Pointwise Expansion of Degenerating Immersions of Finite Total Curvature

Abstract

Generalising classical result of Müller and Šverák (J. Differ. Geom. 42(2), 229-258, 1995), we obtain a pointwise estimate of the conformal factor of sequences of conformal immersions from the unit disk of the complex plane of uniformly bounded total curvature and converging strongly outside of a concentration point towards a branched immersions for which the quantization of energy holds. We show that the multiplicity associated to the conformal parameter becomes eventually constant to an integer equal to the order of the branch point of the limiting branched immersion. Furthermore, we deduce a $C^0$ convergence of the normal unit in the neck regions. Finally, we show that these improved energy quantizations hold for Willmore surfaces of uniformly bounded energy and precompact conformal class, and for Willmore spheres arising as solutions of min-max problems in the viscosity method.

1 Introduction

Let $\Sigma$ be a Riemann surface (not necessarily closed) and $\vec {\Phi }:\Sigma \rightarrow {\mathbb {R}}^n$ be a smooth immersion. Denote by $g=\vec {\Phi }^{*}g_{{\mathbb {R}}^n}$ the induced metric on $\Sigma$. We say that $\vec {\Phi }$ has finite total curvature if

$$\begin{aligned} \int _{\Sigma }|\vec {{\mathbb {I}}}|^2\text {d}{\mathrm {vol}}_{g}<\infty , \end{aligned}$$

where $\vec {{\mathbb {I}}}$ is the second fundamental form of $\vec {\Phi }$. In 1994, T. Toro proved the surprising result that assuming only that $u\in W^{2,2}(\Sigma ,{\mathbb {R}})$, the graph ${\mathscr {S}}={\mathbb {R}}^3\cap \left\{ (x,y): y=u(x)\;\,\text {for some}\;\, x\in \Sigma \right\}$ admits a bi-Lipschitz parametrisation. The following year, Müller and Šverák extended this result and showed that immersed surfaces with finite total curvature are conformally equivalent to a punctured Riemann surface. Furthermore, they proved a pointwise estimate of the conformal parameter of immersions of finite total curvature at the ends. The result can be restated in terms of branched immersions of the disk, and this is the statement due to T. Rivière that we will now state ([29], Lemma A.5). Here, $B_1(0)\subset {\mathbb {C}}$ is the open unit ball of the complex plane.

Theorem

(Müller and Šverák [23], Rivière [29]). Let $n\ge 3$, and $\vec {\Phi }\in W^{2,2}_{{\mathrm {loc}}}(B_1(0),{\mathbb {R}}^n)\cap W^{1,2}(B_1(0),{\mathbb {R}}^n)$ be a conformal immersion of $B_1(0)\setminus \left\{ 0\right\}$ of finite total curvature and assume that

$$\begin{aligned} \lambda =\frac{1}{2}\log |\nabla \vec {\Phi }|\in L^{\infty }_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} ). \end{aligned}$$

Then, $\vec {\Phi }$ can be extended to a Lipschitz conformal immersion of $B_1(0)$, and there exist a positive integer $\theta _0\ge 1$ and $C>0$ such that for all $z\in B_1(0)$

$$\begin{aligned} C(1-o(1))|z|^{\theta _0-1}\le |\partial _{z}\vec {\Phi }|\le C(1+o(1))|z|^{\theta _0-1}. \end{aligned}$$

More precisely, there exists $\mu \in W^{2,1}(B_1(0))$ (so that $\mu \in C^0(B_1(0))$ in particular) and a harmonic function $\nu : B_1(0)\setminus \left\{ 0\right\} \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} \lambda =\mu +\nu , \end{aligned}$$

and

$$\begin{aligned} \nu (z)=(\theta _0-1)\log |z|+h(z), \end{aligned}$$

where $h:B_1(0)\rightarrow {\mathbb {R}}$ is a harmonic function. In particular, we have for some constant $C>0$ depending only on $\vec {\Phi }$

$$\begin{aligned} \left\| \lambda -(\theta _0-1)\log |z|\right\| _{{\mathrm {L}}^{\infty }(B_1(0))}\le C. \end{aligned}$$

In the study of bubbling of sequences of Willmore immersions (or equivalently of the compactness of the moduli space), it is of great interest to understand the pointwise behaviour of degenerations of immersions of uniformly bounded Willmore energy, or equivalently finite total curvature and in the viscosity method (see [16] and [34]).

In the following theorem, we obtain a pointwise expansion of the conformal factor in the full neck region of an arbitrary sequence of immersions (not necessarily Willmore).

The following theorem shows that the multiplicity of weakly converging sequence of immersions becomes eventually constant to an integer. This is a significant improvement of the fundamental work of Müller and Šverák ([23]).

Theorem A

Let $n\ge 3$ be a fixed integer. There exists a universal constant $C_0(n)>0$ with the following property. Let $\{\vec {\Phi }_k\}_{k\in {\mathbb {N}}}$ be a sequence of smooth conformal immersions from the disk $B_1(0)\subset {\mathbb {C}}$ into ${\mathbb {R}}^n$ and $\left\{ \rho _k\right\} _{k\in {\mathbb {N}}}\subset (0,1)$ be such that $\rho _k\underset{k\rightarrow \infty }{\longrightarrow }0$, $\Omega _k=B_1\setminus \overline{B}_{\rho _k}(0)$ and define for all $0<\alpha <1$ the sub-domain $\Omega _k(\alpha )=B_{\alpha }\setminus \overline{B}_{\alpha ^{-1}\rho _k}(0)$. For all $k\in {\mathbb {N}}$, let

$$\begin{aligned} \lambda _k=\log \left( \frac{|\nabla \vec {\Phi }_k|}{\sqrt{2}}\right) \end{aligned}$$

be the conformal factor of $\vec {\Phi }_k$. Assume that

$$\begin{aligned} \sup _{k\in {\mathbb {N}}}\left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k)}<\infty ,\qquad \lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\int _{\Omega _k(\alpha )}|\nabla \vec {n}_k|^2\mathrm{d}x=0 \end{aligned}$$

and that there exists a $W^{2,2}_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} )\cap C^{\infty }(B_1(0)\setminus \left\{ 0\right\} )$ immersion $\vec {\Phi }_{\infty }$ such that

$$\begin{aligned} \log |\nabla \vec {\Phi }_{\infty }|\in L^{\infty }_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} ) \end{aligned}$$

and $\vec {\Phi }_k\underset{k\rightarrow \infty }{\longrightarrow }\vec {\Phi }_{\infty }$ in $C^l_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} )$ (for all $l\in {\mathbb {N}}$). Then, there exists an integer $\theta _0\ge 1$, $\mu _k\in W^{1,(2,1)}(B_1(0))$ such that

$$\begin{aligned} \left\| \nabla \mu _k\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}\le C_0(n)\int _{\Omega _k}|\nabla \vec {n}_k|^2\mathrm{d}x \end{aligned}$$

and a harmonic function $\nu _k$ on $\Omega _k$ such that $\nu _k=\lambda _k$ on $\partial B_1(0)$, $\lambda _k=\mu _k+\nu _k$ on $\Omega _k$ and such that for all $0<\alpha <1$ and for all $k\in {\mathbb {N}}$ sufficiently large, we have

$$\begin{aligned} \left\| \nabla (\nu _k-(\theta _0-1)\log |z|)\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\le C_0(n)\left( \sqrt{\alpha }\left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k)}+\int _{\Omega _k}|\nabla \vec {n}_k|^2\mathrm{d}x\right) . \end{aligned}$$

Finally, we have for all $\rho _k\le r_k\le 1$ and k large enough

$$\begin{aligned} \frac{1}{2\pi }\int _{\partial B_{r _k}}*\, \mathrm{d}\nu _{k}=\theta _0-1. \end{aligned}$$

In particular, there exists a constant $C>0$ independent of $k\in {\mathbb {N}}$ such that for all $k\in {\mathbb {N}}$, and for all $z\in \Omega _k(1)=B_1\setminus \overline{B}_{\rho _k}(0)$

$$\begin{aligned} \frac{1}{C}|z|^{\theta _0-1}\le e^{\lambda _k(z)}\le C|z|^{\theta _0-1}. \end{aligned}$$

Remark

Theorem A corresponds to Theorem 3.1.

This theorem has also been obtained recently by Nicolas Marque in the case of minimal simple bubbling ([17]). It constitutes a fundamental ingredient to show that in this special case, there is an obstruction to the singularity of the limiting Willmore immersion at branch points (it is stated using the second residue, see [1]). As such, this result may be seen as a technical result aimed at providing new applications to the loss compactness of Willmore immersions and in particular an extension of Marque’s main result to arbitrary codimension. This result also constitutes an improvement of Lemma V.3 of Bernard and Rivière ([2]) since it identifies the multiplicity $d_k$ corresponding to $\vec {\Phi }_k$ to be the integer $\theta _0-1\ge 0$ eventually (i.e. for $k\in {\mathbb {N}}$ large enough), which also restricts the possibilities of bubbling of Willmore surfaces. If the limiting branched immersions have a branch point of order, then the bubble that appears at this point must have a branch point of the same order. Since the result also applies to the viscosity method, we expect that it should help shedding some light on the problem to determining the Morse index of branched Willmore spheres realising the min-max sphere eversion (see [18,19,20,21, 34]).

More generally, an $L^{2,1}$ quantization of the energy permits to obtain a pointwise expansion of the conformal parameter by constructing—using by Hélein’s methods ([11]) and their extension to Willmore immersions by T. Rivière ([2, 28])—a controlled $L^{2,1}$ Coulomb frame.

Theorem B

Under the conditions of Theorem A, assume furthermore that the following strong $L^{2,1}$ no-neck energy holds

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\lim _{k\rightarrow \infty }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}=0. \end{aligned}$$

Then, there exists $\alpha _0>0$ such that for all $k\in {\mathbb {N}}$ large enough, there exists a moving frame $(\vec {f}_{k,1},\vec {f}_{k,2})\in W^{1,(2,1)}(B_{\alpha _0}(0))\times W^{1,(2,1)}(B_{\alpha _0}(0))$ and a universal constant $C_{1}(n)$ (independent of k) such that

$$\begin{aligned}&\left\| \nabla \vec {f}_{k,1}\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}+\left\| \nabla \vec {f}_{k,2}\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}\\&\quad \le C_{1}(n)\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}. \end{aligned}$$

Furthermore, there exists a sequence of functions $\mu _k\in W^{2,1}(B_{\alpha _0}(0))$ and a universal constant $C_{2}(n)$ such that

$$\begin{aligned}&\left\| \nabla ^2\mu _k\right\| _{{\mathrm {L}}^{1}(B_{\alpha _0}(0))}+\left\| \nabla \mu _k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}+\left\| \mu _k\right\| _{{\mathrm {L}}^{\infty }(B_{\alpha _0}(0))}\le C_{2}(n)\int _{\Omega _k(\alpha _0)}|\nabla \vec {n}_k|^2\mathrm{d}x \end{aligned}$$

and there exists a sequence of holomorphic functions $\psi _k:B_{\alpha _0}(0)\rightarrow {\mathbb {C}}$ and $\chi _k:B_{\alpha _0}(0)\rightarrow {\mathbb {C}}$ such that $\chi _k(0)=0$, $c\in {\mathbb {C}}$ and $\left\{ c_k\right\} _{k\in {\mathbb {N}}}\subset {\mathbb {C}}$ such that $c_k\underset{k\rightarrow \infty }{\longrightarrow }c$ and

$$\begin{aligned} \psi _k(z)=e^{c_k}z^{\theta _0-1}\left( 1+\chi _k(z)\right) \end{aligned}$$

(1.1)

and

$$\begin{aligned} e^{\lambda _k}=e^{\mu _k}|\psi _k(z)|=e^{{\mathrm {Re}}\,(c_k)}|z|^{\theta _0-1}\left( 1+o(1)\right) ,\qquad \text {for all}\;\, z\in \Omega _k(\alpha ). \end{aligned}$$

(1.2)

Finally, there exists $\vec {A}_0\in {\mathbb {C}}^n$ (satisfying $\langle \vec {A}_0,\vec {A}_0\rangle =0$) and $\{\vec {A}_{k,0}\}_{k\in {\mathbb {N}}}\in {\mathbb {C}}^n$ such that $\vec {A}_{k,0}\underset{k\rightarrow \infty }{\longrightarrow }\vec {A}_0$ and for all $z\in \Omega _k(\alpha _0)$, we have the pointwise identities

$$\begin{aligned} \partial _{z}\vec {\Phi }_k&=\frac{1}{2}e^{c_k+\mu _k(z)}z^{\theta _0-1}\left( 1+\chi _k(z)\right) \left( \vec {f}_{k,1}-i\vec {f}_{k,2}\right) \nonumber \\&=\vec {A}_{k,0}z^{\theta _0-1}+o\left( |z|^{\theta _0-1}\right) . \end{aligned}$$

(1.3)

Remark

Theorem B corresponds to Theorem 3.5 below.

These two theorems have analogues in the case of multiple bubbles but we will not state them here for the sake of simplicity of presentation.

We also prove that this stronger quantization property holds for sequences of Willmore immersions of uniformly bounded Willmore energy and for Willmore spheres arising in min-max constructions in the viscosity method.

Theorem C

Let $\Sigma$ be a closed Riemann surface and assume that $\{\vec {\Phi }_k\}_{k\in {\mathbb {N}}}$ is a sequence of smooth Willmore immersions such that

$$\begin{aligned} \limsup _{k\rightarrow \infty }W(\vec {\Phi }_k)<\infty . \end{aligned}$$

Assume furthermore that the conformal class of $\{\vec {\Phi }_k^{*}g_{{\mathbb {R}}^n}\}_{k\in {\mathbb {N}}}$ lies in a compact subset of the moduli space. Then, for all $0<\alpha <1$, let $\Omega _k(\alpha )=B_{\alpha R_k}\setminus \overline{B}_{\alpha ^{-1}r_k}(0)$ be a neck domain and $\theta _0\in {\mathbb {N}}$ such that (by Theorem 3.1)

$$\begin{aligned} \theta _0-1=\lim \limits _{\alpha \rightarrow 0}\lim \limits _{k\rightarrow \infty }\int _{\partial B_{\alpha ^{-1}r_k}(0)}\partial _{\nu }\lambda _k\,\mathrm{d}{\mathscr {H}}^1, \end{aligned}$$

(1.4)

and define

$$\begin{aligned} \Lambda =\sup _{k\in {\mathbb {N}}}\left( \left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k(1))}+\int _{\Omega _k(1)}|\nabla \vec {n}_k|^2\mathrm{d}x\right) . \end{aligned}$$

Then, there exist a universal constant $C_{3}=C_{3}(n)$, and $\alpha _0=\alpha _0(\{\vec {\Phi }_k\}_{k\in {\mathbb {N}}})>0$ such that for all $0<\alpha <\alpha _0$ and $k\in {\mathbb {N}}$ large enough,

$$\begin{aligned} \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}&\le C_{3}(n)e^{C_{3}(n)\Lambda }\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(4\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(4\alpha ))}. \end{aligned}$$

(1.5)

In particular, we deduce by the $L^{2,1}$ no-neck energy

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}=0. \end{aligned}$$

Remark

Theorem C corresponds to Theorem 4.1 below.

A similar result was proved by Lamm and Sharp ([12]) in the case of conformally invariant problems and in the more general setting introduced by Rivière ([27]) of elliptic systems with antisymmetric potentials, and by Changyou Wang in the case of harmonic maps ([36]).

Finally, we show that this hypothesis is indeed satisfied for sequences of Willmore immersions of precompact conformal class or in the viscosity method for spheres. The proof of such a result builds on the previous work of Rivière ([28, 29]), Bernard and Rivière ([1, 2]) and Laurain and Rivière ([13,14,15]) and on the general philosophy of integration by compensation and geometric analysis on surfaces (including [4, 11, 23, 35]). We refer to Theorems4.1 and  6.2 for the precise (and somewhat technical) statement.

Corollary 1.4

Let $\Sigma$ be a closed Riemann surface and assume that $\{\vec {\Phi }_k\}_{k\in {\mathbb {N}}}$ is a sequence of Willmore immersions from $\Sigma$ into ${\mathbb {R}}^n$ such that

$$\begin{aligned} \limsup _{k\rightarrow \infty }\int _{\Sigma }|\vec {H}_{\vec {\Phi }_k}|^2\mathrm{d}{\mathrm {vol}}_{g_{\vec {\Phi }_k}}<\infty . \end{aligned}$$

Assume furthermore that the conformal class of $\{\vec {\Phi }_k^{*}g_{{\mathbb {R}}^n}\}_{k\in {\mathbb {N}}}$ lies in a compact subset of the moduli space. Then, there exists $\left\{ a_1,\cdots ,a_m\right\} \subset \Sigma$, sequences $\{x_{k}^{i,j}\}_{k\in {\mathbb {N}}}$, $1\le i\le n$, $1\le j\le m_i$ such that $x_{k}^{i,j}\underset{k\rightarrow \infty }{\longrightarrow }a_i$ for all i, j and branched Willmore immersions $\vec {\Phi }_{\infty }:\Sigma \rightarrow {\mathbb {R}}^n$, $\vec {\Phi }_{\infty }^{i,j}:S^2={\mathbb {C}}\cup \left\{ \infty \right\} \rightarrow {\mathbb {R}}^n$ and $\{\rho _k^{i,j}\}_{k\in {\mathbb {N}}}\subset (0,\infty )$ with $\rho _{k}^{i,j}\underset{k\rightarrow \infty }{\longrightarrow }0$ and for all $1\le i\le m$ and $1\le j\ne j'\le m$,

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\max \left\{ \frac{\rho _k^{i,j}}{\rho _{k}^{i,j'}}+\frac{\rho _k^{i,j'}}{\rho _k^{i,j}},\frac{|x_k^{i,j}-x_k^{i,j'}|}{\rho _k^{i,j}+\rho _k^{i,j'}}\right\} =\infty . \end{aligned}$$

such that

$$\begin{aligned} \left\| \nabla \vec {n}_{\vec {\Phi }_k}-\nabla \vec {n}_{\vec {\Phi }_{\infty }}-\sum _{i=1}^{m}\sum _{j=1}^{m_i}\nabla \vec {n}_{\vec {\Phi }_{\infty }^{i,j}}((\rho _k^{i,j})^{-1}(\,\cdot \,-x_k^{i,j}))\right\| _{{\mathrm {L}}^{2,1}(\Sigma )}=0. \end{aligned}$$

(1.6)

In particular, we have

$$\begin{aligned} \left\| \vec {n}_{\vec {\Phi }_k}-\vec {n}_{\vec {\Phi }_{\infty }}-\sum _{i=1}^{m}\sum _{j=1}^{m_i}\left( \vec {n}_{\vec {\Phi }_{\infty }^{i,j}}((\rho _k^{i,j})^{-1}(\,\cdot \,-x_k^{i,j}))-\vec {n}_{\vec {\Phi }_{\infty }^{i,j}}(\infty )\right) \right\| _{{\mathrm {L}}^{\infty }(\Sigma )}=0. \end{aligned}$$

(1.7)

The proof of Corollary is found at the end of Sect. 4.

Remark

• (1) The writing of (1.6) and (1.7), classical in concentration compactness theory, makes use of implicit cutoff functions (see [36]).

• (2) This result is optimal in the $C^{l,\beta }$ topology since the $C^{0,\beta }$ norm for $\beta >0$ is not scaling invariant. For another $C^0$ theory for the blow-up of elliptic equations of order 2, see [12, 24, 36].

More precisely, the $C^0$ energy quantization permits to link the values of the normal of the limiting immersion of the one of bubbles. Let us state the result in the case of a single bubble for simplicity.

Corollary 1.5

Let $\Sigma$ be a closed Riemann surface and assume that $\{\vec {\Phi }_k\}_{k\in {\mathbb {N}}}$ is a sequence of Willmore immersions from $\Sigma$ in ${\mathbb {R}}^n$ such that

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\int _{\Sigma }|\vec {H}_{\vec {\Phi }_k}|^2\mathrm{d}{\mathrm {vol}}_{g_{\vec {\Phi }_k}}<\infty . \end{aligned}$$

Assume furthermore that the conformal class of $\{\vec {\Phi }_k^{*}g_{{\mathbb {R}}^n}\}_{k\in {\mathbb {N}}}$ lies in a compact subset of the moduli space. Following [2], let $\vec {\Phi }_{\infty }:\Sigma \rightarrow {\mathbb {R}}^n$ be such that for some finite collection $\left\{ a_1,\cdots ,a_m\right\} \subset \Sigma$, we have

$$\begin{aligned} \vec {\Phi }_k\underset{k\rightarrow \infty }{\longrightarrow }\vec {\Phi }_{\infty }\qquad \text {in}\;\, C^l_{{\mathrm {loc}}}(\Sigma \setminus \left\{ a_1,\cdots ,a_m\right\} )\;\,\text {for all}\;\, l\in {\mathbb {N}}. \end{aligned}$$

Let $1\le i\le n$ and assume that a single bubble $\vec {\Psi }_{\infty }^i:S^2\rightarrow {\mathbb {R}}^n$ forms at $a_i$. Then, we have

$$\begin{aligned} \vec {n}_{\vec {\Phi }_{\infty }}(a_i)=\vec {n}_{\vec {\Psi }_{\infty }^i}(\infty ). \end{aligned}$$

(1.8)

In the case of bubbles over bubbles, normals at junctions coincide with the value of the normal at $N=\infty \in S^2$ of the bubble. The proof is exactly the same.

2 Uniform Control of the Conformal Factor in Necks

For the definitions related to Lorentz spaces, we refer the reader to the Appendix (Sect. 7.1).

In this section, we obtain a refinement of Lemma V.3 of [2].

Theorem 2.1

There exists a positive real numbers $\varepsilon _1=\varepsilon _1(n)>0$ and $\Gamma _0(n)>0$ with the following property. Let $0<2^6r<R<\infty$ be fixed radii and $\vec {\Phi }:\Omega =B_R\setminus \overline{B}_r(0)\rightarrow {\mathbb {R}}^n$ be a weak immersion of finite total curvature such that

$$\begin{aligned} \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\le \varepsilon _1(n). \end{aligned}$$

(2.1)

Fix some $\left( \dfrac{r}{R}\right) ^{\frac{1}{3}}<\alpha <1$, and define $\Omega _{\alpha }=B_{\alpha R} \setminus \overline{B}_{\alpha ^{-1}r}(0)$. Then, we have

$$\begin{aligned} \left\| \nabla (\lambda -d\log |z|)\right\| _{{\mathrm {L}}^{2,1}(\Omega _{\alpha })}\le \Gamma _0\left( \sqrt{\alpha }\left\| \nabla \lambda \right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) \end{aligned}$$

(2.2)

and for all $r\le \rho <R$, we have

$$\begin{aligned} \left| d-\frac{1}{2\pi }\int _{\partial B_{\rho }}\partial _{\nu }\lambda \,{\mathrm {d}}\mathscr {H}^{1}\right| \le \Gamma _0\left( \int _{B_{\max \left\{ \rho ,2r\right\} }\setminus \overline{B}_r(0)}|\nabla \vec {n}|^2\mathrm{d}x+\frac{1}{\log \left( \frac{R}{\rho }\right) }\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) \end{aligned}$$

(2.3)

In particular, there exists a universal constant $\Gamma _0'=\Gamma _0'(n)$ and $A_{\alpha }\in {\mathbb {R}}$ such that

$$\begin{aligned} \left\| \lambda -d\log |z|-A_{\alpha }\right\| _{{\mathrm {L}}^{\infty }(\Omega _{\alpha })}\le \Gamma _0'\left( \sqrt{\alpha }\left\| \nabla \lambda \right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) . \end{aligned}$$

(2.4)

The proof relies on the strategy developed in [2] (and the lemmas from [13, 15] for the Lemmas 2.2 and 2.3) and the following two lemmas, which will allow us to move from a $L^{2,\infty }$ bound to a $L^{2,1}$ bound in a quantitative way.

Lemma 2.2

Let $u:B_R\setminus \overline{B}_r(0)\rightarrow {\mathbb {R}}$ be a harmonic function such that for some $\rho _0\in (r,R)$

$$\begin{aligned} \int _{\partial B_{\rho _{0}}}\partial _{\nu }u \,\mathrm{d}{\mathscr {H}}^1=0. \end{aligned}$$

Then, there exists a universal constant $\Gamma _1>0$ (independent of $0<4r<R<\infty$) such that for all $\left( \dfrac{r}{R}\right) ^{\frac{1}{2}}<\alpha <\dfrac{1}{2}$, we have

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}\le {\Gamma _1}\left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_{R}\setminus \overline{B}_r(0))}. \end{aligned}$$

Proof

First, we show that for all $\alpha ^{-1}r\le \rho \le \alpha R$, and for all $0<\alpha <\dfrac{1}{2}$, we have

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{\infty }(\partial B_{\rho }(0))}\le \frac{4}{\log (2)}\sqrt{\frac{3}{\pi }}\frac{1}{(1-\alpha )\rho }\left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_{\alpha ^{-1}\rho }\setminus \overline{B}_{\alpha \rho }(0))}. \end{aligned}$$

(2.5)

By a slight abuse of notation, we will write r instead of $\rho$ in the following estimates.

As $0<\alpha <\dfrac{1}{2}$, we have for all $x\in \partial B_r(0)$, the inclusion $B_{(1-\alpha )r}(x)\subset B_{\alpha ^{-1}r}\setminus \overline{B}_{\alpha r}(0)$. Therefore, thanks to the mean value property, we have for all $0<\beta <(1-\alpha )r$

$$\begin{aligned} \nabla u(x)=\frac{1}{2\pi \beta }\int _{\partial B_{\beta }(x)}\nabla u(y)\,\text {d}{\mathscr {H}}^1(y). \end{aligned}$$

(2.6)

Now, thanks to the co-area formula, we have $\Big ($if $I_{\alpha }(r)=\left( \frac{(1-\alpha )}{2}r,(1-\alpha )r\right)$ $\Big )$

$$\begin{aligned}&\int _{B_{(1-\alpha )r}\setminus \overline{B}_{(1-\alpha )r/2}(x)}|\nabla u(y)|\text {d}y=\int _{\frac{(1-\alpha )r}{2}}^{(1-\alpha )r}\left( \int _{\partial B_{\beta }(x)}|\nabla u(y)|\,\text {d}{\mathscr {H}}^1(y)\right) \text {d}\beta \\&\quad \ge \inf _{\beta \in I_{\alpha }(r)}\left( \beta \int _{\partial B_{\beta }(x)}|\nabla u(y)|\text {d}{\mathscr {H}}^1(y)\right) \int _{\frac{(1-\alpha )r}{2}}^{(1-\alpha )r}\frac{\text {d}\beta }{\beta } \\&\quad =\log (2)\inf _{\beta \in I_{\alpha }(r)}\left( \beta \int _{\partial B_{\beta }(x)}|\nabla u(y)|\text {d}{\mathscr {H}}^1(y)\right) \end{aligned}$$

Therefore, there exists $\beta \in \left( \frac{(1-\alpha )r}{2},(1-\alpha )r\right)$ (notice that this shows that the limiting values $\rho =\alpha ^{-1}r$ and $\rho =\alpha R$ are admissible) such that

$$\begin{aligned} \beta \int _{\partial B_{\beta }(x)}|\nabla u(y)|\,\text {d}{\mathscr {H}}^1(y)\le \frac{1}{\log (2)} \int _{B_{(1-\alpha )r}\setminus \overline{B}_{(1-\alpha )r/2}(x)}|\nabla u(y)|\text {d}y \end{aligned}$$

or

$$\begin{aligned} \frac{1}{2\pi \beta }\int _{\partial B_{\beta }(x)}|\nabla u(y)|\text {d}{\mathscr {H}}^1(y)\le \frac{1}{2\pi \log (2)\beta ^2}\int _{B_{(1-\alpha )r}\setminus \overline{B}_{(1-\alpha )r/2}(x)}|\nabla u(y)|\text {d}y. \end{aligned}$$

(2.7)

Now, notice that

$$\begin{aligned} \left\| {\mathrm {1}}_{B_{(1-\alpha )r}\setminus \overline{B}_{(1-\alpha )r/2}(x)}\right\| _{{\mathrm {L}}^{2,1}({\mathbb {R}}^2)}&=4\int _{0}^{\infty }\left( {\mathscr {L}}^2(B_{(1-\alpha )r}\setminus \overline{B}_{(1-\alpha )r/2}(x)\cap \left\{ x: 1>t\right\} \right) ^{\frac{1}{2}}\text {d}t\nonumber \\&=2\sqrt{3\pi }(1-\alpha )r. \end{aligned}$$

(2.8)

Furthermore, as $\beta >\dfrac{(1-\alpha )r}{2}$, we have

$$\begin{aligned} \frac{1}{\beta ^2}\le \frac{4}{(1-\alpha )^2r^2}. \end{aligned}$$

(2.9)

Therefore, we have by the mean value property (2.6), the inequalities (2.7), (2.8), (2.9) and the duality $L^{2,1}/L^{2,\infty }$

$$\begin{aligned} |\nabla u(x)|&\le \frac{1}{2\pi \beta }\int _{\partial B_{\beta }(x)}|\nabla u(y)|\,\text {d}{\mathscr {H}}^1(y)\\&\le \frac{2}{\pi \log (2)(1-\alpha )^2r^2}\left\| {\mathrm {1}}_{B_{(1-\alpha )r}\setminus \overline{B}_{(1-\alpha )r/2}(x)}\right\| _{{\mathrm {L}}^{2,1}({\mathbb {R}}^2)}\left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_{(1-\alpha )r}\setminus \overline{B}_{(1-\alpha )r/2}(x))}\\&\le \frac{4}{\log (2)}\sqrt{\frac{3}{\pi }}\frac{1}{(1-\alpha )r}\left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_{\alpha ^{-1}r}\setminus \overline{B}_{\alpha r}(0))}. \end{aligned}$$

As $x\in \partial B_{r}(0)$ was arbitrary, this proves the inequality (2.5). Now, as u is harmonic, there exists $\left\{ a_n\right\} _{n\in {\mathbb {Z}}}\subset {\mathbb {C}}$ such that

$$\begin{aligned} u(\rho ,\theta )=a_0+d \,\log \rho +\sum _{n\in {\mathbb {Z}}^{*}}^{}\left( a_n\rho ^n+\overline{a_{-n}}\rho ^{-n}\right) e^{in\theta }, \end{aligned}$$

which implies by the hypothesis that

$$\begin{aligned} 0=\int _{\partial B_{\rho _0}}\partial _{\nu }u\,\text {d}{\mathscr {H}}^1=2\pi d \end{aligned}$$

(2.10)

so that for all $r<\rho <R$

$$\begin{aligned} \int _{\partial B_{\rho }}\partial _{\nu }u=0. \end{aligned}$$

Therefore, integrating by parts, we find

$$\begin{aligned} \int _{B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}}|\nabla u(x)|^2\mathrm{d}x&=\int _{\partial B_{\alpha R}}\partial _{\nu } u\,u\,\text {d}{\mathscr {H}}^1-\int _{\partial B_{\alpha ^{-1}r}}\partial _{\nu }u\,u\,\text {d}{\mathscr {H}}^1\nonumber \\&=\int _{\partial B_{\alpha R}}\partial _{\nu }u(u-\overline{u}_{\alpha R})\text {d}{\mathscr {H}}^1-\int _{B_{\alpha ^{-1}r}}\partial _{\nu } u\left( u-\overline{u}_{{\alpha ^{-1}r}}\right) \text {d}{\mathscr {H}}^1 \end{aligned}$$

(2.11)

where is the average of u on $\rho$, for all $r<\rho <R$.

Now, if $\Gamma _2=\Gamma _2(H^{\frac{1}{2}}(S^1),L^1(S^1))$ is the constant of the injection $H^{\frac{1}{2}}(S^1)\hookrightarrow L^1(S^1)$ (for the norm defined by the $L^2$ norm of the harmonic extension), we get by (2.5) for all $r<\rho <R$

$$\begin{aligned} \left| \int _{\partial B_{\rho }}\partial _{\nu }u\left( u-\overline{u}_{\rho }\right) \text {d}{\mathscr {H}}^1\right|&\le \left\| \nabla u\right\| _{{\mathrm {L}}^{\infty }(\partial B_{\rho })}\left\| u-\overline{u}_{\rho }\right\| _{{\mathrm {L}}^{1}(\partial B_{\rho })}\\&\le \frac{4}{\log (2)}\sqrt{\frac{3}{\pi }}\frac{1}{(1-\alpha )\rho }\left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_R\setminus \overline{B}_r(0))}\times \Gamma _2\rho \left\| u\right\| _{{\mathrm {H}}^{\frac{1}{2}}(\partial B_{\rho })}\\&\le \frac{4}{\log (2)}\sqrt{\frac{3}{\pi }}\frac{1}{(1-\alpha )}\Gamma _2\left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_R\setminus \overline{B}_r(0))}\left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))} \end{aligned}$$

which implies by (2.11) that

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r})}\le \frac{8}{\log (2)}\sqrt{\frac{3}{\pi }}\frac{1}{(1-\alpha )}\Gamma _2\left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_R\setminus \overline{B}_r(0))} \end{aligned}$$

and this concludes the proof of the Lemma. $\square$

In the following Lemma, we obtain a slight improvement from [15] and generalise it to a $W^{2,1}$ estimate, that will be used in the proof of Theorem 4.1.

Lemma 2.3

Let $0<4r<R<\infty$ be fixed radii, and $u:\Omega =B_R\setminus \overline{B}_r(0)\rightarrow {\mathbb {R}}$ be a harmonic function such that for some $\rho _0\in (r,R)$

$$\begin{aligned} \int _{\partial B_{\rho _0}}\partial _{\nu }u \,\mathrm{d}{\mathscr {H}}^1=0. \end{aligned}$$

Then for all $\left( \dfrac{r}{R}\right) ^{\frac{1}{2}}<\alpha <1$, we have

$$\begin{aligned}&\left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}\le 32\sqrt{\frac{2}{15}}\frac{\alpha }{1-\alpha } \left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_R\setminus \overline{B}_r(0))},\\&\left\| \nabla ^2u\right\| _{{\mathrm {L}}^{1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}\le 32\sqrt{\frac{\pi }{15}}\frac{\alpha }{1-\alpha }\left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_{R}\setminus \overline{B}_r(0))}. \end{aligned}$$

Proof

As u is harmonic on $B_R\setminus \overline{B}_r(0)$, there exists $\left\{ a_n\right\} _{n\in {\mathbb {Z}}}\subset {\mathbb {C}}$ and $d\in {\mathbb {R}}$ such that

$$\begin{aligned} u(z)=a_0+d\log |z|+2\,{\mathrm {Re}}\,\left( \sum _{n\in {\mathbb {Z}}}a_nz^n\right) . \end{aligned}$$

Thanks to (2.10), we deduce that $d=0$. Furthermore, taking polar coordinates $z=\rho e^{i\theta }$, we have the identity

$$\begin{aligned} |\nabla u|^2=4|\partial _{z}u|^2=4\left| \sum _{n\in {\mathbb {Z}}^{*}}n a_nz^{n-1}\right| ^2=4\sum _{n,m\in {\mathbb {Z}}^{*}}nm\,a_n\overline{a}_m\rho ^{n+m-2}e^{i(n-m)\theta }. \end{aligned}$$

(2.12)

This implies by the inequality $0<4r<R<\infty$ that

$$\begin{aligned} \int _{B_R\setminus \overline{B}_r(0)}|\nabla u(x)|^2\mathrm{d}x&=8\pi \sum _{n\in {\mathbb {Z}}^{*}}^{}\int _{r}^{R}|n|^2|a_n|^2\rho ^{2n-1}\mathrm{d}\rho \nonumber \\&=8\pi \sum _{n\in {\mathbb {Z}}^{*}}|n|^2\left( \frac{1}{2n}|a_n|^2\left( R^{2n}-r^{2n}\right) \right) \nonumber \\&=4\pi \sum _{n\ge 1}^{}|n||a_n|^2R^{2|n|}\left( 1-\left( \frac{r}{R}\right) ^{2|n|}\right) \nonumber \\&\quad +4\pi \sum _{n\le -1}^{}|n||a_{n}|^2\frac{1}{r^{2|n|}}\left( 1-\left( \frac{r}{R}\right) ^{2|n|}\right) \nonumber \\&\ge \frac{15\pi }{4} \sum _{n\ge 1}^{}|n|\left( |a_n|^2R^{2|n|}+|a_{-n}|^2\frac{1}{r^{2|n|}}\right) . \end{aligned}$$

(2.13)

First $L^{2,1}$ estimate. Now, we have

$$\begin{aligned} \left\| 1\right\| _{{\mathrm {L}}^{2,1}(B_R\setminus \overline{B}_r)}=4\sqrt{\pi }\left( R^2-r^2\right) ^{\frac{1}{2}}\le 4\sqrt{\pi }R \end{aligned}$$

while for all $m\ge 1$,

$$\begin{aligned} \left\| |z|^{m}\right\| _{{\mathrm {L}}^{2,1}(B_R\setminus \overline{B}_r(0))}&=4\sqrt{\pi }r^m\left( R^2-r^2\right) ^{\frac{1}{2}}+4\sqrt{\pi }\int _{r^m}^{R^m}(R^2-t^{\frac{2}{m}})^{\frac{1}{2}}\,\text {d}t \\&\le 4\sqrt{\pi }r^{m}R+4\sqrt{\pi }\int _{r^{m}}^{R^m}R\text {d}t =4\sqrt{\pi }R^{m+1}. \end{aligned}$$

Likewise, for all $m\ge 2$

$$\begin{aligned} \left\| \frac{1}{|z|^{m}}\right\| _{{\mathrm {L}}^{2,1}(B_R\setminus \overline{B}_r(0))}&\le 4\sqrt{\pi }\int _{0}^{\frac{1}{r^m}}\left( \frac{1}{t^{\frac{2}{m}}}-r^2\right) ^{\frac{1}{2}}\text {d}t \\&\le 4\sqrt{\pi }\int _{0}^{\frac{1}{r^m}}\frac{\text {d}t}{t^{\frac{1}{m}}} =4\sqrt{\pi }\frac{m}{m-1}\frac{1}{r^{m-1}} \le 8\sqrt{\pi }r^{-m+1}. \end{aligned}$$

By (2.12), we have

$$\begin{aligned} |\nabla u|&\le 2\sum _{n\in {\mathbb {Z}}^{*}}^{}|n||a_n|\rho ^{n-1}, \end{aligned}$$

and the following estimates by Cauchy–Schwarz inequality

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha {-1}r}(0))}&\le 16\sqrt{\pi }\left( \sum _{n\ge 1}^{}|n||a_n|\left( \alpha R\right) ^{|n|}+\sum _{n\ge 1}^{}|n||a_{-n}|\left( \frac{\alpha }{r}\right) ^{|n|}\right) \nonumber \\&\le 16\sqrt{\pi }\left( \sum _{n\in {\mathbb {Z}}^{*}}^{}|n|\alpha ^{2|n|}\right) ^{\frac{1}{2}}\left( \sum _{n\ge 1}^{}|n||a_n|^2R^{2|n|}+|n||a_{-n}|^2\frac{1}{r^{2|n|}}\right) ^{\frac{1}{2}}\nonumber \\&=16\sqrt{2\pi }\frac{\alpha }{1-\alpha ^2}\left( \sum _{n\ge 1}^{}|n||a_n|^2R^{2|n|}+|n||a_{-n}|^2\frac{1}{r^{2|n|}}\right) ^{\frac{1}{2}}. \end{aligned}$$

(2.14)

Combining (2.13) and (2.14) yields

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}&\le \frac{16\sqrt{2\pi }\alpha }{1-\alpha }\times \sqrt{\frac{4}{15\pi }}\left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_R\setminus \overline{B}_r(0))} \\&=32\sqrt{\frac{2}{15}}\frac{\alpha }{1-\alpha }\left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_R\setminus \overline{B}_r(0))}, \end{aligned}$$

which concludes the proof of the first part of the Lemma.

Second $W^{1,1}$ estimate. As $\Delta u=0$, we have $|\nabla ^2 u|=4|\partial _{z}^2u|$, and

$$\begin{aligned} \partial _{z}^2u(z)=\sum _{n\in {\mathbb {Z}}^{*}}n(n-1)z^{n-2}. \end{aligned}$$

Now, for all $m\in {\mathbb {Z}}\setminus \left\{ -2\right\}$, we have

$$\begin{aligned} \left\| |z|^m\right\| _{{\mathrm {L}}^{1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}=2\pi \int _{\alpha ^{-1}r}^{\alpha R}\rho ^{m+1}\text {d}\rho =\frac{2\pi }{m+2}\left( (\alpha R)^{m+2}-(\alpha ^{-1}r)^{m+2}\right) \end{aligned}$$

In particular, we have by the triangle inequality and Cauchy–Schwarz inequality

$$\begin{aligned}&\left\| \partial _{z}^2u\right\| _{{\mathrm {L}}^{1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}\le 2\pi \sum _{n\in {\mathbb {Z}}^{*}}\frac{|n||n-1|}{n}|a_n|\left( \left( \alpha R\right) ^n-\left( \alpha ^{-1}r\right) ^{n}\right) \\&\quad =2\pi \sum _{n\ge 1}|n-1||a_n|(\alpha R)^{|n|}\left( 1-\left( \frac{\alpha ^2r}{R}\right) ^{|n|}\right) \\&\qquad +2\pi \sum _{n\le -1}|n-1||a_n|\left( \frac{\alpha }{r}\right) ^{|n|}\left( 1-\left( \frac{\alpha ^2r}{R}\right) ^{|n|}\right) \\&\quad \le 2\pi \sum _{n\ge 1}{|n-1|}|a_n|(\alpha R)^{|n|}+\sum _{n\ge -1}{|n-1|}|a_n|\left( \frac{\alpha }{r}\right) ^{|n|}\\&\quad \le 2\pi \left( \sum _{n\in {\mathbb {Z}}^{*}}\frac{|n-1|^2}{|n|}\alpha ^{2|n|}\right) ^{\frac{1}{2}}\left( \sum _{n\ge 1}|n||a_n|^2R^{2|n|}+\sum _{n\le -1}|n||a_n|^2\frac{1}{r^{2|n|}}\right) ^{\frac{1}{2}}. \end{aligned}$$

Now, notice that

$$\begin{aligned} \sum _{n\in {\mathbb {Z}}{*}}\frac{|n-1|^2}{|n|}\alpha ^{2|n|} =2\sum _{n\ge 1}\frac{n^2+1}{n}\alpha ^{2n}=\frac{2\alpha ^2}{(1-\alpha ^2)^2}+2 \log \left( \frac{1}{1-\alpha ^2}\right) \le \frac{4\alpha ^2}{(1-\alpha ^2)^2}. \end{aligned}$$

Recalling from (2.3) that

$$\begin{aligned} \int _{B_{R}\setminus \overline{B}_{\alpha ^{-1}r}(0)}|\nabla u(x)|^2\mathrm{d}x\ge \frac{15\pi }{4} \sum _{n\ge 1}^{}|n|\left( |a_n|^2R^{2|n|}+|a_{-n}|^2\frac{1}{r^{2|n|}}\right) , \end{aligned}$$

we deduce that

$$\begin{aligned} \left\| \partial _{z}^2u\right\| _{{\mathrm {L}}^{1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}&\le \frac{4\pi \alpha }{(1-\alpha ^2)}\times \sqrt{\frac{4}{15\pi }}\left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_R\setminus \overline{B}_r(0))}\\&=8\sqrt{\frac{\pi }{15}}\frac{\alpha }{1-\alpha ^2}\left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_R\setminus \overline{B}_r(0))} \end{aligned}$$

which concludes the proof as $|\nabla ^2u|=4|\partial _{z}^2u|$. $\square$

Remark 2.4

Notice that $\left\| \nabla \log |z|\right\| _{{\mathrm {L}}^{2}(B_R\setminus \overline{B}_r(0))}=\sqrt{2\pi }\sqrt{\log \left( \dfrac{R}{r}\right) }$ while

$$\begin{aligned}&\left\| \nabla \log |z|\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}=4\int _{0}^{\frac{1}{\alpha R}}\left( {\mathscr {L}}^2(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))\right) ^{\frac{1}{2}}\text {d}t\\&\quad +4\int _{\frac{1}{\alpha R}}^{\frac{1}{\alpha ^{-1}r}}\left( {\mathscr {L}}^2\left( B_{\frac{1}{t}}\setminus \overline{B}_{\alpha ^{-1}r}(0)\right) \right) ^{\frac{1}{2}}\text {d}t\\&\quad =\frac{4\sqrt{\pi }}{\alpha R}\left( \alpha ^2R^2-\alpha ^{-2}r^2\right) ^{\frac{1}{2}}+4\sqrt{\pi }\int _{\frac{1}{\alpha R}}^{\frac{1}{\alpha ^{-1}r}}\frac{1}{t}\sqrt{1-\frac{r^2t^2}{\alpha ^2}}\text {d}t\\&\quad =4\sqrt{\pi }\left( \log \left( \frac{\alpha ^2R}{r}\right) +\log \left( 1+\sqrt{1-\left( \frac{r}{\alpha ^2R}\right) ^2}\right) \right) . \end{aligned}$$

In particular, for all fixed $0<\alpha <1$, if $\left\{ R_k\right\} _{k\in {\mathbb {N}}},\left\{ r_k\right\} _{k\in {\mathbb {N}}}\subset (0,\infty )$ are sequences chosen such that $\dfrac{R_k}{r_k}\underset{k\rightarrow \infty }{\longrightarrow } \infty$, we have

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\frac{\left\| \nabla \log |z|\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}\setminus \overline{B}_{\alpha ^{-1}r_k}(0))}}{\left\| \nabla \log |z|\right\| _{{\mathrm {L}}^{2}(B_{R_k}\setminus \overline{B}_{r_k}(0))}}=\infty . \end{aligned}$$

If the assumption $4r<R$ does not hold, observe that we get the estimate

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}\le \frac{8\sqrt{2}}{\sqrt{1-\left( \frac{r}{R}\right) ^2}}\frac{\alpha }{1-\alpha ^2}\left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_R\setminus \overline{B}_r(0))}. \end{aligned}$$

Proposition 2.5

Let $0<2^6r<R<\infty$ be fixed radii, and $u:\Omega =B_R\setminus \overline{B}_r(0)\rightarrow {\mathbb {R}}$ be a harmonic function such that for some $\rho _0\in (r,R)$

$$\begin{aligned} \int _{\partial B_{\rho _0}}\partial _{\nu }u \,\mathrm{d}{\mathscr {H}}^1=0. \end{aligned}$$

Then, for all $\left( \dfrac{r}{R}\right) ^{\frac{1}{3}}<\alpha <\dfrac{1}{4}$,

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}\le 24\,\Gamma _1\sqrt{\alpha } \left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_R\setminus \overline{B}_r(0))}, \end{aligned}$$

where $\Gamma _1$ is given in Lemma 2.2.

Proof

Let $\beta =\sqrt{\alpha }$. Then, by Lemma 2.3, we have

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_{\beta ^2R}\setminus \overline{B}_{\beta ^{-2}r}(0))}\le \frac{12\beta }{1-\beta }\left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_{\beta R}\setminus \overline{B}_{\beta ^{-1}r}(0))}. \end{aligned}$$

Furthermore, by Lemma 2.3, we have

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_{\beta R}\setminus \overline{B}_{\beta ^{-1}r}(0))}\le \Gamma _1\left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_R\setminus \overline{B}_{r}(0))} \end{aligned}$$

Therefore, as $\beta =\sqrt{\alpha }<1/2$, we find

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}&\le \frac{12\sqrt{\alpha }}{1-\sqrt{\alpha }}\Gamma _1\left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_{R}\setminus \overline{B}_r(0))} \le 24\,\Gamma _1\sqrt{\alpha }\left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(B_R\setminus \overline{B}_r(0))}, \end{aligned}$$

which concludes the proof of the corollary. $\square$

We will also need a quantitative estimate of the Lorentz–Sobolev embedding $W^{1,(2,1)}(\Omega )\rightarrow C^0(\Omega )$.

Lemma 2.6

Let $n\ge 2$, $\Omega \subset {\mathbb {R}}^n$ be a bounded connected open set and $u\in W^{1,(n,1)}(\Omega )$. Then, $u\in C^0(\Omega )$ and for all $x,y\in \Omega$ such that $B_{2|x-y|}(x)\cup B_{2|x-y|}(y)\subset \Omega$, we have

$$\begin{aligned} |u(x)-u(y)|\le \frac{2^{n+1}}{\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega \cap B_{2|x-y|}(x))}. \end{aligned}$$

(2.15)

Furthermore, if $\Omega$ is a bounded Lipschitz open subset of ${\mathbb {R}}^n$, then there exists a constant $C_{4}=C_{4}(\Omega )$ such that

$$\begin{aligned} \left\| u-\overline{u}_{\Omega }\right\| _{{\mathrm {L}}^{\infty }(\Omega )}\le C_{4}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega )}, \end{aligned}$$

(2.16)

where is the mean of u.

Remarks on the proof

The proof proceeds in a fairly standard way, using an estimate on averages, the $L^{n,1}/L^{\frac{n}{n-1},\infty }$ duality and Lebesgue differentiation theorem on ${\mathbb {R}}^n$. The extension to the case of domains is easily given by extension operators and interpolation theory to obtain a continue linear extension operator $W^{1,(n,1)}(\Omega )\rightarrow W^{1,(n,1)}({\mathbb {R}}^n)$ (using the Stein-Weiss interpolation theorem).

Proof

Let $x\in \Omega$ and $d={\mathrm {dist}}(x,\partial \Omega )>0$. For all $0<r<d$, let

Then, for all $0<r<d$, we have

$$\begin{aligned} u_{x,r}=\frac{1}{\alpha (n)}\int _{B_1(0)}u(x+r(y-x))\text {d}y \end{aligned}$$

so that

(2.17)

Therefore, we have by Fubini theorem and the duality $L^{n,1}/L^{\frac{n}{n-1},\infty }$ (see the estimate (7.8)) for all $0<t\le d$

$$\begin{aligned} \int _{0}^{t}\left| \frac{\mathrm{d}}{\mathrm{d}r}u_{x,r}\right| \text {d}r&\le \frac{1}{\alpha (n)}\int _{0}^t\frac{1}{r^n}\int _{B_r(x)}|\nabla u(y)|\text {d}{\mathscr {L}}^n(y)\mathrm{d}r\\&=\frac{1}{\alpha (n)}\int _{0}^t\int _{B_t(x)}\frac{1}{r^n}|\nabla u(y)|{\mathbf {1}}_{\left\{ |x-y|<r\right\} }\text {d}{\mathscr {L}}^n(y)\text {d}r\\&=\frac{1}{\alpha (n)}\int _{B_t(x)}|\nabla u(y)\left( \int _{|x-y|}^d\frac{\text {d}r}{r^n}\right) \text {d}{\mathscr {L}}^n(y)\\&\le \frac{1}{(n-1)\alpha (n)}\int _{B_t(x)}\frac{|\nabla u(y)|}{|x-y|^{n-1}}\text {d}{\mathscr {L}}^n(y)\\&\le \frac{1}{n^2\alpha (n)}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_t(x))}\left\| \frac{1}{|x-\,\cdot \,|^{n-1}}\right\| _{{\mathrm {L}}^{\frac{n}{n-1},\infty }(B_t(x))} \\&=\frac{1}{n\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_t(x))} \end{aligned}$$

as for all $x\in {\mathbb {R}}^n$

$$\begin{aligned} \left\| \frac{1}{|x-\,\cdot \,|^{n-1}}\right\| _{{\mathrm {L}}^{\frac{n}{n-1},\infty }({\mathbb {R}}^n)}=n\alpha (n)^{\frac{n}{n-1}}. \end{aligned}$$

(2.18)

Therefore, by the Sobolev embedding $W^{1,1}({\mathbb {R}})\subset C^0({\mathbb {R}})$, the function $(0,d]\rightarrow {\mathbb {R}}, r\mapsto u_{x,r}$ is continuous, and for all $0<s<t\le d$, we have

$$\begin{aligned} \left| u_{x,s}-u_{x,t}\right| \le \int _{s}^{t}\left| \frac{d}{\text {d}r}u_{x,r}\right| \text {d}r\le \frac{1}{n\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_t(x))}. \end{aligned}$$

(2.19)

Let $\left\{ r_n\right\} _{n\in {\mathbb {N}}}\subset (0,\infty )$ such that $r_n\underset{n\rightarrow \infty }{\longrightarrow }0$. Then (2.19) implies that

$$\begin{aligned} |u_{x,r_n}-u_{x,r_m}|\le \frac{1}{n\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{\max \left\{ r_n,r_m\right\} }(x))}\underset{n,m\rightarrow \infty }{\longrightarrow }0 \end{aligned}$$

which implies that $\left\{ u_{x,r_n}\right\} _{n\in {\mathbb {N}}}$ is a Cauchy sequence. Now, recall that by the Lebesgue differentiation theorem, for ${\mathscr {L}}^n$ almost all $x\in \Omega$, we have

$$\begin{aligned} u(x)=\lim \limits _{r\rightarrow 0}u_{x,r}. \end{aligned}$$

Therefore, for ${\mathscr {L}}^n$ almost all $x\in \Omega$ and for all $0<r<\text {d}(x)={\mathrm {dist}}(x,\partial \Omega )$, we have

$$\begin{aligned} |u(x)-u_{x,r}|\le \frac{1}{n\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_r(x))}. \end{aligned}$$

(2.20)

To prove that u is continuous, let $x,y\in \Omega$ such that (2.20) holds for x and y (the proof is an adaptation of the Hölder continuous embedding of Campanato spaces of the right indices). Furthermore, without loss of generality, we can assume that $x\ne y$, and $2|x-y|<\max \left\{ \text {d}(x),\text {d}(y)\right\}$, so that

$$\begin{aligned} B_{2|x-y|}(x)\cup B_{2|x-y|}(y)\subset \Omega . \end{aligned}$$

Therefore, if $r=|x-y|$, we have

$$\begin{aligned} \left| u(x)-u(y)\right|&\le |u(x)-u_{x,r}|+|u_{x,r}-u_{y,r}|+|u(y)-u_{y,r}|\nonumber \\&\le \frac{1}{n\alpha (n)^{\frac{1}{n}}}\left( \left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{|x-y|}(x))}+\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{|x-y|}(y))}\right) +|u_{x,r}-u_{y,r}| \end{aligned}$$

(2.21)

so we need only estimate $|u_{x,r}-u_{y,r}|$, as

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{|x-y|}(x))}+\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{|x-y|}(y))}\underset{y\rightarrow x}{\longrightarrow }0. \end{aligned}$$

We have

$$\begin{aligned} u_{x,r}-u_{y,r}&=\frac{1}{\alpha (n)r^n}\int _{B_r(x)}u(z_1)\text {d}{\mathscr {L}}^n(z)_1-\frac{1}{\alpha (n)r^n}\int _{B_r(y)}u(z_2)\text {d}{\mathscr {L}}^n(z_2)\nonumber \\&=\frac{1}{(\alpha (n)r^n)^2}\int _{B_r(x)\times B_r(y)}(u(z_1)-u(z_2))\text {d}{\mathscr {L}}^n(z_1)\mathrm{d}{\mathscr {L}}^n(z_2)\nonumber \\&=\frac{1}{(\alpha (n)r^n)^{2}}\int _{B_r(x)\times B_r(y)}\left( \int _{0}^1\nabla u(z_2+t(z_1-z_2))\cdot (z_1-z_2)\mathrm{d}t\right) \nonumber \\&\qquad \text {d}{\mathscr {L}}^n(z_1)\text {d}{\mathscr {L}}^n(z_2) \end{aligned}$$

(2.22)

Furthermore, for all $t\in [0,1]$ and $(z_1,z_2)\in B_r(x)\times B_r(y)$, we have $z_2+t(z_1-z_2)\in B_{2r}(x)$ and $|z_1-z_2|\le 2r$. Therefore, Fubini’s theorem implies that (by (7.8))

$$\begin{aligned}&\left| \int _{B_r(x)}\left( \int _{0}^{1}\nabla u(z_2+t(z_1-z_2))\cdot (z_1-z_2)\text {d}t\right) \text {d}{\mathscr {L}}^n(z_1)\right| \nonumber \\&\quad \le \int _{0}^{1}\left( \int _{B_r(x)}\frac{|\nabla u(z_2+t(z_1-z_2))|}{|z_1-z_2|^{n-1}}|z_1-z_2|^n\text {d}{\mathscr {L}}^n(z_1)\right) \text {d}t\nonumber \\&\quad \le \frac{1}{n}2^nr^n\int _{0}^1\left\| \nabla u(z_2+t(\,\cdot \,-z_2))\right\| _{{\mathrm {L}}^{n,1}(B_r(x))}\left\| \frac{1}{|\,\cdot \,-z_2|}\right\| _{{\mathrm {L}}^{\frac{n}{n-1},\infty }(B_r(x))}\text {d}t\nonumber \\&\quad \le 2^nr^n\alpha (n)^{\frac{n}{n-1}}\int _{0}^{1}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{2r}(x))}\text {d}t=2^nr^n\alpha (n)^{\frac{n}{n-1}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{2r}(x))}. \end{aligned}$$

(2.23)

Therefore, by (2.22) and (2.23), we find

$$\begin{aligned} |u_{x,r}-u_{y,r}|&\le \frac{1}{\alpha (n)r^n}\int _{B_r(y)} \frac{2^n}{\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{2|x-y|}(x))}\text {d}{\mathscr {L}}^n(z_2)\nonumber \\&=\frac{2^n}{\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{2|x-y|}(x))}. \end{aligned}$$

(2.24)

Furthermore, as the argument is symmetric in x and y notice that

$$\begin{aligned} |u_{x,r}-u_{y,r}|&\le \frac{2^n}{\alpha (n)^{\frac{1}{n}}}\min \left\{ \left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{2|x-y|}(x))},\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{2|x-y|}(y))}\right\} . \end{aligned}$$

Finally, thanks to (2.21) and (2.24), we get

$$\begin{aligned} |u(x)-u(y)|\le \frac{2^{n+1}}{\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{2|x-y|}(x))} \end{aligned}$$

(2.25)

which implies that u is continuous, with modulus of continuity at x

$$\begin{aligned} r\mapsto \frac{2^{n+1}}{\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega \,\cap \, B_{2r}(x))}. \end{aligned}$$

Now, for the $L^{\infty }$ bound, first consider the case $\Omega ={\mathbb {R}}^n$, and let $G:{\mathbb {R}}^n\times {\mathbb {R}}^n\rightarrow {\mathbb {R}}\cup \left\{ \infty \right\}$ be the Green’s function of the Laplacian on ${\mathbb {R}}^n$. Then

$$\begin{aligned} \nabla _y G(x,y)=\frac{1}{n\alpha (n)}\frac{1}{|x-y|^{n-1}}\in L^{\frac{n}{n-1},\infty }({\mathbb {R}}^n) \end{aligned}$$

and we have for all $x\in {\mathbb {R}}^n$

$$\begin{aligned} u(x)=\int _{{\mathbb {R}}^n}\Delta _y G(x,y)u(y)\text {d}y=-\int _{{\mathbb {R}}^n}\nabla _y G(x,y)\cdot \nabla u(y)\text {d}y \end{aligned}$$

and (2.18) implies that

$$\begin{aligned} \left\| u\right\| _{{\mathrm {L}}^{\infty }({\mathbb {R}}^n)}&\le \frac{n-1}{n^2}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}({\mathbb {R}}^n)}\left\| \nabla _y G(x,y)\right\| _{{\mathrm {L}}^{\frac{n}{n-1},\infty }({\mathbb {R}}^n)}\nonumber \\&=\frac{(n-1)}{n^3\alpha (n)}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}({\mathbb {R}}^n)}\left\| \frac{1}{|x-\,\cdot \,|^{n-1}}\right\| _{{\mathrm {L}}^{\frac{n}{n-1},\infty }({\mathbb {R}}^n)}\nonumber \\&=\frac{(n-1)}{n^2\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}({\mathbb {R}}^n)}\le \frac{1}{n\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}({\mathbb {R}}^N)} \end{aligned}$$

(2.26)

Now, (thanks to [3] IX.7) there exists a linear extension operator

$$\begin{aligned} P:\bigcup _{1\le p<\infty }W^{1,p}(\Omega )\rightarrow \bigcup _{1\le p<\infty }W^{1,p}({\mathbb {R}}^n) \end{aligned}$$

such that for $1\le p<\infty$ the restriction $P|W^{1,p}(\Omega )\rightarrow W^{1,p}({\mathbb {R}}^n)$ be a continuous linear operator. Then by identifying $W^{1,p}(\Omega )$ with a closed subset of $L^p({\mathbb {R}}^n)^{n+1}$, the Stein-Weiss interpolation theorem implies that for all P extends as a continuous linear operator $W^{1,(n,1)}(\Omega )$ into $W^{1,(n,1)}({\mathbb {R}}^n)$, as the Sobolev embedding $L^n(\Omega )\hookrightarrow L^q(\Omega )$ for all $1\le q<\infty$ shows that $\nabla u\in L^{n,1}(\Omega )$ implies that $u\in L^{n,1}(\Omega )$. Therefore, by (2.26), for all $u\in W^{1,(n,1)}(\Omega )$, we have

$$\begin{aligned} \left\| u\right\| _{{\mathrm {L}}^{\infty }(\Omega )}\le \left\| \nabla Pu\right\| _{{\mathrm {L}}^{\infty }({\mathbb {R}}^n)}&\le \frac{1}{n\alpha (n)^{\frac{1}{n}}}\left\| P u\right\| _{{\mathrm {L}}^{n,1}({\mathbb {R}}^n)}\le \Gamma _3\left( \left\| u\right\| _{{\mathrm {L}}^{n,1}(\Omega )}+\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega )}\right) \nonumber \\&\le \Gamma _3'(\left\| u\right\| _{{\mathrm {L}}^{n}(\Omega )}+\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega )}), \end{aligned}$$

(2.27)

where we have used in the last line the embedding $W^{1,n}(\Omega )\hookrightarrow L^{n,1}(\Omega )$.

Now, (2.27) implies by the classical Poincaré-Wirtinger inequality and the continuous embedding $L^{n,1}(\Omega )\hookrightarrow L^n(\Omega )$

$$\begin{aligned} \left\| u-\overline{u}_{\Omega }\right\| _{{\mathrm {L}}^{\infty }(\Omega )}&\le \Gamma _3'(\left\| u-\overline{u}_{\Omega }\right\| _{{\mathrm {L}}^{n}(\Omega )}+\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega )})\\&\le \Gamma _3'\left( \Gamma _3''\left\| \nabla u\right\| _{{\mathrm {L}}^{n}(\Omega )}+\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega )}\right) \le C_{4}(\Omega )\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega )} \end{aligned}$$

and this concludes the proof of the Lemma. $\square$

Now, we will need to refine the $L^{\infty }$ bound to obtain an estimate independent of the conformal class (bounded away from $-\infty$) of flat annuli in ${\mathbb {R}}^n$.

Proposition 2.7

Let $0<2r<R<\infty$ and $\Omega =B_R\setminus \overline{B}_r(0)\subset {\mathbb {R}}^n$. Then, there exists a universal constant $\Gamma _4=\Gamma _4(n)$ such that for all $u\in W^{1,(n,1)}(\Omega )$, we have

$$\begin{aligned} \left\| u-\overline{u}_{\Omega }\right\| _{{\mathrm {L}}^{\infty }(\Omega )}\le \Gamma _4(n)\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega )}. \end{aligned}$$

(2.28)

Remarks on the proof

By scaling invariance of the inequality of Lemma 2.6, the constant $C_{4}(\Omega (r))$ inequality (2.16) for annuli $\Omega (r)=B_{2r}\setminus \overline{B}_r(0)$ is independent of $0<r<\infty$, which allows one to introduce a dyadic decomposition of the annulus $\Omega =B_{R}\setminus \overline{B}_{r}(0)$ since the conformal class $\log \left( \frac{R}{r}\right) \ge \log (2)$ is bounded from below. Using once more the $L^{n,1}/L^{\frac{n}{n-1},\infty }$ duality and Fubini’s theorem, we deduce that the various averages can be controlled by the $L^{n,1}$ norm of $\nabla u$ which finally permits after a suitable decomposition to obtain the inequality (2.28).

Proof

First, observe that the $L^{\infty }$ norm and the (n, 1) norm of the gradient $\left\| \nabla \,\cdot \,\right\| _{{\mathrm {L}}^{n,1}(\Omega )}$ are scaling invariant (see (2.40) for the case $n=2$). Therefore, the constant $C_{4}(\Omega )$ in Theorem 2.1 is scaling invariant. In particular, there exists a universal constant $C_{4}'(n)=C_{4}(B_{2}\setminus B_1(0))$ such that for all $0<r<\infty$ and $u\in W^{1,(n,1)}(B_{2r}\setminus \overline{B}_r(0))$, we have

$$\begin{aligned} \left\| u-\overline{u}_{B_{2r}\setminus \overline{B}_r(0)}\right\| _{{\mathrm {L}}^{\infty }(\overline{B}_{2r}\setminus \overline{B}_r(0))}\le C_{4}'(n)\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{2r}\setminus \overline{B}_r(0))}. \end{aligned}$$

(2.29)

Now, as $2r<R$ let $J\in {\mathbb {N}}$ such that

$$\begin{aligned} 2^{J}r<R\le 2^{J+1}r. \end{aligned}$$

Then, we have

$$\begin{aligned} \Omega = B_{R}\setminus B_{\frac{R}{2}}(0)\cup \bigcup _{j=0}^{J-1}B_{2^{j+1}r}\setminus \overline{B}_{2^{j}r}(0). \end{aligned}$$

For the convenience of notation, let us write $\Omega _j=B_{2^{j+1}r}\setminus \overline{B}_{2^{j}r}$ for all $0\le j\le J-1$. Thanks to (2.29) for all $0\le j\le J$, we have

(2.30)

Now define for all $r<t<R$

For all $r<t<R$, thanks to a similar argument as given in (2.17), we have

Furthermore, if $r\le r_1<R$ is a fixed radius, thanks to the co-area formula, we have for ${\mathscr {L}}^1$ almost all $t\in (r_1,R)$

$$\begin{aligned} \int _{\partial B_t}|\nabla u|\,\text {d}{\mathscr {H}}^{n-1}=\frac{\mathrm{d}}{\text {d}t}\int _{r_1}^{t}\left( \int _{\partial B_s}|\nabla u|\text {d}{\mathscr {H}}^{n-1}\right) \text {d}{\mathscr {L}}^1(s)=\frac{\mathrm{d}}{\text {d}t}\int _{B_t\setminus \overline{B}_{r_1}(0)}|\nabla u|\mathrm{d}{\mathscr {L}}^n. \end{aligned}$$

Therefore, we have

$$\begin{aligned}&\int _{r_1}^{r_2}\left| \frac{\text {d}}{\text {d}t}u_t\right| \text {d}t\le \frac{1}{n\alpha (n)}\int _{r_1}^{r_2}\frac{1}{t^{n-1}}\left( \int _{\partial B_t}|\nabla u|\text {d}{\mathscr {H}}^{n-1}\right) \text {d}t\nonumber \\&\quad =\frac{1}{n\alpha (n)}\left[ \frac{1}{t^{n-1}}\int _{B_t\setminus B_{r_1}(0)}|\nabla u|\text {d}{\mathscr {L}}^n\right] _{r_1}^{r_2}+\frac{n-1}{n\alpha (n)}\int _{r_1}^{r_2}\frac{1}{t^n}\left( \int _{B_t\setminus B_{r_1}(0)}|\nabla u|d{\mathscr {L}}^n\right) \text {d}t\nonumber \\&\quad =\frac{1}{n\alpha (n)}\frac{1}{r_2^{n-1}}\int _{B_{r_2}\setminus \overline{B}_{r_1}(0)}|\nabla u|\text {d}{\mathscr {L}}^n+\frac{n-1}{n\alpha (n)}\int _{r_1}^{r_2}\int _{B_{r_2}\setminus \overline{B}_{r_1}(0)}\frac{|\nabla u(x)|}{t^{n}}{\mathrm {1}}_{\left\{ r_1\le |x|\le t\right\} }\text {d}{\mathscr {L}}^n(x) \text {d}t. \end{aligned}$$

(2.31)

Furthermore, observe that

$$\begin{aligned} \int _{B_{r_2}\setminus B_{r_1}(0)}\frac{|\nabla u(x)|}{|x|^{n-1}}\text {d}{\mathscr {L}}^n(x)&\le \frac{1}{n} \left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{r_2}\setminus \overline{B}_{r_1}(0))}\left\| \frac{1}{|x|^{n-1}}\right\| _{{\mathrm {L}}^{\frac{n}{n-1},\infty }(B_{r_2}\setminus \overline{B}_{r_1}(0))}\nonumber \\&\le \alpha (n)^{\frac{n}{n-1}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{r_2}\setminus \overline{B}_{r_1}(0))} \end{aligned}$$

(2.32)

while by Fubini’s theorem

$$\begin{aligned}&\int _{r_1}^{r_2}\int _{B_{r_2}\setminus \overline{B}_{r_1}(0)}\frac{|\nabla u(x)|}{t^{n}}{\mathrm {1}}_{\left\{ r_1\le |x|\le t\right\} }\text {d}{\mathscr {L}}^n(x) \text {d}t\nonumber \\&\quad =\int _{B_{r_2}\setminus \overline{B}_{r_1}(0)}|\nabla u(x)|\left( \int _{|x|}^{r_2}\frac{\text {d}t}{t^{n-1}}\text {d}t\right) \text {d}{\mathscr {L}}^n(x)\nonumber \\&\quad =\frac{1}{n-1}\int _{B_{r_2}\setminus \overline{B}_{r_1}(0)}|\nabla u(x)|\left( \frac{1}{|x|^{n-1}}-\frac{1}{r_2^{n-1}}\right) . \end{aligned}$$

(2.33)

Finally, we get by (2.31), (2.32), (2.33), (2.34) and (2.18)

$$\begin{aligned} \int _{r_1}^{r_2}\left| \frac{d}{\text {d}t}u_t\right| \text {d}t&\le \frac{1}{n\alpha (n)}\frac{1}{r_2^{n-1}}\int _{B_{r_2}\setminus \overline{B}_{r_1}(0)}|\nabla u|\mathrm{d}{\mathscr {L}}^n \nonumber \\&\quad +\frac{n-1}{n\alpha (n)}\int _{r_1}^{r_2}\int _{B_{r_2}\setminus \overline{B}_{r_1}(0)}\frac{|\nabla u(x)|}{t^{n}}{\mathrm {1}}_{\left\{ r_1\le |x|\le t\right\} }\text {d}{\mathscr {L}}^n(x) \text {d}t\nonumber \\&=\frac{1}{n\alpha (n)}\frac{1}{r_2^{n-1}}\int _{B_{r_2}\setminus \overline{B}_{r_1}(0)}|\nabla u|\text {d}{\mathscr {L}}^n\nonumber \\&\quad +\frac{1}{n\alpha (n)}\int _{B_{r_2}\setminus \overline{B}_{r_1}(0)}|\nabla u(x)|\left( \frac{1}{|x|^{n-1}}-\frac{1}{r_{2}}\right) \text {d}\mathscr {L}^{n}(x)\nonumber \\&=\frac{1}{n\alpha (n)}\int _{B_{r_2}\setminus B_{r_1}(0)}\frac{|\nabla u(x)|}{|x|^{n-1}}\text {d}{\mathscr {L}}^n(x) \le \frac{1}{n\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{r_2}\setminus \overline{B}_{r_1}(0))} \end{aligned}$$

(2.34)

Therefore, we have for all $r\le r_1<r_2\le R$

$$\begin{aligned} \left| u_{r_1}-u_{r_2}\right| \le \frac{1}{n\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{r_2}\setminus \overline{B}_{r_1}(0))}. \end{aligned}$$

(2.35)

Furthermore, recalling that $\beta (n)={\mathscr {H}}^{n-1}(S^{n-1})=n\alpha (n)$, we obtain for all $r\le s<t\le R$, thanks to (2.35) that

and the reverse inequality (given by (2.35))

$$\begin{aligned} \int _{\partial B_{\rho }}u\,\text {d}{\mathscr {H}}^{n-1}&\ge \frac{\rho ^{n-1}}{t^{n-1}}\int _{\partial B_t}u\,\text {d}{\mathscr {H}}^{n-1}\\&\quad -\frac{\beta (n)}{n\alpha (n)^{\frac{1}{n}}}\rho ^{n-1}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(B_{t}\setminus \overline{B}_{s}(0))}\quad \text {for all}\;\, s<\rho <t \end{aligned}$$

shows that for all $r\le s<t\le R$

Therefore, by the triangle inequality, we finally obtain that for all $0\le j\le J-1$,

(2.36)

and likewise,

$$\begin{aligned} \left| \overline{u}_{B_R\setminus B_{R/2}(0)}-\overline{u}_{\Omega }\right| \le \frac{3}{n\alpha (n)^{\frac{1}{n}}}\left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega )}. \end{aligned}$$

(2.37)

Finally, thanks to (2.30), (2.36) and (2.37), we have

$$\begin{aligned} \left\| u-\overline{u}_{\Omega }\right\| _{{\mathrm {L}}^{\infty }(\Omega )}\le \left( C_{4}'(n)+\frac{3}{n\alpha (n)^{\frac{1}{n}}}\right) \left\| \nabla u\right\| _{{\mathrm {L}}^{n,1}(\Omega )} \end{aligned}$$

and this concludes the proof of the Proposition. $\square$

We now come back to the proof of Theorem 2.1.

Remarks on the proof

The proof closely follows the one of [2], using the $L^{2,1}$ estimate in lieu of the $L^2$ one, using the previous Lemma (2.5) to prove the inequality (2.2), and Proposition 2.7 for the inequality (2.3).

Proof of Theorem 2.1

Thanks to Lemma IV.1 [2], there exists a universal constant $\Gamma _6=\Gamma _6(n)>0$ and an extension ${\widetilde{\vec {n}}}:B_R(0)\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ of $\vec {n}$ such that

$$\begin{aligned} \left\{ \begin{aligned}&{\widetilde{\vec {n}}}=\vec {n}\quad \text {on}\;\, \Omega =B_R\setminus \overline{B}_r(0)\\&\big \Vert \nabla {\widetilde{\vec {n}}}\big \Vert _{{\mathrm {L}}^{2,\infty }(B_R(0))}\le \Gamma _6(n) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}\right. \end{aligned}$$

(2.38)

Therefore, by Lemma IV.3 of [2], there exists a universal constant $\Gamma _7=\Gamma _7(n)$ and a moving Coulomb frame $(\vec {e}_1,\vec {e}_2)\in W^{1,2}(B_R(0),S^{n-1})\times W^{1,2}(B_R(0),S^{n-1})$ such that

$$\begin{aligned} \left\{ \begin{aligned}&{\widetilde{\vec {n}}}=\star \left( \vec {e}_1\wedge \vec {e}_2\right) \quad {{\,\mathrm{div}\,}}\left( \vec {e}_1\cdot \nabla \vec {e}_2\right) =0\\&\left\| \nabla \vec {e}_1\right\| _{{\mathrm {L}}^{2}(B_R(0))}^2+\left\| \nabla \vec {e}_2\right\| _{{\mathrm {L}}^{2}(B_R(0))}^2\le \Gamma _7(n)\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}^2. \end{aligned}\right. \end{aligned}$$

(2.39)

Furthermore, notice that for all $u\in W^{1,{(2,1)}}_{{\mathrm {loc}}}({\mathbb {R}}^2)$, and for all $\rho >0$, we have

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_{\rho }(0))}&=4\int _{0}^{\infty }\left( {\mathscr {L}}^2(B_{\rho }(0)\cap \left\{ x:|\nabla u(x)|>t\right\} )\right) ^{\frac{1}{2}}\text {d}{\mathscr {L}}^{1}(t)\nonumber \\&=4\int _{0}^{\infty }\left( \int _{B_{\rho }(0)}{\mathrm {1}}_{\left\{ x:|\nabla u(x)|>t\right\} }\text {d}{\mathscr {L}}^2(x)\right) ^{\frac{1}{2}}\text {d}{\mathscr {L}}^{1}(t)\nonumber \\&=4\int _{0}^{\infty }\left( \int _{B_1(0)}{\mathrm {1}}_{\left\{ y:|\nabla (u\circ \varphi _{\rho })(y)|>\rho t\right\} }\rho ^2 \text {d}{\mathscr {L}}^2(y)\right) ^{\frac{1}{2}}\text {d}{\mathscr {L}}^1(t)\nonumber \\&=4\int _{0}^{\infty }\left( \int _{B_1(0)}{\mathrm {1}}_{\left\{ y:|\nabla (u\circ \varphi _{\rho })(y)|>s\right\} }\rho ^2 \text {d}{\mathscr {L}}^2(y)\right) ^{\frac{1}{2}}\rho ^{-1}\, \text {d}{\mathscr {L}}^1(s)\nonumber \\&=\left\| \nabla (u\circ \varphi _{\rho })\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}. \end{aligned}$$

(2.40)

where $\varphi _{\rho }(y)=\rho y$. Now, if $\mu :B_R(0)\rightarrow {\mathbb {R}}$ is the unique solution of the system

$$\begin{aligned} \left\{ \begin{aligned} \Delta \mu&=\nabla ^{\perp }\vec {e}_1\cdot \nabla \vec {e}_2\quad&\text {in}\;\, B_R(0)\\ \mu&=0\quad&\text {on}\;\,\partial B_R(0) \end{aligned}\right. \end{aligned}$$

(2.41)

then ${\widetilde{\mu }}=\mu \circ \varphi _R$ solves (with evident notations)

$$\begin{aligned} \left\{ \begin{aligned} \Delta {\widetilde{\mu }}&=\nabla ^{\perp }\widetilde{\vec {e}_1}\cdot \nabla \widetilde{\vec {e}_2}\quad&\text {in}\;\, B_1(0)\\ {\widetilde{\mu }}&=0\quad&\text {on}\;\, S^1 \end{aligned}\right. \end{aligned}$$

Therefore, the improved Wente inequality ([11], 3.4.1) shows that there exists a universal constant $\Gamma _8>0$ such that

$$\begin{aligned} \left\| \nabla \mu \right\| _{{\mathrm {L}}^{2,1}(B_R(0))}=\left\| \nabla {\widetilde{\mu }}\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}&\le \Gamma _8\left\| \nabla \widetilde{\vec {e}_1}\right\| _{{\mathrm {L}}^{2}(B_1(0))}\left\| \nabla \widetilde{\vec {e}_2}\right\| _{{\mathrm {L}}^{2}(B_1(0))} \nonumber \\&=\Gamma _8\left\| \nabla \vec {e}_1\right\| _{{\mathrm {L}}^{2}(B_R(0))}\left\| \nabla \vec {e}_2\right\| _{{\mathrm {L}}^{2}(B_R(0))}\nonumber \\&\le \frac{1}{2}{\Gamma _7(n)}\Gamma _8\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x. \end{aligned}$$

(2.42)

Furthermore, notice that we also have the optimal inequality

$$\begin{aligned} \left\| \nabla \mu \right\| _{{\mathrm {L}}^{2}(B_R(0))}&\le \frac{1}{4}\sqrt{\frac{3}{\pi }}\left\| \nabla \vec {e}_1\right\| _{{\mathrm {L}}^{2}(B_R(0))}\left\| \nabla \vec {e}_2\right\| _{{\mathrm {L}}^{2}(B_R(0))} \le \frac{1}{8}\sqrt{\frac{3}{\pi }}\Gamma _7(n)\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x. \end{aligned}$$

(2.43)

Now, let $\upsilon =\lambda -\mu$ on $\Omega =B_R\setminus \overline{B}_r(0)$. Then, $\upsilon$ is harmonic on $\Omega$ and $\upsilon =\lambda$ on $\partial B_R(0)$. Then, as $\upsilon$ is harmonic, there exists $d\in {\mathbb {R}}$ and $\left\{ a_k\right\} _{k\in {\mathbb {Z}}}\subset {\mathbb {C}}$ such that

$$\begin{aligned} \upsilon (\rho ,\theta )=a_0+d\,\log \rho +\sum _{k\in {\mathbb {Z}}^{*}}\left( a_k\rho ^k+\overline{a_{-k}}\rho ^{-k}\right) e^{ik\theta }. \end{aligned}$$

Now, noticing that for all $r<\rho <R$

$$\begin{aligned} d=\frac{1}{2\pi }\int _{\partial B_{\rho }}\partial _{\nu }\upsilon , \end{aligned}$$

(2.44)

this implies that $\upsilon -d\log |z|$ satisfies the hypothesis of Proposition 2.5. Therefore, using the identity $\upsilon =\lambda -\mu$, the inequalities (2.43) and $\left\| \,\cdot \,\right\| _{{\mathrm {L}}^{2,\infty }(\,\cdot \,)}\le 2\left\| \,\cdot \,\right\| _{{\mathrm {L}}^{2}(\,\cdot \,)}$, we have for all $\left( \dfrac{r}{R}\right) ^{\frac{1}{3}}<\alpha <\dfrac{1}{4}$

$$\begin{aligned}&\left\| \nabla (\upsilon -d\log |z|)\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r})}\le 24\,\Gamma _1\sqrt{\alpha }\left\| \nabla (\upsilon -d\log |z|)\right\| _{{\mathrm {L}}^{2,\infty }(B_R\setminus \overline{B_r}(0))}\nonumber \\&\quad \le 24\,\Gamma _1\sqrt{\alpha }\left( \left\| \nabla (\lambda -d\log |z|)\right\| _{{\mathrm {L}}^{2,\infty }(B_R\setminus \overline{B}_r(0))}+\left\| \nabla \mu \right\| _{{\mathrm {L}}^{2,\infty }(B_R\setminus \overline{B}_r(0))}\right) \nonumber \\&\quad \le 24\,\Gamma _1\sqrt{\alpha }\left( \left\| \nabla (\lambda -d\log |z|)\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+2\left\| \nabla \mu \right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \nonumber \\&\quad \le 24\,\Gamma _1\sqrt{\alpha }\left( \left\| \nabla (\lambda -d\log |z|)\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\frac{1}{4}\sqrt{\frac{3}{\pi }}\Gamma _7(n)\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) . \end{aligned}$$

(2.45)

Furthermore, notice that by the co-area formula, for all $s\in (r,R)$ such that $2s<R$, we have

$$\begin{aligned} \int _{B_{2s}\setminus \overline{B}_{s}(0)}|\nabla \upsilon (x)|\mathrm{d}x&=\int _{s}^{2s}\left( \rho \int _{\partial B_{\rho }}|\nabla \upsilon |\text {d}{\mathscr {H}}^1\right) \frac{\text {d}\rho }{\rho } \ge \log (2)\inf _{s< \rho < 2s}\left( \rho \int _{\partial B_{\rho }}|\nabla \upsilon |\text {d}{\mathscr {H}}^1\right) . \end{aligned}$$

Therefore, there exists $\rho \in (s,2s)$ such that

$$\begin{aligned} \int _{\partial B_{\rho }}|\nabla \upsilon |\text {d}{\mathscr {H}}^1&\le \frac{1}{\log (2)\rho }\int _{B_{2s}\setminus \overline{B}_s(0)}|\nabla \upsilon (x)|\mathrm{d}x \\&\le \frac{1}{\log (2)\rho }\left\| 1\right\| _{{\mathrm {L}}^{2,1}(B_{2s}\setminus \overline{B}_s)}\left\| \nabla \upsilon \right\| _{{\mathrm {L}}^{2,\infty }(B_{2s}\setminus \overline{B}_s(0))}\\&=\frac{1}{\log (2)\rho }4\sqrt{3\pi }s\left\| \nabla \upsilon \right\| _{{\mathrm {L}}^{2,\infty }(B_{2s}\setminus \overline{B}_s(0))}\\&\le \frac{4\sqrt{3\pi }}{\log (2)}\left( \left\| \nabla \lambda \right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+2\left\| \nabla \mu \right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \\&\le \frac{4\sqrt{3\pi }}{\log (2)}\left( \left\| \nabla \lambda \right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\frac{1}{4}\sqrt{\frac{3}{\pi }}\Gamma _7(n)\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) . \end{aligned}$$

This implies by (2.44) that

$$\begin{aligned} |d|\le \frac{2}{\log (2)}\sqrt{\frac{3}{\pi }}\left( \left\| \nabla \lambda \right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\frac{1}{4}\sqrt{\frac{3}{\pi }}\Gamma _7(n)\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) . \end{aligned}$$

(2.46)

As $\left\| \nabla \log |z|\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}= 2\sqrt{\pi }$, by (2.45) and (2.46) there exists a universal constant $\Gamma _9=\Gamma _9(n)$ such that

$$\begin{aligned} \left\| \nabla (\upsilon -d\log |z|)\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r})}\le \Gamma _9(n)\sqrt{\alpha }\left( \left\| \nabla \lambda \right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) . \end{aligned}$$

(2.47)

Finally, putting together (2.42), (2.47) and recalling that $\lambda =\mu +\upsilon$, we have for all $\left( \dfrac{r}{R}\right) ^{\frac{1}{4}}\le \alpha <\dfrac{1}{4}$

$$\begin{aligned}&\left\| \nabla (\lambda -d\log |z|)\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}\le \left\| \nabla (\upsilon -d\log |z|)\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r})}\nonumber \\&\qquad +\left\| \nabla \mu \right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus B_{\alpha ^{-1}}(0))}\nonumber \\&\quad \le \Gamma _9(n)\sqrt{\alpha }\left( \left\| \nabla \lambda \right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) +\frac{1}{2}\Gamma _7(n)\Gamma _8\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x. \end{aligned}$$

(2.48)

Now, we estimate for $r\le \rho < R$ the following quantity

$$\begin{aligned} \left| d-\frac{1}{2\pi }\int _{\partial B_{\rho }}\partial _{\nu }\lambda \,\text {d}{\mathscr {H}}^1\right| =\left| \frac{1}{2\pi }\int _{\partial B_{\rho }}\partial _{\nu }\mu \,\text {d}{\mathscr {H}}^1\right| . \end{aligned}$$

We have, recalling that $\mu$ is well defined on $B_R(0)$ and satisfies (2.41), we find

$$\begin{aligned} 0&=\int _{B_R\setminus B_{\rho }}\mu (x)\,\Delta \log \left( \frac{|x|}{R}\right) \mathrm{d}x =-\log \left( \frac{R}{\rho }\right) \int _{\partial B_{\rho }}\partial _{\nu }\mu \,\text {d}{\mathscr {H}}^1\nonumber \\&\quad +\int _{B_R\setminus B_{\rho }(0)}\Delta \mu \,\log \left( \frac{|x|}{R}\right) \mathrm{d}x\nonumber \\&=-\log \left( \frac{R}{\rho }\right) \int _{\partial B_{\rho }}\partial _{\nu }\mu \,\text {d}{\mathscr {H}}^1+\int _{B_R(0)}\Delta \mu \log \left( \frac{|x|}{R}\right) \mathrm{d}x\nonumber \\&\quad -\int _{B_\rho (0)}\left( \nabla ^{\perp }\vec {e}_1\cdot \nabla \vec {e}_2\right) \log \left( \frac{|x|}{R}\right) \mathrm{d}x. \end{aligned}$$

(2.49)

First, the previous estimate (2.42) yields

$$\begin{aligned} \left| \int _{B_R(0)}\Delta \mu \log \left( \frac{|x|}{R}\right) \mathrm{d}x\right|&=\left| \int _{B_R(0)}\nabla \mu \cdot \nabla \log |x| \mathrm{d}x\right| \nonumber \\&\le \frac{1}{2}\left\| \nabla \mu \right\| _{{\mathrm {L}}^{2,1}(B_R(0))} \left\| \frac{1}{|x|}\right\| _{{\mathrm {L}}^{2,\infty }(B_R(0))}\nonumber \\&\le \frac{\sqrt{\pi }}{2}\Gamma _7(n)\Gamma _8\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x. \end{aligned}$$

(2.50)

Now, using once more Lemma IV.3 of [2], we see that exists a Coulomb moving frame $(\vec {f_1},\vec {f}_2)\in W^{1,2}(B_{\rho }(0),S^{n-1})\times W^{1,2}(B_{\rho }(0),S^{n-1})$ such that

$$\begin{aligned} {\widetilde{\vec {n}}}=\star (\vec {f}_1\wedge \vec {f}_2) \end{aligned}$$

and using the same inequalities as in (2.38) and (2.39)

$$\begin{aligned}&\left\| \nabla \vec {f}_1\right\| _{{\mathrm {L}}^{2}(B_{\rho }(0))}^2+\left\| \nabla \vec {f}_2\right\| _{{\mathrm {L}}^{2}(B_{\rho }(0))}^2\le \Gamma _7(n)\left\| \nabla {\widetilde{\vec {n}}}\right\| _{{\mathrm {L}}^{2}(B_{\rho }(0))}^2\nonumber \\&\quad =\Gamma _7(n)\int _{B_r(0)}|\nabla {\widetilde{\vec {n}}}|^2\mathrm{d}x+\Gamma _7(n)\int _{B_{\rho }\setminus \overline{B}_r(0)}|\nabla \vec {n}|^2\mathrm{d}x\nonumber \\&\quad \le \Gamma _6(n)\Gamma _7(n)\int _{B_{2r}\setminus \overline{B}_r(0)}|\nabla \vec {n}|^2\mathrm{d}x+\Gamma _7(n)\int _{B_{\rho }\setminus \overline{B}_r(0)}|\nabla \vec {n}|^2\mathrm{d}x \nonumber \\&\quad \le (1+\Gamma _6(n))\Gamma _7(n)\int _{B_{\max \left\{ \rho ,2r\right\} }\setminus \overline{B}_r(0)}|\nabla \vec {n}|^2\mathrm{d}x. \end{aligned}$$

(2.51)

Now, let $\psi$ be the solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta \psi&=\nabla ^{\perp }\vec {f}_1\cdot \nabla \vec {f}_2\quad&\text {in}\;\, B_{\rho }(0)\\ \psi&=0\quad&\text {on}\;\, \partial B_{\rho }(0). \end{aligned}\right. \end{aligned}$$

As in (2.51), we get

$$\begin{aligned} \left\| \nabla \psi \right\| _{{\mathrm {L}}^{2,1}(B_\rho (0))}\le \frac{1}{2}{\Gamma _7(n)}\Gamma _8\int _{B_{\max \left\{ \rho ,2r\right\} }\setminus \overline{B}_r(0)}|\nabla \vec {n}|^2\mathrm{d}x. \end{aligned}$$

(2.52)

Furthermore, we have

$$\begin{aligned}&\int _{B_{\rho }(0)}\left( \nabla ^{\perp }\vec {e}_1\cdot \nabla \vec {e}_2\right) \log \left( \frac{|x|}{R}\right) \mathrm{d}x\nonumber \\&\quad =\int _{B_{\rho }(0)}\left( \nabla ^{\perp }\vec {f}_1\cdot \nabla \vec {f}_2\right) \log \left( \frac{|x|}{R}\right) \mathrm{d}x =\int _{B_{\rho }(0)}\Delta \psi \,\log \left( \frac{|x|}{R}\right) \mathrm{d}x\nonumber \\&\quad =-\log \left( \frac{R}{\rho }\right) \int _{\partial B_{\rho }}\partial _{\nu }\psi \,\text {d}{\mathscr {H}}^1-\int _{B_{\rho }}\nabla \psi \cdot \nabla \log |x|\mathrm{d}x \end{aligned}$$

(2.53)

while by the Cauchy–Schwarz inequality

$$\begin{aligned} \left| \int _{\partial B_{\rho }}\partial _{\nu }\psi \,\text {d}{\mathscr {H}}^1\right|&=\left| \int _{B_{\rho }(0)}\Delta \psi \,\mathrm{d}x\right| =\left| \int _{B_{\rho }(0)}\nabla ^{\perp }\vec {f}_1\cdot \nabla \vec {f}_2\mathrm{d}x\right| \nonumber \\&\quad \le \frac{1}{2}(1+\Gamma _6(n))\Gamma _7(n)\int _{B_{\max \left\{ \rho ,2r\right\} }\setminus B_r(0)}|\nabla \vec {n}|^2\mathrm{d}x. \end{aligned}$$

(2.54)

We estimate as previously by (2.52)

$$\begin{aligned}&\left| \int _{B_{\rho }}\nabla \psi \cdot \nabla \log |x|\mathrm{d}x\right| \le \frac{1}{2} \left\| \nabla \psi \right\| _{{\mathrm {L}}^{2,1}(B_{\rho }(0))}\left\| \frac{1}{|x|}\right\| _{{\mathrm {L}}^{2,\infty }(B_\rho (0))}\nonumber \\&\quad \le \frac{\sqrt{\pi }}{2}\Gamma _7(n)\Gamma _8\int _{B_{\max \left\{ \rho ,2r\right\} }\setminus \overline{B}_r(0)}|\nabla \vec {n}|^2\mathrm{d}x. \end{aligned}$$

(2.55)

Therefore, (2.53), (2.54) and (2.55) yield

$$\begin{aligned}&\left| \int _{B_{\rho }}\left( \nabla ^{\perp }\vec {e}_1\cdot \nabla \vec {e}_2\right) \log \left( \frac{|x|}{R}\right) \mathrm{d}x\right| \nonumber \\&\quad \le \left( \frac{1}{2}(1+\Gamma _6(n))\Gamma _7(n))\log \left( \frac{R}{\rho }\right) +\frac{\sqrt{\pi }}{2}\Gamma _7(n)\Gamma _8\right) \int _{B_{\max \left\{ \rho ,2r\right\} }\setminus \overline{B}_r(0)}|\nabla \vec {n}|^2\mathrm{d}x. \end{aligned}$$

(2.56)

Finally, by (2.49), (2.50) and (2.56), we obtain for some universal constant $\Gamma _0=\Gamma _0(n)$

$$\begin{aligned} \frac{1}{2\pi }\left| \int _{\partial B_{\rho }}\partial _{\nu }\mu \,\text {d}{\mathscr {H}}^1\right| \le \Gamma _0(n)\left( \int _{B_{\max \left\{ \rho ,2r\right\} }\setminus \overline{B}_r(0)}|\nabla \vec {n}|^2\mathrm{d}x+\frac{1}{\log \left( \frac{R}{\rho }\right) }\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) \end{aligned}$$

(2.57)

which completes the proof of the theorem, up to the $L^{\infty }$ estimate which is a direct consequence of the inequality $4r<R$ and of Proposition 2.7. $\square$

3 Pointwise Expansion of the Conformal Factor and of the Immersion

3.1 Case of One Bubbling Domain

In the next Theorem, we obtain an integrality result for the multiplicity of a sequence of weak immersions from annuli converging strongly outside of the origin.

Theorem 3.1

Let $\{\vec {\Phi }_k\}_{k\in {\mathbb {N}}}$ be a sequence of smooth conformal immersions from the disk $B_1(0)\subset {\mathbb {C}}$ into ${\mathbb {R}}^n$, let

$$\begin{aligned} e^{\lambda _k}=\frac{1}{\sqrt{2}}|\nabla \vec {\Phi }_k| \end{aligned}$$

be the conformal factor of $\vec {\Phi }_k$, and $\left\{ \rho _k\right\} _{k\in {\mathbb {N}}}\subset (0,1)$ be such that $\rho _k\underset{k\rightarrow \infty }{\longrightarrow }0$, $\Omega _k=B_1\setminus \overline{B}_{\rho _k}(0)$ and assume that

$$\begin{aligned} \sup _{k\in {\mathbb {N}}}\int _{B_1(0)\setminus \overline{B}_{\rho _k}(0)}|\nabla \vec {n}_k|^2\mathrm{d}x\le \varepsilon _1(n),\qquad \sup _{k\in {\mathbb {N}}}\left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k)}<\infty \end{aligned}$$

where $\varepsilon _1(n)$ is given by the proof of Theorem 2.1. Define for all $0<\alpha <1$ and $k\in {\mathbb {N}}$ large enough $\Omega _k(\alpha )=B_{\alpha }\setminus \overline{B}_{\alpha ^{-1}\rho _k}(0)$, and assume that

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\int _{\Omega _k(\alpha )}|\nabla \vec {n}_k|^2\mathrm{d}x=0 \end{aligned}$$

and that there exists a $W^{2,2}_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} )\cap C^{\infty }(B_1(0)\setminus \left\{ 0\right\} )$ immersion $\vec {\Phi }_{\infty }$ such that

$$\begin{aligned} \log |\nabla \vec {\Phi }_{\infty }|\in L^{\infty }_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} ) \end{aligned}$$

and $\vec {\Phi }_k\underset{k\rightarrow \infty }{\longrightarrow }\vec {\Phi }_{\infty }$ in $C^l_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} )$ (for all $l\in {\mathbb {N}}$). Then, there exists an integer $\theta _0\ge 1$, $\mu _k\in W^{1,(2,1)}(B_1(0))$ such that

$$\begin{aligned} \left\| \nabla \mu _k\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}\le \frac{1}{2}{\Gamma _7(n)}\Gamma _8\int _{\Omega _k}|\nabla \vec {n}_k|^2\mathrm{d}x \end{aligned}$$

and a harmonic function $\nu _k$ on $\Omega _k$ such that $\nu _k=\lambda _k$ on $\partial B_1(0)$, $\lambda _k=\mu _k+\nu _k$ on $\Omega _k$ and such that for all $0<\alpha <1$ and such that for all $k\in {\mathbb {N}}$ sufficiently large

$$\begin{aligned} \left\| \nabla (\nu _k-(\theta _0-1)\log |z|)\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\le \Gamma _{10}\left( \sqrt{\alpha }\left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k)}+\int _{\Omega _k}|\nabla \vec {n}_k|^2\mathrm{d}x\right) \end{aligned}$$

for some universal constant $\Gamma _{10}=\Gamma _{10}(n)$. Furthermore, we have for all $\rho _k\le r_k\le 1$ and k large enough

$$\begin{aligned} \frac{1}{2\pi }\int _{\partial B_{r _k}}*\, \mathrm{d}\nu _k=\theta _0-1. \end{aligned}$$

Remarks on the proof

In Step 1, we first use the classical fact that branch points of Willmore surfaces are positive integers, Theorem 2.1 and the strong convergence outside of 0 to show that the multiplicity $d_k$ converges towards a non-negative integer.

In Step 2, as in [29] (see Lemma A.2, A.3 and A.5), we construct a moving frame that allows us to obtain a precise expansion of $\partial _{z}\vec {\Phi }_k$ in the annular region and show how the existence of a holomorphic function implies in virtue of the first step that for k large enough, the multiplicity must be an integer.

Proof

First, applying Lemma A.5 of [29], we deduce that there exists an integer $\theta _0\ge 1$ and $\vec {A}_0\in {\mathbb {C}}^{n}\setminus \left\{ 0\right\}$ such that

$$\begin{aligned}&\vec {\Phi }_{\infty }(z)={\mathrm {Re}}\,\left( \vec {A}_0z^{\theta _0}\right) +o(|z|^{\theta _0})\nonumber \\&\partial _{z}\vec {\Phi }_{\infty }(z)=\frac{\theta _0}{2}\vec {A}_0z^{\theta _0-1}+o(|z|^{\theta _0-1}). \end{aligned}$$

(3.1)

Step 1: Asymptotic Integrality

First, define $\Omega _k(\alpha )=B_{\alpha }\setminus \overline{B}_{\alpha ^{-1}\rho _k}(0)$ and recall that by Theorem 2.1, we have (applying the inequality on $\Omega _{\alpha }$) for all $\alpha ^{-1} \rho _k<\rho <\alpha$

$$\begin{aligned}&\left| d_k-\frac{1}{2\pi }\int _{\partial B_{\rho }}\partial _{\nu }\lambda \,\text {d}{\mathscr {H}}^1\right| \\&\quad \le \Gamma _0\left( \int _{B_{\max \left\{ \rho ,2\alpha ^{-1}\rho _k\right\} }\setminus B_{\alpha ^{-1}\rho _k}(0)}|\nabla \vec {n}|^2\mathrm{d}x+\frac{1}{\log \left( \frac{\alpha ^2}{\rho _k}\right) }\int _{\Omega _k(\alpha )}|\nabla \vec {n}|^2\mathrm{d}x\right) . \end{aligned}$$

Now, taking $\rho =\alpha ^2$, we get

$$\begin{aligned}&\left| d_k-\frac{1}{2\pi }\int _{\partial B_{\alpha ^2}}\partial _{\nu }\lambda _k\,\text {d}{\mathscr {H}}^1\right| \\&\quad \le \Gamma _0\left( \int _{B_{\alpha ^2}\setminus B_{\alpha ^{-1}\rho _k}}|\nabla \vec {n}_k|^2\mathrm{d}x+\frac{1}{\log \left( \frac{\alpha ^2}{\rho _k}\right) }\int _{\Omega _k(\alpha )}|\nabla \vec {n}_k|^2\mathrm{d}x\right) . \end{aligned}$$

Therefore, the no-neck energy (see [2])

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\int _{\Omega _k(\alpha )}|\nabla \vec {n}_k|^2\mathrm{d}x=0 \end{aligned}$$

implies that

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left| d_k-\frac{1}{2\pi }\int _{\partial B_{\alpha ^2}}\partial _{\nu }\lambda _k\,\text {d}{\mathscr {H}}^1\right| =0. \end{aligned}$$

Furthermore, as $\vec {\Phi }_{\infty }$ has a branch point of order $\theta _0-1\ge 0$ at $z=0$, we have the expansion for some $\beta \in {\mathbb {R}}$

$$\begin{aligned} \lambda _{\infty }(z)=(\theta _0-1)\log |z|+\beta +O(|z|) \end{aligned}$$

we have by the strong convergence

$$\begin{aligned} \frac{1}{2\pi }\int _{\partial B_{\alpha ^2}}\partial _{\nu }\lambda _k\,\text {d}{\mathscr {H}}^1\underset{k\rightarrow \infty }{\longrightarrow }\frac{1}{2\pi }\int _{\partial B_{\alpha ^2}}\partial _{\nu }\lambda _{\infty }\,\text {d}{\mathscr {H}}^1=\theta _0-1+O(\alpha ^2). \end{aligned}$$

Finally, this implies that

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }|d_k-(\theta _0-1)|=0. \end{aligned}$$

(3.2)

Now, recalling that $d_k$ is independent of $\alpha >0$ (as it corresponds to the coefficient in front of the logarithm of the associated harmonic function $\nu _k$ on $B_1\setminus \overline{B}_{\alpha ^{-1}\rho _k}(0)$), we deduce that (3.2) implies that

$$\begin{aligned} d_k\underset{k\rightarrow \infty }{\longrightarrow }\theta _0-1. \end{aligned}$$

(3.3)

Step 2: Moving Frames and Integrality

As in the proof of the forthcoming Theorem 2.1, we introduce an extension of $\widetilde{\vec {n}_k}:B_1(0)\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ of $\vec {n}_k:\Omega _k=B_1\setminus \overline{B}_{\rho _k}(0)\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ such that

$$\begin{aligned} \left\{ \begin{aligned}&\widetilde{\vec {n}_k}=\vec {n}\qquad \text {on}\;\, \Omega _k=B_1\setminus B_{\rho _k}(0)\\&\left\| \nabla \widetilde{\vec {n}_k}\right\| _{{\mathrm {L}}^{2}(B_1(0))}\le \Gamma _6(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k)}. \end{aligned}\right. \end{aligned}$$

Therefore, by Lemma IV.3 of [2], there exists a constant $\Gamma _7(n)$ and a Coulomb moving frame $(\vec {f}_{k,1},\vec {f}_{k,2})\in W^{1,2}(B_1(0),S^{n-1})\times W^{1,2}(B_1(0),S^{n-1})$ of $\widetilde{\vec {n}_k}$ such that

$$\begin{aligned} \left\{ \begin{aligned}&\widetilde{\vec {n}_k}=\star (\vec {f}_{k,1}\wedge \vec {f}_{k,2})\qquad {{\,\mathrm{div}\,}}\left( \vec {f}_{k,1}\cdot \nabla \vec {f}_{k,2}\right) =0\\&\left\| \nabla \vec {f}_{k,1}\right\| _{{\mathrm {L}}^{2}(B_1(0))}+\left\| \nabla \vec {f}_{k,2}\right\| _{{\mathrm {L}}^{2}(B_1(0))}\le \Gamma _7(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k)}. \end{aligned}\right. \end{aligned}$$

(3.4)

Now, define for all $j=1,2$ $\vec {e}_{k,j}=e^{-\lambda _k}\partial _{x_j}\vec {\Phi }_k$. As $\vec {\Phi }_k$ is conformal, $(\vec {e}_{k,1},\vec {e}_{k,2})$ is a Coulomb frame of $\vec {n}_k$ on $\Omega _k$. Furthermore, as $\widetilde{\vec {n}_k}=\vec {n}_k$ on $\Omega _k$, both $(\vec {f}_{k,1},\vec {f}_{k,2})$ and $(\vec {e}_{k,1},\vec {e}_{k,2})$ are Coulomb frames of $\vec {n}_k$ on $\Omega _k$, so there exists a rotation $e^{i\theta _k}$ such that

$$\begin{aligned} (\vec {f}_{k,1}+i\vec {f}_{k,2})=e^{i\theta _k}\left( \vec {e}_{k,1}+i\vec {e}_{k,2}\right) . \end{aligned}$$

(3.5)

Now, we let $f_{k,1},f_{k,2}$ be the vector fields such that

$$\begin{aligned} \text {d}\vec {\Phi }_k(f_{k,j})=\vec {f}_{k,j}\qquad \text {for all}\;\, j=1,2. \end{aligned}$$

(3.6)

Then observe as $\vec {\Phi }_k$ is conformal that

$$\begin{aligned} \delta _{i,j}=\langle \vec {f}_{k,i},\vec {f}_{k,j}\rangle =\langle \text {d}\vec {\Phi }_k(f_{k,i}),\text {d}\vec {\Phi }_k(f_{k,j})\rangle =e^{2\lambda _k}\langle f_{k,i},f_{k,j}\rangle \end{aligned}$$

so we have

$$\begin{aligned} \langle f_{k,i},f_{k,j}\rangle =e^{-2\lambda _k}\delta _{i,j}. \end{aligned}$$

Likewise, if $(f_{k,1}^{*},f_{k,j}^{*})$ is the dual framing, we deduce that

$$\begin{aligned} |f_{k,j}^{*}|=e^{\lambda _k}\qquad \text {for all}\;\, j=1,2. \end{aligned}$$

(3.7)

Now, let $\mu _k$ be the unique solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta \mu _k&=\nabla ^{\perp }\vec {f}_{k,1}\cdot \nabla \vec {f}_{k,2}\qquad&\text {in}\;\, B_1(0)\\ \mu _k&=0\qquad&\text {on}\;\, \partial B_1(0). \end{aligned}\right. \end{aligned}$$

Furthermore, introduce the notation $\nu _k=\lambda _k-\mu _k$. Then, $\nu _k$ is harmonic, and by Step 1, we have

$$\begin{aligned} d_k=\frac{1}{2\pi }\int _{\partial B_{\rho _k}}*\, \text {d}\nu _k\underset{k\rightarrow \infty }{\longrightarrow }\theta _0-1. \end{aligned}$$

As $\vec {f}_{k,1}\cdot \partial _{\nu }\vec {f}_{k,2}=0$ on $\partial B_1(0)$, we also have

$$\begin{aligned} \text {d}\mu _k=*(\vec {f}_{k,1}\cdot \text {d}\vec {f}_{k,2}) \end{aligned}$$

(3.8)

Then, we compute with ${\mathbb {Z}}_2$ indices for all $j\in \left\{ 1,2\right\}$

$$\begin{aligned} \text {d}\mu _k\wedge f_{k,j}^{*}=(*\text {d}\mu _{k})\wedge (*\vec {f}_{k,j}^{*})=(-1)^j(\vec {f}_{k,1}\cdot \text {d}\vec {f}_{k,2})\wedge \vec {f}_{k,{j+1}}. \end{aligned}$$

Likewise, as in [29], we compute

$$\begin{aligned} \text {d}f_{k,j}^{*}=(-1)^j\left( \vec {f}_{k,1}\cdot \text {d}\vec {f}_{k,2}\right) \wedge f_{k,j+1}^{*}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \mathrm{d}\left( e^{-\mu _k}f_{k,j}^{*}\right) =0\qquad \text {in}\;\, \Omega _k\qquad \text {for}\;\, j=1,2. \end{aligned}$$

In particular, by Stokes theorem, we have for all $\rho _k\le r_1<r_2\le 1$

$$\begin{aligned} 0=\int _{B_{r_2}\setminus \overline{B}_{r_1}(0)}\mathrm{d}\left( e^{-\mu _k}f_{k,j}^{*}\right) =\int _{\partial B_{r_2}}e^{-\mu _k}f_{k,j}^{*}-\int _{\partial B_{r_1}}e^{-\mu _k}f_{k,j}^{*}. \end{aligned}$$

Therefore, we introduce the constants $c_j\in {\mathbb {R}}$ defined for all $\rho _k\le \rho \le 1$ by

$$\begin{aligned} c_{k,j}=\int _{\partial B_{\rho }}e^{-\mu _k}f_{k,j}^{*}. \end{aligned}$$

Now, introduce the complex valued 1-forms

$$\begin{aligned} f_{k,z}^{*}=f_{k,1}^{*}+if_{k,2}^{*}\qquad f_{k,\overline{z}}^{*}=f_{k,1}^{*}-if_{k,2}^{*}, \end{aligned}$$

so that

$$\begin{aligned} f_{k,1}^{*}=\frac{1}{2}\left( f_{k,z}^{*}+f_{k,\overline{z}}^{*}\right) \qquad f_{k,2}^{*}=\frac{1}{2i}\left( f_{k,z}^{*}-f_{k,\overline{z}}^{*}\right) . \end{aligned}$$

Notice also that

$$\begin{aligned} *\, \mathrm{d}f_{k,z}^{*}=-if_{k,z}^{*}\qquad \text {and}\;\, *\mathrm{d}f_{k,\overline{z}}^{*}=if_{k,\overline{z}}. \end{aligned}$$

Furthermore, if

$$\begin{aligned} \left\{ \begin{aligned}&f_{k,z}=\frac{1}{2}\left( f_{k,1}-if_{k,2}\right) \\&f_{k,\overline{z}}=\frac{1}{2}\left( f_{k,1}+if_{k,2}\right) , \end{aligned}\right. \end{aligned}$$

(3.9)

then for all smooth function $\varphi :\Omega _k\rightarrow {\mathbb {C}}$, we have

$$\begin{aligned} \text {d}\varphi&=\text {d}\varphi \cdot f_{k,1}\,f_{k,1}^{*}+\text {d}\varphi \cdot f_{k,2}\,f_{k,2}^{*}\\&=\text {d}\varphi \cdot f_{k,\overline{z}}\, f_{k,z}^{*}+\text {d}\varphi \cdot f_{k,z} f_{k,\overline{z}}^{*}. \end{aligned}$$

Now, we introduce the differential form $\alpha \in \Omega ^1({\mathbb {R}}^2\setminus \left\{ 0\right\} )$

$$\begin{aligned} \alpha&=\frac{1}{2\pi }*\, \mathrm{d}\log |z|=\frac{1}{2\pi }*\, \left( \nabla \log |z|\cdot f_{k,z}\,f_{k,z}^{*}+\nabla \log |z|\cdot f_{k,\overline{z}}\,f_{k,\overline{z}}^{*}\right) \\&=\frac{1}{2\pi i}\nabla \log |z|\cdot f_{k,z}\,f_{k,z}^{*}-\frac{1}{2\pi i}\nabla \log |z|\cdot f_{k,\overline{z}}\,f_{k,\overline{z}}^{*}. \end{aligned}$$

In particular, notice that

$$\begin{aligned} \alpha +\frac{1}{2\pi i}\nabla \log |z|\cdot f_{k,\overline{z}} f_{k,\overline{z}}^{*}=\frac{1}{2\pi i}\nabla \log |z|\cdot f_{k,z}f_{k,z}^{*}. \end{aligned}$$

(3.10)

As $\log$ is harmonic on ${\mathbb {R}}^2\setminus \left\{ 0\right\}$, the differential form $\alpha$ is closed on $\Omega _k$ and we deduce that the 1-form

$$\begin{aligned} \omega _{k,j}=e^{-\mu _k}f_{k,j}^{*}-c_{k,j}\alpha \end{aligned}$$

is also closed. Furthermore, as

$$\begin{aligned} \int _{\partial B_{\rho _k}}\omega _{k,j}=0, \end{aligned}$$

we deduce by Poincaré lemma that there exists $(\sigma _{k,1},\sigma _{k,2})\in W^{1,2}(\Omega _k,{\mathbb {R}}^2)$ such that

$$\begin{aligned} \text {d}\sigma _{k,j}=\omega _{k,j}=e^{-\mu _k}f_{k,j}^{*}-c_{k,j}\alpha \qquad \text {for}\;\, j=1,2. \end{aligned}$$

Therefore, we deduce if $c_k=c_{k,1}+ic_{k,2}$ and $\sigma _k=\sigma _{k,1}+i\sigma _{k,2}$ that

$$\begin{aligned} \text {d}\sigma _k&=e^{-\mu _k}\left( f_{k,1}^{*}+if_{k,2}^{*}\right) -c_k\alpha \\&=\left( e^{-\mu _k}-\frac{c_k}{2\pi i}\nabla \log |z|\cdot f_{k,z}\right) \left( f_{k,1}^{*}+f_{k,2}^{*}\right) f_{k,z}^{*}+\frac{c_k}{2\pi i}\nabla \log |z|\cdot f_{k,\overline{z}}\,f_{k,\overline{z}}^{*}. \end{aligned}$$

This implies by (3.10) that

$$\begin{aligned} \mathrm{d}\left( \sigma _k-\frac{c_k}{2\pi i}\log |z|\right) =\left( e^{-\mu _k}-\frac{c_k}{\pi i}\nabla \log |z|\cdot f_{k,z}\right) f_{k,z}^{*}. \end{aligned}$$

(3.11)

Therefore, the function

$$\begin{aligned} \tau _k=\sigma _k-\frac{c_k}{2\pi i}\log |z| \end{aligned}$$

is holomorphic. Now, let $\left( \frac{\partial }{\partial \tau _{k,1}},\frac{\partial }{\partial \tau _{k,2}}\right)$ be the dual basis of $(\tau _{k,1},\tau _{k,2})$, where $\tau _k=\tau _{k,1}+i\tau _{k,2}$. Then, we define

$$\begin{aligned} \varphi =e^{-\mu _k}-\frac{c_k}{\pi i}\nabla \log |z|\cdot f_{k,z}, \end{aligned}$$

(3.12)

and we notice that (3.11) implies that

$$\begin{aligned} \mathrm{d}(\tau _{k,1}+i\tau _{k,2})&=\left( {\mathrm {Re}}\,(\varphi )+i\,{\mathrm {Im}}\,(\varphi )\right) \left( f_{k,1}^{*}+if_{k,2}^{*}\right) \\&=\left( {\mathrm {Re}}\,(\varphi )f_{k,1}^{*}-{\mathrm {Im}}\,(\varphi )f_{k,2}^{*}\right) +i\left( {\mathrm {Im}}\,(\varphi )f_{k,1}^{*}+{\mathrm {Re}}\,(\varphi )f_{k,2}^{*}\right) \end{aligned}$$

Therefore, we deduce that

$$\begin{aligned} \begin{pmatrix} \text {d}\tau _{k,1}\\ \text {d}\tau _{k,2} \end{pmatrix} =\begin{pmatrix} {\mathrm {Re}}\,(\varphi ) &{}-{\mathrm {Im}}\,(\varphi )\\ {\mathrm {Im}}\,(\varphi ) &{} {\mathrm {Re}}\,(\varphi ) \end{pmatrix} \begin{pmatrix} f_{k,1}^{*}\\ f_{k,2}^{*}. \end{pmatrix} \end{aligned}$$

This implies that

$$\begin{aligned} \begin{pmatrix} \frac{\partial }{\partial \tau _{k,1}}\\ \frac{\partial }{\partial \tau _{k,2}} \end{pmatrix} =\frac{1}{|\varphi |^2}\begin{pmatrix} {\mathrm {Re}}\,(\varphi ) &{} {\mathrm {Im}}\,(\varphi )\\ -{\mathrm {Im}}\,(\varphi ) &{} {\mathrm {Re}}\,(\varphi ) \end{pmatrix} \begin{pmatrix} f_{k,1}\\ f_{k,2} \end{pmatrix}. \end{aligned}$$

(3.13)

Now, defining

$$\begin{aligned} \frac{\partial }{\partial \tau _k}=\frac{1}{2}\left( \frac{\partial }{\partial \tau _{k,1}}-i\frac{\partial }{\partial \tau _{k,2}}\right) , \end{aligned}$$

we compute thanks to (3.6) and (3.13)

$$\begin{aligned} \frac{\partial \vec {\Phi }_k}{\partial \tau _k}&=\frac{1}{2|\varphi |^2}\text {d}\vec {\Phi }_k\cdot \left( {\mathrm {Re}}\,(\varphi )f_{k,1}+{\mathrm {Im}}\,(\varphi )f_{k,2}-i\left( {\mathrm {Im}}\,(\varphi )f_{k,1}-{\mathrm {Re}}\,(\varphi )f_{k,2}\right) \right) \\&=\frac{1}{2|\varphi |^2}\left( {\mathrm {Re}}\,(\varphi )\vec {f}_{k,1}+{\mathrm {Im}}\,(\varphi )\vec {f}_{k,2}-i\left( {\mathrm {Im}}\,(\varphi )\vec {f}_{k,1}-{\mathrm {Re}}\,(\varphi )\vec {f}_{k,2}\right) \right) \\&=\frac{1}{2|\varphi |^2}\left( \left( {\mathrm {Re}}\,(\varphi )+i{\mathrm {Im}}\,(\varphi )\right) \vec {f}_{k,1}+\left( {\mathrm {Im}}\,(\varphi )-i{\mathrm {Re}}\,(\varphi )\right) \vec {f}_{k,2}\right) \\&=\frac{\varphi }{2|\varphi |^2}\left( \vec {f}_{k,1}-i\vec {f}_{k,2}\right) =\frac{1}{2\overline{\varphi }}\left( \vec {f}_{k,1}-i\vec {f}_{k,2}\right) . \end{aligned}$$

Therefore, we deduce that

$$\begin{aligned} \frac{e^{\lambda _k}}{2}\left( \vec {e}_{k,1}-i\vec {e}_{k,2}\right) =\partial _{z}\vec {\Phi }_k=\frac{\partial \vec {\Phi }_k}{\partial \tau _k}\frac{\partial \tau _k}{\partial z}=\frac{\tau _k'(z)}{2\overline{\varphi }}\left( \vec {f}_{k,1}-i\vec {f}_{k,2}\right) \end{aligned}$$

(3.14)

Now, recall by (3.5) that there exists a rotation $e^{i\theta _k}$ (beware that the function $\theta _k$ is multi-valued) such that

$$\begin{aligned} \vec {f}_{k,1}+i\vec {f}_{k,2}=e^{i\theta _k}\left( \vec {e}_{k,1}+i\vec {e}_{k,2}\right) . \end{aligned}$$

Therefore, (3.14), and $\lambda _k=\mu _k+\nu _k$ imply that

$$\begin{aligned} e^{\lambda _k}\overline{\varphi }=e^{\nu _k}+\frac{\overline{c_k}e^{\lambda _k}}{\pi i}\nabla \log |z|\cdot f_{k,\overline{z}}=\tau '_{k}(z)e^{-i\theta _k}. \end{aligned}$$

(3.15)

Recalling that

$$\begin{aligned} d_k=\frac{1}{2\pi }\int _{\partial B_{\rho _k}}*\, \text {d}\nu _k\underset{k\rightarrow \infty }{\longrightarrow }\theta _0-1, \end{aligned}$$

we will now show that $d_k=\theta _0-1$ for k large enough. First, recall that there exists a rotation $e^{i\theta _k}$ such that

$$\begin{aligned} (\vec {f}_{k,1}+i\vec {f}_{k,2})=e^{i\theta _k}\left( \vec {e}_{k,1}+i\vec {e}_{k,2}\right) , \end{aligned}$$

(3.16)

and that there exists vector fields $f_{k,1},f_{k,2}$ such that

$$\begin{aligned} \text {d}\vec {\Phi }_k(f_{k,j})=\vec {f}_{k,j}\qquad \text {for all}\;\, j=1,2. \end{aligned}$$

(3.17)

To simplify the notations, we will now delete the subscript k in the following formulas. Now, rewrite (3.16) as

$$\begin{aligned} \vec {f}_{1}+i\vec {f}_{2}=e^{i\theta }(\vec {e}_{1}+i\vec {e}_2)=\cos (\theta )\vec {e}_1-\sin (\theta )\vec {e}_2+i\left( \sin (\theta )\vec {e}_1+\cos (\theta )\vec {e}_2\right) , \end{aligned}$$

so that

$$\begin{aligned} \left\{ \begin{aligned} \vec {f}_1&=\cos (\theta )\vec {e}_1-\sin (\theta )\vec {e}_2\\ \vec {f}_2&=\sin (\theta )\vec {e}_1+\cos (\theta )\vec {e}_2 \end{aligned}\right. \end{aligned}$$

Now, write $f_{1}=(f_1^1,f_1^2)$, $f_2=(f_2^1,f_2^2)$, and observe that

$$\begin{aligned}&\text {d}\vec {\Phi }(f_1)=e^{\lambda }f_1^1\vec {e}_1+e^{\lambda }f_2^2\vec {e}_2=\vec {f}_1=\cos (\theta )\vec {e}_1-\sin (\theta )\vec {e}_2\\&\text {d}\vec {\Phi }(f_2)=e^{\lambda }f_2^1\vec {e}_1+e^{\lambda }f_2^2\vec {e}_2=\vec {f}_2=\sin (\theta )\vec {e}_1+\cos (\theta )e_2 \end{aligned}$$

implies that

$$\begin{aligned} \left\{ \begin{aligned} f_1&=e^{-\lambda }(\cos (\theta ),-\sin (\theta ))\\ f_2&=e^{-\lambda }(\sin (\theta ),\cos (\theta )). \end{aligned}\right. \end{aligned}$$

Therefore, we deduce that

$$\begin{aligned} \left\{ \begin{aligned}&f_1^{*}=e^{\lambda }\cos (\theta )\mathrm{d}x_1-e^{\lambda }\sin (\theta )\mathrm{d}x_2\\&f_2^{*}=e^{\lambda }\sin (\theta )\mathrm{d}x_1+e^{\lambda }\cos (\theta )\mathrm{d}x_2. \end{aligned}\right. \end{aligned}$$

(3.18)

Recall the definitions (from (3.9))

$$\begin{aligned}&c_j=\int _{\partial B_{\rho }}e^{-\mu }f^{*}_j\qquad j=1,2,\qquad f_{\overline{z}}=\frac{1}{2}(f_{1}+if_2). \end{aligned}$$

Introducing

$$\begin{aligned} c=-\frac{1}{2\pi i}(c_1-ic_2), \end{aligned}$$

we have for some holomorphic function $\chi$ on $\Omega _k$ and for all $z\in \Omega _k$ (in the preceding notations, we have $\chi =\tau '_k$ in the previous notations) by (3.15)

$$\begin{aligned} e^{\nu }&=\chi (z)e^{-i\theta }+2c\,e^{\lambda }\nabla \log |z|\cdot f_{\overline{z}} \end{aligned}$$

(3.19)

Notice that $e^{i\theta }=\cos (\theta )+i\sin (\theta )$ implies that

$$\begin{aligned} f_{\overline{z}}&=\frac{e^{-\lambda }}{2}\left( (\cos (\theta ),-\sin (\theta ))+i(\sin (\theta ),\cos (\theta ))\right) \nonumber \\&=\frac{e^{\lambda }}{2}\left( \cos (\theta )+i\sin (\theta ),i\cos (\theta )-\sin (\theta )\right) \nonumber \\&=\frac{e^{-\lambda }}{2}\left( \cos (\theta )+i\sin (\theta ),i(\cos (\theta )+i\sin (\theta ))\right) \nonumber \\&=\frac{e^{-\lambda +i\theta }}{2}\left( 1,i\right) , \end{aligned}$$

(3.20)

Therefore, recalling the notation $z=x_1+ix_2$, (3.19) and (3.20) imply that

$$\begin{aligned} e^{\nu }&=\chi (z)e^{-i\theta }+ce^{i\theta }\nabla \log |z|\cdot (1,i) =\chi (z)e^{-i\theta }+ce^{i\theta }\left( \frac{x_1}{|z|^2},\frac{x_2}{|z|^2}\right) \cdot (1,i)\nonumber \\&=\chi (z)e^{-i\theta }+ce^{i\theta }\frac{x_1+ix_2}{|z|^2}=\chi (z)e^{-i\theta }+ce^{i\theta }\frac{z}{|z|^2}\nonumber \\&=\chi (z)e^{-i\theta }+\frac{ce^{i\theta }}{\overline{z}} \end{aligned}$$

(3.21)

Now, as the left-hand side of (3.21) is real, taking imaginary parts of the right-hand side, we find that

$$\begin{aligned} \chi (z)e^{-i\theta }-\overline{\chi (z)}e^{i\theta }+\frac{ce^{i\theta }}{\overline{z}}-\frac{\overline{c}e^{-i\theta }}{z}=0. \end{aligned}$$

Multiplying this identity by $e^{i\theta }$, we deduce that

$$\begin{aligned} e^{2i\theta }\left( -\overline{\chi (z)}+\frac{c}{\overline{z}}\right) +\chi (z)-\frac{\overline{c}}{z}=0. \end{aligned}$$

This implies that

$$\begin{aligned} e^{2i\theta }=\frac{\chi (z)-\dfrac{\overline{c}}{z}}{\overline{\chi (z)}-\dfrac{c}{\overline{z}}}=\left( \frac{\chi (z)-\dfrac{\overline{c}}{z}}{\left| \chi (z)-\dfrac{\overline{c}}{z}\right| }\right) ^2. \end{aligned}$$

Finally, as $e^{\nu }>0$, we deduce thanks to (3.21) that

$$\begin{aligned} e^{i\theta }=\frac{\chi (z)-\dfrac{\overline{c}}{z}}{\left| \chi (z)-\dfrac{\overline{c}}{z}\right| }. \end{aligned}$$

Letting now $\psi$ be the holomorphic function such that

$$\begin{aligned} \psi (z)=\chi (z)-\frac{\overline{c}}{z}, \end{aligned}$$

we deduce that

$$\begin{aligned} e^{i\theta }=\frac{\psi (z)}{|\psi (z)|}. \end{aligned}$$

(3.22)

This implies readily that

$$\begin{aligned} \text {d}\theta ={\mathrm {Im}}\,\left( \frac{\partial \psi }{\psi }\right) ={\mathrm {Im}}\,\left( \frac{\psi '(z)}{\psi (z)}\text {d}z\right) . \end{aligned}$$

(3.23)

Indeed, we have formally (in other words, the following expression must be understood as the equality of two multi-valued functions, i.e. modulo $2\pi i$)

$$\begin{aligned} i\theta =\log \left( \frac{\psi (z)}{|\psi (z)|}\right) . \end{aligned}$$

Therefore, we have

$$\begin{aligned} i\partial \theta&=\frac{|\psi (z)|}{\psi (z)}\bigg \{\frac{\psi '(z)}{|\psi (z)|}-\frac{1}{2}\psi (z)\psi '(z)\overline{\psi }(z)|\psi (z)|^{-3}\bigg \}\text {d}z=\frac{|\psi (z)|}{\psi (z)}\bigg \{\frac{\psi '(z)}{|\psi (z)|}-\frac{1}{2}\frac{\psi '(z)}{|\psi (z)|}\bigg \}\text {d}z\nonumber \\&=\frac{1}{2}\frac{\psi '(z)}{\psi (z)}\text {d}z=\frac{1}{2}\frac{\partial \psi }{\psi }. \end{aligned}$$

(3.24)

As $\theta$ is real, we deduce that

$$\begin{aligned} i\overline{\partial }\theta =\overline{-i\partial \theta }=-\frac{1}{2}\overline{\left( \frac{\partial \psi }{\psi }\right) }. \end{aligned}$$

(3.25)

Using that $\mathrm{d}=\partial +\overline{\partial }$, we deduce from (3.24) and (3.25) that

$$\begin{aligned} \mathrm{d}\theta =\partial \theta +\overline{\partial }\theta =\frac{1}{2i}\left( \frac{\partial \psi }{\psi }-\overline{\left( \frac{\partial \psi }{\psi }\right) }\right) ={\mathrm {Im}}\,\left( \frac{\partial \psi }{\psi }\right) . \end{aligned}$$

Finally, we deduce from (3.23) that

$$\begin{aligned} \int _{\partial B_{\rho }}\text {d}\theta \in 2\pi {\mathbb {Z}}, \end{aligned}$$

Now, a classical computation shows that

$$\begin{aligned} *\, \text {d}\nu =\text {d}\theta . \end{aligned}$$

This can be directly checked using the Coulomb condition, but as we have already used it to obtain the closedness of $e^{-\mu }f_{1}^{*}$ and $e^{-\mu }f_2^{*}$, we can also check this property with these 1-forms. Recall that thanks to (3.18)

$$\begin{aligned} \left\{ \begin{aligned}&e^{-\mu }f_1^{*}=e^{\nu }\cos (\theta )\mathrm{d}x_1-e^{\nu }\sin (\theta )\mathrm{d}x_2\\&e^{-\mu }f_2^{*}=e^{\nu }\sin (\theta )\mathrm{d}x_1+e^{\nu }\cos (\theta )\mathrm{d}x_2. \end{aligned}\right. \end{aligned}$$

Therefore, that $e^{-\mu }f_1^{*}$ be closed is equivalent to

$$\begin{aligned} 0=\left( \partial _{x_2}\nu \right) e^{\nu }\cos (\theta )-\left( \partial _{x_2}\theta \right) e^{\nu }\sin (\theta )+(\partial _{x_1}\nu )e^{\nu }\sin (\theta )+(\partial _{x_1}\theta )e^{\nu }\cos (\theta ) \end{aligned}$$

or (writing scripts for partial derivatives)

$$\begin{aligned} (\nu _2+\theta _1)\cos (\theta )+\left( \nu _1-\theta _2\right) \sin (\theta )=0. \end{aligned}$$

(3.26)

Likewise, the closedness of $e^{-\mu }f_2^{*}$ is equivalent to

$$\begin{aligned} (-\nu _1+\theta _2)\cos (\theta )+(\nu _2+\theta _1)\sin (\theta )=0. \end{aligned}$$

(3.27)

Therefore, (3.26) and (3.27) are equivalent to the system

$$\begin{aligned} \begin{pmatrix} \cos (\theta )&{} \sin (\theta )\\ \sin (\theta )&{} -\cos (\theta ) \end{pmatrix} \begin{pmatrix} \nu _2+\theta _1\\ \nu _1-\theta _2 \end{pmatrix}=0. \end{aligned}$$

As

$$\begin{aligned} \det \begin{pmatrix} \cos (\theta )&{} \sin (\theta )\\ \sin (\theta )&{} -\cos (\theta ) \end{pmatrix}=-\cos ^2(\theta )-\sin ^2(\theta )=-1\ne 0, \end{aligned}$$

we deduce that

$$\begin{aligned} \left\{ \begin{aligned}&\nu _2+\theta _1=0\\&\nu _1-\theta _2=0 \end{aligned}\right. \end{aligned}$$

(3.28)

In other words, (3.28) is equivalent to $\nabla \nu =\nabla ^{\perp }\theta$, or

$$\begin{aligned} *\,\text {d}\nu =\text {d}\theta . \end{aligned}$$

(3.29)

Therefore, thanks to (3.1) and (3.28), we deduce that for k large enough

$$\begin{aligned} \frac{1}{2\pi }\int _{\partial B_{\rho }}*\, \text {d}\nu _k=\theta _0-1. \end{aligned}$$

This argument concludes the proof of the Proposition. $\square$

We are now going to improve the expansion of the conformal parameter to obtain a pointwise estimate of $\nabla \vec {\Phi }_k$.

We first need an extension lemma which is a refinement of Lemma IV.1 of [2]. For the sake of completeness, we add all details.

Lemma 3.2

Let $0<r<1$ and $\vec {n}\in W^{1,(2,1)}(B_{2r}\setminus \overline{B}_r(0),{\mathscr {G}}_{n-2}({\mathbb {R}}^n))$. There exists $\varepsilon _2(n)>0$ with the following property. Assume that

$$\begin{aligned} \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_{2r}\setminus \overline{B}_r(0))}\le \varepsilon _3(n). \end{aligned}$$

Then, there exists an extension $\widetilde{\vec {n}}\in W^{1,(2,1)}(B_{2r}(0),{\mathscr {G}}_{n-2}({\mathbb {R}}^n))$ such that $\widetilde{\vec {n}}=\vec {n}$ on $B_{2r}\setminus \overline{B}_r(0)$ and a universal constant $C_{5}(n)$ such that

$$\begin{aligned} \left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2,1}(B_{2r})}\le C_{5}(n)\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_{2r}\setminus \overline{B}_r(0))} \end{aligned}$$

Proof

First, as in [5] 3.2.28, we view ${\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ as a submanifold of ${\mathbb {R}}^{N(n)}$ for some (large) N(n). Thanks to the Sobolev embedding $W^{1,(2,1)}(B_{2r}\setminus \overline{B}_r(0))\subset C^0(B_{2r}\setminus \overline{B}_r(0))$ and scaling invariance, there exists $\vec {p}\in {\mathscr {G}}_{n-2}({\mathbb {R}}^n)\subset {\mathbb {R}}^{N(n)}$ and a universal constant $\Gamma _{11}(n)>0$ independent of $r>0$ such that

$$\begin{aligned} \left\| \vec {n}-\vec {p}\right\| _{{\mathrm {L}}^{\infty }(B_{2r}\setminus \overline{B}_r(0))}\le \Gamma _{11}(n)\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_{2r}\setminus \overline{B}_r(0))}\le \Gamma _{11}(n)\varepsilon _3(n). \end{aligned}$$

(3.30)

As ${\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ is a compact smooth submanifold, its injectivity radius is strictly positive, there exists $\varepsilon _3(n)>0$ independent of $\vec {p}\in {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ such that (3.30) implies that $\vec {n}(B_{2r}\setminus \overline{B}_r(0))$ is included in a geodesic ball of ${\mathscr {G}}_{n-2}({\mathbb {R}}^n)$. Therefore, we deduce that there exists $\delta =\delta (n)>0$ such that $\vec {n}(B_{2r}\setminus \overline{B}_r(0))\subset B_{\delta }(\vec {p})$ global coordinates $\varphi :B_{\delta }(\vec {p})\rightarrow \varphi (B_{\delta }(\vec {p}))\subset {\mathbb {R}}^{m(n)}$ (where $m(n)=\dim {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$). Once more, by compactness, we can assume that $\delta =\delta (n)$ has been fixed independently of $\vec {p}$ and such that

$$\begin{aligned} \left\| \nabla \varphi ^{-1}\right\| _{{\mathrm {L}}^{\infty }(\varphi (B_{\delta (n)}(\vec {p})))}<\infty \end{aligned}$$

(3.31)

depends only on n. Furthermore, we can assume without loss of generality that $\varphi (B_{\delta }(\vec {p}))=B_{\delta }^{{\mathbb {R}}^{m(n)}}(0)=B_{\delta }^m(0)$ is the standard geodesics ball in ${\mathbb {R}}^m$ of radius $\delta >0$. Now, apply the extension Theorem 7.2 to the composition $\vec {n}_{\varphi }=\varphi \circ \vec {n}:B_{2r}\setminus \overline{B}_r(0)\rightarrow {\mathbb {R}}^{m(n)}$ to find an extension $\widetilde{\vec {n}}_{\varphi }:B_{2r}(0)\rightarrow {\mathbb {R}}^{m(n)}$ such that

$$\begin{aligned} \left\| \widetilde{\vec {n}}_{\varphi }\right\| _{{\mathrm {W}}^{1,(2,1)}(B_{2r}(0))}\le \Gamma _{12}(n)\left\| \vec {n}_{\varphi }\right\| _{{\mathrm {W}}^{1,(2,1)}(B_{2r}(0))}. \end{aligned}$$

We deduce by the Poincaré-Wirtinger inequality that

$$\begin{aligned} \left\| \nabla \widetilde{\vec {n}}_{\varphi }\right\| _{{\mathrm {L}}^{2,1}(B_{2r}(0))}&\le \Gamma _{12}(n)\left( \left\| \nabla \vec {n}_{\varphi }\right\| _{{\mathrm {L}}^{2,1}(B_{2r}\setminus \overline{B}_r(0))}+\left\| \vec {n}_{\varphi }-\overline{\vec {n}_{\varphi }}_{B_{2r}\setminus \overline{B}_r}\right\| _{{\mathrm {L}}^{2,1}(B_{2r}\setminus \overline{B}_{r}(0))}\right) \\&\le \Gamma _{13}(n)\left( 1+r\right) \left\| \nabla \vec {n}_{\varphi }\right\| _{{\mathrm {L}}^{2,1}(B_{2r}\setminus \overline{B}_r(0))}\\&\le 2\,\Gamma _{13}\left\| \nabla \vec {n}_{\varphi }\right\| _{{\mathrm {L}}^{2,1}(B_{2r}\setminus \overline{B}_r(0))} \end{aligned}$$

Taking $\widetilde{\vec {n}}=\varphi ^{-1}\circ \widetilde{\vec {n}}_{\varphi }$ finishes the proof of the theorem by the previous remark in (3.31). $\square$

The next lemma is an easy consequence of Lemme (5.1.4) of [11] (see also Lemma IV.3 of [2]).

Lemma 3.3

($W^{1,(2,1)}$-controlled Coulomb frame) Let $0<r<\dfrac{1}{2}$ and $\vec {n}\in W^{1,(2,1)}(B_1\setminus \overline{B}_r(0))\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$. Then, there exists $0<\varepsilon _3(n)<\varepsilon _2(n)$ with the following property. Assume that

$$\begin{aligned} \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_1\setminus \overline{B}_r(0))}\le \varepsilon _3(n). \end{aligned}$$

Then, there exists $(\vec {e}_1,\vec {e}_2)\in W^{1,(2,1)}(B_1(0))\times W^{1,(2,1)}(B_1(0))\rightarrow {\mathbb {R}}^n$ which is a Coulomb frame on $B_1\setminus \overline{B}_r(0)$ associated to $\vec {n}$ such that

$$\begin{aligned} \vec {n}=*\left( \vec {e}_1\wedge \vec {e}_2\right) \;\, \text {in}\;\, B_1\setminus \overline{B}_r(0)\qquad \text {and}\qquad \left\{ \begin{aligned} {{\,\mathrm{div}\,}}\left( \vec {e}_1\cdot \nabla \vec {e}_2\right)&=0&\qquad \text {in}\;\, B_1(0)\\ \vec {e}_1\cdot \partial _{\nu }\vec {e}_2&=0&\qquad \text {on}\;\, \partial B_1(0), \end{aligned}\right. \end{aligned}$$

and there exists a universal constant $C_{6}(n)>0$ such that

$$\begin{aligned}&\left\| \nabla \vec {e}_1\right\| _{{\mathrm {L}}^{2}(B_1(0))}^2+\left\| \nabla \vec {e}_2\right\| _{{\mathrm {L}}^{2}(B_1(0))}^2\le \frac{1}{4}C_{5}(n)^2\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_1\setminus \overline{B}_r(0))}^2\nonumber \\&\left\| \nabla \vec {e}_1\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}+\left\| \nabla \vec {e}_2\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}\le C_{6}(n)\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_1\setminus \overline{B}_r(0))}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_1\setminus \overline{B}_r(0))}. \end{aligned}$$

(3.32)

Remark 3.4

Notice that we do not have in general $\vec {e}_1\cdot \partial _{\nu }\vec {e}_2=0$ on $\partial B_r(0)$.

Proof

First, as $\varepsilon _3(n)<\varepsilon _2(n)$, we have

$$\begin{aligned} \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_{2r}\setminus \overline{B}_r(0))}\le \varepsilon _2(n) . \end{aligned}$$

Therefore, by Lemma 3.2, there exists an extension $\widetilde{\vec {n}}:B_1(0)\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ (such that $\widetilde{\vec {n}}=\vec {n}$ on $B_1\setminus \overline{B}_r(0)$) and satisfying (up to replacing $C_{5}(n)$ by $\max \left\{ 1,C_{5}(n)\right\}$ in Lemma 3.2)

$$\begin{aligned} \left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}&\le C_{5}(n)\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_{2r}\setminus \overline{B}_r(0))}+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_1\setminus \overline{B}_{2r}(0))}\nonumber \\&\le C_{5}(n)\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_1\setminus \overline{B}_r(0))} \le C_{5}(n)\varepsilon _3(n). \end{aligned}$$

(3.33)

By the inequality $\Vert \,\cdot \,\Vert _{{\mathrm {L}}^2}\le \dfrac{1}{2\sqrt{2}}\Vert \,\cdot \,\Vert _{{\mathrm {L}}^{2,1}}$ (see the Appendix (7.6)), we deduce by (3.33) that

$$\begin{aligned} \left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2}(B_1(0))}\le \frac{1}{2\sqrt{2}}\left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}\le \frac{1}{2\sqrt{2}}C_{5}(n)\varepsilon _3(n) \end{aligned}$$

so taking

$$\begin{aligned} 0<\varepsilon _3(n)<\frac{2\sqrt{2}}{C_{5}(n)}\frac{8\pi }{3}, \end{aligned}$$

we deduce by Lemme 5.1.4 of [11] that there exists a Coulomb frame $(\vec {e}_1,\vec {e}_2)\in W^{1,2}(B_1(0))\times W^{1,2}(B_1(0))\rightarrow {\mathbb {R}}^n$ such that

$$\begin{aligned} \widetilde{\vec {n}}=*\left( \vec {e}_1\wedge \vec {e}_2\right) \qquad \text {and}\qquad \left\{ \begin{aligned} {{\,\mathrm{div}\,}}\left( \vec {e}_1\cdot \nabla \vec {e}_2\right)&=0&\qquad \text {in}\;\, B_1(0)\\ \vec {e}_1\cdot \partial _{\nu }\vec {e}_2&=0&\qquad \text {on}\;\, \partial B_1(0), \end{aligned}\right. \end{aligned}$$

and (by [11], (5.23, 5.24) p. 244) and the elementary inequality

$$\begin{aligned} 1-\sqrt{1-t}\le t\qquad \text {for all}\;\, t\in [0,1], \end{aligned}$$

we deduce that

$$\begin{aligned} \left\| \nabla \vec {e}_1\right\| _{{\mathrm {L}}^{2}(B_1(0))}^2+\left\| \nabla \vec {e}_2\right\| _{{\mathrm {L}}^{2}(B_1(0))}^2&\le \frac{16\pi }{3}\left( 1-\sqrt{1-\frac{3}{8\pi }\int _{B_1(0)}|\nabla \widetilde{\vec {n}}|^2\mathrm{d}x}\right) \nonumber \\&\le 2\int _{B_1(0)}|\nabla \widetilde{\vec {n}}|^2\mathrm{d}x. \end{aligned}$$

(3.34)

Now, let $\mu :B_1(0)\rightarrow {\mathbb {R}}$ be the unique solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta \mu&=\nabla ^{\perp }\vec {e}_1\cdot \nabla \vec {e}_2 \qquad&\text {in}\;\, B_1(0)\\ \mu&=0\qquad&\text {on}\;\, \partial B_1(0) \end{aligned}\right. \end{aligned}$$

Then, by the generalised Wente inequality (or [4] and the Sobolev embedding $W^{2,1}({\mathbb {R}}^2)\hookrightarrow W^{1,(2,1)}({\mathbb {R}}^n)$), we have

$$\begin{aligned} \left\| \nabla \mu \right\| _{{\mathrm {L}}^{2,1}(B_1(0))}\le 2\Gamma _0\left\| \nabla \vec {e}_1\right\| _{{\mathrm {L}}^{2}(B_1(0))}\left\| \nabla \vec {e}_2\right\| _{{\mathrm {L}}^{2}(B_1(0))} \le \Gamma _0\left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2}(B_1(0))}^2 \end{aligned}$$

(3.35)

Now recall the identity ([11], (5.39), p. 247)

$$\begin{aligned} |\nabla \vec {e}_1|^2+|\nabla \vec {e}_2|^2=2|\nabla \mu |^2+|\nabla \widetilde{\vec {n}}|^2. \end{aligned}$$

(3.36)

Therefore, we have

$$\begin{aligned} |\nabla \vec {e}_1|+|\nabla \vec {e}_2|\le \sqrt{2}\sqrt{|\nabla \vec {e}_1|^2+|\nabla \vec {e}_2|^2}\le 2|\nabla \mu |+\sqrt{2}|\nabla \widetilde{\vec {n}}|. \end{aligned}$$

(3.37)

The identity (3.37) and the estimates (3.33), (3.34) and (3.35) yield

$$\begin{aligned}&\left\| \nabla \vec {e}_1\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}+\left\| \nabla \vec {e}_2\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}\le 2\left\| \nabla \mu \right\| _{{\mathrm {L}}^{2,1}(B_1(0))}+\sqrt{2}\left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}\nonumber \\&\quad \le \Gamma _0\left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2}(B_1(0))}^2+\sqrt{2}\left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}\nonumber \\&\quad \le \frac{1}{8}\Gamma _0\left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}^2+\sqrt{2}\left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2,1}(B_1(0))}\nonumber \\&\quad \le \frac{1}{8}\Gamma _0C_{5}(n)^2\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_1\setminus \overline{B}_r(0))}\nonumber \\&\qquad +\sqrt{2}\,C_{5}(n)\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_1\setminus \overline{B}_r(0))}\nonumber \\&\quad \le C_{6}(n)\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_1\setminus \overline{B}_r(0))}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,1}(B_1\setminus \overline{B}_r(0))}, \end{aligned}$$

(3.38)

where

$$\begin{aligned} C_{6}(n)=\max \left\{ \frac{1}{8}\Gamma _0C_{5}(n)^2,\sqrt{2}\,C_{5}(n)\right\} . \end{aligned}$$

The estimate (3.38) finishes the proof of the lemma. $\square$

We can finally state the precise pointwise estimate.

Theorem 3.5

Under the conditions of Theorem 3.1, assume furthermore that the following strong $L^{2,1}$ no-neck energy holds

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\lim _{k\rightarrow \infty }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}=0. \end{aligned}$$

(3.39)

Then, there exists $\alpha _0>0$ such that for all $k\in {\mathbb {N}}$ large enough, there exists a moving frame $(\vec {f}_{k,1},\vec {f}_{k,2})\in W^{1,(2,1)}(B_{\alpha _0}(0))\times W^{1,(2,1)}(B_{\alpha _0}(0))$ and a universal constant $C_{7}(n)$ (independent of k) such that

$$\begin{aligned}&\left\| \nabla \vec {f}_{k,1}\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}+\left\| \nabla \vec {f}_{k,2}\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}\\&\quad \le C_{7}(n)\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}. \end{aligned}$$

Furthermore, there exists a sequence of functions $\mu _k\in W^{2,1}(B_{\alpha _0}(0))$ and a universal constant $C_{8}(n)$ such that

$$\begin{aligned}&\left\| \nabla ^2\mu _k\right\| _{{\mathrm {L}}^{1}(B_{\alpha _0}(0))}+\left\| \nabla \mu _k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}+\left\| \mu _k\right\| _{{\mathrm {L}}^{\infty }(B_{\alpha _0}(0))}\\&\quad \le C_{8}(n)\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))} \end{aligned}$$

and there exists a sequence of holomorphic functions $\psi _k:B_{\alpha _0}(0)\rightarrow {\mathbb {C}}$ and $\chi _k:B_{\alpha _0}(0)\rightarrow {\mathbb {C}}$ such that $\chi _k(0)=0$, $c\in {\mathbb {C}}$ and $\left\{ c_k\right\} _{k\in {\mathbb {N}}}\subset {\mathbb {C}}$ such that $c_k\underset{k\rightarrow \infty }{\longrightarrow }c$ and

$$\begin{aligned} \psi _k(z)=e^{c_k}z^{\theta _0-1}\left( 1+\chi _k(z)\right) \end{aligned}$$

(3.40)

and

$$\begin{aligned} e^{\lambda _k}=e^{\mu _k}|\psi _k(z)|=e^{{\mathrm {Re}}\,(c_k)}|z|^{\theta _0-1}\left( 1+o(1)\right) ,\qquad \text {for all}\;\, z\in \Omega _k(\alpha ). \end{aligned}$$

(3.41)

Finally, there exists $\vec {A}_0\in {\mathbb {C}}^n$ (such that $\langle \vec {A}_0,\vec {A}_0\rangle =0$) and $\left\{ \vec {A}_{k,0}\right\} _{k\in {\mathbb {N}}}\in {\mathbb {C}}^n$ such that $\vec {A}_{k,0}\underset{k\rightarrow \infty }{\longrightarrow }\vec {A}_0$ and for all $z\in \Omega _k(\alpha _0)$, we have the pointwise identities

$$\begin{aligned} \partial _{z}\vec {\Phi }_k=\frac{1}{2}e^{c_k+\mu _k(z)}z^{\theta _0-1}\left( 1+\chi _k(z)\right) \left( \vec {f}_{k,1}-i\vec {f}_{k,2}\right) =\vec {A}_{k,0}z^{\theta _0}+o(|z|^{\theta _0-1}) \end{aligned}$$

(3.42)

Proof

Step 1: Expansion of $\nabla \vec {\Phi }_k$ in the Neck Region By, fix $\alpha _0>0$ such that for all $k\in {\mathbb {N}}$ large enough

$$\begin{aligned} \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}\le \varepsilon _3(n) \end{aligned}$$

(3.43)

where $\varepsilon _3(n)>0$ is given by Lemma 3.3. Then, we define as in the proof of Theorem 3.1 for all $j=1,2$ $\vec {e}_k,j=e^{-\lambda }_k\partial _{x_j}\vec {\Phi }_k$, and by Lemmas (refextension,  3.3, 3.42 and 3.43), there exists a controlled extension $\widetilde{\vec {n}_k}:B_{\alpha _0}(0)\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ if $\vec {n}_k:\Omega _k(\alpha _0)\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ such that

$$\begin{aligned} \left\{ \begin{aligned}&\widetilde{\vec {n}_k}=\vec {n}_k\qquad \text {in}\;\, \Omega _k(\alpha _0)=B_{\alpha _0}\setminus B_{\alpha _0^{-1}\rho _k}(0)\\&\left\| \nabla \widetilde{\vec {n}_k}\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}\le C_{5}(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k)}. \end{aligned}\right. \end{aligned}$$

and a Coulomb frame $(\vec {f}_{k,1},\vec {f}_{k,2})\in W^{1,(2,1)}(B_{\alpha _0}(0),S^{n-1})\times W^{1,(2,1)}(B_{\alpha _0}(0),S^{n-1})$ associated to $\widetilde{\vec {n}_k}$ such that

$$\begin{aligned}&\widetilde{\vec {n}_k}=\star \,(\vec {f}_{k,1}\wedge \vec {f}_{k,2})\quad \text {in}\;\, B_{\alpha _0}(0)\qquad \text {and}\qquad \left\{ \begin{aligned} {{\,\mathrm{div}\,}}\left( \vec {f}_{k,1}\cdot \nabla \vec {f}_{k,2}\right) =0\quad&\text {in}\;\, B_{\alpha _0}(0)\\ \vec {f}_{k,1}\cdot \partial _{\nu }\vec {f}_{k,2}=0 \quad&\text {on}\;\, \partial B_{\alpha _0}(0) \end{aligned}\right. \end{aligned}$$

(3.44)

and

$$\begin{aligned}&\left\| \nabla \vec {f}_{k,1}\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}+\left\| \nabla \vec {f}_{k,2}\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}\nonumber \\&\quad \le C_{6}(n)\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}. \end{aligned}$$

(3.45)

Finally, we introduce the rotation $\theta _k$ (which is a multi-valued function on $\Omega _k(\alpha )$) such that

$$\begin{aligned} \left( \vec {f}_{k,1}+i\vec {f}_{k,2}\right) =e^{i\theta _k}\left( \vec {e}_{k,1}+i\vec {e}_{k,2}\right) \qquad \text {on}\;\, \Omega _k(\alpha _0) \end{aligned}$$

(3.46)

As previously, let $\mu _k$ the unique solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta \mu _k&=\nabla ^{\perp }\vec {f}_{k,1}\cdot \nabla \vec {f}_{k,2}\qquad&\text {in}\;\, B_{\alpha _0}(0)\\ \mu _k&=0\qquad&\text {on}\;\, \partial B_{\alpha _0}(0). \end{aligned}\right. \end{aligned}$$

Then, we have by the improved Wente inequality $\mu _k\in W^{1,(2,1)}(B_{\alpha _0}(0))\cap C^0(B_{\alpha _0}(0))$ and (3.45) for some universal constant $C_{9}(n)$

$$\begin{aligned}&\left\| \nabla ^2\mu _k\right\| _{{\mathrm {L}}^{1}(B_{\alpha _0}(0))}+\left\| \nabla \mu _k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}+\left\| \mu _k\right\| _{{\mathrm {L}}^{\infty }(B_{\alpha _0}(0))}\nonumber \\&\quad \le C_{9}(n)\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))} \end{aligned}$$

(3.47)

Furthermore, introduce the notation $\nu _k=\lambda _k-\mu _k$. Then, $\nu _k$ is harmonic, and implies that for k large enough

$$\begin{aligned} \frac{1}{2\pi }\int _{\partial B_{\rho _k}}*\, \text {d}\nu _k\underset{k\rightarrow \infty }{\longrightarrow }=\theta _0-1. \end{aligned}$$

Indeed, recall that by the proof of Theorem 3.1, $*\, \text {d}\nu _k=\text {d}\theta _k$ and that there exists a holomorphic function $\psi _k:\Omega _k(\alpha _0)\rightarrow {\mathbb {C}}$ such that

$$\begin{aligned} e^{i\theta _k}=\frac{\psi _k}{|\psi _k|}. \end{aligned}$$

(3.48)

In particular, a computation of the proof of Theorem 3.1 shows that

$$\begin{aligned} \text {d}\theta _k={\mathrm {Im}}\,\left( \frac{\partial \psi _k}{\psi _k}\right) , \end{aligned}$$

(3.49)

so that for all $\alpha _0^{-1}\rho _k<\rho <\alpha _0$

$$\begin{aligned} \frac{1}{2\pi }\int _{\partial B_{\rho }}*\text {d}\nu _k=\frac{1}{2\pi }\int _{\partial B_{\rho }}\text {d}\theta _k=\frac{1}{2\pi }{\mathrm {Im}}\int _{\partial B_{\rho }}\frac{\partial \psi _k}{\psi _k}\in \mathbb {Z}. \end{aligned}$$

As

$$\begin{aligned} d_k=\frac{1}{2\pi }\int _{\partial B_{\rho }}*\text {d}\nu _k\underset{k\rightarrow \infty }{\longrightarrow }\theta _0-1, \end{aligned}$$

we deduce that

$$\begin{aligned} \frac{1}{2\pi }\int _{\partial B_{\rho }}*\text {d}\nu _{k}=\frac{1}{2\pi }\int _{\partial B_{\rho }}\text {d}\theta _k=\theta _0-1 \end{aligned}$$

for all k large enough. In other words, $\nu _k$ satisfies as in (3.28)

$$\begin{aligned} \left\{ \begin{aligned}&\partial _{x_2}\nu _k+\partial _{x_1}\theta _k=0\\&\partial _{x_1}\nu _k-\partial _{x_2}\theta _k=0 \end{aligned}\right. . \end{aligned}$$

(3.50)

Therefore, we deduce by (3.50) that

$$\begin{aligned} \partial _{z}\nu _k=\frac{1}{2}\left( \partial _{x_1}\nu _k-i\partial _{x_2}\nu _k\right) =\frac{1}{2}\left( \partial _{x_2}\theta _k+i\,\partial _{x_1}\theta _k\right) =\frac{i}{2}\left( \partial _{x_1}\theta _k-i\,\partial _{x_2}\theta _k\right) =i\,\partial _{z}\theta _k. \end{aligned}$$

(3.51)

As $\text {d}\theta _k=\partial \theta _k+\overline{\partial }\theta _k$, (3.49) implies that

$$\begin{aligned} i\,\partial \theta _k=\frac{1}{2}\frac{\partial \psi _k}{\psi _k}=\frac{1}{2}\frac{\partial _{z}\psi _k}{\psi _k}\text {d}z=\partial \log |\psi _k|, \end{aligned}$$

(3.52)

as $\partial _{\overline{z}}\psi _k=0$ implies that $\partial _{z}\overline{\psi _k}=\overline{\partial _{\overline{z}}\psi _{k}}=0$ and

$$\begin{aligned} \partial _{z}\log |\psi _k|=\frac{1}{2}\log \left( \psi _k(z)\overline{\psi _k(z)}\right) =\frac{1}{2}\frac{\psi _k'(z)\overline{\psi _k(z)}}{\psi _k(z)\overline{\psi _k(z)}}=\frac{1}{2}\frac{\psi _k'(z)}{\psi _k(z)}. \end{aligned}$$

Therefore, (3.51) and (3.52) show that

$$\begin{aligned} \partial \Big (\nu _k-\log |\psi _k|\Big )=0. \end{aligned}$$

So the function $\nu _k-\log |\psi _k|$ is anti-holomorphic and real, so it must be constant by the maximum principle as $\Omega _k(\alpha _0)=B_{\alpha _0}\setminus \overline{B}_{\alpha _0^{-1}\rho _k}$ is connected. Therefore, there exists $\gamma _k\in {\mathbb {R}}$ such that

$$\begin{aligned} \nu (z)=\gamma _k+\log |\psi _k(z)|, \end{aligned}$$

(3.53)

or

$$\begin{aligned} e^{\nu _k(z)}=e^{\gamma _k}|\psi _k(z)|. \end{aligned}$$

(3.54)

Now, as $\widetilde{\psi _k}=e^{\gamma _k}\psi _k$ is holomorphic and satisfies

$$\begin{aligned} \frac{1}{2\pi }{\mathrm {Im}}\,\int _{\partial B_{\rho }}\frac{\partial \widetilde{\psi _k}}{\widetilde{\psi _k}}=\frac{1}{2\pi }{\mathrm {Im}}\,\int _{\partial B_{\rho }}\frac{\partial \psi _k}{\psi _k}=\theta _0-1, \end{aligned}$$

(3.55)

we can assume without loss of generality that $\gamma _k=0$. Furthermore, (3.55) shows that the holomorphic 1-form $\frac{\partial \psi _k}{\psi _k}$ on $\Omega _k(\alpha _0)$ admits the expansion

$$\begin{aligned} \frac{\partial \psi _k}{\psi _k}=(\theta _0-1)\frac{\text {d}z}{z}+\xi _k(z)\text {d}z, \end{aligned}$$

where $\xi _k$ admits a holomorphic extension on $B_{\alpha _0}(0)$. In particular, $\psi _k$ admits a Laurent series expansion

$$\begin{aligned} \psi _k(z)=\sum _{m= \theta _0-1}^{\infty }a_mz^{m}, \end{aligned}$$

where $a_{\theta _0-1}\ne 0$. Therefore, $\psi _k$ extends holomorphically in $B_{\alpha _0}(0)$, and letting $c_k\in {\mathbb {C}}$ be such that

$$\begin{aligned} e^{c_k}=a_{\theta _0-1}, \end{aligned}$$

there exists a holomorphic function $\chi _k:B_{\alpha _0}(0)\rightarrow {\mathbb {C}}$ such that $\chi _k(0)=0$ and

$$\begin{aligned} \psi _k(z)=e^{c_k}z^{\theta _0-1}\left( 1+\chi _k(z)\right) , \end{aligned}$$

(3.56)

where we have explicitly

$$\begin{aligned} \chi _k(z)=\sum _{m=\theta _0}^{\infty }e^{-c_k}a_mz^{m-(\theta _0-1)}. \end{aligned}$$

Notice in particular as $\lambda _k=\mu _k+\nu _k$ that

$$\begin{aligned} e^{\lambda _k}=e^{\mu _k}|\psi _k(z)|, \end{aligned}$$

(3.57)

where $\psi _k$ is holomorphic and admits the expansion (3.56). Now, we come back to the identity (3.46) to observe that

$$\begin{aligned} \partial _{z}\vec {\Phi }_k=\frac{1}{2}\left( \partial _{x_1}\vec {\Phi }_k-i\partial _{x_2}\vec {\Phi }_k\right) =\frac{1}{2}e^{\lambda _k}\left( \vec {e}_{k,1}-i\,\vec {e}_{k,2}\right) =\frac{1}{2}e^{\lambda _k}e^{i\theta _k}\left( \vec {f}_{k,1}-i\vec {f}_{k,2}\right) . \end{aligned}$$

(3.58)

Now, observe that by (3.48) and (3.57)

$$\begin{aligned} e^{\lambda _k}e^{i\theta _k}=e^{\mu _k}|\psi _k(z)|\times \frac{\psi _k(z)}{|\psi _k(z)|}=e^{\mu _k}\psi _k(z). \end{aligned}$$

(3.59)

Therefore, (3.58), (3.59) and (3.56) finally yield the expansion

$$\begin{aligned} \partial _{z}\vec {\Phi }_k&=\frac{1}{2}e^{\mu _k}\psi _k(z)\left( \vec {f}_{k,1}-i\,\vec {f}_{k,2}\right) \nonumber \\&=\frac{1}{2}e^{c_k+\mu _k}z^{\theta _0-1}\left( 1+\chi _k(z)\right) \left( \vec {f}_{k,1}-i\,\vec {f}_{k,2}\right) . \end{aligned}$$

(3.60)

By (3.45) and (3.47), $e^{\mu _k}\left( \vec {f}_{k,1}-i\,\vec {f}_{k,2}\right) \in W^{1,(2,1)}\cap C^0(B_{\alpha _0}(0),S^{n-1})$ and

$$\begin{aligned}&\left\| e^{\mu _k}\left( \vec {f}_{k,1}-i\,\vec {f}_{k,2}\right) \right\| _{{\mathrm {L}}^{\infty }(B_{\alpha _0}(0))}\le e^{C_{9}(n)\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}}\nonumber \\&\left\| \nabla \left( e^{\mu _k}\left( \vec {f}_{k,1}-i\,\vec {f}_{k,2}\right) \right) \right\| _{{\mathrm {L}}^{2,1}(B_{\alpha _0}(0))}\nonumber \\&\quad \le \left( C_{6}(n)+C_{9}(n)\right) \left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}\right) \nonumber \\&\qquad \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}e^{\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha _0))}} \end{aligned}$$

(3.61)

In particular, if

$$\begin{aligned} \frac{1}{2}e^{c_k+\mu _k(0)}\left( \vec {f}_{k,1}-i\vec {f}_{k,2}\right) (0)=\vec {A}_{k,0}\in {\mathbb {C}}^n\setminus \left\{ 0\right\} \end{aligned}$$

then (notice that $\vec {A}_{k,0}\ne 0$ as $\vec {\Phi }_k$ is an immersion) (3.60) becomes

$$\begin{aligned} \partial _{z}\vec {\Phi }_k=\vec {A}_{k,0}z^{\theta _0-1}+o\left( |z|^{\theta _0-1}\right) \qquad \text {for all}\;\, z\in \Omega _k(\alpha _0). \end{aligned}$$

Furthermore by the strong convergence of $\vec {\Phi }_k$ towards $\vec {\Phi }_{\infty }$ in $C^l_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} )$ (for all $l\in {\mathbb {N}}$) which satisfies

$$\begin{aligned} \partial _{z}\vec {\Phi }_{\infty }=\vec {A}_0z^{\theta _0-1}+o\left( |z|^{\theta _0-1}\right) , \end{aligned}$$

we deduce that

$$\begin{aligned} \vec {A}_{k,0}\underset{k\rightarrow \infty }{\longrightarrow }\vec {A}_{0}. \end{aligned}$$

This concludes the proof of the theorem. $\square$

3.2 General Case

Theorem 3.6

Let $\{\vec {\Phi }_k\}_{k\in {\mathbb {N}}}$ be a sequence of smooth conformal immersions from the disk $B_1(0)\subset {\mathbb {C}}$ into ${\mathbb {R}}^n$. Let $m\in {\mathbb {N}}$, and for all $1\le j\le m$, let $\{a_k^j\}_{k\in {\mathbb {N}}}\subset B_1(0)$, $\{\rho _k^j\}_{k\in {\mathbb {N}}}\subset (0,\infty )$ and define for $0<\alpha <1$ and k large enough

$$\begin{aligned} \Omega _k=B_1(0)\setminus \bigcup _{j=1}^m\overline{B}_{\rho _k^j}(a_{k}^j),\qquad \Omega _k(\alpha )=B_{\alpha }(0)\setminus \bigcup _{j=1}^m\overline{B}_{\alpha ^{-1}\rho _k^j}(a_k^j). \end{aligned}$$

Assume that for all $1\le j\ne j'\le m$, and all $0<\alpha <1$, we have $B_{\alpha ^{-1}\rho _k^j}(a_k^j)\cap B_{\alpha ^{-1}\rho _k^{j'}}(a_k^{j'})=\varnothing$ for k large enough, and

$$\begin{aligned} \rho _k^j\underset{k\rightarrow \infty }{\longrightarrow }0,\qquad a_{k}^j\underset{k\rightarrow \infty }{\longrightarrow }0. \end{aligned}$$

Furthermore, assume that

$$\begin{aligned} \sup _{k\in {\mathbb {N}}}\int _{\Omega _k}|\nabla \vec {n}_k|^2\mathrm{d}x\le \varepsilon _1(n),\qquad \sup _{k\in {\mathbb {N}}}\left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k)}<\infty , \end{aligned}$$

where $\varepsilon _1(n)$ is given by the proof of Theorem 2.1. Finally, assume that

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\int _{\Omega _k(\alpha )}|\nabla \vec {n}_k|^2\mathrm{d}x=0 \end{aligned}$$

and that there exists a $W^{2,2}_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} )\cap C^{\infty }(B_1(0)\setminus \left\{ 0\right\} )$ immersion $\vec {\Phi }_{\infty }$ such that

$$\begin{aligned} \log |\nabla \vec {\Phi }_{\infty }|\in L^{\infty }_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} ) \end{aligned}$$

and $\vec {\Phi }_k\underset{k\rightarrow \infty }{\longrightarrow }\vec {\Phi }_{\infty }$ in $C^l_{{\mathrm {loc}}}(B_1(0)\setminus \left\{ 0\right\} )$. For all $k\in {\mathbb {N}}$, let

$$\begin{aligned} e^{\lambda _k}=\frac{1}{\sqrt{2}}|\nabla \vec {\Phi }_k| \end{aligned}$$

be the conformal factor of $\vec {\Phi }_k$. Then, there exists a positive integer $\theta _0\ge 1$, and for all $k\in {\mathbb {N}}$ integers $\theta _k^1,\cdots ,\theta _k^m\in {\mathbb {N}}$ such that for all $k\in {\mathbb {N}}$ large enough

$$\begin{aligned} \sum _{j=1}^m\theta _k^j=\theta _0-1, \end{aligned}$$

and for all $k\in {\mathbb {N}}$, there exists $1/2<\alpha _k<1$ and $A_k\in {\mathbb {R}}$ such that

$$\begin{aligned} \left\| \lambda _k-\sum _{j=1}^j\theta _k^j\log |z-a_k^j|-A_k\right\| _{{\mathrm {L}}^{\infty }(\Omega _k(\alpha _k))}\le \Gamma _{14}\left( \left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k)}+\int _{\Omega _k}|\nabla \vec {n}_k|^2\mathrm{d}x\right) \end{aligned}$$

(3.62)

for some universal constant $\Gamma _{14}=\Gamma _{14}(n)$. Furthermore, we have for all $0< \rho _k\le 1$ such that

$$\begin{aligned} \bigcup _{j=1}^m\overline{B}_{\rho _k^j}(a_k^j)\subset B_{\rho _k}(0). \end{aligned}$$

and for all $k\in {\mathbb {N}}$ large enough

$$\begin{aligned} \frac{1}{2\pi }\int _{\partial B_{\rho _k}(0)}*\, \mathrm{d}\nu _k=\theta _0-1. \end{aligned}$$

Finally, for all $k\in {\mathbb {N}}$ and $j\in \left\{ 1,\cdots ,m\right\}$, we have

$$\begin{aligned} \frac{1}{2\pi }\int _{\partial B_{\rho _k^j}(a_k^j)}*\, \mathrm{d}\nu _k=\theta _k^j\in {\mathbb {Z}}. \end{aligned}$$

Proof

Indeed, the same argument shows that there exists a holomorphic function $\varphi _k$ on $\Omega _k$ and $c_k^1,\cdots ,c_k^m\in {\mathbb {C}}$ such that

$$\begin{aligned} e^{\nu _k}=\varphi _k(z)e^{-i\theta _k}+\sum _{j=1}^{m}\frac{c_k^je^{i\theta _k}}{\overline{z}-\overline{a_k^j}} \end{aligned}$$

and the same computation shows if

$$\begin{aligned} \psi _k(z)=\varphi _k(z)-\sum _{j=1}^{m}\frac{\overline{c_k^j}}{z-a_k^j} \end{aligned}$$

then

$$\begin{aligned} e^{i\theta _k}=\frac{\psi _k(z )}{|\psi _k(z)|}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \text {d}\theta _k={\mathrm {Im}}\,\left( \frac{\partial \psi _k}{\psi _k}\right) \end{aligned}$$

and for all $1\le j\le m$

$$\begin{aligned} \int _{\partial B_{\rho _k^j}(a_k^j)}\text {d}\theta _k\in 2\pi {\mathbb {Z}}. \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\int _{\partial B_1(0)}\text {d}\theta _k=2\pi (\theta _0-1)\ge 0. \end{aligned}$$

(3.63)

In particular, if $\left\{ \rho _k\right\} _{k\in {\mathbb {N}}}\subset (0,\infty )$ is such that $\rho _k\underset{k\rightarrow \infty }{\longrightarrow } 0$ and

$$\begin{aligned} \bigcup _{j=1}^m\overline{B}_{\rho _k^j}(a_k^j)\subset B_{\alpha ^{-1}\rho _k}(0), \end{aligned}$$

then we also have for $k\in {\mathbb {N}}$ large enough

$$\begin{aligned} \frac{1}{2\pi }{\mathrm {Im}}\,\int _{\partial B_{\rho _k}(0)}\frac{\partial \psi _k}{\psi _k}=\theta _0-1\ge 0, \end{aligned}$$

which implies that $\psi _k$ admits a holomorphic extension on $B_1(0)$. Analytic continuation then implies that for all $1\le j\le m$

$$\begin{aligned} \sum _{j=1}^{m}\theta _k^j=\frac{1}{2\pi }{\mathrm {Im}}\,\int _{\partial B_{\rho _k}(0)}\frac{\partial \psi _k}{\psi _k}=\theta _k^j\ge 0. \end{aligned}$$

Therefore, we have by (3.63) for k large enough

$$\begin{aligned} \frac{1}{2\pi }\sum _{j=1}^m\int _{\partial B_{\rho _k^j}(a_k^j)}\text {d}\theta _k=\theta _0-1. \end{aligned}$$

Then, we deduce by the argument of Lemma V.3 of [2] that there exists a universal constant $\Gamma _{15}(n)=\Gamma _{15}(n)$ such that for all $k\in {\mathbb {N}}$ there exists $1/2<\alpha _k<1$ such that for all $k\in {\mathbb {N}}$ large enough

$$\begin{aligned} \left\| \nu _k-\sum _{j=1}^{m}\theta _k^j\log |z-a_j|-A_k\right\| _{{\mathrm {L}}^{\infty }(\Omega _k(\alpha _k))}\le \Gamma _{15}(n)\left( \left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k)}+\int _{\Omega _k}|\nabla \vec {n}_k|^2\mathrm{d}x\right) , \end{aligned}$$

(3.64)

In particular, as $\mu _k\in L^{\infty }(B_1(0))$, we get the estimate (3.62) from (3.64) and $\left\| \mu _k\right\| _{{\mathrm {L}}^{\infty }(B_1(0))}\le \Gamma _{16}$ for some universal $\Gamma _{16}=\Gamma _{16}(\Lambda ,n)$ (thanks to Wente’s estimate), we deduce that there exists a universal constant $C=C(n,\Lambda )$, where

$$\begin{aligned} \Lambda =\sup _{k\in {\mathbb {N}}}\left( \left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k)}+\int _{\Omega _k}|\nabla \vec {n}_k|^2\mathrm{d}x\right) \end{aligned}$$

such that for all k large enough and $z\in \Omega _k(1/2)$ (noticing that $A_k$ is bounded by the strong convergence outside of 0)

$$\begin{aligned} \frac{1}{C}\le \frac{e^{\lambda _k(z)}}{\displaystyle \prod _{j=1}^{m}|z-a_k^j|^{\theta _k^j}}\le C. \end{aligned}$$

(3.65)

These additional remarks complete the proof of the Proposition. $\square$

Remarks 3.7

• (1) The integers $\theta _k^j$ a priori depend on k, but we will see in the case of interest of bubbling of Willmore immersions, they must stabilise for k large enough.

• (2) The reader will notice that we do not need the limiting immersion to be smooth, but merely $C^{1,\alpha }$ for some $0<\alpha <1$ (this allows one to define branch points, [8]). As in the application we restrict to Willmore immersions, we automatically get the smoothness of the limiting immersion outside of the point of concentration.

Theorem 3.5 also has an analogue in this setting, but we will not state it for the sake of brevity of the paper.

4 Improved Energy Quantization for Willmore Immersions

In this section, we build on [2] to obtain an improved no-neck energy.

Theorem 4.1

Let $\Sigma$ be a closed Riemann surface and assume that $\{\vec {\Phi }_k\}_{k\in {\mathbb {N}}}$ is a sequence of smooth Willmore immersions such that

$$\begin{aligned} \limsup _{k\rightarrow \infty }W(\vec {\Phi }_k)<\infty . \end{aligned}$$

Assume furthermore that the conformal class of $g_k=\vec {\Phi }_k^{*}g_{{\mathbb {R}}^n}$ is precompact in the moduli space. Then for all $0<\alpha <1$ let $\Omega _k(\alpha )=B_{\alpha R_k}\setminus \overline{B}_{\alpha ^{-1}r_k}(0)$ be a neck domain and $\theta _0\in {\mathbb {N}}$ such that (by Theorem 3.1)

$$\begin{aligned} \theta _0-1=\lim \limits _{\alpha \rightarrow 0}\lim \limits _{k\rightarrow \infty }\int _{\partial B_{\alpha ^{-1}r_k}(0)}\partial _{\nu }\lambda _k\,\mathrm{d}{\mathscr {H}}^1, \end{aligned}$$

(4.1)

and define

$$\begin{aligned} \Lambda =\sup _{k\in {\mathbb {N}}}\left( \left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k(1))}+\int _{\Omega _k(1)}|\nabla \vec {n}_k|^2\mathrm{d}x\right) . \end{aligned}$$

Then, there exist a universal constant $\Gamma _{17}=\Gamma _{17}(n)$, and $\alpha _0=\alpha _0(\{\vec {\Phi }_k\}_{k\in {\mathbb {N}}})>0$ such that for all $0<\alpha <\alpha _0$ and $k\in {\mathbb {N}}$ large enough,

$$\begin{aligned} \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}&\le \Gamma _{17}(n)e^{\Gamma _{17}(n)\Lambda }\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(4\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(4\alpha ))}. \end{aligned}$$

(4.2)

In particular, we deduce by the $L^2$ no-neck energy

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}=0. \end{aligned}$$

Proof

Step 1: $L^{2,1}$-quantization of the mean curvature Here, we will prove that

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left( \left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+\left\| e^{\lambda _k}\nabla \vec {H}_k\right\| _{{\mathrm {L}}^{1}(\Omega _k(\alpha ))}\right) =0. \end{aligned}$$

This statement is a consequence of the following lemma.

Theorem 4.2

There exists constants $R_0(n),\varepsilon _4(n)>0$ with the following property. Let $0<100r<R\le R_0(n)$, and $\vec {\Phi }:B_R(0)\rightarrow {\mathbb {R}}^n$ be a weak conformal Willmore immersion of finite total curvature, such that

$$\begin{aligned} \sup _{r<s<R/2}\int _{B_{2s}\setminus \overline{B}_s(0)}|\nabla \vec {n}|^2\mathrm{d}x\le \varepsilon _4(n). \end{aligned}$$

Set $\Omega =B_R\setminus \overline{B}_r(0)$, and

$$\begin{aligned} \Lambda =\left\| \nabla \lambda \right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x, \end{aligned}$$

where $\lambda$ is the conformal parameter of $\vec {\Phi }$. Then there exists a universal constant $\Gamma _{18}=\Gamma _{18}(n)$ such that for all $\left( \dfrac{4r}{5R}\right) ^{\frac{1}{3}}<\alpha <\dfrac{1}{5}$, we have

$$\begin{aligned}&\left\| e^{\lambda }\vec {H}\right\| _{{\mathrm {L}}^{2,1}(\Omega _{\alpha })}+\left\| e^{\lambda }\nabla \vec {H}\right\| _{{\mathrm {L}}^{1}(\Omega _{\alpha })}\nonumber \\&\quad \le \Gamma _{18}(n)\left( 1+\Lambda \right) \,e^{\Gamma _{18}(n)\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

(4.3)

Remarks on the proof

The proof closely follows the proof in [2]. In Step 1, we use the previous results to obtain the $L^{2,1}\cap W^{1,1}$ control for the harmonic parts of tensors, and the Wente inequality for the part with Dirichlet boundary conditions.

In Step 2, we use a structural property of the unit normal $\vec {n}$ to transfer the $L^{2,1}$ control of $e^{\lambda }\vec {H}$ into a $L^{2,1}$ control of $\nabla \vec {n}$. The proof uses other results on moving frames from [11], and the rest follows again by classical Calderón–Zygmund estimates, Wente inequality, and an averaging lemma. The proof is quite lengthy but globally straightforward.

Remark 4.3

Notice that by $L^{2,1}/L^{2,\infty }$ duality, we have

$$\begin{aligned} \left\| \nabla (e^{\lambda }\vec {H})\right\| _{{\mathrm {L}}^{1}(\Omega _{\alpha })}&\le \left\| (\nabla \lambda )e^{\lambda }\vec {H}\right\| _{{\mathrm {L}}^{1}(\Omega _{\alpha })}+\left\| e^{\lambda }\vec {H}\right\| _{{\mathrm {L}}^{1}(\Omega _{\alpha })}\\&\le \left\| \nabla \lambda \right\| _{{\mathrm {L}}^{2,\infty }(\Omega _{\alpha })}\left\| e^{\lambda }\vec {H}\right\| _{{\mathrm {L}}^{2,1}(\Omega _{\alpha })} +\left\| e^{\lambda }\nabla \vec {H}\right\| _{{\mathrm {L}}^{1}(\Omega _{\alpha })} \end{aligned}$$

Proof

Define for all $\left( \dfrac{r}{R}\right) ^{\frac{1}{2}}<\alpha <1$ the open subset $\Omega _{\alpha }=B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}$ of $\Omega$. We follow step by steps the proof of Lemma VI.6 of [2]. First, the pointwise estimate on $\nabla \vec {n}$ is identical and we find that there exists $\Gamma _{19}=\Gamma _{19}(n),\Gamma _{19}'=\Gamma _{19}'(n)>0$ such that for all $z\in B_{4R/5}\setminus \overline{B}_{5r/4}(0)$

$$\begin{aligned} |\nabla \vec {n}(z)|\le \frac{\Gamma _{19}}{|z|^2}\int _{B_{2|z|}\setminus \overline{B}_{|z|/2}(0)}|\nabla \vec {n}|^2\mathrm{d}{\mathscr {L}}^2\le \frac{\Gamma _{19}'\sqrt{\varepsilon _4(n)}}{|z|} \end{aligned}$$

(4.4)

so that

$$\begin{aligned} \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\le \sqrt{\pi }\Gamma _{19}'\sqrt{\varepsilon _4(n)} \end{aligned}$$

and we can choose $\varepsilon _4(n)=\dfrac{\varepsilon _1(n)^2}{\sqrt{\pi }\Gamma _{19}'(n)}.$ Therefore, thanks to Theorem 2.1, there exists $d\in {\mathbb {R}}$ such that

$$\begin{aligned} |d|\le \Gamma _{20}(n)\Lambda . \end{aligned}$$

and for all $\left( \dfrac{5}{4}\right) ^{\frac{2}{3}}\left( \dfrac{r}{R}\right) ^{\frac{1}{3}}=\left( \dfrac{\frac{5r}{4}}{\frac{4R}{5}}\right) ^{\frac{1}{3}}<\dfrac{5\alpha }{4}<\dfrac{1}{4}$, there exists $A_{\alpha }\in {\mathbb {R}}$ such that

$$\begin{aligned} \left\| \lambda -d\log |z|-A_{\alpha }\right\| _{{\mathrm {L}}^{\infty }(\Omega _{\alpha })}\le \Gamma _0'(n)\sqrt{\frac{5\alpha }{4}}\Lambda +\Gamma _0'(n)\le \Gamma _0''(n)\left( \sqrt{\alpha }\Lambda +\int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) . \end{aligned}$$

(4.5)

As $\vec {\Phi }$, is Willmore, the following 1-form is closed :

$$\begin{aligned} \vec {\alpha }={\mathrm {Im}}\,\left( \partial \vec {H}+|\vec {H}|^2\partial \vec {\Phi }+g^{-1}\otimes \langle \vec {H},\vec {h}_0\rangle \otimes \overline{\partial }\vec {\Phi }\right) . \end{aligned}$$

As $\vec {\Phi }$ is well defined on $B_R(0)$, the Poincaré lemma implies that there exists $\vec {L}:B_R(0)\rightarrow {\mathbb {R}}^n$ such that

$$\begin{aligned} 2i\,\partial \vec {L}=\partial \vec {H}+|\vec {H}|^2\partial \vec {\Phi }+g^{-1}\otimes \langle \vec {H},\vec {h}_0\rangle \otimes \overline{\partial }\vec {\Phi }. \end{aligned}$$

(4.6)

Now, introduce for $0<s<R/2$

$$\begin{aligned} \delta (s)=\left( \frac{1}{s^2}\int _{B_{2s}\setminus \overline{B}_{s/2}(0)}|\nabla \vec {n}|^2\mathrm{d}x\right) ^{\frac{1}{2}}. \end{aligned}$$

Then, we have trivially for all $2r<s<R/2$

$$\begin{aligned} s\delta (s)\le \left( \int _{\Omega }|\nabla \vec {n}|^2\mathrm{d}x\right) ^{\frac{1}{2}}=\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )} \end{aligned}$$

(4.7)

and Fubini’s theorem implies that for all $r\le r_1<r_2\le R/2$

$$\begin{aligned} \int _{r_1}^{r_2}s\delta (s)^2\text {d}s&=\int _{r_1}^{r_2}\frac{1}{s}\left( \int _{B_{2s}\setminus \overline{B}_{s/2}(0)}|\nabla \vec {n}(x)|^2\mathrm{d}x\right) \text {d}s\nonumber \\&=\int _{r_1}^{r_2}\int _{B_{2r_2}\setminus \overline{B}_{r_1/2}(0)}\frac{|\nabla \vec {n}(x)|^2}{s}\varvec{1}_{\left\{ s/2\le |x|\le 2s\right\} }\mathrm{d}x\text {d}s\nonumber \\&=\log (4)\int _{B_{2r_2}\setminus \overline{B}_{r_1}(0)}|\nabla \vec {n}|^2\mathrm{d}x. \end{aligned}$$

(4.8)

Now, (4.4) shows that for some $C_{10}=C_{10}(n)$

$$\begin{aligned} \max \left\{ e^{\lambda (z)}|\vec {H}(z)|,e^{\lambda (z)}|\vec {H}_0(z)|\right\} \le |\nabla \vec {n}(z)|\le C_{10}\delta (|z|). \end{aligned}$$

(4.9)

Furthermore, the same argument of [2] (see [1] for more details) using a Theorem from [7] implies that there exists a constant $C_{11}=C_{11}(n)$ such that

$$\begin{aligned} e^{\lambda (z)}|\nabla \vec {H}(z)|\le C_{11}\frac{\delta (|z|)}{|z|}\qquad \text {for all}\;\, z\in \Omega _{1/2} \end{aligned}$$

(4.10)

Therefore, we have thanks to (4.6), (4.9) and (4.10)

$$\begin{aligned} |\nabla \vec {L}(z)|=2|\partial \vec {L}(z)|\le e^{-\lambda (z)}\left( C_{11}\frac{\delta (z)}{|z|}+2C_{10}\delta (z)^2\right) . \end{aligned}$$

(4.11)

Now assume for simplicity that $\alpha =1/2$ (then we do not need to use the precised form (4.5) and we can use instead Lemma V.3 from [2]). Denoting for all $r<s<R$

we deduce from (4.10) that for all $z\in \Omega _{1/2}$ (taking $\alpha =1/2$ in (4.5))

$$\begin{aligned} |\vec {L}(z)-\vec {L}_{|z|}|&\le \int _{\partial B_{|z|}}|\nabla \vec {L}|\text {d}{\mathscr {H}}^1\le 2\pi e^{2\Gamma _1\Lambda }e^{-\lambda (|z|)}\left( C_{11}\delta (|z|)+2C_{10}|z|\delta (|z|)\cdot \delta (|z|)\right) \nonumber \\&\le 2\pi e^{2\Gamma _1\Lambda }e^{-\lambda (|z|)}\left( C_{11}+2C_{10}\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \delta (|z|). \end{aligned}$$

(4.12)

Then, we get

$$\begin{aligned}&\int _{\Omega _{1/2}}|\vec {L}(z)-\vec {L}_{|z|}|^2e^{2\lambda (z)}|\text {d}z|^2\nonumber \\&\quad \le 2\pi e^{2\Gamma _1\Lambda }\left( C_{11}+2C_{10}\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \int _{\Omega }\delta (|z|)^2|\text {d}z|^2\nonumber \\&\quad =4\pi ^2e^{2\Gamma _1\Lambda }\left( C_{11}+2C_{10}\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \int _{2r}^{R/2}s\delta ^2(s)\text {d}s\nonumber \\&\quad =4\pi ^2\log (4)e^{2\Gamma _1\Lambda }\left( C_{11}+2C_{10}\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

(4.13)

Now, we continue the proof in an exact same way to obtain the pointwise estimate (for some universal constant $C_{12}=C_{12}(n)$)

$$\begin{aligned} e^{\lambda (z)}|z||\vec {L}_{|z|}|\le C_{12}e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

(4.14)

Therefore, we get

$$\begin{aligned} \left\| e^{\lambda (z)}\vec {L}_{|z|}\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _{1/2})}\le 2\sqrt{\pi }C_{12}e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

(4.15)

Combining (4.13) and (4.15) implies as $\left\| \,\cdot \,\right\| _{{\mathrm {L}}^{2,\infty }(\,\cdot \,)}\le \left\| \,\cdot \,\right\| _{{\mathrm {L}}^{2}(\,\cdot \,)}$ that

$$\begin{aligned} \left\| e^{\lambda }\vec {L}\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _{1/2})}&\le \left( 8\pi ^2\log (4)+2\sqrt{\pi }C_{12}\right) e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\\&=C_{13}(n)e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

The estimates (4.12), (4.14) and (4.7) imply that for all $z\in \Omega _{1/2}$

$$\begin{aligned} e^{\lambda (z)}|\vec {L}(z)|&\le \left( 2\pi \max \left\{ C_{11}(n),2C_{10}(n)\right\} \right. \nonumber \\&\quad \left. +C_{13}(n)\right) e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left( \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}+|z|\delta (|z|)\right) |z|^{-1}\nonumber \\&\le 2\left( 2\pi \max \left\{ C_{11}(n),2C_{10}(n)\right\} +C_{13}(n)\right) \nonumber \\&\quad \times e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\,|z|^{-1}\nonumber \\&=C_{14}(n)e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\frac{1}{|z|}. \end{aligned}$$

(4.16)

Now, recall that there exists $S:B_R(0)\rightarrow {\mathbb {R}}$ and $\vec {R}:B_R(0)\rightarrow \Lambda ^2{\mathbb {R}}^n$ such that

$$\begin{aligned} \left\{ \begin{aligned}&\nabla S=\vec {L}\cdot \nabla \vec {\Phi }\\&\nabla \vec {R}=\vec {L}\wedge \nabla \vec {\Phi }+2\,\vec {H}\wedge \nabla ^{\perp }\vec {\Phi }, \end{aligned}\right. \end{aligned}$$

we trivially obtain from the pointwise inequality (4.16), (4.9) and (4.7) for all $z\in \Omega _{1/2}$

$$\begin{aligned} |\nabla S(z)|\le 2C_{14}(n)e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\frac{1}{|z|} \end{aligned}$$

and

$$\begin{aligned} |\nabla \vec {R}(z)|&\le 2C_{14}(n)e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\frac{1}{|z|}+4C_{10}(n)\delta (|z|)\\&\le 2\left( C_{14}(n)+2C_{10}(n)\right) e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\frac{1}{|z|}. \end{aligned}$$

Therefore, if $C_{15}(n)=4\sqrt{\pi }(C_{14}(n)+2C_{10}(n))>0$, we deduce that

$$\begin{aligned}&\left\| \nabla S\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _{1/2})}\le C_{15}(n)e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\nonumber \\&\left\| \nabla \vec {R}\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _{1/2})}\le C_{15}(n) e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

(4.17)

Now, define for all $2r\le \rho <\dfrac{R}{2}$

Following the exact same steps as [2], we find that for some universal constant $C_{16}=C_{16}(n)$

$$\begin{aligned} \left| \frac{\text {d}S_\rho }{\text {d}\rho }\right| +\left| \frac{\text {d}\vec {R}_{\rho }}{\text {d}\rho }\right| \le C_{16}(n)e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\delta (\rho ). \end{aligned}$$

(4.18)

Therefore, (4.8) and (4.18) imply that

$$\begin{aligned} \int _{2r}^{\frac{R}{2}}\left( \left| \frac{\text {d} S_{\rho }}{\text {d}\rho }\right| ^2+\left| \frac{\text {d}\vec {R}_{\rho }}{\text {d}\rho }\right| ^2\right) \rho \,\text {d}{\mathscr {L}}^1(\rho )\le C_{16}(n)^2e^{4\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) ^2\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}^3. \end{aligned}$$

(4.19)

We will now use a precised version of Lemma VI.2 of [2] (proved in [13], see also [15]).

Lemma 4.4

There exists a universal constant $R_0>0$ with the following property. Let $0<4r<R<R_0$, $\Omega =B_R\setminus \overline{B}_r(0)\rightarrow {\mathbb {R}}$, $a,b:\Omega \rightarrow {\mathbb {R}}$ such that $\nabla a\in L^{2,\infty }(\Omega )$ and $\nabla b\in L^2(\Omega )$, and $u:\Omega \rightarrow {\mathbb {R}}$ be a solution of

$$\begin{aligned} \Delta \varphi =\nabla a\cdot \nabla ^{\perp }b\qquad \text {in}\;\ \Omega . \end{aligned}$$

Furthermore, define for $r\le \rho \le R$

Then $\nabla \varphi \in L^2(\Omega )$, and there exists a positive constant $\Gamma _{20}>0$ independent of $0<4r<R$ such that for all $\left( \dfrac{r}{R}\right) ^{\frac{1}{2}}<\alpha <\dfrac{1}{2}$

$$\begin{aligned} \left\| \nabla \varphi \right\| _{{\mathrm {L}}^{2}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r})}\le \Gamma _{20}\left( \left\| \nabla a\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\left\| \nabla b\right\| _{{\mathrm {L}}^{2}(\Omega )}+\left\| \nabla \overline{\varphi }_r\right\| _{{\mathrm {L}}^{2}(\Omega )}+\left\| \nabla \varphi \right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\right) . \end{aligned}$$

Proof

Let $\widetilde{a}:B_R(0)\rightarrow {\mathbb {R}}$ and $\widetilde{b}:B_R(0)\rightarrow {\mathbb {R}}$ the extensions of a and b given by Theorem 7.2. As $0<4r<R$ and scaling invariance of the $L^{2,\infty }$ and the $L^{2}$ norm of the gradient, we deduce that there exists a universal constant $\Gamma _{20}>>0$ such that

$$\begin{aligned}&\left\| \nabla \widetilde{a}\right\| _{{\mathrm {L}}^{2,\infty }(B_R(0))}\le \Gamma _{20}\left( \left\| \nabla a\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\left\| a\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\right) \\&\left\| \nabla \widetilde{b}\right\| _{{\mathrm {L}}^{2}(B_R(0))}\le \Gamma _{20}\left( \left\| \nabla b\right\| _{{\mathrm {L}}^{2}(\Omega )}+\left\| b\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) . \end{aligned}$$

Thanks to Poincaré-Wirtinger inequality, and as $\widetilde{a}=a$ and $\widetilde{b}=b$ on $\Omega$, we deduce that

$$\begin{aligned} \left\| \nabla \widetilde{a}\right\| _{{\mathrm {L}}^{2,\infty }(B_R(0))}&\le \Gamma _{20}\left( \left\| \nabla a\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\left\| a-\overline{\widetilde{a}}_{B_R(0)}\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\right) \\&=\Gamma _{20}\left( \left\| \nabla a\right\| _{{\mathrm {L}}^{2,\infty }(B_R(0))}+\left\| \widetilde{a}-\overline{\widetilde{a}}_{B_R(0)}\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\right) \\&\le \Gamma _{20}\left( \left\| \nabla a\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\left\| \widetilde{a}-\overline{\widetilde{a}}_{B_R(0)}\right\| _{{\mathrm {L}}^{2,\infty }(B_R(0))}\right) \\&\le \Gamma _{20}\left\| \nabla a\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}+\Gamma _{20}C_{PW}(L^{2,\infty })R\left\| \nabla \widetilde{a}\right\| _{{\mathrm {L}}^{2,\infty }(B_R(0))}. \end{aligned}$$

Therefore, if $\Gamma _{20}C_{PW}(L^{2,\infty })R_0\le \dfrac{1}{2}$, we find

$$\begin{aligned} \left\| \nabla \widetilde{a}\right\| _{{\mathrm {L}}^{2,\infty }(B_R(0))}\le 2\Gamma _{20}\left\| \nabla a\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}, \end{aligned}$$

and likewise, provided $\Gamma _{20}C_{PW}(L^2)R_0\le \dfrac{1}{2}$, we find

$$\begin{aligned} \left\| \nabla \widetilde{b}\right\| _{{\mathrm {L}}^{2}(B_R(0))}\le 2\Gamma _{20}\left\| \nabla b\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

Now, let $u:B_R(0)\rightarrow {\mathbb {R}}$ be the solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta u&=\nabla \widetilde{a}\cdot \nabla ^{\perp }\widetilde{b}\qquad&\text {in}\;\, B_R(0)\\ u&=0\qquad&\text {on}\;\, \partial B_R(0). \end{aligned}\right. \end{aligned}$$

Then, the improved Wente inequality of Bethuel ([11], 3.3.6) and the scaling invariance show that there exists a universal constant $\Gamma _{21}>>0$ such that

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_R(0))}&\le \Gamma _{21}\left\| \nabla \widetilde{a}\right\| _{{\mathrm {L}}^{2,\infty }(B_R(0))}\left\| \nabla \widetilde{b}\right\| _{{\mathrm {L}}^{2}(B_R(0))}\\&\le 4 \Gamma _{20}^2\Gamma _{21}\left\| \nabla a\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\left\| \nabla b\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

Now, let $v=\varphi -u-\overline{(\varphi -u)}_r$. Then, v is a harmonic function such that for all $r<\rho <R$

$$\begin{aligned} \int _{\partial B_{\rho }}\partial _{\nu }v\,\mathrm{d} {\mathscr {H}}^1=0. \end{aligned}$$

Therefore, Lemma 2.2 implies that

$$\begin{aligned} \left\| \nabla v\right\| _{{\mathrm {L}}^{2}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}&\le \Gamma _1\left\| \nabla v\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )} \\&\le \Gamma _1'\left( \left\| \nabla a\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\left\| \nabla b\right\| _{{\mathrm {L}}^{2}(\Omega )}+\left\| \nabla \varphi _r\right\| _{{\mathrm {L}}^{2}(\Omega )}+\left\| \nabla \varphi \right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\right) \end{aligned}$$

which concludes the proof. $\square$

Now, recall that the following system holds

First, thanks to Lemma IV.1 of [2], we extend the restriction $\vec {n}:B_R\setminus \overline{B}_r(0)\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ to a map $\widetilde{\vec {n}}:B_R(0) \rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ such that

$$\begin{aligned} \left\| \nabla \widetilde{\vec {n}}\right\| _{{\mathrm {L}}^{2}(B_R(0))}\le C_0(n)\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

In particular, we have

(4.20)

Therefore, applying the proof of Lemma 4.4 by using the already constructed extension of $\vec {n}$, we deduce thanks to (4.17) and (4.19) that

$$\begin{aligned} \left\| \nabla S\right\| _{{\mathrm {L}}^{2}(\Omega _{1/4})}+\left\| \nabla \vec {R}\right\| _{{\mathrm {L}}^{2}(\Omega _{1/4})}\le C_{17}(n)e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

As in [2], we obtain readily

$$\begin{aligned} \left\| \nabla S\right\| _{{\mathrm {L}}^{2}(\Omega _{1/2})}+\left\| \nabla \vec {R}\right\| _{{\mathrm {L}}^{2}(\Omega _{1/2})}&\le C_{18}(n)e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\nonumber \\ \left\| S\right\| _{{\mathrm {L}}^{\infty }(\Omega _{1/2})}+\left\| \vec {R}\right\| _{{\mathrm {L}}^{\infty }(\Omega _{1/2})}&\le C_{18}(n)e^{2\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

(4.21)

Now, introduce the following slight variant from a Lemma of [13].

Lemma 4.5

Let $R_0>0$ be the constant of Lemma 4.4. Let $0<16r<R<R_0$, $\Omega =B_R\setminus \overline{B}_r(0)\rightarrow {\mathbb {R}}$, $a,b:\Omega \rightarrow {\mathbb {R}}$ such that $\nabla a\in L^{2}(\Omega )$ and $\nabla b\in L^2(\Omega )$, and $\varphi :\Omega \rightarrow {\mathbb {R}}$ be a solution of

$$\begin{aligned} \Delta \varphi =\nabla a\cdot \nabla ^{\perp }b\quad \text {in}\;\ \Omega . \end{aligned}$$

Assume that $\left\| \varphi \right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}<\infty$. Then there exists a universal constant $\Gamma _{22}>0$ such that for all $\left( \dfrac{r}{R}\right) ^{\frac{1}{2}}<\alpha <\dfrac{1}{4}$,

$$\begin{aligned}&\left\| \varphi \right\| _{{\mathrm {L}}^{\infty }(\Omega )}+\left\| \nabla \varphi \right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}+\left\| \nabla ^2\varphi \right\| _{{\mathrm {L}}^{1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}\\&\quad \le \Gamma _{22}\left( \left\| \nabla a\right\| _{{\mathrm {L}}^{2}(\Omega )}\left\| \nabla b\right\| _{{\mathrm {L}}^{2}(\Omega )}+\left\| \varphi \right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}\right) . \end{aligned}$$

Proof

As in the proof of Lemma 4.4, introduce extensions $\widetilde{a}:B_R(0)\rightarrow {\mathbb {R}}$ and $\widetilde{b}:B_R(0)\rightarrow {\mathbb {R}}$ of a and b, such that

$$\begin{aligned}&\Vert \nabla \widetilde{a}\Vert _{{\mathrm {L}}^{2}(B_R(0))}\le 2\Gamma _{20}\left\| \nabla a\right\| _{{\mathrm {L}}^{2}(\Omega )}\\&\Vert \nabla \widetilde{b}\Vert _{{\mathrm {L}}^{2}(B_R(0))}\le 2\Gamma _{20}\left\| \nabla b\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

Now, let $v:B_R(0)\rightarrow {\mathbb {R}}$ be the solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta v&=\nabla \widetilde{a}\cdot \nabla ^{\perp }\widetilde{b}\qquad&\text {in}\;\, B_R(0)\\ v&=0\qquad&\text {on}\;\, \partial B_R(0). \end{aligned}\right. \end{aligned}$$

Then, the improved Wente inequality and the Coifman-Lions-Meyer-Semmes estimate [4] show (by scaling invariance of the different norms considered) that

$$\begin{aligned} \left\| v\right\| _{{\mathrm {L}}^{\infty }(B_R(0))}+\left\| \nabla v\right\| _{{\mathrm {L}}^{2,1}(B_R(0))}+\left\| \nabla ^2v\right\| _{{\mathrm {L}}^{1}(B_R(0))}\le \Gamma _{22}\left\| \nabla a\right\| _{{\mathrm {L}}^{2}(\Omega )}\left\| \nabla b\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

(4.22)

Now let $u=\varphi -v$. Then, u is harmonic, and let $d\in {\mathbb {R}}$, $\left\{ a_n\right\} _{n\in {\mathbb {Z}}}\subset {\mathbb {C}}$ such that

$$\begin{aligned} u(z)=a_0+d\log |z|+{\mathrm {Re}}\,\left( \sum _{n\in {\mathbb {Z}}^{*}}a_nz^n\right) . \end{aligned}$$

Then, we have by the maximum principle for all $r\le \rho \le R$

$$\begin{aligned} |a_0+d\log \rho |=\left| \frac{1}{2\pi }\int _{0}^{2\pi }u(\rho e^{i\theta })\text {d}\theta \right| \le \left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} |d|\log \left( \frac{R}{r}\right)&=\left| a_0+d\log R-\left( a_0+d\log r\right) \right| \nonumber \\&\le |a_0+d\log R|+|a_0+d\log r|\le 2\left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}. \end{aligned}$$

(4.23)

Now, recall that

$$\begin{aligned}&\left\| \nabla \log |z|\right\| _{{\mathrm {L}}^{2,1}(B_{R}\setminus \overline{B}_{r}(0))}=4\sqrt{\pi }\left( \log \left( \frac{R}{r}\right) +\log \left( 1+\sqrt{1-\left( \frac{r}{R}\right) ^2}\right) \right) \nonumber \\&\left\| \nabla ^2\log |z|\right\| _{{\mathrm {L}}^{1}(B_R\setminus \overline{B}_{r}(0))}=4\left\| \partial _{z}^2\log |z|\right\| _{{\mathrm {L}}^{1}(B_R\setminus \overline{B}_r(0))}=4\pi \log \left( \frac{R}{r}\right) . \end{aligned}$$

(4.24)

Therefore, as $R>4r$, (4.23) and (4.24) imply that

$$\begin{aligned}&\left\| \nabla \left( d\log |z|\right) \right\| _{{\mathrm {L}}^{2,1}(B_R\setminus \overline{B}_r(0))}\le 4\sqrt{\pi }\left( \log \left( \frac{R}{r}\right) +\log (2)\right) |d|\le 16\sqrt{\pi }\left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}\nonumber \\&\left\| \nabla ^2\left( d\log |z|\right) \right\| _{{\mathrm {L}}^{1}(B_R\setminus \overline{B}_r(0))}\le 8\pi \left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}. \end{aligned}$$

(4.25)

These estimates (4.23) imply by Lemmas 2.3 and 4.5 imply that

$$\begin{aligned}&\left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}+\left\| \nabla ^2u\right\| _{{\mathrm {L}}^{1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}\nonumber \\&\quad \le 64\frac{\sqrt{2}+\sqrt{\pi }}{\sqrt{15}}(2\alpha ) \left\| \nabla \left( u-d\log |z|\right) \right\| _{{\mathrm {L}}^{2}(B_{R/2}\setminus \overline{B}_{r/2}(0))}\nonumber \\&\qquad +24\pi \left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}\nonumber \\&\quad \le 128\frac{\sqrt{2}+\sqrt{\pi }}{\sqrt{15}}\alpha \left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_{R/2}\setminus \overline{B}_{2r}(0))}+\left( 24\pi +256\pi \frac{\sqrt{2}+\sqrt{\pi }}{\sqrt{15}}\alpha \right) \left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}. \end{aligned}$$

(4.26)

Now, recall that the mean value formula and the maximum principle ([9] 1.10) imply that for all $x\in B_{R}\setminus \overline{B}_r(0)$, and $0<\rho <{\mathrm {dist}}(x,\partial \Omega )$,

$$\begin{aligned} |\nabla u(x)|\le \frac{2}{\rho }\left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial B_{\rho }(x))}\le \frac{2}{\rho }\left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}. \end{aligned}$$

(4.27)

As

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_{R/2}^2\setminus \overline{B}_{2r}(0))}^2&=\int _{\partial B_{R/2}(0)} u\,\partial _{\nu }u\,\text {d}{\mathscr {H}}^1-\int _{\partial B_{2r}(0)}u\,\partial _{\nu }u\,\text {d}{\mathscr {H}}^1\nonumber \\&\le \left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}\left( \int _{\partial B_{R/2}(0)}|\nabla u|\text {d}{\mathscr {H}}^1+\int _{\partial B_{2r}(0)}|\nabla u|\text {d}{\mathscr {H}}^1\right) \end{aligned}$$

(4.28)

the estimate (4.27) shows that

$$\begin{aligned} \int _{\partial B_{R/2}(0)}|\nabla u|\text {d}{\mathscr {H}}^1+\int _{\partial B_{2r}(0)}|\nabla u|\text {d}{\mathscr {H}}^1&\le 4\int _{\partial B_{R/2}(0)}\frac{\left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}}{R}\text {d}{\mathscr {H}}^1\nonumber \\&\quad +\int _{\partial B_{2r}(0)}\frac{\left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}}{r}\text {d}{\mathscr {H}}^1\nonumber \\&=8\pi \left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}. \end{aligned}$$

(4.29)

Therefore, we have by (4.28) and (4.29)

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2}(B_{R/2}\setminus \overline{B}_{2r}(0))}\le 2\sqrt{2\pi }\left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}. \end{aligned}$$

(4.30)

Combining (4.26) and (4.30) shows that

$$\begin{aligned}&\left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}+\left\| \nabla ^2u\right\| _{{\mathrm {L}}^{1}(B_{\alpha R}\setminus \overline{B}_{\alpha ^{-1}r}(0))}\nonumber \\&\quad \le \left( 256\frac{2\sqrt{\pi }+\pi \sqrt{2}}{\sqrt{15}}\alpha +24\pi +256\pi \frac{\sqrt{2}+\sqrt{\pi }}{\sqrt{15}}\alpha \right) \left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}\nonumber \\&\quad \le \left( 24\pi +512\pi \frac{\sqrt{2}+\sqrt{\pi }}{\sqrt{15}}\alpha \right) \left\| u\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega )}. \end{aligned}$$

(4.31)

Combining the maximum principle and inequalities (4.22), (4.31) yields the expected estimate. $\square$

Now, apply Lemma 4.5 to the estimates (4.21) shows by using the previous extension $\widetilde{\vec {n}}$ of $\vec {n}$ that

$$\begin{aligned}&\left\| \nabla S\right\| _{{\mathrm {L}}^{2,1}(\Omega _{1/2})}+\left\| \nabla \vec {R}\right\| _{{\mathrm {L}}^{2,1}(\Omega _{1/2})}\le C_{19}(n)e^{4\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\nonumber \\&\left\| \nabla ^2S\right\| _{{\mathrm {L}}^{1}(\Omega _{1/2})}+\left\| \nabla ^2\vec {R}\right\| _{{\mathrm {L}}^{1}(\Omega _{1/2})}\le C_{19}(n)e^{4\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

(4.32)

As (see [28] for the definition of the restriction operator between a 2-vector and a vector)

(4.33)

we trivially have

$$\begin{aligned} \left\| e^{\lambda }\vec {H}\right\| _{{\mathrm {L}}^{2,1}(\Omega _{1/2})}&\le \left\| \nabla S\right\| _{{\mathrm {L}}^{2,1}(\Omega _{1/2})}+\left\| \nabla \vec {R}\right\| _{{\mathrm {L}}^{2,1}(\Omega _{1/2})}\nonumber \\&\le 2C_{19}(n)e^{4\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}. \end{aligned}$$

(4.34)

Now, (4.33) implies that

so that

As $\nabla \lambda \in L^{2,\infty }$, $e^{\lambda }\vec {H}\in L^{2,1}$ and $e^{-\lambda }\nabla ^2\vec {\Phi }\in L^{2,\infty }$, we deduce by (4.32) and (4.34) that

$$\begin{aligned} \left\| e^{\lambda }\nabla \vec {H}\right\| _{{\mathrm {L}}^{1}(\Omega _{1/2})}\le C_{20}(n)\left( 1+\Lambda \right) e^{4\Gamma _1\Lambda }\left( 1+\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}\right) \left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega )}, \end{aligned}$$

and this concludes the proof of the Theorem. $\square$

For all neck region of the form $\Omega _k=B_{R_k}\setminus \overline{B}_{r_k}(0)$, define for all $0<\alpha <1$

$$\begin{aligned} \Omega _k(\alpha )=B_{\alpha R_k}\setminus \overline{B}_{\alpha ^{-1}r_k}(0). \end{aligned}$$

The estimate (4.3) implies that

$$\begin{aligned}&\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+\left\| e^{\lambda _k}\nabla \vec {H}_k\right\| _{{\mathrm {L}}^{1}(\Omega _k(\alpha ))}\nonumber \\&\quad \le C_{21}(n)\left( 1+\Lambda \right) e^{C_{21}\Lambda }\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}, \end{aligned}$$

(4.35)

where

$$\begin{aligned} \Lambda =\sup _{k\in {\mathbb {N}}}\left( \left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k)}+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k)}\right) <\infty , \end{aligned}$$

is finite by hypothesis. Therefore, the no-neck energy

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}=0 \end{aligned}$$

(4.36)

implies by (4.35) that

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left( \left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+\left\| e^{\lambda _k}\nabla \vec {H}_k\right\| _{{\mathrm {L}}^{1}(\Omega _k(\alpha ))}\right) =0. \end{aligned}$$

Step 2: $L^{2,1}$-quantization of the Weingarten tensor The proof relies on an algebraic computation first given in [28] (II.10). We will give its easy derivation in codimension 1.

Algebraic identity in codimension 1. Let $\vec {\Phi }:B_1(0)\rightarrow {\mathbb {R}}^3$ be a conformal immersion, and $\vec {n}:B_1(0)\rightarrow S^2$ be its unit normal. If $e^{\lambda }=\dfrac{1}{\sqrt{2}}|\nabla \vec {\Phi }|$ is the associated conformal parameter and $\vec {e}_j=e^{-\lambda }\partial _{x_j}\vec {\Phi }$ for $j=1,2$, we have by definition

$$\begin{aligned} \vec {n}=\vec {e}_1\times \vec {e}_2, \end{aligned}$$

where $\times$ is the vector product. Recall the Grassmann identity valid for all $\vec {u},\vec {v},\vec {w}\in {\mathbb {R}}^3$

$$\begin{aligned} \left( \vec {u}\times \vec {v}\right) \times \vec {w}=\langle \vec {u},\vec {w}\rangle \vec {v}-\langle \vec {v},\vec {w}\rangle \vec {u}. \end{aligned}$$

Therefore, we deduce that

$$\begin{aligned} \left\{ \begin{aligned} \vec {n}\times \vec {e}_1&=\left( \vec {e}_1\times \vec {e}_2\right) \times \vec {e}_1=\langle \vec {e}_1,\vec {e}_1\rangle \vec {e}_2-\langle \vec {e}_2,\vec {e}_1\rangle \vec {e}_1=\vec {e}_2\\ \vec {n}\times \vec {e}_2&=-\vec {e}_1. \end{aligned}\right. \end{aligned}$$

(4.37)

As $|\vec {n}|=1$, we have for all $j=1,2$

$$\begin{aligned} \partial _{x_j}\vec {n}&=\langle \nabla _{\partial _{x_j}}\vec {n},\vec {e}_1\rangle \vec {e}_1+\langle \nabla _{\partial _{x_j}}\vec {n},\vec {e}_2\rangle \vec {e}_2 =-{\mathbb {I}}_{1,j}\vec {e}_1-{\mathbb {I}}_{2,j}\vec {e}_2. \end{aligned}$$

This implies that

$$\begin{aligned} \nabla \vec {n}=(-{\mathbb {I}}_{1,1}\vec {e}_1-{\mathbb {I}}_{1,2}\vec {e}_2,-{\mathbb {I}}_{1,2}\vec {e}_1-{\mathbb {I}}_{2,2}\vec {e}_2) \end{aligned}$$

(4.38)

and (4.37) combined with the identity $\vec {u}\times \vec {v}=-(\vec {v}\times \vec {u})$ (valid for all $\vec {u},\vec {v}\in {\mathbb {R}}^3$) yield

$$\begin{aligned} \partial _{x_j}\vec {n}\times \vec {n}=-{\mathbb {I}}_{1,j}\vec {e}_1\times \vec {n}-{\mathbb {I}}_{2,j}\vec {e}_2\times \vec {n}={\mathbb {I}}_{1,j}\vec {e}_2-{\mathbb {I}}_{2,j}\vec {e}_1. \end{aligned}$$

Therefore, we deduce that

$$\begin{aligned} \nabla ^{\perp }\vec {n}\times \vec {n}=\left( \partial _{x_2}\vec {n}\times \vec {n},-\partial _{x_1}\vec {n}\times \vec {n}\right) =(-{\mathbb {I}}_{2,2}\vec {e}_1+{\mathbb {I}}_{1,2}\vec {e}_2,-{\mathbb {I}}_{1,1}\vec {e}_2+{\mathbb {I}}_{1,2}\vec {e}_1). \end{aligned}$$

(4.39)

As

$$\begin{aligned} e^{\lambda }H=\frac{1}{2}\left( {\mathbb {I}}_{1,1}+{\mathbb {I}}_{2,2}\right) , \end{aligned}$$

(4.40)

the identities (4.39) and (4.40) show that

$$\begin{aligned} \nabla ^{\perp }\vec {n}\times \vec {n}+2H\nabla \vec {\Phi }&=({\mathbb {I}}_{1,1}\vec {e}_1+{\mathbb {I}}_{1,2}\vec {e}_2,{\mathbb {I}}_{1,2}\vec {e}_1+{\mathbb {I}}_{2,2}\vec {e}_2). \end{aligned}$$

(4.41)

Comparing (4.41) and (4.38), we deduce that

$$\begin{aligned} \nabla \vec {n}=\vec {n}\times \nabla ^{\perp }\vec {n}-2H\,\nabla \vec {\Phi }. \end{aligned}$$

(4.42)

Taking the divergence of this equation, we find

$$\begin{aligned} \Delta \vec {n}=\nabla \vec {n}\times \nabla ^{\perp }\vec {n}-2\,{{\,\mathrm{div}\,}}(H\nabla \vec {\Phi }). \end{aligned}$$

Argument in arbitrary codimension Then, we can find a trivialisation of $\vec {n}$ such that $\vec {n}=\vec {n}_1\wedge \vec {n}_2\wedge \cdots \wedge \vec {n}_{n-2}$ satisfying the Coulomb condition

$$\begin{aligned} {{\,\mathrm{div}\,}}\left( \nabla \vec {n}_{\beta }\cdot \vec {n}_{\gamma }\right) =0\qquad \text {for all}\;\, 1\le \beta ,\gamma \le n-2. \end{aligned}$$

(4.43)

Furthermore, recall that for all $1\le \beta \le n-2$, [28] implies that (using (4.43) for the second condition)

$$\begin{aligned} \nabla \vec {n}_{\beta }&=-*\left( \vec {n}\wedge \nabla ^{\perp }\vec {n}_{\beta }\right) +\sum _{\gamma =1}^{n-2}\langle \nabla \vec {n}_{\beta },\vec {n}_{\gamma }\rangle \cdot \vec {n}_{\gamma }-2\,H_{\beta }\nabla \vec {\Phi }\end{aligned}$$

(4.44)

Taking the divergence of this equation yields by the Coulomb condition (4.43)

$$\begin{aligned} \Delta \vec {n}_{\beta }=-*\left( \nabla \vec {n}\wedge \nabla ^{\perp }\vec {n}_{\beta }\right) +\sum _{\gamma =1}^{n-2}\langle \nabla \vec {n}_{\beta },\vec {n}_{\gamma }\rangle \cdot \nabla \vec {n}_{\gamma }-2\,{{\,\mathrm{div}\,}}\left( H_{\beta }\nabla \vec {\Phi }\right) . \end{aligned}$$

(4.45)

Now, as in (2.39) (recall that this comes from Lemma IV.1. in [2]), construct for small enough $\alpha >0$ and k large enough (thanks to the no-neck property) an extension $\widetilde{\vec {n}}_k:B_{\alpha R_k}(0)\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ of $\vec {n}_k:\Omega _k(\alpha )=B_{\alpha R_k}(0)\setminus \overline{B}_{\alpha ^{-1}r_k}(0)\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ such that for some universal constant $C_{22}=C_{22}(n)>0$

$$\begin{aligned} \left\| \nabla \widetilde{\vec {n}}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\le C_{22}\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}. \end{aligned}$$

(4.46)

Furthermore, as in Lemma IV.1 of [2] (see also [11] 4.1.3–4.1.7), we can construct extensions $\widetilde{\vec {n}}_{k}^{\beta }$ of $\vec {n}_k^{\beta }$ on $B_{\alpha R_k}(0)$ such that

$$\begin{aligned} \widetilde{\vec {n}}_k=\widetilde{\vec {n}}_k^{1}\wedge \cdots \wedge \widetilde{\vec {n}}_k^{n-2}\qquad \text {on}\;\, B_{\alpha R_k}(0) \end{aligned}$$

satisfying the Coulomb condition for all $1\le \beta ,\gamma \le n-2$

$$\begin{aligned} \left\{ \begin{aligned} {{\,\mathrm{div}\,}}\left( \nabla \widetilde{\vec {n}}_k^{\beta }\cdot \widetilde{\vec {n}}_k^{\gamma }\right)&=0\qquad&\text {for all}\;\, B_{\alpha R_k}(0)\\ \partial _{\nu }\widetilde{\vec {n}}_k^{\beta }\cdot \widetilde{\vec {n}}_k^{\gamma }&=0\qquad&\text {on}\;\,\text {on}\;\, \partial B_{\alpha R_k}(0) \end{aligned}\right. \end{aligned}$$

(4.47)

and for all $1\le \beta \le n-2$ (by (4.46) for the second inequality)

$$\begin{aligned} \left\| \nabla \widetilde{\vec {n}}_k^{\beta }\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\le \widetilde{C_{22}}(n)\left\| \nabla \widetilde{\vec {n}}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\le C_{22}'(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}. \end{aligned}$$

(4.48)

Furthermore, using [11] 4.1.7, we have the estimate for all $1\le \beta \le n-2$

$$\begin{aligned} \left\| \nabla \widetilde{\vec {n}}_{k}^{\beta }\cdot \widetilde{\vec {n}}_k^{\gamma }\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\le C_{22}''(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2. \end{aligned}$$

(4.49)

Let us recall the argument for this crucial step. By (4.47), there exists $\vec {A}_{\beta ,\gamma }:B_{\alpha R_k}(0)\rightarrow {\mathbb {R}}^n$ such that

$$\begin{aligned} \nabla ^{\perp }\vec {A}_{\beta ,\gamma }=\nabla \widetilde{\vec {n}}_{k}^{\beta }\cdot \widetilde{\vec {n}_k}^{\gamma }. \end{aligned}$$

(4.50)

Furthermore, the boundary conditions of (4.47) imply that we can choose $\vec {A}_{\beta \gamma }$ such that $\vec {A}_{\beta ,\gamma }=0$ on $\partial B_{\alpha R_k}(0)$. Therefore, we have

$$\begin{aligned} \left\{ \begin{aligned} \Delta \vec {A}_{\beta ,\gamma }&=\nabla \widetilde{\vec {n}}_{k}^{\beta }\cdot \nabla ^{\perp }\widetilde{\vec {n}}_k^{\gamma }\qquad&\text {in}\;\, B_{\alpha R_k}(0)\\ \vec {A}_{\beta ,\gamma }&=0\qquad&\text {on}\;\, \partial B_{\alpha R_k}(0) \end{aligned} \right. \end{aligned}$$

(4.51)

Therefore, we get by the improved Wente estimate and (4.48)

$$\begin{aligned}&\left\| \nabla \vec {A}_{\beta ,\gamma }\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\le C_{22}''(n)\left\| \nabla \widetilde{\vec {n}}_k^{\beta }\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\left\| \nabla \widetilde{\vec {n}}_k^{\gamma }\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\nonumber \\&\quad \le C_{22}''(n)C_{22}'(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2. \end{aligned}$$

(4.52)

Combining the pointwise identity (4.50) with (4.52) yields (4.49).

Now fix some $1\le \beta \le n-2$ and let $\vec {u}_k:B_{\alpha R_k}(0)\rightarrow {\mathbb {R}}^n$ be the unique solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta \vec {u}_k&=-*\left( \nabla \widetilde{\vec {n}}_k\wedge \nabla ^{\perp }\widetilde{\vec {n}}_k^{\beta }\right) +\sum _{\gamma =1}^{n-2}\langle \nabla \widetilde{\vec {n}}_k^{\beta },\widetilde{\vec {n}}_k^{\gamma }\rangle \cdot \nabla \widetilde{\vec {n}}_k^{\beta }\qquad&\text {in}\;\, B_{\alpha R_k}(0)\\ \vec {u}_k&=0\qquad&\text {in}\;\, \partial B_{\alpha R_k}(0). \end{aligned}\right. \end{aligned}$$

(4.53)

Now, thanks to (4.43), we can apply [4], scaling invariance and (4.46) to find that there exists $C_{23}=C_{23}(n)>0$ such that

$$\begin{aligned} \left\| \nabla ^2\vec {u}_k\right\| _{{\mathrm {L}}^{1}(B_{\alpha R_k}(0))}\le C_1\left\| \nabla \widetilde{\vec {n}}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))} ^2\le C_{22}^2C_{23}\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2. \end{aligned}$$

Furthermore, as $\vec {u}_k=0$ on $\partial B_{\alpha R_k}(0)$, and scaling invariance (of $\left\| u_k\right\| _{{\mathrm {L}}^{\infty }(B_{\alpha R_k}(0))}$, $\left\| \nabla \vec {u}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}$ and $\left\| \nabla ^2u_k\right\| _{{\mathrm {L}}^{1}(B_{\alpha R_k}(0))}$) and Sobolev embedding, there exists $C_{24}=C_{24}(n)>0$ such that

$$\begin{aligned} \left\| \vec {u}_k\right\| _{{\mathrm {L}}^{\infty }(B_{\alpha R_k}(0))}+\left\| \nabla \vec {u}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}+\left\| \nabla ^2\vec {u}_k\right\| _{{\mathrm {L}}^{1}(B_{\alpha R_k}(0))}\le C_{24}\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2. \end{aligned}$$

(4.54)

Now, by Theorem (4.2), $H_k^{\beta }\nabla \vec {\Phi }_k\in L^{2,1}(\Omega _k(\alpha ))$. Furthermore, as

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{R_k}{r_k}=\infty ,\qquad \limsup _{k\rightarrow \infty }R_k<\infty , \end{aligned}$$

there exists by Theorem 7.2 an extension $\vec {F}_k:B_{\alpha R_k}(0)\rightarrow {\mathbb {R}}^n$ of $H_k^{\beta }\nabla \vec {\Phi }_k$ such that for all k large enough

$$\begin{aligned} \left\| \vec {F}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\le C_{25}(n)\left\| H_{k}^{\beta }\nabla \vec {\Phi }_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}, \end{aligned}$$

where $C_{25}(n)>0$ is independent of k large enough and $0<\alpha <\alpha _0(n)$ fixed (small enough with respect to some $\alpha _0(n)>0$). Now, let $\vec {v}_k:\Omega _k(\alpha )\rightarrow {\mathbb {R}}^n$ be the solution of the system

$$\begin{aligned} \left\{ \begin{aligned} \Delta \vec {v}_k&=-2{{\,\mathrm{div}\,}}\left( \vec {F}_k\right) \qquad&\text {in}\;\, B_{\alpha R_k}(0)\\ \vec {v}_k&=0\qquad&\text {on}\;\, \partial B_{\alpha R_k}(0). \end{aligned}\right. \end{aligned}$$

As we trivially have

$$\begin{aligned} \left\| {{\,\mathrm{div}\,}}(\vec {F}_k)\right\| _{{\mathrm {W}}^{-1,(2,1)}(B_{\alpha R_k}(0))}\le \left\| \vec {F}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}, \end{aligned}$$

scaling invariance and standard Calderón–Zygmund estimates show that there exits a universal constant $C_{26}=C_{26}(n)$ such that

$$\begin{aligned} \left\| \nabla \vec {v}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}&\le C_{26}(n)\left\| \vec {F}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\le C_{25}(n)C_{26}(n)\left\| H_{k}^{\beta }\nabla \vec {\Phi }_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\nonumber \\&\le 2C_{25}(n)C_{26}(n)\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}. \end{aligned}$$

(4.55)

Furthermore, the Sobolev embedding shows that for some universal constant $\Gamma _{23}>0$

$$\begin{aligned} \left\| \vec {v}_k\right\| _{{\mathrm {L}}^{\infty }(B_{\alpha R_k}(0))}\le \Gamma _{23}\left\| \nabla \vec {v}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\le \,\Gamma _{23}C_{25}(n)C_{26}(n)\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}. \end{aligned}$$

(4.56)

Finally, let $\vec {\varphi }_k=\vec {n}_k^{\beta }-\vec {u}_k-\vec {v}_k$. The $\vec {\varphi }_k:\Omega _k(\alpha )\rightarrow {\mathbb {R}}^n$ is harmonic and

$$\begin{aligned} \left\{ \begin{aligned} \Delta \vec {\varphi }_k&=0\qquad&\text {in}\;\, \Omega _k(\alpha )\\ \varphi _k&=\vec {n}_k^{\beta }\qquad&\text {on}\;\, \partial B_{\alpha R_k}(0)\\ \varphi _k&=\vec {n}_k^{\beta }-\vec {u}_k-\vec {v}_k\qquad&\text {on}\;\, \partial B_{\alpha ^{-1}r_k}(0). \end{aligned}\right. \end{aligned}$$

In particular, as $\vec {u}_k,\vec {v}_k,\vec {n}_k^{\beta }\in L^{\infty }(\Omega _k(\alpha ))$ (as $|\vec {n}_k^{\alpha }|=1$ and using the bounds (4.54) and (4.69)), if $\vec {d}_k\in {\mathbb {R}}$ and $\left\{ \vec {a}_n\right\} _{n\in {\mathbb {Z}}}\subset {\mathbb {C}}^n$ are such that

$$\begin{aligned} \vec {\varphi }_k(z)=\vec {a}_0+\vec {d_k}\log |z|+{\mathrm {Re}}\,\left( \sum _{n\in {\mathbb {Z}}^{*}}\vec {a}_nz^n\right) , \end{aligned}$$

then

$$\begin{aligned} |\vec {d}_k|&\le \frac{\left\| \vec {\varphi }_k\right\| _{{\mathrm {L}}^{\infty }(\partial \Omega _k(\alpha ))}}{\log \left( \frac{\alpha ^2R_k}{r_k}\right) } \nonumber \\&\le \frac{2}{\log \left( \frac{\alpha ^2R_k}{r_k}\right) }\left( 1+C_{27}(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2+C_{27}(n)\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\right) , \end{aligned}$$

(4.57)

so that by the proof of Lemma 4.5

$$\begin{aligned}&\left\| \nabla \vec {\varphi }_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha /2))} \nonumber \\&\quad \le 16\sqrt{\pi }+C_{28}(n)\left( \left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}+\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\right) \nonumber \\&\left\| \nabla ^2\vec {\varphi }_k\right\| _{{\mathrm {L}}^{1}(\Omega _k(\alpha /2))}\nonumber \\&\quad \le 8\pi +C_{28}(n)\left( \left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}+\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\right) . \end{aligned}$$

(4.58)

Finally, we have by (4.54), (4.55), (4.58) and Theorem 4.2 for some $C_{29}(n)>0$

$$\begin{aligned} \left\| \nabla \vec {n}_k^{\beta }\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha /2))}&\le \left\| \nabla \varphi _k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ^2))}+\left\| \nabla \vec {u}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ^2))}+\left\| \nabla \vec {v}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ^2))}\nonumber \\&\le 16\sqrt{\pi }+C_{29}(n)\left( 1+\Lambda \right) \left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}. \end{aligned}$$

(4.59)

Therefore, the no-neck energy yields for all $1\le \beta \le n-2$

$$\begin{aligned} \limsup _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \nabla \vec {n}_k^{\beta }\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\le 16\sqrt{\pi }. \end{aligned}$$

(4.60)

Now, as

$$\begin{aligned} |\nabla \vec {n}_k|&=\left| \sum _{\beta =1}^{n-2}\vec {n}_{k}\wedge \cdots \wedge \nabla \vec {n}_{k}^{\beta }\wedge \cdots \wedge \vec {n}_k^{n-2}\right| \le \sum _{\beta =1}^{n-2}|\nabla \vec {n}_k^{\beta }|, \end{aligned}$$

(4.61)

we deduce from (4.60) that

$$\begin{aligned} \limsup _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \nabla \vec {n}_{k}\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\le 16\sqrt{\pi }(n-2)<\infty \end{aligned}$$

(4.62)

Now, define $\overline{\vec {n}}_{k}:B_{\alpha R_k}(0)\setminus \overline{B}_{\alpha ^{-1}r_k}(0)$ such that for all $z\in \Omega _k(\alpha )$ such that $|z|=r$

We will prove that for certain universal constants $C_{30}(n)$

$$\begin{aligned} \left\| \nabla \overline{\vec {n}}_{k}\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\le C_{30}(n)e^{\Gamma _2(n)\Lambda }\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}, \end{aligned}$$

(4.63)

and this will finish the proof of the Theorem by using Lemmas 4.4, 4.5. Indeed, notice that the following lemma implies by (4.54) and (4.55) that

$$\begin{aligned}&\left\| \nabla \overline{\vec {u}_k}\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\le C_{31}(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2\nonumber \\&\left\| \nabla \overline{\vec {v}_k}\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\le C_{31}(n)\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}. \end{aligned}$$

(4.64)

Lemma 4.6

Let $n\ge 2$, $0<r<R<\infty$, $\Omega =B_R\setminus \overline{B}_r(0)\subset {\mathbb {R}}^n$, $1\le p<\infty$ and assume that $u\in W^{1,p}(B_R\setminus \overline{B}_r(0))$. Define $\overline{u}:\Omega \rightarrow {\mathbb {R}}$ to be the radial function such that for all $r<t<R$ if $t=|x|$, then

Then, $\overline{u}\in W^{1,p}(\Omega )$ and

$$\begin{aligned} \left\| \nabla \overline{u}\right\| _{{\mathrm {L}}^{p}(\Omega )}\le \left\| \nabla u\right\| _{{\mathrm {L}}^{p}(\Omega )}. \end{aligned}$$

Furthermore, for all $1<p<\infty$, and $1\le q\le \infty$, there exists a constant C(p, q) independent of $0<r<R<\infty$ such that for all $u\in W^{1,(p,q)}(\Omega )$, $\overline{u}\in W^{1,(p,q)}(\Omega )$ and

$$\begin{aligned} \left\| \nabla \overline{u}\right\| _{{\mathrm {L}}^{p,q}(\Omega )}\le C(p,q)\left\| \nabla u\right\| _{{\mathrm {L}}^{p,q}(\Omega )}. \end{aligned}$$

Proof

First, assume that $u\in W^{1,p}(\Omega )$ for some $1\le p<\infty$. Recall that by the proof of Proposition 2.7, we have

(4.65)

Therefore, as $\overline{u}$ is radial, we have by the co-area formula

$$\begin{aligned} \left\| \nabla \overline{u}\right\| _{{\mathrm {L}}^{p}(\Omega )}^p=\beta (n)\int _{r}^{R}\left| \frac{\mathrm{d}}{\text {d}t}u_t\right| ^pt^{n-1}\text {d}t. \end{aligned}$$

(4.66)

Furthermore, by Hölder’s inequality and (4.65),

$$\begin{aligned} \left| \frac{d}{\text {d}t}u_t\right| ^p\le & {} \frac{1}{\left( \beta (n)t^{n-1}\right) ^p}\left| \int _{\partial B_t(0)}|\nabla u|\text {d}{\mathscr {H}}^{n-1}\right| ^p \nonumber \\\le & {} \frac{1}{(\beta (n)t^{n-1})^p}\int _{\partial B_t(0)}|\nabla u|^p\,\text {d}{\mathscr {H}}^{n-1}\left( \beta (n)t^{n-1}\right) ^{\frac{p}{p'}}\end{aligned}$$

(4.67)

$$\begin{aligned}= & {} \frac{1}{\beta (n)t^{n-1}}\int _{\partial B_t(0)}|\nabla u|^p\,\text {d}{\mathscr {H}}^{n-1}. \end{aligned}$$

(4.68)

Putting together (4.66) and (4.67), we find by a new application of the co-area formula

$$\begin{aligned} \left\| \nabla \overline{u}\right\| _{{\mathrm {L}}^{p}(\Omega )}^p&\le \int _{r}^R\left( \int _{\partial B_t(0)}|\nabla u|^p\text {d}{\mathscr {H}}^{n-1}\right) \mathrm{d}t=\int _{B_R\setminus \overline{B}_r(0)}|\nabla u|^p\mathrm{d}{\mathscr {L}}^n\\&=\left\| \nabla u\right\| _{{\mathrm {L}}^{p}(\Omega )}^p. \end{aligned}$$

The last statement comes from the Stein-Weiss Interpolation Theorem ([11], 3.3.3). $\square$

Now, in order to obtain (4.63), recall the algebraic equation on $\Omega _k(\alpha )$ from (4.44)

$$\begin{aligned} \nabla \vec {n}_{k}^{\beta }=-*\left( \vec {n}_k\wedge \nabla ^{\perp }\vec {n}_k^{\beta }\right) +\sum _{\gamma =1}^{n-2}\langle \nabla \vec {n}_k^{\beta },\vec {n}_k^{\gamma }\rangle -2\,H_k^{\beta }\nabla \vec {\Phi }_k. \end{aligned}$$

To simplify notations, let

$$\begin{aligned} \vec {G}_k=\sum _{\beta =1}^{n-2}\langle \nabla \vec {n}_k^{\beta },\vec {n}_k^{\gamma }\rangle -2\,H_k^{\beta }\nabla \vec {\Phi }_k. \end{aligned}$$

Then, (4.49) implies that

$$\begin{aligned} \left\| \vec {G}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\le C_{32}(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2+4\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}. \end{aligned}$$

(4.69)

We have

(4.70)

Furthermore, by (4.69) and Lemma 4.6, we have (as $\overline{\vec {G}}_k$ is radial)

$$\begin{aligned} \left\| \frac{d}{\text {d}t}{\vec {G}}_{k,t}\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}&=\left\| \nabla \overline{\vec {G}}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\nonumber \\&\le C_{33}(n)\left( \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2+\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\right) . \end{aligned}$$

(4.71)

Now, the $\varepsilon$-regularity ([28], I.5) combined with the small $L^2$ norm of $\nabla \vec {n}_k$ in $\Omega _k(2\alpha )$ implies that there exists a universal constant $C_{34}(n)$ such that

$$\begin{aligned} \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{\infty }(\partial B_t)}\le \frac{C_{34}(n)}{t}\left( \int _{B_{2t}\setminus \overline{B}_{t/2}(0)}|\nabla \vec {n}_k|^2\mathrm{d}x\right) ^{\frac{1}{2}} \end{aligned}$$

so that

$$\begin{aligned} \left\| \vec {n}_k-\overline{\vec {n}}_{k,t}\right\| _{{\mathrm {L}}^{\infty }(\partial B_t(0))}\le \int _{\partial B_t(0)}|\nabla \vec {n}_k|\,\text {d}{\mathscr {H}}^1\le 2\pi C_{34}(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(B_{2t}\setminus \overline{B}_{t/2}(0))}. \end{aligned}$$

Therefore,

(4.72)

The proof of Lemma 4.6 now implies by (4.72) that

(4.73)

Finally, thanks to (4.70), (4.71) and (4.73), we find

$$\begin{aligned} \left\| \nabla \overline{\vec {n}}_{k}^{\beta }\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}&\le C_{33}(n)\left( \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2+\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\right) \nonumber \\&\quad +C_{35}(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\left\| \nabla \vec {n}_k^{\beta }\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}. \end{aligned}$$

(4.74)

Therefore, (4.59) and (4.74) imply that

$$\begin{aligned}&\left\| \nabla \overline{\vec {n}}_k^{\beta }\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))} \nonumber \\&\quad \le C_{33}(n)\left( \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2+\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\right) \nonumber \\&\qquad +C_{35}(n)\left( 16\sqrt{\pi }+C_9(n)\left( 1+\Lambda \right) \left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\nonumber \\&\quad \le C_{36}(n)\left( 1+\Lambda \right) ^2e^{4\Gamma _1(n)\Lambda }\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}. \end{aligned}$$

(4.75)

Therefore, (4.64) and (4.75) imply that

$$\begin{aligned}&\left\| \nabla \overline{\vec {\varphi }_k}\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))} \nonumber \\&\quad \le \left\| \nabla \overline{\vec {n}_k}\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+\left\| \nabla \overline{\vec {u}_k}\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+\left\| \nabla \overline{\vec {v}}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}\nonumber \\&\quad \le C_{37}(n)e^{\Gamma _2(n)\Lambda }\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))} \end{aligned}$$

(4.76)

We can now use Lemma 2.3 (or equivalently Proposition 2.5) and Lemma 4.6 to get for all $0<\beta <1$

$$\begin{aligned}&\left\| \nabla \left( \vec {\varphi }_k-\overline{\vec {\varphi _k}}\right) \right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\beta \alpha ))}\le 24\beta \left\| \nabla \left( \vec {\varphi _k}-\overline{\vec {\varphi }_k}\right) \right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))} \le 48\beta \left\| \nabla \vec {\varphi }_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}\nonumber \\&\quad \le 48\beta \left( \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}+\left\| \nabla \vec {u}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}+\left\| \nabla \vec {v}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}\right) \nonumber \\&\quad \le C_{38}(n)\beta \, e^{\Gamma _2(n)\Lambda }\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}. \end{aligned}$$

(4.77)

Therefore, taking $\beta =1/2$ in (4.77), we get by (4.76) and (4.77) show that

$$\begin{aligned} \left\| \nabla \vec {\varphi }_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha /2))}\le C_{39}(n)\, e^{\Gamma _2(n)\Lambda }\left( 1+\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}\right) \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}. \end{aligned}$$

(4.78)

Finally, by (4.54), (4.55) and (4.56), we obtain the expected estimate for $\vec {n}_k^{\beta }=\vec {u}_k+\vec {v}_k+\vec {\varphi }_k$ on $\Omega _k(\alpha /2)$, and for $\vec {n}_k$ by the algebraic inequality (4.61). $\square$

Remark 4.7

Observe that for the mean curvature, we have the improved (because of the Sobolev embedding $W^{1,1}({\mathbb {R}}^2)\hookrightarrow L^{2,1}({\mathbb {R}}^2)$) no-neck energy

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| e^{\lambda _k}\nabla \vec {H}_k\right\| _{{\mathrm {L}}^{1}(\Omega _k(\alpha ))}=0 \end{aligned}$$

but this is not completely clear if this also holds for $\nabla ^2\vec {n}_k$ (here, $\Omega _k(\alpha )=B_{\alpha R_k}\setminus \overline{B}_{\alpha ^{-1}r_k}(0)$). However, notice that (4.51) implies that

$$\begin{aligned} \left\| \nabla ^2\vec {A}_{\beta ,\gamma }\right\| _{{\mathrm {L}}^{1}(B_{\alpha R_k}(0))}\le C(n)\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2 \end{aligned}$$

and as $\nabla ^{\perp }\vec {A}_{\beta ,\gamma }=\nabla \widetilde{\vec {n}}_k^{\beta }\cdot \widetilde{\vec {n}}_k^{\gamma }$, we deduce that

$$\begin{aligned} \nabla ^2\widetilde{\vec {n}}_k^{\beta }\cdot \widetilde{\vec {n}}_k^{\gamma }+\nabla \widetilde{\vec {n}}_{k}^{\beta }\cdot \nabla \widetilde{\vec {n}}_{k}^{\gamma }\in L^1(B_{\alpha R_k}(0)), \end{aligned}$$

and by the Cauchy–Schwarz inequality, this implies that for all $1\le \beta ,\gamma \le n-2$

$$\begin{aligned} \left\| \nabla ^2\widetilde{\vec {n}}_k^{\beta }\cdot \widetilde{\vec {n}}_{k}^{\gamma }\right\| _{{\mathrm {L}}^{1}(B_{\alpha R_k}(0))}\le C'(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}^2. \end{aligned}$$

Therefore, we deduce as $\widetilde{\vec {n}}_k^{\beta }=\vec {n}_k^{\beta }$ on $\Omega _k(\alpha )$ that

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \pi _{\vec {n}_k}(\nabla ^2{\vec {n}}_k)\right\| _{{\mathrm {L}}^{1}(\Omega _k(\alpha ))}=0, \end{aligned}$$

(where $\pi _{\vec {n}_k}$ is the projection on the normal bundle) but this is not completely clear how one may obtain the same result for the tangential part of $\nabla ^2\vec {n}_k$.

We finish this section by the proof of Corollary 1.4.

Proof of Corollary 1.4

Introduce for all $\alpha >0$ small enough the domain decomposition of [2]:

$$\begin{aligned} \Sigma =\left( \Sigma \setminus \bigcup _{i=1}^m\overline{B}_{\alpha }(a_i)\right) \cup \Omega _k(\alpha )\cup \sum _{i=1}^m\sum _{j=1}^{m_i}B(i,j,\alpha ,k), \end{aligned}$$

where

$$\begin{aligned} \Omega _k(\alpha )&=\left( \bigcup _{i=1}^mB_{\alpha }(a_i)\setminus \bigcup _{j=1}^{m_i}B_{\alpha ^{-1}\rho _k^{i,j}(x_k^{i,j})}\right) \bigcup _{i=1}^m\bigcup _{j=1}^{m_i}\bigcup _{j'\in I^{i,j}}\\&\qquad \left( B_{\alpha \rho _k^{i,j}}(x_k^{i,j'})\setminus \bigcup _{j''\in I^{i,j}}B_{\alpha ^{-1}\rho _k^{i,j''}}(x_k^{i,j''})\right) \\&=\bigcup _{i=1}^m\Omega _k^i(\alpha )\bigcup _{i=1}^m\bigcup _{j=1}^{m_i}\bigcup _{j'\in I^{i,j}}\Omega _k^{i,j,j'}(\alpha ) \end{aligned}$$

and

$$\begin{aligned} B(i,j,\alpha ,k)=B_{\alpha ^{-1}\rho _{k}^{i,j}}(x_k^{i,j})\setminus \bigcup _{j'\in I^{i,j}}B_{\alpha \rho _k^{i,j}}(x_k^{i,j'}) \end{aligned}$$

and for all $1\le i\le m$, for all $1\le j\le m_i$, we have

$$\begin{aligned} \frac{\rho _k^{i,j}}{\rho _{k}^{i,j'}}\underset{k\rightarrow \infty }{\longrightarrow }\infty . \end{aligned}$$

Thanks to the no-neck property, we have

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}=0. \end{aligned}$$

By the strong convergence, for all $0<\alpha \le \alpha _0$, we have

$$\begin{aligned} \limsup _{k\rightarrow \infty }\left\| \vec {n}_k-\vec {n}_{\infty }\right\| _{{\mathrm {L}}^{\infty }(\Sigma _{\alpha })}=0, \end{aligned}$$

where $\Sigma _{\alpha }=\Sigma \setminus \bigcup _{i=1}^m\overline{B}_{\alpha }(a_i)$. This implies by Proposition 2.7 that there exist sequences of constants $\{\vec {c}_k^{\,i}(\alpha )\}_{k\in {\mathbb {N}}}, \{\vec {c}_k^{\,i,j,j'}(\alpha )\}_{k\in {\mathbb {N}}}\subset \Lambda ^{n-2}{\mathbb {R}}^n$ such that for all i, j

$$\begin{aligned}&\lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \vec {n}_k-\vec {c}_k^{\,i}(\alpha )\right\| _{{\mathrm {L}}^{\infty }(\Omega _k^i(\alpha ))}=0\\&\lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \vec {n}_k-\vec {c}_k^{\,i,j,j'}(\alpha )\right\| _{{\mathrm {L}}^{\infty }(\Omega _k^{i,j,j'}(\alpha ))}=0. \end{aligned}$$

Since $|\vec {n}_k|=1$, we deduce that up to a subsequence $\vec {c}_k^{\,i}(\alpha _0)\underset{k\rightarrow \infty }{\longrightarrow }\vec {c}_{\infty }^{\,i}(\alpha _0)$ such that $|\vec {c}_{\infty }^{\,i}(\alpha _0)|=1$. Likewise, there exists $\left\{ \alpha _k\right\} _{k\in {\mathbb {N}}}\subset (0,\infty )$ such that $\vec {c}_{\infty }^{\,i}(\alpha _k)\rightarrow \vec {c}_{\infty }^{\,i}$ where $|\vec {c}_{\infty }^{\,i}|=1$. Therefore, we deduce that there exists $\vec {c}_{\infty }^{\,i},\vec {c}_{\infty }^{\,i,j,j'}\in S^{n-1}$ such that

$$\begin{aligned}&\lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \vec {n}_k-\vec {c}_{\infty }^{\,i}\right\| _{{\mathrm {L}}^{\infty }(\Omega _k^i(\alpha ))}=0\\&\lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \vec {n}_k-\vec {c}_{\infty }^{\,i,j,j'}\right\| _{{\mathrm {L}}^{\infty }(\Omega _k^{i,j,j'}(\alpha ))}=0. \end{aligned}$$

Finally, in a bubble domain $B(i,j,\alpha ,k)$, there exists a sequence $\{\mu _k^{i,j}\}_{k\in {\mathbb {N}}}\subset {\mathbb {R}}$ such that the function

$$\begin{aligned} \vec {\Phi }_{k}^{i,j}:B_{\alpha ^{-1}}(x_k^{i,j'})\setminus \bigcup _{j''\in I^{i,j}}B_{\alpha }(x_k^{i,j''})&\rightarrow {\mathbb {R}}^n\\ z&\mapsto e^{-\mu _{k}^{i,j}}\left( \vec {\Phi }_k(\rho _k^{i,j}z)-\vec {\Phi }_k(x_k^{i,j'})\right) \end{aligned}$$

converges smoothly towards to the branched Willmore sphere $\vec {\Phi }_{\infty }^{i,j}:{\mathbb {C}}\rightarrow {\mathbb {R}}^n$. Since

$$\begin{aligned} \vec {n}_{\vec {\Phi }_k^{i,j}}(z)=\vec {n}_{\vec {\Phi }_k}(\rho _kz), \end{aligned}$$

we deduce that for all $0<\alpha <\alpha _0$,

$$\begin{aligned} \left\| \nabla \vec {n}_k-\nabla \vec {n}_{\vec {\Phi }_{\infty }^{i,j}}((\rho _k^{i,j})^{-1}\,\cdot \,)\right\| _{{\mathrm {L}}^{2,1}(B(i,j,\alpha ,k))}=0. \end{aligned}$$

This implies that there exists $\{\vec {d}_k^{\,i,j}(\alpha )\}_{k\in {\mathbb {N}}}\subset \Lambda ^{n-2}{\mathbb {R}}^{n}$ such that

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \vec {n}_k-\vec {n}_{\vec {\Phi }_{\infty }^{\,i,j}}((\rho _k^{i,j})^{-1}\,\cdot \,))-\vec {d}_k^{\,i,j}(\alpha )\right\| _{{\mathrm {L}}^{\infty }(B(i,j,\alpha ,k))}=0 \end{aligned}$$

Since $\vec {n}_k$ and $\vec {n}_{\vec {\Phi }_{\infty }^{\,i,j}}$ are unitary, we deduce that

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\lim \limits _{k\rightarrow \infty }|\vec {d}_k^{\,i,j}(\alpha )|=0, \end{aligned}$$

so that

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \vec {n}_k-\vec {n}_{\vec {\Phi }_{\infty }^{\,i,j}}((\rho _k^{i,j})^{-1}\,\cdot \,)\right\| _{{\mathrm {L}}^{\infty }(B(i,j,\alpha ,k))}=0. \end{aligned}$$

Notice that the function $\vec {n}_k-\vec {n}_{\vec {\Phi }_{\infty }^{\,i,j}}((\rho _k^{i,j})^{-1}\,\cdot \,)$ is independent of $\alpha$ and that $B(i,j,\alpha ,k)\subset B(i,j,\beta ,k)$ for $\alpha <\beta$, which implies that

$$\begin{aligned} \lim _{k\rightarrow \infty }\left\| \vec {n}_k-\vec {n}_{\vec {\Phi }_{\infty }^{\,i,j}}((\rho _k^{i,j})^{-1}\,\cdot \,)\right\| _{{\mathrm {L}}^{\infty }(B(i,j,\alpha _0,k))}=0. \end{aligned}$$

Now, using the proof of Proposition 2.7, we deduce that we can take

and since $\vec {\Phi }_{\infty }^{i,1}:{\mathbb {C}}\rightarrow {\mathbb {R}}^n$ extends to an immersion $S^2\rightarrow {\mathbb {R}}^n$, the normal has a continuous extension and identifying $N=(0,0,1)\in S^2$ and $\infty \in {\mathbb {C}}\cup \left\{ \infty \right\}$, we deduce that

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\lim \limits _{k\rightarrow \infty }\left\| \vec {n}_k-\vec {n}_{\vec {\Phi }_{\infty }^{i,1}}(N)\right\| _{{\mathrm {L}}^{\infty }(\Omega _k^i(\alpha ))}=0, \end{aligned}$$

and likewise for all $1\le i\le m$ and $1\le j\le m_i$, we have

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \vec {n}_k-\vec {n}_{\vec {\Phi }_{\infty }^{\,i,j}}(N)\right\| _{{\mathrm {L}}^{\infty }(\Omega _k^{i,j}(\alpha ))}=0, \end{aligned}$$

which completes the proof of the theorem. $\square$

In the next section, we recall basic facts on the viscosity method for the Willmore energy, and then in the following section, we show the improved $L^{2,1}$ quantization in this setting.

5 The Viscosity Method for the Willmore Energy

We first introduce for all weak immersion $\vec {\Phi }:S^2\rightarrow {\mathbb {R}}^n$ of finite total curvature the associated metric $g=\vec {\Phi }^{*}g_{{\mathbb {R}}^n}$ on $S^2$. By the uniformisation theorem, there exists a function $\omega :S^2\rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} g=e^{2\omega }g_0, \end{aligned}$$

where $g_0$ is a metric of constant Gauss curvature $4\pi$ and unit volume on $S^2$. Furthermore, in all fixed chart $\varphi :B_1(0)\rightarrow S^2$, we define $\mu :B_1(0)\rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} \lambda =\omega +\mu , \end{aligned}$$

where in the given chart

$$\begin{aligned} g=e^{2\lambda }|\text {d}z|^2. \end{aligned}$$

For technical reasons, we will have to make a peculiar choice of $\omega$ (see [34], Definition III.2).

Definition 5.1

Under the preceding notations, we say that a choice $(\omega ,\varphi )$ of a map $\omega :S^2\rightarrow {\mathbb {R}}$ and of a diffeomorphism $\varphi :S^2\rightarrow S^2$ is an Aubin gauge if

$$\begin{aligned} \varphi ^{*}g_0=\frac{1}{4\pi }g_{S^2}\qquad \text {and}\qquad \int _{S^2} x_j e^{2\omega \circ \varphi (x)}\text {d}{\mathrm {vol}}_{g_{S^2}}(x)=0\qquad \text {for all}\;\, j=1,2,3, \end{aligned}$$

where $g_{S^2}$ is the standard metric on $S^2$.

We also recall that the limiting maps arise from a sequence of critical point of the following regularisation of the Willmore energy (see [34] for more details) :

$$\begin{aligned} W_{\sigma }(\vec {\Phi })&=W(\vec {\Phi })+\sigma ^2\int _{S^2}\left( 1+|\vec {H}|^2\right) ^2\text {d}{\mathrm {vol}}_g\\&\quad +\frac{1}{\log \left( \frac{1}{\sigma }\right) }\left( \frac{1}{2}\int _{S^2}|\text {d}\omega |_g^2\text {d}{\mathrm {vol}}_g+4\pi \int _{S^2}\omega \, e^{-2\omega }\text {d}{\mathrm {vol}}_g-2\pi \log \int _{S^2}\text {d}{\mathrm {vol}}_g\right) \end{aligned}$$

where $\omega :S^2\rightarrow {\mathbb {R}}$ is as above.

We need a refinement of a standard estimate (see [11], 3.3.6).

Lemma 5.2

Let $\Omega$ be a open subset of ${\mathbb {R}}^2$ whose boundary is a finite union of $C^1$ Jordan curves. Let $f\in L^1(\Omega )$ and let u be the solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta u&=f\qquad&\text {in}\;\, \Omega \\ u&=0\qquad&\text {on}\;\, \partial \Omega . \end{aligned} \right. \end{aligned}$$

(5.1)

Then, $\nabla u\in L^{2,\infty }(\Omega )$, and

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\le 3\sqrt{\frac{2}{\pi }}\left\| f\right\| _{{\mathrm {L}}^{1}(\Omega )}. \end{aligned}$$

Remark 5.3

We need an estimate independent of the domain for a sequence of annuli of conformal class diverging to $\infty$, but the argument applies to a general domain (although some regularity conditions seem to be necessary).

Proof

First assume that $f\in C^{0,\alpha }(\overline{\Omega })$ for some $0<\alpha <1$. Then by Schauder theory, $u\in C^{2,\alpha }(\overline{\Omega })$, and by Stokes theorem ([10], 1.2.1), we find as $u=0$ on $\partial \Omega$ that for all $z\in \Omega$

$$\begin{aligned} \partial _{z}u(z)=\frac{1}{2\pi i}\int _{\Omega }\frac{\partial _{\overline{z}}\left( \partial _{z}u(\zeta )\right) }{\zeta -z}\text {d}\zeta \wedge \text {d}\overline{\zeta }. \end{aligned}$$

(5.2)

As $\Delta u=4\,\partial _{z\overline{z}}^2u$ and $|\mathrm{d}\zeta |^2=\dfrac{\text {d}\overline{\zeta }\wedge \text {d}\zeta }{2i}$, the pointwise estimate (5.2) implies that

$$\begin{aligned} \partial _{z}u(z)=-\frac{1}{4\pi }\int _{\Omega }\frac{\Delta u(\zeta )}{\zeta -z}|\text {d}\zeta |^2=-\frac{1}{4\pi }\int _{\Omega }\frac{f(\zeta )}{\zeta -z}|\text {d}\zeta |^2. \end{aligned}$$

(5.3)

Now, define $\overline{f}\in L^1({\mathbb {R}}^2)$ by

$$\begin{aligned} \overline{f}(z)&=\left\{ \begin{aligned}&f(z)\qquad&\text {for all}\;\, z\in \Omega \\&0\qquad&\text {for all}\;\, z\in {\mathbb {R}}^2\setminus \Omega . \end{aligned}\right. \end{aligned}$$

and $U:{\mathbb {R}}^2\rightarrow {\mathbb {C}}$ by

$$\begin{aligned} U(z)=-\frac{1}{4\pi }\int _{{\mathbb {R}}^2}\frac{\overline{f}(\zeta )}{\zeta -z}|\text {d}\zeta |^2=-\frac{1}{4\pi }\left( \left( \zeta \mapsto \frac{1}{\zeta }\right) *\overline{f}\right) (z), \end{aligned}$$

(5.4)

where $*$ indicates the convolution on ${\mathbb {R}}^2$. Now, recall that for all $1\le p<\infty$ and $g\in L^p({\mathbb {R}}^2,{\mathbb {C}})$, we have

$$\begin{aligned} \left\| \overline{f}*g\right\| _{{\mathrm {L}}^{p}({\mathbb {R}}^2)}\le \left\| \overline{f}\right\| _{{\mathrm {L}}^{1}({\mathbb {R}}^2)}\left\| g\right\| _{{\mathrm {L}}^{p}({\mathbb {R}}^2)}. \end{aligned}$$

Interpolating between $L^1$ and $L^p$ for all $p>2$ shows by the Stein-Weiss interpolation theorem ([11], 3.3.3) that for all $g\in L^{2,\infty }({\mathbb {R}}^2,{\mathbb {C}})$

$$\begin{aligned} \left\| \overline{f}*g\right\| _{{\mathrm {L}}^{2,\infty }({\mathbb {R}}^2)}&\le \sqrt{2}\left( \frac{2\cdot 1}{2-1}+\frac{p\cdot 1}{p-2}\right) \left\| \overline{f}\right\| _{{\mathrm {L}}^{1}({\mathbb {R}}^2)}\left\| g\right\| _{{\mathrm {L}}^{2,\infty }({\mathbb {R}}^2)} \\&=\sqrt{2}\left( 2+\frac{p}{p-2}\right) \left\| \overline{f}\right\| _{{\mathrm {L}}^{1}({\mathbb {R}}^2)}\left\| g\right\| _{{\mathrm {L}}^{2,\infty }({\mathbb {R}}^2)}. \end{aligned}$$

Taking the infimum in $p>2$ (that is, $p\rightarrow \infty$) shows that for all $g\in L^{2,\infty }({\mathbb {R}}^2)$,

$$\begin{aligned} \left\| \overline{f}*g\right\| _{{\mathrm {L}}^{2,\infty }({\mathbb {R}}^2)}\le 3\sqrt{2}\left\| \overline{f}\right\| _{{\mathrm {L}}^{1}({\mathbb {R}}^2)}\left\| g\right\| _{{\mathrm {L}}^{2,\infty }({\mathbb {R}}^2)}. \end{aligned}$$

(5.5)

Therefore, we deduce from (5.3) and (5.5) that

$$\begin{aligned} \left\| U\right\| _{{\mathrm {L}}^{2,\infty }({\mathbb {R}}^2)}\le \frac{ 3\sqrt{2}}{4\pi }\left\| \overline{f}\right\| _{{\mathrm {L}}^{1}({\mathbb {R}}^2)}\left\| \frac{1}{|\,\cdot \,|}\right\| _{{\mathrm {L}}^{2,\infty }({\mathbb {R}}^2)}=\frac{3}{\sqrt{2\pi }}\left\| f\right\| _{{\mathrm {L}}^{1}(\Omega )}. \end{aligned}$$

Now, as $U=\partial _{z}u$ on $\Omega$ and $2|\partial _{z}u|=|\nabla u|$, we finally deduce that

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\le 3\sqrt{\frac{2}{\pi }}\left\| f\right\| _{{\mathrm {L}}^{1}(\Omega )}. \end{aligned}$$

(5.6)

In the general case $f\in L^1(\Omega )$, by density of $C^{\infty }_c(\Omega )$ in $L^1(\Omega )$, let $\left\{ f_k\right\} _{k\in {\mathbb {N}}}\subset C_{c}^{\infty }(\Omega )$ such that

$$\begin{aligned} \left\| f_k-f\right\| _{{\mathrm {L}}^{1}(\Omega )}\underset{k\rightarrow \infty }{\longrightarrow }0. \end{aligned}$$

(5.7)

Then, $u_k\in C^{\infty }(\overline{\Omega })$ (defined to be the solution of the system (5.1) with f replaced by $f_k$ and the same boundary conditions) so for all $k\in {\mathbb {N}}$, $\nabla u_k\in L^{2,\infty }(\Omega )$ and

$$\begin{aligned} \left\| \nabla u_k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\le 3\sqrt{\frac{2}{\pi }}\left\| f_k\right\| _{{\mathrm {L}}^{1}(\Omega )}. \end{aligned}$$

(5.8)

As $\left\{ \left\| f_k\right\| _{{\mathrm {L}}^{1}(\Omega )}\right\} _{k\in {\mathbb {N}}}$ is bounded, up to a subsequence $u_k\underset{k\rightarrow \infty }{\rightharpoonup } u_{\infty }$ in the weak topology of $W^{1,(2,\infty )}(\Omega )$. Therefore, (5.7) and (5.8) yield

$$\begin{aligned} \left\| \nabla u_{\infty }\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\le \liminf _{k\rightarrow \infty }\left\| \nabla u_k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}\le 3\sqrt{\frac{2}{\pi }}\left\| f\right\| _{{\mathrm {L}}^{1}(\Omega )}. \end{aligned}$$

Furthermore, as $f_k\underset{k\rightarrow \infty }{\longrightarrow }f$ in $L^1(\Omega )$, we have $\Delta u_{\infty }=f$ in ${\mathscr {D}}'(\Omega )$, so we deduce that $u_{\infty }=u$ and this concludes the proof of the lemma. $\square$

Finally, recall the following Lemma from [2] (se also [6]).

Lemma 5.4

Let $\Omega$ be a Lipschitz bounded open subset of ${\mathbb {R}}^2$, $1<p<\infty$ and $1\le q\le \infty$, and $(a,b)\in W^{1,(p,q)}(B_1(0))\times W^{1,(2,\infty )}(B_1(0))$. Let $u:B_1(0)\rightarrow {\mathbb {R}}$ be the solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta u&=\nabla a\cdot \nabla ^{\perp }b\qquad&\text {in}\;\, \Omega \\ u&=0\qquad&\text {on}\;\, \partial \Omega . \end{aligned} \right. \end{aligned}$$

Then, there exists a constant $C_{p,q}(\Omega )>0$ such that

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{p,q}(\Omega )}\le C_{p,q}(\Omega )\left\| \nabla a\right\| _{{\mathrm {L}}^{p,q}(\Omega )}\left\| \nabla b\right\| _{{\mathrm {L}}^{2,\infty }(\Omega )}. \end{aligned}$$

Remark 5.5

Notice that by scaling invariance, we have for all $R>0$ if $\Omega _R=B_R(0)$

$$\begin{aligned} \left\| \nabla u\right\| _{{\mathrm {L}}^{2,1}(B_R(0))}\le C_{2,1}(B_1(0))\left\| \nabla a\right\| _{{\mathrm {L}}^{2,1}(B_R(0))}\left\| \nabla b\right\| _{{\mathrm {L}}^{2,\infty }(B_R(0))}. \end{aligned}$$

6 Improved Energy Quantization in the Viscosity Method

The viscosity method [21, 22, 25, 26, 31,32,33,34] developed by the T. Rivière and collaborators aims at constructing solutions of min-max problems for functionals that do not satisfy the Palais–Smale condition or defined on spaces that are not Banach manifolds. Here, we will be focussing on the viscosity method for Willmore surfaces [34]. Let us recall a couple of definitions

Definition 6.1

Let ${\mathcal {M}}=W^{2,4}_{\iota }(S^2,{\mathbb {R}}^n)$ be the space of $W^{2,4}$ immersions from the sphere $S^2$ into ${\mathbb {R}}^n$. We say that a family ${\mathcal {A}}\subset {\mathcal {P}}({\mathcal {M}})$ is an admissible family if for every homeomorphism $\Psi$ of ${\mathscr {M}}$ isotopic to the identity, we have

$$\begin{aligned} \forall A\in {\mathcal {A}},\quad \Psi (A)\in {\mathcal {A}}. \end{aligned}$$

Now fix some admissible family ${\mathscr {A}}\subset {\mathcal {P}}(W^{2,4}_{\iota }(S^2,{\mathbb {R}}^n))$ and define

$$\begin{aligned} \beta _0=\inf _{A\in {\mathscr {A}}}\sup W(A). \end{aligned}$$

For all $\sigma >0$ and all smooth immersion $\vec {\Phi }:S^2\rightarrow {\mathbb {R}}^n$, recall the definition

$$\begin{aligned} W_{\sigma }(\vec {\Phi })=W(\vec {\Phi })+\sigma ^2\int _{\Sigma }\left( 1+|\vec {H}|^2\right) ^2\text {d}{\mathrm {vol}}_g+\frac{1}{\log \left( \frac{1}{\sigma }\right) }{\mathscr {O}}(\vec {\Phi }), \end{aligned}$$

where ${\mathscr {O}}$ is the Onofri energy (see above or [34] for more details), and define

$$\begin{aligned} \beta (\sigma )=\inf _{A\in {\mathscr {A}}}\sup W_{\sigma }(A). \end{aligned}$$

We can now introduce the main result of this section.

Theorem 6.2

Let $\left\{ \sigma _k\right\} _{k\in {\mathbb {N}}}\subset (0,\infty )$ be such that $\sigma _k\underset{k\rightarrow \infty }{\longrightarrow } 0$ and let $\{\vec {\Phi }_k\}_{k\in {\mathbb {N}}}:S^2\rightarrow {\mathbb {R}}^n$ be a sequence of critical points associated to $W_{\sigma _k}$ such that

$$\begin{aligned} \left\{ \begin{aligned}&W_{\sigma _k}(\vec {\Phi }_k)=\beta (\sigma _k)\underset{k\rightarrow \infty }{\longrightarrow }\beta _0\\&W_{\sigma _k}(\vec {\Phi }_k)-W(\vec {\Phi }_k)=o\left( \frac{1}{\log \left( \frac{1}{\sigma _k}\right) \log \log \left( \frac{1}{\sigma _k}\right) }\right) . \end{aligned} \right. \end{aligned}$$

(6.1)

Let $\left\{ R_k\right\} _{k\in {\mathbb {N}}}, \left\{ r_k\right\} _{k\in {\mathbb {N}}}\subset (0,\infty )$ be such that

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{R_k}{r_k}=0,\qquad \limsup _{k\rightarrow \infty }R_k<\infty , \end{aligned}$$

and for all $0<\alpha <1$ and $k\in {\mathbb {N}}$, let $\Omega _k(\alpha )=B_{\alpha R_k}\setminus \overline{B}_{\alpha ^{-1}r_k}(0)$ be a neck region, i.e. such that

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\lim _{k\rightarrow \infty }\sup _{2\alpha ^{-1}r_k<s<\alpha R_k/2}\int _{B_{2s}\setminus \overline{B}_{s/2}(0)}|\nabla \vec {n}_k|^2\mathrm{d}x=0. \end{aligned}$$

Then, we have

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}=0. \end{aligned}$$

Remarks on the proof

The proof is in the same spirit of the proof of the no-neck energy for the $L^{2,1}$ norm in the case of Willmore immersions, up to the need to introduce more conversation laws and derive more estimates to obtain the $L^{2,1}$ estimates.

Proof

As in [34], we give the proof in the special case $n=3$. By Theorem 4.1, this is not restrictive.

$$\begin{aligned} \Lambda =\sup _{k\in {\mathbb {N}}}\left( \left\| \nabla \lambda _k\right\| _{{\mathrm {L}}^{2,\infty }(B_1(0))}+\int _{B_1(0)}|\nabla \vec {n}_k|^2\mathrm{d}x\right) <\infty \end{aligned}$$

and

$$\begin{aligned} l(\sigma _k)=\frac{1}{\log \left( \frac{1}{\sigma _k}\right) },\qquad \widetilde{l}(\sigma _k)=\frac{1}{\log \log \left( \frac{1}{\sigma _k}\right) }. \end{aligned}$$

Furthermore, the entropy condition (6.1) and the improved Onofri inequality show (see [2], III.2)

$$\begin{aligned}&\frac{1}{\log \left( \frac{1}{\sigma _k}\right) }\left\| \omega _k\right\| _{{\mathrm {L}}^{\infty }(B_1(0))}=o\left( \frac{1}{\log \log \left( \frac{1}{\sigma _k}\right) }\right) \nonumber \\&\frac{1}{\log \left( \frac{1}{\sigma _k}\right) }\int _{S^2}|\text {d}\omega _k|_{g_k}^2\text {d}{\mathrm {vol}}_{g_k}=o\left( \frac{1}{\log \log \left( \frac{1}{\sigma _k}\right) }\right) \nonumber \\&\frac{1}{\log \left( \frac{1}{\sigma _k}\right) }\left( \frac{1}{2}\int _{S^2}|\text {d}\omega _k|_{g_k}^2\text {d}{\mathrm {vol}}_{g_k}+4\pi \int _{S^2}\omega _ke^{-2\omega _k}\text {d}{\mathrm {vol}}_{g_k}-2\pi \log \int _{S^2}\mathrm{d}{\mathrm {vol}}_{g_k}\right) \nonumber \\&\quad =o\left( \frac{1}{\log \log \left( \frac{1}{\sigma _k}\right) }\right) . \end{aligned}$$

(6.2)

Thanks to [34], we already have

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}=0. \end{aligned}$$

Therefore, as in Lemma IV.1 in [2] (and using the same argument as in Lemma 4.4), there exists a controlled extension $\widetilde{\vec {n}}_k:B_{\alpha R_k}(0)\rightarrow {\mathscr {G}}_{n-2}({\mathbb {R}}^n)$ such that $\widetilde{\vec {n}}_k=\vec {n}_k$ on $\Omega _k(\alpha )=B_{\alpha R_k}(0)\setminus \overline{B}_{\alpha ^{-1}r_k}(0)$ and

$$\begin{aligned}&\left\| \nabla \widetilde{\vec {n}}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\le \kappa _0(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}\nonumber \\&\left\| \nabla \widetilde{\vec {n}}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\le \kappa _0(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}, \end{aligned}$$

(6.3)

in all equations involving $\vec {n}_k$ on $B_{\alpha R_k}(0)$, we replace $\vec {n}_k$ by $\widetilde{\vec {n}}_k$ as one need only obtain estimates on $\Omega _k(\alpha )$, where $\widetilde{\vec {n}}_k=\vec {n}_k$. Likewise, $\vec {H}_k$ can be replaced by a controlled extension using Lemma B.4 in [15] (see also the Appendix).

Now, by [34], let $\vec {L}_k:B_1(0)\rightarrow {\mathbb {R}}^3$ be such that

(6.4)

Then, following [34], we have

$$\begin{aligned} e^{\lambda _k(z)}|\vec {L}_k(z)|\le \left( \kappa _1(n)\left( 1+\Lambda \right) e^{\kappa _1(n)\Lambda }\left\| \nabla \vec {n}\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}+\widetilde{l}(\sigma _k)\right) \frac{1}{|z|}\qquad \text {for all}\;\, z\in \Omega _k(\alpha /2), \end{aligned}$$

so that

$$\begin{aligned} \left\| e^{\lambda _k}\vec {L}_k\right\| _{{\mathrm {L}}^{2,\infty }(\Omega _k(\alpha /2))}\le 2\sqrt{\pi }\left( \kappa _1(n)\left( 1+\Lambda \right) e^{\kappa _1(n)\Lambda }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}+\widetilde{l}(\sigma _k)\right) . \end{aligned}$$

Now let $Y_k:B_{\alpha R_k}(0)\rightarrow {\mathbb {R}}$ (see [34], VI.21) be the solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta Y_k&=-4e^{2\lambda _k}\sigma _k^2\left( 1-H_k^4\right) -2l(\sigma _k)K_{g_0}\omega _k e^{2\mu _k}+8\pi \,l(\sigma _k)e^{2\lambda _k}{\mathrm {Area}}(\vec {\Phi }(S^2))^{-1}\qquad&\text {in}\;\, B_{\alpha R_k}(0)\\ Y_k&=0\qquad&\text {on}\;\, \partial B_{\alpha R_k}(0). \end{aligned} \right. \end{aligned}$$

(6.5)

Then, we have (recall that $K_{g_0}=4\pi$ by the chosen normalisation in Definition 5.1)

$$\begin{aligned} \left\| \Delta Y_k\right\| _{{\mathrm {L}}^{1}(B_{\alpha R_k}(0))}&\le 4\sigma _k^2\int _{B_{\alpha R_k}(0)}\left( 1+H_k^4\right) \text {d}{\mathrm {vol}}_{g_k}+8\pi \,l(\sigma _k)\left\| \omega _k\right\| _{{\mathrm {L}}^{\infty }(B_{\alpha _k}(0))} \int _{B_{\alpha R_k}(0)}e^{2\mu _k}\mathrm{d}x\nonumber \\&\quad +8\pi \,l(\sigma _k)\frac{{\mathrm {Area}}(\vec {\Phi }_k(B_{\alpha R_k}(0)))}{{\mathrm {Area}}(\vec {\Phi }_k(S^2))}=o(\widetilde{l}(\sigma _k)). \end{aligned}$$

(6.6)

Therefore, Lemma 5.2 implies by (6.6) that

$$\begin{aligned} \left\| \nabla Y_k\right\| _{{\mathrm {L}}^{2,\infty }(B_{\alpha R_k}(0))}\le 3\sqrt{\frac{2}{\pi }}\left\| \Delta Y_k\right\| _{{\mathrm {L}}^{1}(B_{\alpha R_k}(0))}=o(\widetilde{l}(\sigma _k))\le \widetilde{l}(\sigma _k) \end{aligned}$$

(6.7)

for k large enough. Now, let $\vec {v}_k:B_{\alpha R_k}(0)\rightarrow {\mathbb {R}}^3$ be the solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta \vec {v}_k&=\nabla \widetilde{\vec {n}}_k\cdot \nabla ^{\perp }Y_k\qquad&\text {in}\;\, B_{\alpha R_k}(0)\\ \vec {v}_k&=0\qquad&\text {on}\;\, \partial B_{\alpha R_k}(0). \end{aligned}\right. \end{aligned}$$

(6.8)

By scaling invariance and the inequality of Lemma 5.4, we deduce by (6.7) that for some universal constant $\kappa _2>0$

$$\begin{aligned}&\left\| \nabla \vec {v}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\le \kappa _2\left\| \nabla \widetilde{\vec {n}}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\left\| \nabla Y_k\right\| _{{\mathrm {L}}^{2,\infty }(B_{\alpha R_k}(0))}\nonumber \\&\quad \le \kappa _2\kappa _0(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\left\| \nabla Y_k\right\| _{{\mathrm {L}}^{2,\infty }(B_{\alpha R_k}(0))}\le \widetilde{l}(\sigma _k)\,\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}. \end{aligned}$$

(6.9)

Furthermore, we have by Lemma 5.4 and scaling invariance

$$\begin{aligned} \left\| \nabla \vec {v}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}&\le \kappa _3\left\| \nabla \widetilde{\vec {n}}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\left\| \nabla Y_k\right\| _{{\mathrm {L}}^{2,\infty }(B_{\alpha R_k}(0))}\nonumber \\&\le \kappa _3\kappa _0(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}o(\widetilde{l}(\sigma _k))\nonumber \\&\le \widetilde{l}(\sigma _k)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))} \end{aligned}$$

(6.10)

Now, recall that the Codazzi identity ([34], III.58) implies that

$$\begin{aligned} {{\,\mathrm{div}\,}}\left( e^{-2\lambda _k}\sum _{j=1}^2{\mathbb {I}}_{2,j}\partial _{x_j}\vec {\Phi }_k,-e^{-2\lambda _k}\sum _{j=1}^2{\mathbb {I}}_{1,j}\partial _{x_j}\vec {\Phi }_k\right) =0\qquad \text {in}\;\, B_{\alpha R_k}(0) \end{aligned}$$

(6.11)

Therefore, by the Poincaré Lemma, there exists $\vec {D}_k:B_{\alpha R_k}(0)\underset{k\rightarrow \infty }{\longrightarrow } {\mathbb {R}}^3$ such that

$$\begin{aligned} \nabla \vec {D}_k=\left( e^{-2\lambda _k}\sum _{j=1}^2{\mathbb {I}}_{1,j}\partial _{x_j}\vec {\Phi }_k,e^{-2\lambda _k}\sum _{j=1}^2{\mathbb {I}}_{2,j}\partial _{x_j}\vec {\Phi }_k\right) . \end{aligned}$$

Notice that we have the trivial estimate

$$\begin{aligned} \left\| \nabla \vec {D}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha _k}(0))}\le 2\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha _k}(0))}\le 2\sqrt{\Lambda }. \end{aligned}$$

(6.12)

Furthermore,

$$\begin{aligned} l(\sigma _k)\left\| \nabla \vec {D}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\le 2\,l(\sigma _k)\,\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}. \end{aligned}$$

(6.13)

Now, let $\vec {E}_k:B_{\alpha R_k}(0)\rightarrow {\mathbb {R}}^3$ be the solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta \vec {E}_k&=2\,\nabla (l(\sigma _k)\omega _k)\cdot \nabla ^{\perp }\vec {D}_k\qquad&\text {in}\;\, B_{\alpha R_k}(0)\\ \vec {E}_k&=0\qquad&\text {on}\;\, \partial B_{\alpha R_k}(0). \end{aligned}. \right. \end{aligned}$$

(6.14)

The improved Wente estimate, the scaling invariance and the estimates (6.1) and (6.12) imply that

$$\begin{aligned} \left\| \nabla \vec {E}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}&\le 2\kappa _0\,l(\sigma _k)\left\| \nabla \omega _k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\left\| \nabla \vec {D}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\nonumber \\&\le 4\kappa _0\sqrt{\Lambda }\,o(\sqrt{l(\sigma _k)})\le \sqrt{l(\sigma _k)}\nonumber \\ \left\| \nabla \vec {E}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}&\le \frac{1}{2}\sqrt{\frac{3}{\pi }}\,l(\sigma _k)\left\| \nabla \omega _k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\left\| \nabla \vec {D}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\le \sqrt{l(\sigma _k)}. \end{aligned}$$

(6.15)

Now, let $\vec {F}_k:B_{\alpha R_k}(0)\rightarrow {\mathbb {R}}^3$ be such that

$$\begin{aligned} 2\omega _k\,l(\sigma _k)\,\nabla ^{\perp }\vec {D}_k=\nabla ^{\perp }\vec {F}_k+\nabla \vec {E}_k. \end{aligned}$$

Combining (6.13), (6.15), and recalling that $l(\sigma _k)\left\| \omega _k\right\| _{{\mathrm {L}}^{\infty }(B_{\alpha R_k}(0))}=o(\widetilde{l}(\sigma _k))$ (by (6.1)), we deduce that

$$\begin{aligned} \left\| \vec {F}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}&\le 2\,l(\sigma _k)\left\| \omega _k\right\| _{{\mathrm {L}}^{\infty }(\Omega _k(\alpha ))}\left\| \nabla \vec {D}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+\left\| \nabla \vec {E}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}\nonumber \\&\le \widetilde{l}(\sigma _k)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+\sqrt{l(\sigma _k)}. \end{aligned}$$

(6.16)

Finally, let $\vec {w}_k:B_{\alpha R_k}(0)\rightarrow {\mathbb {R}}^3$ be the solution of

$$\begin{aligned} \left\{ \begin{aligned} \Delta \vec {w}_k&=\nabla \widetilde{\vec {n}}_k\cdot \nabla ^{\perp }\left( \vec {v}_k-\vec {E}_k\right) \qquad&\text {in}\;\, B_{\alpha R_k}(0)\\ \vec {w}_k&=0\qquad&\text {on}\;\, \partial B_{\alpha R_k}(0). \end{aligned} \right. \end{aligned}$$

As previously, the improved Wente implies that

$$\begin{aligned} \left\| \nabla \vec {w}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0))}&\le \kappa _0\left\| \nabla \widetilde{\vec {n}}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\left\| \nabla (\vec {v}_k-\vec {E}_k)\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\nonumber \\&\le \kappa _0\left\| \nabla \widetilde{\vec {n}}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}\left( \left\| \nabla \vec {v}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}+\left\| \nabla \vec {E}_k\right\| _{{\mathrm {L}}^{2}(B_{\alpha R_k}(0))}\right) \nonumber \\&\le \kappa _0\kappa (n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}\left( \widetilde{l}(\sigma _k)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}+\sqrt{l(\sigma _k)}\right) \nonumber \\&\le \kappa _0\kappa (n)\sqrt{\Lambda }\left( \widetilde{l}(\sigma _k)\sqrt{\Lambda }+\sqrt{l(\sigma _k)}\right) \le \widetilde{l}(\sigma _k) \end{aligned}$$

(6.17)

for k large enough. Finally, if $\vec {Z}_k:\Omega _k(\alpha )\rightarrow {\mathbb {R}}^3$ satisfies

$$\begin{aligned} \nabla ^{\perp }\vec {Z}_k=\vec {n}_k\times \nabla ^{\perp }\left( \vec {v}_k-\vec {E}_k\right) -\nabla \vec {w}_k, \end{aligned}$$

the estimates (6.9), (6.15), (6.17) show that (as $\widetilde{\vec {n}}_k=\vec {n}_k$ on $\Omega _k(\alpha )$)

$$\begin{aligned} \left\| \nabla \vec {Z}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}&\le \widetilde{l}(\sigma _k)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+\sqrt{l(\sigma _k)}+\widetilde{l}(\sigma _k). \end{aligned}$$

(6.18)

Finally, following constants and using the controlled extension $\widetilde{\vec {n}}_k$ of $\vec {n}_k$, we deduce as in [34] (see (VI.75)) that

$$\begin{aligned}&\left\| 2\left( 1+2\sigma _k^2\left( 1+H_k^2\right) -l(\sigma _k)\omega _k\right) e^{\lambda _k}\vec {H}_k+\left( \nabla \vec {v}_k+\nabla ^{\perp }\left( \vec {F}_k+\vec {Z}_k\right) \right) \right. \nonumber \\&\qquad \left. \times \nabla \vec {\Phi }_k\,e^{-\lambda _k} +l(\sigma _k)\nabla ^{\perp }\vec {D}_k\cdot \nabla \vec {\Phi }_k\,e^{-\lambda _k}\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\nonumber \\&\quad \le \kappa _4(n)e^{\kappa _4(n)\Lambda }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}. \end{aligned}$$

(6.19)

Furthermore, as $l(\sigma _k)\left\| \omega _k\right\| _{{\mathrm {L}}^{\infty }(\Omega _k(\alpha ))}=o(\widetilde{l}(\sigma _k))$, we have $2(1+2\sigma _k^2(1+H_k^2)-l(\sigma _k)\omega _k)\ge 1$ for k large enough and by the estimates (6.9), (6.13), (6.16), (6.18), (6.19), we deduce that

$$\begin{aligned}&\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))} \le \kappa _4(n)e^{\kappa _4(n)\Lambda }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))} +\widetilde{l}(\sigma _k)\,\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\nonumber \\&\qquad +\widetilde{l}(\sigma _k)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+\sqrt{l(\sigma _k)} +\widetilde{l}(\sigma _k)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\nonumber \\&\qquad +\sqrt{l(\sigma _k)}+\widetilde{l}(\sigma _k) +2\,l(\sigma _k)\,\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\nonumber \\&\quad \le \kappa _4(n)e^{\kappa _4(n)\Lambda }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}+5\,\widetilde{l}(\sigma _k)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+3\,\widetilde{l}(\sigma _k). \end{aligned}$$

(6.20)

Thanks to the proof of Theorem 3.1 and (6.20), we have

$$\begin{aligned} \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha /2))}&\le \kappa _5(n)e^{\kappa _5(n)\Lambda }\left( \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(\alpha ))}+\left\| e^{\lambda _k}\vec {H}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\right) \nonumber \\&\le \kappa _6(n)e^{\kappa _6(n)\Lambda }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}+5\,\widetilde{l}(\sigma _k)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+3\,\widetilde{l}(\sigma _k). \end{aligned}$$

(6.21)

Furthermore, thanks to the $\varepsilon$-regularity ([28]), we obtain

$$\begin{aligned}&\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(B_{\alpha R_k}(0)\setminus \overline{B}_{\alpha R_k/2}(0))}\le \kappa _7(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(B_{2\alpha R_k}\setminus \overline{B}_{\alpha R_k/4}(0))}\nonumber \\&\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(B_{2\alpha ^{-1}r_k}\setminus \overline{B}_{\alpha ^{-1}r_k}(0))}\le \kappa _7(n)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(B_{4\alpha ^{-1}r_k}\setminus \overline{B}_{\alpha ^{-1} r_k/2}(0))}. \end{aligned}$$

(6.22)

Finally, by (6.21) and (6.22), we have

$$\begin{aligned} \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\le \kappa _8(n)e^{\kappa _8(n)\Lambda }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))}+5\,\widetilde{l}(\sigma _k)\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}+3\,\widetilde{l}(\sigma _k), \end{aligned}$$

which directly implies as $\widetilde{l}(\sigma _k)\underset{k\rightarrow \infty }{\longrightarrow }0$ that for k large enough

$$\begin{aligned} \left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}\le 2\kappa _8(n)e^{\kappa _8(n)\Lambda }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2}(\Omega _k(2\alpha ))} \end{aligned}$$

and the improved no-neck energy

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\limsup _{k\rightarrow \infty }\left\| \nabla \vec {n}_k\right\| _{{\mathrm {L}}^{2,1}(\Omega _k(\alpha ))}=0. \end{aligned}$$

This concludes the proof of the Theorem. $\square$

We close this article with a short appendix concerning Lorentz spaces.

Appendix

1.1 Some Basic Properties of Lorentz Spaces

Fix a measured space $(X,\mu )$. Define for all $0<t<\infty$ the measurable function $f_{*}$ on $(0,\infty )$ by

$$\begin{aligned} f_{*}(t)=\inf \left\{ \lambda>0: \mu (X\cap \left\{ x:|f(x)|>\lambda \right\} )\le t\right\} \end{aligned}$$

and recall that for all $\lambda >0$

$$\begin{aligned} {\mathscr {L}}^1\left( (0,\infty )\cap \left\{ t:f_{*}(t)>\lambda \right\} \right) =\mu \left( X\cap \left\{ x:|f(x)|>\lambda \right\} \right) . \end{aligned}$$

In particular, using twice the usual slicing formula (valid for an arbitrary measure $\mu$ that need not be $\sigma$-finite), we find

$$\begin{aligned} \left\| f\right\| _{{\mathrm {L}}^{p}(X,\mu )}&=p\int _{0}^{\infty }\lambda ^{p}\mu \left( X\cap \left\{ x:|f(x)|>\lambda \right\} \right) \frac{\text {d}\lambda }{\lambda }\\&=p\int _{0}^{\infty }\lambda ^p{\mathscr {L}}^1\left( (0,\infty )\cap \left\{ t:f_{*}(t)>\lambda \right\} \right) \frac{\text {d}\lambda }{\lambda }\\&=\int _{0}^{\infty }f_{*}^p(t)\text {d}t =\left\| f_{*}\right\| _{{\mathrm {L}}^{p}((0,\infty ),{\mathscr {L}}^1)}. \end{aligned}$$

To simplify notations we will often remove the reference to the measure $\mu$. This motivates the introduction of the following quasi-norm for $1<p<\infty$ and $1\le q<\infty$

$$\begin{aligned} \left| f\right| _{{\mathrm {L}}^{p,q}(X)}=\left( \int _{0}^{\infty }t^{\frac{q}{p}}f_{*}^q(t)\frac{\text {d}t}{t}\right) ^{\frac{1}{q}}. \end{aligned}$$

(7.1)

If we define $\displaystyle f_{**}(t)=\dfrac{1}{t}\int _{0}^{t}f_{*}(s)\text {d}s,$ then the associated norm to $L^{p,q}$ is

$$\begin{aligned} \left\| f\right\| _{{\mathrm {L}}^{p,q}(X)}=\left( \int _{0}^{\infty }t^{\frac{q}{p}}f_{**}^q(t)\frac{\text {d}t}{t}\right) ^{\frac{1}{q}}, \end{aligned}$$

(7.2)

and $(L^{p,q}(X,\mu ),\left\| \,\cdot \,\right\| _{{\mathrm {L}}^{p,q}(X)})$ is a Banach space for all $1<p<\infty$ and $1\le q< \infty$. Now, we have by Fubini’s theorem for all $f\in L^{p,q}(X,\mu )$

$$\begin{aligned} \left\| f\right\| _{{\mathrm {L}}^{p,1}(X)}&=\int _{0}^{\infty }t^{\frac{1}{p}}f_{**}(t)\frac{\text {d}t}{t}=\int _{0}^{\infty }\int _{0}^{\infty }t^{\frac{1}{p}-2}f_{*}(s){\mathbf {1}}_{\left\{ 0<s<t\right\} }\text {d}s\text {d}t \\&=\int _{0}^{\infty }f_{*}(s)\left( \int _{0}^{\infty }t^{\frac{1}{p}-2}{\mathbf {1}}_{\left\{ 0<s<t\right\} }\text {d}t\right) \text {d}s\\&=\int _{0}^{\infty }f_{*}(s)\left( \int _{s}^{\infty }t^{\frac{1}{p}-2}\text {d}t\right) \text {d}s=\frac{p}{p-1}\int _{0}^{\infty }s^{\frac{1}{p}-1}f_{*}(s)\text {d}s\\&=\frac{p}{p-1}\left| f\right| _{{\mathrm {L}}^{p,1}(X)}. \end{aligned}$$

Therefore, $\left| \,\cdot \,\right| _{{\mathrm {L}}^{p,1}(X)}$ is a norm for all $1<p<\infty$. Furthermore, notice that Fubini’s theorem also shows ([30]) that

$$\begin{aligned} \left| f\right| _{{\mathrm {L}}^{p,q}(X)}=p^{\frac{1}{q}}\left( \int _{0}^{\infty }\lambda ^{q}\mu \left( X\cap \left\{ x:|f(x)|>\lambda \right\} \right) ^{\frac{q}{p}}\frac{\text {d}\lambda }{\lambda }\right) ^{\frac{1}{q}}. \end{aligned}$$

(7.3)

In particular, for $q=1$ each of the quantities (7.1), (7.2) and (7.3) defines a norm on $L^{p,1}(X,\mu )$. Finally, for $q=\infty$, we define the quasi-norm

$$\begin{aligned} \left| f\right| _{{\mathrm {L}}^{p,\infty }(X)}=\sup _{\lambda>0}\,t\left( \mu \left( X\cap \left\{ x:|f(x)|>\lambda \right\} \right) \right) ^{\frac{1}{p}}=\sup _{t>0}\,t^{\frac{1}{p}}f_{*}(t) \end{aligned}$$

and the norm

$$\begin{aligned} \left\| f\right\| _{{\mathrm {L}}^{p,\infty }(X)}= \sup _{t>0}\,t^{\frac{1}{p}}f_{**}(t) \end{aligned}$$

makes $(L^{p,\infty }(X),\left\| \,\cdot \,\right\| _{{\mathrm {L}}^{p,\infty }(X)})$ a Banach space (they are the classical Marcinkiewicz weak $L^p$ spaces). Notice however that $L^{1,\infty }$ is not a Banach space. We have the general inequality for all $1<p<\infty$

$$\begin{aligned} \left| f\right| _{{\mathrm {L}}^{p,\infty }(X)}\le \left\| f\right\| _{{\mathrm {L}}^{p,\infty }(X)}\le \frac{p}{p-1}\left| f\right| _{{\mathrm {L}}^{p,\infty }(X)}. \end{aligned}$$

The norms are related as follows (see [30]).

Lemma 7.1

For all $1<p<\infty$ and $1\le q\le r<\infty$, and for all $f\in L^{p,q}(X,\mu )$, we have

$$\begin{aligned}&\left| f\right| _{{\mathrm {L}}^{p,\infty }(X)}\le \left( \frac{q}{p}\right) ^{\frac{1}{q}}\left| f\right| _{{\mathrm {L}}^{p,q}(X)}\\&\left| f\right| _{{\mathrm {L}}^{p,r}(X)}\le \left( \frac{q}{p}\right) ^{\frac{r-q}{rq}}\left| f\right| _{{\mathrm {L}}^{p,q}(X)}. \end{aligned}$$

Proof

As $f_{*}$ is decreasing, we have for all $0<t<\infty$

$$\begin{aligned} t^{\frac{1}{p}}f_{*}(t)&=\left( \frac{q}{p}\int _{0}^ts^{\frac{q}{p}}f_{*}^q(t)\frac{\text {d}s}{s}\right) ^{\frac{1}{q}}\le \left( \frac{q}{p}\int _{0}^ts^{\frac{q}{p}}f_{*}^q(s)\frac{\text {d}s}{s}\right) ^{\frac{1}{q}}\\&\le \left( \frac{q}{p}\right) ^{\frac{1}{q}}\left| f\right| _{{\mathrm {L}}^{p,q}(X)}, \end{aligned}$$

which implies that for all $1\le q<\infty$

$$\begin{aligned} \left\| f\right\| _{{\mathrm {L}}^{p,\infty }(X)}\le \left( \frac{q}{p}\right) ^{\frac{1}{q}}\left| f\right| _{{\mathrm {L}}^{p,q}(X)}. \end{aligned}$$

(7.4)

Now, assume that $1\le q<r<\infty$. Then, (7.4) implies that

$$\begin{aligned} \left| f\right| _{{\mathrm {L}}^{p,r}(X)}&=\left( \int _{0}^{\infty }t^{\frac{r}{p}}f_{*}^r(t)\frac{\text {d}t}{t}\right) ^{\frac{1}{r}}=\left( \int _{0}^{\infty }t^{\frac{q}{p}}f_{*}^q(t)t^{\frac{r-q}{p}}f_{*}(t)^{r-q}\frac{\text {d}t}{t}\right) ^{\frac{1}{r}}\nonumber \\&\le \left| f\right| _{{\mathrm {L}}^{p,\infty }(X)}^{\frac{r-q}{r}}\left| f\right| _{{\mathrm {L}}^{p,q}(X)}^{\frac{q}{r}}\nonumber \\&\le \left( \frac{q}{p}\right) ^{\frac{r-q}{rq}}\left| f\right| _{{\mathrm {L}}^{p,q}(X)}. \end{aligned}$$

(7.5)

This concludes the proof of the Lemma. $\square$

In particular, if $q=1$, as $\left\| f\right\| _{{\mathrm {L}}^{p,1}(X)}=\frac{p}{p-1}\left| f\right| _{{\mathrm {L}}^{p,1}(X)}$, we deduce by (7.5) that

$$\begin{aligned} \left| f\right| _{{\mathrm {L}}^{p,r}(X)}\le \frac{p-1}{p}\left( \frac{1}{p}\right) ^{\frac{r-1}{r}}\left\| f\right\| _{{\mathrm {L}}^{p,1}(X)}=\frac{p-1}{p^{2-\frac{1}{r}}}\left\| f\right\| _{{\mathrm {L}}^{p,1}(X)}\le \left\| f\right\| _{{\mathrm {L}}^{p,1}(X)}. \end{aligned}$$

In particular as $\left| \,\cdot \,\right| _{{\mathrm {L}}^{p,p}(X)}=\left\| \,\cdot \,\right\| _{{\mathrm {L}}^{p}(X)}$, we have

$$\begin{aligned} \left\| f\right\| _{{\mathrm {L}}^{p}(X)}\le \frac{p-1}{p^{2-\frac{1}{p}}}\left\| f\right\| _{{\mathrm {L}}^{p,1}(X)} \end{aligned}$$

Notice that for $p=2$. this yields

$$\begin{aligned} \left\| f\right\| _{{\mathrm {L}}^{2}(X)}\le \frac{1}{2\sqrt{2}}\left\| f\right\| _{{\mathrm {L}}^{2,1}(X)}. \end{aligned}$$

(7.6)

Finally, recall the inequality

$$\begin{aligned} \left| \int _{X}fg\text {d}\mu \right| \le \int _{0}^{\infty }f_{*}(t)g_{*}(t)\text {d}t. \end{aligned}$$

It implies that for all $1<p<\infty$

$$\begin{aligned} \int _{0}^{\infty }f_{*}(t)g_{*}(t)\text {d}t&=\int _{0}^{\infty }t^{\frac{1}{p}}f_{*}(t)t^{p'}g_{*}(t)\frac{\text {d}t}{t}\le \left| g\right| _{{\mathrm {L}}^{p',\infty }(X)}\int _{0}^{\infty }t^{\frac{1}{p}}f_{*}(t)\frac{\text {d}t}{t}\\&=\left| f\right| _{{\mathrm {L}}^{p,1}(X)}\left| g\right| _{{\mathrm {L}}^{p',\infty }(X)}, \end{aligned}$$

while for all $1<p<\infty$ and $1\le q< \infty$, we have by Hölder’s inequality (applied to the Haar measure $\nu =\dfrac{\text {d}t}{t}$ on $(0,\infty )$)

$$\begin{aligned} \int _{0}^{\infty }f_{*}(t)g_{*}(t)\text {d}t&=\int _{0}^{\infty }t^{\frac{1}{p}}f_{*}(t)t^{\frac{1}{p'}}g_{*}(t)\frac{\text {d}t}{t}\\&\le \left( \int _{0}^{\infty }t^{\frac{p}{q}}f_{*}^q(t)\frac{\text {d}t}{t}\right) ^{\frac{1}{q}}\left( \int _{0}^{\infty }t^{\frac{q'}{p'}}g_{*}^{q'}(t)\text {d}t\right) ^{\frac{1}{q'}}\\&=\left| f\right| _{{\mathrm {L}}^{p,q}(X)}\left| g\right| _{{\mathrm {L}}^{p',q'}(X)}. \end{aligned}$$

Therefore, we have for all $1<p<\infty$ and $1\le q\le \infty$

$$\begin{aligned} \left| \int _{X}fg\, \text {d}\mu \right| \le \left| f\right| _{{\mathrm {L}}^{p,q}(X)}\left| g\right| _{{\mathrm {L}}^{p',q'}(X)}\le \left\| f\right\| _{{\mathrm {L}}^{p,q}(X)}\left\| g\right\| _{{\mathrm {L}}^{p',q'}(X)} \end{aligned}$$

(7.7)

and one shows that for all $1<p<\infty$ and $1\le q< \infty$, the dual space of $L^{p,q}(X,\mu )$ is $L^{p',q'}(X,\mu )$. In particular, (7.7) implies that for all $1<p<\infty$

$$\begin{aligned} \left\| f\right\| _{{\mathrm {L}}^{p,1}(X)}=\frac{p^2}{p-1}\int _{0}^{\infty }\mu \left( X\cap \left\{ x:|f(x)|>t\right\} \right) ^{\frac{1}{p}}\text {d}t \end{aligned}$$

The main case of interest in this article is the $L^{2,1}$ norm, which now can be defined as

$$\begin{aligned} \left\| f\right\| _{{\mathrm {L}}^{2,1}(X)}=4\int _{0}^{\infty }\left( \mu \left( X\cap \left\{ x:|f(x)|>t\right\} \right) \right) ^{\frac{1}{2}}\text {d}t, \end{aligned}$$

and

$$\begin{aligned} \left| \int _{X}fg\,\text {d}\mu \right| \le \frac{1}{2}\left\| f\right\| _{{\mathrm {L}}^{2,1}(X)}\left| g\right| _{{\mathrm {L}}^{2,\infty }(X)}\le \left\| f\right\| _{{\mathrm {L}}^{2,1}(X)}\left\| g\right\| _{{\mathrm {L}}^{2,\infty }(X)}. \end{aligned}$$

As

$$\begin{aligned} \frac{1}{n}\left\| \frac{1}{|x-y|^{n-1}}\right\| _{{\mathrm {L}}^{\frac{n}{n-1},\infty }({\mathbb {R}}^n,{\mathscr {L}}^n)}=\left| \frac{1}{|x-y|^{n-1}}\right| _{{\mathrm {L}}^{\frac{n}{n-1},\infty }({\mathbb {R}}^n,{\mathscr {L}}^n)}=\alpha (n)^{\frac{n}{n-1}}, \end{aligned}$$

we have for all open subset $\Omega \subset {\mathbb {R}}^n$ and $f\in L^{n,1}(\Omega )$, for all $y\in {\mathbb {R}}^n$

$$\begin{aligned} \int _{\Omega }\frac{|f(x)|}{|x-y|^{n-1}}\text {d}{\mathscr {L}}^n(x)&\le \frac{n-1}{n^2}\left\| f\right\| _{{\mathrm {L}}^{2,1}(\Omega )}\left\| \frac{1}{|x-y|^{n-1}}\right\| _{{\mathrm {L}}^{\frac{n}{n-1},\infty }({\mathbb {R}}^n)}\nonumber \\&= \frac{n-1}{n}\alpha (n)^{\frac{n-1}{n}}\left\| f\right\| _{{\mathrm {L}}^{n,1}(\Omega )}. \end{aligned}$$

(7.8)

In particular, if $\Omega \subset {\mathbb {R}}^2$, we have

$$\begin{aligned} \int _{\Omega }\frac{|f(x)|}{|x-y|}\text {d}{\mathscr {L}}^2(x)\le \frac{\sqrt{\pi }}{2}\left\| f\right\| _{{\mathrm {L}}^{2,1}(\Omega )}. \end{aligned}$$

(7.9)

1.2 Extension Operators on Annuli

The following result was used in [2] and [15].

Lemma 7.2

Let $n\ge 2$, $\varepsilon >0$ and $1+\varepsilon<R<\infty$ and $\Omega _{R}=B_R\setminus \overline{B}_1(0)$ be the associated annulus. Then, there exists a linear extension operator

$$\begin{aligned} T:\bigcup _{1\le p<\infty }W^{1,p}(\Omega _{R})\rightarrow \bigcup _{1\le p<\infty }W^{1,p}(B_R(0)) \end{aligned}$$

such that for all $1\le p<\infty$, there exists a universal constant $C_1(n,\varepsilon )>0$ (independent of $R>1+\varepsilon$) such that for all $1\le p<\infty$

$$\begin{aligned} \left\| T u\right\| _{{\mathrm {W}}^{1,p}(B_R(0))}\le C_1(n,\varepsilon )\left\| u\right\| _{{\mathrm {W}}^{1,p}(\Omega _{R})}. \end{aligned}$$

Furthermore, for all $1<p<\infty$ and $1\le q\le \infty$, T extends as a linear operator $W^{1,(p,q)}(\Omega _R)\rightarrow W^{1,(p,q)}(B_R(0))$ such that for some universal constant $C_2(n,p,q,\varepsilon )$

$$\begin{aligned} \left\| T u\right\| _{{\mathrm {W}}^{1,(p,q)}(B_R(0))}\le C_2(n,p,q,\varepsilon )\left\| u\right\| _{{\mathrm {W}}^{1,(p,q)}(\Omega _{R})}. \end{aligned}$$

Proof

The second assertion follows directly from the Stein-Weiss interpolation theorem ([11], 3.3.3). For the first part, construct by [3], IX.7 a linear extension operator $\widetilde{T}$ such that for all $u \in W^{1,p}(B_{1+\varepsilon }\setminus \overline{B}_1(0))$, $\widetilde{T} u\in W^{1,p}(B_{1+\varepsilon }(0))$ and such that

$$\begin{aligned} \left\| \widetilde{T}u\right\| _{{\mathrm {W}}^{1,p}(B_{1+\varepsilon }(0))}\le C(n,\varepsilon )\left\| u\right\| _{{\mathrm {W}}^{1,p}(B_{1+\varepsilon }(0))}. \end{aligned}$$

(7.10)

Now, if $u\in W^{1,p}(B_R(0))$, just consider the restriction $\overline{u}|B_{1+\varepsilon }(0)\setminus \overline{B}_1(0)$, and define

$$\begin{aligned} Tu(x)=\left\{ \begin{aligned}&u(x)\qquad&\text {if}\;\, x\in B_R\setminus \overline{B}_{1+\varepsilon }(0)\\&\widetilde{T}u(x)\qquad&\text {if}\;\, x\in B_{1+\varepsilon }(0). \end{aligned} \right. \end{aligned}$$

As $\widetilde{T}u=u$ on $B_{1+\varepsilon }\setminus \overline{B}_1(0)$, T satisfies the claimed properties by (7.10). $\square$

Remark 7.3

Although the norm of the operator $T:W^{1,p}(\Omega _R)\rightarrow W^{1,p}(B_R(0))$ does not depend on $1<p<\infty$, the norm of $T:W^{1,(p,q)}(\Omega _R)\rightarrow W^{1,(p,q)}(B_R(0))$ depends a priori on $1<p<\infty$ and $1\le q\le \infty$, as the constant of the Stein-Weiss interpolation theorem depends on these parameters.