The Helmholtz decomposition of a space of vector fields with bounded mean oscillation in a bounded domain

Abstract

We introduce a space of vector fields with bounded mean oscillation whose “tangential” and “normal” components to the boundary behave differently. We establish its Helmholtz decomposition when the domain is bounded. This substantially extends the authors’ earlier result for a half space.

1 Introduction

The Helmholtz decomposition of a vector field is a fundamental tool to analyze the Stokes and the Navier–Stokes equations. It is formally a decomposition of a vector field $v=(v^1,\ldots ,v^n)$ in a domain $\Omega$ of $\mathbf {R}^n$ into

$$\begin{aligned} v = v_{0} + \nabla q; \end{aligned}$$

(1)

here $v_{0}$ is a divergence free vector field satisfying supplemental conditions like boundary condition and $\nabla q$ denotes the gradient of a function (scalar field) q. If v is in $L^{p}$ ($1<p<\infty$) in $\Omega$, such a decomposition is well-studied. For example, a topological direct sum decomposition

$$\begin{aligned} \left( L^p(\Omega ) \right) ^n = L^p_\sigma (\Omega ) \oplus G^p(\Omega ) \end{aligned}$$

holds for various domains including $\Omega =\mathbf {R}^n$, a half space $\mathbf {R}^n_+$, a bounded smooth domain [8]; see e.g. Galdi [9]. Here, $L^p_\sigma (\Omega )$ denotes the $L^p$-closure of the space of all div-free vector fields compactly supported in $\Omega$ and $G^p(\Omega )$ denotes the totality of $L^p$ gradient fields. It is impossible to extend this Helmholtz decomposition to $L^\infty$ even if $\Omega =\mathbf {R}^n$ since the projection $v\mapsto \nabla q$ is a composite of the Riesz operators which is not bounded in $L^\infty$. We have to replace $L^\infty$ with a class of functions of bounded mean oscillation. However, if the vector field is of bounded mean oscillation (BMO for short), such a problem is only studied when $\Omega$ is a half space $\mathbf {R}^n_+$ [10], where the boundary is flat.

Our goal is to establish the Helmholtz decomposition of BMO vector fields in a smooth bounded domain in $\mathbf {R}^n$, which is a typical example of a domain with curved boundary. Although the space of BMO functions in $\mathbf {R}^n$ is well studied, the situation is less clear when one considers such a space in a domain, because there are several possible definitions. One should be careful about the behavior of a function near the boundary $\Gamma = \partial \Omega$. In this paper we study a space of BMO vector fields introduced in [11] and establish its Helmholtz decomposition when $\Omega$ is a bounded $C^3$ domain.

Let us recall the space $vBMO(\Omega )$ introduced in [11]. We first recall the BMO seminorm for $\mu \in (0,\infty ]$. For a locally integrable function f, i.e., $f \in L^1_\mathrm {loc}(\Omega )$ we define

$$\begin{aligned} {[}f]_{BMO^\mu (\Omega )} := \sup \left\{ \frac{1}{\left| B_r(x)\right| } \int _{B_r(x)} \left| f(y) - f_{B_r(x)} \right| \, dy \biggm | B_r(x) \subset \Omega ,\ r < \mu \right\} , \end{aligned}$$

where $f_B$ denotes the average over B, i.e.,

$$\begin{aligned} f_B := \frac{1}{|B|} \int _B f(y) \, dy \end{aligned}$$

and $B_r(x)$ denotes the closed ball of radius r centered at x and |B| denotes the Lebesgue measure of B. The space $BMO^\mu (\Omega )$ is defined as

$$\begin{aligned} BMO^\mu (\Omega ) := \left\{ f \in L^1_\mathrm {loc}(\Omega ) \bigm | [f]_{BMO^\mu } < \infty \right\} . \end{aligned}$$

This space may not agree with the space of restrictions $r_\Omega f$ of $f \in BMO^\mu (\mathbf {R}^n)$. As in [1,2,3,4] we introduce a seminorm controlling the boundary behavior. For $\nu \in (0,\infty ]$, we set

$$\begin{aligned}{}[f]_{b^\nu } := \sup \left\{ r^{-n} \int _{\Omega \cap B_r(x)} \left| f(y) \right| \, dy \biggm | x \in \Gamma ,\ 0<r<\nu \right\} . \end{aligned}$$

In these papers, the space

$$\begin{aligned} BMO^{\mu ,\nu }_b(\Omega ) := \left\{ f \in BMO^\mu (\Omega ) \bigm | [f]_{b^\nu } < \infty \right\} \end{aligned}$$

is considered. Note that this space $BMO^{\infty ,\infty }_b(\Omega )$ is identified with Miyachi’s BMO introduced by [21] if $\Omega$ is a bounded Lipschitz domain or a Lipschitz half space as proved in [4]. However, unfortunately, it turns out such a boundary control for whole components of vector fields is too strict to have the Helmholtz decomposition. We separate tangential and normal components. Let $d_\Gamma (x)$ denote the distance from the boundary $\Gamma$, i.e.,

$$\begin{aligned} d_\Gamma (x) := \inf \left\{ |x - y|,\ y \in \Gamma \right\} . \end{aligned}$$

For vector fields, we consider

$$\begin{aligned} vBMO^{\mu ,\nu }(\Omega ) := \left\{ v \in \left( BMO^\mu (\Omega )\right) ^n \bigm | [\nabla d_\Gamma \cdot v]_{b^\nu } < \infty \right\} , \end{aligned}$$

where $\, \cdot \,$ denotes the standard inner product in $\mathbf {R}^n$. The quantity $(\nabla d_\Gamma \cdot v)\nabla d_\Gamma$ on $\Gamma$ is the component of v normal to the boundary $\Gamma$. We set

$$\begin{aligned}{}[v]_{vBMO^{\mu ,\nu }(\Omega )} := [v]_{BMO^\mu (\Omega )} + [\nabla d_\Gamma \cdot v]_{b^\nu }. \end{aligned}$$

If $\Omega$ is the half space, this is not a norm but a seminorm. However, if it has a fully curved part in the sense of [11,  Definition 7], then this becomes a norm [11,  Lemma 8]. In particular, when $\Omega$ is a bounded $C^2$ domain, this is a norm. Roughly speaking, the boundary behavior of a vector field v is controlled for only normal part of v if $v \in vBMO^{\mu ,\nu }(\Omega )$. For a bounded domain, this norm is equivalent no matter how $\mu$ and $\nu$ are taken; in other words, $vBMO^{\mu ,\nu }(\Omega )=vBMO^{\infty ,\infty }(\Omega )$. This is because $vBMO^{\mu ,\nu }(\Omega )\subset L^1(\Omega )$ when $\Omega$ is bounded, which follows from the characterization of $vBMO^{\mu ,\nu }(\Omega )$ in [11,  Theorem 9]. We shall simply write $vBMO^{\mu ,\nu }(\Omega )$ as $vBMO(\Omega )$. We are now in a position to state our main result.

Theorem 1

Let $\Omega$ be a bounded $C^3$ domain in $\mathbf {R}^n$. Then the topological direct sum decomposition

$$\begin{aligned} vBMO(\Omega ) = vBMO_\sigma (\Omega ) \oplus GvBMO(\Omega ) \end{aligned}$$

(2)

holds with

$$\begin{aligned} vBMO_\sigma (\Omega )&:= \left\{ v \in vBMO(\Omega ) \bigm | {\text {div}}v = 0\ \text {in}\ \Omega ,\ v \cdot \mathbf {n} = 0\ \text {on}\ \Gamma \right\} , \\ GvBMO(\Omega )&:= \left\{ \nabla q \in vBMO(\Omega ) \bigm | q \in L^1_\mathrm {loc}(\Omega ) \right\} , \end{aligned}$$

where $\mathbf {n}$ denotes the exterior unit normal vector field. In other words, for $v\in vBMO(\Omega )$, there are unique $v_0\in vBMO_\sigma (\Omega )$ and $\nabla q\in GvBMO(\Omega )$ satisfying $v=v_0+\nabla q$. Moreover, the mappings $v\mapsto v_0$, $v\mapsto \nabla q$ are bounded in $vBMO(\Omega )$.

As shown in [11], the normal trace $v \cdot \mathbf {n}$ is well defined as an element of $L^\infty (\Gamma )$ for $v \in vBMO(\Omega )$ with ${\text {div}}v = 0$. So far, the Helmholtz decomposition of BMO type space in a domain is only known for $vBMO^{\infty ,\infty }$ when $\Omega$ is the half space

$$\begin{aligned} \mathbf {R}^n_+ = \left\{ x = (x_1, \ldots , x_n) \in \mathbf {R}^n \bigm | x_n > 0 \right\} \end{aligned}$$

as shown in [10], where the normal trace is taken in locally $H^{-1/2}$ sense.

Here is our strategy to show Theorem 1. For a vector field v, we construct a linear map $v\longmapsto q_1$ such that $q_1$ satisfies

$$\begin{aligned} - \Delta q_1 = {\text {div}}v \quad \text {in}\quad \Omega , \end{aligned}$$

where the divergence is taken in the sense of distribution. There are many ways to construct such a map because there is no boundary condition. A naive way is to extend v in a suitable way to a function $\overline{v}$ on $\mathbf {R}^n$ so that $v\longmapsto \overline{v}$ is linear. We next consider the volume potential of ${\text {div}}\overline{v}$, i.e.,

$$\begin{aligned} q_0(x) := \int _{\mathbf {R}^n} E(x-y) {\text {div}} \overline{v}(y) \, dy = E * {\text {div}} \overline{v}, \end{aligned}$$

where E is the fundamental solution of $-\Delta$ in $\mathbf {R}^n$, i.e.,

$$\begin{aligned} E(x) := {\left\{ \begin{array}{ll} - \log |x| / 2 \pi &{} \quad (n = 2) \\ |x|^{2-n} / \left( n(n-2) \alpha (n) \right) &{} \quad (n \ge 3), \end{array}\right. } \end{aligned}$$

where $\alpha (n)$ denotes the volume of the unit ball $B_1(0)$ of $\mathbf {R}^n$. By the famous BMO-BMO estimate due to Fefferman and Stein [7], we have

$$\begin{aligned}{}[\nabla q_0]_{BMO^\infty (\mathbf {R}^n)} \le C_0[\overline{v}]_{BMO^\infty (\mathbf {R}^n)} \end{aligned}$$

with $C_0>0$ independent of $\overline{v}$. However, it is difficult to control $[\nabla d_\Gamma \cdot \nabla q_0]_{b^\nu }$ so we construct another function $q_1$ instead of $q_0$.

Although BMO space does not allow the standard cut-off procedure, our space is in $L^1$, so we are able to decompose v into two parts $v=v_1+ v_2$ such that the support of $v_2$ is close to $\Gamma$ while the support of $v_1$ is away from $\Gamma$; see Proposition 1. For $v_1$ we just set

$$\begin{aligned} q^1_1 = E * {\text {div}}v_1 \end{aligned}$$

by extending $v_1$ as zero outside its support. Then, the $L^\infty$ bound for $\nabla q^1_1$ is well controlled near $\Gamma$, which yields a bound for $b^\nu$ semi-norm. To estimate $v_2$, we use a normal coordinate system near $\Gamma$ and reduce the problem to the half space. Let d denote the signed distance function where $d=d_\Gamma$ in $\Omega$ and $d = - d_\Gamma$ outside $\Omega$. We extend $v_2$ to $\mathbf {R}^n$ so that the normal part $(\nabla d\cdot \overline{v}_2) \nabla d$ is odd and the tangential part $\overline{v_2} - (\nabla d \cdot \overline{v_2}) \nabla d$ is even in the direction of $\nabla d$ with respect to $\Gamma$. In such type of coordinate system, the minus Laplacian can be transformed as

$$\begin{aligned} L = A - B + \text {lower order terms},\ A = -\Delta _\eta , \ B = \sum _{1\le i,j\le n-1} \partial _{\eta _i} b_{ij} \partial _{\eta _j}, \end{aligned}$$

where $\eta _n$ is the normal direction to the boundary so that $\{ \eta _n>0\}$ is the half space. By choosing a suitable coordinate system to represent $\Gamma$ locally, we are able to arrange $b_{ij}=0$ at one point of the boundary of the local coordinate system. We use a freezing coefficient method to construct volume potential $q_1^2$ and $q^3_1$, which corresponds to the contribution from the tangential part $\overline{v_2}^{\mathrm {tan}}$ and the normal part $\overline{v_2}^{\mathrm {nor}}$ respectively. Since the leading term of ${\text {div}} \overline{v_2}^{\mathrm {nor}}$ in normal coordinate consists of the differential of $\eta _n$ only, if we extend the coefficient $b_{ij}$ even in $\eta _n$, $q_1^3$ is constructed so that the leading term of $\nabla d \cdot \nabla q^3_1$ is odd in the direction of $\nabla d$. On the other hand, as the leading term of ${\text {div}} \overline{v_2}^{\mathrm {tan}}$ in normal coordinate consists of the differential of $\eta ' = (\eta _1, ... , \eta _{n-1})$ only, the even extension of $b_{ij}$ in $\eta _n$ gives rise to $q_1^2$ so that the leading term of $\nabla d \cdot \nabla q_1^2$ is also odd in the direction of $\nabla d$. Disregarding lower order terms and localization procedure, we set $q_1^2$ and $q^3_1$ of the form

$$\begin{aligned} q_1^2&= -L^{-1} {\text {div}}\overline{v}^{\mathrm {tan}}_2 = -A^{-1} (I-BA^{-1})^{-1} {\text {div}}\overline{v}^{\mathrm {tan}}_2, \\ q^3_1&= -L^{-1} {\text {div}}\overline{v}^{\mathrm {nor}}_2 = -A^{-1} (I-BA^{-1})^{-1} {\text {div}}\overline{v}^{\mathrm {nor}}_2. \end{aligned}$$

One is able to arrange $BA^{-1}$ small by taking a small neighborhood of a boundary point. Then $(I-BA^{-1})^{-1}$ is given as the Neumann series $\sum ^\infty _{m=0}(BA^{-1})^m$. We are able to establish BMO-BMO estimate for $\nabla q_1^2$ and $\nabla q^3_1$, i.e.

$$\begin{aligned} \left[ \nabla q_1^2 \right] _{BMO(\mathbf {R}^n)} \le C'_0 \left[ \overline{v}^{\mathrm {tan}}_2 \right] _{BMO(\mathbf {R}^n)}, \; \left[ \nabla q^3_1 \right] _{BMO(\mathbf {R}^n)} \le C'_0 \left[ \overline{v}^{\mathrm {nor}}_2 \right] _{BMO(\mathbf {R}^n)} \end{aligned}$$

with some constant $C'_0$ independent of $\overline{v_2}$. Since the leading term of $\nabla d \cdot (\nabla q_1^2 + \nabla q^3_1)$ is odd in the direction of $\nabla d$ with respect to $\Gamma$, the BMO bound implies $b^\nu$ bound. Note that $\left[ \overline{v_2}^{\mathrm {nor}}\right] _{BMO(\mathbf {R}^n)}$ is controlled by $[v_2]_{b^\nu }$ and $[v_2]_{BMO(\Omega )}$ since $\overline{v_2}^{\mathrm {nor}}$ is odd in the direction of $\nabla d$ with respect to $\Gamma$. By the procedure sketched above, we are able to construct a suitable operator by setting $q_1=q^1_1+q^2_1+q^3_1$.

Theorem 2

(Construction of a suitable volume potential) Let $\Omega$ be a bounded $C^3$ domain in $\mathbf {R}^n$. Then, there exists a linear operator $v\longmapsto q_1$ from $vBMO(\Omega )$ to $L^\infty (\Omega )$ such that

$$\begin{aligned} - \Delta q_1 = {\text {div}}v \quad \text {in}\quad \Omega \end{aligned}$$

and that there exists a constant $C_1=C_1(\Omega )$ satisfying

$$\begin{aligned} \Vert \nabla q_1\Vert _{vBMO(\Omega )} \le C_1 \Vert v\Vert _{vBMO(\Omega )}. \end{aligned}$$

In particular, the operator $v\longmapsto \nabla q_1$ is a bounded linear operator in $vBMO(\Omega )$.

By this operator, we observe that $w=v-\nabla q_1$ is divergence free in $\Omega$. Unfortunately, this w may not fulfill the trace condition $w\cdot \mathbf {n}=0$ on the boundary $\Gamma$. We construct another potential $q_2$ by solving the Neumann problem

$$\begin{aligned} \Delta q_2&= 0 \quad \quad \;\;\,\text {in}\quad \Omega \\ \frac{\partial q_2}{\partial \mathbf {n}}&= w\cdot \mathbf {n} \quad \text {on}\quad \Gamma . \end{aligned}$$

We then set $q=q_1+q_2$. Since $\partial q_2/\partial \mathbf {n}=\nabla q_2\cdot \mathbf {n}$, $v_0=v-\nabla q$ gives the Helmholtz decomposition (1). To complete the proof of Theorem 1, it suffices to prove that $\Vert \nabla q_2\Vert _{vBMO(\Omega )}$ is bounded by a constant multiply of $\Vert v\Vert _{vBMO(\Omega )}$.

Lemma 1

(Estimate of the normal trace) Let $\Omega$ be a bounded $C^{2+\kappa }$ domain in $\mathbf {R}^n$ with $\kappa \in (0,1)$. Then there is a constant $C_2 = C_2(\Omega )$ such that

$$\begin{aligned} \Vert w \cdot \mathbf {n} \Vert _{L^\infty (\Gamma )} \le C_2 \Vert w \Vert _{vBMO(\Omega )} \end{aligned}$$

for all $w \in vBMO(\Omega )$ with ${\text {div}}w = 0$.

This is a special case of the trace theorem established in [11]. We finally need the estimate for the Neumann problem.

Lemma 2

(Estimate for the Neumann problem) Let $\Omega$ be a bounded $C^2$ domain. For $g \in L^\infty (\Gamma )$ satisfying $\int _\Gamma g \, d\mathcal {H}^{n-1}=0$, there exists a unique (up to constant) solution u to the Neumann problem

$$\begin{aligned}&\begin{aligned} \Delta u&= 0 \quad \text {in}\quad \Omega \\ \frac{\partial u}{\partial \mathbf {n}}&= g \quad \text {on}\quad \Gamma \end{aligned} \end{aligned}$$

(3)

such that the operator $g \longmapsto u$ is linear and that there exists a constant $C_3 = C_3(\Omega )$ such that

$$\begin{aligned} \Vert \nabla u \Vert _{vBMO(\Omega )} \le C_3 \Vert g\Vert _{L^\infty (\Gamma )}. \end{aligned}$$

Combining these two lemmas, Theorem 2 yields

$$\begin{aligned} \Vert \nabla q_2\Vert _{vBMO(\Omega )}&\le C_3 C_2 \Vert v-\nabla q_1\Vert _{vBMO(\Omega )} \\&\le C_3 C_2 (1 + C_1) \Vert v\Vert _{vBMO(\Omega )}. \end{aligned}$$

Setting $q=q_1+q_2$ and $v_0=v-\nabla q$, we now observe that the projections $v\longmapsto v_0$, $v\longmapsto \nabla q$ are bounded in $vBMO(\Omega )$, which yields (2) in Theorem 1.

To show Lemma 2 let N(x, y) be the Neumann Green function. Then a solution of (3) is given by $\int _{\Gamma } N(x,y) g(y) \, \mathrm {d} \mathcal {H}^{n-1}$. It is well-known (see e.g. [12,  Appendix]) that leading part of N is $E(x-y)$. We have to estimate

$$\begin{aligned} \left\| \nabla E *(\delta _{\Gamma } \otimes g)\right\| _{vBMO^{\infty ,\nu }(\Omega )}. \end{aligned}$$

Here $\delta _{\Gamma }$ denotes the delta function supported on $\Gamma$, i.e.,

$$\begin{aligned} \delta _{\Gamma } : \psi \mapsto \int _{\Gamma } \psi \, \mathrm {d} \mathcal {H}^{n-1} \end{aligned}$$

for $\psi \in C_{c}^{\infty }(\mathbb {R}^n)$. We take a $C^{2}$ cutoff function $\theta \ge 0$ such that $\theta (\sigma ) = 1$ for $\sigma \le 1$, $\theta (\sigma ) = 0$ for $\sigma \ge 2$. We take $\delta$ small so that $2\delta$ is smaller than the reach of $\Gamma$, for the definition of the reach of $\Gamma$ please refer to Sect. 2.1. By this choice, $\theta _d = \theta (d/\delta )$ is $C^{2}$ in $\mathbf {R}^n$, where d denotes the signed distance function from $\Gamma$ so that $\nabla d = - \mathbf {n}$ on $\Gamma$. For $g \in L^{\infty }(\Gamma )$, we extend g so that $\nabla d \cdot \nabla g = 0$ near the $2\delta$-neighborhood of $\Gamma$. Let $g_{e}$ denote this extension and set $g_{e,c} = \theta _d g_e$. A key observation is that

$$\begin{aligned}&\delta _{\Gamma } \otimes g = ( \nabla 1_{\Omega } \cdot \nabla d ) g_{e,c} = {\text {div}} \, (g_{e,c} 1_{\Omega } \nabla d) - 1_{\Omega } {\text {div}} \, (g_{e,c} \nabla d) \\&{\text {div}} \, (g_{e,c} \nabla d) = g_{e,c} \Delta d + \nabla d \cdot \nabla g_{e,c} = g_{e,c} \Delta d + \frac{\theta '(d/\delta )}{\delta } g_e, \end{aligned}$$

where $1_{\Omega }$ is the characteristic function of $\Omega$. The leading (singular) part of $\nabla E *(\delta _{\Gamma } \otimes g)$ is the term involving ${\text {div}} \, (g_{e,c} 1_{\Omega } \nabla d)$. The famous $L^\infty$-BMO estimate for the singular integral operator $\nabla E*{\text {div}}$ yields

$$\begin{aligned} \left\| \nabla E* {\text {div}}\,(g_{e,c} 1_{\Omega } \nabla d)\right\| _{BMO(\mathbf {R}^n)} \le C \Vert g_{e,c} \nabla d\Vert _{L^{\infty }(\Omega )} \le C' \Vert g\Vert _{L^{\infty }(\Gamma )} \end{aligned}$$

with C and $C'$ independent of g. All other terms can be estimated easily since the integral kernel is integrable. A direct calculation gives an $L^\infty$ estimate near $\Gamma$ for $\nabla d\cdot \nabla E*(\delta _\Gamma \otimes g)$ which yields

$$\begin{aligned} \left[ \nabla d \cdot \nabla E *(\delta _\Gamma \otimes g) \right] _{b^\nu } \le C_4 \Vert g \Vert _{L^\infty (\Gamma )} \end{aligned}$$

with $C_4$ independent of g, but it is impossible to estimate $b^\nu$-seminorm of the tangential part. This is the main reason why we use vBMO instead of $BMO_b$-type space where $b^\nu$-boundedness of ALL components of vector fields is imposed; see the end of Sect. 4.2.

To extend our results to a more general domain it seems to be reasonable to consider $vBMO\cap L^2$. This is because $L^p\cap L^2\ (p>2)$ admits the Helmholtz decomposition for arbitrary uniformly $C^2$ domains as proved in [5, 6].

Our approach in this paper is to derive the boundedness of the operator $v\mapsto \nabla q$ by a potential-theoretic approach. In $L^p$ setting there is a variational approach based on duality introduced by [23]; see also [5]. The key estimate is

$$\begin{aligned} \Vert \nabla q\Vert _{L^p(\Omega )} \le C_5 \sup \left\{ \int _\Omega \nabla q\cdot \nabla \varphi \, dx \Bigm | \Vert \nabla \varphi \Vert _{L^{p'}(\Omega )} \le 1 \right\} \end{aligned}$$

with $C_5$ independent of q, where $1/p+1/p'=1$, $1<p<\infty$. Formally, this estimate yields the desired bound $\Vert \nabla q\Vert _{L^p(\Omega )}\le C_5\Vert v\Vert _{L^p(\Omega )}$ since $\int _\Omega \nabla q\cdot \nabla \varphi \, dx = \int _\Omega v\cdot \nabla \varphi \, dx$. At this moment, it is not clear that similar estimate holds if one replaces $L^p(\Omega )$ by vBMO since the predual space of vBMO is not clear.

For $BMO_b$ type solution, it is known that the Stokes semigroup is analytic [1, 3]. However, it is nontrivial to extend to the space vBMO since in the half space the Stokes operator with Dirichlet boundary condition does not generate a semigroup because $\left[ u(t)\right] _{vBMO}$ for the solution u(t) may be non-zero for $t>0$ for initial data $u_0$ with $[u_0]_{vBMO}=0$ so that $u^{\tan }_0$ may be a non-zero constant [1,  Example 6.5].

This paper is organized as follows. In Sect. 2, to construct a volume potential of ${\text {div}}v$, we localize the problem and reduce the problem to small neighborhoods of points on the boundary. In Sect. 3, we construct a leading part of the volume potential by a perturbation method called the freezing coefficient method. In these two sections, we complete the proof of Theorem 2. In Sect. 4, we prove Lemma 2 by estimating the single layer potential.

2 Construction of volume potentials

For $v \in vBMO(\Omega )$, we shall construct a suitable potential $q_1$ so that $v\longmapsto \nabla q_1$ is a bounded linear operator in vBMO as stated in Theorem 2. In this section, as a preliminary, we reduce the problem to the case that the support of v is contained in a small neighborhood of a point of the boundary and it consists of only normal part.

2.1 Localization procedure

Let $\Omega$ be a uniformly $C^k$ domain in $\mathbf {R}^n$ ($k\ge 1$). In other words, there exist $r_*, \delta _*>0$ such that for each $z_0 \in \Gamma$, up to translation and rotation, there exists a function $h_{z_0}$ which is $C^k$ in a closed ball $B_{r_*}(0')$ of radius $r_*$ centered at the origin $0'$ of $\mathbf {R}^{n-1}$ satisfying following properties:

• (i) $K_\Gamma := \sup _{B_{r_*}(0')} \left| (\nabla ')^s h_{z_0} \right| < \infty$ for $s=0,1,2,\ldots ,k$, where $\nabla '$ denotes the gradient in $x'\in \mathbf {R}^{n-1}$; $\nabla ' h_{z_0}(0')=0$, $h_{z_0}(0')=0$,

• (ii) $\Omega \cap U_{r_*,\delta _*, h_{z_0}}(z_0)=\left\{ (x',x_n)\in \mathbf {R}^n \bigm | h_{z_0}(x')<x_n<h_{z_0}(x')+\delta _*,\ |x'|<r_* \right\}$ for

  $$\begin{aligned} U_{r_*,\delta _*, h_{z_0}}(z_0) := \left\{ (x',x_n)\in \mathbf {R}^n \bigm | h_{z_0}(x')-\delta _*< x_n< h_{z_0}(x')+\delta _*,\ |x'|<r_* \right\} , \end{aligned}$$

• (ii) $\Gamma \cap U_{r_*,\delta _*, h_{z_0}}(z_0)=\left\{ (x',x_n)\in \mathbf {R}^n \bigm | x_n = h_{z_0}(x'),\ |x'|<r_* \right\}$.

A bounded $C^k$ domain is, of course, a uniformly $C^k$ domain.

Let d denote the signed distance function from $\Gamma$ which is defined by

$$\begin{aligned} d(x) := \left\{ \begin{array}{ll} \displaystyle \inf _{y\in \Gamma }|x-y| &{}\quad \text {for}\quad x\in \Omega , \\ \displaystyle -\inf _{y\in \Gamma }|x-y| &{}\quad \text {for}\quad x\notin \Omega \end{array} \right. \end{aligned}$$

(4)

so that $d(x)=d_\Gamma (x)$ for $x\in \Omega$. If $\Omega$ is a bounded $C^2$ domain, then there is $R^\Omega >0$ such that for $x \in \Omega$ with $\left| d(x)\right| <R^\Omega$, there is a unique point $\pi x \in \Gamma$ such that $|x-\pi x|=\left| d(x)\right|$. The supremum of such $R^\Omega$ is called the reach of $\Gamma$ in $\Omega$, we denote this supremum by $R_*^\Omega$. Let $\Omega ^{\mathrm {c}}$ be the complement of $\Omega$ in $\mathbf {R}^n$. Similarly, there is $R^{\Omega ^{\mathrm {c}}}>0$ such that for $x \in \Omega ^{\mathrm {c}}$ with $\left| d(x)\right| < R^{\Omega ^{\mathrm {c}}}$, we can also find a unique point $\pi x \in \Gamma$ such that $|x-\pi x|=\left| d(x)\right|$. The supremum of such $R^{\Omega ^{\mathrm {c}}}$ is called the reach of $\Gamma$ in $\Omega ^{\mathrm {c}}$, we denote this supremum by $R_*^{\Omega ^{\mathrm {c}}}$. We then define

$$\begin{aligned} R_* := \min \left( R^\Omega _*, R^{\Omega ^{\mathrm {c}}}_* \right) , \end{aligned}$$

which we call it the reach of $\Gamma$. Moreover, d is $C^2$ in the $R_*$-neighborhood of $\Gamma$, i.e., $d \in C^2\left( \Gamma ^{\mathbf {R}^n}_{R_*}\right)$ with

$$\begin{aligned} \Gamma ^{\mathbf {R}^n}_{R_*} := \left\{ x \in \mathbf {R}^n \bigm | \left| d(x) \right| < R_*\right\} ; \end{aligned}$$

see [13,  Chap. 14, Appendix], [19,  §4.4]. Let $K_\Gamma ^*:= \mathrm {max} \, \{ K_\Gamma , 1 \}$. There exists $0< \rho _0 < \min (r_*, \delta _*, \frac{R_*}{2}, \frac{1}{2n K_\Gamma ^*})$ such that

$$\begin{aligned} U_\rho (z_0) := \left\{ x \in \mathbf {R}^n \bigm | (\pi x)' \in {\text {int}} B_\rho (0'),\ \left| d(x) \right| < \rho \right\} \end{aligned}$$

is contained in the coordinate chart $U_{r_*,\delta _*, h_{z_0}}(z_0)$ for any $\rho \le \rho _0$.

We always take $\rho <\rho _0$. Since $\Omega$ is bounded and

$$\begin{aligned} \bigcup _{z \in \Gamma } U_\rho (z) \end{aligned}$$

covers the compact set $K=\mathrm {cl}\left( \Gamma ^{\mathbf {R}^n}_{\rho /2}\right)$, there exists a finite subcover $\left\{ U_\rho (z_j) \right\} ^m_{j=1}$ of K, where the number m depends on $\rho$. For $\sigma > 0$, we denote that

$$\begin{aligned} \Omega ^\sigma = \Omega \backslash \Gamma ^{\mathbf {R}^n}_\sigma , \; \, U_{\sigma ,j} := U_\sigma (z_j). \end{aligned}$$

Observe that

$$\begin{aligned} \overline{\Omega } \subset \bigcup ^m_{j=1} U_{\rho ,j} \cup \Omega ^{\rho /2}. \end{aligned}$$

Let $\{ \varphi _j \}^m_{j=0}$ be a partition of the unity associated with $\{ U_{\rho ,j}\} \cup \{\Omega ^{\rho /2}\}$ in the sense that

$$\begin{aligned}&\varphi _j \in C^\infty _c (U_{\rho ,j} \cap \overline{\Omega }), \; \; 0 \le \varphi _j \le 1 \quad \text {for} \quad j=1, \ldots , m, \\&\varphi _0 \in C^\infty _c (\Omega ^{\rho /2}), \; \; 0 \le \varphi _0 \le 1, \; \; \varphi _0 = 1 \quad \text {in} \quad \Omega ^{\rho } \end{aligned}$$

and

$$\begin{aligned} \sum ^m_{j=0} \varphi _j = 1 \quad \text {in} \quad \overline{\Omega }. \end{aligned}$$

Here $C^\infty _c(W)$ denotes the space of all smooth function in W whose support is compact in W.

Throughout this paper, unless otherwise specified, the symbol C in an inequality represents a positive constant independent of quantities that appeared in the inequality. For a fixed $\rho >0$, $C_\rho$ represents a constant depending only on $\rho$. $C_n$ represents a constant depending only on n and $C_{\Omega ,n}$ represents a constant depending only on $\Omega$ and n.

2.2 Cut-off and extension

In general, multiplication by a smooth function to BMO is not bounded in BMO. Fortunately, our space is closed by multiplication.

Proposition 1

(Multiplication) Let $\Omega$ be a bounded $C^2$ domain in $\mathbf {R}^n$. Let $\varphi \in C^\gamma (\Omega )$, $\gamma \in (0,1)$. For each $v \in vBMO(\Omega )$, the function $\varphi v \in vBMO(\Omega )$ satisfies

$$\begin{aligned} \Vert \varphi v \Vert _{vBMO(\Omega )} \le C\Vert \varphi \Vert _{C^\gamma (\Omega )} \Vert v\Vert _{vBMO(\Omega )} \end{aligned}$$

with C independent of $\varphi$ and v.

Proof

Since

$$\begin{aligned} \left[ \nabla d\cdot \varphi v \right] _{b^\nu } \le \Vert \varphi \Vert _{L^\infty (\Omega )} \left[ \nabla d\cdot v \right] _{b^\nu }, \end{aligned}$$

it suffices to establish the estimate

$$\begin{aligned} \left[ \varphi v \right] _{BMO^\infty (\Omega )} \le c_0 \Vert \varphi \Vert _{C^\gamma (\Omega )} \Vert v\Vert _{vBMO(\Omega )} \end{aligned}$$

(5)

with $c_0$ independent of $\varphi$ and v. Since a bounded Lipschitz domain is a uniform domain, we are able to apply [11,  Theorem 13] to get

$$\begin{aligned} \left[ \varphi v \right] _{BMO^\infty (\Omega )} \le c_1 \Vert \varphi \Vert _{C^\gamma (\Omega )} \big ( [v]_{BMO^\infty (\Omega )} + \Vert v\Vert _{L^1(\Omega )} \big ). \end{aligned}$$

This is based on the product estimate of a Hölder function and a function in $bmo_\infty ^\infty (\Omega ) := BMO^\infty (\Omega )\cap L^1_\mathrm {ul}(\Omega )$ where

$$\begin{aligned} L^1_{\mathrm {ul}}(\Omega ) := \bigg \{ f \in L^1_{\mathrm {loc}} (\Omega ) \biggm | \Vert f \Vert _{L^1_{\mathrm {ul}}(\Omega )} := \sup _{x \in \mathbf {R}^n} \int _{B_1(x) \cap \Omega } \bigl | f(y) \bigr | \, dy < \infty \bigg \}. \end{aligned}$$

The space $bmo_\infty ^\infty (\Omega )$ is equipped with the norm

$$\begin{aligned} \Vert f \Vert _{bmo_\infty ^\infty (\Omega )}:= [f]_{BMO^\infty (\Omega )} + \Vert f \Vert _{L_{\mathrm {ul}}^1(\Omega )} \end{aligned}$$

for $f \in bmo_\infty ^\infty (\Omega )$. For local BMO spaces defined in a domain, this space was firstly introduced in [11,  Section 2]. In the case where $\Omega$ is the whole space $\mathbf {R}^n$, for simplicity we omit the infinity subscript and superscript and denote this space by $bmo(\mathbf {R}^n)$. The product estimate for $bmo(\mathbf {R}^n)$ follows from a similar result for a local Hardy space $h^1(\mathbf {R}^n) = F^0_{1,2}(\mathbf {R}^n)$ [22,  Remark 4.4] and duality $bmo(\mathbf {R}^n) = \big ( h^1(\mathbf {R}^n) \big )'$ [22,  Theorem 3.26]. To handle a function in $\Omega$, we need an extension to conclude [11,  Theorem 13]. Fortunately, by the characterization of vBMO for a bounded $C^2$ domain [11,  Theorem 9],

$$\begin{aligned} \Vert v\Vert _{L^1(\Omega )} \le c_2 \Vert v\Vert _{vBMO(\Omega )}. \end{aligned}$$

Here $c_j$ denotes a constant independent of v and $\varphi$ for $j=1,2$. Combining these two estimates, we obtain (5) with $c_0 = c_1(1+c_2)$. This yields Proposition 1. $\square$

For a bounded $C^3$ domain, we next consider an extension based on the normal coordinate in $U_\rho (z_0)$ for $\rho \le \rho _0$ of the form

$$\begin{aligned} \left\{ \begin{array}{lcl} x' &{}=&{} \eta ' + \eta _n \nabla 'd ( \eta ', h_{z_0}(\eta ') ); \\ x_n &{}=&{} h_{z_0}(\eta ') + \eta _n \partial _{x_n} d ( \eta ', h_{z_0}(\eta ') ). \end{array} \right. \end{aligned}$$

(6)

Let $V_{\sigma } := B_\sigma (0') \times (- \sigma , \sigma )$ for $\sigma \in (0,\rho _0)$. We shall write this coordinate change by $x=\psi (\eta )$ with $\psi \in C^2(V_{\rho _0})$ and

$$\begin{aligned} x = \pi x - d(x) \mathbf {n}(\pi x), \quad \mathbf {n}(\pi x) = -\nabla d(\pi x). \end{aligned}$$

We consider the projection to the direction to $\nabla d$. For $x \in \Gamma _{\rho _0}^{\mathbf {R}^n}$, we set

$$\begin{aligned} P(x) = \nabla d(\pi x) \otimes \nabla d(\pi x) = \mathbf {n}(\pi x) \otimes \mathbf {n}(\pi x). \end{aligned}$$

For later convenience, we set $Q(x) = I - P(x)$ which is the tangential projection for $x \in \Gamma _{\rho _0}^{\mathbf {R}^n}$. For a function f in $\Gamma _\rho ^{\mathbf {R}^n} \cap \overline{\Omega }$, let $f_{\mathrm {even}}$ (resp. $f_{\mathrm {odd}}$) denote its even (odd) extension to $\Gamma _\rho ^{\mathbf {R}^n}$ defined by

$$\begin{aligned} f_{\mathrm {even}} \left( \pi x + d(x)\mathbf {n}(\pi x) \right)&= f \left( \pi x - d(x)\mathbf {n}(\pi x) \right)&\text {for}\quad x \in \Gamma _\rho ^{\mathbf {R}^n} \backslash \overline{\Omega }, \\ f_{\mathrm {odd}} \left( \pi x + d(x)\mathbf {n}(\pi x) \right)&= -f \left( \pi x - d(x)\mathbf {n}(\pi x) \right)&\text {for}\quad x \in \Gamma _\rho ^{\mathbf {R}^n} \backslash \overline{\Omega }. \end{aligned}$$

We denote $r_W$ to be the restriction in W for any subset $W \subset \mathbf {R}^n$. Let f be a function (or a vector field) defined in $V_\sigma$ for some $\sigma \in (0,\infty ]$. We set $E_{\mathrm {even}} f$ to be the even extension of f in $V_\sigma \cap \mathbf {R}_+^n$ to $V_\sigma$ with respect to the n-th variable, i.e.,

$$\begin{aligned} E_{\mathrm {even}} f (\eta ', - \eta _n) = f(\eta ', \eta _n) \end{aligned}$$

for any $(\eta ', \eta _n) \in V_\sigma \cap \mathbf {R}_+^n$.

For $v\in vBMO(\Omega )$ with ${\text {supp}} \, v \subset U_\rho (z_0) \cap \overline{\Omega }$, let $\overline{v}$ be its extension of the form

$$\begin{aligned} \overline{v}(x) := (P v_{\mathrm {odd}}) (x) + (Q v_{\mathrm {even}}) (x) \end{aligned}$$

(7)

for $x \in U_{\rho }(z_0)$. Notice that ${\text {supp}} \, \overline{v} \subset U_{\rho }(z_0)$, $\overline{v}$ is indeed defined in $\mathbf {R}^n$ with $\overline{v}(x) = 0$ for any $x \in U_\rho (z_0)^{\mathrm {c}}$. Define

$$\begin{aligned} L_*:= \underset{z_0 \in \Gamma , \, \rho \le \rho _0}{\sup } \, \mathrm {max} \, \{ \Vert \nabla \psi \Vert _{L^\infty (V_\rho )} + \Vert \nabla \psi ^{-1} \Vert _{L^\infty (U_{\rho }(z_0))}, 1 \}. \end{aligned}$$

Since the boundary $\Gamma$ is uniformly $C^3$, $L_*$ is finite that depends on $\Omega$ only. We set $\rho _{0,*} = \rho _0/12L_*$.

Proposition 2

Let $\Omega \subset \mathbf {R}^n$ be a bounded $C^2$ domain, $z_0 \in \Gamma$ and $\rho \in (0, \rho _{0,*})$. There exists a constant $C_\rho$, which depends on $\rho$ only, such that

$$\begin{aligned}{}[\overline{v}]_{BMO\left( \mathbf {R}^n \right) }&\le C_\rho \Vert v\Vert _{vBMO(\Omega )}, \\ [\nabla d\cdot \overline{v}]_{b^\nu (\Gamma )}&\le C_\rho \Vert v\Vert _{vBMO(\Omega )} \end{aligned}$$

for all $v \in vBMO(\Omega )$ with ${\text {supp}} \, v \subset U_\rho (z_0) \cap \overline{\Omega }$ and $\nu >0$.

In the normal coordinate, $P\overline{v} = P v_{\mathrm {odd}}$ is odd in $\eta _n$ and $Q\overline{v} = Q v_{\mathrm {even}}$ is even in $\eta _n$. The key idea of proving this proposition is to reduce the problem to the case where the boundary is locally flat by invoking the normal coordinate.

Proof

Since $vBMO(\Omega ) \subset L^1(\Omega )$, see e.g. [11,  Theorem 9], by considering the normal coordinate change $y = \psi (\eta )$ in $U_{\rho }(z_0)$ we can deduce that $v_{\mathrm {even}}, v_{\mathrm {odd}} \in L^1(\mathbf {R}^n)$ satisfying

$$\begin{aligned} \Vert v_{\mathrm {even}} \Vert _{L^1(\mathbf {R}^n)} = \Vert v_{\mathrm {odd}} \Vert _{L^1(\mathbf {R}^n)} \le 2 L_*^2 \Vert v \Vert _{L^1(\Omega )}. \end{aligned}$$

Hence $\overline{v} \in L^1(\mathbf {R}^n)$ satisfies the estimate $\Vert \overline{v} \Vert _{L^1(\mathbf {R}^n)} \le C_{\Omega ,n} \Vert v \Vert _{L^1(\Omega )}$. Since $\Omega$ is a uniform domain, by [17,  Theorem 1] there exists $v_J \in BMO(\mathbf {R}^n)$ with $r_\Omega v_J = v$ and

$$\begin{aligned}{}[v_J]_{BMO(\mathbf {R}^n)} \le C_{\Omega ,n} [v]_{BMO^\infty (\Omega )}. \end{aligned}$$

Suppose that $B_r(\zeta ) \subset V_{4 \rho L_*}^+ := V_{4 \rho L_*} \cap \mathbf {R}_+^n$. The mean value theorem implies that $\psi (B_r(\zeta )) \subset B_{L_*r}(x)$ with $x = \psi (\zeta )$. By change of variables $y = \psi (\eta )$ in $U_{4 \rho L_*}(z_0)$, we see that

$$\begin{aligned} \frac{1}{|B_r(\zeta )|} \int _{B_r(\zeta )} | v \circ \psi (\eta ) - c | \, d \eta&\le L_*\cdot \frac{1}{|B_r(\zeta )|} \int _{\psi (B_r(\zeta ))} | v(y) - c | \, dy \\&\le C_n L_*^{n+1} \cdot \frac{1}{|B_{L_*r}(x)|} \int _{B_{L_*r}(x)} | v_J (y) -c | \, dy \end{aligned}$$

for any constant vector $c \in \mathbf {R}^n$. By considering an equivalent definition of the BMO-seminorm, see e.g. [15,  Proposition 3.1.2], we deduce that

$$\begin{aligned}{}[v \circ \psi ]_{BMO^\infty (V_{4 \rho L_*}^+)} \le C_{\Omega ,n} [v]_{BMO^\infty (\Omega )}. \end{aligned}$$

By recalling the results concerning the even extension of BMO functions in the half space, see [10,  Lemma 3.2 and Lemma 3.4], we can deduce that

$$\begin{aligned}{}[ v_{\mathrm {even}} \circ \psi ]_{BMO^\infty (V_{4 \rho L_*})} \le C_{\Omega ,n} [v]_{BMO^\infty (\Omega )}. \end{aligned}$$

(8)

Next, we shall estimate the BMO-seminorm of $v_{\mathrm {even}}$. Let $B_r(x)$ be a ball with radius $r \le \frac{\rho }{2}$. If either $B_r(x) \cap U_\rho (z_0) = \emptyset$ or $B_r(x) \subset \Omega$, there is nothing to prove. It is sufficient to consider $B_r(x)$ that intersects both $U_\rho (z_0)$ and $\Omega ^{\mathrm {c}}$. In this case we can find $x_0 \in B_r(x) \cap U_\rho (z_0)$. Since $B_r(x) \subset B_{2r}(x_0) \subset B_{4 \rho }(z_0) \subset U_{8 \rho }(z_0)$, by considering the change of variables $y = \psi (\eta )$ in $U_{8 \rho }(x_0)$, we have that

$$\begin{aligned} \frac{1}{|B_r(x)|} \int _{B_r(x)} | v_{\mathrm {even}} (y) - c | \, dy \le \frac{L_*}{|B_r(x)|} \int _{\psi ^{-1}(B_r(x))} | v_{\mathrm {even}} \circ \psi (\eta ) - c | \, d\eta . \end{aligned}$$

For any $y \in B_r(x)$, we have that $| y - z_0 | < 4 \rho$. Hence $\psi ^{-1}(B_r(x)) \subset B_{L_*r}(\zeta ) \subset B_{4 \rho L_*}(0) \subset V_{4 \rho L_*}$. By (8), we deduce that

$$\begin{aligned} \frac{1}{|B_r(x)|} \int _{B_r(x)} | v_{\mathrm {even}} (y) - (v_{\mathrm {even}})_{B_r(x)} | \, dy \le C_{\Omega ,n} [v]_{BMO^\infty (\Omega )}. \end{aligned}$$

Thus, we obtain that

$$\begin{aligned}{}[v_{\mathrm {even}}]_{BMO^{\frac{\rho }{2}}(\mathbf {R}^n)} \le C_{\Omega ,n} [v]_{BMO^\infty (\Omega )}. \end{aligned}$$

For a ball B with radius $r(B) > \frac{\rho }{2}$, a simple triangle inequality implies that

$$\begin{aligned} \frac{1}{|B|} \int _B | v_{\mathrm {even}} (y) - (v_{\mathrm {even}})_B | \, dy \le \frac{2}{|B|} \int _B | v_{\mathrm {even}} (y) | \, dy \le \frac{C_n}{\rho ^n} \Vert v_{\mathrm {even}} \Vert _{L^1(\mathbf {R}^n)}. \end{aligned}$$

Therefore, we obtain the BMO estimate for $v_{\mathrm {even}}$, i.e.,

$$\begin{aligned}{}[v_{\mathrm {even}}]_{BMO(\mathbf {R}^n)} \le \frac{C_{\Omega ,n}}{\rho ^n} \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

We shall then give the BMO estimate for $P v_{\mathrm {odd}}$. Since $\nabla d \in C^1(\Gamma _{\rho _0}^{\mathbf {R}^n})$, there exists $D_e \in C^1(\mathbf {R}^n)$ such that $\Vert D_e \Vert _{C^1(\mathbf {R}^n)} \le \Vert \nabla d \Vert _{C^1(\Gamma _{\rho _0}^{\mathbf {R}^n})}$ and $r_{\Gamma _{\rho _0}^{\mathbf {R}^n}} D_e = \nabla d$, see the proof of [11,  Theorem 13]. By the multiplication rule for bmo functions, we have that $(P v)_E := (D_e \cdot v_{\mathrm {even}} ) D_e \in bmo(\mathbf {R}^n)$, see also [11,  Theorem 13]. Consider the normal coordinate change in $U_{4 \rho L_*}(z_0)$. Since $(P v)_E = Pv$ in $U_{4 \rho L_*}(z_0) \cap \Omega$, the same argument in the second paragraph implies that

$$\begin{aligned}{}[P v \circ \psi ]_{BMO^\infty (V_{4 \rho L_*}^+)} \le C_{\Omega ,n} \Vert (P v)_E \Vert _{bmo(\mathbb {R}^n)} \le \frac{C_{\Omega ,n}}{\rho ^n} \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

Let $\zeta \in V_{12 \rho L_*} = \psi ^{-1}(U_{12 \rho L_*}(z_0))$ with $\zeta _n = 0$. Let $B_r(\zeta ) \subset V_{12 \rho L_*}$ and $x = \psi (\zeta )$. Since $F(B_r(\zeta ) \cap V_{12 \rho L_*}^+) \subset B_{L_*r}(x) \cap \Omega$, by considering the change of variables $y = \psi (\eta )$ in $U_{12 \rho L_*}(z_0)$, we can deduce that

$$\begin{aligned} \frac{1}{|B_r(\zeta )|} \int _{B_r(\zeta ) \cap V_{12 \rho L_*}^+} | P v_{\mathrm {odd}} \circ \psi (\eta ) | \, d\eta \le L_*^{n+1} [\nabla d \cdot v]_{b^\nu }. \end{aligned}$$

(9)

Recall the results concerning the odd extension of BMO functions in the half space, see [10,  Lemma 3.1], we have the estimate

$$\begin{aligned}{}[P v_{\mathrm {odd}} \circ \psi ]_{BMO^\infty (V_{4 \rho L_*})} \le \frac{C_{\Omega ,n}}{\rho ^n} \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

(10)

By considering (10) and the fact that $P v_{\mathrm {odd}} = (P v)_E$ in $\Omega$, the same argument in the third paragraph implies the BMO estimate for $P v_{\mathrm {odd}}$, i.e.,

$$\begin{aligned}{}[P v_{\mathrm {odd}}]_{BMO(\mathbf {R}^n)} \le \frac{C_{\Omega ,n}}{\rho ^n} \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

Combining the BMO estimates for $v_{\mathrm {even}}$ and $P v_{\mathrm {odd}}$, we have that

$$\begin{aligned}{}[\overline{v}]_{BMO(\mathbf {R}^n)} \le \frac{C_{\Omega ,n}}{\rho ^n} \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

Notice that $\nabla d \cdot \overline{v} = v_{\mathrm {odd}} \cdot \nabla d$ in $\mathbf {R}^n$. Let $x \in \Gamma$ and $r \le \frac{\rho }{L_*}$. If $B_r(x) \cap U_\rho (z_0) = \emptyset$, then $v_{\mathrm {odd}} = 0$ in $B_r(x)$. Suppose that $B_r(x) \cap U_\rho (z_0) \ne \emptyset$. Then we can find $x_0 \in B_r(x) \cap U_\rho (z_0) \cap \Gamma$. Let $\zeta _0 = \psi ^{-1}(x_0)$, we have that $\psi ^{-1}(B_r(x)) \subset B_{2 L_*r}(\zeta _0) \subset V_{12 \rho L_*}$. Hence,

$$\begin{aligned} r^{-n} \int _{B_r(x)} | v_{\mathrm {odd}} \cdot \nabla d | \, dy&\le \frac{2 L_*}{r^n} \int _{B_{2 L_*r}(\zeta _0) \cap V_{12 \rho L_*}^+} | ( v \cdot \nabla d ) \circ \psi | \, d\eta \\&\le \frac{2 L_*^2}{r^n} \int _{B_{2 L_*^2 r}(x_0) \cap \Omega } | \nabla d \cdot v | \, dy \le C_{\Omega ,n} [\nabla d \cdot v]_{b^\nu }. \end{aligned}$$

For $r > \frac{\rho }{L_*}$, we simply have that

$$\begin{aligned} r^{-n} \int _{B_r(x)} | v_{\mathrm {odd}} \cdot \nabla d | \, dy \le \frac{C_{\Omega ,n}}{\rho ^n} \Vert v_{\mathrm {odd}} \Vert _{L^1(\mathbf {R}^n)} \le \frac{C_{\Omega ,n}}{\rho ^n} \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

$\square$

2.3 Volume potentials

To construct mapping $v\mapsto q_1$ in Theorem 2, for some $\rho _*$ to be determined later in the next section, we localize v by using the partition of the unity $\{\varphi _j\}^m_{j=0}$ associated with the covering

$$\begin{aligned} \{ U_{\rho ,j} \}^m_{j=1} \cup \Omega ^{\rho /2} \end{aligned}$$

as in Sect. 2.1, where $\rho$ is always assumed to satisfy $\rho \le \rho _*/2$. Here and hereafter we always assumed that $\Omega$ is a bounded $C^3$ domain in $\mathbf {R}^n$.

Proposition 3

There exists a constant $C_\rho$, which depends on $\rho$ only, such that

$$\begin{aligned}{}[ \nabla q^1_1 ]_{BMO^\infty (\mathbf {R}^n)}&\le C_\rho \Vert v\Vert _{vBMO(\Omega )}, \\ \Vert \nabla q^1_1(x) \Vert _{L^\infty (\Gamma _{\rho /4}^{\mathbf {R}^n})}&\le C_\rho \Vert v\Vert _{vBMO(\Omega )} \end{aligned}$$

for $q^1_1= E*{\text {div}} \, (\varphi _0 v)$ and $v \in vBMO(\Omega )$. In particular,

$$\begin{aligned} \left[ \nabla q^1_1 \right] _{b^\nu (\Gamma )} \le C_\rho \Vert v \Vert _{vBMO(\Omega )} \end{aligned}$$

for $\nu < \rho /4$.

Proof

By the BMO-BMO estimate [7], we have the estimate

$$\begin{aligned} \left[ \nabla q^1_1 \right] _{BMO(\mathbf {R}^n)} \le C [\varphi _0 v]_{BMO(\mathbf {R}^n)}. \end{aligned}$$

Consider $x \in \Gamma _{\rho /4}^{\mathrm {R}^n}$. Since $\nabla q_1^1$ is harmonic in $\Gamma _{\rho /2}^{\mathrm {R}^n}$ and $B_{\frac{\rho }{4}}(x) \subset \Gamma _{\rho /2}^{\mathrm {R}^n}$, the mean value property for harmonic functions implies that

$$\begin{aligned} \nabla q_1^1 (x) = \frac{C_n}{\rho ^n} \int _{B_{\frac{\rho }{4}}(x)} \nabla q_1^1 (y) \, dy. \end{aligned}$$

By H$\ddot{\text {o}}$lder’s inequality, we can estimate $| \nabla q_1^1 (x) |$ by $\frac{C_n}{\rho ^{n/2}} \Vert \nabla q_1^1 \Vert _{L^2(\mathbf {R}^n)}$. Since the convolution with $\nabla ^2 E$ is bounded in $L^p$ for any $1<p<\infty$, see e.g. [14,  Theorem 5.2.7 and Theorem 5.2.10], an interpolation inequality (cf. [4,  Lemma 5]) implies that

$$\begin{aligned} \Vert \nabla q_1^1 \Vert _{L^2(\mathbf {R}^n)} \le C \Vert \varphi _0 v \Vert _{L^2(\mathbf {R}^n)} \le C \Vert \varphi _0 v \Vert _{L^1(\mathbf {R}^n)}^{\frac{1}{2}} [ \varphi _0 v ]_{BMO(\mathbf {R}^n)}^{\frac{1}{2}}. \end{aligned}$$

View $\varphi _0 v$ as the extension of $\varphi _0 v$ from $\Omega$ to $\mathbf {R}^n$. By the extension theorem for bmo functions [11,  Theorem 12], we estimate $[ \varphi _0 v ]_{BMO(\mathbf {R}^n)}$ by $C_\rho [ \varphi _0 v ]_{BMO^\infty (\Omega )}$. Since $vBMO(\Omega ) \subset L^1(\Omega )$, see [11,  Theorem 9], Proposition 1 implies that

$$\begin{aligned} | \nabla q^1_1 (x) | \le C_\rho \Vert v\Vert _{vBMO(\Omega )} \end{aligned}$$

for any $x \in \Gamma _{\rho /4}^{\mathrm {R}^n}$. $\square$

We next set $v_1 := \varphi _0 v$ and $v_2 := 1-v_1$. For each $\varphi _j v_2$ ($j=1,.,m$), we extend as in Proposition 2 to get $\overline{\varphi _j v_2}$ and set

$$\begin{aligned} \overline{v_2} = \sum ^m_{j=1} \overline{\varphi _j v_2}. \end{aligned}$$

Indeed, this extension is independent of the choice $\varphi _j$’s but we do not use this fact. We next set

$$\begin{aligned} \overline{v_2}^{\tan } := Q \, \overline{v_2} = \sum _{j=1}^m Q \, (\varphi _j v_2)_{\mathrm {even}}. \end{aligned}$$

For $1 \le j \le m$, $\varphi _j \in C_{\mathrm {c}}^\infty (U_{\rho ,j} \cap \overline{\Omega })$ implies that the even extension of $\varphi _j$ in $U_{\rho ,j}$ with respect to $\Gamma$ is H$\ddot{\text {o}}$lder continuous in the sense that $(\varphi _j)_{\mathrm {even}} \in C^{0,1}(U_{\rho ,j})$. Moreover, we have that $(\varphi _j)_{\mathrm {even}} \in C^{0,1}(\mathbf {R}^n)$ satisifes

$$\begin{aligned} \Vert (\varphi _j)_{\mathrm {even}} \Vert _{C^{0,1}(\mathbf {R}^n)} \le C_\rho \Vert (\varphi _j)_{\mathrm {even}} \Vert _{C^{0,1}(U_{\rho ,j})}. \end{aligned}$$

For simplicity of notations, we denote $Q \, (\varphi _j v_2)_{\mathrm {even}}$ by $w_j^{\mathrm {tan}}$ for every $1 \le j \le m$. Now, we are ready to construct the suitable potential corresponding to $\overline{v_2}^{\mathrm {tan}}$.

Proposition 4

There exists $\rho _*> 0$ such that if $\rho <\rho _*/2$, then for every $1 \le j \le m$, there exists a linear operator $v \longmapsto p_j^{\mathrm {tan}}$ from $vBMO(\Omega )$ to $L^\infty (\mathbf {R}^n)$ such that

$$\begin{aligned} - \Delta p_j^{\mathrm {tan}} = {\text {div}} w_j^{\mathrm {tan}} \; \; \text {in} \; \; U_{2 \rho ,j} \cap \Omega \end{aligned}$$

and that there exists a constant $C_\rho$, independent of v, such that

$$\begin{aligned}{}[\nabla p_j^{\mathrm {tan}}]_{BMO(\mathbf {R}^n)}&\le C_{\rho } \Vert v\Vert _{vBMO(\Omega )}, \\ \sup _{x\in \Gamma , r<\rho } \, \frac{1}{r^n} \int _{B_r(x)} |\nabla d \cdot \nabla p_j^{\mathrm {tan}}| \, dy&\le C_{\rho } \Vert v\Vert _{vBMO(\Omega )}. \end{aligned}$$

Having the estimate for the volume potential near the boundary regarding its tangential component, we are left to handle the contribution from $\overline{v}^{\mathrm {nor}}_2:=\overline{v}_2-\overline{v}^{\tan }_2$. We recall its decomposition

$$\begin{aligned} \overline{v}^{\mathrm {nor}}_2 = \sum ^m_{j=1} P \, ( \varphi _j v_2)_{\mathrm {odd}}. \end{aligned}$$

For simplicity of notations, we denote $P \, (\varphi _j v_2)_{\mathrm {odd}}$ by $w_j^{\mathrm {nor}}$ for every $1 \le j \le m$. With a similar idea of proof, we can establish the suitable potential corresponding to $\overline{v}^{\mathrm {nor}}_2$.

Proposition 5

There exists $\rho _*> 0$ such that if $\rho <\rho _*/2$, then for every $1 \le j \le m$, there exists a linear operator $v\longmapsto p_j^{\mathrm {nor}}$ from $vBMO(\Omega )$ to $L^\infty (\mathbf {R}^n)$ such that

$$\begin{aligned} - \Delta p_j^{\mathrm {nor}} = {\text {div}} w_j^{\mathrm {nor}} \; \; \text {in} \; \; U_{2 \rho ,j} \cap \Omega \end{aligned}$$

and that there exists a constant $C_{\rho }$, independent of v, such that

$$\begin{aligned}{}[\nabla p_j^{\mathrm {nor}}]_{BMO(\mathbf {R}^n)}&\le C_{\rho } \Vert v\Vert _{vBMO(\Omega )}, \\ \sup _{x\in \Gamma , r<\rho } \, \frac{1}{r^n} \int _{B_r(x)} |\nabla d \cdot \nabla p_j^{\mathrm {nor}}| \, dy&\le C_{\rho } \Vert v\Vert _{vBMO(\Omega )}. \end{aligned}$$

Once these two propositions are proved, we are able to prove Theorem 2.

Proof

(Theorem 2 admitting Proposition 4 and 5) Fix $1 \le j \le m$. Let us first consider the contribution from the tangential part. We take a cut-off function $\theta _j\in C^\infty _c(U_{2\rho ,j})$ such that $\theta _j=1$ on $U_{\rho ,j}$ and $0 \le \theta _j \le 1$. We next set

$$\begin{aligned} q^{\mathrm {tan}}_{1,j} := \theta _j p_j^{\mathrm {tan}} + E * \left( p_j^{\mathrm {tan}} \Delta \theta _j + 2\nabla \theta _j \cdot \nabla p_j^{\mathrm {tan}} \right) . \end{aligned}$$

By definition, Proposition 4 says that

$$\begin{aligned} - \Delta q^{\mathrm {tan}}_{1,j}&= - \Delta (\theta _j p_j^{\mathrm {tan}}) + p_j^{\mathrm {tan}} \Delta \theta _j + 2\nabla \theta _j \cdot \nabla p_j^{\mathrm {tan}} \\&= \theta _j {\text {div}} w_j^{\mathrm {tan}} = {\text {div}} w_j^{\mathrm {tan}} \end{aligned}$$

in $\Omega$ as ${\text {supp}}w_j^{\mathrm {tan}} \subset U_{\rho ,j}$. By interpolation as in the proof of Proposition 4, we observe that $\Vert p_j^{\mathrm {tan}}\Vert _{L^\infty (\mathbf {R}^n)}$, $\Vert \nabla p_j^{\mathrm {tan}}\Vert _{L^p(\mathbf {R}^n)}$ are controlled by $\Vert v\Vert _{BMO(\Omega )}$. Since $\nabla E$ is in $L^{p'}(B_R)$ for $p'<n/(n-1)$ where $R = {\text {diam}}\Omega + 4\rho$, it follows that

$$\begin{aligned} \sup _{\mathbf {R}^n} | \nabla E* ( p_j^{\mathrm {tan}} \Delta \theta _j + 2 \nabla \theta _j \cdot \nabla p_j^{\mathrm {tan}} ) | \le C_\rho \Vert v\Vert _{vBMO(\Omega )}. \end{aligned}$$

Thus, by Proposition 4, we conclude that the restriction of $q^{\mathrm {tan}}_{1,j}$ on $\Omega$, which is still denoted by $q^{\mathrm {tan}}_{1,j}$, fulfills

$$\begin{aligned} \Vert \nabla q^{\mathrm {tan}}_{1,j} \Vert _{vBMO(\Omega )} \le C_\rho \Vert v\Vert _{vBMO(\Omega )}. \end{aligned}$$

(11)

By Proposition 5, a similar argument yields an estimate of type (11) for

$$\begin{aligned} q^{\mathrm {nor}}_{1,j} := \theta _j p_j^{\mathrm {nor}} + E * (p_j^{\mathrm {nor}} \Delta \theta _j + 2\nabla \theta _j \cdot \nabla p_j^{\mathrm {nor}} ). \end{aligned}$$

Set

$$\begin{aligned} q^2_1 = \sum ^m_{j=1} q^{\mathrm {tan}}_{1,j}, \; q_1^3 = \sum _{j=1}^m q_{1,j}^{\mathrm {nor}}, \; q_1 = q_1^1 + q_1^2 + q_1^3. \end{aligned}$$

Observe that $q_1^2$ and $q^3_1$ satisfy the desired estimates in Theorem 2. Moreover, by construction we have that

$$\begin{aligned} - \Delta q_1&= - \Delta q^1_1 - \Delta q^2_1 - \Delta q^3_1 \\&= {\text {div}}v_1 + \sum _{j=1}^m {\text {div}} w_j^{\mathrm {tan}} + \sum ^m_{j=1} {\text {div}} w_j^{\mathrm {nor}} \\&= {\text {div}} (v_1 + v_2) = {\text {div}} v \end{aligned}$$

in $\Omega$. $\square$

3 Volume potentials based on normal coordinates

Our goal in this section is to prove Propositions 4 and 5. We write the Laplace operator by a normal coordinate system and construct a volume potential keeping the parity of functions with respect to the boundary. For this purpose, we adjust a perturbation method called a freezing coefficient method which is often used to construct a fundamental solution to an operator with variable coefficients.

3.1 A perturbation method keeping parity

We consider an elliptic operator of the form

$$\begin{aligned} L_0 = A-B, \quad A = -\Delta _\eta , \quad B = \sum _{1\le i,j \le n-1} \partial _{\eta _i} b_{ij}(\eta ) \partial _{\eta _j} \end{aligned}$$

in a cylinder $V_{4 \rho }$. We assume that

• (B1) (Regularity) $b_{ij} \in {\text {Lip}}(V_{4 \rho })$ ($1\le i,j \le n-1$),

• (B2) (Parity) $b_{ij}$ is even in $\eta _n$, i.e., $b_{ij}(\eta ', \eta _n)=b_{ij}(\eta ', - \eta _n)$ for $\eta \in V_{4 \rho }$,

• (B3) (Smallness) $b_{ij}(0)=0$ ($1\le i,j\le n-1$).

For $\rho > 0$, let $Y_\rho$ denote the space

$$\begin{aligned} \left\{ g \in BMO(\mathbf {R}^n) \cap L^2(\mathbf {R}^n) \bigm | {\text {supp}} g \subset V_\rho ,\ g(\eta ', \eta _n) = g(\eta ', -\eta _n)\ \text {for}\ \eta \in V_\rho \right\} , \end{aligned}$$

whereas $Z_\rho$ denotes the space

$$\begin{aligned} \left\{ f \in BMO(\mathbf {R}^n) \bigm | {\text {supp}} f \subset V_\rho ,\ f(\eta ', \eta _n) = -f(\eta ', -\eta _n)\ \text {for}\ \eta \in V_\rho \right\} . \end{aligned}$$

The oddness condition in $Z_\rho$ guarantees that

$$\begin{aligned} \frac{1}{r^n} \int _{B_r(\eta ',0)} f \, d\eta = 0 \end{aligned}$$

for any $r > 0$ and $\eta ' \in \mathbf {R}^{n-1}$, which implies that

$$\begin{aligned} \frac{1}{r^n} \int _{B_r(\eta ',0)} |f| \, d\eta \le [f]_{BMO(\mathbf {R}^n)} \end{aligned}$$

for any $r > 0$ and $\eta ' \in \mathbf {R}^{n-1}$. Hence f is $L^1$ in $\mathbf {R}^n$.

Lemma 3

Assume that (B1)–(B3). Then, there exists $\rho _*>0$ depending only on n and b such that the following property holds provided that $\rho \in (0,\rho _*)$. There exists a bounded linear operator $f \mapsto q_o$ from $Z_\rho$ to $L^\infty (\mathbf {R}^n)$ such that

• (i)
  $$\begin{aligned}&[ \nabla _\eta q_o ]_{BMO(\mathbf {R}^n)} \le C[f]_{BMO(\mathbf {R}^n)} \quad \text {for all}\quad f\in Z_\rho \end{aligned}$$

  with some C independent of f;

• (ii)
  $$\begin{aligned}&L_0 q_o = \partial _{\eta _n} f \quad \text {in}\quad V_{2 \rho }; \end{aligned}$$

• (iii) $q_o$ is even in $\mathbf {R}^n$ with respect to $\eta _n$, i.e. $q_o(\eta ',\eta _n)=q_o(\eta ',-\eta _n) \; \forall \, \eta \in \mathbf {R}^n$;

• (iv)
  $$\begin{aligned} \sup \left\{ \frac{1}{r^n} \int _{B_r(\eta ',0)} \left| \partial _{\eta _n} q_o \right| \, d\eta \biggm | 0< r < \infty ,\, \eta ' \in \mathbf {R}^{n-1} \right\} \le C[f]_{BMO(\mathbf {R}^n)}. \end{aligned}$$

Proof

By (B3) and (B1), we observe that

$$\begin{aligned} \varlimsup _{\rho \downarrow 0} \, \Vert b_{ij}\Vert _{C^\gamma (V_{4 \rho })} \bigm / \rho ^{1-\gamma } < \infty \end{aligned}$$

for any $\gamma \in (0,1)$ and $1 \le i,j \le n-1$. Indeed, for $1 \le i,j \le n-1$, (B1) and (B3) imply that

$$\begin{aligned} \Vert b_{ij}\Vert _{L^\infty (V_{4 \rho })}&\le 8 L\rho , \\ [b_{ij}]_{C^\gamma (V_{4 \rho })}&:= \sup \left\{ | b_{ij} (\eta ) - b_{ij}(\zeta ) | \bigm / | \eta - \zeta |^\gamma \Bigm | \eta ,\zeta \in V_{4 \rho } \right\} \\&\le L(16 \rho )^{1-\gamma }, \end{aligned}$$

where L is the maximum of Lipschitz bound for $b_{ij}$ for all $1\le i,j\le n-1$. We next take a cut-off function. We take $\theta \in C^\infty _c(V_4)$ such that $\theta =1$ on $V_2$ and $0\le \theta \le 1$ in $V_4$, we may assume $\theta$ is radial so that $\theta$ is even in $\eta _n$. We rescale $\theta$ by setting

$$\begin{aligned} \theta _\rho (\eta ) = \theta (\eta /\rho ) \end{aligned}$$

so that $\theta _\rho =1$ on $V_{2 \rho }$. Since $\Vert \nabla \theta _\rho \Vert _\infty \rho$ is bounded as $\rho \rightarrow 0$, we see that

$$\begin{aligned} \varlimsup _{\rho \downarrow 0} \, [\theta _\rho ]_{C^\gamma (V_{4 \rho })} \rho ^\gamma < \infty . \end{aligned}$$

Hence, the estimate

$$\begin{aligned}{}[\theta _\rho b_{ij}]_{C^\gamma (V_{4 \rho })} \le [\theta _\rho ]_{C^\gamma (V_{4 \rho })} \Vert b_{ij}\Vert _{L^\infty (V_{4 \rho })} + [b_{ij}]_{C^\gamma (V_{4 \rho })} \Vert \theta _\rho \Vert _{L^\infty (V_{4 \rho })} \end{aligned}$$

implies that

$$\begin{aligned} \varlimsup _{\rho \downarrow 0} \, \Vert \theta _\rho b_{ij} \Vert _{C^\gamma (V_{4 \rho })} \bigm / \rho ^{1-\gamma } < \infty . \end{aligned}$$

We then set

$$\begin{aligned} L_1 = A - B_1, \quad B_1 = \sum _{1\le i,j \le n-1} \partial _{\eta _i} b^\rho _{ij} \partial _{\eta _j}, \quad b^\rho _{ij} = b_{ij} \theta _\rho . \end{aligned}$$

For $1 \le i,j \le n-1$, notice that $b^\rho _{ij}$ satisfies the same property of $b_{ij}$ in (B1)–(B3). Moreover,

$$\begin{aligned} {\text {supp}} \, b^\rho _{ij} \subset V_{4 \rho } \quad \text {and}\quad \left\| b^\rho _{ij} \right\| _{C^\gamma (V_{4 \rho })} \le c_b \rho ^{1-\gamma },\ \rho > 0 \end{aligned}$$

with some $c_b$ independent of $\rho$. Since ${\text {supp}} \, b^\rho _{ij} \subset V_{4 \rho }$, we actually have that $b_{ij}^\rho \in C^\gamma (\mathbf {R}^n)$ together with the estimate

$$\begin{aligned} \Vert b_{ij}^\rho \Vert _{C^\gamma (\mathbf {R}^n)} \le \Vert b_{ij}^\rho \Vert _{C^\gamma (V_{4 \rho })}. \end{aligned}$$

For a given $f \in Z_\rho$, we define $q_o$ by

$$\begin{aligned} q_o := \sum ^\infty _{k=0} A^{-1} \left( B_1 A^{-1} \right) ^k \partial _{\eta _n} f, \end{aligned}$$

where formally for a function h we mean $A^{-1} h$ by $E*h$. The parity condition (iii) is clear once $q_o$ is well defined as a function. Since

$$\begin{aligned} L_1 q_o = \sum ^\infty _{k=0} \left( B_1 A^{-1} \right) ^k \partial _{\eta _n} f - \sum ^\infty _{k=1} \left( B_1 A^{-1} \right) ^k \partial _{\eta _n} f = \partial _{\eta _n}f \end{aligned}$$

in $\mathbf {R}^n$, the property (ii) then follows since $L_1=L_0$ in $V_{2 \rho }$.

It remains to prove the convergence of $q_o$ as well as (i). For this purpose, we reinterpret $q_o$ in a different way. We rewrite

$$\begin{aligned} B _1 = {\text {div}}' \cdot \nabla '_B \quad \text {with}\quad \nabla '_B = \left( \sum ^{n-1}_{j=1} b^\rho _{ij} \partial _{\eta _j} \right) _{1\le i \le n-1} \end{aligned}$$

and observe that

$$\begin{aligned} q_o= & {} \sum ^\infty _{k=0} A^{-1} {\text {div}}' \cdot G^k \cdot \nabla '_B A^{-1} \partial _{\eta _n} f + A^{-1} \partial _{\eta _n} f, \\ G:= & {} \nabla '_B A^{-1} {\text {div}}'. \end{aligned}$$

Denote

$$\begin{aligned} b^\rho := \left( b_{ij}^\rho \right) _{1 \le i,j \le n-1}. \end{aligned}$$

Since $\partial _{\eta _\alpha } A^{-1}\partial _{\eta _\beta }$ is bounded in BMO [7] and also in $L^p$ ($1<p<\infty$) for all $\alpha , \beta =1,\ldots ,n$, see e.g. [14,  Theorem 5.2.7 and Theorem 5.2.10], by a multiplication theorem we can deduce the estimates

$$\begin{aligned} \Vert Gh \Vert _{L^p(\mathbf {R}^n)}&\le C_p \Vert b^\rho \Vert _{L^\infty (\mathbf {R}^n)} \Vert h \Vert _{L^p(\mathbf {R}^n)}, \end{aligned}$$

(12)

$$\begin{aligned} _{BMO(\mathbf {R}^n)}&\le C'_\infty \Vert b^\rho \Vert _{C^\gamma (\mathbf {R}^n)} \left( [h]_{BMO(\mathbf {R}^n)} + \Vert h \Vert _{L^1(\mathbf {R}^n)} \right) \end{aligned}$$

(13)

provided that ${\text {supp}} \, h \subset V_{4 \rho }$ and $\rho <1$. Here $C_p$ and $C'_\infty$ are independent of $\rho$ and h. Similar estimate holds for $\nabla '_B A^{-1}\partial _{\eta _n}$. Since $\Vert f\Vert _{L^1(\mathbf {R}^n)} \le C_\rho [f]_{BMO(\mathbf {R}^n)}$ for $f\in Z_\rho$, by an interpolation (cf. [4,  Lemma 5]) we see that the $L^p$ norm of f is also controlled, i.e., $\Vert f\Vert _{L^p(\mathbf {R}^n)} \le C_\rho [f]_{BMO(\mathbf {R}^n)}$ for any $1\le p<\infty$. For h supported within $V_{4 \rho }$ with $\rho <1$, since $A^{-1}$ is of a convolution type we can see that $A^{-1}{\text {div}}' h$ and $A^{-1} \partial _{\eta _n} h$ are indeed ${\text {div}}' A^{-1} h$ and $\partial _{\eta _n} A^{-1} h$, respectively. Since the convolution kernel of the operator $\partial _{\eta _i} A^{-1}$ is dominated by a constant multiple of $\left| \eta \right| ^{-(n-1)}$ for any $1 \le i \le n$, H$\ddot{\text {o}}$lder’s inequality with any $p>n$ implies the estimate

$$\begin{aligned} \Vert \partial _{\eta _i} A^{-1} h \Vert _{L^\infty (\mathbf {R}^n)} \le K \Vert h \Vert _{L^p(\mathbf {R}^n)} \end{aligned}$$

with some constant K independent of $\rho <1$. Hence, we deduce that

$$\begin{aligned} \Vert q_o \Vert _{L^\infty (\mathbf {R}^n)}&\le K \left( \left\| \sum ^\infty _{k=0} G^k \nabla '_B A^{-1}\partial _{\eta _n} f \right\| _{L^p(\mathbf {R}^n)} + \Vert f\Vert _{L^p(\mathbf {R}^n)} \right) \\&\le K \left( \sum ^\infty _{k=0} C^{k+1}_p \Vert b^\rho \Vert ^{k+1}_{L^\infty (\mathbf {R}^n)} \Vert f\Vert _{L^p(\mathbf {R}^n)} + \Vert f\Vert _{L^p(\mathbf {R}^n)} \right) , \quad p>n. \end{aligned}$$

If $\rho$ is taken small so that

$$\begin{aligned} \sum ^\infty _{k=0} (C_p \cdot 8 L \rho )^{k+1} < \infty , \end{aligned}$$

then $q_o$ converges uniformly in $\mathbf {R}^n$ and $\Vert q_o \Vert _{L^\infty (\mathbf {R}^n)} \le C_\rho [f]_{BMO(\mathbf {R}^n)}$ with some $C_\rho$ independent of f.

Set

$$\begin{aligned} \Vert h \Vert _{BMOL^p(\mathbf {R}^n)} := [h]_{BMO(\mathbf {R}^n)} + \Vert h \Vert _{L^p(\mathbf {R}^n)}. \end{aligned}$$

By estimates (12) and (13), we observe that

$$\begin{aligned} \Vert G h \Vert _{BMOL^p(\mathbf {R}^n)} \le C_*\Vert b^\rho \Vert _{C^\gamma (\mathbf {R}^n)} \Vert h \Vert _{BMOL^p(\mathbf {R}^n)}, \quad 1<p<\infty , \end{aligned}$$

where $C_*= C_p + C'_\infty \cdot C_n$ with $C_n$ independent of $\rho$ and h. We next estimate $\nabla q_o$. By the similar estimate for $\nabla '_B A^{-1}{\text {div}}'$ and $\nabla '_B A^{-1} \partial _{\eta _n}$, we have that

$$\begin{aligned} \Vert \nabla q_o \Vert _{BMOL^p(\mathbf {R}^n)} \le \left( \sum ^\infty _{k=0} C^{k+1}_*\Vert b^\rho \Vert ^{k+1}_{C^\gamma (\mathbf {R}^n)} + C_* \Vert b^\rho \Vert _{C^\gamma (\mathbf {R}^n)} \right) \Vert f\Vert _{BMOL^p(\mathbf {R}^n)}. \end{aligned}$$

We fix $p>n$ and take $\rho < \frac{1}{8 L C_p}$ sufficiently small so that

$$\begin{aligned} \sum ^\infty _{k=0} \left( C_* \cdot c_b \rho ^{1-\gamma } \right) ^{k+1} < \infty . \end{aligned}$$

Then we get our desired estimate

$$\begin{aligned} \Vert \nabla q_o \Vert _{BMOL^p (\mathbf {R}^n)} \le C_\rho \Vert f\Vert _{BMOL^p(\mathbf {R}^n)} \le C_\rho [f]_{BMO(\mathbf {R}^n)} \end{aligned}$$

for $f \in Z_\rho$. This completes the proof of (i).

Since $\partial _{\eta _n} q_o$ is odd in $\eta _n$ so that

$$\begin{aligned} \frac{1}{r^n} \int _{B_r(\eta ',0)} \partial _{\eta _n} q_o \, d\eta = 0 \end{aligned}$$

for any $\eta ' \in \mathbf {R}^{n-1}$, the left-hand side of (iv) is estimated by a constant multiple of $[\partial _{\eta _n} q_o]_{BMO(\mathbf {R}^n)}$, which is estimated by a constant multiple of $[f]_{BMO(\mathbf {R}^n)}$. The proof of (iv) is now complete. $\square$

Similarly, we are able to establish the following which corresponds to a version of Lemma 3 for the space $Y_\rho$.

Lemma 4

Assume that (B1)–(B3). Then, there exists $\rho _*>0$ depending only on n and b such that the following property holds provided that $\rho \in (0,\rho _*)$. For each $1 \le i \le n-1$, there exists a bounded linear operator $g \mapsto q_{e,i}$ from $Y_\rho$ to $L^\infty (\mathbf {R}^n)$ such that

• (i)
  $$\begin{aligned}&[ \nabla q_{e,i} ]_{BMO(\mathbf {R}^n)} \le C \Vert g\Vert _{BMOL^2(\mathbf {R}^n)} \quad \text {for all} \quad g \in Y_\rho \end{aligned}$$

  with some C independent of f;

• (ii)
  $$\begin{aligned}&L_0 q_{e,i} = \partial _{\eta _i} g \quad \text {in} \quad V_{2 \rho }; \end{aligned}$$

• (iii) $q_{e,i}$ is even in $\mathbf {R}^n$ with respect to $\eta _n$, i.e. $q_{e,i}(\eta ',\eta _n)=q_{e,i}(\eta ',-\eta _n) \; \forall \, \eta \in \mathbf {R}^n$;

• (iv)
  $$\begin{aligned} \sup \left\{ \frac{1}{r^n} \int _{B_r(\eta ',0)} \left| \partial _{\eta _n} q_{e,i} \right| \, d\eta \biggm | 0< r < \infty ,\, \eta ' \in \mathbf {R}^{n-1} \right\} \le C \Vert g\Vert _{BMOL^2(\mathbf {R}^n)}. \end{aligned}$$

Proof

Fix $1 \le i \le n-1$. Since g is even in $\mathbf {R}^n$ with respect to $\eta _n$, $\partial _{\eta _i} g$ is also even in $\mathbf {R}^n$ with respect to $\eta _n$. This means that $\partial _{\eta _i} g$ has the same parity with $\partial _{\eta _n} f$ in Lemma 3. By considering

$$\begin{aligned} q_{e,i} := \sum _{k=0}^\infty A^{-1} (B_1 A^{-1})^k \partial _{\eta _i} g, \end{aligned}$$

exactly the same arguments of the proof of Lemma 3 finish the rest of the work. $\square$

We take $\rho _*$ in Lemmas 3 and 4 to be

$$\begin{aligned} \rho _*:= \mathrm {min} \, \left\{ \rho _{0,*}, \; \frac{1}{8 L C_p}, \; \bigg ( \frac{1}{C_*\cdot c_b} \bigg )^{\frac{1}{1-\gamma }} \right\} . \end{aligned}$$

3.2 Laplacian in a normal coordinate system

Take $z_0 \in \Gamma$. Let us recall the normal coordinate system (6) introduced in Sect. 2.1, i.e.,

$$\begin{aligned} \left\{ \begin{array}{lcl} x' &{}=&{} \eta ' + \eta _n \nabla ' d ( \eta ', h_{z_0}(\eta ') ); \\ x_n &{}=&{} h_{z_0}(\eta ') + \eta _n \partial _{\eta _n} d ( \eta ', h_{z_0}(\eta ') ) \end{array} \right. \end{aligned}$$

in $U_{\rho _0}(z_0)$ with $\nabla ' h_{z_0}(0')=0$, $h_{z_0}(0')=0$ up to translation and rotation such that $z_0=0$ and

$$\begin{aligned} -\mathbf {n} \left( \eta ', h_{z_0}(\eta ') \right) = \left( -\nabla ' h_{z_0}(\eta '), 1 \right) \Bigm / \left( 1 + \left| \nabla '_{z_0} h( \eta ' )\right| ^2 \right) ^{1/2}, \quad \eta ' \in B_{\rho _0}. \end{aligned}$$

Since $\Gamma$ is $C^3$, the mapping $x=\psi (\eta ) \in C^2(V_{\rho _0})$ in $U_{\rho _0}(z_0)$, it is a local $C^2$-diffeomorphism. Moreover, its Jacobi matrix $D\psi$ is the identity at 0, i.e.,

$$\begin{aligned} \nabla \psi (0) = I = \nabla \psi ^{-1} (0). \end{aligned}$$

A direct calculation shows that in $U_{\rho _0}(z_0) \cap \Omega$,

$$\begin{aligned} -\Delta _x&= -\Delta _{\eta } - \left\{ \sum _{\begin{array}{c} 1\le i,j \le n-1 \\ i\ne j \end{array}} \gamma _{ij} \partial _{\eta _i} \partial _{\eta _j} + \sum ^{n-1}_{j=1} (\gamma _{jj}-1) \partial ^2_{\eta _j} \right\} - \sum _{1\le i,j \le n} \frac{\partial ^2 \eta _j}{\partial x^2_i} \partial _{\eta _j}, \\ \gamma _{ij}&= \sum ^n_{k=1} \frac{\partial \eta _j}{\partial x_k} \frac{\partial \eta _i}{\partial x_k}. \end{aligned}$$

Note that $\gamma _{jj}(0)=1$ while $\gamma _{ij}(0)=0$ if $i \ne j$. Changing order of multiplication and differentiation, we conclude that

$$\begin{aligned} -\Delta _x&= {\tilde{L}}_0 + {\tilde{M}}, \\ {\tilde{L}}_0&:= A - {\tilde{B}},\quad A := -\Delta _\eta ,\quad {\tilde{B}} := \sum _{1\le i,j \le n-1} \partial _{\eta _i} {\tilde{b}}_{ij} (\eta ) \partial _{\eta _j}, \\ {\tilde{M}}&:= \sum ^n_{j=1} {\tilde{c}}_j (\eta ) \partial _{\eta _j} \end{aligned}$$

with ${\tilde{b}}_{ij}=\gamma _{ij}-\delta _{ij}$, ${\tilde{c}}_j=-\sum ^n_{i=1} \frac{\partial ^2 \eta _j}{\partial x^2_i} + \sum ^n_{i=1} \partial _{\eta _i} \gamma _{ij}$. Note that if $\Gamma =\partial \Omega$ is $C^3$, ${\tilde{b}}_{ij} \in C^1(V_{\rho _0})$ and $\tilde{c_j} \in C(V_{\rho _0})$. We restrict ${\tilde{b}}_{ij}$, ${\tilde{c}}_j$ in $V_{\rho _0} \cap \mathbf {R}_+^n$ and extend to $V_{\rho _0}$ so that the extended function $b_{ij}$, $c_j$’s are even in $V_{\rho _0}$ with respect to $\eta _n$, i.e., we set $b_{ij} = E_{\mathrm {even}} \; r_{V_{\rho _0} \cap \mathbf {R}_+^n} \, \tilde{b_{ij}}$ and $c_j = E_{\mathrm {even}} \; r_{V_{\rho _0} \cap \mathbf {R}_+^n} \, \tilde{c_j}$. By this extension, $b_{ij}$ may not be in $C^1$ but still Lipschitz and $c_j\in C (V_{\rho _0})$. We set

$$\begin{aligned} B&:= \sum _{1\le i,j \le n-1} \partial _{\eta _i} b_{ij} (\eta ) \partial _{\eta _j}, \\ M&:= \sum ^n_{j=1} c_j (\eta ) \partial _{\eta _j} \end{aligned}$$

and

$$\begin{aligned} L := L_0 + M,\quad L_0 = A - B. \end{aligned}$$

The operator L may not agree with $-\Delta _x$ outside $U_{\rho _0}(z_0) \cap \Omega$. We summarize what we observe so far.

Proposition 6

Let $\Gamma =\partial \Omega$ be $C^3$ and $\rho _0$ be chosen as in Sect. 2.1. For $z_0 \in \Gamma$, $L_0$ satisfies (B1)–(B3). Moreover, $-\Delta _x=L$ in $U_{\rho _0}(z_0) \cap \Omega$ and the coefficient of M is in $C(V_{\rho _0})$.

Although we do not use the explicit formula of $\Delta$ in normal coordinates, we give it for $n=2$ when we take the arc length parameter s to represent $\Gamma$. The coordinate transform is of the form

$$\begin{aligned} x_1&= \phi _1(x) + r\phi '_2 (s) \\ x_2&= \phi _2(x) - r\phi '_1 (s) \end{aligned}$$

with $\phi '^2_1 + \phi '^2_2=1$ and $r=d(x)$. A direct calculation yields

$$\begin{aligned} -\Delta _x = -\Delta _{s,r} -\partial _s \left( \frac{1}{J^2}-1 \right) \partial _s - \frac{\partial _s J}{J^3} \partial _s - \frac{1}{r} \left( 1-\frac{1}{J} \right) \partial _r, \end{aligned}$$

where $J=1+r\kappa$ and $\kappa$ is the curvature. We see that that the even extension of coefficient does not agree with $-\Delta _x$ outside $\Omega$.

3.3 bmo invariant under local $C^1$-diffeomorphism

Before we give the proofs to Propositions 4 and 5, we shall first establish the fact that the bmo estimate of a compactly supported function is preserved under a local $C^1$-diffeomorphism. Let $V,U \subset \mathbf {R}^n$ be two domains, we consider a local $C^1$-diffeomorphism $\psi : V \mapsto U$. Suppose that

$$\begin{aligned} \Vert \nabla _\eta \psi \Vert _{L^\infty (V)} + \Vert \nabla _x \psi ^{-1} \Vert _{L^\infty (U)} < \infty . \end{aligned}$$

For simplicity of notations, from now on we denote $h_\psi := h \circ \psi$ for a scalar function or a vector field h that is defined in U. If h is a vector field, we further denote $h_{\psi ,i} := h_i \circ \psi$ for $1 \le i \le n$. Vice versa, if h is defined in V, then $h_{\psi ^{-1}}$ and $h_{\psi ^{-1},i}$ for $1 \le i \le n$ would mean compositions with the coordinate change $\eta =\psi ^{-1}(x)$.

Let $\rho >0$. Assume that there exist two bounded subdomains $V_\rho \subset V, U_\rho \subset U$ such that $\psi :V_\rho \mapsto U_\rho$ is also a local $C^1$-diffeomorphism. Set

$$\begin{aligned} K_*:= \mathrm {max} \left\{ 1, \Vert \nabla _\eta \psi \Vert _{L^\infty (V)} + \Vert \nabla _x \psi ^{-1} \Vert _{L^\infty (U)} \right\} . \end{aligned}$$

We assume further that there exists a constant $c_0$ such that for some $\eta _0 \in V_\rho$,

$$\begin{aligned} V_\rho \subset B_{c_0 \rho }(\eta _0) \subset B_{K_*(c_0+3) \rho }(\eta _0) \subset V, \; \; U_\rho \subset B_{c_0 \rho }(x_0) \subset B_{K_*(c_0+3) \rho }(x_0) \subset U \end{aligned}$$

where $x_0 = \psi (\eta _0)$.

Proposition 7

Let $f \in bmo(\mathbf {R}^n)$ with ${\text {supp}} f \subset V_\rho$, then $f_{\psi ^{-1}} \in bmo(\mathbf {R}^n)$ satisfies

$$\begin{aligned} \Vert f_{\psi ^{-1}} \Vert _{bmo(\mathbf {R}^n)} \le C_\rho \Vert f \Vert _{bmo(\mathbf {R}^n)}. \end{aligned}$$

Proof

Since ${\text {supp}} f_{\psi ^{-1}} \subset U_\rho$, we can treat $f_{\psi ^{-1}}$ as a function in $\mathbf {R}^n$ with value zero outside $U_\rho$. The compactness of $V_\rho$ in $\mathbf {R}^n$ implies that $\Vert f \Vert _{bmo(\mathbf {R}^n)} = \Vert f \Vert _{BMOL^1(\mathbf {R}^n)}$. Thus, the $L^1$ estimate

$$\begin{aligned} \Vert f_{\psi ^{-1}} \Vert _{L^1(\mathbf {R}^n)} \le C \Vert f \Vert _{L^1(\mathbf {R}^n)} \end{aligned}$$

is obvious. Since $\psi \in C^1(V_\rho )$, an equivalent definition of the BMO-seminorm (cf. [14,  Proposition 3.1.2]) implies that

$$\begin{aligned}{}[f_{\psi ^{-1}}]_{BMO^\infty (B_{( c_0+1) \rho }(x_0))} \le \Vert \nabla _x \psi ^{-1} \Vert _{L^\infty (U)}^n \cdot \Vert \nabla _\eta \psi \Vert _{L^\infty (V)} \cdot [f]_{BMO(\mathbf {R}^n)}. \end{aligned}$$

As $U_\rho \subset B_{c_0 \rho }(x_0)$, by the extension theorem of bmo functions [11,  Theorem 12], we obtain that

$$\begin{aligned} \Vert f_{\psi ^{-1}} \Vert _{bmo(\mathbf {R}^n)} \le C_\rho \Vert f_{\psi ^{-1}} \Vert _{bmo_\infty ^\infty (B_{(c_0+1) \rho }(x_0))} \le C_\rho \Vert f \Vert _{bmo(\mathbf {R}^n)}. \end{aligned}$$

$\square$

Similarly, if $g \in bmo(\mathbf {R}^n)$ with ${\text {supp}} g \subset U_\rho$, then we have that $g_\psi \in bmo(\mathbf {R}^n)$ satisfying

$$\begin{aligned} \Vert g_\psi \Vert _{bmo(\mathbf {R}^n)} \le C_\rho \Vert g \Vert _{bmo(\mathbf {R}^n)}. \end{aligned}$$

Even if we are considering vector fields instead of scalar functions, similar results hold.

Proposition 8

Let $\nabla _\eta f \in bmo(\mathbf {R}^n)$ with ${\text {supp}} \nabla _\eta f \subset V_\rho$, then $\nabla _x f_{\psi ^{-1}} \in bmo(\mathbf {R}^n)$ satisfying

$$\begin{aligned} \Vert \nabla _x f_{\psi ^{-1}} \Vert _{bmo(\mathbf {R}^n)} \le C_\rho \Vert \nabla _\eta f \Vert _{bmo(\mathbf {R}^n)}. \end{aligned}$$

Proof

Since $\nabla _\eta f$ is compactly supported, the $L^1$ estimate

$$\begin{aligned} \Vert \nabla _x f_{\psi ^{-1}} \Vert _{L^1(\mathbf {R}^n)} \le C \Vert \nabla _\eta f \Vert _{L^1(\mathbf {R}^n)} \end{aligned}$$

is obvious. Pick a cut-off function $\theta _{*,\rho } \in C_c^\infty (B_{K_*(c_0+3) \rho }(\eta _0))$ such that $\theta _{*,\rho } = 1$ in $B_{K_*(c_0+2) \rho }(\eta _0)$. Consider $B_r(x) \subset B_{(c_0+1) \rho }(x_0)$ with $r < \rho$. Let $\eta = \psi ^{-1}(x)$. Since $\psi ^{-1}(B_r(x)) \subset B_{K_*(c_0+2) \rho }(\eta _0)$, we have that

$$\begin{aligned} \frac{1}{r^n} \int _{B_r(x)} | \partial _{x_i} f_{\psi ^{-1}} - c | \, dy&\le \frac{K_*}{r^n} \int _{\psi ^{-1}(B_r(x))} \bigg | \sum _{1 \le l \le n} \theta _{*,\rho } \bigg ( \frac{\partial \eta _l}{\partial x_i} \bigg )_\psi \partial _{\eta _l} f - c \bigg | \, d\eta \end{aligned}$$

for any $c \in \mathbf {R}^n$, $1 \le i \le n$. By considering an equivalent definition of the BMO-seminorm, see e.g. [14,  Proposition 3.1.2], we deduce that

$$\begin{aligned}{}[\nabla _x f_{\psi ^{-1}}]_{BMO^\infty (B_{(c_0+1) \rho } (x_0))}&\le K_*^{n+1} \bigg [ \sum _{1 \le i,l \le n} \theta _{*,\rho } \bigg ( \frac{\partial \eta _l}{\partial x_i} \bigg )_\psi \partial _{\eta _l} f \bigg ]_{BMO(\mathbf {R}^n)} \\&\le C_\rho \Vert \nabla _\eta f \Vert _{bmo(\mathbf {R}^n)}. \end{aligned}$$

As $U_\rho \subset B_{c_0 \rho }(x_0)$, by the extension theorem of bmo functions [11,  Theorem 12], we obtain that

$$\begin{aligned} \Vert \nabla _x f_{\psi ^{-1}} \Vert _{bmo(\mathbf {R}^n)} \le C_\rho \Vert \nabla _x f_{\psi ^{-1}} \Vert _{bmo_\infty ^\infty (B_{(c_0+1) \rho }(x_0))} \le C_\rho \Vert \nabla _\eta f \Vert _{bmo(\mathbf {R}^n)}. \end{aligned}$$

$\square$

If $\nabla _x g \in bmo(\mathbf {R}^n)$ with ${\text {supp}} \nabla _x g \subset U_\rho$, the same proof of Proposition 8 shows that $\nabla _\eta g_\psi \in bmo(\mathbf {R}^n)$ satisfying

$$\begin{aligned} \Vert \nabla _\eta g_\psi \Vert _{bmo(\mathbf {R}^n)} \le C_\rho \Vert \nabla _x g \Vert _{bmo(\mathbf {R}^n)}. \end{aligned}$$

3.4 Volume potential for tangential component

Let $\rho \in (0,\rho _*/2)$ and fix $1 \le j \le m$. Since $\varphi _j v_2 \in vBMO(\Omega )$ with ${\text {supp}} \,\varphi _j v_2 \subset U_{\rho ,j} \cap \overline{\Omega }$, Proposition 2 implies that $(\varphi _j v_2)_{\mathrm {even}} \in BMOL^1(\mathbf {R}^n)$. By the product estimate for bmo functions [11,  Theorem 13], we see that $w_j^{\mathrm {tan}} = Q (\varphi _j v_2)_{\mathrm {even}} \in BMOL^1(\mathbf {R}^n)$ with ${\text {supp}} w_j^{\mathrm {tan}} \subset U_{\rho ,j}$. For simplicity of notations, we set $v_{2,j} := (\varphi _j v_2)_{\mathrm {even}}$.

Let $\psi : V_{4 \rho } \mapsto U_{4 \rho ,j}$ be the normal coordinate change defined by (6) in Sect. 2.1. Since $\rho < \rho _*/2$, we have that

$$\begin{aligned} V_{4 \rho } \subset B_{12 \rho }(0) \subset B_{24 L_*\rho }(0) \subset V_{\rho _0}, \; \; U_{4 \rho ,j} \subset B_{12 \rho }(z_j) \subset B_{24 L_*\rho }(z_j) \subset U_{\rho _0,j}. \end{aligned}$$

By Propositions 7 and 8, we see that $\psi$, in this case, is a local $C^2$-diffeomorphism that preserves bmo estimates for functions or vector fields compactly supported in $V_{4 \rho }$. As a result, $(v_{2,j})_\psi \in BMOL^1(\mathbf {R}^n)$ satisfies the estimate

$$\begin{aligned} \Vert (v_{2,j})_\psi \Vert _{BMOL^1(\mathbf {R}^n)} \le C_\rho \Vert v_{2,j} \Vert _{BMOL^1(\mathbf {R}^n)}. \end{aligned}$$

Note that similar conclusions hold if we consider $\psi ^{-1}: U_{4 \rho ,j} \mapsto V_{4 \rho }$ instead.

Proof

(Proposition 4) For $1 \le i \le n$ and $1 \le k \le n-1$, we define

$$\begin{aligned} \bigg (\frac{\partial \eta _k}{\partial x_i} \bigg )_*:= E_{\mathrm {even}} \; r_{V_{4 \rho } \cap \mathbf {R}_{+}^n} \; \bigg ( \frac{\partial \eta _k}{\partial x_i} \bigg )_\psi \, \; \text { and } \, \; g_{i,k} := \bigg ( \frac{\partial \eta _k}{\partial x_i} \bigg )_*\cdot (v_{2,j})_{\psi ,i}. \end{aligned}$$

We consider

$$\begin{aligned} ( {\text {div}}_x w_j^{\mathrm {tan}} )_{\psi ,*}:= & {} \underset{1 \le k \le n-1}{\sum _{1 \le i \le n,}} \left\{ \partial _{\eta _k} g_{i,k} - \partial _{\eta _k} \bigg ( \frac{\partial \eta _k}{\partial x_i} \bigg )_\psi \cdot (v_{2,j})_{\psi ,i} \right\} \\&\quad - \underset{1 \le k \le n-1}{\sum _{1 \le i \le n,}} \bigg ( \frac{\partial \eta _k}{\partial x_i} \bigg )_\psi \cdot \left( \sum _{1 \le l \le n} (v_{2,j})_{\psi ,l} \cdot \bigg ( \frac{\partial \eta _n}{\partial x_l} \bigg )_\psi \right) \cdot \frac{\partial ^2 x_i}{\partial \eta _k \partial \eta _n} \end{aligned}$$

in $V_{4 \rho } = \psi ^{-1}(U_{4 \rho ,j})$. Let $L = L_0 + M$ be the operator in Proposition 6 and $L_0^{-1}$ be the operator in Lemma 4. Let $1 \le i \le n$ and $1 \le k \le n-1$. We set

$$\begin{aligned} q_{j,1,\psi }^{i,k} := - \theta _\rho L_0^{-1} \partial _{\eta _k} g_{i,k} \end{aligned}$$

where $\theta _\rho$ is the cut-off function defined in the proof of Lemma 3. There exists $\overline{( \frac{\partial \eta _k}{\partial x_i} )_*} \in C^{0,1}(\mathbf {R}^n)$, see e.g. [11,  Theorem 13], such that the restriction of $\overline{( \frac{\partial \eta _k}{\partial x_i} )_*}$ in $V_{4 \rho }$ equals $( \frac{\partial \eta _k}{\partial x_i} )_*$ and $\Vert \overline{( \frac{\partial \eta _k}{\partial x_i} )_*} \Vert _{C^{0,1}(\mathbf {R}^n)} \le \Vert ( \frac{\partial \eta _k}{\partial x_i} )_*\Vert _{C^{0,1}(V_{4 \rho })}$. By viewing $g_{i,k}$ as $\overline{( \frac{\partial \eta _k}{\partial x_i} )_*} \cdot (v_{2,j})_{\psi ,i}$, we see that $g_{i,k} \in BMOL^1(\mathbf {R}^n)$. Hence, $q_{j,1,\psi }^{i,k} \in L^\infty (\mathbf {R}^n)$ is well-defined which satisfies all conditions in Lemma 4. Let $f_{j,1,\psi }^{i,k} := M \theta _\rho L_0^{-1} \partial _{\eta _k} g_{i,k}$. We can define

$$\begin{aligned} q_{j,1}^{i,k} := q_{j,1,\psi }^{i,k} \circ \psi ^{-1}, f_{j,1}^{i,k} := f_{j,1,\psi }^{i,k} \circ \psi ^{-1} \end{aligned}$$

in $U_{\rho _0,j}$. Notice that ${\text {supp}} q_{j,1}^{i,k}, {\text {supp}} f_{j,1}^{i,k} \subset U_{4 \rho ,j}$, we can indeed treat $q_{j,1}^{i,k}, f_{j,1}^{i,k}$ as functions defined in $\mathbf {R}^n$ where their values outside $U_{4 \rho ,j}$ equal zero. Proposition 8 shows that $\nabla _x q_{j,1}^{i,k} \in BMO(\mathbf {R}^n)$ satisfies the estimate

$$\begin{aligned}{}[\nabla _x q_{j,1}^{i,k}]_{BMO(\mathbf {R}^n)} \le C_\rho \Vert \nabla _\eta q_{j,1,\psi }^{i,k} \Vert _{BMOL^2(\mathbf {R}^n)} \le C_\rho \Vert g_{i,k}\Vert _{BMOL^2(\mathbf {R}^n)}. \end{aligned}$$

Let $p_{j,1}^{i,k} := E *f_{j,1}^{i,k}$. By Lemma 4 again, we can prove that

$$\begin{aligned} \Vert p_{j,1}^{i,k} \Vert _{L^\infty (\mathbf {R}^n)} + \Vert \nabla _x p_{j,1}^{i,k} \Vert _{L^\infty (\mathbf {R}^n)} \le C_\rho \Vert f_{j,1,\psi }^{i,k} \Vert _{L^p(V_{2 \rho })} \le C_\rho \Vert g_{i,k} \Vert _{L^p(\mathbf {R}^n)} \end{aligned}$$

with some $p > n$. Thus, $p_{j,1}^{i,k}$ is well-defined. By Proposition 2, we have that

$$\begin{aligned} \Vert g_{i,k} \Vert _{BMOL^1(\mathbf {R}^n)}&\le C_\rho \Vert v_{2,j} \Vert _{BMOL^1(\mathbf {R}^n)} \le C_\rho \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

Hence, by an interpolation (cf. [4,  Lemma 5]),

$$\begin{aligned} \Vert g_{i,k} \Vert _{L^p(\mathbf {R}^n)} \le C_\rho \Vert v \Vert _{vBMO(\Omega )} \end{aligned}$$

for any $1<p<\infty$.

For lower order term $q_{j,2,\psi }^{i,k} := \partial _{\eta _k} ( \frac{\partial \eta _k}{\partial x_i} )_\psi \cdot (v_{2,j})_{\psi ,i}$, we set $q_{j,2}^{i,k} := q_{j,2,\psi }^{i,k} \circ \psi ^{-1}$ in $U_{\rho _0,j}$. Similar as $q_{j,1}^{i,k}$, we can treat $q_{j,2}^{i,k}$ as a function in $\mathbf {R}^n$ with value zero outside $U_{\rho ,j}$ since ${\text {supp}} q_{j,2}^{i,k} \subset U_{\rho ,j}$. Define $p_{j,2}^{i,k} := E *q_{j,2}^{i,k}$. Since E and $\nabla _x E$ are locally integrable, we have that

$$\begin{aligned} \Vert p_{j,2}^{i,k} \Vert _{L^\infty (\mathbf {R}^n)} + \Vert \nabla _x p_{j,2}^{i,k} \Vert _{L^\infty (\mathbf {R}^n)} \le C_\rho \Vert q_{j,2,\psi }^{i,k} \Vert _{L^p(V_{\rho })} \le C_\rho \Vert v_{2,j} \Vert _{L^p(U_{\rho ,j})} \end{aligned}$$

for some $p>n$. By an interpolation (cf. [4,  Lemma 5]) again, we deduce that

$$\begin{aligned} \Vert p_{j,2}^{i,k} \Vert _{L^\infty (\mathbf {R}^n)} + \Vert \nabla _x p_{j,2}^{i,k} \Vert _{L^\infty (\mathbf {R}^n)} \le C_\rho \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

This argument also holds for lower order term

$$\begin{aligned} q_{j,3,\psi }^{i,k} := \bigg ( \frac{\partial \eta _k}{\partial x_i} \bigg )_\psi \cdot \left( \sum _{1 \le l \le n} (v_{2,j})_{\psi ,l} \cdot \bigg ( \frac{\partial \eta _n}{\partial x_l} \bigg )_\psi \right) \cdot \frac{\partial ^2 x_i}{\partial \eta _k \partial \eta _n}. \end{aligned}$$

By letting $q_{j,3}^{i,k} := q_{j,3,\psi }^{i,k} \circ \psi ^{-1}$ in $U_{\rho _0,j}$ and $p_{j,3}^{i,k} := E *q_{j,3}^{i,k}$, we can show that

$$\begin{aligned} \Vert p_{j,3}^{i,k} \Vert _{L^\infty (\mathbf {R}^n)} + \Vert \nabla _x p_{j,3}^{i,k} \Vert _{L^\infty (\mathbf {R}^n)} \le C_\rho \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

Set

$$\begin{aligned} p_j^{\mathrm {tan}} := \underset{1 \le k \le n-1}{\sum _{1 \le i \le n,}} \big ( q_{j,1}^{i,k} + p_{j,1}^{i,k} + p_{j,2}^{i,k} + p_{j,3}^{i,k} \big ). \end{aligned}$$

Since a direct calculation implies that

$$\begin{aligned} ({\text {div}}_x w_j^{\mathrm {tan}})_{\psi }&= \underset{1 \le k \le n-1}{\sum _{1 \le i \le n,}} \bigg (\frac{\partial \eta _k}{\partial x_i} \bigg )_\psi \cdot \partial _{\eta _k} (v_{2,j})_{\psi ,i} \\&\quad - \underset{1 \le k \le n-1}{\sum _{1 \le i \le n,}} \bigg ( \frac{\partial \eta _k}{\partial x_i} \bigg )_\psi \cdot \left( \sum _{1 \le l \le n} (v_{2,j})_{\psi ,l} \cdot \bigg ( \frac{\partial \eta _n}{\partial x_l} \bigg )_\psi \right) \cdot \frac{\partial ^2 x_i}{\partial \eta _k \partial \eta _n} \end{aligned}$$

in normal coordinate in $V_{4 \rho } = \psi ^{-1}(U_{4 \rho ,j})$, it is easy to see that

$$\begin{aligned} - \Delta _x p_j^{\mathrm {tan}} = {\text {div}} w_j^{\mathrm {tan}} \end{aligned}$$

in $U_{2 \rho ,j}\cap \Omega$. Calculations above ensures that

$$\begin{aligned}{}[\nabla _x p_j^{\mathrm {tan}}]_{BMO(\mathbf {R}^n)} \le C_\rho \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

Since ${\text {supp}} q_{j,1}^{i,k} \subset U_{4 \rho ,j}$, we consider $x \in \Gamma$ and $r < \rho$ such that $B_r(x) \cap U_{4 \rho ,j} \ne \emptyset$. By change of variables $y = \psi (\eta )$ in $U_{4 \rho ,j}$, we deduce that

$$\begin{aligned} \int _{B_r(x) \cap U_{4 \rho ,j}} | \nabla _y q_{j,1}^{i,k} \cdot \nabla _y d| \, dy \le C \int _{B_{L_*r}(\zeta )} | \partial _{\eta _n} q_{j,1,\psi }^{i,k} | \, d\eta \end{aligned}$$

where $\zeta = \psi ^{-1}(x)$ and $\zeta _n = 0$. By Lemma 4, we see that

$$\begin{aligned} \int _{B_{L_*r}(\zeta )} | \partial _{\eta _n} q_{j,1,\psi }^{i,k} | \, d\eta \le r^n C_\rho \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

Since $\nabla _x p_{j,l}^{i,k} \in L^\infty (\mathbf {R}^n)$ for $l = 1,2,3$, we finally obtain that

$$\begin{aligned} \frac{1}{r^n} \int _{B_r(x)} | \nabla _y p_j^{\mathrm {tan}} \cdot \nabla _y d | \, dy \le C_\rho \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

$\square$

3.5 Volume potential for normal component

Consider $\rho \in (0,\rho _*/2)$ and $1 \le j \le m$. We let $g_j := \nabla d \cdot (\varphi _j v_2)_{\mathrm {odd}}$. Since $\varphi _j v_2 \in vBMO(\Omega )$ with ${\text {supp}} \varphi _j v_2 \subset U_{\rho ,j} \cap \overline{\Omega }$, by Proposition 2 we see that $g_j \in BMO(\mathbf {R}^n) \cap b^\nu (\Gamma )$. In particular, we have the estimate

$$\begin{aligned}{}[g_j]_{BMO(\mathbf {R}^n)} + [g_j]_{b^\nu (\Gamma )} \le C_\rho \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

Considering the normal coordinate in $U_{4 \rho ,j}$, $g_j$ is odd in $\eta _n$. Note that $w_j^{\mathrm {nor}} = g_j \nabla d$.

Proof

(Proposition 5) Since $\nabla d \in C^1(U_{\rho _0,j})$, by Proposition 2 we have that

$$\begin{aligned}{}[w_j^{\mathrm {nor}}]_{BMO(\mathbf {R}^n)} \le C \Vert \nabla d\Vert _{C^\gamma (U_{\rho _0,j})} \Vert g_j\Vert _{BMOL^1(\mathbf {R}^n)} \le C_\rho \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

We note that

$$\begin{aligned} {\text {div}}_x w_j^{\mathrm {nor}} = \nabla _x g_j \cdot \nabla _x d + g_j \Delta _x d. \end{aligned}$$

Let $g_{j,\psi } := g_j \circ \psi$ in $U_{\rho _0,j}$. We may treat $g_{j,\psi }$ as a function in $\mathbf {R}^n$ with value zero outside $V_\rho$. By Proposition 7, we have that

$$\begin{aligned}{}[g_{j,\psi }]_{BMO(\mathbf {R}^n)} \le C_\rho \Vert g_j \Vert _{BMOL^1(\mathbf {R}^n)}. \end{aligned}$$

In the normal coordinate, $\nabla _x g_j \cdot \nabla _x d=\partial _{\eta _n} g_{j,\psi }$. We introduce the operator $L=L_0+M$ in Proposition 6. Since $g_{j,\psi } \in Z_\rho$, we set

$$\begin{aligned} p_{1,j,\psi } := \theta _\rho L^{-1}_0 \partial _{\eta _n} g_{j,\psi } \end{aligned}$$

where $\theta _\rho$ is the cut-off function of $V_{2 \rho }$ in the proof of Lemma 3. $p_{1,j,\psi }$ satisfies all conditions in Lemma 3. Set $f_{j,\psi } := - M\theta _\rho L^{-1}_0 \partial _{\eta _n} g_{j,\psi }$. We define

$$\begin{aligned} p_{1,j} := p_{1,j,\psi } \circ \psi ^{-1}, f_j := f_{j,\psi } \circ \psi ^{-1} \end{aligned}$$

in $U_{\rho _0,j}$. Notice that $p_{1,j} \in L^\infty (\mathbf {R}^n)$ and $f_j \in L^p(\mathbf {R}^n)$ with some $p>n$. By Proposition 8,

$$\begin{aligned}{}[\nabla _x p_{1,j}]_{BMO(\mathbf {R}^n)} \le C_\rho [\nabla _\eta p_{1,j,\psi }]_{BMO(\mathbf {R}^n)} \le C_\rho [g_{j,\psi }]_{BMO(\mathbf {R}^n)}. \end{aligned}$$

Set

$$\begin{aligned} p_j^{\mathrm {nor}} = p_{1,j} + p_{2,j} + p_{3,j} \end{aligned}$$

with $p_{2,j} = E *f_j$ and $p_{3,j}=E *(g_j \Delta _x d)$. This $p_j^{\mathrm {nor}}$ satisfies all desired properties required. For lower order terms $p_{2,j}$ and $p_{3,j}$, we have that

$$\begin{aligned} \Vert p_2 \Vert _{L^\infty (\mathbf {R}^n)} + \Vert \nabla p_2 \Vert _{L^\infty (\mathbf {R}^n)} + \Vert \nabla p_3 \Vert _{L^\infty (\mathbf {R}^n)} + \Vert p_3 \Vert _{L^\infty (\mathbf {R}^n)} \le C_\rho \Vert g_j \Vert _{L^p(\mathbf {R}^n)} \end{aligned}$$

as E and $\nabla _x E$ are both locally integrable. By an interpolation (cf. [4,  Lemma 5]), we obtain that

$$\begin{aligned}{}[\nabla _x p_j^{\mathrm {nor}}]_{BMO(\mathbf {R}^n)} \le C_\rho \Vert g_j \Vert _{BMOL^1(\mathbf {R}^n)} \le C_\rho \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

Since ${\text {supp}} p_{1,j} \subset U_{\rho ,j}$, we consider $x \in \Gamma$ and $r<\rho$ such that $B_r(x) \cap U_{\rho ,j} \ne \emptyset$. Set $\zeta = \psi ^{-1}(x)$ with $\zeta _n = 0$. Consider the change of variable $y = \psi (\eta )$ in $U_{4 \rho ,j}$, by Lemma 3 we see that

$$\begin{aligned} \int _{B_r(x) \cap U_{\rho ,j}} | \nabla _y d \cdot \nabla _y p_{1,j} | \, dy \le C \int _{B_{L_*r}(\zeta )} | \partial _{\eta _n} p_{1,j,\psi } | \, d\eta \le C_\rho [g_{j,\psi }]_{BMO(\mathbf {R}^n)}. \end{aligned}$$

By the $L^\infty$-estimates of $\nabla _y p_2$ and $\nabla _y p_3$, we get that

$$\begin{aligned} \frac{1}{r^n} \int _{B_r(x)} |\nabla _y d \cdot \nabla _y p_j^{\mathrm {nor}}| \, dy \le C_\rho \Vert v \Vert _{vBMO(\Omega )}. \end{aligned}$$

Finally, a simple substitution shows that

$$\begin{aligned} - \Delta _x p_j^{\mathrm {nor}} = \nabla _x d \cdot \nabla _x g_j - f_j + f_j + g_j \Delta _x d = {\text {div}}_x w_j^{\mathrm {nor}} \end{aligned}$$

in $U_{2 \rho }(z_0)\cap \Omega$. $\square$

4 Neumann problem with bounded data

We consider the Neumann problem for the Laplace equation problem (3) for the Laplace equation. If $\Omega$ is a smooth bounded domain, as well-known, for $g\in H^{-1/2}(\Gamma )$, there is a unique (up to constant) weak solution $u \in H^1(\Omega )$ provided that g fulfills the compatibility condition

$$\begin{aligned} \int _\Gamma g \, d \mathcal {H}^{n-1} = 0; \end{aligned}$$

(14)

see e.g. [20]. The main goal of this section is to prove that $\nabla u$ belongs to $vBMO^{\infty ,\infty }(\Omega )$ provided that $g \in L^\infty (\Gamma )$. In other words, we prove Lemma 2.

To prove Lemma 2, we represent the solution by using the Neumann-Green function. Let N(x, y) be the Green function, i.e., a solution v of

$$\begin{aligned} -\Delta _x v&= \delta (x-y) - |\Omega |^{-1}&\text {in}&\quad \Omega \\ \frac{\partial v}{\partial \mathbf {n}_x}&= 0&\text {on}&\quad \partial \Omega \end{aligned}$$

for $y\in \Omega$. It is easy to see that the solution u of (3) satisfying $\int _\Omega u \, dx=0$ is given as

$$\begin{aligned} u(x) = \int _\Gamma N(x, y) g(y) \, d\mathcal {H}^{n-1} (y). \end{aligned}$$

The function N is decomposed as

$$\begin{aligned} N(x, y) = E(x - y) + h(x, y), \end{aligned}$$

where $h\in C^\infty (\Omega \times \Omega )$ is a milder part. We recall $h(x,y)=h(y,x)$ and

$$\begin{aligned} \sup _{x\in \Omega } \int _\Omega \left| \nabla ^k_y h(x, y) \right| ^{1+\delta } \, dy < \infty \end{aligned}$$

for $k=0,1,2$ with some $\delta >0$; see [12,  Lemma 3.1]. In particular, by applying the standard $L^p$ estimate for the Neumann problem in the proof of [12,  Lemma 3.1] to $\nabla _y h( \cdot ,y)$, we can deduce that

$$\begin{aligned} \sup _{x \in \Omega } \int _\Omega \left| \nabla _x \nabla _y h(x,y)\right| ^{1+\delta } \, dy<\infty . \end{aligned}$$

Hence, we see that $\nabla _x h(x,\cdot ) \in W^{1,1+\delta }(\Omega _y)$. By the trace theorem for Sobolev space $W^{1,1+\delta }(\Omega _y)$, this yields

$$\begin{aligned} M_0 := \sup _{x\in \Omega } \int _\Gamma \left| \nabla _x h(x,y) \right| ^{1+\delta } \, d\mathcal {H}^{n-1}(y) < \infty . \end{aligned}$$

(15)

We decompose u as

$$\begin{aligned} u(x) = E * (\delta _\Gamma \otimes g) + \int _\Gamma h(x,y) g(y) \, d\mathcal {H}^{n-1}(y) = I + I\!\!I. \end{aligned}$$

The estimate (15) yields

$$\begin{aligned} \Vert \nabla I\!\!I \Vert _{L^\infty (\Omega )} \le M_0 \Vert g\Vert _{L^\infty (\Gamma )}, \end{aligned}$$

so to prove Lemma 2 it suffices to estimate $\nabla I$. In other words, Lemma 2 follows from the next lemma.

Lemma 5

Let $\Omega$ be a bounded domain in $\mathbf {R}^n$ with $C^2$ boundary $\Gamma =\partial \Omega$.

• (i) (BMO estimate) There exists a constant $C_1$ such that

  $$\begin{aligned} \left[ \nabla \left( E * (\delta _\Gamma \otimes g) \right) \right] _{BMO(\mathbf {R}^n)} \le C_1 \Vert g\Vert _{L^\infty (\Gamma )} \end{aligned}$$

  (16)

  for all $g\in L^\infty (\Gamma )$.

• (ii) $(L^\infty$ estimate for normal component) There exists a constant $C_2$ such that

  $$\begin{aligned} \left\| \nabla d \cdot \nabla \left( E * (\delta _\Gamma \otimes g) \right) \right\| _{L^\infty (\Gamma _{\rho _0}^{\mathbf {R}^n} \cap \Omega )} \le C_2 \Vert g\Vert _{L^\infty (\Gamma )} \end{aligned}$$

  (17)

  for all $g\in L^\infty (\Gamma )$.

Here $E*(\delta _\Gamma \otimes g)$ is defined as $E*(\delta _\Gamma \otimes g)(x) := \int _\Gamma E(x-y)g(y) \, d\mathcal {H}^{n-1}(y)$ for a function g on $\Gamma$. We shall prove Lemma 5 in following subsections.

4.1 BMO estimate

To see the idea, we shall prove (16) when $\Gamma$ is flat. Let $\Gamma =\partial \mathbf {R}^n_+$ and $\mathbf {R}^n_+=\left\{ (x_1,\ldots ,x_n)\mid x_n>0 \right\}$. In this case,

$$\begin{aligned} \nabla \left( E*(\delta _\Gamma \otimes g) \right) = \nabla \partial _{x_n} E*1_{\mathbf {R}^n_+} {\widetilde{g}} \end{aligned}$$

where ${\widetilde{g}} \in L^\infty (\mathbf {R}^n)$ is defined by ${\widetilde{g}} (x',x_n) := g(x',0)$ for any $x \in \mathbf {R}^n$. By the $L^\infty$-BMO estimate for the singular integral operator [15,  Theorem 4.2.7], we obtain (16) when $\Gamma =\partial \mathbf {R}^n_+$.

Proof

(Lemma 5 (i)) Note that the signed distance function d is $C^2$ in $\Gamma _{\rho _0}^{\mathbf {R}^n}$, see [13,  Section 14.6]. Let $\delta \in (0,\rho _0/2)$. We take a $C^2$ cut-off function $\theta \ge 0$ such that $\theta (\sigma )=1$ for $\sigma \le 1$ and $\theta (\sigma )=0$ for $\sigma \ge 2$. By the choice of $\delta$, we see that $\theta _d=\theta (d/\delta )$ is $C^2$ in $\mathbf {R}^n$. We extend $g\in L^\infty (\Gamma )$ to $g_e \in L^\infty (\Gamma _{2 \delta }^{\mathbf {R}^n})$ by setting

$$\begin{aligned} g_e(x) := g(\pi x) \end{aligned}$$

for any $x \in \Gamma _{2 \delta }^{\mathbf {R}^n}$ with $\pi x$ denoting the projection of x on $\Gamma$. For $x \in \Gamma _{2 \delta }^{\mathbf {R}^n}$, by considering the normal coordinate $x = \psi (\eta )$ in $U_{2 \delta }(\pi x)$, we have that

$$\begin{aligned} (\nabla _x d)_\psi \cdot (\nabla _x g_e)_\psi = \partial _{\eta _n} (g_e)_\psi = 0 \end{aligned}$$

as $(g_e)_\psi (\eta ',\alpha ) = (g_e)_\psi (\eta ',\beta )$ for any $|\eta '| < 2 \delta$ and $\alpha ,\beta \in (-2 \delta ,2 \delta )$. Hence, we see that $\nabla d\cdot \nabla g_e=0$ in $\Gamma _{2 \delta }^{\mathbf {R}^n}$.

Let us consider $g_{e,c} := \theta _d g_e$. A key observation is that

$$\begin{aligned} \delta _\Gamma \otimes g&= (\nabla 1_\Omega \cdot \nabla d) g_{e,c} \\&= {\text {div}} (g_{e,c} 1_\Omega \nabla d) - 1_\Omega {\text {div}} (g_{e,c} \nabla d), \\ {\text {div}} (g_{e,c} \nabla d)&= g_{e,c} \Delta d + \nabla d \cdot \nabla g_{e,c} = g_{e,c} \Delta d + \frac{\theta '(d/\delta )}{\delta } g_e. \end{aligned}$$

Thus

$$\begin{aligned} \nabla E*(\delta _\Gamma \otimes g) = \nabla {\text {div}} \left( E*( g_{e,c} 1_\Omega \nabla d) \right) -\nabla E* \left( 1_\Omega g_e f_{\theta ,\delta } \right) = I_1 + I_2 \end{aligned}$$

where $f_{\theta ,\delta } := \theta _d \Delta d + \frac{\theta '(d/\delta )}{\delta }$. By the $L^\infty$-BMO estimate for the singular integral operator [15,  Theorem 4.2.7], the first term is estimated as

$$\begin{aligned}{}[I_1]_{BMO(\mathbf {R}^n)} \le C \Vert g_{e,c} \nabla d \Vert _{L^\infty (\Omega )} \le C \Vert g\Vert _{L^\infty (\Gamma )}. \end{aligned}$$

Since

$$\begin{aligned} A = \sup _{x \in \mathbf {R}^n \setminus \{0\}} |x|^{n-1} \left| \nabla E(x) \right| < \infty , \end{aligned}$$

for $x \in \mathbf {R}^n$ with $d(x,\Omega ) = \inf _{y \in \Omega } |x-y| < 1$ we have that

$$\begin{aligned} \left| I_2(x) \right| \le A \int _\Omega \frac{1}{|x-y|^{n-1}} \, dy \Vert f_{\theta ,\delta } \Vert _{L^\infty (\Gamma _{2 \delta }^{\mathbf {R}^n})} \Vert g_{e,c} \Vert _{L^\infty (\Gamma _{2 \delta }^{\mathbf {R}^n})} \le C_{\Omega ,\delta } \Vert g\Vert _{L^\infty (\Gamma )} \end{aligned}$$

with $C_{\Omega ,\delta }$ depending only on $\Omega$ and $\delta$. For $x \in \mathbf {R}^n$ with $d(x,\Omega ) = \inf _{y \in \Omega } |x-y| \ge 1$, the above estimate is trivial as $|x-y|^{-(n-1)} \le 1$ for any $y \in \Omega$. The proof of (i) is now complete. $\square$

4.2 Estimate for normal derivative

We shall estimate normal derivative of E.

Lemma 6

Let $\Omega$ be a bounded domain in $\mathbf {R}^n$ with $C^2$ boundary $\Gamma$. Then

• (i)
  $$\begin{aligned} \int _\Gamma \frac{\partial E}{\partial \mathbf {n}_y} (x-y) \, d\mathcal {H}^{n-1}(y) = -1 \quad \text {for}\quad x \in \Omega , \end{aligned}$$
• (ii)
  $$\begin{aligned} \sup _{x\in \Omega } \int _\Gamma \left| \frac{\partial E}{\partial \mathbf {n}_y} (x-y) \right| \, d\mathcal {H}^{n-1}(y) < \infty . \end{aligned}$$

Proof

• (i) This follows from the Gauss divergence theorem. We observe that

  $$\begin{aligned} \int _\Gamma \frac{\partial E}{\partial \mathbf {n}_y} (x-y) \, d\mathcal {H}^{n-1}(y) = \int _\Omega \Delta _y E(x-y) \, dy. \end{aligned}$$

  Since $\Delta _y E(x-y) = -\delta (x-y)$, we obtain

  $$\begin{aligned} \int _\Gamma \frac{\partial E}{\partial \mathbf {n}_y} (x-y) \, d\mathcal {H}^{n-1}(y) = -1 \end{aligned}$$

  for $x\in \Omega$.

• (ii) We recall our local coordinate patches $\{U_i\}^m_{i=1}$ with $U_i=U_{\rho ,i}$ as in Sect. 2.1. For $x \in \Omega ^\rho$ and $y \in \Gamma$, obviously $| \nabla E (x-y) | \le C \rho ^{-(n-1)}$. Let $x \in \Gamma _\rho ^{\mathbf {R}^n} \cap \Omega$. If $d(x, U_i \cap \Gamma ) \ge \rho$, similarly $| \nabla E (x-y) | \le C \rho ^{-(n-1)}$ for $y \in U_i \cap \Gamma$. Hence, it is sufficient to consider $U_i$ such that $d(x, U_i \cap \Gamma ) < \rho$, i.e., it suffices to prove

  $$\begin{aligned} \int _{U_i\cap \Gamma } \left| \frac{\partial E}{\partial \mathbf {n}_y} (x-y) \right| d\mathcal {H}^{n-1}(y) < \infty . \end{aligned}$$

  for $U_i$ such that $d(x, U_i \cap \Gamma ) < \rho$. Since $-\partial E/\partial \mathbf {n}_y(x-y)$ is invariant under translations and rotations, we can write $-\partial E/\partial \mathbf {n}_y(x-y)$ in the local coordinate. Let $U_i$ be such that $d(x, U_i \cap \Gamma ) < \rho$ and denote $h_{z_i}$ by $h_i$ for simplicity. Let us observe that

  $$\begin{aligned} -\mathbf {n} \left( y', h_i(y') \right) = \left( -\nabla ' h_i(y'), 1 \right) / \omega _i(y') \end{aligned}$$

  with $\omega _i(y')=\left( 1+ |\nabla ' h_i(y')|^2 \right) ^{1/2}$, where $\nabla '$ is the gradient in $y'$ variables. This implies that

  $$\begin{aligned} -n\alpha (n)\frac{\partial E}{\partial \mathbf {n}_y}(x-y) = \frac{\sigma _i(y')}{\omega _i(y') \left( |x'-y'|^2 + \left( x_n- h_i(y') \right) ^2 \right) ^{n/2}} \end{aligned}$$

  for $y \in \Gamma _i$ with

  $$\begin{aligned} \sigma _i(y') := -\nabla ' h_i(y) \cdot (x'-y') + \left( x_n - h_i(y') \right) \; \; \text {where} \; \; x_n > h_i(x'),\ x' \in B_{3 \rho }(0'). \end{aligned}$$

  We set

  $$\begin{aligned} K_i(x', y', x_n) = \frac{\sigma _i(y')}{\left( |x'-y'|^2 + \left( x_n - h_i(y') \right) ^2 \right) ^{n/2}}. \end{aligned}$$

  By the Taylor expansion

  $$\begin{aligned} h_i(x') = h_i(y') + \nabla ' h_i(y') \cdot (x'-y') + r_i(x',y') \end{aligned}$$

  with

  $$\begin{aligned} r_i(x',y') = (x'-y')^{\mathrm {T}} \cdot \int ^1_0 (1-\theta ) \nabla '^2 h_i \left( \theta x'+(1-\theta )y' \right) d\theta \cdot (x'-y'), \end{aligned}$$

  we obtain

  $$\begin{aligned} \sigma _i(y') = x_n - h_i(x') + r_i(x',y') \end{aligned}$$

  with an estimate

  $$\begin{aligned} \left| r_i(x',y') \right| \le \Vert \nabla '^2 h_i \Vert _{L^\infty (B_{3 \rho }(0'))} |x'-y'|^2. \end{aligned}$$

  (18)

  We decompose $K_i$ into a leading term and a remainder term

  $$\begin{aligned} K_i(x', y', x_n) = K_0^i(x', y', x_n) + R_i(x', y', x_n) \end{aligned}$$

  with

  $$\begin{aligned} K_0^i(x', y', x_n)&:= \frac{x_n - h_i(x')}{\left( |x'-y'|^2 + \left( x_n - h_i(y') \right) ^2 \right) ^{n/2}} \\ R_i(x', y', x_n)&:= \frac{r_i(x, y)}{\left( |x'-y'|^2 + \left( x_n - h_i(y') \right) ^2 \right) ^{n/2}}. \end{aligned}$$

  The term $K_0^i$ is very singular but it is positive. The term $R_i$ is estimated as

  $$\begin{aligned} \left| R_i(x', y', x_n) \right| \le \Vert \nabla '^2 h_i \Vert _{L^\infty (B_{3 \rho }(0'))} |x'-y'|^{2-n} \end{aligned}$$

  by the estimate (18). Hence,

  $$\begin{aligned} \int _{\Gamma \cap U_i} \left| \frac{R_i(x',y',x_n)}{\omega _i(y')} \right| \, d\mathcal {H}^{n-1}(y) \le C \int _{B_\rho (0')} \frac{1}{|x'-y'|^{n-2}} \, dy' \le C \rho \end{aligned}$$

  with C independent of $\rho$ and i. By (i), we observe that

  $$\begin{aligned} n \alpha (n)&= \sum _{i : d(x, U_i \cap \Gamma ) < \rho } \int _{B_\rho (0')} \frac{K_i(x', y', x_n)}{\omega _i(y')} \, dy' \\&\quad - n \alpha (n) \sum _{j : d(x, U_j \cap \Gamma ) \ge \rho } \int _{U_j \cap \Gamma } \frac{\partial E}{\partial \mathbf {n}_y} (x-y) \, d\mathcal {H}^{n-1}(y). \end{aligned}$$

  Since $K_0^i$ is positive for any i such that $d(x,U_i \cap \Gamma ) < \rho$,

  $$\begin{aligned} \sum _{i : d(x, U_i \cap \Gamma ) < \rho } \int _{B_\rho (0')} \frac{K_0^i(x', y', x_n)}{\omega _i(y')} \, dy' \le n \alpha (n) \cdot (1+\frac{m \cdot C \cdot S(\Gamma )}{\rho ^{n-1}}) + m \cdot C \cdot \rho \end{aligned}$$

  where $S(\Gamma )$ denotes the surface area of $\Gamma$, which is bounded. Thus, the estimate

  $$\begin{aligned} \int _{U_i\cap \Gamma } \left| \frac{\partial E}{\partial \mathbf {n}_y}(x-y) \right| \, d\mathcal {H}^{n-1}(y) \le \frac{1}{n \alpha (n)} \int _{B_\rho (0')} \frac{K_0^i + |R_i|}{\omega _i(y')} \, dy' < \infty \end{aligned}$$

  holds for any $U_i$ such that $d(x, U_i \cap \Gamma ) < \rho$. The proof of (ii) is now complete.

$\square$

Based on Lemma 6, we are able to prove Lemma 5 (ii).

Proof

(Lemma 5 (ii)) We decompose

$$\begin{aligned} \nabla d (x) \cdot \nabla \left( E*(\delta _\Gamma \otimes g) \right) (x) = \int _\Gamma \left( \nabla d(x)-\nabla d(y) \right) \cdot \nabla E(x-y) g(y) \, d\mathcal {H}^{n-1}(y) \\ + \int _\Gamma \frac{\partial E}{\partial \mathbf {n}_y}(x-y)g(y) \, d\mathcal {H}^{n-1}(y) = I_1 + I_2. \end{aligned}$$

Let $x \in \Gamma _{\rho _0}^{\mathbf {R}^n}$ and $\pi x$ be the projection of x on $\Gamma$. For $y \in U_{\rho _0}(\pi x)$, there exists a constant $L'$, independent of x and y, such that

$$\begin{aligned} \left| \nabla d(x) - \nabla d(y) \right| \le L' |x-y|. \end{aligned}$$

For $y \in \Gamma _{\rho _0}^{\mathbf {R}^n} \setminus U_{\rho _0}(\pi x)$, we have that $|x-y| \ge \frac{\rho _0}{2}$. Since $\overline{\Gamma _{\rho _0/2}^{\mathbf {R}^n}}$ is compact in $\mathbf {R}^n$, by considering a finite subcover of $\cup _{z \in \Gamma } U_{\rho _0}(z)$ we are able to show that there exists $M>0$ such that the estimate

$$\begin{aligned} |\nabla d (x) - \nabla d (y)| \le M |x-y| \end{aligned}$$

holds for any $x,y \in \Gamma _{\rho _0}^{\mathbf {R}^n}$. Thus,

$$\begin{aligned} H(x,y) = \left( \nabla d(x) - \nabla d(y) \right) \cdot \nabla E(x-y) \end{aligned}$$

is estimated as

$$\begin{aligned} \left| H(x,y) \right| \le \frac{M}{|x-y|^{n-2}} \end{aligned}$$

in $\Gamma _{\rho _0}^{\mathbf {R}^n} \times \Gamma _{\rho _0}^{\mathbf {R}^n}$. We observe that

$$\begin{aligned} \sup _{x \in \Gamma _{\rho _0}^{\mathbf {R}^n} \cap \Omega } \left| I_1(x) \right|&\le \sup _{x \in \Gamma _{\rho _0}^{\mathbf {R}^n} \cap \Omega } \int _\Gamma H(x,y) \, d\mathcal {H}^{n-1}(y) \Vert g\Vert _{L^\infty (\Gamma )} \\&\le M\sup _{x \in \Gamma _{\rho _0}^{\mathbf {R}^n} \cap \Omega } \int _\Gamma \frac{d\mathcal {H}^{n-1}(y)}{|x-y|^{n-2}} \Vert g\Vert _{L^\infty (\Gamma )}. \end{aligned}$$

Since

$$\begin{aligned} \sup _{x \in \Gamma _{\rho _0}^{\mathbf {R}^n} \cap \Omega } \left| I_2(x) \right| \le \sup _{x \in \Gamma _{\rho _0}^{\mathbf {R}^n} \cap \Omega } \int _\Gamma \left| \frac{\partial E}{\partial \mathbf {n}_y}(x-y) \right| \, d\mathcal {H}^{n-1}(y) \Vert g\Vert _{L^\infty (\Gamma )}, \end{aligned}$$

Lemma 6 (ii) now yields (17). The proof is now complete. $\square$

We wonder whether the tangential component of $\nabla E*(\delta _\Gamma \otimes g)$ satisfies the same estimate. Unfortunately, the estimate

$$\begin{aligned} \left\| \nabla \left( E*(\delta _\Gamma \otimes g)\right) \right\| _{L^\infty (\Gamma _{\rho _0}^{\mathbf {R}^n} \cap \Omega )} \le C \Vert g\Vert _{L^\infty (\Gamma )} \end{aligned}$$

should not hold even if $\Gamma$ is flat. Even weaker estimate

$$\begin{aligned} \left[ \nabla \left( E*(\delta _\Gamma \otimes g)\right) \right] _{b^\nu (\Gamma )} \le C \Vert g\Vert _{L^\infty (\Gamma )} \end{aligned}$$

should not hold in general.

To illustrate the problem, we consider the case that $\Gamma$ is flat. We may assume $\Gamma =\partial \mathbf {R}^n_+$, $\mathbf {R}^n_+ = \{x_n>0 \}$.

Lemma 7

The estimate

$$\begin{aligned} \left\| \partial _{x_n} \left( E*(\delta _\Gamma \otimes g) \right) \right\| _{L^\infty (\mathbf {R}_+^n)} \le \frac{1}{2}\Vert g\Vert _{L^\infty (\mathbf {R}^{n-1})} \end{aligned}$$

holds for $g \in L^\infty (\mathbf {R}^{n-1})$.

Proof

This is because $-\partial _{x_n} \left( E*(\delta _\Gamma \otimes g) \right)$ is the half of the Poisson integral, i.e.,

$$\begin{aligned} -\partial _{x_n} \left( E*(\delta _\Gamma \otimes g) \right) (x) = \frac{1}{2} \int _{\mathbf {R}^{n-1}} P_{x_n} (x'-y') g(y')dy', \end{aligned}$$

where $P_{x_n}$ denotes the Poisson kernel. Thus the desired $L^\infty$ estimate follows from the maximum principle of the Dirichlet problem for the Laplacian or from the property that $\int _{\mathbf {R}^{n-1}} P_{x_n} (x')dx'=1$ and $P_{x_n} \ge 0$. $\square$

Theorem 3

There is a bounded sequence of smooth functions $\{g_\ell \}_{\ell \in \mathbf {N}} \subset L^\infty (\mathbf {R}^{n-1})$ such that

$$\begin{aligned} \lim _{\ell \rightarrow \infty } \left[ \partial _{x'} \left( E*(\delta _\Gamma \otimes g_\ell ) \right) \right] _{b^\nu } = \infty \end{aligned}$$

for any $\nu >0$.

Proof

If g is smooth, $E*(\delta _\Gamma \otimes g)$ is smooth up to the boundary. In this case, if $\left[ \partial _{x'}\left( E*(\delta _\Gamma \otimes g)\right) \right] _{b^\nu }$ is bounded by $C\Vert g\Vert _{L^\infty (\mathbf {R}^{n-1})}$, $\left\| \partial _{x'} \left( E*(\delta _\Gamma \otimes g) \right) \right\| _{L^\infty (\Gamma )}$ is also bounded by $c_0 C\Vert g\Vert _{L^\infty (\mathbf {R}^{n-1})}$ with a constant $c_0$ depending only on n since the mean value over r-ball around x converges to its value at x as $r\rightarrow 0$.

We consider the Neumann problem

$$\begin{aligned} \Delta u = 0&\quad \text {in}\quad \mathbf {R}^n_+, \\ \frac{\partial u}{\partial \mathbf {n}} = g&\quad \text {on}\quad \Gamma = \partial \mathbf {R}^n_+. \end{aligned}$$

By using the tangential Fourier transform, we see that

$$\begin{aligned} u(x,t) = \Lambda ^{-1} \exp (-x_n \Lambda ) g \end{aligned}$$

where $\Lambda =(-\Delta ')^{1/2}$. If $\Vert \nabla ' u\Vert _{L^\infty (\Gamma )} \le C\Vert g\Vert _{L^\infty (\mathbf {R}^{n-1})}$ were true, sending $x_n>0$ to zero would imply $L^\infty$ boundedness of the Riesz operator $\nabla ' \Lambda ^{-1}$, which is absurd.

The operator $E*(\delta _\Gamma \otimes g)$ is the half of the solution operator of the Neumann problem, so $L^\infty$ bound for $\nabla ' E*(\delta _\Gamma \otimes g)$ should not hold even if it is restricted to smooth functions. $\square$

Corollary 1

Assume that $\Omega =\mathbf {R}^n_+$. Let $v\mapsto \nabla q$ be the Helmholtz projection to a gradient field. Then, this projection is unbounded from $\left( L^\infty (\Omega )\right) ^n$ to $\left( BMO^{\mu ,\nu }_b(\Omega )\right) ^n$ for any $\mu ,\nu >0$.

Proof

We consider

$$\begin{aligned} v = \left( 0, \ldots , 0, v_n(x')\right) \end{aligned}$$

with $v_n \in L^\infty (\mathbf {R}^{n-1})$. This evidently solves ${\text {div}}v=0$. The normal trace equals $-v_n(x')$. If

$$\begin{aligned} {[}\nabla q]_{b^\nu } \le C \Vert v_n\Vert _{L^\infty (\mathbf {R}^{n-1})} \end{aligned}$$

for all $v_n \in L^\infty (\mathbf {R}^{n-1})$ with C independent of v, then this would contradict Theorem 3. $\square$