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\(C^{1,\alpha }\)-regularity for surfaces with \(H\in L^p\)

Abstract

In this paper, we prove several results on the geometry of surfaces immersed in \(\mathbb {R}^3\) with small or bounded \(L^2\) norm of \(|A|\). For instance, we prove that if the \(L^2\) norm of \(|A|\) and the \(L^p\) norm of \(H\), \(p>2\), are sufficiently small, then such a surface is graphical away from its boundary. We also prove that given an embedded disk with bounded \(L^2\) norm of \(|A|\), not necessarily small, then such a disk is graphical away from its boundary, provided that the \(L^p\) norm of \(H\) is sufficiently small, \(p>2\). These results are related to previous work of Schoen–Simon (Surfaces with quasiconformal Gauss map. Princeton University Press, Princeton, vol 103, pp 127–146, 1983) and Colding–Minicozzi (Ann Math 160:69–92, 2004).

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Correspondence to Giuseppe Tinaglia.

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G. Tinaglia is partially supported by EPSRC Grant No. EP/L003163/1.

Appendix

Appendix

For the sake of completeness, in this appendix, we prove two results in differential geometry that are used throughout the paper. In Remark 5.3, we also discuss the \(C^{1,\alpha }\) regularity.

Let

$$\begin{aligned} \Omega :=\{(x_1,x_2)\in \mathbb {R}^2| a^2<x_1^2+x_2^2<b^2\} \end{aligned}$$

for certain \(b>a>0\) and let \(\mathcal A\) denote the graph above \(\Omega \) of a smooth function \(u\). That is \(u\in C^\infty (\Omega )\) and \({{\mathrm{graph}}}u=\mathcal A\).

Lemma 5.1

Assume that

$$\begin{aligned} |Du|\le r\le 1 \text { and} \int _\mathcal A|A|^2\,\mathrm{d}\mathcal {H}^2\le \varepsilon \end{aligned}$$

Then, there exists \(\rho \in (a,b)\) for which

$$\begin{aligned} \int _{u(S_\rho )}k \,\mathrm{d}s\le 2\pi \left( 1+r\sqrt{2}+ \left( \frac{2\varepsilon b}{b-a}\right) ^\frac{1}{2}\right) , \end{aligned}$$

where \(k\) is the curvature of the curve \({{\mathrm{graph}}}u|_{S_\rho }\) and \(S_\rho =\{(x_1, x_2): x_1^2+x_2^2=\rho ^2\}\).

Proof

Recall that in graphical coordinates

$$\begin{aligned} A_{ij}(x, u(x))=\left( \frac{\partial ^2}{\partial x_i \partial x_j}\left( x_1, x_2, u(x_1, x_2)\right) \right) ^\bot =\left( 0,0, D_{ij} u\right) \cdot \nu = \frac{D_{ij} u}{\sqrt{1+|Du|^2}}, \end{aligned}$$

where \(A_{ij}\), \(i=1,2\), are the coefficients of the second fundamental form, and also that

$$\begin{aligned} |A|^2=A_{ij}A_{kl}g^{ik}g^{il}, \end{aligned}$$

where \(g\) is the induced metric. We have that

$$\begin{aligned} |D^2 u|^2=\sum _{i,j=1}^2|D_{ij} u|^2 \le |A|^2 (1+|Du|^2)^3 \end{aligned}$$

(see for example [5]). On \(\Omega \), we are assuming that

$$\begin{aligned} |Du(x)|\le r \,, \quad \forall x\in \Omega \end{aligned}$$

and this, together with the area formula, gives

$$\begin{aligned} \int _\Omega |D^2u(x)|^2\,\mathrm{d}x&= \int _\Omega \frac{|D^2u|^2}{(1+|Du|^2)^3} (1+|Du|^2)^3 \mathrm{d}x\\&\le (1+r^2)^\frac{5}{2} \int _\Omega \frac{|D^2u|^2}{(1+|Du|^2)^3} \sqrt{1+|Du|^2} \mathrm{d}x\\&= (1+r^2)^\frac{5}{2}\int _{\mathcal A}|A|^2\,\mathrm{d}\mathcal {H}^2\le 2(1+r)\varepsilon . \end{aligned}$$

By the coarea formula, we can pick \(\rho \in (a, b)\), so that

$$\begin{aligned} \int _{S_\rho }|D^2u|^2\,\mathrm{d}x\le \frac{2(1+r)\varepsilon }{b-a}. \end{aligned}$$

Let \(\Gamma = {{\mathrm{graph}}}u|_{S_\rho }\). \(\Gamma \) is a closed curve in \(\mathcal A\) and we want to compute

$$\begin{aligned} \int _\Gamma k\,\mathrm{d}s. \end{aligned}$$

Let \(\gamma :[0,1]\rightarrow S_\rho \) be the following parametrization of \(S_\rho \):

$$\begin{aligned} \gamma (t)=(\rho \cos 2\pi t, \rho \sin 2\pi t) \end{aligned}$$

and consider the parametrization of \(\Gamma \) given by

$$\begin{aligned} f(t)=(\gamma (t), u(\gamma (t)))\,,\,\,t\in [0,1]. \end{aligned}$$

Recall that

$$\begin{aligned} \int _\Gamma k\,\mathrm{d}s=\int _0^1k(f(t)) |f'(t)|\,\mathrm{d}t\le \int _0^1\frac{|f''|}{|f'|}\mathrm{d}t, \end{aligned}$$

since

$$\begin{aligned} k=\frac{|f'\times f''|}{|f'|^3}\implies k\le \frac{|f''|}{|f'|^2}. \end{aligned}$$

Furthermore,

$$\begin{aligned} |f'|^2&= |\gamma '(t)|^2+\left| \frac{\text{ d }}{{\text{ d }}t}u(\gamma (t))\right| ^2\implies \\ (2\pi \rho )^2&= |\gamma '(t)|^2\le |\gamma '(t)|^2+|Du|^2|\gamma '(t)|^2= |f'|^2 \end{aligned}$$

and

$$\begin{aligned} |f''|&= \left| \left( \gamma ''(t),\frac{\mathrm{d}^2}{\mathrm{d}t^2}u(\gamma (t))\right) \right| \le \left| \gamma ''(t)\right| +\left| \frac{\mathrm{d}^2}{\mathrm{d}t^2}u(\gamma (t))\right| \\&\le \rho (2\pi )^2+\sqrt{2}|D^2u||\gamma '(t)|^2+\sqrt{2}|Du||\gamma ''(t)|\\&\le \rho (2\pi )^2+\sqrt{2}|D^2u|(2\pi \rho )^2+\sqrt{2}r \rho (2\pi )^2, \end{aligned}$$

where we have used the computation:

$$\begin{aligned} \left| \frac{{\text{ d }}^2}{{\text{ d }}t^2}u(\gamma )\right|&= \left| \frac{\text{ d }}{{\text{ d }}t}(Du\cdot \gamma ')\right| \le |\gamma '|^2\left( \sum _{i,j=1}^2 |D_{ij} u|\right) +|Du|\left( |\gamma _1''|+|\gamma _2''|\right) \\&\le |\gamma '|^2\sqrt{2}\left( \sum _{i,j=1}^2 |D_{ij} u|^2\right) ^\frac{1}{2}+|Du|\sqrt{2}|\gamma ''|\\&= |\gamma '|^2\sqrt{2}|D^2u|+|Du|\sqrt{2}|\gamma ''|. \end{aligned}$$

Hence,

$$\begin{aligned} \int _\Gamma k\,\mathrm{d}s&\le \int _0^1\frac{\rho (2\pi )^2+\sqrt{2}|D^2u|(2\pi \rho )^2+ r\rho \sqrt{2} (2\pi )^2}{2\pi \rho }\mathrm{d}t\\&= 2\pi (1+\sqrt{2}r)+2\sqrt{2}\pi \rho \int _0^1|D^2u(\gamma (t))|\mathrm{d}t. \end{aligned}$$

Using again the area formula, we have

$$\begin{aligned} \int _0^1 |D^2u(\gamma (t))|\mathrm{d}t&= \int _0^1 \frac{|D^2u(\gamma (t))|}{|\gamma '(t)|}|\gamma '(t)|\mathrm{d}t=\frac{1}{2\pi \rho } \int _0^1 |D^2u(\gamma (t))||\gamma '(t)|\mathrm{d}t\\&= \frac{1}{2\pi \rho }\int _{S_\rho }|D^2u(x)|\mathrm{d}x\le \frac{1}{2\pi \rho }\left( \int _{S_\rho }|D^2u(x)|^2\mathrm{d}x\right) ^\frac{1}{2}|S_\rho |^\frac{1}{2}\\&\le \frac{1}{2\pi \rho } \left( \frac{4\pi \rho (1+r)\varepsilon }{ b-a}\right) ^\frac{1}{2}= \left( \frac{(1+r)\varepsilon }{\pi \rho ( b-a)}\right) ^\frac{1}{2}. \end{aligned}$$

Hence,

$$\begin{aligned} \int _\Gamma k\, \mathrm{d}s&\le 2\pi (1+r\sqrt{2})+2\sqrt{2}\pi \rho \left( \frac{(1+r)\varepsilon }{\pi \rho ( b-a)}\right) ^\frac{1}{2}=2\pi \left( 1+r\sqrt{2}+\left( \frac{2(1+r)\varepsilon \rho }{\pi (b-a)}\right) ^\frac{1}{2}\right) \\&\le 2\pi \left( 1+r\sqrt{2}+\left( \frac{2\varepsilon b}{b-a}\right) ^\frac{1}{2}\right) . \end{aligned}$$

\(\square \)

Lemma 5.2

Let \(\mathcal {B}_R:=\mathcal {B}_R(x_0) \subset M\setminus \partial M\) and assume that

$$\begin{aligned} g(x)=\frac{1}{\sqrt{2}}|\nu (x)- e_3|\le r\,,\,\forall x\in \mathcal {B}_R, \end{aligned}$$

for some \(r\in \left[ 0,\frac{1}{\sqrt{2}}\right] \). Then, \(\mathcal {B}_R\) is locally graphical over the plane \(\{x_3=0\}\) with gradient bounded by \(3r\). Moreover, \(\mathcal {B}_R\) contains a graph of a function \(u\) over the disk in the plane \(\{x_3=0\}\) centered at \(\Pi (x_0)\) and of radius \(\rho =\frac{R}{\sqrt{1+(3r)^2}}\); where \(\Pi \) denotes the projection on the plane \(\{x_3=0\}\).

Proof

Since \(g(x)\le 1\) for all \(x\in \mathcal {B}_R\), we have that \(\mathcal {B}_R\) is locally a graph over the plane \(\{x_3=0\}\) and at each point \(x=(x_1,x_2, u(x_1, x_2))\in \mathcal {B}_R\), we have

$$\begin{aligned} \nu (x)=\left( -\frac{D_1u}{\sqrt{1+|Du|^2}}, -\frac{D_2u}{\sqrt{1+|Du|^2}},\frac{1}{\sqrt{1+|Du|^2}}\right) \end{aligned}$$
(30)

where \(\nu \) is the upward pointing unit normal. We estimate now \(|Du|^2=|D_1u|^2+|D_2u|^2\) using the estimate for \(g\) as follows: Note first that

$$\begin{aligned} g(x)&= \frac{1}{\sqrt{2}}|\nu (x)-e_3|=\frac{1}{\sqrt{2}}\left( \frac{|D_1u|^2}{1+|Du|^2}+\frac{|D_2 u|^2}{1+|Du|^2}+\left( 1-\frac{1}{\sqrt{1+|Du|^2}}\right) ^2\right) ^\frac{1}{2}\\&= \frac{1}{\sqrt{2}}\left( \frac{|Du|^2}{1+|Du|^2}+\frac{1}{1+|Du|^2}+1-2\frac{1}{\sqrt{1+|Du|^2}}\right) ^\frac{1}{2}\\&= \frac{1}{\sqrt{2}}\left( 2-2\frac{1}{\sqrt{1+|Du|^2}}\right) ^\frac{1}{2}=\left( 1-\frac{1}{\sqrt{1+|Du|^2}}\right) ^\frac{1}{2}. \end{aligned}$$

Hence, since \(g(x)\le r\), we get

$$\begin{aligned} g(x)=\left( 1-\frac{1}{\sqrt{1+|Du|^2}}\right) ^\frac{1}{2}&\le r\implies 1-\frac{1}{\sqrt{1+|Du|^2}}\le r^2\implies \\ \sqrt{1+|Du|^2}&\le \frac{1}{1-r^2}\le 1+2 r^2, \end{aligned}$$

with the last inequality being true since \(r\le \frac{1}{\sqrt{2}}\implies r^2-2 r^4\ge 0\). Squaring both sides, we obtain

$$\begin{aligned} 1+|Du|^2\le 1+4r^4+4 r^2\implies |Du|^2\le 9 r^2\implies |Du|\le 3 r. \end{aligned}$$

This finishes the proof of the first part of the lemma.

By the previous discussion, \(\mathcal {B}_R\) is a graph of a function \(u\) around the point \(x_0\). Let \(\rho \) be such that \(u\) is defined on the disk centered at \(\Pi (x_0)\) of radius \(\rho \) in the plane \(\{x_3=0\}\), \(D_\rho (\Pi (x_0))\). Without loss of generality, let \(x_0=0\). We will prove a lower estimate for the radius \(\rho \) of the disk where the function \(u\) is defined. To do this, let \(\rho \) be the maximum such radius. Then, there exists a point \((x_1,x_2)\in \partial D_\rho (0)\), for which \((x_1, x_2, u(x_1, x_2))\in \partial \mathcal {B}_R\), else \(u\) maps \(\partial D_\rho (0)\) in the interior of \(\mathcal {B}_R\) and since \(\mathcal {B}_R\) is locally a graph over the plane \(\{x_3=0\}\) we could increase \(\rho \). Let \(\gamma (t)\) be the path in \(\mathcal {B}_R\) defined by

$$\begin{aligned} \gamma :[0,1]\rightarrow \mathcal {B}_R, \quad \gamma (t)=(tx_1,tx_2,u(t(x_1,x_2))). \end{aligned}$$

The path \(\gamma \) joins \(0\), that is the center of \(\mathcal {B}_R\), with \(x\in \partial \mathcal {B}_R\), therefore it must have length at least \(R\), from which we get

$$\begin{aligned} R\le {{\mathrm{Length}}}(\gamma )=\int _0^1|\dot{\gamma }|\mathrm{d}t\le \int _0^1 \rho \sqrt{1+|Du|^2}\mathrm{d}t \le \rho \sqrt{1+(3r)^2} \end{aligned}$$

which implies that

$$\begin{aligned} \rho \ge \frac{R}{\sqrt{1+(3r)^2}} \end{aligned}$$

and this finishes the proof of the lemma. \(\square \)

Remark 5.3

Standard PDE theory implies that under the hypotheses of Lemma 5.2 and if in addition \(r\le \frac{1}{\sqrt{3}}\) and \(H\in L^p(\mathcal {B}_R)\), \(p>2\), we obtain \(C^{1,\alpha }\) estimates in \(\mathcal {B}_{\frac{R}{2}}\); namely, there exist constants \(\alpha \in (0,1)\) and \(C\), depending on \(r, R^{1-\frac{2}{p}}\int _{\mathcal {B}_R}|H|^p \mathrm{d}\mathcal {H}^2\), and \(p\), such that

$$\begin{aligned} \frac{|\nu (x)-\nu (y)|}{|x-y|^\alpha }\le R^{-\alpha }C,\quad \forall x,y\in \mathcal {B}_{R/2} \end{aligned}$$

To see why the above remark is true, we can assume without loss of generality that \(x_0=0\). Note that by Lemma 5.2, \(\mathcal {B}_R\) contains a graph of a function \(u\) over \(\Omega :=\{(x_1, x_2): x_1^2+x_2^2\le \rho ^2\}\) with \(|Du|\le 3r\) and with

$$\begin{aligned} \rho =\frac{R}{\sqrt{1+ (3r)^2}}\ge \frac{R}{\sqrt{10}}\ge \frac{R}{2}. \end{aligned}$$

Thus, \(\mathcal {B}_{\frac{R}{2}}\subset {{\mathrm{graph}}}u\). Furthermore, \(u\) satisfies the equation

$$\begin{aligned} \sum _{i=1}^2 D_i\left( \frac{D_iu(x)}{\sqrt{1+|Du(x)|^2}}\right) = H(x, u(x)). \end{aligned}$$

By differentiating the above equation, we obtain that \(w= D_ku\), for \(k=1,2\), is a solution to the equation

$$\begin{aligned} \sum _{i,j=1}^2D_i(a^{ij} D_jw)= D_k H \end{aligned}$$

(cf. [7, pages 319–320]) with

$$\begin{aligned} a^{ij}=\frac{\delta _{ij}}{\sqrt{1+|Du|^2}}- \frac{D_iu D_ju}{\sqrt{1+|Du|^2}}. \end{aligned}$$

Note that since \(|Du|\) is bounded, we have that \(H\in L^p(\mathcal {B}_R)\implies H\in L^p (\Omega )\). We can then apply Theorem 8.22 in [7] to obtain that

$$\begin{aligned} {\mathop {\mathrm{sup}}_{x, y\in \Omega }}\frac{|Du(x)- Du(y)|}{|x-y|^\alpha }\le \rho ^{-\alpha } C \end{aligned}$$

for some \(\alpha \in (0,1)\) and with \(C\) and \(\alpha \) depending on \(\mathrm{sup}_{\Omega }|Du|\), \(\rho ^{1-\frac{2}{p}}\Vert H\Vert _{L^p(\Omega )}\) and \(p\), namely on \(r\), \(R^{1-\frac{2}{p}}\Vert H\Vert _{L^p(\mathcal {B}_R)}\) and \(p\), and \(\rho \ge \frac{R}{2}\). Using the formula for \(\nu \) as in (30), we get the required estimate.

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Bourni, T., Tinaglia, G. \(C^{1,\alpha }\)-regularity for surfaces with \(H\in L^p\) . Ann Glob Anal Geom 46, 159–186 (2014). https://doi.org/10.1007/s10455-014-9417-1

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Keywords

  • Embedded Disk
  • Geodesic Ball
  • Gauss Bonnet Theorem
  • Mean Curvature
  • Versus Extrinsic