$C^{1,\alpha }$-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).

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.