The deformation of symplectic critical surfaces in a Kähler surface-II—compactness

Abstract

In this paper we consider the compactness of $\beta$-symplectic critical surfaces in a Kähler surface. Let M be a compact Kähler surface and $\Sigma _i\subset M$ be a sequence of closed $\beta _i$-symplectic critical surfaces with $\beta _i\rightarrow \beta _0\in (0,\infty )$. Suppose the quantity $\int _{\Sigma _i}\frac{1}{\cos ^q\alpha _i}d\mu _i$ (for some $q>4$) and the genus of $\Sigma _{i}$ are bounded, then there exists a finite set of points ${{\mathcal {S}}}\subset M$ and a subsequence $\Sigma _{i'}$ which converges uniformly in the $C^l$ topology (for any $l<\infty$) on compact subsets of $M\backslash {{\mathcal {S}}}$ to a $\beta _0$-symplectic critical surface $\Sigma \subset M$, each connected component of $\Sigma \setminus {{\mathcal {S}}}$ can be extended smoothly across ${{\mathcal {S}}}$.

Appendix: Monotonicity formula

In this appendix, we will prove the following monotonicity formula for submanifolds in a Riemannian manifold, which has been used in the proof of the main theorem. It is known to experts. For the convenience of the readers, we provide all the details here.

Theorem 4.2

Let $(M^n,\bar{g})$ be a Riemannian manifold with sectional curvature bounded by $K_0$ (i.e., $|K_M|\le K_0$) and injectivity radius bounded below by $i_0>0$. Suppose that $\Sigma ^k\subset M^n$ is a smooth submanifold and $x_0\in M$, if f is a nonnegative function on $\Sigma$, then for any $0<s_1<s_2< \min \{i_0,\frac{1}{\sqrt{K_0}}\}$,

$$\begin{aligned}&e^{k\sqrt{K_0}s_2}\frac{\int _{B_{s_2}(x_0)\cap \Sigma }fd\mu }{s_2^k}-e^{k\sqrt{K_0}s_1}\frac{\int _{B_{s_1}(x_0)\cap \Sigma }fd\mu }{s_1^k}\nonumber \\&\quad \ge \int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }e^{k\sqrt{K_0}r}\frac{|\nabla ^{\perp }r|^2}{r^k}f d\mu \nonumber \\&\qquad +\int _{s_1}^{s_2}e^{k\sqrt{K_0}s}\frac{1}{2s^{k+1}}\int _{B_s(x_0)\cap \Sigma } \mathbf{H (r^2)}fd\mu ds\nonumber \\&\qquad +\int _{s_1}^{s_2}e^{k\sqrt{K_0}s}\frac{1}{2s^{k+1}}\int _{B_s(x_0)\cap \Sigma } (s^2-r^2)\Delta fd\mu ds. \end{aligned}$$

(4.13)

In particular, by taking $f\equiv 1$, we have:

$$\begin{aligned} e^{k\sqrt{K_0}s_2}\frac{Area(B_{s_2}(x_0)\cap \Sigma )}{s_2^k}\ge & {} e^{k\sqrt{K_0}s_1}\frac{Area(B_{s_1}(x_0)\cap \Sigma )}{s_1^k}\nonumber \\&+\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }e^{k\sqrt{K_0}r}\frac{|\nabla ^{\perp }r|^2}{r^k}d\mu \nonumber \\&+\int _{s_1}^{s_2}e^{k\sqrt{K_0}s}\frac{1}{2s^{k+1}}\int _{B_s(x_0)\cap \Sigma } \mathbf{H (r^2)}d\mu ds. \end{aligned}$$

(4.14)

The proof of the monotonicity formula needs the following estimate, which is a consequence of the standard hessian comparison theorem (see Lemma 5.1 of [6]).

Lemma 4.3

Let $(M^n,\bar{g})$ be a complete n-dimensional Riemannian manifold with sectional curvature bounded by $K_0$ (i.e., $|K_M|\le K_0$) and injectivity radius bounded below by $i_0>0$. Then for $r<\min \{i_0,\frac{1}{\sqrt{K_0}}\}$ and any vector X with $|X|=1$,

$$\begin{aligned} \left| Hess_r(X,X)-\frac{1}{r}\langle X-\langle X,Dr\rangle Dr,X-\langle X,Dr\rangle Dr\rangle \right| \le \sqrt{K_0}, \end{aligned}$$

where r is the distance function from a fixed point on M.

Proof

Since $-K_0\le K_M\le K_0$, by Hessian Comparison Theorem, we have for any $|X|=1$, $X_1\in T\bar{M}(K_0)$, $X_2\in T\bar{M}(-K_0)$ with

$$\begin{aligned} |X_1|_1=|X_2|_2=|X|=1, \ \ \langle X,Dr\rangle _M=\langle X_1,Dr_1\rangle _{\bar{M}(K_0)}=\langle X_2,Dr_2\rangle _{\bar{M}(-K_0)}, \end{aligned}$$

we have

$$\begin{aligned} Hess^1_{r_1}(X_1,X_1)\le Hess_r(X,X)\le Hess^2_{r_2}(X_2,X_2). \end{aligned}$$

(4.15)

Here, $\bar{M}(K_0)$, $\bar{M}(K_0)$ denote the space form of constant curvatures $K_0$, $-K_0$, $r_1$ and $r_2$ are the distance functions on $\bar{M}(K_0)$ and $\bar{M}(-K_0)$, and $Hess^1$ and $Hess^2$ are the Hessians of $\bar{M}(K_0)$ and $\bar{M}(-K_0)$, respectively.

Recall that, if we set

$$\begin{aligned} f(r)=\left\{ \begin{array}{ll} \frac{\sin (\sqrt{K_0}r)}{\sqrt{K_0}},&{} \quad K_0>0, \quad \ r<\frac{\pi }{\sqrt{K_0}}, \\ r, &{}\quad K_0=0, \\ \frac{\sinh (\sqrt{-K_0}r)}{\sqrt{-K_0}},&{} \quad K_0<0, \end{array}\right. \end{aligned}$$

then the hessian of the distance function r on the space form $\bar{M}(K_0)$ is given by

$$\begin{aligned} Hess_r=\frac{f'(r)}{f(r)}(\bar{g}-dr\otimes dr). \end{aligned}$$

Therefore, we have

$$\begin{aligned} Hess^1_{r_1}(X_1,X_1)= & {} \frac{\sqrt{K_0}\cos (\sqrt{K_0}r)}{\sin (\sqrt{K_0}r)}(|X_1|_1^2-\langle X_1,Dr_1\rangle _{\bar{M}(K_0)}\langle X_1,Dr_1\rangle _{\bar{M}(K_0)})\\= & {} \frac{\sqrt{K_0}\cos (\sqrt{K_0}r)}{\sin (\sqrt{K_0}r)}(|X|^2-\langle X,Dr\rangle \langle X,Dr\rangle )\\= & {} \frac{\sqrt{K_0}\cos (\sqrt{K_0}r)}{\sin (\sqrt{K_0}r)}\langle X-\langle X,Dr\rangle Dr,X-\langle X,Dr\rangle Dr\rangle . \end{aligned}$$

Similarly, we have

$$\begin{aligned} Hess^2_{r_2}(X_2,X_2)=\frac{\sqrt{K_0}\cosh (\sqrt{K_0}r)}{\sinh (\sqrt{K_0}r)}\langle X-\langle X,Dr\rangle Dr,X-\langle X,Dr\rangle Dr\rangle . \end{aligned}$$

Then, by (4.15), we can easily see that

$$\begin{aligned}&\frac{\sqrt{K_0}r\cos (\sqrt{K_0}r)-\sin (\sqrt{K_0}r)}{\sqrt{K_0}r\sin (\sqrt{K_0}r)} \langle X-\langle X,Dr\rangle Dr,X-\langle X,Dr\rangle Dr\rangle \sqrt{K_0}\\&\quad \le Hess_r(X,X)-\frac{1}{r}\langle X-\langle X,Dr\rangle Dr,X-\langle X,Dr\rangle Dr\rangle \\&\quad \le \frac{\sqrt{K_0}r\cosh (\sqrt{K_0}r)- \sinh (\sqrt{K_0}r)}{\sqrt{K_0}r\sinh (\sqrt{K_0}r)} \langle X-\langle X,Dr\rangle Dr,X-\langle X,Dr\rangle Dr\rangle \sqrt{K_0}. \end{aligned}$$

Note that

$$\begin{aligned} 0\le \langle X-\langle X,Dr\rangle Dr,X-\langle X,Dr\rangle Dr\rangle =|X|^2-\langle X,Dr\rangle ^2\le 1. \end{aligned}$$

An elementary computation shows that

$$\begin{aligned} \frac{x\cosh x-\sinh x}{x\sinh x}\le 1, \ \ \forall x>0; \ \ \ \frac{x\cos x-\sin x}{x\sin x}\ge -1, \ \ \forall x\in (0,1]. \end{aligned}$$

Then the conclusion follows easily. $\square$

Proof of Theorem 4.2

In the proof, we will denote r the extrinsic distance from a fixed point $x_0$ on M, Dr the gradient of r with respect to the Rimammian metric $\bar{g}$. Then $Dr=\nabla r+\nabla ^{\perp }r$, where $\nabla r$ and $\nabla ^{\perp } r$ denote the projections of Dr on the tangent bundle $T\Sigma$ and normal bundle $N\Sigma$, respectively. Also, we denote D and $\nabla$ the Levi-Civita connection on $(M,\bar{g})$ and the induced connection on $(\Sigma ,g)$, respectively, where g denotes the induced metric on $\Sigma$ from $(M,\bar{g})$.

We choose local orthonormal frame $\{e_1,\ldots ,e_k,N_1,\ldots ,N_{n-k}\}$ so that: $\{e_1,\ldots ,e_k\}$ spans $T\Sigma$ and $\{N_1,\ldots ,N_{n-k}\}$ spans $N\Sigma$. Then for any $C^2$ function f on $\Sigma$, we have

$$\begin{aligned} \Delta f= & {} \sum _{i=1}^{k}[e_i(e_i(f))-(\nabla _{e_i}e_i) f]\\= & {} \sum _{i=1}^{k}[e_i(e_i(f))-(D_{e_i}e_i) f+\mathbf A (e_i,e_i)f]\\= & {} \sum _{i=1}^{k}Hess_f(e_i,e_i)+\mathbf H (f), \end{aligned}$$

where $\mathbf{A }$ denotes the second fundamental form of $\Sigma$ in M, $\mathbf H$ is the mean curvature vector of $\Sigma$ in M, and Hess is the Hessian of $(M,\bar{g})$. Therefore, we have

$$\begin{aligned} \Delta r^2= & {} \sum _{i=1}^{k}Hess_{r^2}(e_i,e_i)+\mathbf H (r^2)\\= & {} \sum _{i=1}^{k}[2rHess_r(e_i,e_i)+2(dr\otimes dr)(e_i,e_i)]+\mathbf H (r^2)\\= & {} \sum _{i=1}^{k}[2rHess_r(e_i,e_i)+2\langle e_i,Dr\rangle \langle e_i,Dr\rangle ]+\mathbf H (r^2)\\= & {} 2r\sum _{i=1}^{k}\left[ Hess_r(e_i,e_i)+\frac{1}{r}\langle e_i,Dr\rangle \langle e_i,Dr\rangle -\frac{1}{r}\langle e_i,e_i\rangle +\frac{1}{r}\right] +\mathbf H (r^2)\\= & {} 2r\sum _{i=1}^{k}\left[ Hess_r(e_i,e_i)-\frac{1}{r}\langle e_i-\langle e_i,Dr\rangle Dr,e_i-\langle e_i,Dr\rangle Dr\rangle \right] +2k+\mathbf H (r^2). \end{aligned}$$

Applying Lemma 4.3, we have

$$\begin{aligned} |\Delta r^2-2k-\mathbf H (r^2)|\le 2k\sqrt{K_0}r. \end{aligned}$$

(4.16)

Denote $B_s(x_0)$ the geodesic ball of radius s centered at $x_0$ on M. For $x_0\in M$,we see that the unit outer normal vector of $\partial B_s(x_0)$ is given by Dr and the unit outer normal $\partial B_s(x_0)\cap \Sigma$ is given by $\frac{\nabla r}{|\nabla r|}$.

The coarea formula gives us that

$$\begin{aligned} \int _{B_s(x_0)\cap \Sigma } fd\mu =\int _0^s\left( \int _{\partial B_{\tau }(x_0)\cap \Sigma }\frac{f}{|\nabla r|}d\sigma \right) d\tau , \end{aligned}$$

(4.17)

which implies that

$$\begin{aligned} \frac{d}{ds}\int _{B_s(x_0)\cap \Sigma } fd\mu =\int _{\partial B_{s}(x_0)\cap \Sigma }\frac{f}{|\nabla r|}d\sigma . \end{aligned}$$

(4.18)

On the other hand, by divergence theorem, we have

$$\begin{aligned} \int _{B_s(x_0)\cap \Sigma } f\Delta r^2d\mu= & {} \int _{\partial B_s(x_0)\cap \Sigma } f\frac{\partial r^2}{\partial \nu }d\mu -\int _{B_s(x_0)\cap \Sigma } \langle \nabla f,\nabla r^2\rangle d\mu \\= & {} \int _{\partial B_s(x_0)\cap \Sigma } 2rf\langle \nabla r,\frac{\nabla r}{|\nabla r|}\rangle d\mu -\int _{\partial B_s(x_0)\cap \Sigma } \frac{\partial f}{\partial \nu }r^2d\sigma \\&+\int _{B_s(x_0)\cap \Sigma }r^2\Delta fd\mu \\= & {} \int _{\partial B_s(x_0)\cap \Sigma } 2rf|\nabla r|d\mu -s^2\int _{B_s(x_0)\cap \Sigma } \Delta fd\mu +\int _{B_s(x_0)\cap \Sigma }r^2\Delta fd\mu \\= & {} 2s\int _{\partial B_s(x_0)\cap \Sigma } f|\nabla r|d\mu -\int _{B_s(x_0)\cap \Sigma } (s^2-r^2)\Delta fd\mu . \end{aligned}$$

Combining with (4.16) and (4.18), we get that

$$\begin{aligned} \frac{d}{ds}\left( \frac{\int _{B_s(x_0)\cap \Sigma }fd\mu }{s^k}\right)= & {} \frac{1}{s^{k+1}}\left( s\int _{\partial B_{s}(x_0)\cap \Sigma }\frac{f}{|\nabla r|}d\sigma -k\int _{B_s(x_0)\cap \Sigma }fd\mu \right) \\= & {} \frac{1}{s^{k+1}}\left( s\int _{\partial B_{s}(x_0)\cap \Sigma }\frac{f}{|\nabla r|}d\sigma -s\int _{\partial B_s(x_0)\cap \Sigma } f|\nabla r|d\mu \right. \\&\left. +\int _{B_s(x_0)\cap \Sigma } f\left( \frac{1}{2}\Delta r^2-k\right) d\mu +\frac{1}{2}\int _{B_s(x_0)\cap \Sigma } (s^2-r^2)\Delta fd\mu \right) \\\ge & {} \frac{1}{s^{k}}\int _{\partial B_{s}(x_0)\cap \Sigma }\frac{|\nabla ^{\perp }r|^2}{|\nabla r|}f d\sigma +\frac{1}{2s^{k+1}}\int _{B_s(x_0)\cap \Sigma } \mathbf{H (r^2)}fd\mu \\&-k\sqrt{K_0}\frac{\int _{B_s(x_0)\cap \Sigma }fd\mu }{s^k}+\frac{1}{2s^{k+1}}\int _{B_s(x_0)\cap \Sigma } (s^2-r^2)\Delta fd\mu , \end{aligned}$$

which implies that

$$\begin{aligned} \frac{d}{ds}\left( e^{k\sqrt{K_0}s}\frac{\int _{B_s(x_0)\cap \Sigma }fd\mu }{s^k}\right)\ge & {} e^{k\sqrt{K_0}s}\frac{1}{s^{k}}\int _{\partial B_{s}(x_0)\cap \Sigma }\frac{|\nabla ^{\perp }r|^2}{|\nabla r|}f d\sigma \\&+\,e^{k\sqrt{K_0}s}\frac{1}{2s^{k+1}}\int _{B_s(x_0)\cap \Sigma } \mathbf{H (r^2)}fd\mu \\&+\,e^{k\sqrt{K_0}s}\frac{1}{2s^{k+1}}\int _{B_s(x_0)\cap \Sigma } (s^2-r^2)\Delta fd\mu \\= & {} \frac{d}{ds} \left( \int _{B_{s}(x_0)\cap \Sigma }e^{k\sqrt{K_0}r}\frac{|\nabla ^{\perp }r|^2}{r^k}f d\sigma \right) \\&+\,e^{k\sqrt{K_0}s}\frac{1}{2s^{k+1}}\int _{B_s(x_0)\cap \Sigma } \mathbf{H (r^2)}fd\mu \\&+\,e^{k\sqrt{K_0}s}\frac{1}{2s^{k+1}}\int _{B_s(x_0)\cap \Sigma } (s^2-r^2)\Delta fd\mu . \end{aligned}$$

Integration from $s_1$ to $s_2$ gives (4.13). $\square$

The following corollary is a generalization of (1.3) in [15].

Corollary 4.4

Let $(M^n,\bar{g})$ be a closed Riemannian manifold with sectional curvature bounded by $K_0$ (i.e., $|K_M|\le K_0$) and injectivity radius bounded below by $i_0>0$. Suppose that $\Sigma ^2\subset M^n$ is a smooth surface and $x_0\in M$, then for any $0<s_1<s_2< \min \{i_0,\frac{1}{\sqrt{K_0}}\}$

$$\begin{aligned}&e^{4\sqrt{K_0}s_1}\left( \frac{Area(B_{s_1}(x_0)\cap \Sigma )}{s_1^2}+\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }\frac{|\nabla ^{\perp }r|^2}{r^2}d\mu \right) \nonumber \\&\quad \le 2e^{4\sqrt{K_0}s_2}\left( \frac{Area(B_{s_2}(x_0)\cap \Sigma )}{s_2^2}+\int _{B_{s_2}(x_0)\cap \Sigma }|\mathbf H |^2d\mu \right) . \end{aligned}$$

(4.19)

Proof

First note that (4.14) (with $k=2$) can be written as

$$\begin{aligned} e^{2\sqrt{K_0}s_1}\frac{Area(B_{s_1}(x_0)\cap \Sigma )}{s_1^2}\le & {} e^{2\sqrt{K_0}s_2}\frac{Area(B_{s_2}(x_0)\cap \Sigma )}{s_2^2}\nonumber \\&-\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }e^{2\sqrt{K_0}r}\frac{|\nabla ^{\perp }r|^2}{r^2}d\mu \nonumber \\&-\int _{s_1}^{s_2}e^{2\sqrt{K_0}s}\frac{1}{2s^{3}}\int _{B_s(x_0)\cap \Sigma } \mathbf{H (r^2)}d\mu ds\nonumber \\\le & {} e^{2\sqrt{K_0}s_2}\frac{Area(B_{s_2}(x_0)\cap \Sigma )}{s_2^2}\nonumber \\&-e^{2\sqrt{K_0}s_1}\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }\frac{|\nabla ^{\perp }r|^2}{r^2}d\mu \nonumber \\&-\int _{s_1}^{s_2}e^{2\sqrt{K_0}s}\frac{1}{2s^{3}}\int _{B_s(x_0)\cap \Sigma } \mathbf{H (r^2)}d\mu ds. \end{aligned}$$

(4.20)

We need to estimate the last term. By coarea formula and Fubini Theorem, we compute

$$\begin{aligned}&\int _{s_1}^{s_2}e^{2\sqrt{K_0}s}\frac{1}{2s^{3}}\int _{B_s(x_0)\cap \Sigma } {\mathbf{H }(r^2)}d\mu ds\\&\quad = \int _{s_1}^{s_2}\int _{0}^{s}e^{2\sqrt{K_0}s}\frac{1}{2s^{3}}\int _{(\partial B_{\tau }(x_0))\cap \Sigma } \frac{\mathbf{H }(r^2)}{|\nabla r|}d\sigma d\tau ds\\&\quad = \int ^{s_1}_{0}\int _{s_1}^{s_2}e^{2\sqrt{K_0}s}\frac{1}{2s^{3}}\int _{(\partial B_{\tau }(x_0))\cap \Sigma } \frac{\mathbf{H }(r^2)}{|\nabla r|}d\sigma ds d\tau \\&\qquad +\int _{s_1}^{s_2}\int _{\tau }^{s_2}e^{2\sqrt{K_0}s}\frac{1}{2s^{3}}\int _{(\partial B_{\tau }(x_0))\cap \Sigma } \frac{\mathbf{H }(r^2)}{|\nabla r|}d\sigma ds d\tau \\&\quad = \int _{s_1}^{s_2}e^{2\sqrt{K_0}s}\frac{1}{2s^{3}}ds\int _{B_{s_1}(x_0)\cap \Sigma } \mathbf{H }(r^2)d\mu \\&\qquad +\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }\left( \int _{r}^{s_2}e^{2\sqrt{K_0}s}\frac{1}{2s^{3}}ds\right) \mathbf{H }(r^2)d\mu . \end{aligned}$$

Since $\mathbf H (r^2)=2r\langle \mathbf H ,\nabla ^{\perp }r\rangle$, we have

$$\begin{aligned}&\left| -\int _{s_1}^{s_2}e^{2\sqrt{K_0}s}\frac{1}{2s^{3}}\int _{B_s(x_0)\cap \Sigma } \mathbf{H (r^2)}d\mu ds\right| \\&\quad \le e^{2\sqrt{K_0}s_2}\int _{s_1}^{s_2}\frac{1}{s^{3}}ds\int _{B_{s_1}(x_0)\cap \Sigma } r|\mathbf H ||\nabla ^{\perp }r|d\mu \\&\qquad +\,e^{2\sqrt{K_0}s_2}\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }\left( \int _{r}^{s_2}\frac{1}{s^{3}}ds\right) r|\mathbf H ||\nabla ^{\perp }r|d\mu \\&\quad = e^{2\sqrt{K_0}s_2}\frac{1}{2s_1^{2}}\int _{B_{s_1}(x_0)\cap \Sigma } r|\mathbf H ||\nabla ^{\perp }r|d\mu -e^{2\sqrt{K_0}s_2}\frac{1}{2s_2^{2}}\int _{B_{s_2}(x_0)\cap \Sigma } r|\mathbf H ||\nabla ^{\perp }r|d\mu \\&\qquad +\,\frac{1}{2}e^{2\sqrt{K_0}s_2}\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }\frac{1}{r}|\mathbf H ||\nabla ^{\perp }r|d\mu \\&\quad \le e^{2\sqrt{K_0}s_2}\frac{1}{2s_1}\int _{B_{s_1}(x_0)\cap \Sigma } |\mathbf H |d\mu +\frac{1}{2}e^{2\sqrt{K_0}s_2}\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }\frac{1}{r}|\mathbf H ||\nabla ^{\perp }r|d\mu \\&\quad \le \frac{1}{2}e^{2\sqrt{K_0}s_1}\frac{Area(B_{s_1}(x_0)\cap \Sigma )}{s_1^2}+\frac{1}{8}e^{4\sqrt{K_0}s_2-2\sqrt{K_0}s_1}\int _{B_{s_1}(x_0)\cap \Sigma } |\mathbf H |^2d\mu \\&\qquad +\frac{1}{2}e^{2\sqrt{K_0}s_1}\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }\frac{|\nabla ^{\perp }r|^2}{r^2}d\mu \\&\qquad +\frac{1}{8}e^{4\sqrt{K_0}s_2-2\sqrt{K_0}s_1}\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }|\mathbf H |^2d\mu \\&\quad = \frac{1}{2}e^{2\sqrt{K_0}s_1}\frac{Area(B_{s_1}(x_0)\cap \Sigma )}{s_1^2}+\frac{1}{2}e^{2\sqrt{K_0}s_1}\int _{(B_{s_2}(x_0)\backslash B_{s_1}(x_0))\cap \Sigma }\frac{|\nabla ^{\perp }r|^2}{r^2}d\mu \\&\qquad +\,\frac{1}{8}e^{4\sqrt{K_0}s_2-2\sqrt{K_0}s_1}\int _{B_{s_2}(x_0)\cap \Sigma }|\mathbf H |^2d\mu . \end{aligned}$$

Putting this inequality into (4.20) yields the desired estimate. $\square$