A monotonicity approach to Pogorelov's Hessian estimates for Monge- Ampère equation

Yu Yuan; Yu Yuan

doi:10.3934/mine.2023037

Mathematics in Engineering

2023, Volume 5, Issue 2: 1-6. doi: 10.3934/mine.2023037

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A monotonicity approach to Pogorelov's Hessian estimates for Monge- Ampère equation

Yu Yuan ^,

Department of Mathematics, Box 354350, University of Washington, Seattle, WA 98195, USA

Received: 21 April 2022 Revised: 25 May 2022 Accepted: 25 May 2022 Published: 02 June 2022

We present an integral approach to Pogorelov's Hessian estimates for the Monge-Ampère equation, originally obtained via a pointwise argument.

Keywords:

Monge-Ampère equation

Citation: Yu Yuan. A monotonicity approach to Pogorelov's Hessian estimates for Monge- Ampère equation[J]. Mathematics in Engineering, 2023, 5(2): 1-6. doi: 10.3934/mine.2023037

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Abstract

We present an integral approach to Pogorelov's Hessian estimates for the Monge-Ampère equation, originally obtained via a pointwise argument.

Dedicated to Neil S. Trudinger on the occasion of his 80th birthday.

In this note, we present a mean value inequality approach to Pogorelov's Hessian estimates for the Monge-Ampère equation, derived via a pointwise argument ^[3].

Theorem 0.1. Let $u$ be a smooth convex solution to $\det D^{2}u = 1$ with $Du\left(0\right) = 0$ on $D_{\tau} = \left\{ x\in\mathbb{R}^{n}:x\cdot u_{x}\leq\tau^{2}\right\}$ $.$ Then

$\begin{equation} \left\vert D^{2}u\left( 0\right) \right\vert \leq\left[ 2\left\vert B_{1}\right\vert \frac{\tau^{n}}{\left\vert D_{\tau}\right\vert } \frac{\left\vert \partial D_{\tau}\right\vert }{\left\vert D_{\tau}\right\vert }\left\Vert Du\right\Vert _{L^{\infty}\left( D_{\tau}\right) }\right] ^{2n}. \end{equation}$

(0.1)

The Hessian estimates for the (dual) potential equation of minimal Lagrangian surfaces, including the two dimensional Monge-Ampère equation $\det D^{2}u = 1,$ obtained in recent years, originate in Trudinger's classic mean value inequality proof of the gradient estimates for the minimal hypersurface equation, by Bombieri-De Giorgi-Miranda ^[2].

0.1. Monotonicity on maximal surface

Taking the gradient of the both sides of the Monge-Ampère equation

$\begin{equation} \ln\det D^{2}u = 0, \end{equation}$

(0.2)

we have

$\begin{equation} { \sum\limits_{i, j = 1}^{n}} g^{ij}\partial_{ij}\left( x, Du\left( x\right) \right) = 0, \end{equation}$

(0.3)

where $\left(g^{ij}\right)$ is the inverse of the induced metric $g = \left(g_{ij}\right) = D^{2}u$ on the Lagrangian graph $M = \left(x, Du\left(x\right) \right) \subset\left(\mathbb{R}^{n}\times \mathbb{R}^{n}, 2dxdy\right)$ (for simplicity of notation, we drop the $2$ in $g = 2D^{2}u$ ). Because of (0.2) and (0.3), the Laplace-Beltrami operator of the metric $g$ also takes the non-divergence form $\bigtriangleup _{g} = { \sum\limits_{i, j = 1}^{n}} g^{ij}\partial_{ij}.$ Denote the extrinsic distance of the position vector $\left(x, Du\right)$ to the origin by

$z = \left( x_{1}, \cdots, x_{n}\right) \cdot Du = x\cdot u_{x}\overset{u_{x} \left( 0\right) = 0}{ = }x\cdot\left( u_{x}\left( x\right) -u_{x}\left( 0\right) \right) \overset{u\ \text{convex}}{\geq}0.$

Then

$\begin{gather*} \left\vert \nabla_{g}z\right\vert ^{2} = { \sum\limits_{i, j = 1}^{n}} g^{ij}\partial_{i}z\partial_{j}z = { \sum\limits_{i, j, k = 1}^{n}} g^{ij}\left( u_{i}+x_{k}u_{ki}\right) \left( u_{j}+x_{k}u_{kj}\right) \\ \overset{p}{ = } { \sum\limits_{i = 1}^{n}} g^{ii}\left( u_{i}^{2}+x_{i}^{2}u_{ii}^{2}+2x_{i}u_{i}u_{ii}\right) \geq4x\cdot u_{x}, \\ \bigtriangleup_{g}z = x\cdot\bigtriangleup_{g}u_{x}+u_{x}\cdot\bigtriangleup _{g}x+2\langle \nabla_{g}x, \nabla_{g}u_{x}\rangle _{g} = 2 { \sum\limits_{i, j, k = 1}^{n}} g^{ij}\partial_{i}x_{k}\partial_{j}u_{k}\\ \overset{p}{ = }2 { \sum\limits_{i = 1}^{n}} g^{ii}u_{ii} = 2n\leq\frac{n}{2}\frac{\left\vert \nabla_{g}z\right\vert ^{2}} {z}, \end{gather*}$

where at any fixed point $p,$ we assume that $D^{2}u$ is diagonalized, and we use (0.3) for $\bigtriangleup_{g}z.$ In terms of $s = \sqrt{z},$ we have

$\begin{equation} \left\vert \bigtriangledown_{g}s\right\vert \geq1\ \text{and\ }\bigtriangleup _{g}s\leq\left( n-1\right) \left\vert \bigtriangledown_{g}s\right\vert ^{2}/s. \end{equation}$

(0.4)

Following [2, p.392], set

$\psi\left( s\right) \overset{\chi = \chi_{\left[ 0, 1\right] }}{ = }\int _{s}^{\infty}t\chi\left( t/\rho\right) dt = \left\{ \begin{array} [c]{l} \frac{1}{2}\left( \rho^{2}-s^{2}\right) \ \ \ 0\leq s\leq\rho\\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s > \rho \end{array} \right. ,$

actually in the following, $\chi$ is taken as a nonnegative smooth approximation of the characteristic function of $(-\infty, 1)\subset\left(-\infty, \infty\right)$ with support in $(-\infty, 1).$ We have

$\begin{align*} \bigtriangleup_{g}\psi\left( s\right) & = \psi^{\prime}\bigtriangleup _{g}s+\psi^{\prime\prime}\left\vert \bigtriangledown_{g}s\right\vert ^{2}\\ & = -s\chi\left( s/\rho\right) \bigtriangleup_{g}s-\left[ \chi\left( s/\rho\right) +\frac{s}{\rho}\chi^{\prime}\left( s/\rho\right) \right] \left\vert \bigtriangledown_{g}s\right\vert ^{2}\\ & \geq-\left[ n\chi\left( s/\rho\right) +\frac{s}{\rho}\chi^{\prime }\left( s/\rho\right) \right] \left\vert \bigtriangledown_{g}s\right\vert ^{2}\\ & = \rho^{n+1}\frac{d}{d\rho}\left[ \rho^{-n}\chi\left( s/\rho\right) \right] \left\vert \bigtriangledown_{g}s\right\vert ^{2}, \end{align*}$

where we use (0.4) in the above inequality. Multiply both sides by any nonnegative superharmonic quantity $q:$ $q\geq0$ and $\bigtriangleup _{g}q\leq0,$ then integrate over the whole maximal surface $M,$ one has

$0\geq\int_{M}\psi\bigtriangleup_{g}qdv_{g} = \int_{M}q\bigtriangleup_{g}\psi dv_{g}\geq\rho^{n+1}\frac{d}{d\rho}\left[ \int_{M}q\rho^{-n}\chi\left( s/\rho\right) \left\vert \bigtriangledown_{g}s\right\vert ^{2}dv_{g}\right] .$

Note $1\leq\left\vert \bigtriangledown_{g}s\right\vert \overset{x\rightarrow 0}{\rightarrow }1$ by tedious asymptotic analysis and $dv_{g} = dx,$ after taking limit in the smooth approximation of the characteristic function, we obtain

$\begin{equation} \left\vert B_{1}\right\vert q\left( 0\right) \geq\tau^{-n}\int_{D_{\tau} }q\left\vert \bigtriangledown_{g}s\right\vert ^{2}dv_{g}\geq\tau^{-n} \int_{D_{\tau}}qdx. \end{equation}$

(0.5)

0.2. Superharmonic quantity

Lemma 0.1. Suppose $u$ is a smooth convex solution to $\det D^{2}u = 1.$ Then

$\begin{equation} \bigtriangleup_{g}\ln\det\left[ I+D^{2}u\left( x\right) \right] \geq \frac{1}{2n}\ \left\vert \bigtriangledown_{g}\ln\det\left[ I+D^{2}u\left( x\right) \right] \right\vert ^{2}, \end{equation}$

(0.6)

or equivalently for $q\left(x\right) = \left\{ \det\left[I+D^{2}u\left(x\right) \right] \right\} ^{\frac{-1}{2n}}$

$\begin{equation} \bigtriangleup_{g}q\leq0. \end{equation}$

(0.7)

To begin the proof of Lemma 0.1, we first denote $b\left(x\right) = \ln\det\left[I+D^{2}u\left(x\right) \right]$ and rewrite $\bigtriangleup_{g}b$ only in terms of the second and third order derivatives of $u,$ relying on the following equations for the first and second order derivatives of $u:$

$\begin{gather} 0 = \partial_{\alpha}\ln\det D^{2}u = { \sum\limits_{i, j = 1}^{n}} g^{ij}\partial_{ij}u_{\alpha}\overset{p}{ = } { \sum\limits_{i = 1}^{n}} g^{ii}u_{ii\alpha}, \end{gather}$

(0.8)

$\begin{gather} 0 = { \sum\limits_{i, j = 1}^{n}} \partial_{\beta}\left( g^{ij}\partial_{ij}u_{\alpha}\right) = { \sum\limits_{i, j = 1}^{n}} g^{ij}\partial_{ij}u_{\alpha\beta}- { \sum\limits_{i, j, k, l = 1}^{n}} g^{ik}\partial_{\beta}g_{kl}g^{lj}\partial_{ij}u_{\alpha}, \\ \bigtriangleup_{g}u_{\alpha\beta} = { \sum\limits_{i, j = 1}^{n}} g^{ij}\partial_{ij}u_{\alpha\beta}\overset{p}{ = } { \sum\limits_{k, l = 1}^{n}} g^{kk}g^{ll}u_{kl\alpha}u_{kl\beta}, \end{gather}$

(0.9)

where at any fixed point $p,$ we assume that $D^{2}u$ is diagonalized. The first and second derivatives of $b$ are

$\begin{align*} \partial_{\alpha}b & = { \sum\limits_{i, j = 1}^{n}} \left( I+g\right) ^{ij}u_{ij\alpha}\\ \partial_{\alpha\beta}b & = { \sum\limits_{i, j = 1}^{n}} \left( I+g\right) ^{ij}\partial_{\alpha\beta}u_{ij}- { \sum\limits_{i, j, k, l = 1}^{n}} \left( I+g\right) ^{ik}\partial_{\beta}\left( \delta_{kl}+g_{kl}\right) \left( I+g\right) ^{lj}u_{ij\alpha}\\ & \overset{p}{ = } { \sum\limits_{i = 1}^{n}} \left( 1+u_{ii}\right) ^{-1}\partial_{\alpha\beta}u_{ii}- { \sum\limits_{k, l = 1}^{n}} \left( 1+u_{kk}\right) ^{-1}\left( 1+u_{ll}\right) ^{-1}u_{kl\alpha }u_{kl\beta}, \end{align*}$

where $\left(\left(I+g\right) ^{ij}\right) = \left(I+g\right) ^{-1}.$ Coupled with (0.9), we arrive at

$\begin{align} \bigtriangleup_{g}b & = { \sum\limits_{\alpha, \beta = 1}^{n}} g^{\alpha\beta}\partial_{\alpha\beta}b\overset{p}{ = } { \sum\limits_{\alpha = 1}^{n}} g^{\alpha\alpha}\partial_{\alpha\alpha}b\\ & = { \sum\limits_{i = 1}^{n}} \left( 1+u_{ii}\right) ^{-1}\bigtriangleup_{g}u_{ii}- { \sum\limits_{\alpha, k, l = 1}^{n}} g^{\alpha\alpha}\left( 1+u_{kk}\right) ^{-1}\left( 1+u_{ll}\right) ^{-1}u_{kl\alpha}^{2}\\ & = { \sum\limits_{i, k, l = 1}^{n}} \left( 1+\lambda_{i}\right) ^{-1}g^{kk}g^{ll}u_{kli}^{2}- { \sum\limits_{\alpha, k, l = 1}^{n}} g^{\alpha\alpha}\left( 1+\lambda_{k}\right) ^{-1}\left( 1+\lambda _{l}\right) ^{-1}u_{kl\alpha}^{2}\\ & = { \sum\limits_{i, j, k = 1}^{n}} \left[ \left( 1+\lambda_{i}\right) ^{-1}g^{jj}g^{kk}u_{ijk}^{2}-\left( 1+\lambda_{i}\right) ^{-1}\left( 1+\lambda_{k}\right) ^{-1}g^{jj} u_{ijk}^{2}\right] \\ & = { \sum\limits_{i, j, k = 1}^{n}} \lambda_{i}\left( 1+\lambda_{i}\right) ^{-1}\left( 1+\lambda_{k}\right) ^{-1}g^{ii}g^{jj}g^{kk}u_{ijk}^{2} \\ & = { \sum\limits_{i, j, k = 1}^{n}} \lambda_{i}\left( 1+\lambda_{i}\right) ^{-1}\left( 1+\lambda_{k}\right) ^{-1}h_{ijk}^{2}, \end{align}$

(0.10)

where we denote (the second fundamental form) $\sqrt{g^{ii}g^{jj}g^{kk} }u_{ijk}$ by $h_{ijk}.$ Let $\mu_{i} = \frac{\lambda_{i}-1}{\lambda_{i}+1} \in\left(-1, 1\right),$ and regrouping those terms $h_{ijk}$ with three repeated indices, two repeated ones, and none, we have

$\begin{align*} \bigtriangleup_{g}b & = \frac{1}{4} { \sum\limits_{i, j, k = 1}^{n}} \left( 1+\mu_{i}\right) \left( 1-\mu_{k}\right) h_{ijk}^{2}\\ & = \left\{ \begin{array} [c]{c} \frac{1}{4}\sum\limits_{i}\left[ \left( 1-\mu_{i}^{2}\right) h_{iii}^{2} +\sum\limits_{j\neq i}\left( 3-\mu_{j}^{2}-2\mu_{i}\mu_{j}\right) h_{ijj} ^{2}\right] \\ +\frac{1}{2}\sum\limits_{i > j > k}\left( 3-\mu_{i}\mu_{j}-\mu_{j}\mu_{k}-\mu_{k}\mu _{i}\right) h_{ijk}^{2} \end{array} \right\} \geq0. \end{align*}$

Accordingly at $p,$ we have

$\begin{align*} \left\vert \bigtriangledown_{g}b\right\vert ^{2} & = { \sum\limits_{\alpha, \beta = 1}^{n}} g^{\alpha\beta}\partial_{\alpha}b\partial_{\beta}b\overset{p}{ = } { \sum\limits_{\alpha = 1}^{n}} g^{\alpha\alpha}\left[ { \sum\limits_{j = 1}^{n}} \left( 1+\lambda_{j}\right) ^{-1}u_{jj\alpha}\right] ^{2}\\ & = { \sum\limits_{\alpha = 1}^{n}} \left[ { \sum\limits_{j = 1}^{n}} \left( 1+\lambda_{j}\right) ^{-1}\lambda_{j}g^{jj}\sqrt{g^{\alpha\alpha} }u_{jj\alpha}\right] ^{2}\\ & = \frac{1}{4} { \sum\limits_{i = 1}^{n}} \left[ { \sum\limits_{j = 1}^{n}} \left( 1+\mu_{j}\right) h_{ijj}\right] ^{2} = \frac{1}{4} { \sum\limits_{i = 1}^{n}} \left[ { \sum\limits_{j = 1}^{n}} \left( 1-\mu_{j}\right) h_{ijj}\right] ^{2}, \end{align*}$

where the last equality follows from (0.8) or $\sum_{j = 1}^{n}h_{ijj} = 0,$ and the corresponding expressions with $\left(1+\mu_{j}\right)$ and $\left(1-\mu_{j}\right)$ for each $\mu_{i} < 0$ and $\mu_{i}\geq0$ respectively are used to justify the Jacobi inequality (0.6) in the following.

For each fixed $i,$ case $\mu_{i}\geq0$ :

$\begin{align*} \frac{1}{2n}\left( { \sum\limits_{j = 1}^{n}} \left( 1-\mu_{j}\right) h_{ijj}\right) ^{2} & \leq\frac{1}{2}\left( 1-\mu_{i}\right) ^{2}h_{iii}^{2}+ { \sum\limits_{j\neq i}} \frac{1}{2}\left( 1-\mu_{j}\right) ^{2}h_{ijj}^{2}\\ & \leq\left( 1+\mu_{i}\right) \left( 1-\mu_{i}\right) h_{iii}^{2}+ { \sum\limits_{j\neq i}} \left[ 1-\mu_{j}^{2}+2\left( 1-\mu_{i}\mu_{j}\right) \right] h_{ijj}^{2}, \end{align*}$

where in the last inequality we used

$\frac{1}{2}\left( 1-\mu_{j}\right) ^{2}\leq\left\{ \begin{array} [c]{l} 1-\mu_{j}^{2}\ \ \ \ \ \ \ \ \ \ \text{for }~~\mu_{j}\in\lbrack0, 1)\\ 2\left( 1-\mu_{i}\mu_{j}\right) \ \ \ \text{for }~~\mu_{j}\in(-1, 0)\ \text{and }~~\mu_{i}\geq0 \end{array} \right. ;$

case $\mu_{i}\in(-1, 0)$ : Symmetrically we have

$\frac{1}{2n}\left( { \sum\limits_{j = 1}^{n}} \left( 1+\mu_{j}\right) h_{ijj}\right) ^{2}\leq\left( 1+\mu_{i}\right) \left( 1-\mu_{i}\right) h_{iii}^{2}+ { \sum\limits_{j\neq i}} \left[ 1-\mu_{j}^{2}+2\left( 1-\mu_{i}\mu_{j}\right) \right] h_{ijj}^{2}.$

We have proved the Jacobi inequality (0.6) in Lemma 0.1.

0.3. Divergence of $\bigtriangleup u$

Plug in the superharmonic quantity from (0.7) to (0.5), we get

$\left\{ \det\left[ I+D^{2}u\left( 0\right) \right] \right\} ^{\frac {1}{2n}} = q^{-1}\left( 0\right) \leq\left\vert B_{1}\right\vert \tau^{n} \frac{1}{\int_{D_{\tau}}qdx}.$

From

$\left\vert D_{\tau}\right\vert ^{2} = \left( \int_{D_{\tau}}q^{1/2} q^{-1/2}dx\right) ^{2}\leq\int_{D_{\tau}}qdx\int_{D_{\tau}}q^{-1}dx,$

we have

$\frac{1}{\int_{D_{\tau}}qdx}\leq\frac{1}{\left\vert D_{\tau}\right\vert ^{2} }\int_{D_{\tau}}q^{-1}dx.$

Now

$\begin{gather} \int_{D_{\tau}}q^{-1}dx = \int_{D_{\tau}}\left[ \left( 1+\lambda_{1}\right) \cdots\left( 1+\lambda_{n}\right) \right] ^{\frac{1}{2n}}dx < \int_{D_{\tau} }\left( 1+\lambda_{\max}\right) dx \\ \overset{1\leq\lambda_{\max}}{\leq}\int_{D_{\tau}}2\lambda_{\max}dx\leq 2\int_{D_{\tau}}\bigtriangleup udx = 2\int_{\partial D_{\tau}}u_{\gamma} dA\leq2\left\vert \partial D_{\tau}\right\vert \left\Vert Du\right\Vert _{L^{\infty}\left( D_{\tau}\right) }. \end{gather}$

(0.11)

Therefore, we arrive at the claimed estimate in Theorem 0.1.

$\left\vert D^{2}u\left( 0\right) \right\vert < \det\left[ I+D^{2}u\left( 0\right) \right] \leq\left[ 2\left\vert B_{1}\right\vert \frac{\tau^{n} }{\left\vert D_{\tau}\right\vert }\frac{\left\vert \partial D_{\tau }\right\vert }{\left\vert D_{\tau}\right\vert }\left\Vert Du\right\Vert _{L^{\infty}\left( D_{\tau}\right) }\right] ^{2n}.$

Remark 0.1. Relying on a "rougher" superharmonic quantity $q = \lambda_{\max }^{-1/\left(n-1\right) }$ satisfying $\bigtriangleup_{g}q\leq0,$ repeat the above arguments, in particular, with $\left(1+\lambda_{\max}\right)$ in (0.11) replaced by $\lambda_{\max},$ we have a sharper estimate

$\begin{equation} \left\vert D^{2}u\left( 0\right) \right\vert = \lambda_{\max}\left( 0\right) \leq\left[ \left\vert B_{1}\right\vert \frac{\tau^{n}}{\left\vert D_{\tau}\right\vert }\frac{\left\vert \partial D_{\tau}\right\vert }{\left\vert D_{\tau}\right\vert }\left\Vert Du\right\Vert _{L^{\infty}\left( D_{\tau}\right) }\right] ^{n-1}. \end{equation}$

(0.12)

Remark 0.2. In addition to the conditions in Theorem 0.1, assuming $u\left(0\right) = 0,$ and the solution $u\left(x\right)$ exists on $\left\{ x\in\mathbb{R}^{n}:u\left(x\right) \leq\tau ^{2}\right\},$ then we have

$\Gamma_{\tau} = \left\{ x\in\mathbb{R}^{n}:u\left( x\right) \leq \varepsilon\left( n\right) \tau^{2}\right\} \subset D_{\tau} = \left\{ x\in\mathbb{R}^{n}:x\cdot u_{x}\leq\tau^{2}\right\} \subset\Gamma_{\tau /\sqrt{\varepsilon\left( n\right) }}$

for a small dimensional constant $\varepsilon\left(n\right),$ where the second inclusion follows from $0\leq u_{r} = \left(ru_{r}\right) _{r} -ru_{rr}\leq\left(ru_{r}\right) _{r}$ for the convex function $u;$ and the first inclusion follows from the fact that the gradient $Du$ is small at low enough level set of $u,$ which can be derived from the "separation" Corollary 1 in [, p.40], of lower level set of the convex solution $u$ from the boundary of the upper level set of $u,$ combined with the invariance of the "extrinsic distance" $x\cdot u_{x}\left(x\right)$ and the equation $\det D^{2}u\left(x\right) = 1$ under affine transform $v\left(x\right) = u\left(Ax\right)$ with $\det A = 1:$ $x\cdot v_{x}\left(x\right) = Ax\cdot u_{x}\left(Ax\right),$ $\det D^{2}v\left(x\right) = 1,$ and the invariance of the equation $\det D^{2}u\left(x\right)$ under scaling $v\left(x\right) = u\left(\tau x\right) /\tau^{2}:$ $\det D^{2}v\left(x\right) = 1.$

We claim

$\begin{equation} \left\vert D^{2}u\left( 0\right) \right\vert = \lambda_{\max}\left( 0\right) \leq\left[ C\left( n\right) \frac{\left\vert \partial\Gamma _{\tau}\right\vert }{\left\vert \Gamma_{\tau}\right\vert }\left\Vert Du\right\Vert _{L^{\infty}\left( \Gamma_{\tau}\right) }\right] ^{n-1}, \end{equation}$

(0.13)

or a weaker estimate

$\begin{equation} \left\vert D^{2}u\left( 0\right) \right\vert \leq\left[ C\left( n\right) \frac{\left\vert \partial\Gamma_{\tau}\right\vert }{\left\vert \Gamma_{\tau }\right\vert }\left\Vert Du\right\Vert _{L^{\infty}\left( \Gamma_{\tau }\right) }\right] ^{2n}. \end{equation}$

(0.14)

In fact, going with the sharper superharmonic quantity $q = \lambda _{\max}^{-1/\left(n-1\right) },$ (0.5) becomes

$\left\vert B_{1}\right\vert q\left( 0\right) \geq\tau^{-n}\int_{D_{\tau} }qdx\geq\tau^{-n}\int_{\Gamma_{\tau}}qdx.$

Repeating Step 0.3 Divergence of $\bigtriangleup u,$ with $D_{\tau}$ replaced by $\Gamma_{\tau},$ we have

$\left\vert D^{2}u\left( 0\right) \right\vert = \lambda_{\max}\left( 0\right) \leq\left[ \left\vert B_{1}\right\vert \frac{\tau^{n}}{\left\vert \Gamma_{\tau}\right\vert }\frac{\left\vert \partial\Gamma_{\tau}\right\vert }{\left\vert \Gamma_{\tau}\right\vert }\left\Vert Du\right\Vert _{L^{\infty }\left( \Gamma_{\tau}\right) }\right] ^{n-1}.$

By John's lemma, there exists an ellipsoid $E$ such that the convex set $\Gamma_{\tau}$ satisfies $E\subset\Gamma_{\tau}\subset nE.$ Alexandrov estimate and simple barrier argument combined with the equation $\det D^{2}u = 1$ on $\Gamma_{\tau}$ and $E$ respectively, lead to $c\left(n\right) \tau^{n}\leq\left\vert \Gamma_{\tau}\right\vert \leq C\left(n\right) \tau^{n}.$

Consequently, we arrive at the sharper Hessian estimate (0.13) in terms of the level set of solution $u.$

For the weaker Hessian estimate (0.14) in terms of the level set $u,$ just repeat the above argument with the weaker superharmonic quantity $q = \left\{ \det\left[I+D^{2}u\right] \right\} ^{-\frac{1}{2n}}.$

Acknowledgments

This work is partially supported by an NSF grant.

Conflict of interest

The author declares no conflict of interest.

References

[1]	L. A. Caffarelli, A priori estimates and the geometry of the Monge Ampère equation, In: Nonlinear partial differential equations in differential geometry, Providence, RI: Amer. Math. Soc., 1996, 5–63.
[2]	D. Gilbarg, N. S. Trudinger, Elliptic partial differential equations of second order, Berlin, Heidelberg: Springer, 2001. https://doi.org/10.1007/978-3-642-61798-0
[3]	A. V. Pogorelov, The Minkowski multidimensional problem, New York-Toronto-London: Halsted Press [John Wiley & Sons], 1978.

This article has been cited by:

1.	Julie Clutterbuck, Jiakun Liu, Preface to the Special Issue: Nonlinear PDEs and geometric analysis – Dedicated to Neil Trudinger on the occasion of his 80th birthday, 2023, 5, 2640-3501, 1, 10.3934/mine.2023095
2.	Ravi Shankar, Yu Yuan, Regularity for the Monge–Ampère equation by doubling, 2024, 307, 0025-5874, 10.1007/s00209-024-03508-6

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Mathematics in Engineering

A monotonicity approach to Pogorelov's Hessian estimates for Monge- Ampère equation

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Abstract

0.1. Monotonicity on maximal surface

0.2. Superharmonic quantity

0.3. Divergence of $\bigtriangleup u$

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通讯作者: 陈斌, bchen63@163.com

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Mathematics in Engineering

A monotonicity approach to Pogorelov's Hessian estimates for Monge- Ampère equation

Related Papers:

Abstract

0.1. Monotonicity on maximal surface

0.2. Superharmonic quantity

0.3. Divergence of △u \bigtriangleup u

Acknowledgments

Conflict of interest

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

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0.3. Divergence of $\bigtriangleup u$