## Improvement of Aleksandrov maximum principle

$\mathbf{Problem:}$ For $u\in C^2(\Omega)\cap C^0(\overline{\Omega})$ we have

$\displaystyle \sup\limits_{\Omega}u\leq \sup\limits_{\partial \Omega}u+\frac{d}{2w_n^{\frac{1}{n}}}\left(\int_{\Gamma^+}|det D^2u|\right)^{\frac{1}{n}}$

where $d=diam \Omega$.

There is an easy proof when the RHS is $\displaystyle \frac{d}{w_n^{\frac{1}{n}}}$. We will adjust the original proof.

$\mathbf{Proof:}$ By replacing $u$ with $\displaystyle u-\sup\limits_{\partial \Omega}u$, it suffices to assume $u\leq 0$ on $\partial \Omega$. Suppose $0\in\Omega$ and $u$ achieves maximum $M$ at 0.

Firstly, $\displaystyle |\chi(\Gamma^+)|\leq \int_{\Gamma^+}|det D^2u|$. Let $v$ be the function whose graph is the cone $K$ with vertex $(0,M)$ and base $\partial \Omega$. Since $diam(\Omega)=d$, there exists a ball $B_\frac{d}{2}(x_0)$ such that $\Omega\subset B_\frac{d}{2}(x_0)$. Let $\bar{v}$ be the function whose graph is cone $\bar{K}$ with vertex $(0,M)$ and base $B_\frac{d}{2}(x_0)$. Then

$\chi_{\bar{v}}(B_\frac{d}{2}(x_0))\subset \chi_{v}(\Omega)\subset \chi(\Omega)$

Define $d(y)$ is the length of the ray in direction $y$ tha lies in $B_\frac{d}{2}(x_0)$,

$\displaystyle \chi_{\bar{v}}(y)=\begin{cases}-\frac{M}{d(y)}\frac{y}{|y|}\quad y\neq 0, y\in B_\frac{d}{2}(x_0)\\ E\quad\quad\quad y=0\end{cases}$

$E$ is defined as the region bounded by $-\frac{M}{d(y)}\frac{y}{|y|}$, $y\in \partial B_\frac{d}{2}(x_0)$. This is because $p\in\mathbb{R}^n$ such that $\bar{v}(y)\leq \bar{v}(0)+p\cdot y$ means that

$\displaystyle M-\frac{M|y|}{d(y)}\leq M+p\cdot y\quad \forall\, y\in B_\frac{d}{2}(x_0)$

$\displaystyle -\frac{M|y|}{d(y)}\leq p\cdot y\quad \forall\, y\in B_\frac{d}{2}(x_0).$

Suppose a bounded region $G$ in $\mathbb{R}^n$ with boundary $(r(\omega), \omega)$ , $\omega\in(\mathbb{R}^{n-1})$,  then the volume of $\displaystyle G=\frac{1}{n}\int_{\mathbb{S}^{n-1}}r^n(\omega)d\sigma(\omega)$. With this fact in hand,

$\displaystyle |\chi_{\bar{v}}(B_\frac{d}{2}(x_0))|=\frac{1}{n}\int_{\mathbb{S}^{n-1}}\left(\frac{M}{d(\omega)}\right)^nd\sigma(\omega)$ and $\displaystyle |B_\frac{d}{2}(x_0)|=\frac{1}{n}\int_{\mathbb{S}^{n-1}}d(\omega)^nd\sigma(\omega)$

By the holder inequality

$\displaystyle |\chi(B_\frac{d}{2}(x_0))||B_\frac{d}{2}(x_0)|\geq \left(\frac{1}{n}\int_{\mathbb{S}^{n-1}}\sqrt{M^n}d\sigma(\omega)\right)^2=M|B_1(0)|^2$.

So $\displaystyle |\chi(B_\frac{d}{2}(x_0))|\geq w_n\left(\frac{2M}{d}\right)^n$

$\displaystyle M\leq \frac{d}{2w_n^{\frac{1}{n}}}\left(\int_{\Gamma^+}|det D^2u|\right)^{\frac{1}{n}}$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$Gilbarg Trudinger’s book chapter 9. 9.1