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

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: