Subharmonic function equivalent non-negative laplacian

Question

I want to ask for a proof that if $v(x,y)$ is $C^2$ and is subharmonic [here, define as satisfyingthen $\Delta v \geq 0$ where $\Delta v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2}$ is the Laplacian operator. I think it has something to do with geometric interpretation of $\Delta v$ but I fail to see what kind of interpretation it can be.

Answer 1

We will show the more general characterization of subharmonic function in $\mathbb{R}^{n}$ :

Let $u\in\mathcal{C}^{2}\left(\Omega;\mathbb{R}\right)$ where $\Omega\subset\mathbb{R}^{n}$ is smooth open and bounded. Then the following assertions are equivalent :

1. $u$ is subharmonic in $\Omega$,
2. $u$ satisfies $$u(x)\leq \frac{1}{|\partial B_x(r)|}\int_{\partial B_x(r)}u(y)\mathrm{d}\sigma(y)\quad\quad\forall x\in\Omega,\forall r>0;B_x(r)\Subset\Omega,$$
3. $\Delta u(x)\geq0\quad\quad\forall x\in\Omega$.

Definitions and prerequisite :

$B_x(r)$ is the ball $\{z\in\mathbb{R}^{n}\mid|z-x|<r\}$ and $B_x(r)\Subset\Omega$ means that $\bar{B}_x(r)\subset\Omega$. $\mathrm{d}\sigma$ is the "surface" measure.

A subharmonic function is a function $u\in\mathcal{C}^{0}\left(\Omega;\mathbb{R}\right)$ such that for all $B_x(r)\Subset\Omega$ and for all harmonic function $\psi$ in $B_x(r)$, we have $$\left(u\leq\psi\text{ on $\partial B_x(r)$}\right)\quad\Longrightarrow\quad\left(u\leq\psi\text{ in $B_x(r)$}\right).$$ Sometimes, $u$ is asked to be upper semi-continuous ; we will not require it here.

An important result to know is the

Theorem : For all $\varphi\in\mathcal{C}^{0}\left(\bar{B}_x(r);\mathbb{R}\right)$, there is one and only one solution to the Laplace equation with Dirichlet's conditions on the boundary \begin{cases} \Delta u=0 & \text{in $B_x(r)$} \\ u=\varphi & \text{on $\partial B_x(r)$} \end{cases}

See Evans - "Partial differential equations" for a proof.

The answer :

We show $1.\Longrightarrow 2.$ Let $\bar{u}$ the harmonic extension of $u$ in $B_x(r)\Subset\Omega$, that is \begin{cases} \Delta \bar{u}=0 & \text{in $B_x(r)$} \\ \bar{u}=u & \text{on $\partial B_x(r)$} \end{cases} We assume known the fact that any harmonic function satisfies the mean value property. Then $$u(x)\leq\bar{u}(x)=\frac{1}{|\partial B_x(r)|}\int_{\partial B_x(r)}\bar{u}(y)\mathrm{d}\sigma(y)=\frac{1}{|\partial B_x(r)|}\int_{\partial B_x(r)}u(y)\mathrm{d}\sigma(y)$$ because $\bar{u}=u$ on $\partial B_x(r)$.

We show $2.\Longrightarrow 1.$ Assume $\psi$ is harmonic in $B_x(r)\Subset\Omega$ and is such that $u\leq\psi$ on $\partial B_x(r)$. We must show that $u\leq\psi$ holds everywhere in $B_x(r)$. To see this, we consider the set $$A=\{z\in B_x(r)\mid (u-\psi)(z)=M\}$$ where $M:=\max_{\bar{B}_x(r)}(u-\psi)$. $A$ is close because $u-\psi$ is continuous, and if $A$ where not open, then there would be an element $z_0\in A$ such that $$B_{z_0}(R)\nsubseteq A\quad\quad\forall R>0.$$ But from $2.$ it follows that $$M=(u-\psi)(y_0)\leq\frac{1}{|\partial B_{y_0}(R)|}\int_{\partial B_{y_0}(R)}(u-\psi)\mathrm{d}\sigma\leq M$$ with equality iff $(u-\psi)=M$ on $\partial B_{y_0}(R)$. This holds for any $R>0$ whence $(u-\psi)=M$ on a ball $B_{y_0}(R_0)$. Hence, $A$ is open, a contradiction. Now there are two cases : if $A\neq\emptyset$, then by connexity of $B_x(r)$, we have $u-\psi=M$ everywhere, hence $u=\psi+M\leq\psi$ in $B_x(r)$ because of the boundary condition $u\le\psi$ (i.e. $M\leq0$); now if $A=\emptyset$, then $\max_{\bar{B}_x(r)}(u-\psi)$ is realized on $\partial B_x(r)$, whence $M\leq0$ because of the boundary condition $u\leq\psi$, and then $u-\psi\leq M\leq0$ in $B_x(r)$. Whatever the situation is, we do end with the conclusion that $u$ is subharmonic.

We show $3.\Longrightarrow 2.$ Let us define $$\mu(r):=\frac{1}{|\partial B_x(r)|}\int_{\partial B_x(r)}u(y)\mathrm{d}\sigma(y).$$ Let us compute \begin{align} \frac{\mathrm{d}}{\mathrm{d}r}\mu(r) & = \frac{\mathrm{d}}{\mathrm{d}r}\frac{1}{|\partial B_x(r)|}\int_{\partial B_x(r)}u(y)\mathrm{d}\sigma(y) \\ & = \frac{\mathrm{d}}{\mathrm{d}r}\frac{1}{|\partial B_0(1)|}\int_{\partial B_0(1)}u(x+ry)\mathrm{d}\sigma(y) \\ & = \frac{1}{|\partial B_0(1)|}\int_{\partial B_0(1)}\nabla u(x+ry)\cdot y\mathrm{d}\sigma(y) \\ & = \frac{1}{|\partial B_x(r)|}\int_{\partial B_x(r)}\nabla u(y)\cdot \frac{y-x}{r}\mathrm{d}\sigma(y). \end{align} Now we remark that $(y-x)/r=\nu$ is the unitary outward normal vector of $\partial B_x(r)$. Then, applying Green's theorem, we find \begin{align} \frac{\mathrm{d}}{\mathrm{d}r}\mu(r) & = \frac{1}{|\partial B_x(r)|}\int_{\partial B_x(r)}\frac{\partial u}{\partial\nu}(y)\mathrm{d}\sigma(y) \\ & = \frac{1}{|\partial B_x(r)|}\int_{B_x(r)}\Delta u(y)\mathrm{d}y \\ & \geq0 \end{align} whence $\mu$ is non-decreasing. Next, we establish that $\lim\mu(r)=u(x)$ as $r\to0$ : indeed, as $u$ is continuous on $\partial B_x(r)\subsetneq\Omega$, we have for $r$ very small and $y\in\partial B_x(r)$ $$|u(y)-u(x)|<\varepsilon$$ whence $$|\mu(r)-u(x)|\leq\frac{1}{|\partial B_x(r)|}\int_{\partial B_x(r)}|u(y)-u(x)|\mathrm{d}\sigma(y)<\frac{1}{|\partial B_x(r)|}\int_{\partial B_x(r)}\varepsilon\mathrm{d}\sigma(y)=\varepsilon.$$ Therefore, we have $$\frac{1}{|\partial B_x(r)|}\int_{\partial B_x(r)}u(y)\mathrm{d}\sigma(y)=\mu(r)\geq\lim_{r\to0}\mu(r)=u(x)$$ and $2.$ follows.

We show $2.\Longrightarrow 3.$ Assume there is $x_0\in B_x(r)\Subset\Omega$ such that $\Delta u(x_0)<0$. Then if $u\in\mathcal{C}^{2}\left(\Omega;\mathbb{R}\right)$, $\Delta u$ is continuous in $\Omega$ and then there is a radius $r_0\in\left(0,\mathrm{dist}\left(x_0,\partial B_x(r)\right)\right)$ such that $\Delta u<0$ in $B_{x_0}(R)$ for all $R\in\left(0,r_0\right)$. But then, we compute as above that the function $\mu$ is decreasing and hence $$u(x_0)>\frac{1}{|\partial B_{x_0}(r)|}\int_{\partial B_{x_0}(r)}u(y)\mathrm{d}\sigma(y)$$ which shows that $u$ can not be subharmonic in $\Omega$.