I want to show that $u$ is a harmonic function in a domain $\Omega\subset \mathbb{R}^{n}$ if and only if $u$ is continuous in $\Omega$ and the following two conditions hold:
(1) if $u-\varphi$ has a local maximum at $x_{0}\in\Omega$ and $\varphi \in C^{2}(\Omega)$, then $\Delta \varphi(x_{0})\geq 0$ and
(2) if $u-\varphi$ has a local minimum at $x_{0}\in\Omega$ and $\varphi \in C^{2}(\Omega)$, then $\Delta \varphi(x_{0})\leq 0$.
To show that a harmonic function satisfies these conditions is easy to check, but I have problems showing the other implication. Can someone help?
Consider a non-harmonic function $u.$ There is a ball $B=B(z,r)$ such that $u(z)$ is not the average of $u$ on $\partial B.$ There's a harmonic function $\psi$ that is equal to $u$ on $\partial B$ (Dirichlet boundary conditions). Replacing $u$ by $u-\psi,$ we can assume $u(x)=0$ for $x\in\partial B,$ and replacing $u$ by $-u$ if necessary we can assume $u(z)>0.$ Pick a positive $\epsilon<u(z)/r^2$ and let $\varphi(x)=(r^2-|x-z|^2)\epsilon.$ Note that $u(x)-\varphi(x)=0$ for $x\in\partial B,$ but $u(z)-\varphi(z)>0.$ Let $x_0$ be a point in $B$ maximizing $u(x_0)-\phi(x_0).$ Then $x_0$ is a local maximum (because it's not in $\partial B$), which means condition (1) is violated for $u,x_0,\varphi.$