Let $ u\in C(B(0,1)) $ is a viscosity solution of Laplacian, that is for if for any point $ x_{0}\in \Omega $, any $ C^{2} $ function $ \phi $, such that $ \phi (x_{0})=u(x_{0}) $ and $ \phi \geq u $ in a neighborhood of $ x_{0} $, we have $ \Delta\phi(x_0)\geq 0 $ and any $ C^{2} $ function $ \psi $, such that $ \psi (x_{0})=u(x_{0}) $ and $ \psi \leq u $ in a neighborhood of $ x_{0} $, we have $ \Delta\psi(x_0)\leq 0 $. Is $ u $ actually a harmonic function in $ B(0,1) $?
Here is my try, I want to use the mean value property to prove that $ u $ is harmonic, but I can not use the information of the $ \Delta \phi(x_0) $ or $ \Delta \psi(x_0) $. Can you give me some hints or references?
Best Answer
Here's a sketch of the proof.
Let $v$ be the unique harmonic function on $B(0,1)$ satisfying $u=v$ on $\partial B(0,1)$. Then $v$ is smooth on $B(0,1)$ and you can use it as a test function in the definition of viscosity solution.
For $\delta>0$, define $\phi_\delta(x) = v(x) - \delta|x|^2 + \delta$. Show that $\Delta \phi_\delta < 0$, and so the viscosity property for $u$ implies that $u-\phi_\delta$ attains its maximum value over $B(0,1)$ on the boundary $\partial B(0,1)$, which implies that $u \leq \phi_\delta$ on $B(0,1)$. Send $\delta\to 0$ to get $u\leq v$. Repeat in the other direction to get that $u=v$.
If you don't want to assume $u$ is continuous up to the boundary, apply the argument above on any slightly smaller ball $B(0,r)$ for $r<1$.