# if $\Delta u \geq c$ for some $c>0$ then $u$ has a max on the boundary

Let $D=\{(x,y): \vert(x,y)\vert \leq 1\}$ and let $u:D\rightarrow \mathbb R$ be continuous function with three continous derivatives in the interior of $D$.

Show that if there is a number $c>0$ such that $\Delta u \geq c$ then $u$ attains its max on the boundary of $D$.

I know that this is somehow related to the maximum principals for harmonic functions.But I can't seem to prove this.

Here is what I thought of: Suppose $u$ has a max on the interior of $D$. Let $(x_0,y_0)$ be this point. Let $g(t)=u(x_0+t,y_0)$ and $g(t)=u(x_0,y_0+t)$. Is there anyway I can use the regular second derivative test to prove what I am looking for?

This is the maximum-principle for subharmonic functions (infact $c=0$ is true too). – AD. Jan 11 '13 at 7:28
Suppose $u$ has a max in the interior of $D$. A maximum is a critical point, so $\nabla u = 0$ there. We look at its Hessian matrix $\nabla^2 u$ at that point. The Hessian matrix is symmetric, so can be diagonalized. That $\lambda_1 > \lambda_2$ be the eigenvalues in decreasing order. Let $e_1$ be the unit eigenvector corresponding to $\lambda_1$. If $\lambda_1 > 0$, then applying the usual second derivative test to the function $f(t) = u((x_0,y_0) + te_1)$ shows that $f''(t) > 0$ and so $f(0)$ cannot be a local maximum. So we have that $\lambda_1 \leq 0$. This implies that the Hessian matrix of $u$ at the critical point must be negative semi-definite. Hence its trace $\leq 0$. But the trace of the Hessian is just $\triangle u$, which by assumption is strictly positive, so we reached a contradiction.