Le be an open ball of $\mathbb{R}^2$, :$\mathbb{R}^2→\mathbb{R}$ a function and $\Delta$ the Laplacian operator. $m_1$ is a local maximum of $f$ in $D$.
Recall that $\forall h \in \mathbb{R}^2, d^2_{m_1}(h,h)\leq 0$ where $d^2_{m_1}$ is the second-order differential of $f$ at $m_1$.
Show that if $\Delta f >0$ on , then $\forall m \in D$, $$f(m) < \sup_{m' \in \partial D} f(m').$$
I think I have found the answer.
Since $d^2_{m_1}f(h,h) \leq 0$, the Hessian matrix of $f$ is negative in a local interior maximum of $f$ in $D$. Therefore,
$$\frac{\partial^2 f}{\partial x^2} \leq 0 \text{ and } \frac{\partial^2 f}{\partial y^2} \leq 0$$
So at a local interior maximum: $$\Delta f \leq 0$$
More generally, for all interior maximum of $f$ in $D$, $$\Delta f \leq 0$$
Hence, if $\Delta f > 0$, then $f$ can only reach its maximum can't reach its maximum on an interior point of $D$.
Finally,
$$\forall m \in D, f(m) < \sup_{m \in \partial D} f(m')$$