Exercise on maximum principle for PDE

Question

Let $f \in\;\mathcal{C^3(U)\cup C(\bar U)}$ when $\mathcal U$ is an open and bounded subset of $\mathbb R^2$.
Assume that $\det\mathbf D^2f(x,y)\le 0\; for\; every\;(x,y) \in \mathcal U.$

Prove that $\max_{\bar U} f=\max_{\partial U} f$ and $ \min_{\bar U} f=\min_{\mathcal \partial U} f.$ By $\mathbf D^2f(x,y)$ we denote $\mathbf D^2f(x,y)\;=\;\begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \\ \end{pmatrix}$ where $\;f_{xx} \; f_{xy}\; f_{yx}\; f_{yy}$ are partial derivatives of $f$.

I recently started to study about the maximum principle for PDE with the guidance of my professor. The above is an exercise which is assigned to me. The exercise also comes with some hints, which are the following:

1. Proof by contradiction. Let $m=\max_{\partial U} f \lt \max_{\bar U} f=M=f(x_0,y_0)$. Then we denote $g(x,y)=M-\varepsilon [(x-x_0)^2 + (y-y_0)^2]\;\;for\;\varepsilon \gt 0 $ and $ φ(x,y)=f(x,y)-g(x,y)$. In addition we set $μ=\max φ=φ(x_1,y_1)$
2. Show that $g(x,y) \gt m$ at the boundary of $U$ for $\varepsilon \gt 0$ small enough and conclude that $g(x,y) \gt f(x,y)$ for every $(x,y) \in \partial U$
3. Finally, show that $\mathbf D^2f(x_1,y_1)$ is strictly negative definite.

I 've been stuck on step 3. It's easy to see $\mathbf D^2f(x_1,y_1)=\mathbf D^2 φ(x_1,y_1) +\mathbf D^2g(x_1,y_1)$ and furthermore that $\det \mathbf D^2g(x_1,y_1)=4\varepsilon^2\gt 0$.Although step 2 seems quite trivial , I cannot see the connection with step 3. Since $φ$ attains its maximum at $(x_1,y_1)$, can't we conclude that $\det \mathbf D^2 φ(x_1,y_1) \ge 0$? It can't be that simple, I feel like I'm missing something.

Could somebody make that clear for me or suggest another solution?

I 'm really new to these kind of exercises so I apologize in advance if the above questions are silly. I would appreciate any help!

Thanks a lot!!

Answer 1

I have to apologize for not beeing able to give you the whole answer. First I think you have to correct your statement to one of the following kind: If $\text{det}D^2f\geq 0$ on $U$ then the maximum of $f$ is not attained on $U$ but on the boundary.

I will only deal with the (most instructive) case $\text{det}D^2f>0$. For assume that the maximum is attained in the interior then $D^2f$ is surely negative semidefinite there contradicting $\text{det}D^2f>0$.

If you only have $\text{det}D^2f\geq 0$ then you have to perturb $f$ a bit as in Gilbarg Trudinger Thm. 3.1. If you want a more general statement than yours consult [GT] Thm 17.1.