Second derivative test with non zero result

Question

Suppose function f(x,y) is of $C^2$, and we know that $f_{xx}f_{yy}-f_{xy}^2$ does not equal to 0. Furthermore, $f_{xx}+f_{yy} \ge 0$. Show that f does not have a strict local max.

So the second derivative test tells us that it is either a saddle point, min, or max since it does not equal to 0. And we know that $f_{xy}^2 \ge 0$. If $f_{xx}f_{yy} > f_{xy}^2 $, then we know $f_{xx}f_{yy}-f_{xy}^2 > 0$, then if $f_{xx}>0$, local min. But it can be a local max if $f_{xx}<0$.

If $f_{xx}f_{yy} < f_{xy}^2 $, then we know $f_{xx}f_{yy}-f_{xy}^2 < 0$, then saddle point.

How can I show $f_{xx}<0$ is not possible through $f_{xx}+f_{yy} \ge 0$?

Answer 1

The Hessian matrix of $f$ at $(0,0)$ has two real eigenvalues, $\lambda$ and $\mu$. You know that $\lambda\mu\neq0$ and that $\lambda+\mu\geqslant0$. Therefore, either $\lambda$ and $\mu$ have opposite signs, in which case $(0,0)$ is a saddle point, or $\lambda,\mu>0$, in which case $f$ has a local minmum at $(0,0)$. In either case, $(0,0)$ is not a local maximum.

Answer 2

Case 1: $$f_{xx}f_{yy}-f_{xy}^2>0 \Rightarrow \\ f_{xx}f_{yy}>f_{xy}^2\ge 0 \Rightarrow \\ f_{xx}f_{yy}>0 \Rightarrow \\ \{f_{xx}>0,f_{yy}>0\} \ \text{or} \ \{f_{xx}<0,f_{yy}<0\}$$ Since $f_{xx}+f_{yy}>0$, then: $f_{xx}>0,f_{yy}>0$. So, it is the local minimum point.

Case 2: $$f_{xx}f_{yy}-f_{xy}^2<0$$ It is a saddle point.