Suppose that there exists $(a,b) \in S$ such that $f(x,y) \leq f(a,b)$ on $S$, then show that $(\Delta f) (a,b) \leq 0$

Asked Dec 09 '20 at 13:23

Active Dec 09 '20 at 13:38

Viewed 34 times

Let $S$ be open subset of $R^2$ and let $f$ be twice differentiable real valued function on $S$. Suppose that there exists $(a,b) \in S$ such that $f(x,y) \leq f(a,b)$ on $S$, then show that $(\Delta f) (a,b) \leq 0$ where $\Delta f = f_{xx} + f_{yy}$

I have proved a similar result for functions define on $R$. If $g$ is a function defined on an open interval $I$ such that $a\in I$ and $g(x) \leq g(a) $ on $I$, then $g''(a) \leq 0$

I proved the above using the fact that $g''(a) =\lim_{h\to 0} \frac{g(a+h) + g(a-h) - 2g(a)}{h^2}$ and by observing that $g(a+h) + g(a-h) - 2g(a) \leq 0$

Can I use this somehow to arrive at the required result?

edited Dec 09 '20 at 13:38

asked Dec 09 '20 at 13:23

So Lo

1,567

1

$\nabla f$ usually denotes the gradient of $f$. Do you mean $\Delta f = f_{xx} + f_{yy}$? – Martin R Dec 09 '20 at 13:27
corrected, thanks. – So Lo Dec 09 '20 at 13:38

Suppose that there exists $(a,b) \in S$ such that $f(x,y) \leq f(a,b)$ on $S$, then show that $(\Delta f) (a,b) \leq 0$

0 Answers0