Maximum of non differentiable function

Question

Suppose that $F:[a,b]\rightarrow ℝ$ is continous and $F(a)=F(b)$, and there exists $\epsilon \geq 0$ such that:

$\limsup\limits_{h\rightarrow0}\frac{F(x+h)-2F(x)+F(x-h)}{h^{2}} \geq \epsilon$

for all $x \in (a,b)$.

Show F cannot attain maximum at any $x \in (a,b)$.

Answer 1

Hints: (i) $F(x_h)-2F(X)+F(x+h)=(F(x-h)-F(x))+(F(x+h)-F(x))$.

(ii) If $a\le0$ and $b\le0$ then $a+b\le0$. So if $a+b>0$ then either $a>0$ or $b>0$.

Answer 2

Suppose that $c\in (a,b)$ is the maximum of $F$, there exists $h>0$ such that $F(c+h)+F(c-h)-2F(c)>h^2\epsilon>0$, but $F(c+h)+F(c-h)-2F(c)=F(c+h)-F(c)+F(c-h)-F(c)$,

since $c$ is the maximum, $F(c-h)-F(c)\leq 0, F(c+h)-F(c)\leq 0$, this implies that $F(c+h)-F(c)+F(c-h)-F(c)=F(c+h)+F(c-h)-2F(c)\leq 0$, contradiction.