Suppose $A \subset \mathbb{R}^n$ is a compact set, and $f:A\rightarrow \mathbb{R}$ is a continuous function with a unique point of maximum $x_0$.
I want to understand what are the minimal assumptions on $f$ for which there exists an open set $U$ containing $x_0$, and some $\alpha>0$ such that:
- $\forall x \in U$, $f(x) > f(x_0) - \alpha$
- $\forall x \in A\setminus \bar{U}$, $f(x) < f(x_0) - \alpha$
For example, if we assume that the function $f$ is smooth, then we can do a Taylor's series expansion around $x_0$ to get:
$ f(x_0 + th) = f(x_0) + \frac{t^2}{2}h^T(\nabla^2f(x_0))h + o(t^2)$ which means that for form $t_0$ small enough, we have for all $t \leq t_0$,
$f(x_0 + th) - f(x_0) \leq -\lambda_0t^2/2$
I want to know whether we can obtain similar result without the assumption of the existence of the Hessian.
Under the condition that $x_0$ is the only maximum point (local and global).
Take any $\alpha$ such that $\max\{ \min f, ~ 0 \} < \alpha < f(x_0) .$
Now set $U:= \{ x \in A ~ | f(x) > f(x_0) - \alpha \}$ . Then $U$ and $V= A \setminus \bar{U}$ are two open set in $A$ satisfying $1,2$
P.S under our assumption we made here one can show that $$ V= \{ x \in A ~ | f(x) < f(x_0) - \alpha \} $$