I'm stuck trying to prove the following equivalence :
let $V$ be normed vector space, $U$ an open subset of $V$, $C$ a convex subset of $U$, and $f : U \to \mathbb{R}$ a convex function on $C$.
then a local minima of $f$ on $C$ is a global minima of $f$ on $C$ and vice versa.
in other words : $$f(\alpha) = \min_{x\in C} f(x) \iff \alpha \in C \; \text{&} \; \exists \epsilon>0, \; f(\alpha) =\inf_{x \in C \cap \overline{B}(\alpha, \epsilon)} f(x)$$ here's my attempt at proving the 1st implication :
$\forall \epsilon > 0, \; C \cap \overline{B}(\alpha, \epsilon) \subset C$ and $\alpha \in C$
then $\forall \epsilon > 0 ,\; f(\alpha) \leq \inf_{x \in C \cap \overline{B}(\alpha, \epsilon)} f(x) $ which entails that $f(\alpha) = \inf_{x \in C \cap \overline{B}(\alpha, \epsilon)} f(x) $
concerning the second implication I don't even know where to begin.
any help would be appreciated. thanks !
A global minimum is always a local minimum, so the first implication is trivial.
The second part can be shown using a proof by contradiction.
Assume $C$ convex and $f$ convex on $C$, and that $x^*$ is a local but not global mininum of $f$ on $C$. Then there exists $\bar{x} \in C$ such that $f(\bar{x})<f(x^*)$. Let $\lambda\in[0,1].$
By convexity of $C$, $\lambda \bar{x}+(1-\lambda)x^*\in C$. By convexity of $f$, we have that
$f(\lambda \bar{x}+(1-\lambda)x^*)\leq\lambda f(\bar{x})+(1-\lambda)f(x^*)<\lambda f(x^*)+(1-\lambda)f(x^*)=f(x^*)$.
For small $\lambda$, this contradicts our assumption about $x^*$ being a local minimum.