local minimum of convex function is necessarily global

Theorem 1.

A local minimum (resp. local maximum) of a convex function(resp. concave function) on aconvex subset of a topological vector space, is always a global extremum.

Proof.

Let $f:S\to ℝ$ be a convex functionon a convex set $S$ in a topological vector space.

Suppose $x$ is a local minimum for $f$;that is, there is an open neighborhood $U$ of $x$where $f(x)\le f(\xi )$ for all $\xi \in U$.We prove $f(x)\le f(y)$ for arbitrary $y\in S$.

Consider the convex combination $(1-t)x+ty$ for $0\le t\le 1$:

Since scalar multiplication and vector addition are, by definition,continuous in a topological vector space, the convex combination approaches$x$ as $t\to 0$. Therefore for small enough $t$,$(1-t)x+ty$ is in the neighborhood $U$.Then

$f(x)$$\le f\left(ty+(1-t)x\right)$for small $t>0$

$\le tf(y)+(1-t)f(x)$since $f$ is convex.

Rearranging , we have $f(x)\le f(y)$.

To show the analogous situation for a concave function $f$,the above reasoning can be applied after replacing $f$ with $-f$.∎