# local minimum of convex function is necessarily global

###### Proof.

Let $f:S\to\mathbb{R}$ be a convex function on 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 $f(x)\leq f(\xi)$ for all $\xi\in U$. We prove $f(x)\leq f(y)$ for arbitrary $y\in S$.

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

 $\displaystyle f(x)$ $\displaystyle\leq f\bigl{(}ty+(1-t)x\bigr{)}$ for small $t>0$ $\displaystyle\leq t\,f(y)+(1-t)f(x)$ since $f$ is convex.

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

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

Title local minimum of convex function is necessarily global LocalMinimumOfConvexFunctionIsNecessarilyGlobal 2013-03-22 13:33:37 2013-03-22 13:33:37 stevecheng (10074) stevecheng (10074) 13 stevecheng (10074) Theorem msc 26B25 extremal value of convex/concave functions local maximum of concave function is necessarily global