local minimum of convex function is necessarily global

Theorem 1.

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

Proof.

Let $f:S\to ℝ$ 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 $f(x)\le f(\xi )$ 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 $-f$. ∎

Title	local minimum of convex function is necessarily global
Canonical name	LocalMinimumOfConvexFunctionIsNecessarilyGlobal
Date of creation	2013-03-22 13:33:37
Last modified on	2013-03-22 13:33:37
Owner	stevecheng (10074)
Last modified by	stevecheng (10074)
Numerical id	13
Author	stevecheng (10074)
Entry type	Theorem
Classification	msc 26B25
Synonym	extremal value of convex/concave functions
Synonym	local maximum of concave function is necessarily global