# 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 $f:S\to \mathbb{R}$ 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 $f(x)\le f(\xi )$ 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 |