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A proof about maximum point, critical point and differentiation

ianchenmu

Member
Feb 3, 2013
74
Let $E\subset\mathbb{R}^n$ and $f: E\rightarrow\mathbb{R}$ be a continuous function. Prove that if $a$ is a local maximum point for $f$, then either $f$ is differentiable at $x = a$ and $Df(a) = 0$ or $f$ is not differentiable at $x = a$. Deduce that if $f$ is differentiable on $E^o$, then a global maximum point of f is either a critical point of f or an element of $∂E$.





It's a little bit about optimization but stil analysis. Well I have no idea about this question and I think I need a proof. Thank you!
 
Last edited:

jakncoke

Active member
Jan 11, 2013
68
If a is the local maximum then

either f is differentiable at a, or its not.

if f is differentiable at a.

Then for a neighborhood $U$ containing a, f(x) = f(a) + Df(a)(x-a)

Since by def of local max, for any $\epsilon > 0$, |x-a|<$\epsilon$ implies
f(x) $\leq$ f(a).

f(a) + Df(a)|(x-a)| $\leq$ f(a) for a small neighborhood around a.

so Df(a)|x-a| $\leq $ 0, or Df(a) = 0.

If f is differentiable on the interior of a set $E$, and if there exists an open ball located in $E$ which contains the global max a, a type of local max, then i showed earlier Df(a) = 0 (thus a critical point). If there exists no such open ball which contains point a, it is in the closure of E.