I have that $X$ is a normed vector spaces over $\mathbb{R}$, $U\subseteq X$ is an open set and $F:U\rightarrow \mathbb{R}$ is a differentiable function. Need to show that each local max/min is also a critical point. I figured one could somehow attempt to combine/input the critical point, $F'(a)=\mathbf{0}\hspace{1mm}\forall\hspace{1mm}a\in U$ with the local minimum $F(a)=min_{x\in V}F(x)\hspace{1mm}\forall\hspace{1mm}a\in U$ where $V$ is a neighbourhood of $a$ and then simply attempt to solve/simplify this. Any tips on how to approach this?
Show that each local max/min point is a critical point
normed-spacesreal-analysis
Best Answer
(I'm going to assume the derivative is Gateaux.)
Suppose $a \in U$ is a local extreme point. By replacing $F$ by $-F$ as necessary, we may assume without loss of generality that $a$ is a local maximum. Let $r > 0$ be such that $a$ is the maximum of $F$ on $B[a; r] \subseteq U$.
Let $\phi \in X^*$ be the Gateaux derivative of $F$. That is, for all $h \in X$, we have $$\lim_{t \to 0} \frac{F(a + th) - F(a)}{t} = \phi(h).$$ If $\phi \neq 0$, then there exists some $h \in X$ such that $\phi(h) \neq 0$. By replacing $h$ with $-h$ as necessary, we may assume $\phi(h) > 0$. It follows therefore that $$\lim_{t \to 0} \frac{F(a + th) - F(a)}{t} = \phi(h) > 0.$$ Using the $\varepsilon$-$\delta$ definition of continuity, there must exist some $\delta > 0$ corresponding to $\varepsilon = \phi(h)$. That is, $\delta$ satisfies \begin{align*} 0 < |t| < \delta &\implies \left|\frac{F(a + th) - F(a)}{t} - \phi(h)\right| < \phi(h) \\ &\implies \frac{F(a + th) - F(a)}{t} > 0 \\ &\implies F(a + th) > F(a). \end{align*} Choose $t$ small enough so that $a + th \in B[a; r]$, and we also get $F(a + th) \le F(a)$. This presents a contradiction; $\phi = 0$.