Exercise 9.8

Question

Exercise 8: 
    Suppose that $f$ is a differentiable real function in an open set $E\subset\R n$, and that $f$ has a local maximum at a point $\v x\in E$. Prove that $f'(\v x)=\v0$.

analambanomenos · Accepted Answer

You can use Theorem 9.17 to express $f'$ as a sum of the partial derivatives and easily reduce the problem to the the single-variable case, Theorem 5.8. However, I thought I'd use the new
    definition of derivative (commonly called a Fr\'echet derivative, by the way) instead.

By Exercise 5, there is a $\v y\in\R n$ such that $f'(\v x)=\v y\cdot\v x$. Let $\v h=h\v y/|\v y|$, and take the limit in the definition of derivative as $h$ approaches 0 through positive
    numbers. Then we get 
    \begin{align*}
        0 &= \lim_{\v hightarrow \v 0}\frac{\big|f(\v x+\v h)-f(\v x)-f'(\v x)\v h\big|}{|\v h|} \
        &= \lim_{hightarrow 0+}\frac{\big|f(\v x+\v h)-f(\v x)-\v y\cdot(h\v y)/|\v y|\big|}{h|\v y|/|\v y|} \
        &= \lim_{hightarrow 0+}\frac{\big|f(\v x+\v h)-f(\v x)-h|\v y|\big|}{h} & \hbox{(since $\v y\cdot\v y=|\v y|^2$)} \
        &= \lim_{hightarrow 0+}\frac{f(\v x)-f(\v x+\v h)+h|\v y|}{h} & \hbox{(since $f(\v x+\v h)-f(\v x)$ and $-h|\v y|$ are $\le0$)} \
        &= |\v y| + \lim_{hightarrow 0+}\frac{f(\v x)-f(\v x+\v h)}{h}
    \end{align*}
    Since the term in the limit is non-negative for small $h$, this forces $|\v y|\le 0$, or $\v y=\v 0$. Hence $f'(\v x)=0$.

Exercise 9.8

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Exercise 9.8

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