A question from my Analysis list that I could not have any idea. Any help would be great. I don't want a complete solution, just a little hint, because I need to do at least one part alone.

Let $f: U \longrightarrow \mathbb{R}$ differentiable in open $U \subset \mathbb{R}^{m}$. Suppose that $df(a) \neq 0$ for $a \in U$ and unitary vector $u \in \mathbb{R}^{m}$ such that $df(a)u = \max \lbrace df(a)h\,|\,|h|=1 \rbrace$. If $v \in \mathbb{R}^{m}$ is such that $df(a)v=0$, show that $v$ is perpendicular to $u$.

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    $\begingroup$ You should ask the first question here, but delete the second one and ask that under separate cover. $\endgroup$ – zhw. Apr 16 '18 at 22:48

Small hints

The question has nothing to do with real anlysis: if $A \neq 0$ is a $1\times m$ matrix and $u$ such that: $$A u =\max_{|h|=1} A h$$ and if $A v=0$ then $\langle u,v \rangle =0$.

(What can be said of $A\frac{u+t v}{|u+tv|} $?)

  • $\begingroup$ $\frac{u+tv}{|u+tv|}$ is an unitary vector, so $A\frac{u+tv}{|u+tv|} = Au$ for $t \in \mathbb{R}$? $\endgroup$ – Lucas Corrêa Apr 16 '18 at 23:11
  • $\begingroup$ Not necessarily but as it is a unitary vector and by definition of $u$ you have $A \frac{u+tv}{|u+tv|} \leq Au$ i.e the function of $t$ as a maximum at $t=0$ $\endgroup$ – Delta-u Apr 16 '18 at 23:13
  • $\begingroup$ $\varphi(t) = A\frac{u+tv}{|u+tv|}$ and $\varphi'(0) = 0$? How does this help? $\endgroup$ – Lucas Corrêa Apr 16 '18 at 23:18
  • $\begingroup$ Exactly :-). You can then compute $\varphi'(t)$ to show that $\varphi'(0)= A \frac{v}{|u|}-A u \frac{\langle u, v\rangle}{|u|^\frac{3}{2}}$ and as $A v=0$ you obtain $\langle u, v\rangle=0$. $\endgroup$ – Delta-u Apr 16 '18 at 23:24

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