1
$\begingroup$

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 $u$.

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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|} $?)

$\endgroup$
  • $\begingroup$ $\frac{u+tv}{|u+tv|}$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.