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Seja $f: \mathbb{R}^n \to \mathbb{R}$ a continuous function that have all directional derivates in any point of $\mathbb{R^n}$. If $f'(u;u)>0$ exist for all $u \in S^{n-1}$, prove that exist $a \in \mathbb{R}^n$ such that $f'(a;v)=0$ for all $v \in \mathbb{R^n}$.

OBS: $f'(v;u)$ is directional derivate in $v$. Direction $u$.

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    Directional derivative at $u$ in the direction $u$? (Not sure if this right)2012-10-15

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Your function is continuous in $$B^{n}=\{x\in\mathbb{R}^{n}: \|x\|\leq 1\}$$

Hence $f$ must attain a maximum and a minimum in $B^{n}$. All we have to show is that there exist a maximum or a minimum in $int (B^{n})$, where $int$ is interior.

Indeed, $f$ cannot assume a minimum in a point $u\in S^{n-1}$ because by hypothesis, the function $f$ restricted to the set $\{tu:\ t\in\mathbb{R}\}$ is increasing in a neighbourhood of $u$. Then the minimum is attained in a point $u\in int(B{^{n}})$.

We can define a function $\phi: \mathbb{R} \to \mathbb{R}$ where $ \phi (t)=f(a+tv)$, whith $f(a)$ a "local" minimum point. As there is all directional derivates $\phi$ is derivable and $\phi(0)$ is minimum point. Then we have $\phi'(0)=0$ and $\phi(0)=f'(a;v)$.

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    where you using $f$ is continuous and $D_if$ exist?2012-10-16
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    Im using the continuity of $f$ when i say that $f$ assume a minimum or maximum in $B^{n}$ (Bolzano–Weierstrass theorem). Im using the partial derivatives in the gradient of $f$.2012-10-16
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    I changed differentiability by partial derivatives. For instance we dont know if the function is differentiable.2012-10-16
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    ok... another question... in the question we have conclude that exist $f'(a;v)=0$ FOR ALL $v \in \mathbb{R}^n$...in your argument we are considering the canonical basis $(0,..1,...,0)$ (partial derivates), thanks....2012-10-16
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    "If" $f'(a:v)$ exist then $$f'(a:v)=\sum_{i=1}^{n}D_{i}f(a)v_{i}$$ where $v=(v_{1},...,v_{n})$. Hence it is sufficient to prove only to the partial derivatives.2012-10-16
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    ok... but this equality is valid if only if $f$ is differentible2012-10-16
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    Yes. Note that i said "If $f'(a:v)$ exist". In fact, the postulate of your exercise is wrong. You have to put something like this: If $f'(a:v)$ exist then it is $0$.2012-10-16
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    ???????????? We don't use the equality because for instance we dont know if the function is differentiable.2012-10-16
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    Ok you are right, let me think a little more, then i will finish it.2012-10-16
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    Ok. I think you need more hipothesis. I found this same exercise here in a book of mine, but the author admits that the directional derivatives exist in every point. I think it is possible to construct an example by using partition of unity (for example) such that there is directions where the derivative is not zero.2012-10-16
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    you're right is not partial but directional. I'll fix..2012-10-16
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    Well ok, so i think your problem is solved. Ok?2012-10-16
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    We can define a function $\phi: \mathbb{R} \to \mathbb{R}$ where $ \phi (t)=f(a+tv)$, whith $f(a)$ the minimum point. As there is all directional derivates $\phi$ is derivable and $\phi(0)$ is minimum point. Then we have $\phi'(0)=0$ and $\phi(0)=f'(a;v)$. I think this concludes the question2012-10-16
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    Yes. Thats the conclusion.2012-10-16