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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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    Yes. Thats the conclusion.2012-10-16