Let $f:\mathbb{R}^m\rightarrow\mathbb{R}$ be continuous such that all the directional derivatives exist in $\mathbb{R}^m$. If $\frac{\partial f}{\partial u}(u)>0$ for every $u\in S^{m-1}$, then there exists $a\in \mathbb{R}^m$ such that $\frac{\partial f}{\partial v}(a)=0\,\,\forall v.$
Tried using MVT for the straight lines through zero, but we don't have that the partial derivatives are continuous… Any tips?
Best Answer
Denote by $m$ the least value of (a continuous) $f$ on the (compact) unit ball $B^m$. $m$ is attained somewhere on $B^m$.
Suppose to the contrary that there is $u \in S^{m - 1}$ such that $f(u) = m$. From the assumption it follows that $f(u - h u) < m$ for $h > 0$ sufficiently small. But $u - h u \in B^m$, which contradicts the choice of $m$.
Therefore, there is $a \in B^m \setminus S^{m - 1}$ such that $f(a) = m$. So $f$ has a (local) minimum at $a$. All directional derivatives of $f$ at $a$ must thus be zero.