Let $f:\mathbb{R}^{m} \rightarrow \mathbb{R}$ be a continuous function such that all the directional derivatives exist for every point in $\mathbb{R}^{m}$. Suppose $\frac{\partial f}{\partial u}(u)>0$ for all $u \in S^{m-1}$. Prove there exists a point $a \in \mathbb{R}^{m}$ such that $\frac{\partial f}{\partial v}(a)=0$ for all $v \in \mathbb{R}^{m}$.
So far I've given that I'll need to use that a continuous function attains its maximum and minimum value in the unit ball, but I don't know how to go from there; any books referenced or tips would be welcome.
Best Answer
Your function, being continuous on the closed unit ball $B_1(0)=\{x\in R^m: \|x\|\le 1\}$, should attain its minimum there (the closed ball is compact). It does not attain the min on the boundary of the ball $S^{m-1}$ for, suppose it does attain its min at $a\in S^{m-1}$. Since at all those points the radial derivative $\partial f/\partial u>0$, $f(\lambda a)<f(a)$ for $\lambda<1$, sufficiently close to $1$. The min is thus achieved inside the unit ball. At that point, all directional derivatives have to be equal to zero (otherwise $f$ would decrease in some direction, a contradiction).