$F: B(0,1)\to R$ is differentiable, $|F|\leq 1$. show $\exists\ \xi\in B(0,1)$, $|\nabla F(\xi)|\leq 2$.
Here $B(0,1)$ is the unit ball in $\Bbb R^d$.
If I just use Lagrange intermediate value theorem on two points of the boundary, I could just deduce some $\xi$ exists, such that one directional deriative has absolute value $\leq 1$...What idea for the problem?
Suppose $\|\nabla F(x)\| > 2$ for all $x \in B(0,1)$. Then starting at the origin, flow in the direction of $\nabla F(x)$. That is, let ${\bf r}(t)$ be a curve such that $\|{\bf r}'(t)\| = 1$ and $\partial_t F({\bf r}(t)) = \|\nabla F({\bf r}(t))\|$.
Then ${\bf r}(t)$ is defined on $(-1,1)$ and $\partial_t (F({\bf r}(t))) > 2$. Thus the range of $F({\bf r}(t))$ is an interval of length greater than $2$, contradicting that $|F| \leq 1$ on $B(0,1)$.
