Let $B_1(0) := \{(x,y) \in \mathbb{R}^2: x^2 + y^2 \leq 1\}$ be the unit disc in $\mathbb{R}^2$. What is the infimum of all $k$ such that for all $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ which are (not necessarily continuously) differentiable and satisfy $\left| f(x,y) \right| \leq 1$ for all $(x,y) \in B_1(0)$, there necessarily exists a point $(x_0,y_0) \in B_1(0)$ with $|\nabla f(x_0,y_0)| \leq k$?
The backstory is that friends of mine had to prove that for every such function, there is a point $(x_0,y_0)$ with $|\nabla f(x_0,y_0)|\leq 4$, i.e. that $\inf\{k\} \leq 4$. You do this by considering $f(x,y)-2x^2-2y^2$, which has $|f(\partial B_1(0))|<-1$ and $f(0,0)>-1$, thus its maximum is at some $(x_0,y_0)$ inside $B_1(0)$. Therefore, $\nabla f(x_0,y_0) = 4(x_0,y_0)^T$, proving the claim.
Trading $2x^2+2y^2$ for a smooth approximations of $2|(x,y)^T|$ we can push this down all the way to $\inf\{k\}\leq 2$.
On the other hand, if we assume $\mathcal{C}^{1,1}$-regularity, a simple gradient flow argument yields $\inf\{k\}=1$, which is clearly optimal, as $f(x,y) = x$ shows.
This gap between the optimal bound for $\mathcal{C}^{1,1}$ regularity and the much weaker bound we could get for mere differentiable functions is interesting, since it implies that a possible counterexample would have to be really weird.
I just found out that the optimal bound holds also in the differentiable case.
Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a differentiable function such that $f(B_1(0)) \in [-1,1]$. Then $\inf\{|\nabla f(x,y)|: (x,y)\in B_1(0)\} \leq 1$.
Let $\epsilon > 0$. We will prove that there exist $(x_0,y_0)$ with $|\nabla f(x_0,y_0)|\leq 1+\epsilon$. For this, consider a smooth approximation $g$ to $|(x,y)^T|$ which satisfies $g(0,0,) = 0$, $g(\partial B_1(0)) = 1$ and $|\nabla g(x,y)| \leq 1+\epsilon$. If $f+g$ has a minimum at $(x_0,y_0)$ in $B_1(0)^\circ$, we are done, since there $|\nabla f(x_0,y_0)| = |-\nabla g(x_0,y_0)| \leq 1+\epsilon$ holds.
Suppose now that $f+g$ reaches its minimum on $\partial B_1(0)$. From $(f+g)(\partial B_1(0)) \in [0,2]$ we infer $(f+g)(0,0) = f(0,0) \geq 0$.
Since $(f-g)(\partial B_1(0)) \in [-2,0]$, and $(f-g)(0,0) \geq 0$, $f-g$ has to reach its maximum in $B_1(0)^\circ$ (either at $(0,0)$ if the maximum is $0$ or at some other point, if it's larger). Thus, there exists a point $(x_0,y_0)$ with $|\nabla f(x_0,y_0)| = |\nabla g(x_0,y_0)| \leq 1+\epsilon$.