$\sup_{f} \inf_{z\in D} [f_x^2(z)+f_y^2(z)]$ for $|f|\leq1$ on a unit disk

Question

Let $f:\mathbb{R^2}\mapsto\mathbb{R}$ be continuous and have partial derivatives in $D=\{(x,y):x^2+y^2\leq1\}$, and let $\mathscr{H}$ the set of such functions for which $\sup_D |f|\leq1$.

Could someone calculate $$ \sup\limits_{f \in \mathscr{H}} \inf\limits_{z\in D} [f_x^2(z)+f_y^2(z)]\;? $$

Current work:

I have an rigorous proof that it is less than 16.

Proof as follows:

Define $F(x,y)=f(x,y)+2(x^2+y^2)$. We have $F(0,0)\leq1$ and $F|_{\partial D}\geq1$, which mean that minimun can be obtained in the interior. Therefore, we have an interior point $(x_0,y_0)$ s.t. $$F_x(x_0,y_0)=F_y(x_0,y_0)=0. $$ Then $f_x(x_0,y_0)=-4x_0,f_y(x_0,y_0)=-4y_0$ $$f_x^2(x_0,y_0)+f_y^2(x_0,y_0)=16(x_0^2+y_0^2)<16.$$

Q.E.D.

Intuition:

However, the upper bound may be too rough. We could always find some point where $f_x^2+f_y^2$ is small.

If there is some place slopes fast, then the restriction of $f$ means there is always somewhere plain.

Examples like $f(x,y)=x, y $ or $\displaystyle\frac{x+y}{\sqrt{2}}$ all have $f_x^2+f_y^2$ with 1.

Answer 1

The supremum is indeed equal to $1$.

Suppose that $F$ is continuously differentiable and let $\Gamma$ be a maximal gradient line of $F$ in the unit disk (that is, $\Gamma$ is tangent to $\nabla F$ at each point of $\Gamma$, and the endpoints of $\Gamma$ either lie on the boundary of the disk or are critical points of $F$). Let $|\Gamma|$ be the length of $\Gamma$. Since $F$ takes values in $[-1, 1]$, by the mean value theorem there is a point $(x_0,y_0)$ on $\Gamma$ such that $|\nabla F(x_0,y_0)| \le 2 / |\Gamma|$.

If $F$ has a critical point, then there is $(x_0, y_0)$ such that $|\nabla F(x_0, y_0)| = 0$. Otherwise, all gradient lines begin and end at the boundary, and therefore there must be a gradient line of length at least $2$. It follows that there always is a point $(x_0, y_0)$ such that $|\nabla F(x_0, y_0)| \le 1$, as desired.

(This question seems to be more appropriate for Math.SE, though).