# Show that there is $(a,b)$ s.t. $f_x(a,b)^2+f_y(a,b)^2<4$

Let $f:D\to\mathbb{R}$ where $D:=\{(x,y):x^2+y^2\leq 1\}$. Suppose all partial derivatives of $f$ are continuous and $|f|\leq 1$. Show that there exists $(a,b)\in Int_D$ s.t.:

$$\left[\frac{\partial f}{\partial x}(a,b)\right]^2+\left[\frac{\partial f}{\partial y}(a,b)\right]^2<4$$

Here $Int_D$ means the interior of $D$, that is , the open unit disk.

I try to reduce to the one dimension case, that is: there exists $a\in(-\sqrt{1-y^2},\sqrt{1-y^2})$ s.t $\left[\frac{\partial f}{\partial x}(a,y)\right]^2\leq 2$.

Proof by contradiction Assume $||\nabla f||\geq 2$ for each point in $D$ and define a vector field $v$ on $D$ such that $$\frac{dx}{dt}=\nabla f$$ Consider the solution curve $\gamma(t)$ passing through the origin at time 0. Since $||\nabla f||>0$ for each point in $D$, the curve cannot terminate in the region $D$. Then the arc length of $\gamma(t)$ from $t=t_1<0$ to $t=t_2>0$ must be no less than 2 (the diameter of $D$), where $\gamma(t_1),\gamma(t_2)\in\partial D$ belongs to the boundary of $D$. Then we have $$f(\gamma(t_2))-f(\gamma(t_1))=\int_{\gamma(t_1)}^{\gamma(t_2)}df=\int_{\gamma(t_1)}^{\gamma(t_2)}\nabla f\cdot dx=\int_{\gamma(t_1)}^{\gamma(t_2)}||\nabla f||~dx\geq4$$ On the other hand, $$f(\gamma(t_2))-f(\gamma(t_1))\leq||f(\gamma(t_2))-f(\gamma(t_1))||\leq||f(\gamma(t_2))||+||f(\gamma(t_1))||\leq 2$$ which leads to contradiction. Hence there must be some point $x\in D$ such that $||\nabla f(x)||^2<4$.
Suppose $\|\nabla f(z)\|\ge2$ for all $z\in D$. Note that $f$ cannot attain its minimum in the interior of $D$, since $\nabla f(z)\ne0$ for all $z\in D$. Hence $\epsilon := \frac14 (f(0)+1)\in(0,\frac12]$. We will reach a contradiction by following the direction field $\nabla f(z)$ starting from $z_0=0$.
Proof 1: (Like Shuchang's argument.) Suppose we can solve the ODE system $\dot z(t)=\nabla f(z(t))$, $z(0)=0$, on some maximal time interval $[0,T)$, with either $T=\infty$ or $\|z(t)\|\to1$ as $t\to T$. Then for any $t\in[0,T)$, since $\nabla f(z)\dot z = \|\nabla f(z)\|^2$, $$2-4\epsilon \ge f(z(t))-f(0) = \int_0^t \|\nabla f(z(s))\|^2\,ds \ge 2\int_0^t \|\dot z(s)\|\,ds \ge 2\|z(t)\|,$$ whence $\|z(t)\|\le 1-2\epsilon<1$. We infer $T=\infty$. However, $$f(z(t))-f(0) = \int_0^t \|\nabla f(z(s))\|^2\,ds \ge 4t >2$$ for $t>\frac12$, which yields a contradiction.
Proof 2: We can avoid invoking ODE existence and continuation theory by using a discrete argument, choosing a sufficiently small $h>0$ and defining $$z_{j+1} = z_{j} + h\nabla f(z_{j})$$ for $j=0,1,\ldots$, as long as $z_{j}\in D$.
We will find that the hypotheses imply that $|z_j|\le 1-\epsilon$ for all $j>0$ but also $f(z_j)-f(0)\ge 2hj$, which contradicts the bound $|f|\le1$ when $j>1/h$.
The gradient is uniformly continuous in $D$, hence there exists $\delta>0$ such that $\|\hat z\|<\delta$ implies $$\| \nabla f(z+\hat z)-\nabla f(z)\| <\epsilon.$$ Let $h=\min(\frac12, \delta/M, \epsilon/M)$ where $M=\sup_D \|\nabla f(z)\|$. Note that as long as $\|z_j\|\le 1-\epsilon$ we have $$2h \le \| h\nabla f(z_j)\| = \|z_{j+1}-z_j\| \le hM\le \min(\delta, \epsilon),$$ hence $\|z_{j+1}\|\le1$. Now we can write $$f(z_{j+1})-f(z_j) = \int_0^1 \nabla f(z_j+s(z_{j+1}-z_j)) (z_{j+1}-z_j)\,ds = \nabla f(z_j)(z_{j+1}-z_j) + E_j,$$ where $$E_j = \int_0^1 (\nabla f(z_j+s(z_{j+1}-z_j))-\nabla f(z_j)) (z_{j+1}-z_j)\,ds.$$ This satisfies the bound $$| E_j |\le \epsilon \|z_{j+1}-z_j\|.$$
Consequently, since $\nabla f(z_j)(z_{j+1}-z_j)=\|\nabla f(z_j)\|\|z_{j+1}-z_j\|\ge 2\|z_{j+1}-z_{j}\|$, $$f(z_{j+1})-f(z_j) \ge \|z_{j+1}-z_j\| (2-\epsilon ) \ge 2h.$$ Summing, however (recall $z_0=0$), we infer that $$\|z_{j+1}\| \le \sum_{k=0}^{j} \|z_{k+1}-z_k\| \le \frac{ f(z_{j+1})-f(0)}{2-\epsilon } \le \frac{ 2- 4\epsilon}{2-\epsilon }\le 1-\epsilon.$$ Thus $\|z_j\|\le 1-\epsilon$ for all $j$. However, $$f(z_{j+1})-f(0) = \sum_{k=0}^{j} (f(z_{k+1})-f(z_k)) \ge 2h(j+1)$$ and this contradicts the bound $|f|\le 1$ for large $j$.