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 $\left[\frac{\partial f}{\partial x}(a,y)\right]^2\leq 2$.
 A: 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 $||\nabla f(x)||^2<4$.
A: 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 $j$.
