TITLE: How prove there exist $(a,b)$ such $f'^2_{x}(a,b)+f'^2_{y}(a,b)-4(a^2+b^2)=0$
QUESTION [6 upvotes]: Question:

let $D=\{(x,y):x^2+y^2<1\}$,and $f\in C^{1}(D)$,if
  $$|f(x,y)|\le 1 ,((x,y)\in D)$$
show that:
  $\exists (a,b)\in D,$$$f'^2_{x}(a,b)+f'^2_{y}(a,b)-4(a^2+b^2)=0$$

I only solve this problem:
let $D=\{(x,y):x^2+y^2\le 1\}$,and $f\in C^{1}(D)$,if
$$|f(x,y)|\le 1 ,((x,y)\in D)$$

show that:
  $\exists (a,b)\in D,$$$f'^2_{x}(a,b)+f'^2_{y}(a,b)-16(a^2+b^2)=0$$
proof: let $$g(x,y)=f(x,y)+2(x^2+y^2)$$
if $$x^2+y^2=2,\Longrightarrow |g(x,y)|\ge 1$$
  since $g(0,0)\le 1$
  so
  $g(x,y)$ is minimum when $x^2+y^2<1$
  so

$$g'_{x}(a)=0,g'_{y}(b)=0$$
so
$$f'^2_{x}(a,b)+f'^2_{y}(a,b)-16(a^2+b^2)=0$$
But for the coefficient  is 4(and D is also different) ,so I 
 let
$$g(x,y)=f(x,y)+(x^2+y^2)$$
if $x^2+y^2=1$. then
$$0\le g(x,y)\le 2$$
then I can't works

REPLY [3 votes]: If $f(0,0)\ge0$ then let $g(x,y)=x^2+y^2-f(x,y)$; otherwise, if 
$f(0,0)<0$ then let $g(x,y)=x^2+y^2+f(x,y)$. In all cases, $g(0,0)\le0$.
We show that $g$ has a minimal value. This solves the problem: if the minimum of $g$ is attained at $(a,b)$ (which must lie in the interior) then $g'_x(a,b)=g'_y(a,b)=0$, so $f'_x(a,b)=\pm 2a$ and $f'_y(a,b)=\pm 2b$.
Case 1: $g\ge0$ everywhere. The property $g(0,0)\le0$ enforces $g(0,0)=0$ and this is the minimum of $g$.
Case 2: $g$ attains a negative value somewhere, say $g(p,q)=-c$ with $(p,q)\in D$ and $c>0$. Chose a radius $0<r<1$ such that $p^2+q^2\le r^2$ and $r^2>1-c$.
On the compact set $x^2+y^2\le r^2$, the function $g$ has a minimum value $m$. Since the point $(p,q)$ is located in this set, we have $m\le g(p,q)=-c$.
By the definition of $m$, for $x^2+y^2\le r^2$ we have $g(x,y)\ge m$.
For $x^2+y^2>r^2$ we have
$g(x,y) \ge x^2+y^2-|f(x,y)| \ge r^2-1 > -c \ge m$. Hence, $m$ is the minimum of $g$ on the entire disc.