Vtyjkyuk

Let $f:mathbbR^ntomathbbR^n$, let $AsubseteqmathbbR^n$ be compact and suppose $nabla f=0$ on $A$. Show that there is a constant $B$ such that $|f(x)-f(z)|leq B|x-z|^2$ for all $x,zin A$.

My attempt: Cover $A$ by finitely many open balls $B_delta(x_1),...B_delta(x_q)$ of radius $delta$. Since open balls are convex, if $x,zin B_delta(x_j)cap A$ for some $j$, then $|f(x)-f(z)|=0$ by the mean value theorem. There is an $epsilon$ such that if $x,z$ are not in the same ball, then $|x-z|geqepsilon$. Then $|f(x)-f(z)|leq||f(A)||=||f(A)||frac1epsilon^2|x-z|^2$, where $|f(A)|$ is the diameter of $f(A)$. So setting $B=|f(A)|frac1epsilon^2$ gives the result.

Does this proof look correct?

No, I don't agree with your proof. You can cover $A$ by balls as you say, but in general $A$ will be properly contained in the balls. So although the ball $B_delta(x_j)$ is convex, the set $B_delta(x_j) cap A$ need not be, nor even connected. Hence you cannot conclude that $f$ is constant on $B_delta(x_j) cap A$.

In fact, I think this claim is false. Take $n=1$ and start with the function $f(x) = |x|^3/2$, which is differentiable at $x=0$ with $f'(0)=0$. Consider the sequence $x_n = 1/n$ and choose neighborhoods $U_n$ of the $x_n$ which are disjoint. Now modify $f$ inside each $U_n$ so that it is constant on a smaller neighborhood of $x_n$, without changing the value of $f(x_n) = |x_n|^3/2$. We can do this in such a way that $|f(x)| le 2|x|^3/2$ everywhere, and this will imply that $f$ is still differentiable at $x=0$ with $f'(0)=0$. Now $f'$ vanishes on the set $A = x_1, x_2, x_3, dots, 0$ which is compact, but $f(x) = |x|^3/2$ for every $x in A$, and this is not 2-Holder continuous.