TITLE: $C^{2} ( \mathbb{R}^{2}) $ function-proof of inequality
QUESTION [7 upvotes]: Let $f\in C^{2}( \mathbb{R}^{2} )$. Suppose that $\triangledown f=0
$ on a compact set $A\subseteq \mathbb{R}^{2}$. I want to prove that there is a strictly positive constant $\lambda > 0$ such that: $$\left | f\left ( x \right )-f\left ( y \right ) \right |\leq \lambda \left | x-y \right |^{2}$$ for all $x,y$ in $A$.
What I had in mind is that if the gradient of f is zero on a compact set $A$, then the function $f$ is constant and the inequality is obvious.
Any suggestions to solve this problem?

REPLY [5 votes]: One of the formulations of the Taylor theorem tells you that for any $x,y$ you can approximate:
$$
f(y) = f(x) + \nabla f(x) \cdot (y-x) + O_x(|x-y|^2)
$$
where the $O_x$ notation is taken close to $x$. In fact, the constant inside $O_x$ can be approximated by the norm of the second derivative of $f$ on $[x,y]$ (times $1/2$, actually, and it is Taylor again). In partiular, if you take $\lambda$ to be maximal value of $\frac{1}{2}|| f''||$ on the convex hull $\mathrm{conv} A$, and have in whole generality:
$$
|f(y) - f(x) - \nabla f(x) \cdot (y-x)| \leq \lambda \cdot|x-y|^2
$$
Now, the claim follows from $\nabla f(x) = 0$.
Sorry about being a little vague. I'm sure you can find an appropriate version of the Taylor's theorem in a notation you find convenient  with a little search.