Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
The argument has to be more involved than what you write. Think about $A=\{0\}\cup\{1/n;n\in\mathbb N\}$ in dimension $1$. – Did Mar 2 '12 at 6:31
up vote 4 down vote accepted

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.

share|cite|improve this answer

Part of the problem here is that the set $A$ need not have any `interior' points. Back in the one dimensional case, think of a quadratic polynomial on the real line- generally speaking $A$ will be a two point set containing the $x$-values of the local max and min.

Anyway it seems to me that if $f$ is continuous, then, in particular $f$ is continuous on $A$ and therefore achieves a max and a min, so if $x$ and $y$ are not close together then your inequality is obvious. Now what happens if $x$ and $y$ are close together?

Update: Basically you can just use Feanor's suggestion for the case when $x$ is close to $y$. Here is one way you might think of it: for any $x_0 \in A$, Taylor's theorem tells us that (since $\nabla f(x_0) = 0$) $$f(x) = f(x_0) + h^2g(x)$$ for some function $g(x)$ that is bounded in a small neighborhood of radius $\epsilon$ about $x_0$. Here $h = |x - x_0|$ and we assume that $x$ is in the $\epsilon$-neighborhood about $x_0$. This gives you the result you want for all $x$ that are close to $x_0$, by choosing $\lambda$ to be an upper bound for $|g(x)|$. You can then cover $A$ by a finite number of such balls of radius $\epsilon$, and you can therefore deduce that if $x$ and $y$ lie in any one of these balls then you have your estimate. If you happen to pick numbers $x$ and $y$ that do not lie in a single one of the covering balls then you know that $|x -y| \geq \epsilon$ and so the argument I mentioned at the top of the page works in this case.

share|cite|improve this answer
If $x^*$ and $x_*$ are the maximizer and the minimizer respectively, then for $x,y\in A$, we have $|f(x)-f(y)|\le |f(x^*)-f(x_*)|:=M$. So if $|x-y|\ge \epsilon$, then $|f(x)-f(y)|\le \frac{M}{\epsilon^2}|x-y|^2$. Can you explain the $|x-y|<\epsilon$ part? – Ashok Mar 5 '12 at 6:05
I updated my answer to address your question. – treble Mar 5 '12 at 23:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.