# Thread: A variation of the Mean Value Theorem

1. ## A variation of the Mean Value Theorem

Hi, I need help proving this:

Let U be an open set in R^n, f:U--->R be of class C^2 and Gradient of f(a) = 0, a is an element of S contained in U where S is compact and convex.

Prove that there exists a constant M such that:
|f(x)-f(a)| <= M* |x-a|^2 for all x in S.

Thanks a lot.

2. Use Taylor theorem for approaching with a degree 1 polynomial. There exists $\displaystyle 0\leq t\leq 1$ such that

$\displaystyle f(x)=f(a)+\sum_{|\alpha|=1}D^{\alpha}f(a)(x-a)^\alpha+ \sum_{|\alpha|=2}\frac{D^{\alpha}f(a+t(x-a))}{\alpha!}(x-a)^\alpha$

Here $\displaystyle \alpha$ are multiinedex. For $\displaystyle |\alpha|=1$ all the terms are 0 by the hypothesis. For $\displaystyle |\alpha|=2$ use the hypothesis that $\displaystyle S$ is convex and compact for finding a bound $\displaystyle M$ for each $\displaystyle |D^{\alpha}f(a+t(x-a))|$ for $\displaystyle |\alpha|=2$, and clearly each $\displaystyle |(x-a)^\alpha|\leq |x-a|^2$ for a multiindex $\displaystyle \alpha$ such that $\displaystyle |\alpha|=2$.

3. How do I use the fact that S is convex and compact to find an M such that$\displaystyle |D^{\alpha}f(a+t(x-a))|$?

4. $\displaystyle a+t(x-a)$ is in $\displaystyle S$ because $\displaystyle S$ is convex. Since $\displaystyle S$ is also compact and $\displaystyle f$ is $\displaystyle C^2$ all the derivatives of order 2 are uniformly bounded on $\displaystyle S$.

5. Yes, but what exactly is the bound for the constant M?

6. OK, there are a finite number of order 2 derivatives, all of them are continuous on the compact set $\displaystyle S$.Take $\displaystyle M$ as the biggest bound for the modulus of each one of all these derivatives!