Holder condition.

• Jun 5th 2008, 09:31 AM
aurora
Holder condition.
Hello!
I've been asked to find all the functions f: R^n--->R^m that satisfies Holder's condition in a certain D, for alpha>1 and M>0
(meaning: for all x,y in D, ||f(x)-f(y)||<= M||x-y||^alpha

I have absoultely no idea how to do that :=\
• Jun 5th 2008, 12:40 PM
Opalg
Quote:

Originally Posted by aurora
Hello!
I've been asked to find all the functions f: R^n--->R^m that satisfies Holder's condition in a certain D, for alpha>1 and M>0
(meaning: for all x,y in D, ||f(x)-f(y)||<= M||x-y||^alpha

I have absoultely no idea how to do that :=\

It's always a good idea to look at simple cases first, so let's see what happens when n=m=1. Then we have $f:\mathbb{R}\to\mathbb{R}$ and $|f(y)-f(x)|\leqslant|y-x|^\alpha.$ Divide by |y–x|, to get $\Bigl|\frac{f(y)-f(x)}{y-x}\Bigr|\leqslant|y-x|^{\alpha-1}.$ Now let y→x and you see that f'(x)=0 (for all x). So f has to be constant on each component of D.

Now do the same sort of thing for the general case.
• Jun 5th 2008, 02:37 PM
ThePerfectHacker
Quote:

Originally Posted by Opalg
Now do the same sort of thing for the general case.

I been curious to see how the argument would go, let us not go to crazy and look at $f:\mathbb{R}^n\mapsto \mathbb{R}$. Tell me if I have the right idea. Let $\bold{x}$ be some point. The first step is to show that the partials $\partial_i f(\bold{x})$ exist in the neighborhood of the point for $i=1,2,...,n$ by using the argument you used above with one variable at a time. Next we use the fact that the Holder condition implies (uniform) continuity. Thus, since the partials $\partial_i f(\bold{x})$ exist in neighborhood of $\bold{x}$ and $f$ is continous at $\bold{x}$ it follows $f$ is differenciable at $\bold{x}$. Furthermore, $\nabla f(\bold{x}) = (\partial_1 f(\bold{x}) , ... , \partial_ n f(\bold{x}) ) = \bold{0}$. Thus, the function is constant (if $D$ is connected).
• Jun 6th 2008, 03:46 PM
aurora
I didn't see any need to describe the fact that the partial derivatives of F are continous. I have used the straight forward definition of Diffrentiabillity. I have shown such a "Diffrential" exists (zero, or the zero matrix, in this case), and therefore shown the function can be differentiated.

Anyway, thank you guys very, very much!