# Thread: Holder condition.

1. ## 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 :=\

2. 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.

3. 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).

4. 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!