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 :=\

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- June 5th 2008, 08:31 AMauroraHolder 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 :=\ - June 5th 2008, 11:40 AMOpalg
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 and Divide by |y–x|, to get 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. - June 5th 2008, 01:37 PMThePerfectHacker
I been curious to see how the argument would go, let us not go to crazy and look at . Tell me if I have the right idea. Let be some point. The first step is to show that the partials exist in the neighborhood of the point for 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 exist in neighborhood of and is continous at it follows is differenciable at . Furthermore, . Thus, the function is constant (if is connected).

- June 6th 2008, 02:46 PMaurora
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!