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Math Help - Holder condition.

  1. #1
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    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 :=\
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  2. #2
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    Quote Originally Posted by aurora View Post
    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 |yx|, 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.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    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).
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  4. #4
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    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!
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