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 :=\
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 $\displaystyle f:\mathbb{R}\to\mathbb{R}$ and $\displaystyle |f(y)-f(x)|\leqslant|y-x|^\alpha.$ Divide by |y–x|, to get $\displaystyle \Bigl|\frac{f(y)-f(x)}{y-x}\Bigr|\leqslant|y-x|^{\alpha-1}.$ 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.
I been curious to see how the argument would go, let us not go to crazy and look at $\displaystyle f:\mathbb{R}^n\mapsto \mathbb{R}$. Tell me if I have the right idea. Let $\displaystyle \bold{x}$ be some point. The first step is to show that the partials $\displaystyle \partial_i f(\bold{x})$ exist in the neighborhood of the point for $\displaystyle i=1,2,...,n$ 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 $\displaystyle \partial_i f(\bold{x})$ exist in neighborhood of $\displaystyle \bold{x}$ and $\displaystyle f$ is continous at $\displaystyle \bold{x}$ it follows $\displaystyle f$ is differenciable at $\displaystyle \bold{x}$. Furthermore, $\displaystyle \nabla f(\bold{x}) = (\partial_1 f(\bold{x}) , ... , \partial_ n f(\bold{x}) ) = \bold{0}$. Thus, the function is constant (if $\displaystyle D$ is connected).
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!