# Proving that Holder Continuity with a>1 implies a constant function

1. Sep 11, 2010

### cinlef

1. The problem statement, all variables and given/known data
More or less the thread title:
Given f: Rn -> Rm, and f is both differentiable and satisfies the condition:
$$\left| f(x) - f(y) \right| \leq C \left| x-y \right|^{\alpha}.$$ for all x,y in Rn, and alpha > 1, prove that f is a constant function.

2. Relevant equations

3. The attempt at a solution
I've stared at this for a couple hours already and can't seem to figure out how to set the problem up. It seems to me that I'd need to show that the matrix of partial derivatives of f is 0... I would assume by squeezing it between $$-C \left| x-y \right|^{\alpha}$$ and $$C\left| x-y \right|^{\alpha}$$, and taking a derivative, then showing that both sides are 0 if alpha>1. But, I haven't the foggiest what I would take the derivative w.r.t., or how I would justify such a choice, since both x and y are points in Rn, and I would assume that they are both constant.

Any help would be appreciated.
Thanks,
Cinlef

2. Sep 12, 2010

### snipez90

Perhaps show |f'(x)| = 0 directly from the definition of the derivative.