# If f is holder continuous with a > 1, then f is constant?

• Apr 29th 2011, 05:37 AM
moses
If f is holder continuous with a > 1, then f is constant?
Hello

I'm trying to prove that if f satisfies the holder condition, i.e.
there exists a constan K such that for all x,y in R : |f(x) - f(y)| <= K(|x - y|^a)
and a > 1
then f is a constant function.

I assume that the right direction is proving that the derivative of f is constantly zero, but I don't see why f has to be differentiable in the first place...and even assuming that f is differentiable I don't see why the derivative must be zero..
• Apr 29th 2011, 06:18 AM
girdav
Let $\displaystyle x_0\in\mathbb R$. We have $\displaystyle \left|\frac{f(x)-f(x_0)}{x-x_0}\right|\leq K|x-x_0|$ if $\displaystyle x\neq x_0$. Now you can see what the derivative in $\displaystyle x_0$ is and conclude.
• Apr 30th 2011, 02:23 AM
moses
So would it be correct to take the limit of both sides of the inequality as X goes to Xo, and then we get that the absolute value of the derivative at Xo is smaller or equal to zero?
• Apr 30th 2011, 05:02 AM
girdav
Yes, it's correct.