Is Holder Continuity with Alpha Greater Than 1 Sufficient for Constant Function?

[tom_rylex]

Homework Statement

Prove that if f(x) is Holder continuous, i.e,

\sup_{a<x , y<b} \frac{\abs{f(x) - f(y)}}{\abs{x-y}^\alpha} = K^f_\alpha<\inf
with \alpha > 1, then f(x) is a constant function

Homework Equations

The Attempt at a Solution

I've been staring at this for a while, but I'm unsure of where to start. I'm guessing that I'm supposed to show that the derivative of something is zero. I've seen hints elsewhere about applying Taylor's theorem, but I am unsure on how to apply it.

 

[morphism]

First let's fix up your LaTeX: f(x) is Holder continuous (on (a,b)) if

\sup_{a < x < y < b} \frac{\left| f(x) - f(y) \right|}{\left| x-y \right|^{\alpha}} < \infty,

where we're assuming that \alpha>1. We can restate this as: there exists a fixed C such that for all x & y in (a,b)

\left| f(x) - f(y) \right| \leq C \left| x-y \right|^{\alpha}.

Your idea of proving that f'(x)=0 is a good one, but we don't know a priori that f is differentiable. So let's resort to the epsilon-delta definition. Given e>0, can we find a d>0 such that if |h|<d then

\left| \frac{f(x+h) - f(x)}{h} \right| &lt; \epsilon?

Try to use the Holder continuity inequality to find a suitable one.