Prove that a Hölder function is derivable.

Question

It is said that a order $\gamma > 0$ continuous function is named Hölder when

$|f(x)-f(y)| <= C|x-y|^\gamma$

being $C$ a constant, such as $C > 0$.

A) Prove that if $f$ is a order $\gamma > 1$ continous Hölder function, then, $f$ is derivable. Show that, in fact, $f$ must be constant.

(also, see https://math.stackexchange.com/questions/361400/function-on-a-b-that-satisifies-a-hölder-condition-of-order-alpha-1-is and https://math.stackexchange.com/questions/1851983/if-f-is-holder-continuous-for-alpha-1-then-f-is-constant?noredirect=1&lq=1, among others) — Clement C., Dec 03 '19 at 21:15

score 0 · Answer 1 · answered Dec 03 '19 at 23:32

$|\frac{f(a+h)-f(a)}{h}|=\frac{|f(a+h)-f(a)|}{|h|} \le \frac{C|a+h -a|^\gamma}{|h|} = \frac{C|h|^\gamma}{|h|} = C|h|^{\gamma-1}$

When $\gamma > 1$, $\lim\limits_{h \to 0}C|h|^{\gamma-1} = 0.0$

Thus, $\lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h} = 0.0 $

By definition, f is derivable and its derivative is 0.0 (i.e. f is constant)

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