TITLE: How to show that every $\alpha$-Hölder function, with $\alpha>1$, is constant?
QUESTION [13 upvotes]: Suppose $f:(a,b) \to \mathbb{R} $ satisfy $|f(x) - f(y) | \le M |x-y|^\alpha$ for some  $\alpha >1$
and all $x,y \in (a,b) $. Prove that $f$ is constant on $(a,b)$. 
I'm not sure which theorem should I look to prove this question. Can you guys give me a bit of hint? First of all how to prove some function $f(x)$ is constant on $(a,b)$? Just show $f'(x) = 0$?

REPLY [20 votes]: Derivatives are not needed here (which is a good thing since proofs based on them can lead to erroneous arguments, see the accepted answer for an example)...

Choose $y\gt x$ in the interval $(a,b)$. For every $n\geqslant1$, divide the interval $(x,y)$ into $n$ subintervals $(x_i,x_{i+1})$ of length $x_{i+1}-x_i=(y-x)/n$. By hypothesis, for every $i$, $$|f(x_i)-f(x_{i+1})|\leqslant M(x_{i+1}-x_i)^\alpha=M(y-x)^\alpha n^{-\alpha},$$ hence, by the triangular inequality, $$|f(x)-f(y)|\leqslant \sum\limits_{i=1}^n|f(x_i)-f(x_{i+1})|\leqslant M(y-x)^\alpha n^{1-\alpha}.$$ If $\alpha\gt1$, the RHS goes to zero when $n\to\infty$ because $n^{1-\alpha}\to0$, hence $f(x)=f(y)$, QED.