Is a function that has Holder order bigger than one constant?

Question

I see from the Wikipedia that if a function $f$ over $[a,b]$ is Holder continuous with order strictly bigger than one, i.e. $$|f(x) - f(y)| < K |x-y|^\alpha$$ for some constant $K$ and $\alpha>1$ over $[a,b]$, then $f$ is a constant over this interval. My question is how should we prove this? Thanks!

Answer 1

Hint: we can rewrite this as $$ \left|\frac{f(x) - f(y)}{x-y}\right| \leq K|x-y|^{\alpha-1} $$

Answer 2

Hint: Pick any $x,y \in [a,b]$ with $x<y$, and subdivide the interval $[x,y]$ into $N$ equal parts. Applying Hölder continuity to each subinterval gives a new bound on $|f(x) - f(y)|$ depending on $N$. What happens as $N \to \infty$?