Let be $h: \mathbb{R} \rightarrow \mathbb{R}$ and c a positive constant, if $\forall a, b \in \mathbb{R}$ we have that:
$$\frac{|h(a)-h(b)|}{a-b}\leq |a-b|^c$$
Prove that $h$ is constant.
I tried to use the definition of derivative or the triangle inequality but I didn't find the solution.
On
One approach mentioned above is to show that the derivative exists and is equal to zero. Let's try to do something more elementary.
Note that by assumption $$|h(b)-h(a)|\le |b-a|^{c+1}.$$
Fix a point $x$ and let $n$ be a positive integer. Write $h(x)-h(0)$ as a telescopic sum and take absolute value (first line), then use triangle inequality (second line), and finally use the assumption, and take limit as $n\to\infty$ (third line):
\begin{align*} |h(x) - h(0)| &= |\sum_{j=1}^n h(\frac{jx}{n})-h(\frac{(j-1)x}{n}) |\\& \le \sum_{j=1}^n |h(\frac{jx}{n}) - h(\frac{(j-1)x}{n})| \\ & \le n|\frac{1}{n} x|^{c+1}=n^{-c}|x|^{c+1}\underset{n\to\infty}{\to} 0 \end{align*}
Therefore $h(x)=h(0)$ for any $x$.
Take $$\underset{h \to 0}{\lim} \frac{|h(x+h)-h(x)|}{h}$$
We have for all $h >0$: $$ 0 \le \frac{|h(x+h)-h(x)|}{h} \le h^c$$
Thus, we have $$0 \le \underset{h \to 0}{\lim}\frac{|h(x+h)-h(x)|}{h} \le \underset{h \to 0}{\lim} h^c =0$$
It follows by the sandwich/squeeze theorem that $$\underset{h \to 0}{\lim}\frac{|h(x+h)-h(x)|}{h} =0 $$
Hence $h$ is differentiable everywhere, and has derivative of 0 everywhere. It follows that it must be a constant function.