Consider a function $$f:\Bbb R \rightarrow \Bbb R$$ such that

$$|f(x)-f(y)| \leq |x-y|^\alpha\tag{1}\label{eq447:1}$$

for all $$x,y\in \Bbb R$$ and for $$\alpha >1$$. Then $$f(x)$$ is constant.

Proof.

Using triangular inequality and then hypothesis \eqref{eq447:1} yields

\begin{eqnarray}\left| f(x) – f(y) \right| &=& \left| f(x) – f\left(\frac{x+y}{2}\right)+f\left(\frac{x+y}{2}\right)- f(y) \right| \leq\\
&\leq& \left| f(x) – f\left(\frac{x+y}{2}\right)\right| + \left|f\left(\frac{x+y}{2}\right)- f(y) \right|\leq \\
&\leq& \left|x-\frac{x+y}{2}\right|^\alpha + \left|\frac{x+y}{2}-y\right|^\alpha=\\
&=& \frac{|x-y|^\alpha}{2^{\alpha-1}}.\end{eqnarray}

The previous steps can be generalized to obtain what follows. If $$\forall x,y \in \Bbb R$$

$|f(x) – f(y)| \leq \frac{|x-y|^\alpha}{2^{(\alpha-1)n}},$

then

|f(x) – f(y)| \leq \frac{|x-y|^\alpha}{2^{(\alpha-1)(n+1)}}.\tag{2}\label{eq447:2}

Thus by induction \eqref{eq447:2} is true for all $$n \in \Bbb Z^+$$. Consequently, since the RHS of \eqref{eq447:2} can be made arbitrarily small, it must be

$|f(x)-f(y)| = 0$

and $$f(x)$$ is constant.