Consider a function \(f:\Bbb R \rightarrow \Bbb R\) such that

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

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

\begin{equation}

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

\end{equation}

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.