A condition for a function to be constant I need to proof this result:
Let $\alpha >1$ and $c\in\mathbb{R}$. If $f:U\subset\mathbb{R}^m\rightarrow\mathbb{R}^n$, U open, satisfies $|f(x)-f(y)|\leq c|x-y|^\alpha$ for every $x$, $y$ $\in U$, then $f$ is constant in every component of $U$.
I just didn't have any idea on how to start it, I'm doing my first multivariable analysis course now!

We show that $f$ is locally constant. Let $x_0\in U$ that I assume open, and let $r$ such that $B(x_0,r)\subset U$. Then for $y\in B(x_0,r)$ and $n\geq 1$
\begin{align*}
|f(x_0)-f(y)|&\leq \sum_{k=0}^{n-1}\left|f\left(x_0+\frac{k+1}n(y-x_0)\right)-f\left(x_0+\frac kn(y-x_0)\right)\right|\\
&\leq \sum_{k=0}^{n-1}\left|x_0+\frac{k+1}n(y-x_0)-\left(x_0+\frac kn(y-x_0)\right)\right|^{\alpha}\\
&=\sum_{k=0}^{n-1}n^{-\alpha}|y-x_0|^{\alpha}\\
&\leq r^{\alpha}n^{1-\alpha}
\end{align*}
and since $n$ is arbitrary, $f(x_0)=f(y)$ for all $y\in B(x_0,r)$.
