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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!

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Could you please explicitly state what your question is. – Sasha Mar 25 '12 at 13:51
I need to prove this result. I have no idea how to start solving this. – Marra Mar 25 '12 at 13:54
Ok, I'll do it! – Marra Mar 25 '12 at 14:12
I retracted that downvote and I'll delete the comment. – user21436 Mar 25 '12 at 14:52
up vote 3 down vote accepted

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 $y\in B(x_0,r)$.

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I think you meant that $r$ is arbitrary. – Marra Mar 25 '12 at 13:58
@GustavoMarra No, that's $n$ since after I take the limit $n\to \infty$. – Davide Giraudo Mar 25 '12 at 13:59
But $n^{1-\alpha}$ does not converge to zero when $n\rightarrow\infty$. – Marra Mar 25 '12 at 14:06
After bounding, we don't have a sum anymore, and since $1-\alpha<0$, $n^{1-\alpha}$ does converge to $0$. – Davide Giraudo Mar 25 '12 at 14:07
Whoops, just saw my misunderstanding. Thanks! – Marra Mar 25 '12 at 14:11

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