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Let $\beta>1$ and $c\in \mathbb R$. If $f:U\to \mathbb R^n$, defined in the open subset $U\subset \mathbb R^m$.

I'm trying to prove that if $f$ holds the condition $|f(x)-f(y)|\le c|x-y|^{\beta}$ for any $x,y\in U$, then $f$ is constant in each component of $U$.

The only thing I could discover was $f$ must be continuous because of the inequality above (the way to prove is similar to prove every Lipschitz function is continuous).

The book I'm studying has only explained so far basic facts of several variables vector-valued functions. So the tools I have in my pocket are the only ones when either $n=1$ or $m=1$.

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Hint:

Take two points $x,y$ on the same component of $U$ such that the segment joining them is in $U$. Consider the segment $\gamma$ from $x$ to $y$. Evaluate $$\lim_{h\to 0}\frac{f(\gamma(t+h))-f(\gamma(t))}h$$

Can any two points in the same component of $U$ be joined by a finite sequence of segments, all of them in $U$? (Hint for this: you can join the points with a continuous path $\gamma:[0,1]\to U$ and $[0,1]$ is compact).

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    Why is there such a $\gamma$? the open set $U$ is not necessarily path-connected. Thank you for your answer – 2017-01-06
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    @user42912 Because the end points are in the same component of $U$. – 2017-01-06
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    Yes, but is this component path-connected? – 2017-01-06
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It helps to prove it first in the case of $n=1$ and you will notice there is nothing special about this case. The same argument works just as well for higher dimensions. The idea is to show that $f$ is differentiable at every point and its derivative is zero so it must be constant. Toward this, fix $x \in \mathbb{R}$ then by assumption $$\frac{|f(x) - f(y)|}{|x-y|} \leq c|x-y|^{\beta - 1} = c|x-y|^{\epsilon}$$ where $\epsilon > 0$. Now taking a limit as $y \rightarrow x$ on the left is nothing more than the derivative of $f$ which exists and is equal to 0 by the squeeze theorem. Hence, $f'(x) = 0$ for any $x \in \mathbb{R}$ proving $f$ is constant.