A Lipschitz-type function on a connected open subset of $\mathbb{R}^m$ is constant Let $1<\theta<\infty$ and $f:U \to \mathbb{R}^m$, where $U \subset \mathbb{R}^n$ is open connected. Suppose that
$$ | f(x)-f(y)  | \leq K  |x-y |^{\theta} $$
for all $x,y \in U$, where $K>0$. Show that $f$ is constant in $U$.
I tried this: Let $a \in U$, define the sets $A=\{ x \in U  |f(x)=f(a)\}$ and $B=\{ x \in U  |f(x)\neq f(a) \}$, and show that A and B are open disjoint. I could show that $B$ is open, and for showing that A is open, let $y \in A \subset U$, then exists $\delta >0$ such that $B(y,\delta) \subset U$, if $ |v|<\delta$ then $y+v \in U$, so using the hypothesis
$$ | f(y)-f(y+v)  | \leq K  |v |^{\theta} $$
I would like to show that the inequality is zero and thus $y+v \in A$, by the connectedness of $U$, $A=U$.
Any help or hint would be greatly appreciated.

If $m=n=1$ then note that
$$\left|{f(x) - f(y) \over x-y}\right| \leq |x-y|^{\theta-1}$$
Letting $y$ go to $x$, we see that the first term goes to zero, which means that $f$ has a derivative and that derivative is zero. So $f$ is locally constant and since its domain of definition is connected, it is constant everywhere.
If $m$ is greater than 1, I guess we can go back to the one-dimensional case by working on the segment connecting $x$ and $y$.
