Showing that if a function between metric spaces has Hölder exponent $\alpha > 1$, this function must be constant

Question

Let us recall the definition of a Hölder continuous function in the most general setting:

Let $(M,d)$ and $(M',d')$ be two metric spaces. A function $f : M \to M'$ is said to be Hölder continuous with exponent $\alpha$, $0 < \alpha \leq 1$, if there exists a real constant $C$ s.t. $$d(f(x),f(y)) \leq C d(x,y)^\alpha$$ holds for all $x,y \in M$.

Now the question is why do we only consider $0 < \alpha \leq 1$. As I heard in todays lecture, this is because if $\alpha > 1$, then the only functions satisfying the above criterion are the constant ones. So naturally I wanted to prove this, but I got stuck. I mean, I think the best way of proving this is by contradiction: Hence assume that $f(x) \neq f(y)$ for some $x,y \in M$. Moreover, the above condition implies that $f$ is continuous.

My question (since I've searched for answers already, but only found the case where $M$ is an interval): How does one prove that if $\alpha > 1$, then $f$ must be constant? Any hint is appreciated.

Answer 1

Answer is: this is wrong. You can look in On Jordan Arcs and Lipschitz functions defined on them by Besicovitch and Schoenberg.

Regarding your question, the important parts are the theorem 1 and the beginning of page $116$ : there exists an arc $J \subset \mathbb{C}$ and an injective function $f : J \mapsto \mathbb{R}$ such that for all $x,y \in J$, $|f(x)-f(y)| \le |x-y|^{\frac{3}{2}}$.

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Usually I prefer to look at their result the other way around : for $0 \le \varepsilon<\frac{1}{2}$, there exists $g : [0,1] \mapsto \mathbb{C}$ a continuous function such that for all $x,y \in [0,1]$, $|g(x)-g(y)| \ge |x-y|^{\varepsilon}$.

(You get to this by considering reciprocal functions : for $\varepsilon=\frac{2}{3}$, you just have to rescale $f$ as defined in the previous paragraph and take $g=f^{-1}$).

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Note that "arc" is to be understood simply as the image of $[0,1]$ by a continuous injective function from $[0,1]$ to $\mathbb{C}$.

Answer 2

This is false. Consider $(M,d)=(M',d')=(\{a,b\},d)$ where $d(a,b) = 1$. Let $f: M \to M, f(a) = a, f(b) = b$, and $C = 1$. Then the Hölder condition is satisfied for any $\alpha > 0$ (I guess any $\alpha \in \mathbb{R}$).

What was probably meant in the lecture is that in the case that $(M',d) = \mathbb{R}$ and $(M,d) = \mathbb{R}^n$ (for some $n$), if a function satisfies a Hölder $\alpha$ condition with $\alpha > 1$, then it must be constant. Note the main reason this is true is the dilation structure of $\mathbb{R}$, which is not present in my counterexample above ($d(x,y)$ is never between $0$ and $1$, so raising it to the $\alpha > 1$ won't make it very small).