Hölder exponents greater than 1 imply function to be constant? I have been having trouble with Hölder exponents. The definition of Hölder continuity tells me that a function $f$ between metric spaces must satisfy 
$d(f(x),f(y)) \leq C \cdot d(x,y)^\alpha$ for some exponent $\alpha > 0$. 
The Wikipedia article however states that for an exponent $\alpha >1$ this condition implies that the function $f$ is constant. I've been mulling it over but just can't see why this is the case. Let's assume that $\alpha>1$. Then I have two (interesting) cases: 
Case 1: $d(x,y) > 1$. Here the Hölder condition tells me that the function values are allowed to be even further apart than the input values, which doesn't seem to enforce $f$ to be constant. 
Case 2: $d(x,y) < 1$. This time the Hölder condition tells me that if the input values lie close together, then the function values have to be even tighter together. To me it is plausible that this is precisely what yields the continuity of Hölder-continuous functions, however again saying that $f$ needs to be constant seems to me to still be a strong conclusion. 
I've seen similar questions being posted, however they all make (at least indirect) use of some differentiability assumption on $f$ which I do not want to make use of. 
Can anyone enlighten me please? 
I appreciate your answers ;). 
 A: This is not true for general metric spaces - if you have a function whose domain is a two-point space, then this function is $\alpha$-Hölder for every $\alpha>0$. $\alpha$-Hölder property for $\alpha>1$ implies the function being constant only in special spaces, like $\mathbb R^n$. Let me just focus on functions $f:\mathbb R\to\mathbb R$.
You mention "[the proofs] all make (at least indirect) use of some differentiability assumption on $f$". This is not quite correct - you don't need to assume differentiability, since $\alpha$-Hölder condition for $\alpha>1$ implies that the derivative exists - indeed, we just consider
$$\left|\frac{f(x+h)-f(x)}{h}\right|\leq\frac{C|h|^\alpha}{|h|}\to 0,$$
so the limit exists and is zero everywhere. That way or another, we can prove that $f$ is constant directly as well: take any two points $x<y$ and let $x_0=x,x_1=x+\frac{y-x}{n},x_2=x+2\frac{y-x}{n},\dots,x_n=x+n\frac{y-x}{n}=y$. Then we have
$$|f(x)-f(y)|\leq|f(x_0)-f(x_1)|+|f(x_1)-f(x_2)|+\dots+|f(x_{n-1})-f(x_n)|\\
\leq C|x_0-x_1|^\alpha+C|x_1-x_2|^\alpha+\dots+C|x_{n-1}-x_n|^\alpha\\
=n\cdot C\left|\frac{x-y}{n}\right|^\alpha=n^{1-\alpha}C|x-y|^\alpha\to 0,$$
so that $f(x)=f(y)$.
A: I doubt that the statement holds in all generality. Consider a set $X$ endowed with the discrete metric and let $f$ be the identity on $X$. Then $f$ satisfies Hölder continuity (with $C=1$ and any $\alpha>0$) but it is not constant.
