What can we say about the function $|f(x)-f(y)|\leq |x-y|^c$ when $c\geq 1$

Question

When $c=1$ we know that it is Lipschitz, hence the function $f(x)$ is uniformly continuous. But if $c>1$ then what are the possibilities may come out?

Note- Function is defined on real numbers.

Answer 1

The function is constant: take $y=x+h$, then $$ \lvert f(x+h)-f(x) \rvert \leqslant h^c, $$ and dividing both sides by $h$ and taking the limit as $h \to 0$ shows that $f'(x)=0$. But this is true for any $x$, and the only continuous functions with zero derivative everywhere are constant.

Answer 2

These functions are called Hölder functions. If the exponent is $> 1$ and the space isn't too weird, only constant functions satisfy the condition.

Hint of proof: if $f(x) \neq f(y)$, you can find a chain of small intervals which connect $x$ to $y$ (if the distance between $x$ and $y$ is $d$, you will roughly need $n$ intervals of size $d/n$). But the Hölder condition implies that these intervals will be sent to intervals of size $\leq (d/n)^c$, do the distance between $f(x)$ and $f(y)$ has to be $\leq n \times \left(\dfrac dn\right)^c$. As this quantity converges to $0$ when $n$ grows, you have a contradiction.