Problem in solving a question concerning real analysis. The question is :
Does there exist any function $f : \mathbb R \longrightarrow \mathbb R$ such that $f(1) = 1$, $f(-1) = -1$ and $|f(x) - f(y)| \leq |x - y|^{\frac {3} {2}}$?
It is clear that $f$ is continuous over $\mathbb R$ by the given condition and hence it attains all the values between $-1$ and $1$ in $(-1,1)$.Now how can I proceed?Please help me.
Thank you in advance.

Notice that if $|f(x) - f(y)| \leq {|x - y|}^{\frac {3} {2}}$, then
$$\left|\frac{f(x)-f(y)}{x-y}\right|\leq {|x - y|}^{\frac {1} {2}}$$
and hence $f$ is differentiable... but $f'=0$ everywhere. In other words, $f$ is constant, so one may not have $f(1)\neq f(-1)$.
