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.

Another way:
$|f(x)-f(y)| \le \sum_{k=0}^{n-1} | f(y+ {k+1 \over n} (x-y))-f(y+ {k \over n} (x-y)) | \le n ({1 \over n}(x-y))^{3 \over 2} = {1 \over \sqrt{n} } (x-y)^{3 \over 2}$,
and since $n$ is arbitrary, we see that $f(x) = f(y)$ for any $x,y$.
