Let $\displaystyle f: \Re \mapsto \Re $ satisfy a Lipshitz condition of order p. Is f differentiable everywhere when p>1 and p=1?

a) p > 1

b) p = 1; Let $\displaystyle x,y \in \Re \ , \ |f(x)-f(y)| \leq C|x-y|^{p}$, then we have $\displaystyle \frac { |f(x)-f(y)| } {|x-y|} \leq C \ \ \ \ \ \ C > 0$

I tend to think that f is differentiable when p=1, but not p > 1, but I'm stuck on actually proving it here.

Thank you.