Problem 1.2.2 Suppose that $f$ maps compact interval $I$ into itself and that $$\mid f(x)-f(y) \mid < \mid x-y \mid$$ for all $x, y \in I, x\neq y$. Can one conclude that there is some constant $M<1$ such that, for all $x, y \in I$, $$\mid f(x)-f(y) \mid \leq M\mid x-y \mid$$?
If such function exists it is not continuously differentiable, otherwise $f'$ will attain supremum.
I tried to define $g: I \times I \rightarrow I$ s.t. $g(x,y)=\frac{\mid f(x)-f(y) \mid}{\mid x-y \mid}$ which is continuous and attains supremum. But $g$ is not defined for $x=y$
Please give hint. Please do not give solution. Thanks!