give a counter example that $T^n$ is contraction will not imply that $T$ is contraction.

Question

Let $T$ be a contraction map then $T^n$ is contraction.. We can prove this result by induction on n.. But the converse is not true... help me to give a counter example that $T^n$ is contraction will not imply that $T$ is contraction.

Answer 1

Let $f : \mathbb{R} \rightarrow \mathbb{R}$, $$f : x \mapsto \begin{cases} -x & \text{ if } x \ge 0 \\ \frac{x}{2} & \text{ if } x < 0 \end{cases}.$$ Then $f$ maintains distances between positive numbers so it isn't a contraction. On the other hand, you can easily verify that $f \circ f$ is a contraction with Lipschitz constant $\frac{1}{2}$.