n-th iteration of T is contraction mapping, must m-th iteration (m>n) of T be a contraction mapping?

Question

$T$ is a self-map on a metric space $M$. $M$ is not necessarily complete.

$T$ itself is not a contraction mapping, but $n$-th iteration of $T$ is a contraction mapping.

Must $T^m$ ($m>n$) be a contraction mapping as well?

The intuition is no. But I cannot find a nice counterexample.

Answer 1

It is possible to come up with an example of linear contraction on the euclidean plane. Let $$A=\begin{pmatrix} 2^{-1/2} & (2N)^{-1} \\ N & -2^{-1/2}\end{pmatrix}$$ Then $A^2=I.$ For $T=2^{-1}A$ the mapping $T^2=4^{-1}I$ is a contraction on $\mathbb{R}^2.$ However $T^3=8^{-1}A$ is not a contraction for $N>8$ as $$d(T^3e_1,0)=8^{-1}d(Ae_1,0)>8^{-1}N>1=d(e_1,0)$$ where $e_1,e_2$ denote the elements of the standard basis. Thus $T$ is not a contraction either.

Answer 2

Let us consider $M = \{0, 2,3,6,8\}$ (equipped with the Euclidean distance) and the map $T$ defined via $$ T(0) = T(2) = 0, \quad T(3) = 2, \quad T(6) = 3, \quad T(8) = 6. $$ Then, it can be checked that $$ d(T(6),T(8)) = d(3,6) = 3 > 2 = d(6,8) $$ and $$ d(T^3(6), T^3(8)) = d(0,2) = 2 = d(6, 8),$$ i.e., $T$ and $T^3$ are not contractive. However, by looking at $$ T^2(0) = T^2(2) = T^2(3) = 0, \quad T^2(6) = 2, \quad T^2(8) = 3$$ we see that $T^2$ is a contraction.