Let $C$ be a nonempty subset of a Banach space $X$. A mapping $T:C\to C$ is said to be asymptotically nonexpansive if for all $n\in \mathbf{N},$ there exists a positive constant $k_n\geq1$ such that $\lim\limits_{n\to \infty}k_n=1$ $$\|T^nx-T^ny\|\leq k_n\|x-y\| \quad \text{ for all } x, y\in C.$$ I think that if $T$ is a nonexpansive mapping ($\|Tx-Ty\|\leq \|x-y\|$), then it will be asymptotically nonexpansive. But I cannot prove that there is an asymptotically nonexpansive mapping which is not nonexpansive.
Thank you in advance!
Take $X=C=\mathbb R^2$ (or let $C$ be a subspace of dimension two). Let $T$ be the linear map given by the matrix $\begin{pmatrix}0&17\\0&0\end{pmatrix}$. The optimal $k_1$ is now 17. But $T^2=0$, so $T^nx-T^ny=0$ whenever $n\geq2$, so the estimate you desire holds for every choice of $k_n$. Thus if you choose $k_1=17$ and $k_n=1$ for $n\geq2$, $$ \|T^nx-T^ny\|\leq k_n\|x-y\| \quad \text{ for all } x, y\in C $$ and $k_n\to1$ as $n\to\infty$, but $T$ is not nonexpansive.