First we recall the following result (see also this post):
Weissinger's fixed point theorem: Let $X$ be a complete metric space and assume that $f:X\longrightarrow X$ satisfies the following
$$
d(f^{i}(x),f^{i}(y))\leq \alpha_{i} d(x,y), \quad (*)
$$
for all $i\geq 1$, where $f^{i}$ stands for the composition of $f$ with itself $i$-times, and $\alpha_{i}$ is a sequence of non-negative numbers such that $\sum_{i\geq 1}\alpha_{i}<\infty$.
I am looking for an example (if any) of a continuous and not compact mapping $f:X\longrightarrow X$, $X$ being the closed unit ball of an infinite dimensional (real) Banach space such that:
(1) $f$ does not satisfy (*)
(2) $f^{i}$ is compact (i.e., $f^{i}$ maps bounded subsets into precompact ones) for some $i\geq 2$.
Of course, above $X$ can be replaced for any other convex and closed subset.
Many thanks in advance for your comments.
Thanks!
Best Answer
In the spirit of the comment by Matthias Klupsch, you can also take something like $$ Tx := (x_1, x_3, 0, x_5, 0, x_7, 0, \ldots ) $$ for $x \in \ell^2$. Then, $T$ is not compact, but $T^i x = (x_1, 0, \ldots)$ for $i \ge 2$ which is compact. Moreover, the Lipschitz constant of $T^i$ is always $1$, hence, $\alpha_i \ge 1$.