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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!

user123043
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  • Take any discontinuous map such that $f^2$ is constant. This can even be done with linear maps. – MaoWao Aug 12 '21 at 11:19
  • Thanks @MaoWao, but if $f$ is discontinuous, it does not sastify condition (*) with $i=1$. – user123043 Aug 12 '21 at 12:27
  • But that is exactly what you want in (1), isn't it? In any case, I just saw you want a continuous $f$ ... – MaoWao Aug 12 '21 at 12:38
  • I haven't given this much thought, but what happens if you take $f$ to be a projection from an infinite dimensional Banach space to a finite-dimensional non-trivial subspace (and restrict to the closed unit ball, if you want)? I mean $d(f^i(x),f^i(y)) = d(x,y)$ for $x$ and $y$ in this subspace, so (*) is not satisfied, yet $f^i = f$ maps bounded sets to bounded sets of a finite dimensional space and these are definitely precompact. Am I missing something here? – Matthias Klupsch Aug 18 '21 at 08:59
  • Thanks for your response, but such $f$ is, itself, compact, and I am looking for a non-compact (but $f^{i}$ compact for some $i\geq 2$). I edit this detail in the post. Also, I am not sure that, in your context, $d(x,y)=d(f^{i}(x),f^{i}(y)))$. However, I will revise your comment, because it seems interesting. – user123043 Aug 18 '21 at 09:44

1 Answers1

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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, $\alpha_i \ge 1$.

gerw
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