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It's well known that:

  1. If $X$ is a finite-dimensional normed space, $C$ is a closed and bounded subset of $X$ and $f:C\subset X\to X$ is continuous, then $f(C)$ is closed and bounded.

  2. If $X$ is any normed space, $C$ is a compact subset of $X$ and $f:C\subset X\to X$ is continuous, then $f(C)$ is compact.

In the finite-dimensional case, "compact" is the same as "closed and bounded". Therefore, Item 1 is a particular case of Item 2.

Question:

Item 2 does not hold with "compact" replaced by "closed and bounded", right? What are the standard counterexamples? More precisely:

  • What is an example of a Banach space $X$, a closed and bounded subset $C$ of $X$ and a continuous function $f:C\subset X\to X$ such that $f(C)$ is not bounded?

  • What is an example of a Banach space $X$, a closed and bounded subset $C$ of $X$ and a continuous function $f:C\subset X\to X$ such that $f(C)$ is not closed?

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  • $\begingroup$ There aren't too many well-known examples of the first one, since such a map is necessarily non-linear. For the second, James's Theorem guarantees the existence of continuous linear functionals on non-reflexive spaces which map the closed unit ball to an open (and not closed) interval in $\Bbb{R}$. $\endgroup$ – Theo Bendit May 12 '19 at 15:12
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    $\begingroup$ See this for the first question. $\endgroup$ – David Mitra May 12 '19 at 15:22
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Let $X = \ell^2$, and $C$ an orthonormal basis. Then $C$ is closed and bounded, but it is a discrete set, so any function on it is continuous. In particular, you can map it continuously to any sequence, which may be unbounded or non-closed.

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  • $\begingroup$ And by Tietze-Urysohn you can extend any such function to a continuous function on all of $\ell^2$. $\endgroup$ – Jochen May 13 '19 at 9:59

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