Let $r: X \to P$ be a retraction, i.e., continuous and $r(x)=x$ for all $x\in P$. Here, $X$ is a Banach space and $P$ is a closed convex subset of $X$ or more specifically $P$ is a positive cone.
(1) I was wondering if $r$ is a bounded map, i.e., for a bounded subset $A$, $r(A)$ is bounded.
The above question is related to the following result:
(2) If $T: P \to P$ is a continuous and compact operator, i.e., for a bounded subset $A$, $T(A)$ is relatively compact, then $T\circ r$ is a compact operator.
Actually, I’d like to prove $(2)$. If $(1)$ is true, then $(2)$ follows. $(1)$ may not be true generally.
Please let me know if you have any idea or comment for this question.
Thanks in advance!
Unfortunately both are false, at least in an infinite dimensional case.
For (1) let $X=l^\infty$ be the Banach space of all real bounded sequences with $sup$ norm.
Consider $e_i=(0,0,\ldots,0,1,0,\ldots)$ with $1$ on $i$-th position. Note that $\lVert e_i-e_j\lVert=1$ for $i\neq j$ and $\lVert e_i\rVert=1$. Furthermore let $P=\{(r,0,0,\ldots)\ |\ r\in\mathbb{R}\}$. Now let $A=P\cup\{e_i\}_{i=1}^\infty$ and note that $A$ is closed in $X$. Consider the function
$$f:A\to P$$ $$f(v)=\begin{cases} v &\text{if }v\in P \\ (i,0,0,\ldots) &\text{if }v=e_i \end{cases}$$
Note that $f$ is continuous because $A$ is (topologically) a disjoint union of $P, \{e_1\}, \{e_2\},\ldots$ By the Tietze Extension Theorem $f$ can be extended to $r:X\to P$ which is a retraction by the construction.
However $r(\{e_i\}_{i=1}^\infty)=f(\{e_i\}_{i=1}^\infty)$ is not bounded even though $\{e_i\}_{i=1}^\infty$ is.
This example can be used in (2) to show that $(T\circ r)(\{e_i\})$ is unbounded and thus not relatively compact, e.g. take $T:P\to P$ to be the identity (note that the identity is a compact operator on finite dimensional spaces).
So assumptions about $T$ and $r$ being linear are essential here as noted by @Paul Frost.