Is every retraction a bounded map? 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!
 A: 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.
A: Partial answer
(2) is true. 
Note that (2) only makes sense if $r$ is linear. In that case $P$ must be a closed linear subspace of $X$ (since the image of a linear map is always a linear subspace).
Now (2) is due to the fact the the composition $T \circ L$ is compact if $L$ is linear and continuous and $T$ is compact.
