Let $\mathcal X$ be the Banach space of $L^1$ functions on some probability space, $\mathcal Y$ be some other Banach space, $T:\mathcal X\to \mathcal Y$ be some surjective continuous linear map, $\mathcal X\_+$ be the set of all elements of $\mathcal X$ with a nonnegative version (a closed convex cone), and $\mathcal Y\_+:=T(\mathcal X\_+)$ (a convex cone).


Question:


* Is $\mathcal Y\_+$ necessarily closed in $\mathcal Y$?
* If not, are there nice, easily verifiable conditions on $\mathcal Y\_+$ that are sufficient for it to be closed?
* Do either of these answers change if I'm willing to assume that $\mathcal Y$ is itself a closed subspace of some $L^1$ space, and $T$ is positive?



Closedness of linear image of positive L1 functions
Take $\mathcal X = L^1(\Omega,\mu)$ where $\Omega = \{1,2,3,\dots\}$ and measure $\mu(\{k\}) = p\_k$ with $p\_k > 0$,
$\sum p\_k = 1$. The norm in $\mathcal X$ is
$$
\|f\|\_{\mathcal X} = \sum\_k |f(k)|p\_k .
$$
Let $\mathcal Y = \mathbb R^2$ with norm
$$
\|(x,y)\|\_{\mathcal Y} = \frac{1}{2}|x|+\frac{1}{2}|y|.
$$
Thus $\mathcal Y$ is also $L^1$ of a probability space.


Let $(t\_k)\_{k=1}^\infty$ be a sequence or reals in $(0,1)$, so that $t\_k \to 0$. We may assume $t\_1 = 2/3, t\_2 = 1/3$. Define $T : \mathcal X \to \mathcal Y$ as follows.
Suppose $f \in \mathcal X$; that is $\sum\_k |f(k)| p\_k < \infty$. Then define
$$
T(f) = \left(\sum\_k f(k)p\_k t\_k , \sum\_k f(k)p\_k (1-t\_k)\right) \in \mathcal Y.
$$
Then:


$\bullet\quad$ $T$ is linear


$\bullet\quad$ $T$ is bounded


$\bullet\quad$ $T$ is positive


$\bullet\quad$ for each $k \in \mathbb N$, we have $(t\_k,1-t\_k) \in T(\mathcal X\_+)$


$\bullet\quad$ $(2/3,1/3),(1/3,2/3) \in T(\mathcal X)$, so $T(\mathcal X) = \mathcal Y$


$\bullet\quad$ $(0,1) \notin T(\mathcal X\_+)$


Thus the convex cone $T(\mathcal X\_+)$ is not closed in $\mathcal Y$.


