Closedness of linear image of positive L1 functions 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?

 A: 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 $\mathcal Y$.
