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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?
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    $\begingroup$ $T$ is bounded? $\endgroup$ Nov 8, 2020 at 21:00
  • $\begingroup$ Yes, thank you! I added the word "continuous" accordingly. $\endgroup$ Nov 8, 2020 at 21:28

1 Answer 1

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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 $\mathcal Y$.

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