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

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 $$\mathcal Y$$.