Let $E$ be a real ordered Banach space with the positive cone $P$ of $E$. It is easy to see that the positive cone $P$ is the closed convex cone of $E$.

We assume that $E_0 := P - P $ is a dense linear subspace of $E$. This means that $P$ is a total cone of $E$ (i.e., $\overline{P-P} = E$).

We know that if there is a bounded linear operator $A \colon E_0 \to E_0$, by Continuous Linear extension theorem, then the operator $A$ has a unique extension $\bar{A} \colon E \to E$ such that $\bar{A}$ is also a bounded linear operator with $\bar{A} v = Av$ for each $v \in E_0$ and $\| \bar{A} \| = \| A \|$.

My question is that if we suppose further that the operator $A$ is positive (i.e., $A(P) \subset P$), then is its extension $\bar{A}$ still positive (i.e., $\bar{A}(P) \subset P$)? If so, how to prove it?

Any ideas or suggestions are most welcome! Thanks in advance:)

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    $\begingroup$ Its not necessarily true unless you have some further restrictions. For example let $E$ be $\ell^2(\Bbb N)$ and $P$ be the cone of such sequences with only entries $≥0$ and infinitely many non-zero entries. Take $E_0$ to be the dense subspace of such sequences with only finitely many non-zero elements, then $E_0\cap P= 0$ and any linear map sends $E_0\cap P$ into $P$. It does not follow that the extension sends $P$ to $P$. $\endgroup$ – s.harp Nov 27 '17 at 23:18
  • $\begingroup$ You should want at least $E_0\cap P$ to be dense in $P$ and possibly for $P$ to be closed, in that case the result is easy. $\endgroup$ – s.harp Nov 27 '17 at 23:20
  • $\begingroup$ Thank you so much @s.harp. You are right, I have edited my question again. In that case, it is clear that $E_0 \cap P = P$ with $P$ being closed. Thus, for any fixed $v \in P$, by positive property of $A$, $Av \in P$, which in turn yields that $\bar{A} v = Av \in P$, by the construction of the unique extension $\bar{A}$. This means that $\bar{A}$ is indeed positive as desired. To show this result, I haven't used the closeness of the cone $P$. So, I'm wondering that is the closeness of $P$ necessary here? $\endgroup$ – Paradiesvogel Nov 28 '17 at 0:56
  • $\begingroup$ Since $P \subset E_0$ we have $A|_{P} = \bar A|_{P}$. Thus, $A(P) = \bar A(P)$. $\endgroup$ – gerw Nov 28 '17 at 7:41
  • $\begingroup$ Thanks @gerw. I agree with you. May I ask you one more question please? As $E_0$ is a dense subspace of $E$, we know that the well-known BLT (bounded linear transformation) extension theorem is talking about the existence of a unique extension of a bounded linear operator $A \colon E_0 \to E$ rather than $A \colon E_0 \to E_0$ I wrote above, I'm not 100% sure that does BLT extension theorem apply to $A \colon E_0 \to E_0$? Is it correct? Many thanks again:) $\endgroup$ – Paradiesvogel Nov 28 '17 at 8:04

To answer the question asked in the OP:

Let $P$ be a cone in a Banach space $E$ so that $E_0=\mathrm{span}(P)$ is dense in $E$, if a continuous linear operator $A:E_0\to E$ is positive, does it follow that its closure $\overline{A}: E\to E$ is positive?

This is true for these special conditions, namely since $P\subset E_0$ we have $\overline A(P) = \overline A\lvert_{E_0}(P) = A(P)\subset P$ by the positivity of $A$.

Perhaps more natural is the question what happens when $P\not\subset E_0$. But we cannot be too general, for example if $E_0$ and $P$ have intersection $\{0\}$ then any map $E_0\to E$ is positive, even the restriction of non-positive maps will be so.

I think the clearest demands to make are that $E_0\cap P$ is dense in $P$ and that $P$ is a closed cone. Then for any $p\in P$ we have a sequence $p_n\in E_0\cap P$ with $p_n\to p$. So $$\overline A(p) =\lim_n \overline A(p_n) =\lim_n\, A(p_n).$$ Now by positivity of $A$ we have that $A(p_n)$ is in $P$ and by closedness of $P$ the limit of this must lie in $P$. Thus $\overline A(p)\in P$. So $\overline A$ is a positive linear map.


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