I'm trying to investigate the interplay between the norm and cone of positive elements in ordered Banach spaces. In particular, I would like a nice characterisation of when the norm of a positive operator can be obtained as a supremum of norms over all norm 1 positive elements (see 1. below). It is easy to show that this always holds for Banach lattices, but gets tricky when trying to prove this more generally.

To be a bit more detailed, let $X$ be a Banach space with a cone (or wedge) of 'positive' elements, denoted by $X_{+}$; $X$ is then called an ordered Banach space and $x\leq y$ means $y-x\in X_{+}$. I will call $X_{+}$ generating if $X=X_{+}-X_{-}$; proper if $X_{+}\cap(-X_{+})=\{0\}$; and normal if $0\leq x\leq y$ implies $\|x\|\leq\|y\|$.

I have a conjecture that something like the following holds:

Let $X$ be an ordered Banach space with a closed, proper, generating, normal cone. Then (1) below, is equivalent with (2) AND (3) simultaneously. I.e., (1)$\Leftrightarrow$(2)$\wedge$(3).

1. For any bounded, positive operator $T:X\to X$, $\|T\|=\sup_{x\in X_{+},\|x\|=1}\|Tx\|$
2. For all $x\in X$ there exist $X\ni x_{1},x_{2}\geq0$ such that $x=x_{1}-x_{2}$ and $\|x_{j}\|\leq\|x\|$ for $j=1,2$.
3. For all $x\in X$ and $X\ni a,b\geq0$, $-a\leq x\leq b$ implies $\|x\|\leq\max \{ \|a\|,\|b\| \}$.

That (2) and (3) together imply (1) is easy. That (1) implies (2) also holds, one can prove the contrapositive statement by constructing a positive operator that doesn't satisfy (1), when one assumes that (2) is false.

That (1) implies (3) remains just out of reach. It could be that I'm missing some essential assumption on the norm of $X$. (I've played around with also assuming (what I've sometimes seen called) a Fremlin norm on $X$, meaning that for all $x\in X$, $\|x\|=\inf \{ \|y\|:y\geq0;-y\leq x\leq y \}$, but with not much luck)

EDIT: By "$T:X\to X$ a positive operator", I of course mean: $x\geq 0$ implies $Tx\geq 0$

I'm trying to investigate the interplay between the norm and cone of positive elements in ordered Banach spaces. In particular, I would like a nice characterization of when the norm of a positive operator can be obtained as a supremum of norms over all norm 1 positive elements (see 1. below). It is easy to show that this always holds for Banach lattices, but gets tricky when trying to prove this more generally.

To be a bit more detailed, let $X$ be a Banach space with a cone (or wedge) of 'positive' elements, denoted by $X_{+}$; $X$ is then called an ordered Banach space and $x\leq y$ means $y-x\in X_{+}$. I will call $X_{+}$ generating if $X=X_{+}-X_{-}$; proper if $X_{+}\cap(-X_{+})=\{0\}$; and normal if $0\leq x\leq y$ implies $\|x\|\leq\|y\|$. A bounded linear operator $T:X\to X$ is called positive, if $x\geq 0$ implies $Tx\geq 0$. An ordered Banach space with the property that every positive operator satisfies $$\|T\|=\sup\{\|Tx\|:x\in X_{+},\|x\|=1\}$$ I will say has the positive operator property, i.e., the norms of positive operators are completely determined by their behavior on the cone.

My question is:

Let $X$ be an ordered Banach space with a closed, proper, generating, normal cone. Is there a characterization of the positive operator property in terms of the cone-norm interaction?

Thus far I have found a few sufficient conditions:

2. For all $x\in X$ there exist $X\ni x_{1},x_{2}\geq0$ such that $x=x_{1}-x_{2}$ and $\|x_{j}\|\leq\|x\|$ for $j=1,2$.
3. For all $x\in X$ and $X\ni a,b\geq0$, $-a\leq x\leq b$ implies $\|x\|\leq\max \{ \|a\|,\|b\| \}$.

together imply the positive operator property.

Another sufficient condition is having the property that for any $x\in X$ it holds that $\|x\| = \inf \{\|z_+ + z_-\|:z_\pm \geq 0;x=z_+ -z_-\}$. Being a Banach lattice is sufficient to have this property (by invoking the property $\|x\|=\||x|\|$), but seems to be a bit more general since $\mathbb{R}^3$ with the Euclidean norm and the `ice-cream cone' $\{(x_1,x_2,x_3):x_1\geq (x_2^2+x_3^2)^{1/2} \}$ (which is not a lattice) also has this property (so being a lattice is not necessary for the positive operator property).

I've been unable to prove that any of these conditions are necessary. The last one had been my best bet so far, but I'm beginning to doubt the existence of a nice list of conditions on the norm and cone that together are necessary and sufficient.

UPDATE: Cleaned it up a bit, and posted some new information I found since first posting. I found no mention in any literature to this question. I also asked around a bit with no hits from people who are a bit in the know, so it seems to be wide open. Gets the 'open-problem' tag.

I'm trying to investigate the interplay between the norm and cone of positive elements in ordered Banach spaces. In particular, I would like a nice characterisation of when the norm of a positive operator can be obtained as a supremum of norms over all norm 1 positive elements (see 1. below). It is easy to show that this always holds for Banach lattices, but gets tricky when trying to prove this more generally.

To be a bit more detailed, let $X$ be a Banach space with a cone (or wedge) of 'positive' elements, denoted by $X_{+}$; $X$ is then called an ordered Banach space and $x\leq y$ means $y-x\in X_{+}$. I will call $X_{+}$ generating if $X=X_{+}-X_{-}$; proper if $X_{+}\cap(-X_{+})=\{0\}$; and normal if $0\leq x\leq y$ implies $\|x\|\leq\|y\|$.

I have a conjecture that something like the following holds:

Let $X$ be an ordered Banach space with a closed, proper, generating, normal cone. Then (1) below, is equivalent with (2) AND (3) simultaneously. I.e., (1)$\Leftrightarrow$(2)$\wedge$(3).

1. For any bounded, positive operator $T:X\to X$, $\|T\|=\sup_{x\in X_{+},\|x\|=1}\|Tx\|$
2. For all $x\in X$ there exist $X\ni x_{1},x_{2}\geq0$ such that $x=x_{1}-x_{2}$ and $\|x_{j}\|\leq\|x\|$ for $j=1,2$.
3. For all $x\in X$ and $X\ni a,b\geq0$, $-a\leq x\leq b$ implies $\|x\|\leq\max \{ \|a\|,\|b\| \}$.

That (2) and (3) together imply (1) is easy. That (1) implies (2) also holds, one can prove the contrapositive statement by constructing a positive operator that doesn't satisfy (1), when one assumes that (2) is false.

That (1) implies (3) remains just out of reach. It could be that I'm missing some essential assumption on the norm of $X$. (I've played around with also assuming (what I've sometimes seen called) a Fremlin norm on $X$, meaning that for all $x\in X$, $\|x\|=\inf \{ \|y\|:y\geq0;-y\leq x\leq y \}$, but with not much luck)

Any suggestions, counterexamples or pointers to the literature that anyone may have will be greatly appreciated.

I'm trying to investigate the interplay between the norm and cone of positive elements in ordered Banach spaces. In particular, I would like a nice characterisation of when the norm of a positive operator can be obtained as a supremum of norms over all norm 1 positive elements (see 1. below). It is easy to show that this always holds for Banach lattices, but gets tricky when trying to prove this more generally.

To be a bit more detailed, let $X$ be a Banach space with a cone (or wedge) of 'positive' elements, denoted by $X_{+}$; $X$ is then called an ordered Banach space and $x\leq y$ means $y-x\in X_{+}$. I will call $X_{+}$ generating if $X=X_{+}-X_{-}$; proper if $X_{+}\cap(-X_{+})=\{0\}$; and normal if $0\leq x\leq y$ implies $\|x\|\leq\|y\|$.

I have a conjecture that something like the following holds:

Let $X$ be an ordered Banach space with a closed, proper, generating, normal cone. Then (1) below, is equivalent with (2) AND (3) simultaneously. I.e., (1)$\Leftrightarrow$(2)$\wedge$(3).

1. For any bounded, positive operator $T:X\to X$, $\|T\|=\sup_{x\in X_{+},\|x\|=1}\|Tx\|$
2. For all $x\in X$ there exist $X\ni x_{1},x_{2}\geq0$ such that $x=x_{1}-x_{2}$ and $\|x_{j}\|\leq\|x\|$ for $j=1,2$.
3. For all $x\in X$ and $X\ni a,b\geq0$, $-a\leq x\leq b$ implies $\|x\|\leq\max \{ \|a\|,\|b\| \}$.

That (2) and (3) together imply (1) is easy. That (1) implies (2) also holds, one can prove the contrapositive statement by constructing a positive operator that doesn't satisfy (1), when one assumes that (2) is false.

That (1) implies (3) remains just out of reach. It could be that I'm missing some essential assumption on the norm of $X$. (I've played around with also assuming (what I've sometimes seen called) a Fremlin norm on $X$, meaning that for all $x\in X$, $\|x\|=\inf \{ \|y\|:y\geq0;-y\leq x\leq y \}$, but with not much luck)

Any suggestions, counterexamples or pointers to the literature that anyone may have will be greatly appreciated.

EDIT: By "$T:X\to X$ a positive operator", I of course mean: $x\geq 0$ implies $Tx\geq 0$