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:
- 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 $j=1,2$.
- 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.
Any suggestions, counterexamples or pointers to the literature that anyone
may have will be greatly appreciated.
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.
Best Answer
Here are examples to show that neither of the two conditions 1. and 2. that you give as jointly sufficient can just be dropped.
Take $X$ to be two-dimensional real space with the pointwise order. Norm it by taking the closed unit ball to be the convex hull of $\{(-3,3),(0,2),(2,0),(3,-3),(0,-2),(-2,0),(-3,3)\}$. On the positive cone, $\|(x,y)\|=(x+y)/2$. Clearly, $(-1,-1)\le (-1,1)\le (1,1)$ and $\|(-1,-1)\|=\|(1,1)\|=1$ whilst $\|(-1,1)\|=1/3$.
Any attempt to write $(-1,1)=u-v$ with $u$ and $v$ positive will have to have $\|u\|,\|v\|\ge 1/2$, so your condition 1 fails. Condition 2 does hold. Define $T(x,y)=(y,y)$, which is certainly positive. For $(x,y)\ge 0$, $\|T(x,y)\|=y\le 2\|(x,y)\|$. On the other hand $\|T(-1,1)\|=\|(1,1)\|=1$ whilst $\|(-1,1)\|=1/3$ so that $\|T\|\ge 3$.
For the second example, take $Y$ to be two-dimensional real space with the pointwise order and normed by $\|(x,y)\|=\max\{\|(x,y)\|_\infty, |x-y|\}$. Take $Z$ to be two-dimensional real space with the pointwise order and supremum norm. For $(x,y)\ge (0,0)$ the two norms are equal. $Y$ has your property 1 but not 2. The identity map from $Z$ into $Y$ has norm at least 2 as $\|(-1,1)\|=2$ whilst $\|(-1,1)\|_\infty=1$. On the positive cone all norms of images are equal. To get an example from a space into itself, take $X=Y\oplus Z$ and the operator $(y,z)\mapsto (z,0)$.