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 $Tx\geq 0$