I've got my hands over Aliprantis and Burkinshaw's Positive operators (@t.b.: thank you for hints!) and read something on Riesz spaces and Banach lattices. A Banach lattice is a triple $(E, \le , \lVert \cdot \rVert )$, where
- $E$ is a real vector space.
- $\le$ is an ordering on $E$ compatible with vector space operations and s.t. for every $f, g\in E$ there exist $f \vee g , f \wedge g \in E$. We then define $\lvert f \rvert= f \vee (-f)$.
- $\lVert \cdot \rVert$ is a Banach norm on $E$ s.t. $\lvert f \rvert \le \lvert g \rvert \Rightarrow \lVert f \rVert \le \lVert g \rVert$.
A positive operator on $E$ is then a linear operator that is order preserving. It is not obvious that every positive linear operator is continuous (cfr. Aliprantis & Burkinshaw, §4.3), hence bounded.
Now let $T$ be a positive operator on $E$. By definition
$\lVert T \rVert= \sup_{\lVert f\rVert \le 1} \lVert Tf \rVert.$
We claim that, being $T$ positive, it suffices to take the supremum over positive $f$ (that is, over all $f$ that are greater than the origin):
$\lVert T \rVert = \sup_{\lVert f \rVert \le 1, f \ge 0} \lVert Tf \rVert.$
In fact, it is clear that the second supremum is lesser than the first. To prove the reverse inequality take $\lVert f \rVert \le 1$ and observe that $ \lVert \lvert f \rvert \rVert \le 1$ also. We have $Tf \le T\lvert f \rvert$ and so $\lVert Tf \rVert \le \lVert T \lvert f \rvert \rVert$.