How does positivity affect operator norms?

Question

Take a measure space $\Omega$, an exponent $1 \le p < \infty$ and a linear operator $T \colon L^p(\Omega) \to L^p(\Omega)$. Suppose that $T$ is positive, that is $Tf \ge 0$ a.e. if $f \ge 0$ almost everywhere. Then

$$\lVert T \rVert = \sup_{f \ge 0, \int_{\Omega} f^p \le 1} \lVert Tf \rVert_p;$$

that is, it suffices to take the supremum over positive functions.

I was wondering if there is something more general here. For example, does this hold if $T$ is an operator on $C(K)$ for compact $K$? Is this property true in some abstract class of Banach spaces?

Answer 1

I've got my hands over Aliprantis and Burkinshaw's Positive operators (@Theo Buehler: 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, by positivity, it suffices to take the supremum over positive (that is, greater than the origin) $f$:

$$\lVert T \rVert = \sup_{\lVert f \rVert \le 1, f \ge 0} \lVert Tf \rVert.$$

In fact, clearly the second supremum is lesser than the first; now 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 $\lVert Tf \rVert \le \lVert T \lvert f \rvert \rVert$. This proves the other inequality.