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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?

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In $C(K)$ you can simply use that every function in the unit ball satisfies $-1_{K} \leq f \leq 1_{K}$, where $1_{K}(x) = 1$ is the constant function. Using this you can even show that a positive operator is continuous (that's e.g. why measures on $K$ are the same thing as positive linear functionals on $C(K)$). The general theory behind this is the theory of Riesz spaces and Banach lattices. However, I find it hard to find a good exposition on that. If I'm not completely mistaken Yosida's book contains some stuff that might be of interest to you. – t.b. May 19 '11 at 19:20
My remark on measures was maybe a bit of an overstatement. But that positivity implies continuity is a general theme in Riesz theory and often very useful. The order structure on Banach spaces is, if it interacts well with the norm, an extremely useful tool. – t.b. May 19 '11 at 19:36
up vote 3 down vote accepted

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 $\lVert Tf \rVert \le \lVert T \lvert f \rvert \rVert$.

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That's it, exactly! It wasn't that hard, was it? – t.b. May 21 '11 at 19:29
No, not hard at all! The only difficulty was to wade through all of those new definitions and properties. But in doing so I discovered a very nice book! – Giuseppe Negro May 21 '11 at 19:37
I'll be sure to have a look at it (I guess you mean Aliprantis and Burkinshaw) because I don't know it, thanks a lot for this pointer! I agree that these definitions aren't that easy and fun to go through because they seem a bit unmotivated at first. – t.b. May 21 '11 at 19:39

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