# Takesaki lemma 1.16 (volume II, chapter VII)

I am trying to understand the proof of the implication $$(i)\implies (ii)$$ in Takesaki's book "Theory of operator algebras II", chapter VII, which says the following:

The relevant setting and notations are described in the following picture:

And finally, here is Takesaki's proof:

There are several things about this proof that I don't understand, and I strongly suspect that it contains some typos/mistakes.

For example, I cannot understand the proof of the equality $$(F-A_+)^\circ = F^{\hat{}}$$. Rather, I think that we have $$F^{\hat{}}= (F-A_+)^\circ \cap A_+^*.$$

I can see that by the Hahn-Banach separation theorem and the hypothesis, we have $$F= (F-A_+)^{\circ\circ }\cap A^+.$$

I think I can finish if I can show that $$((F-A_+)^\circ\cap A_+^*)^\circ \cap A_+ = (F-A_+)^{\circ\circ}\cap A_+.$$ The inclusion $$\supseteq$$ is obvious, however I struggle to prove the other inclusion. Under the extra assumption that every functional in $$A^*$$ is a difference of two positive ones, i.e. $$A^*= A_+^* - A_+^*$$, I think I can also prove the other inclusion. But I think this assumption need not hold in general. Any help in the matter is highly appreciated!

$$\newcommand{\con}{\operatorname{conv}}$$[Edit: This has turned out to be more involved than I thought, and my argument is now somewhat different from the book.] Firstly, some facts about $$A_+$$. Let $$a,b\in A_+$$ so $$a,b\geq0$$ and so $$a\geq 0 \implies a+b\geq b \implies a+b\geq 0$$. Also $$t\in\mathbb R$$ with $$t\geq 0$$ implies that $$tA_+ \subseteq A_+$$. It follows that $$A_+$$ is convex, and so also $$-A_+$$ is convex.

We are proving (i)$$\implies$$(ii) so by assumption $$F$$ is a hereditary convex closed set. Set $$F_0 = \{ tf : 0\leq t<1, f\in F \}$$. As $$F$$ is hereditary, if $$F$$ is non-empty there is $$f\in F$$ so $$0\leq0\leq f\implies 0\in F$$. Thus $$F_0\subseteq F$$, and notice that given any $$f\in F_0$$ there is $$0 and $$f_0\in F_0$$ with $$f = t_0f_0$$. $$F_0$$ is convex.

By definition, $$F-A_+ = \{ f-a : f\in F, a\in A_+\}$$. As $$F_0$$ and $$A_+$$ are convex, \begin{align*} \con(F_0\cup(-A_+)) &= \{ tf + s(-a) : f\in F_0,a\in A_+, s,t\geq0, s+t=1\} \\ &= \{ tf - a : f\in F_0, a\in A_+, 0\leq t<1 \} \cup F_0, \end{align*} as $$A_+$$ is invariant under scaling by a positive number. We union with $$F_0$$ to correspond to the case $$t=1$$, but as $$0\in A_+$$, we can write $$F_0\ni f = f + 0$$. Given $$f-a\in F_0-A_+$$, pick $$t_0,f_0$$ with $$f = t_0 f_0$$ so $$f-a = t_0f_0 + (1-t_0)(-(1-t_0)^{-1}a)$$. Hence $$\con(F_0\cup(-A_+)) = F_0 - A_+.$$

For any subsets $$X,Y$$ we have $$(X\cup Y)^\circ = X^\circ \cap Y^\circ$$, just from the definition. So $$(F_0 - A_+)^\circ = (F_0\cup(-A_+))^\circ = F_0^\circ \cap (-A_+)^\circ.$$ We first show that $$(-A_+)^\circ = A_+^*$$. If $$\omega\in (-A_+)^\circ$$ then $$\omega(-ta) \leq 1$$ for all $$t\geq0,a\in A_+$$ so $$\omega(a)\geq -1/t$$ for all $$t>0$$ so $$\omega(a)\geq 0$$. Conversely, if $$\omega(A_+)\subseteq[0,\infty)$$ then $$\omega(-a)\leq 0\leq 1$$ for all $$a\in A_+$$. Now consider \begin{align*} F_0^\circ &= \{ \omega : t\omega(f)\leq 1 \ (0\leq t<1, f\in F) \} \\ &= \{ \omega : \omega(f)\leq 1 \ (f\in F) \} = F^\circ. \end{align*} So $$(F_0 - A_+)^\circ = F^\circ \cap A_+^* = F^\wedge$$.

We now continue; we finally get to use condition (i)! Hahn-Banach shows that $$X^{\circ\circ}$$ is the closed convex hull of $$X$$, for any $$X$$ which contains $$0$$. So $$(F_0 - A_+)^{\bar{}} = (F_0 - A_+)^{\circ\circ} = (F^\wedge)^\circ.$$ As $$F_0 \subseteq F$$ and is dense, we have $$(F - A_+)^{\bar{}} \subseteq (F_0 - A_+)^{\bar{}} \subseteq (F - A_+)^{\bar{}}$$ and so we have equality. Thus $$(F - A_+)^{\bar{}} = (F^\wedge)^\circ$$. Now use that (i) holds, so $$F = (F-A_+)^{\bar{}} \cap A_+$$ so $$F = (F^\wedge)^\circ \cap A_+ = F^{\wedge\wedge}$$, that is, (ii).

• You are right. I hope I have now fixed these problems. This technique of using $(0,1)F$ instead of $F$ occurs elsewhere, when you want to get rid of "boundary" problems. May 14 at 11:02
• I think the calculation of the convex hull can be left out. Indeed, the inclusion $\operatorname{conv}(F_0\cup(-A_+))\subseteq F_0-A_+$ is obvious (the right hand side is convex and contains $F_0$ and $-A_+$), the other one is clear by the reasoning in your answer (which does not use these set equalities). yesterday