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Related to Jon's question, I have two questions. Let $\mathcal{A}$ be a concrete $C^*$-algebra on a Hilbert space $\mathcal{H}$. For any selfadjoint subset $S$ of $\mathbb{B}(\mathcal{H})$, let $S^m$ denote the set of elements of $\mathbb{B}(\mathcal{H})_{sa}$ that can be obtained as the strong limits of monotone increasing nets from $S$.

Question 1. Is $((\mathcal{A}_{sa})^m)^m=(\mathcal{A}_{sa})^m$? (Maybe this is very basic.)
Question 2. Does the $C^*$-algebra $C^*((\mathcal{A}_{sa})^m)$ generated by $(\mathcal{A}_{sa})^m$ in $\mathbb{B}(\mathcal{H})$ coincide with the strong closure of $\mathcal{A}$ in $\mathbb{B}(\mathcal{H})$?

For Question 2, I have been thinking that Pedersen's up-down-up theorem [Theorem 2 in American Journal of Mathematics 94 (1972), 955-962] might be useful, but I couldn't figure out.

flag	asked Aug 3 at 12:56 Masayoshi Kaneda 145●6

Nik Weaver · Accepted Answer · 2012-08-04 15:20:25Z UTC

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Edited. As Masayoshi points out, my reading of Hamana's paper was incorrect. I'm quite sure question 1 is false in general but I don't have a reference. (Masayoshi, did you look in Pedersen's book? I feel the answer may be there but I don't have access to it right now.)

I guess I'd better be more explicit about question 2. For example, take $A = C[0,1]$ acting by multiplication on $l^2[0,1]$. Then the strong closure of $A$ equals $l^\infty[0,1]$, but $A_{sa}^m$ is contained in e.g. the set of bounded Borel measurable functions, so the C*-algebra it generates is also contained in that class.

edited Aug 4 at 15:20

answered Aug 3 at 15:36

Nik Weaver
4,810●1●8●21

	Thank you Nik for your answer. Please allow me to ask further questions following your answer. For Question 1: Yes, I know that paper by Hamana. In fact, I previously highlighted the sentence `$B_{sa}$ is itself a unique real linear subspace $S$ of $B_{sa}$ such that $A_{sa}\subset S$ and $S^m=S$'' in Introduction. I think that you are referring this part. In this statement, $S$ is assumed to be a real linear subspace of $B_{sa}$. However, in my question, $(A_{sa})^m$ is just a selfadjoint `subset'' of $B(H)_{sa}$ and not a linear subspace in general. Could you please clarify this point? – Masayoshi Kaneda Aug 4 at 10:45
	For Question 2: If you don't mind, could you please provide me with your counterexample in the abelian case? Thank you. P.S. My Question 1 seems exactly what Jon is asking. – Masayoshi Kaneda Aug 4 at 11:14
	Nik, thank you again for your response. For Question 2, I see, the point is to consider $l^{\infty}[0,1]$ instead of $L^{\infty}[0,1]$. Thank you for a nice example. For Question 1, of course, I looked over Pedersen’s book, but I couldn’t find the answer. This is why I am placing the question on this board. – Masayoshi Kaneda Aug 5 at 13:05
	(Continuation) In Pedersen’s book (Theorem 2.4.4), Kadison’s other (but related) result “A $C^*$-subalgebra $M$ of $B(H)$ is a von Neumann algebra if and only if $(M_{sa})^m= M_{sa}$” (Lemma 1 of [Kadison, Operator algebras with a faithful weakly-closed representation, Ann. of Math. (2) 64 (1956), 175-181]) is presented. But the proof in Pedersen’s book is not Kadison’s original one, but one using his own lemma which was used to prove his “up-down” theorem. – Masayoshi Kaneda Aug 5 at 13:07
	(Continuation) Kadison’s result cited in Hamana’s paper can be found on pp. 316-317 of [Kadison, Unitary invariants for representations of operator algebras, Ann. of Math. (2) 66 (1957), 304-379], though it is not presented as a theorem or lemma. (Note that in this paper, $A_1^m$ is used in a different meaning from the usual one. It is the union of the monotone closures of $A_1$ taken “transfinitely recursively”. Also the monotone closures are taken in both directions, i.e., both increasing and decreasing.) – Masayoshi Kaneda Aug 5 at 13:11

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