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Consider the following fragment from Takesaki's book "Theory of operator algebra I":

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Why is the boxed sentence true? It looks like they replace $A$ by its strong$^*$-closure. Is this correct? If so, why is this allowed? It looks like I would have to show that if $B$ is the strong$^*$-closure of $A$, then $$\overline{A_1}^{\operatorname{strong}^*}= \overline{B_1}^{\operatorname{strong}^*}.$$ Obviously the inclusion $\subseteq$ holds, and to see the converse inclusion it suffices to show that $B_1 \subseteq \overline{A_1}^{\operatorname{strong^*}}$ but I think this is not easy to show, so probably there is soem other reasoning why we can assume $\operatorname{id}_H \in A$. Any help will be greatly appreciated!

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  • $\begingroup$ Of course Mathew's answer works nicely. But if you read Takesaki's book carefully, it is already done in Theorem 3.9 in Chapter II. The existence of the multiplicative identity for any strong-$*$ closed self-adjoint algebra is in turn dealt with in the study of extreme points of the closed unit ball of $C^*$-algebras (section 10 of Chapter I) and the Krein-Milman theorem. $\endgroup$
    – Hua Wang
    Jan 9, 2022 at 15:42

1 Answer 1

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No, it is not the case that $A$ is being replaced by its strong closure.

That paragraph performs a whole series of reductions:

  • We may assume that $A$ is norm closed (i.e. is a $C^*$-algebra)
  • We may assume that $A$ acts non-degenerately, so the weak closure of $A$, say $M$, is a unital (so is a von Neumann algebra).

Next, it is proved that:

  • the strong$^*$-closure of the unit ball of $A$ contains the identity operator $1$.

We are trying to prove that the unit ball of $A$ is strong$^*$-dense in the unit ball of $M$. Let $X$ be the strong$^*$-closure of the unit ball of $A$, let $A_1$ be the unitisation of $A$ (so the linear span of $A$ and $1$) and let $X_1$ be the strong$^*$-closure of the unit ball of $A_1$. So by definition, $X\subseteq X_1$. (We know that $1 \in X$, but I cannot see how to directly use this.)

Let $x = 1+a\in A_1$ with $\|x\|\leq 1$. As in the paragraph, let $(u_i)$ be an approximate identity for $A$, so we know that $(u_i)\rightarrow 1$ strong$^*$. For $\xi\in H$, $$ \| u_ix\xi - x\xi\| \rightarrow 0 \quad\implies\quad \|(u_i+u_ia)\xi - x\xi\|\rightarrow 0, $$ and $$ \| (u_i+u_ia)^*\xi - x^*\xi\| = \|x^*(u_i\xi-\xi)\|\leq \|x\| \|u_i\xi-\xi\|\rightarrow 0, $$ and so $(u_i + u_ia) \rightarrow x$ strong$^*$. As $\|u_i+u_ia\| \leq \|x\| \leq 1$, it follows that $x\in X$. So the unit ball of $A_1$ is contained in $X$ and hence taking strong$^*$-closures shows that $X = X_1$.

Hence, $X$ equals the ball of $M$ if and only if $X_1$ equals the ball of $M$, and so we can (without loss of generality) replace $A$ by $A_1$, and so we may assume that $A$ is unital.

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  • $\begingroup$ Ah of course, replace $A$ by $A + \mathbb{C}1$ makes much more sense...Thanks for your answer! $\endgroup$
    – Andromeda
    Jan 5, 2022 at 21:12

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