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Let $M$ be a von Neumann algebra and $A\subseteq M$ a $\sigma$-weakly dense $*$-subalgebra of $M$. Does there exist an increasing net $\{a_i\}_{i\in I}\subseteq A\cap M^+$ such that $a_i\to 1$ in the $\sigma$-weak topology?

The following two questions are related:

(1) This earlier post motivates this question.

(2) If we replace $1$ by a positive element in $M$, such approximation is no longer possible.

Thanks in advance for your assistance.

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  • $\begingroup$ Is $A \cap M^+$ in general equal to $A^+$? $\endgroup$
    – Matthew Daws
    Commented Mar 28, 2023 at 8:49
  • $\begingroup$ @MatthewDaws How do you define $A^+$? For me, $A^+ = A\cap M^+$ is a definition. $\endgroup$
    – Andromeda
    Commented Mar 28, 2023 at 8:50
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    $\begingroup$ Probably naively, I'd define $A^+ = \{a^*a : a\in A \}$... $\endgroup$
    – Matthew Daws
    Commented Mar 28, 2023 at 8:57
  • $\begingroup$ @MatthewDaws If $A$ is just a $*$-algebra, you need to take linear spans or the sum of two positive elements need not be positive. $\endgroup$
    – Andromeda
    Commented Mar 28, 2023 at 9:01
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    $\begingroup$ You are asking if any $\ast$-algebra $A$ has an increasing approximate unit. Although I don't know the answer, I doubt it's true. However it's probably hard to come up with a counterexample, because it's true if $A$ has a commuting approximate unit: (1) For $a\in A_+$ with $\| a\|<1$, one has $1-(1-a)^n\nearrow\operatorname{supp}a$. (2) If $a\in A_+$ with $\| a \| <1$, then for any $b\in A_+$ with $\|b\|<1$, one has $$a \le (1-\frac{1}{n})(1-(1-\frac{1}{2}a)^n) \le (1-\frac{1}{n})(1-(1-\frac{1}{2}(a+b))^n)$$ for $n$ large enough. $\endgroup$
    – Narutaka OZAWA
    Commented Mar 29, 2023 at 5:52

1 Answer 1

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In his comment, Narutaka Ozawa already gave a positive answer in the abelian case. The following argument gives a positive answer in the separable case, or more generally when $M$ is countably decomposable.

Proposition. Let $M \subset B(H)$ be a countably decomposable von Neumann algebra and $A \subset M$ a $\sigma$-weakly dense $*$-subalgebra of $M$. Then there exists a sequence $a_n \in A \cap M^+$ such that $a_n \leq a_{n+1}$ in $M$ for all $n \in \mathbb{N}$ and $a_n \to 1$ strongly.

Proof. Choose a faithful normal state $\omega$ on $M$. Choose a net $(a_i)_{i \in I}$ in $A$ such that $\|a_i\| \leq 1$ for all $i \in I$ and $a_i \to 1$ strongly. Then $\omega(a_i^* a_i) \to 1$ and we can choose a subsequence $b_n = a_{i_n}^* a_{i_n}$ such that $\omega(b_n) \to 1$. Since $0 \leq b_n \leq 1$ for all $n$, we get that $b_n \to 1$ strongly. Defining $p_n \in M$ as the orthogonal projection onto $\bigcap_{k=1}^n \text{Ker}\ b_k$, we get that $p_n$ is a decreasing sequence of projections and $p_n \to 0$ strongly.

Whenever $k_n$ is a sequence in $\mathbb{N}$, we can inductively define the sequence $a_n \in A$ by the formula $a_1 = 1-(1-b_1)^{k_1}$ and $$a_n = 1 - \bigl( (1-a_{n-1}) (1-b_n) (1-a_{n-1}) \bigr)^{k_n}$$ for all $n \geq 2$. We always have that $a_n \in A \cap M^+$. Because $$\bigl( (1-a_{n-1}) (1-b_n) (1-a_{n-1}) \bigr)^{k_n} = (1-a_{n-1}) d_n (1-a_{n-1}) \leq 1-a_{n-1}$$ for some $0 \leq d_n \leq 1$, we get that $a_n \geq a_{n-1}$.

Also note that the projection onto the kernel of $a_n$ equals $p_n$. Having chosen $k_1,\ldots,k_{n-1}$, note that the sequence $$\bigl( (1-a_{n-1}) (1-b_n) (1-a_{n-1}) \bigr)^k$$ decreases to $p_n$. We choose $k_n$ large enough such that $\omega(a_n) \geq \omega(1-p_n) - 1/n$.

By construction, the sequence $a_n$ is increasing and $0 \leq a_n \leq 1$ for all $n$. Also, $\omega(a_n) \to 1$. So, $a_n \to 1$ strongly.

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