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Let $\varepsilon>0$. If we consider a sequence $\{f_n\}$ in $L_\infty(0,1)$, then there exists a very small subset $A$ of $(0,1)$ with $m(A)<\varepsilon$ such that $$\|f_n \chi_A\|_\infty =\|f_n \|_\infty $$ for all $n$. My question is, do we have an analogue of this result in the von Neuamnn algebra setting? Precisely, let $M$ be an atomless von Neumann algebra and let $\{x_n\}$ be a sequence in $M$. Can we find a projection $p$ in $M$ which is very small (say, for a semifinite faithful normal weight, $\omega(p)<\varepsilon$) such that $$\|x_n p\|_\infty =\|x_n \|_\infty $$ for all $n$.

For the case of a semifinite von Neumann algebra, it is true because we may take $p_n$ to be very small such that $\|x_n p_n\|_\infty =\|x_n\|_\infty$ and let $p:=\vee p_n$. Since $\tau(p)\le \sum_{n\ge 1}\tau(p_n)$, we may choose suitable $p_n$ such that $\tau(p)<\varepsilon$. Moreover, $$\|x_n p\|_\infty =\|x_n \|_\infty $$ for all $n$. However, for the type III case, it seems to be rather difficult.

I already asked this question in stackexchange https://math.stackexchange.com/questions/4804057/projections-in-atomless-von-neumann-algebras

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2 Answers 2

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Also with respect to a normal semifinite weight on a non-atomic von Neumann algebra, the result is true. But the proof that I found is much less elegant than the proof of Narutaka Ozawa for the case of a state.

Proposition 1. Let $M$ be a non-atomic von Neumann algebra and let $\varphi$ be a normal semifinite weight. Let $x_n \in M$ be a sequence of elements. For every $\varepsilon > 0$, there exists a projection $p \in M$ such that $\varphi(p) < \varepsilon$ and $\|x_n p\| = \|x_n\|$ for all $n$.

Adding to $\varphi$ a faithful normal semifinite weight, we may assume that $\varphi$ is faithful. We fix $M$ and $\varphi$. We view $M$ as acting on a Hilbert space $H$. We need a few lemmas and the first one is the key lemma.

Lemma 2. Let $p,q \in M$ be projections with $\varphi(p) < \infty$ and $\varphi(q) < \infty$ and $q \neq 0$. There exists a sequence of nonzero projections $q_n \leq q$ such that $\varphi(p \vee q_n) < \infty$ and $\varphi(p \vee q_n - p) \to 0$.

Proof. If $q \leq p$, we take $q_n = q$ for all $n$. So we may assume that $q \not\leq p$. We use Theorem 1.3 in the article of Raeburn and Sinclair (https://www.jstor.org/stable/24491975) to model the position of the two projections $p$ and $q$. So we define the C$^*$-algebra

$$A = \bigl\{F \in C([0,1],M_2(\mathbb{C})) \bigm| F(0) \;\text{is diagonal and}\; F(1) \;\text{is a multiple of}\; \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & 0\end{smallmatrix}\bigr)\bigr\}$$

with projections $P,Q \in A$ given by

$$P(t) = \begin{pmatrix} t & \sqrt{t(1-t)} \\ \sqrt{t(1-t)} & 1-t \end{pmatrix} \;\;\text{and}\;\; Q(t) = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$

By Theorem 1.3 in the above mentioned paper, there is a unique $*$-homomorphism $\pi : A \to M$ satisfying $\pi(P) = p$ and $\pi(Q) = q$. Denoting by $S(M_2(\mathbb{C}))$ the state space of $M_2(\mathbb{C})$, we find a measure $\mu$ on the Borel $\sigma$-algebra of $[0,1]$ and a Borel map $[0,1] \to S(M_2(\mathbb{C})) : t \mapsto \rho_t$ such that

$$\varphi(\pi(F)) = \int_0^1 \rho_t(F(t)) \, d\mu(t)$$

for all $F \in A^+$. Note that $\mu([0,t_0]) < \infty$ for all $t_0 \in [0,1)$. Define the Borel set ${\mathcal{U}} = \{t \in [0,1] \mid \rho_t(Q(t)) > 0\}$. Since $q \neq 0$, we have that $\mu({\mathcal{U}}) > 0$. We distinguish three cases.

Case 1. The measure $\mu|_{\mathcal{U}}$ has a nonatomic part. We can then find $t_0 \in [0,1)$ and a decreasing sequence ${\mathcal{U}}_n \subset {\mathcal{U}} \cap (0,t_0)$ of Borel sets with empty intersection such that $\mu({\mathcal{U}}_n) > 0$ for all $n$. To every Borel set ${\mathcal{U}}_n$ corresponds (by properly extending $\pi$) a projection $z_n \in M$ that commutes with $p$ and $q$ and that satisfies $\varphi(z_n) < \infty$ and $q z_n \neq 0$ for all $n$, and $\varphi(z_n) \to 0$. Write $q_n = q z_n$. Since $z_n$ commutes with $p$ and $q$, we have that

$$(p \vee q_n)(1-z_n) = p(1-z_n) \vee q_n(1-z_n) = p (1-z_n) \; .$$

So $p \vee q_n - p \leq z_n$ and the lemma follows.

Case 2. The measure $\mu|_{\mathcal{U}}$ has an atom in $0$. To this atom corresponds a projection $z \in M$ that commutes with $p$ and $q$ and that satisfies the following properties: $\varphi(z) < \infty$, $qz \neq 0$ and $pz = (1-q)z$. Since $M$ is non-atomic, we can choose a sequence of nonzero projections $q_n \leq qz$ such that $q_n \to 0$ strongly. Since $\varphi(qz) < \infty$, we have that $\varphi(q_n) \to 0$. We again find that $(p \vee q_n - p)(1-z) = 0$, while

$$(p \vee q_n - p)z = pz \vee q_n - pz = q_n$$

because $pz = (1-q)z$ is orthogonal to $q_n$. Again the lemma follows.

Case 3. The measure $\mu|_{\mathcal{U}}$ has an atom in $t \in (0,1)$. To this atom corresponds a projection $z \in M$ that commutes with $p$ and $q$ and that satisfies the following properties: $\varphi(z) < \infty$, $qz \neq 0$ and there exists a unital embedding $\theta : M_2(\mathbb{C}) \to z M z$ such that

$$\theta\begin{pmatrix} t & \sqrt{t(1-t)} \\ \sqrt{t(1-t)} & 1-t \end{pmatrix} = pz \;\;\text{and}\;\; \theta \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = q z \; .$$

Define the partial isometry $v \in z M z$ by

$$v = \theta \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \; .$$

Since $M$ is non-atomic, we can choose a sequence of nonzero projections $q_n \leq qz$ such that $q_n \to 0$ strongly. Define the projections $e_n \in z M z$ by $e_n = q_n + v q_n v^*$. By construction, $e_n$ commutes with $p$ and $q$ and $e_n \to 0$ strongly. Since $\varphi(z) < \infty$, also $\varphi(e_n) \to 0$. As before, $(p \vee q_n - p)(1-e_n) = 0$, so that $p \vee q_n - p \leq e_n$, so that the lemma follows.

The remaining case would be that $\mu|_{\mathcal{U}}$ is a Dirac measure at $1$. But that would imply that $q \leq p$, which we ruled out at the start. So the lemma is proven.

Lemma 3. Let $e_n$ be a sequence of nonzero projections in $M$ with $\varphi(e_n) < \infty$ for all $n$. For every $\varepsilon > 0$, there exists a projection $p \in M$ with $\varphi(p) \leq \varepsilon$ and $\|e_n p\| = 1$ for all $n$.

Proof. We make the following inductive construction of a sequence of nonzero projections $p_n \leq e_n$ such that $\varphi(p_1 \vee \cdots \vee p_n) \leq \varepsilon(1-2^{-n})$. For $n=1$, since $M$ is non-atomic and $\varphi(e_1) < \infty$, we can choose a nonzero projection $p_1 \leq e_1$ such that $\varphi(p_1) \leq \varepsilon / 2$. Having chosen $p_1,\ldots,p_n$, we then apply Lemma 2 to $p_1 \vee \cdots \vee p_n$ and $p_{n+1}$.

Define $p$ as the join of all the projections $p_n$. By construction, $\varphi(p) \leq \varepsilon$. Since $p \geq p_n$, also $\|e_n p\| \geq \|e_n p_n\| = \|p_n\| = 1$. So, $\|e_n p\| = 1$ for all $n$ and the lemma is proven.

Lemma 4. For every $x \in M$, there exists a sequence of projections $p_n \in M$ such that $\varphi(p_n) < \infty$ and $\|x p_n\| \to \|x\|$.

Proof. Choose unit vectors $\xi_n \in H$ such that $\|x \xi_n\| \to \|x\|$. Since $\varphi$ is semifinite, we can choose projections $p_n \in M$ such that $\varphi(p_n) < \infty$ and $\|p_n \xi_n - \xi_n\| < 1/n$. Then also $\|x p_n \xi_n\| \to \|x\|$ and the lemma is proven.

Lemma 5. Assume that $x \in M$ and that $p_n \in M$ is a sequence of projections such that $\|x p_n\| \to \|x\|$. If $p \in M$ is a projection satisfying $\|x p_n p\| = \|x p_n\|$ for all $n$, then also $\|x p\| = \|x\|$.

Proof. Since $\|xp_n\| \to \|x\|$ and $\|x p_n p\| = \|x p_n\|$ for all $n$, we can choose a sequence of unit vectors $\xi_n \in H$ such that $\|x p_n p \xi_n\| \to \|x\|$. Since $\|p_n p \xi_n\| \leq 1$ for all $n$, it follows in particular that $\|p_n p \xi_n\| \to 1$. This implies that $\|p_n p \xi_n - p \xi_n\| \to 0$. Thus, $\|x p_n p \xi_n - x p \xi_n\| \to 0$, so that $\|x p \xi_n\| \to \|x\|$. Hence, $\|xp\| = \|x\|$.

We are now ready to prove Proposition 1. Fix $\varepsilon > 0$. By Lemma 4, we can choose projections $p_{n,k} \in M$ such that $\varphi(p_{n,k}) < \infty$ for all $n,k$ and $\lim_k \|x_n p_{n,k}\| = \|x_n\|$ for all $n$. By Lemma 5, it suffices to find a projection $p \in M$ with $\varphi(p) \leq \varepsilon$ and $\|x_n p_{n,k} p\| = \|x_n p_{n,k}\|$ for all $n,k$. Replacing $x_n$ by the countable family $x_n p_{n,k}$, we might thus assume from the start that $\varphi(|x_n|) < \infty$ for all $n$. Replacing $x_n$ by a multiple, we may also assume that $\|x_n\| = 1$ for all $n$.

Denote by $e_{n,k}$ the spectral projection

$$e_{n,k} = 1_{[1-1/k,1]}(|x_n|) \; .$$

Since $\|x_n\| = 1$, we have that all $e_{n,k}$ are nonzero. Since $\varphi(|x_n|) < \infty$, we also have that $\varphi(e_{n,k}) < \infty$ for all $n,k$. By Lemma 3, we find a projection $p \in M$ with $\varphi(p) \leq \varepsilon$ and $\|e_{n,k} p\| = 1$ for all $n,k$. Fix $n$. For every $k$, we have

$$\|x_n p\|^2 = \| p |x_n|^2 p \| \geq (1-1/k)^2 \|p e_{n,k} p\| = (1-1/k)^2 \|e_{n,k} p\|^2 = (1-1/k)^2 \; .$$

Since this holds for all $k$, it follows that $\|x_n p\| = 1$ for all $n$ and the proposition is proven.

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  • $\begingroup$ beautiful proof. Thank you very much! $\endgroup$
    – user92646
    Commented Nov 13, 2023 at 1:57
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It is not true as is stated, even for a single $x$ and the faithful normal trace $\tau$ on $B(\ell_2)$; e.g., if $x=\mathrm{diag}_n(1-n^{-1})$, then any projection $p$ such that $\|xp\|=1$ is infinite.

On the other hand, the answer is yes if it is requested that $p_i\searrow0$ and $\|x_np_i\|=\|x_n\|$ for all $n$ and $i$.

Proof. For each $n$, there is a pure state $\phi_n$ on $M$ that satisfies $\phi_n(x_n^*x_n)=\| x_n\|^2$. Then, since $M$ is non-atomic, all $\phi_n$ are singular and so is $\phi:=\sum 2^{-n}\phi_n$. Hence there is a net of projections $p_i$ such that $p_i\searrow 0$ and $\phi(p_i)=1$ for all $i$. Then $\phi_n(p_i)=1$ for every $n$ and $i$, implying $\| x_n p_i\| = \|x_n\|$.

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    $\begingroup$ Isn’t $B(l^2)$ atomic though? $\endgroup$
    – David Gao
    Commented Nov 12, 2023 at 2:13
  • $\begingroup$ Dear Professor Ozawa, I would like to understand why you need to mention that \phi_n's are singular. Where did you use it? $\endgroup$
    – user92646
    Commented Nov 12, 2023 at 2:38
  • $\begingroup$ @user92646 The existence of $p_i \searrow 0$ with $\phi(p_i) = 1$ for all $i$ is equivalent to $\phi$ being singular. $\endgroup$
    – David Gao
    Commented Nov 12, 2023 at 2:50
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    $\begingroup$ @David Gao: Yes, I have to take back the first paragraph, and I no longer see if snf weight case is impossible. $\endgroup$ Commented Nov 12, 2023 at 3:32
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    $\begingroup$ @NarutakaOZAWA: I think that also the case of a normal semifinite weight works, but my argument is quite involved. I have posted it as a separate answer. $\endgroup$ Commented Nov 12, 2023 at 12:12

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