Projections in atomless von Neumann algebras

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.

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.

• beautiful proof. Thank you very much! Commented Nov 13, 2023 at 1:57

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 $$\| x_n p_i\| = \|x_n\|$$.

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