# Self-adjoint bounded operator, resolution of the identity, def. of the diagonal

Let $A$ be a self adjoint bounded linear operator with a continuous spectrum $\sigma(A)=[a,b]$ which acts on a separable Hilbert space. Let $E_\lambda$ be its resolution of the identity.

For example, $A$ is an operator on $L_2(0,1)$ which acts by multiplication $$A(x(t)):= t \cdot x(t), \ \ \ \sigma(A)=[0,1].$$

Let $\theta=\{\lambda_0=a, \lambda_1,...,\lambda_n=b\}$ be a decomposition of an interval $[a,b]$ and let $B$ be an arbitrary bounded self adjoint operator. Let's consider the sum $$Diag_\theta(B):=\sum_{k=1}^nE_{\Delta_k}\cdot B \cdot E_{\Delta_k}, \ \ \text{where} \ \ \Delta_k=\lambda_k-\lambda_{k-1}.$$

The question: what are the conditions on $B$ for the following sum to exist? $$Diag(B) = \lim_{m(\theta) \to 0}Diag_\theta(B), \ \ \text{where} \ \ m(\theta)=\max_{1\leq k\leq n} \Delta_k \$$ Meaning, when the diagonal'' of $B$ with respect to $A$ is well-defined?

Clearly if $A$ and $B$ commute $Diag(B)=B$. Also, if $B$ is compact $Diag(B)=0$. What is known about a general case? It would suffice to know the answer when $B$ is a projector on an infinite-dimensional subspace.

Below is the same question in Russian.

Пусть $A$ -- самосопряженный ограниченный оператор с чисто непрерывным спектром $\sigma(A)=[a,b]$, действующий на сепарабельном гильбертовом пространстве. Пусть $E_\lambda$ -- его разложение единицы.

Например, $A$ -- это оператор в $L_2(0,1)$ умножения на аргумент: $$A(x(t)):= t \cdot x(t), \ \ \ \sigma(A)=[0,1].$$

Пусть $\theta=\{\lambda_0=a, \lambda_1,...,\lambda_n=b\}$ -- произвольное разбиение отрезка $[a,b]$.

Пусть $B$ -- произвольный ограниченный самосопряженный оператор.

Рассмотрим сумму $$Diag_\theta(B):=\sum_{k=1}^nE_{\Delta_k}\cdot B \cdot E_{\Delta_k}, \ \ \text{где} \ \ \Delta_k=\lambda_k-\lambda_{k-1}.$$

Вопрос: для любого ли самосопряженного ограниченного оператора $B$ существует предел $$Diag(B) = \lim_{m(\theta) \to 0}Diag_\theta(B), \ \ \text{где} \ \ m(\theta)=\max_{1\leq k\leq n} \Delta_k \ \text{?}$$

Речь идет о существовании "диагонали" \ оператора $B$ относительно оператора $A$.

Очевидно, если операторы $A$ и $B$ коммутируют, то $Diag(B)=B$.

Если $B$ компактный оператор, то $Diag(B)=0$.

Достаточно ответить на поставленный вопрос для проектора $B$ на бесконечномерное подпространство.

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Thanks for the english version! –  sisn Jul 28 '11 at 20:53
Yakov, I bet all Russian speaking users of MO are OK with English. –  AlB Jul 28 '11 at 21:06
I don't think that $Diag(B)=0$ when $B$ is compact. What happens if $B$ is compact and commutes with $A$? –  Martin Argerami Jul 30 '11 at 18:49
Martin: I do not see any contradiction. This would just prove that (under the appropriate hypotheses) the only compact operator commuting with $A$ is $0$. –  Mikael de la Salle Aug 1 '11 at 11:22
Thanks, Mikael, I had not paid attention to the word "continuous" in the question. Now it's obvious: if $A$ generates a diffuse masa, then this masa cannot contain any finite-dimensional projections. –  Martin Argerami Aug 1 '11 at 22:09

I think the answer to the question, as it is formulated, is "no" (the explanation is below).
However, there may be replacements for your map $Diag$, which we'll discuss at the end.

Let us say that $A$ has spectral multiplicity identically equal to $1$ (i.e., the von Neumann algebra $\mathfrak{A}$ generated by $A$ in the algebra of all bounded operators $B(H)$ on $H$ is maximal abelian).

Suppose for a contradiction that $Diag(B)$ were defined by your limit procedure for any $B$ in $B(H)$. Note that $Diag_\theta(B)$ commutes with the projections $E_{\Delta_k}$ coming from the partition $\theta$; using this, I think you can prove that $Diag(B)$ commutes with $\mathfrak{A}$ and thus lies in $\mathfrak{A}$. You can also note that $Diag$ is a linear map from $B(H)$ to $\mathfrak{A}$ which is actually $\mathfrak{A}$-bilinear: $Diag(X B Y)=X Diag(B) Y$ for any $X,Y\in \mathfrak{A}$ and $B\in B(H)$. Finally, I think it will be the case that $Diag$ is a normal map, i.e., if $B_i$ is a sequence of uniformly bounded operators, converging weakly to some $B$, then also $Diag(B_i)$ will converge weakly to $Diag(B)$.

This means that $Diag$ is a normal conditional expectation from $B(H)$ onto $\mathfrak{A}$, which is impossible. Indeed, for any measure $\mu$ on $[a,b]$ which is abs. continuous wrt the spectral measure of $A$, the formula $$\tau_\mu (B)=\int Diag(B) d\mu$$ defines a state on $B(H)$ which is (by assumption on normality of $Diag$) normal and thus has the form $\tau_\mu(B)=Tr(B D_\mu)$ for some trace-class operator $D_\mu$. Because of the properties of $Diag$, $\mathfrak{A}$ is in the centralizer of $\tau_\mu$: $$\tau_\mu(X B)=\tau_\mu (B X),\quad \forall X\in \mathfrak{A},\ B\in B(H).$$ Thus $\tau_\mu$ is what is known as a hypertrace. But this entails $D_\mu \in \mathfrak{A}'$ which is impossible (since no trace-class (even compact) operators can commute with a diffuse abelian algebra). This is a contradiction.

On the positive side, there are (many) non-normal conditional expectations from $B(H)$ onto the algebra $\mathfrak{A}$; so if you don't exactly care for the formula in the definition of your $Diag$, then such "diagonals" exist. In fact, you can choose $E:B(H)\to\mathfrak{A}$ so that, for some choice of unitaries $u_k\in \mathfrak{A}$ going to zero weakly, $E(B)\in \overline{co}(\{u_k B u_k^*\})$ (i.e., $E(B)$ is computed as a kind of "average" over the unitary group of $\mathfrak{A}$. This is because $\mathfrak{A}$ is abelian and thus amenable. In fact, $E(B)$ may be called the "virtual diagonal" of $B$ in some literature.

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Dima: what makes you think that, if it exists, the limit will be normal? Are you assuming that the limit is in the norm topology? Because in the point-weak* topology, the net $Diag_\theta$ is relatively compact (because the $Diag_\theta$'s are unital completely positive), and hence has (at least) one cluster point. What is clear is that such a cluster point will be a norm $1$ map with values in $\mathfrak A$, which is the identity on $\mathfrak A$. It is therefore a conditional expectation. Not normal by your proof. –  Mikael de la Salle Aug 1 '11 at 11:37
Perhaps you are right, in any case it was not very clear what the original question wants in terms of convergence. The original question stated that working out what $Diag(B)$ means for $B$ a spectral projection is enough, which makes me think that the original question required some form of convergence that would make $Diag$ a normal map. –  Dima Shlyakhtenko Aug 2 '11 at 2:12