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In linear algebra it is true that: a) if $\mathcal{A}$ is a set of unitarily diagonalizable matrices (in $\mathbb{C}$, i.e. normal matrices) that commute with each other then they are simultaneously diagonalizable. Moreover, b) if another matrix $D$ commutes with all matrices in $\mathcal{A}$ and at least one of those in $\mathcal{A}$ has all distinct eigenvalues then it's true that $D$ can be diagonalized simultaneously with all of $\mathcal{A}$ (this diagonalization being unique mod permutations of basis elements). Moreover, $D$ is in the norm-closure of the algebra of matrices generated by $\mathcal{A}$.

Now, when dealing with infinite-dimensional Hilbert spaces, if I have a) $\mathcal{A}$, a Banach *-subalgebra of normal operators in $\mathcal{B}(H)$ then it's true that $\mathcal{A}$ is simultaneously diagonalizable (that is, there's a unitary operator $V$ from $H$ to some $L^2(X,\mu)$ such that $VTV^*(g)=f_T\cdot g$ for some function $f_T$ on $X$ and all operators $T$ in $\mathcal{A}$, functions $g\in L^2(X, \mu)$.

My question is then the infinite dimensional analogue of b) above, that is, let us assume that the measures of the resolution of identity $\{E_{\phi,\phi}\}_{\phi\in H}$ are distinct for distinct $\phi$ (this is the condition I deem equivalent to the distinct eigenvalues above, just to be clear, I mean the measures on $\Delta$, the maximal ideal space of $\mathcal{A}$, given by the Spectral Theorem for normal *-subalgebras as $\langle T\phi,\psi\rangle=\int_\Delta \hat{T}(\chi)dE_{\phi,\psi}(\chi)$ for $T\in\mathcal{A}$ and $\hat{T}(\chi)=\chi(T)$ the Gelfand transform). If I then have a (bounded) operator $D$ that commutes with all $\mathcal{A}$, does it follow that $D$ can be diagonalized in the same diagonalization we have for $\mathcal{A}$ (that is, $VDV^*$ is a multiplication operator in $L^2(X,\mu)$)? And if true, does it follow that $D$ must be on the WOT-closure of $\mathcal{A}$? (I assume the SOT-closure won't work)

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The phenomena you are looking for may be connected with the notion of a Maximal Abelian SubAlgebra (MASA) of a von Neumann algebra. Let $M\subset B(H)$ be a von Neumann algebra (i.e., a weak-operator-closed $*$-subalgebra of $B(H)$, one example being $M=B(H)$), and let $A\subset M$ be an abelian von Neumann subalgebra (i.e. $A$ is also weakly closed and is a unital subalgebra). Let $$A' = \{ T\in B(H) : Ta=aT\ \forall a\in A\}$$ be the commutant of $A$. One says that $A$ is a MASA if $A'\cap M = A$.

For $M=B(H)$, the following are standard examples of MASAs: (a) for $H=\ell^2(S)$ for $S$ a discrete set, $A = \ell^\infty(S)$ acting on $\ell^2(S)$ as multiplication operators; for $S$ finite this is the situation you were referring to in finite dimension; (b) for $H=L^2(X,\mu)$, $A=L^\infty(X,\mu)$ acting by multiplication operators on $L^2(X,\mu)$.

Let now $T\in B(H)$ be a normal operator. You may wish to ask when the von Neumann algebra $A$ generated by $T$ (= all operators of the form $f(T)$ for $f$ a bounded function defined on the spectrum of $T$ and measurable wrt the spectral measure of $T$) is a MASA. The answer comes from a version of the spectral theorem that gives the complete invariant (up to unitary conjugacy) for a normal operator $T\in B(H)$. It states: there exists a projection-valued spectral measure $\mu : \sigma(T)\to B(H)$ and a multiplicity function $n : \sigma(T)\to B(H)$, so that for some measure $\nu$ equivalent to $\mu$, $H\cong L^2(\sigma(T),\nu, \mathbb{C}^n)$ and $T$ acts as multiplication by $z$ on that space. Here $ L^2(\sigma(T),\nu, \mathbb{C}^n) $ means the subspace of $L^2(\sigma(t),\nu,\ell^2(\mathbb{N}))$ consisting of functions for which $f(x)$ lies in the subspace spanned by the first $n(x)$ basis vectors of $\ell^2(\mathbb{N})$. The function $n$ is referred to as the multiplicity function.

In this case, it is not hard to figure out what the commutant $A'\cap B(H)$ is. It consists of functions $f: \sigma(T)\to M_{n(x)\times n(x)}(\mathbb{C})$ (defined in a suitable way) which are measurable and essentially bounded on $(\sigma(T),\mu)$. (A more precise description requires the notion of a measurable field of von Neumann algebras, but I hope that you get the idea).

In particular, $A$ is a MASA iff $n(x)=1$ identically.

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