My question concersns operators on Hilbert spaces.
It is well-known that any family of mutually commuting compact self-adjoint operators can be simultaneously diagonalized. Is the same true for any family of mutually commuting bounded self-adjoint operators?
The answer depends on the definition of "diagonalized" operator and "simultaneously diagonalized" family of operators. If you mean the existence of a basis in which all operators take diagonal form, then the answer is negative, because even for one self-adjoint bounded operator such a basis not always exists.
In some very abstract form something like diagonalization follows from Gelfand-Neimark theorem on commutative C*-algebras: the subalgebra of algebra of operators generated by your commutative family of self-adjoint operators is a C*-algebra, so it is isomorphic to a $C(K)$-algebra.
Thank you for the reply. Let me explain the question.
By von Neumann's theorem, two bounded self-adjoint operators $A$ and $B$ commute if and only if for the corresponding projection-valued measure $m_A$ and $m_B$ we have following: every value of $m_A$ commutes with all values of $m_B$. Now, let $F$ be a family of mutually commuting operators and let $X$ be the set formed by all values of the projective-valued measures of operators from $F$.
The question is the following: Is there an orthonormal basis such that the range of every projection from $X$ is spanned by a subset of this basis?
Dear Mark Pankov ,
I doubt that the answer can be positive even in the case of one operator.
Namely, consider the operator $T: L_2[0, 1] \to L_2[0, 1]$ that acts by the rule $(Tf)(t) = t f(t)$. This is a self-adjoint operator, its spectrum is $[0, 1]$ and its projection-valued measure $m_T$ is easy to calculate: for every Borel subset $D \subset [0, 1]$ $m_T(D)$ is the orthoprojector onto the subspace $L_2(D)$ of those $f \in L_2[0, 1]$ whose support lies (up to a subset of Lebesgue measure 0) in D.
So, for this $T$ you want to have an orthonormal basis $E= (e_k)_{k \in \mathbb N}$ such that for every Borel subset $D \subset [0, 1]$ of positive measure the intersection of $E$ and $L_2(D)$ forms a basis in $L_2(D)$. But this is impossible, because for the sequence $A_n$ of supports of $e_n$ one can find a Borel set $D$ of positive measure such that for each $n$ the set $[0, 1] \setminus D$ intersects $A_n$ by a set of positive measure, which means that for this special $D$ the intersection of $E$ and $L_2(D)$ is empty.
Best, Vladimir
Dear Vladimir, thank you very much! So, there is a family $F$ of mutually commuting projections with the following property: there is no orthonormal basis $B$ such that the range of every projection from $F$ is spanned by a subset of $B$. With best regards, Mark
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