Exercise 7, § 5

Let A be a set of closed operators with dense domain on a Hilbert space E. Let $\lambda \in \mathbf{C}$. One says that A is $\lambda$-commutative if $\lambda$ belongs to the resolvent set of $u$ for every $u\in A$, and if the endomorphisms $R(u, \lambda )$ of E, for $u\in A$, are pairwise permutable.

One says that A is commutative in the sense of the resolvent if there exists $\lambda \in \mathbf{C}$ such that A is $\lambda$-commutative.

A set reduced to a single closed operator with dense domain $u$ is commutative if, and only if, the resolvent set of $u$ is non-empty.

a) If $A\subset \mathscr{L}$ (E), then A is commutative in the sense of the resolvent if, and only if, the union of the spectra of the elements of A is not equal to $\mathbf{C}$, and if A is commutative.

b) Let A be a set of closed operators with dense domain on E which is $\lambda$-commutative. Then A is $\mu$-commutative for every complex number $\mu$ belonging to the resolvent set of every $u\in A$.

c) Let A be a set of closed operators with dense domain on E, commutative in the sense of the resolvent, and such that $u^*\in A$ for every $u\in A$. There exist a locally compact topological space X, a positive measure $\mu$ on X, an isometric isomorphism $v$ of $L^2(X, \mu)$ onto E, and a mapping $u\mapsto g_u$ of A into the set of $\mu$-measurable functions such that $u=v\circ m_{g_u}\circ v^{-1}$ for every $u\in A$.