Direct Integral - Decomposable Operators

Decomposable Operators

In our running example, any bounded linear operator T on

is given by an infinite matrix

\begin{bmatrix} T_{1 1} & T_{1 2} & \cdots & T_{1 n} & \cdots \\ T_{2 1} & T_{2 2} & \cdots & T_{2 n} & \cdots \\ \vdots & \vdots & \ddots & \vdots & \cdots \\ T_{n 1} & T_{n 2} & \cdots & T_{n n} & \cdots \\ \vdots & \vdots & \cdots & \vdots & \ddots \end{bmatrix}.

Let us consider operators that are block diagonal, that is all entries off the diagonal are zero. We call these operators decomposable. These operators can be characterized as those that commute with diagonal matrices:

\begin{bmatrix} \lambda_{1} & 0 & \cdots & 0 & \cdots \\ 0 & \lambda_{2} & \cdots & 0 & \cdots \\ \vdots & \vdots & \ddots & \vdots & \cdots \\ 0 & 0 & \cdots & \lambda_{n} & \cdots \\ \vdots & \vdots & \cdots & \vdots & \ddots \end{bmatrix}.

We now proceed to the general definition: A family of bounded operators {Tx}x∈ X with Tx ∈ L(Hx) is said to be strongly measurable if and only if its restriction to each Xn is strongly measurable. This makes sense because Hx is constant on Xn.

Measurable families of operators with an essentially bounded norm, that is

define bounded linear operators

acting in a pointwise fashion, that is

Such operators are said to be decomposable.

Examples of decomposable operators are those defined by scalar-valued (i.e. C-valued) measurable functions λ on X. In fact,

Theorem. The mapping

given by

is an involutive algebraic isomorphism onto its image.

For this reason we will identify L∞μ(X) with the image of φ.

Theorem Decomposable operators are precisely those that are in the operator commutant of the abelian algebra L∞μ(X).

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