# Functional calculus for direct integrals

Suppose I have a direct integral of Hilbert spaces $H = \int^\oplus H_x dx$, and suppose I have an operator $T: H \to H$ which is decomposable, and so it can be written as $T = \int^\oplus T_x$ for some measurable field of operators $T_x$. Suppose furthermore that every $T_x$ is self-adjoint (and so also $T$ is self-adjoint), and let $f$ be a bounded measurable function on $\mathbb R$.

Under what conditions $f(T)$ is decomposable (I guess always) and equal to the integral of the field $f(T_x)$ ?

One paper which says something about this problem is Chow, Gilfeather, "Functions of direct integrals of operators". It actually states that the only necessary condition is that $T_x$ are contractions. But unfortunately I don't understand this paper, since it doesn't state its assumptions very precisely - for example, it doesn't seem to be assumed that the operator $T$ (or operators $T_x$) is (are) normal, and so I don't what kind of functional calculus is considered.

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(1) I guess you mean every T_x is self-adjoint. (2) The Chow and Gilfeather paper tells us that they are using the functional calculus developed by Schreiber in "A functional calculus for general operators in Hilbert space." The rough idea appears to be compression of functional calculus for unitary dilations of contractions. – Jonas Meyer Jan 18 '10 at 21:39
Normal generalizes self-adjoint. en.wikipedia.org/wiki/Normal_operator – Steve Huntsman Jan 18 '10 at 21:45
Jonas: (1) I edited the question accordingly, (2) Thanks for this clarification. I will look in the paper of Schreiber then, but probably this isn't what I'm looking for, since, for example, Chow-Gilfeather deal with contractions which I don't know how could be relevant to the question I asked. – Łukasz Grabowski Jan 19 '10 at 16:10

Your guess that it is always decomposable is correct. Here is a way to see this without verifying the expected formula: Borel functional calculus keeps you inside the von Neumann algebra generated by $T$, and the set of decomposable operators on $H$ is the von Neumann algebra of operators that commute with the diagonal operators on $H$ (Kadison-Ringrose 5.2.8, Takesaki 8.16; see also K-R 14.1.10 which has no Google preview).

(However, I don't know a reference (or have a proof) that the expected formula is correct. I think it should follow by Fubination once the case of characteristic functions is known.)

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what do you mean by fubination? – Martin Brandenburg Jan 20 '10 at 1:13
I mean using Fubini's theorem. There are 2 integrals going on in the Borel functional calculus of a decomposable operator: the decomposition as mentioned in the question; and the functional calculus itself, which uses the spectral decomposition of a normal operator. The case of characteristic functions is just making sure that the spectral decompositions of the T_x's behave as one would want, and my feeling was that if this were established the rest would follow by a form (or analogue) of Fubini's Theorem. It was left vague here and above because I have not put much thought into it yet. – Jonas Meyer Jan 20 '10 at 2:27

If you want just to give a brief argument with possible references to known results, you can proceed in the following way: one picks a sequence $p_n$ of polynomials converging to $f$ in the weak-measure topology on the Borel functions; then $p_n(T)$ converges to $f(T)$ even strongly (see e.g. Helemski. Lectures and exercises on functional analysis, p. 388). As $p_n(T)$ commuted with every diagonal operator, $f(T)$ does commute as well, and therefore is decomposable (Dixmier. Les algèbres d'opérateurs dans l'espace hilbertien, Thm. II.2.5.1), say, as $\int^\oplus S_x d\nu(x)$. Now, there is a subsequence $p_{n_k}$ such that $p_{n_k}(T_x)$ converges strongly to $S_x$ $x$-almost everywhere (Dixmier, Prop. II.2.3.4), so $S_x=f(T_x)$ almost everywhere.

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How do you know that there exists a sequence of polynomials converging to $f$ in the weak-measure topology? You can certainly find a sequence of polynomials for a given operator $T$ such that $p_n(T)$ converges to $f(T)$, but I'm not sure how to get such convergence for all operators $T$. For example, you can't just take a sequence $p_n$ which converges to $f$ point-wise because such a sequence might not exist. – Łukasz Grabowski Feb 10 '10 at 23:35

I wanted to make it a comment to Jonas' answer, but the system didn't allow me (because it's too long?)

I might be forced to write down my own proof of the expected formula. What do you think about the following sketch? The statement is clear for polynomials. Then take a sequence of polynomials $p_n$ converging somehow to $f$ (How?). This should imply that $p_n(T)$ converges weakly to $f(T)$ and similarly for $p_n(T_x)$ for a.e. $x$. Now one needs to check that $f(T_x)$ is a measurable field, i.e. whether $lim_n (p_n(T_x)v_x,w_x)$ converges to a measurable function whenever $v_x, w_x$ are measurable vecotr fields. But this limit is the same as $lim_n ([p_n(T)]_xv_x,w_x) = ([f(T)]_xv_x,w_x)$, because by your argument we know that $f(T)$ is decomposable (there is some argument needed here). The expected formula then holds because of the same reasoning and uniqueness of the weak limit.

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