Functional calculus for direct integrals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T01:46:33Z http://mathoverflow.net/feeds/question/12232 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12232/functional-calculus-for-direct-integrals Functional calculus for direct integrals Łukasz Grabowski 2010-01-18T20:44:43Z 2010-02-10T14:08:37Z <p>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 $\mathbb R$.</p> <p>Under what conditions $f(T)$ is decomposable (I guess always) and equal to the integral of the field $f(T_x)$ ?</p> <p>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. </p> http://mathoverflow.net/questions/12232/functional-calculus-for-direct-integrals/12258#12258 Answer by Jonas Meyer for Functional calculus for direct integrals Jonas Meyer 2010-01-18T23:35:38Z 2010-01-18T23:35:38Z <p>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 $H$ (<a href="http://books.google.com/books?id=y8t50fVBHfYC&amp;lpg=PA322&amp;dq=5.2.8%20kadison%20ringrose&amp;client=firefox-a&amp;pg=PA321#v=onepage&amp;q=&amp;f=false" rel="nofollow">Kadison-Ringrose 5.2.8</a>, <a href="http://books.google.com/books?id=dTnq4hjjtgMC&amp;lpg=PA273&amp;dq=takesaki%20operator%20algebras%208.16&amp;client=firefox-a&amp;pg=PA273#v=onepage&amp;q=&amp;f=false" rel="nofollow">Takesaki 8.16</a>; see also K-R 14.1.10 which has no Google preview). </p> <p>(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.)</p> http://mathoverflow.net/questions/12232/functional-calculus-for-direct-integrals/12320#12320 Answer by Łukasz Grabowski for Functional calculus for direct integrals Łukasz Grabowski 2010-01-19T16:43:03Z 2010-01-19T16:43:03Z <p>I wanted to make it a comment to Jonas' answer, but the system didn't allow me (because it's too long?)</p> <p>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.</p> http://mathoverflow.net/questions/12232/functional-calculus-for-direct-integrals/14902#14902 Answer by Vadim Alekseev for Functional calculus for direct integrals Vadim Alekseev 2010-02-10T14:08:37Z 2010-02-10T14:08:37Z <p>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. <i>Helemski. Lectures and exercises on functional analysis</i>, p. 388). As $p_n(T)$ commuted with every diagonal operator, $f(T)$ does commute as well, and therefore is decomposable (<i>Dixmier. Les algèbres d'opérateurs dans l'espace hilbertien</i>, 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.</p>