Talk:Borel functional calculus

message: By considering Borel functions on the complex plane, the requirement of the operator T being self-adjoint will be changed to being normal.
:the normal case is more general anyhow, but there's no real-valued restriction for the self-adjoint case. we simply consider the real and imaginary parts of a function of a real variable. given an operator T, the Borel calculus gives a spectral measure, or the resolution of the identity, {EB} indexed by the Borel sets in σ(T). we define f(T) by <h, f(T) h> = ∫ f d <h, EB(T) h>. f can be complex-valued. Mct mht 14:55, 16 August 2006 (UTC)
replies: text: in the special case that f is real valued, f(T) would also be self adjoint. Mct mht 14:58, 16 August 2006 (UTC)
thread_title:  Complex valued functions 

message: I don't believe the most recent edit concerning the domain of the functional calculus is correct. The functional calculus structly speaking is a function of 2 arguments a function h and an operator T. 
* For the continuous functional calculus, h varies over continuous functions and T over self-adjoint operators (or maybe normal ones depending)
* For the Borel functional calculus, h varies over Borel functions and T over self-adjoint operators (or maybe normal ones depending)
What sense does it make to say the domain of the calculus is an algebra of operators?
Perhaps what the article means to say is that for a fixed T, the range of the funcal calculus is a C*-algebra or a von Neuman algebra.--CSTAR 14:13, 13 December 2006 (UTC)
replies: text: i was thinking of C(σ(T)) and L∞(σ(T)), the domains. and yes i took "functional calculus" to mean the map that assigns f to f(T), for a fixed T. in that case, seems to me that the domain/range wording is not an issue. Mct mht 17:40, 13 December 2006 (UTC)
thread_title:  Domain of functional calculus 

message: In the first sentence of the lead, "... a functional calculus (that is, an assignment of operators to functions defined on the real line) ...", there is a parenthetical expression which is supposed to clarify what "functional calculus" means. However, to me at least, it is still unclear. You need to be more explicit about what an operator is and how it is related to the function from the reals to the reals. Are you saying that functions from the reals to the reals can be extended to functions from Hermitian matrices to Hermitian matrices in a systematic way? By applying the function to the real eigenvalues of the matrices? And that operations on these real functions (such as adding two of them together) are also extended to operations on the Hermitian matrices? JRSpriggs (talk) 17:46, 1 December 2010 (UTC)
replies: text: It's really only a non-trivial remark when there is a continuous spectrum. If there's a discrete spectrum and we say "square the operator by taking the same eigenvectors and square the eigenvalues" it hardly looks like heavy lifting. When the eigenvalues are varying continuously, you'd have to think "which functions?" rather than "any function" to get a theory that is of any value. You can always do polynomials. Holomorphic functions being in a sense a limit of polynomials (on compact sets), there is an analytical challenge. For general Borel functions (i.e. about as bad as you'd want to contemplate), much more so. But it is pretty standard spectral theory from about von Neumann's, I think. Charles Matthews (talk) 21:13, 1 December 2010 (UTC)

text: I agree: at least the phrasing is awkward, since the purpose of a mathematical field is to study various objects, their relations to each other, and the associated problems, not to assign them. — Kallikanzaridtalk 10:59, 19 December 2010 (UTC)
thread_title:  First sentence 

message: At the end of the section The bounded functional calculus, currently there is a comment:
Given an operator T, the range of the continuous functional calculus h → h(T) is the (abelian) C*-algebra C(T) generated by T. The Borel functional calculus has a larger range, that is the closure of C(T) in the weak operator topology, a (still abelian) von Neumann algebra.
As far as I know, this is true only when the Hilbert space is separable. For example, if we consider \ell^{2}[0,1] and an operator T:e_{t}\mapsto te_{t} where e_{t} is the basis vector corresponding to t\in[0,1], then \sigma(T)=[0,1] so the Borel functional calculus gives the algebra of bounded Borel measurable functions on [0,1]. This is of course not a von Neumann algebra because the base measure here is the counting measure, not the Lebesgue measure. (The true von Neumann algebra in this case is the algebra of every bounded function on [0,1].) How can we fix the above comment?  — Preceding unsigned comment added by 122.36.103.214 (talk) 17:00, 19 January 2019 (UTC)
thread_title:  Borel functional calculus may not give a von Neumann algebra