Talk:Borel functional calculus

Resolution of Identity

This part makes no fucking sense. I gave up and read it in one of the referenced sources, and it was crystal clear. I don't feel confident enough to change it though...

Complex valued functions

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, {E_B} indexed by the Borel sets in σ(T). we define f(T) by <h, f(T) h> = ∫ f d <h, E_B(T) h>. f can be complex-valued. Mct mht 14:55, 16 August 2006 (UTC)

in the special case that f is real valued, f(T) would also be self adjoint. Mct mht 14:58, 16 August 2006 (UTC)

Domain of functional calculus

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)

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)

CSTAR, good edit. Thanks. Mct mht 05:48, 7 April 2006 (UTC) —Preceding unsigned comment added by 160.39.152.32 (talk)

Borel functions ?? Huh?

The current article states:

More precisely, the Borel functional calculus allows us to apply an arbitrary Borel function to a self-adjoint operator, in a way which generalizes applying a polynomial function.

I find this hard to believe, or more precisely, it seems flat-out wrong to me. The current link Borel function links to an article about measurable functions and integrability; is that really the intent here? If so, then this article is misleading ... e.g. The lead talks about taking the square-root or the exponential of an operator; square-root and exponential are nice, well-behaved analytic functions, they have nothing to do with integrability, as "most" integrable functions are continuous-nowhere and unbounded on any open interval! Giving examples using holomorphic functions while trying to imply that the results extend to integrable functions is not believable. Perhaps something else is meant by 'Borel function' ? Maybe square-integrable function i.e. L_2, which is a whole lot tamer than L_1, which is tamer than the space of measurable functions ? Extending holomorphic functional calculus to something defined on L_2 is intuitively "doable", extending it to measurable functions seems absurd.

Other absurities: This formula:

$h(T)e_{k}=h(\lambda _{k})e_{k}$

For a "typical" measurable function h, and a fixed value x, h(x) will be either zero, or plus/minus infinity. Defining an operator this way is non-sense, as measurablility is a statistical property, really.

Oh, wait, more then half-way into the article we get talk of "bounded measurable functions" -- OK, well, the space of bounded measureable functions is far more tame tame than the space of measureable functions; indeed, h(x) will be finite almost everywhere, unless you are so unlucky that the spectrum of the operator lands on the discontinuities of h. Hmmm.

Then farther down we get this example:

$U_{t}=e^{itA}\quad t\in \mathbb {R}$

True, but this is a textbook example from quantum mechanics, exp is entire, it lies in the domain of holomorphic functional calculus, and no need of measure theory is needed for this. Perhaps this formula, and the comments about quantum, should be ripped out, as they have nothing to do with anything??

linas (talk) 06:18, 29 November 2010 (UTC)

Hi Linas, I had a quick glance to the article and everything seems ok. Note that here "Borel" refers in any case to Borel $\scriptstyle \mathbb {C}$-valued functions, so $h(x)=+\infty$ can never occur. You can find the construction of the Borel functional calculus in any book of Functional Analysis. Just to give you an idea, for a bounded self-adjoint operator A on H , the set {f(A) : f continuous function on spec(A)} is exactly the norm-closed algebra generated by A. The larger set {f(A) : f bounded Borel function on spec(A)} is exactly the w* closed algebra generated by A (recall that the space of bounded operators on the Hilbert space is a second dual). You are right that some more examples should be good.--pma 09:06, 29 November 2010 (UTC)

Not sure what you mean by "Borel $\scriptstyle \mathbb {C}$-valued functions"; last I looked, Borel functions were defined with respect to a sigma-algebra. Thus, although the integral of the function is always bounded and valued in some field e.g. C, the point value may be infinite, e.g. the Dirac delta function, for which $\delta (0)=+\infty$ not only occurs, but is in some sense the "typical" situation. Right? The Dirac delta is a Borel function, isn't it? If not, then I am really misunderstanding something. So, many Borel functions are sums of delta functions and their derivatives, but even those are "well-behaved"; there are even worse out there. If I have $h(x)=\delta (x-1/2)$ and ${\frac {1}{2}}\in \operatorname {spec} T$, what is h(T)? What if $h(x)=\delta (x)$ and T has a sequence of eigenvalues with an accumulation point at zero? (My old/out-of-date books on functional analysis do not cover this topic; any suggestions?). linas (talk) 15:03, 29 November 2010 (UTC)

OK, a function $f:X\to Y$ between two topological is Borel measurable if for any Borel subset S of Y the pre-image $f^{-1}(S)$ is a Borel set of X. In our case X is a compact of $\mathbb {C}$ and Y is $\mathbb {C}$. A Dirac delta (say supported at in 0) is not a function; talking of "a function taking value $+\infty$ at 0, 0 elsewhere, with unit Lebesgue integral" is a picturesque description that appeals somehow to intuition, but it's not a formal definition, and it's of no practical use ( note that the Lebesgue integral of such a function should be 0) . A Dirac delta is a distribution, and in fact it's a Borel measure concentrated on a point. It has weak derivatives, that are still distributions, but not measures. Whatever sense we give to $\delta (T)$, it is not within the Borel functional calculus. Here is a positive example that should clarify the picture: consider the Hilbert space of square integrable functions $L^{2}(X).$ Choose a function $a\in L^{\infty }(X)$. It defines a bounded linear operator $M_{a}$ on $L^{2}(X)$ by multiplication: for any $L^{2}$ function f, $M_{a}f$ is the $L^{2}$ function $a(x)f(x)$. It turns out that the spectrum S of $M_{a}$ is exactly the essential range of $a.$ Moreover, $M_{a}$ is a normal operator, and it is a self-adjoint operator iff $a(x)$ is real-valued. In any case it has a Borel functional calculus, and for any bounded Borel function $\phi :S\to \mathbb {C}$ the operator $\phi (M_{a})$ is still a multiplication operator: namely, the multiplication by the composition function $\phi \circ a$. The relevant feature of this example is that for any Hilbert space H and for any bounded normal operator T on H, there is a representation of H with a space $L^{2}(X)$ such that T becomes a multiplication operator. Unbounded operators may be treated as well.--pma 18:37, 29 November 2010 (UTC)

OK, we are talking past each other. Wayyyy past each other. The current wikilink Borel function redirects to measurable function. We both agree that the dirac delta is not a function from the reals to the reals in any normal sense. It seems we agree that it's a 'measurable function'. We both seem to understand that the correct definition of the Dirac delta is that it is an indicator function: it has a value of exactly 1 on any (Borel) set containing zero. To me, I think this means that the Dirac delta is a 'Borel function', and I see nothing on the page Borel function that would contradict my belief. So -- am I wrong in believing that the Dirac delta is a "Borel function"?

If I am wrong in believing that the Dirac delta is a "Borel function", then it seems that the article Borel function should be modified to clarify this. But I'm really pretty sure I'm not wrong. If I'm right, then this article has several non-sensical statements, as pointed out above: so, for example,

$h(T)e_{k}=h(\lambda _{k})e_{k}$

is nonsense when h is the dirac delta. Worse, it uses bogus notation: a Borel function is defined on Borel sets; I don't understand how to interpret $\lambda _{k}$ as a Borel set; to me, $\lambda _{k}$ is a number.

In your rebuttal, you give an L^2 example. I don't see what this has to do with anything, since, in general, Borel functions do not belong to L^2, and L^2 is just a "small" subset of all Borel functions. In particular, L^2 is a lot simpler, and tamer, and much better behaved than the set of measurable functions! L^2 has a lot of nice, elegant properties that make it easy to use; these properties do not hold in general for arbitrary measurable functions.

Again, let me point out: the lead states:

More precisely, the Borel functional calculus allows us to apply an arbitrary Borel function to a self-adjoint operator, in a way which generalizes applying a polynomial function.

It does not say ...apply an arbitrary L^2 function .... So: can we define a functional calculus for Borel functions, or are all examples limited to the much simpler case of L^2? Meanwhile, I'll try to find & read the two books cited, but this will take some weeks or more. My gut impression is that this article does not come even close to describing what Borel functional calculus is, and/or that it has serious, fatal misrepresentations. linas (talk) 03:32, 30 November 2010 (UTC)

The delta function is not a function. In particular it is not a Borel function. See the second paragraph of the article Dirac delta function for an explanation of why this is so.

L² is relevant because the operator h(T) is not a bounded operator, as the article notes.

I am removing the expert tag. Ozob (talk) 12:03, 30 November 2010 (UTC)

Given the article's contents, Linas' misgivings are understandable, in particular at least the lead of the article doesn't quite do enough to distinguish the various functional calculi. Unfortunately, there also seems to be some confusion over what a Borel function is. It is just a complex-valued function in the true sense (such that...see above). In particular, the Dirac delta function is not a Borel function since it is not an ordinary function. But also Borel functions are not a statistical notion: we don't identify Borel functions if they agree almost everywhere (since there is no background measure with which to do this). Also, at least for the purposes of this article, even unbounded Borel functions must still have a finite value everywhere. PMajer's example is a good one; please look at it again. The operator $T=M_{a}$ is a self adjoint operator from Hilbert space to itself, and φ is an arbitrary Borel function. In general, $\phi (T)$ will be unbounded (densely-defined). Sławomir Biały (talk) 13:37, 30 November 2010 (UTC)

Bull-hockey. The Dirac delta is a classic, prototypical example of a Borel function. Is is even a bounded Borel function: It has a value of exactly 1 on every Borel set that contains the origin, and it has a value of zero on every Borel set that does NOT contain the origin. See indicator function. Dirac delta and indicator functions are exactly one and the same thing on Borel sets! Take a look at the article Borel function, or open any book on measure theory. The proper measure-theoretic presentation of the Dirac delta is usually given within the first 10 pages or so. linas (talk) 15:08, 30 November 2010 (UTC)

The Delta function is not a function f : R → R. It is a Borel measure. That is, it is a mapping from the sigma algebra of Borel sets to the reals satisfying certain axioms. This is not what is meant by Borel function. Also, there is a big difference between the indicator function supported at {0} and the delta function. Sławomir Biały (talk) 15:17, 30 November 2010 (UTC)

Argh: let me review: let ${\mathcal {B}}$ be a collection of Borel sets, say, for example, covering the complex plane or some portion thereof, whatever. A Borel function f is a function $f:{\mathcal {B}}\to {\mathcal {E}}$ where ${\mathcal {E}}$ is another collection of Borel sets, say, over the reals. That is, f maps elements $B\in {\mathcal {B}}$ to elements $E\in {\mathcal {E}}$.

Consider the Borel function $1_{x}:{\mathcal {B}}\to {\mathcal {E}}$ that is equal to:

$1_{x}(B)={\begin{cases}1&{\mbox{ if }}x\in B\\0&{\mbox{ if }}x\notin B\end{cases}}$

This function $1_{x}:{\mathcal {B}}\to {\mathcal {E}}$ is a Borel function. It is called the indicator function for x. If you think about it, you will realize that its the Dirac delta. To use colloquial, olde-fashioned notation, one writes

$1_{x}(B)=\int _{B}\delta (x-z)dz$

The above is the proper measure theoretic definition of the Dirac delta. This is how this material is presentedd in textbooks on measure theory. Whatever it is that you guys think is a "Borel function", I don't know, but it appears that this article is not using the standard text-book definition of a "Borel function", which I just gave above! linas (talk) 15:32, 30 November 2010 (UTC)

You seem to be using different conventions than most of measure theory, which could be a source of your confusion. Usually, the indicator function as it is defined in most treatments of the subject (also called the characteristic function) fixes a subset B, and then $1_{B}:\mathbb {R} \to \mathbb {R}$ is the function that returns x if $x\in B$ and zero otherwise (see our article indicator function, for instance). This is an actual point function. What you are calling the indicator function is what others would call a unit mass or Dirac measure (iirc, Rudin calls it the unit mass). This is a set function, a special case of what is known as a Borel measure. However, it is not a Borel function, since Borel functions are ordinary point functions. Whether it is a Borel measure or Borel function depends on whether one thinks of the set as fixed (indicator function) or the point as fixed (unit mass measure). Sławomir Biały (talk) 16:14, 30 November 2010 (UTC)

Please consider that saying "huh?" isn't actually an argument. As Oliver Cromwell once wrote, think it possible that you may be mistaken. Charles Matthews (talk) 15:36, 30 November 2010 (UTC)

Charles, I find this entire conversation frustrating, please be helpful. Where, exactly, does the article Borel function state that "Borel functions are ordinary point functions"? I suppose they are; I just don't see where it says so. There seems to be an additional axiom that I am missing. Let me guess wildly; please correct me where I am wrong: a Borel function is a function f defined as:

1. $f:{\mathcal {B}}\to {\mathcal {E}}$ where ${\mathcal {B}}$ and ${\mathcal {E}}$ are Borel sets; that is, f maps elements of ${\mathcal {B}}$ to elements of ${\mathcal {E}}$.
2. If $A,B\in {\mathcal {B}}$ and $A\subset B$, then $f(A)\subset f(B)$

Is that right? Possibly the second axiom should use f^-1 instead? From what I can tell, this second axiom is missing from the article Borel function, yet, it seems that everyone else here is implicitly assuming it (or something similar). So, now I think I understand: yes, clearly, the Dirac measure satisfies #1 but fails to satisfy axiom #2. But axiom #2, or whatever it's correct statement is, is missing from the article Borel function. linas (talk) 02:25, 1 December 2010 (UTC)

From measurable function:

If $(X,\Sigma )$ and $(Y,\mathrm {T} )$ are Borel spaces, a measurable function $f:(X,\Sigma )\rightarrow (Y,\mathrm {T} )$ is also called a Borel function.

It is f⁻¹ that gives the map of sigma-algebras. Ozob (talk) 02:54, 1 December 2010 (UTC)

Yes, OK. I just now dug into a book that explained this nicely. The problem is that the article measurable function never actually states that f⁻¹ gives the map of the sigma-algebras. It was this minor little 'fact' that I had forgotten, and that lead to the morass of confusion above. Could someone please fix measurable function so that it correctly defines what a measurable function actually is? Viz. that it adds the above crucial fact? Thanks. linas (talk) 03:33, 1 December 2010 (UTC)

The notation $f:(X,\Sigma )\rightarrow (Y,\mathrm {T} )$ leaves something to be desired. Sławomir Biały (talk) 14:53, 1 December 2010 (UTC)

First sentence

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)

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)

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. — Kallikanzarid^talk 10:59, 19 December 2010 (UTC)

Borel functional calculus may not give a von Neumann algebra

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)