18
$\begingroup$

This is another history question.

Hilbert phrased the spectral theorem in terms of resolutions of the identity.

While this remained the form of Stone and von Neumann, they did also have the functional calculus and that became more common after the Banach algebra revolution of the 40's.

Circa 1960 Dunford-Schwartz were still mainly resolutions of the identity.

By the time reed and I did Reed-Simon in the early 1970, we preferred to talk about multiplication operators emphasizing the individual spectral measures which was in the air when we wrote. Who first using that language and who made it popular?

$\endgroup$

4 Answers 4

7
$\begingroup$

I do not know who used it first, but I claim that Halmos made it popular in his 1963 Monthly paper.

There he makes the connection to the diagonalization of hermitian matrices and mentions that people usually called "multiciplicity theroy" this part and not the spectral theorem.

Multiplicity theory is described in great detail in his 1951 Hilbert space book.

Finally, Dieudonne claims in his History of functional analysis book that this connection was maid by von Neumann after the development of the Gelfand-Najmark theory in 1943, but he does not give an explicit reference.

ADDED: After reading the answer of Francois Ziegler and Igor Khavkine, let me mention the following. NEumann in his 1931 Ann. Math. paper had all what you need to formulate the spectral theorem in this form. Stone in his famous book, Theorem 7.10 formulates this statement with proof. Thank you for following this up.

$\endgroup$
2
  • $\begingroup$ I believe this is the passage you're referring to in Dieudonné -- but then, I don't see an attribution (of the diagonalization idea) to von Neumann there. $\endgroup$ May 4, 2014 at 18:43
  • $\begingroup$ @FrancoisZiegler: Well, you are right, I might have been misled by the "modern version of Riesz-von Neumann decomposition of unity" expression. $\endgroup$ May 4, 2014 at 19:01
6
$\begingroup$

Following András's lead: Halmos quotes Segal (1951) where the multiplication operator version appears as Lemma 4.2. Segal in turn refers to Plesner-Rokhlin (1946) and Wecken (1939).

Plesner-Rokhlin have it as Theorem 1, page 137: "Any cyclic operator $A$ is isomorphic to $K_\rho$", where they defined $K_\rho F(\xi)=\xi F(\xi)$ on $L^2(\rho)$ (unattributed).

Wecken on the other hand only quotes (a version of) it inside a proof, page 432, with attribution to Stone (1932), Theorem 7.10. There the buck seems to stop.

$\endgroup$
3
$\begingroup$

Dieudonne records the following in Chapter VII of History of Functional Analysis:

Following the method of von Neumann, it is then easy to extend the homomorphism $f \mapsto f(N)$ to the algebra $\mathcal{U}(\sigma(N))$ of all universally measurable bounded functions in $\sigma(N)$.

Finally, by adapting the arguments of von Neumann, Hellinger and Hahn, one arrives at the modern description of the Riesz-von Neumann "decomposition of unit" and of the "multiplicity theory" of Hellinger-Hahn:

Dieudonne then describes the multiplication version of Spectral Theory in 4 enumerated points, after which he writes:

One says that this description is a diagonalization of the normal operator $N$.

So Dieudonne describes diagonalization as a natural consequence of, application of, or interpretation of, the general theory up to that point. That would explain why there is no milestone to be found.

$\endgroup$
3
  • $\begingroup$ See also this part of my answer and comment Ziegler afterwards where he links to this passage of Dieudonne. $\endgroup$ May 13, 2014 at 22:04
  • $\begingroup$ @Andras Batkai: I don't see "that this connection was maid by von Neumann" in Dieudonne. Did I miss a statement of that kind? It would not surprise me that von Neumann put the pieces together, but I didn't see such a reference. $\endgroup$ May 13, 2014 at 23:21
  • 1
    $\begingroup$ It seems at the moment that Stone is the earliest reference for this statement in 1932, see other answers and the added part to my answer. $\endgroup$ May 14, 2014 at 6:49
2
$\begingroup$

A note (to big to fit into a comment) about Francois Ziegler's investigation. I must say that the Plesner-Rokhlin (1946) article is exceptionally bad at citing references. I have only noted 2 references in the first 70+ pages of the text. Though, that may be understandable since it essentially consists of the pedagogical notes for a course that had been given by Plesner. Here's a translation of a footnote to the title, explaining the origin of these notes:

The first part was published in issue IX of Uspekhi (old series). [Uspekhi Mat. Nauk, 1941:9, 3–125]

The current, second part records, with insignificant changes, the content of the lectures that I read in the spring semester of 1939 at the Faculty of Mechanics and Mathematics, MSU [Moscow State University]. The typesetting of this part of my course was mostly finished by V. A. Rokhlin in the summer of 1941, when the war interrupted him in this task. Thereafter, I myself added only §22 (Hellinger types) part 3, §26 and parts of the Appendix (parts 3 and 5), though I also omitted parts of the treatment of real operators. However, an addition was made (§28) of the theory of generalized functions of a hermitian operator, which I developed in 1942. In relation to this, some rearrangements were made in the preceding chapters and §29 suffered some revisions. --- A. Plesner (A short exposition of the basics necessary for the understanding of part II of the article by A. I. Plesner and V. A. Rokhlin can be found in A. I. Plesner's article Fundamental concepts of the spectral theory of Hermitian operators within the current issue of Uspekhi.

For reference, the part with the theorem cited by Francois Ziegler is in §23.1. The two works that Plesner-Rokhlin article does cite by that point are Stone (Ann. Math. 33, 1932) and von Neumann (Ann. Math. 31, 1931). It seems that the content of Plesner's lectures was fixed a few years before the work of Gelfand-Naimark (1943), so Dieudonné's remark does not seem consistent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.