Talk:Riesz representation theorem

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[edit] Split up article

This page should be split up into three articles. If there is no objection, I will do this soon.CSTAR 02:23, 24 Dec 2004 (UTC)

I agree. What would you call the first two sections? The third one has a canonical name, Riesz-Markov theorem. Dbenbenn 03:42, 24 Dec 2004 (UTC)

How about Riesz representation theorem and Extension of Radon measures? There could also be a disambiguation page called Riesz theorems or something.CSTAR 03:53, 24 Dec 2004 (UTC)

Er, not so happy about that. Rudin (Real and Complex Analysis) presents the second version as the Riesz representation theorem. Another textbook of mine, Funcional Analysis by Michael Reed and Barry Simon, calls the first version the "Riesz lemma". (Disambiguation page is good, though.) Dbenbenn 04:27, 24 Dec 2004 (UTC)

Hmmm. "Riesz lemma" makes the result sound pretty lame (Riesz layma). We could give it a really long name. Riesz representation theorem for linear functionals on Hilbert spaces or Riesz representation theorem on Hilbert space. CSTAR 04:35, 24 Dec 2004 (UTC)

Those sound fine to me. And the page can always be moved later if you find a canonical name. (Lemma doesn't sound lame to me, but I didn't grow up speaking Spanish.) Dbenbenn 05:50, 24 Dec 2004 (UTC)

[edit] Introduction

The first paragraph says: "The theorem is the justification for the bra-ket notation popular in the mathematical treatment of quantum mechanics."... What? I would hardly call it the "justification" for that notation (which can be easily explained in terms of inner products!); nor do I think that some imagined connection to physics is a helpful way to start the article. If anything, the sentence should be reworded and moved to the very bottom. A5 16:34, 17 February 2006 (UTC)

Everything, more or less, about Hilbert space can be 'explained in terms of inner products'. I don't see that that invalidates what is said. Charles Matthews 16:49, 17 February 2006 (UTC)

Should I have used the phrase "trivially explained"? I.e. . What do the operator and the vector depend on the Riesz representation theorem for? Obviously more needs to be said by way of explanation. A5 18:36, 25 February 2006 (UTC)

Hmm. This article, as currently written, is clear enough if one already knows the subject matter, and one's interest in reading it to "refresh one's memory". But if one does not know Hilbert spaces well, then it appears confusing and poorly structured. I note that User:A5 is a new editor for WP math articles: if you think you know this material well, and can improve it, then please do so. Just be careful not to wreck the article in the process: us "old timers" find many well-meaning edits to be of questionable value, often muddying or obscuring the subject matter. linas 16:15, 26 February 2006 (UTC)

I think I understand now what the comment is attempting to say. I will move the comment lower down (because the theorem has applications outside of QM), and I will try to clarify it. If there is a problem, please fix rather than just reverting. A5 20:13, 27 February 2006 (UTC)

By the way, I'm not really a new editor of math articles. I'm doing it more frequently now because I'm revising for exams, but also it hasn't been until recently that my browser would keep around the Wikipedia cookie for more than a few days, so most of my edits are anonymous. A5 20:44, 27 February 2006 (UTC)

Thanks. The edits looked good. I made a minor change. Also, my browser has a "remember me" button which keeps me logged in for weeks at a time. linas 01:18, 28 February 2006 (UTC)