Talk:Resolvent formalism

Boundedness

On 1 August 2006,User:Mct mht removed the following text:

When A is a compact operator, the Fredholm alternative states that the resolvent is a bounded operator.

with the comment: (that's not right. at least misleading. the resolvent is by definition bounded, including for unbounded operators.). I don't understand the objection. If the spectrum is discrete, then yes, I suppose the resolvent is always bounded. If the spectrum is continuous, then the resolvent is not bounded, since one can construct vectors that are arbitrarily close to pole. Right? Or is just that its late at night, and I'm missing something? linas 03:43, 27 October 2006 (UTC)[reply]

hm, AFAIK, only bounded resolvents are considered, including when the operator itself is unbounded. seems to me a large part of the utility of the resolvent is that it allows one to use techniques from the bounded case. if you have a reference where unbounded resolvents are considered in some detail, i'd be interested to know. and re: If the spectrum is continuous, then the resolvent is not bounded, since one can construct vectors that are arbitrarily close to pole...: well the spectrum is closed no matter what, so looks like the operator (T - λ)^(-1) is gonna be bounded if λ is not in the spectrum with the 1/(distance between lambda and spectrum of T) being the upper bound for the operator norm, no? Mct mht 03:59, 27 October 2006 (UTC)[reply]

No idea

I have to mention that whoever wrote this article has no idea about the topic. The resolvent operator (if it exists) is always bounded. On the other hand, calling it "formalism" shows the ignorance concerning Spectral Theory. —Preceding unsigned comment added by Glickglock (talk • contribs) 19:58, 1 April 2008 (UTC)[reply]

That would be me. When I wrote this, I had no idea about the topic; what little I knew, I knew from physics texts, which rarely/never treat the topic with mathematical rigor. I still am unaware of any text that treats this subject matter in any sort of detail, although I note that Dunford and Shcwatrz have a few pages on it. linas (talk) 03:05, 10 November 2008 (UTC)[reply]

Erroneous formula

If I compare the integral formula for the eigenspace projection operator with the formulas in the holomorphic functional calculus article they seem to disagree.. a $\zeta$ factor is maybe missing ? —Preceding unsigned comment added by 91.66.113.1 (talk) 08:01, 11 May 2008 (UTC)[reply]

Move

I think this article should be renamed "resolvent function", with resolvent set merged into spectrum (functional analysis). Anyone have opinions?Quietbritishjim (talk) 14:04, 22 June 2009 (UTC)[reply]

Resolvent identity

LFEnilec (talk) 18:13, 26 January 2016 (UTC)[reply]
I have to say that I disagree with the second resolvent identity. In my opinion this is false, the simpler way to see it, is to consult Theorem 5.10.4 from the book by Hille and Phillips cited in the article. More precisely, the formula to look at is (5.10.6).

Last comments go to the bottom of the page, not the top. Hille & Phillips, as you can see, use the backwards notation, like Dunforth and Schwarz, as indicated a few lines above: so their R is −R defined in the first formula of the article. But you may check the whole thing yourself, for noncommutative operators:

R_{A}-R_{B}=1/(A-zI)-1/(B-zI)=1/(A-zI)~~(1-(A-zI)~1/(B-zI))

=1/(A-zI)~~((B-zI)-(A-zI))~~1/(B-zI)=1/(A-zI)~~(B-A)~~1/(B-zI)

as I expressly invited you, no?? Cuzkatzimhut (talk) 19:03, 26 January 2016 (UTC)[reply]

You mean Dunford and Schwartz. 2601:200:C000:1A0:B150:8229:678:CF53 (talk) 17:37, 9 September 2022 (UTC)[reply]

As I can see the courtesy is not really the force of wikipedia user, at least for you. Not a good experience. Anyway, the mixing of the two notations in the article lead to be confused, I confesses my error.

I apologize if I inadvertently offended you. I assumed you had checked the commutative limit (one-dimensional Hilbert space!), and seen the sign from the high-school algebra involved. But there really is only one notational convention in the article, not two: The cautionary remark is precisely for people who prefer external sources to the definition provided. Cuzkatzimhut (talk) 19:25, 26 January 2016 (UTC)[reply]

OK you are right for the notation, I do not take enough care when I read it. I won't bother you on the article anymore. Thank you ! LFEnilec (talk) 19:39, 26 January 2016 (UTC)[reply]

Too difficult for beginners - Can someone remove the jargon?

As a person attempting to learn about this topic at graduate level in mechanical engineering I honestly cannot even start reading this page. There is way too much unnecessary jargon. Pretty much every term that shows as a hyperlink I do not know anything about, and I'm pretty sure I don't need to know everything to understand the overall workings of this machinery. Can someone knowledgeable about the subject perhaps translate this page to English? I mean, a simpler "TL;DR" paragraph that explains what the resolvent operator is even doing?

(Please no hate, I know math is difficult and builds over several ideas, but this is unnecessarily complicated in my opinion. I'm sure someone with basic knowledge of linear algebra should be able to grasp what this operator is doing!) — Preceding unsigned comment added by Fzigunov (talk • contribs) 19:12, 5 November 2019 (UTC)[reply]

No hate, as you put it; I am a physicist myself. While your request is sensible, why are you here? I mean, the basic identity provided is manifest, if you think of matrices, which you almost certainly do in engineering. The article dots the is and crosses the ts, so to speak, and steers you to subtleties if you were interested in them. The brutal facts are self-evident. Most physics and engineering texts don't even use these terms--they just invert matrices. Cuzkatzimhut (talk) 20:04, 5 November 2019 (UTC)[reply]

Bad writing

The article begins as follows:

"In mathematics, the resolvent formalism is a technique for applying concepts from complex analysis to the study of the spectrum of operators on Banach spaces and more general spaces. Formal justification for the manipulations can be found in the framework of holomorphic functional calculus.

"The resolvent captures the spectral properties of an operator in the analytic structure of the functional. Given an operator $A$ , the resolvent may be defined as

R(z;A)=(A-zI)^{-1}~.

"

Here are three problems, all serious, with this passage:

1. The phrase "the functional" is never explained. What functional?

2. The expression given for $R(z;A)$ is meaningless in case $A-zI$ happens to not be invertible. But the article never mentions for which z this definition makes sense.

3. But most important by far: Regardless of the correct expression for the resolvent $R(z;A)$ , the article never states what kind of object the resolvent is.

This is a fatal flaw for any Wikipedia article, especially on on a topic of advanced mathematics.

Presumably, the resolvent $R(z;A)$ is a certain kind of function, having a domain and a codomain.

It would be an excellent idea for someone knowledgeable about this topic to include these pieces of information in the article, preferably near the beginning.

Readers need to know what a resolvent is. 2601:200:C000:1A0:B150:8229:678:CF53 (talk) 17:35, 9 September 2022 (UTC)[reply]

The resolvent is an operator as defined, a function of the operator A; in may have singular points, explorable from context. This is a summary, not a tutorial ! Cuzkatzimhut (talk) 17:51, 9 September 2022 (UTC)[reply]

Hille-Yoshida theorem

In the book of Hille and Phillips, there is no mention of A needing to be Hermitian. I suppose we should remove this condition. — Preceding unsigned comment added by Csoler (talk • contribs) 10:24, 23 November 2022 (UTC)[reply]