Talk:Resolvent formalism

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[edit] 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)

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)

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)

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 ζ factor is maybe missing ? —Preceding unsigned comment added by 91.66.113.1 (talk) 08:01, 11 May 2008 (UTC)