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I have corrected the statement of the spectral theorem. It read
The spectral theory for compact operators in the abstract was worked out by Frigyes Riesz. It shows that a compact operator K has a discrete spectrum, with finite multiplicities (so that K − λI has a finite-dimensional kernel for all complex λ).
Which is close, but case where the spectrum has 0 as a limit point is not a discrete subset of C. Moreover, 0 need not be an eigenvector even though it is always in the spectrum (e.g.
Volterra operator) and if 0 is an eigenvector it may have infinite multiplicity (e.g. 0 operator)
It might be worth expanding on volterra operator, either here or in a new page, but I don't have time now.
--
AndrewKepert 07:57, 7 Apr 2004 (UTC)
OK, fine. 'Discrete spectrum' as opposed to 'continuous spectrum' is sort of lax terminology, I guess.
In the following, X,Y,Z,W are Banach spaces, B(X,Y) is space of bounded operators from X to Y. K(X,Y) is space of compact operators from X to Y. B(X)=B(X,X), K(X)=K(X,X). is the unit ball in X.
A
bounded operator is compact if and only if any of the following is true
there exists a neighbourhood of 0, , and compact set such that .
is relatively compact
Image of any bounded set under T is relatively compact
ok, Fredholm operators have closed range and Fredholm-ness is preserved by homotopy, in the set of Fredholm operators. seems to me that the latter fact is not entirely trivial. just wanted to note that the property can also be shown directly.
Mct mht03:51, 5 April 2007 (UTC)reply
A question: The article says "A
bounded operator is compact if and only if any of the following is true". This seems to suggest boundedness is necessary for the following conditions to be equivalent to compactness. Is this true? —Preceding
unsigned comment added by
130.207.197.164 (
talk)
17:40, 25 October 2010 (UTC)reply
Finite spectrum
Article said:
It shows that a compact operator K on an infinite-dimensional Banach space has spectrum that is either a finite subset of C which includes 0 (in that case, the operator has finite rank)
This is wrong. A compact operator may have spectral radius 0, hence finite spectrum without being of finite rank. Consider for example the integration operator
There is something wrong here. As it stands it would imply the identity is compact. Surely it needs to say the singular values tend to zero. I will look for a good reference then fix it.
Billlion (
talk)
08:06, 2 May 2013 (UTC)reply
No, the text is OK, even if not formulated in the clearest way. It's OK because you need infinitely many orthonormal vectors to represent the identity in an infinite dimensional Hilbert space, and then the will accumulate at a non zero value, which has been excluded in the text.
Bdmy (
talk)
08:19, 2 May 2013 (UTC)reply
Examples
By Riesz's lemma, the identity operator is a compact operator if and only if the space is finite-dimensional.
It seems that this is wrong, because because if X is rational numbers, closed unit ball is not compact, so image of closed unit ball is not relatively compact, therefore identity on rational numbers is not compact. I think that the equivalence holds only on complete spaces.