Talk:Decomposition of spectrum (functional analysis)

Some incorrect statements made in the article were changed, perhaps someone can elaborate on the details.

above unsigned comment was added by me. about two months(?) ago. Mct mht 22:54, 5 May 2006 (UTC)

spectrum of multiplication operator

I think there's an error in the proof that the spectrum of $T_{h}$ equals the essential range of h. the $f_{n}$ are not in $L^{p}$ unless we find a subset of $S_{n}$ of finite positive measure and take the characteristic funktion of this subset. we have to require that the measure $\mu$ is locally finite, i.e. every measurable set of positive measure contains a finite measurable set of positive measure.

for an counterexample, consider $S=\{x_{1},x_{2}\}$, where $x_{1}$ has finite and $x_{2}$ has infinite measure. then $L^{p}(\mu )$ consits of those functions which vanish in $x_{2}$. let $h$ be the characteristic function of $\{x_{2}\}$. then the essential range of $h$ is $\{0,1\}$, but $T_{h}=0$ and hence the spectrum is $\{0\}$. ---oo- 23:58, 14 February 2007 (UTC)

yes, you're right. article should be modified. Mct mht 00:06, 15 February 2007 (UTC)

well, i am gonna change it to σ-finite. Mct mht 08:23, 6 March 2007 (UTC)

Another error seems to be in the phrase "isolated point of the spectrum", say $g\in C[0,1]$ (real-valued just for kicks), then by continuity, the image of g is the same as the essential image, and since [0,1] is compact and connected, the image is a compact, connected subset of $\mathbb {R}$, and an isolated point would be if the image were a single point, in which case, yes that point would be in $\sigma _{p}(T_{g})$, however, say that the support of $g$ is not all of [0,1] and say g is not identically zero, examples of such functions are readily available. Then take $f:[0,1]\to \mathbb {R}$ with disjoint support, which is possible because the support of g is a compact subset of [0,1], and hence a disjoint union of closed intervals (finite), then gf=0 contradicts $T_{g}$ 1-1, so 0 is in the point spectrum even though g may not be constant, i.e. a point spectrum value may not be an isolated point. As a result, I'm going to remove the language about being isolated and just leave the part that explains what it really means. — Preceding unsigned comment added by Guardian of Light (talk • contribs) 16:23, 2 November 2011 (UTC)

suggest merge and expand

the article discrete spectrum could be merged with this one. mathematically, continuous spectrum can mean several things (similar but possibly distinct, depending on the particular operator). maybe the article can be expanded to include couple typical break-ups of the spectrum that one might encounter in the literature. Mct mht 12:15, 14 August 2006 (UTC)

also, non-mathematical meanings should have its own page. i restored some remarks from that point of view without attempting to integrate with existing text. Mct mht 07:33, 15 August 2006 (UTC)

i forked those remarks to continuous spectrum. Mct mht 08:08, 15 August 2006 (UTC)

Thanks very much for your work. The various kinds of spectrum have always confused me, and perhaps next time I need to look into this, I can use this page for some guidance (hint: please add references!). I wonder though why you used the title Decomposition of spectrum (functional analysis) instead of Decomposition of spectrum. Do you expect other pages with the title "Decomposition of spectrum (something)"? Otherwise, disambiguation is not necessary. -- Jitse Niesen (talk) 12:38, 15 August 2006 (UTC)

no particular reason why i used that title, maybe you're right. i just thought since physicists speak of spectra, in non-rigorous fashion, all the time, better to specify. some references are Dunford & Schwartz, and Reed & Simon, will add them now. Mct mht 12:54, 15 August 2006 (UTC)

some discussion of relationship of discrete spectrum to localization?

It would be nice to have some explicit discussion of the discreteness of the spectrum to localization of the corresponding eigenmodes for Hermitian operators. For example, under some fairly weak conditions on the Hermitian operator, if the eigenmodes have compact support then the eigenvalue spectrum is discrete. (And conversely, continuous spectra arise only for infinitely extended modes.) There's a proof of this, I think restricted to Sturm-Liouville problems, in Courant & Hilbert, if I recall correctly.

This has such far-reaching consequences that I was surprised not to see it mentioned. For example, it leads to the discrete energies of bound electron states in quantum mechanics, to the discrete harmonics of a piano string, and to the discrete solution curves in an electromagnetic waveguide.

I don't remember the exact conditions on the operator required by Courant & Hilbert. At least continuity, and a lower bound for the eigenvalues, I think. A reference for a more general form of the theorem would be welcome.

—Steven G. Johnson 18:15, 24 August 2006 (UTC)

yeah, sure the article can be expanded. not quite sure what you mean by eigenmodes there. perhaps you mean eigenvectors. (physicists speak of "eigenvectors" not in the Hilbert space, but mathematically that requires the rigged Hilbert space. the article does not do this and spectrum is characterized by the corresponding spectral measures. there is no great difference and IMHO, the rigged space formulation offers no advantage.)

if T is a compact operator, every nonzero number in the spectrum of T is an eigenvalue. so compact operators only have point, or discrete, spectrum. in particular, the inverse of the Sturm-Liouville operator is compact and self adjoint. pure point, or discrete spectrum is have their spectral measures being Dirac point measures. the article does mention, although very briefly, they correspond to the bound states in physics. additional physical perspective would be nice, although i would suggest rigor of presentation not be compromised.

Re:I don't remember the exact conditions on the operator required by Courant & Hilbert. At least continuity, and a lower bound for the eigenvalues, I think. continuity would seem to be far too restrictive there. being bounded below is reasonable for a sufficient condition. a particular case is the Sturm-Liouville operator, which is not continuous.Mct mht 18:49, 24 August 2006 (UTC)

also, small oscillations and normal modes is a finite dimensional problem, perhaps the relevant stuff can be added to a page devoted to the matricial case, i.e. eigenvalues, spetral theorem for normal matrices, etc. Mct mht 18:57, 24 August 2006 (UTC)

I wasn't suggesting that rigor of presentation be compromised, and I'd love to see the most general correct statement of the theorem relating discreteness to localization. (You're right that continuity/boundedness is too strong, sorry. By "eigenmode" I do indeed mean eigenvector, possibly in the rigged Hilbert space depending on the problem...physicists tend to stick "eigen-" in front of whatever they are dealing with (state, field, mode, etc.) instead of "vector". I hadn't thought about just looking at the spectrum of the inverse operator to make it compact, by the way; that's a nice trick. Obviously, functional analysis is not my field.) I noticed that the article mentions that the bound states correspond to a discrete spectrum, but what it doesn't say (clearly) is that this is not accidental: the very fact that they are "bound" (i.e. localized) is what leads them to have a discrete spectrum. —Steven G. Johnson

the spectral measure being Dirac point measures alludes to this. the support of a Dirac point measure consists of a single point. we could elaborate a little. let's say T is self adjoint operator with discrete spectrum {x_n}. then by spectral theorem, T is unitarily equivalent to mutiplication by x on L²({x_n}). the eigenvector corresponding to x_m in then the function that is 1 at x_m and zero elsewhere. this perhaps offers a mathematical justification of the "localization" you speak of? in contrast, we can not do this for absolutely continuous spectrum. for instance, the position operator of a free spinless particle is mutiplication by x on the real line and has no eigenvalues. Mct mht 04:40, 25 August 2006 (UTC)

That might do it...since it's unitary, this means that the eigenfunctions in the original space are square-integrable as well, which says that they cannot be infinitely extended (i.e. they must decay sufficiently fast outside some bounded region). —Steven G. Johnson 16:18, 25 August 2006 (UTC)

gave it a shot there and added a short paragraph, feel free to add to it. Mct mht 10:03, 26 August 2006 (UTC)

I'm not sure what you mean about small oscillations necessarily being a finite-dimensional problem. The vibrational modes of a continuous medium such as a string etc. live in an infinite-dimensional vector space. —Steven G. Johnson 19:26, 24 August 2006 (UTC)

interesting, didn't know that. Mct mht 04:40, 25 August 2006 (UTC)

Redirect

Why was this redirected to from "Continuous Spectrum" if they're separate articles? —The preceding unsigned comment was added by 199.197.111.212 (talk) 17:01, 24 April 2007 (UTC).

sry, i moved the article but not the talk page. will be fixed shortly. Mct mht 08:24, 2 May 2007 (UTC)

Some Issues

I have some issues relating to the discussion of self-adjoint operators on Hilbert spaces. Usually, one means by continuous spectrum, the union of the absolutely continuous spectrum and the singular continuous spectrum. On the same note by singular spectrum the union of the pure point spectrum and the singular continuous spectrum. So the notation with H_{cont} for the subspace of the AC spectrum is a little bit confusing. The same is true for the notation H_{sing} for the singular continuous spectrum.

Furthermore, the discrete spectrum corresponds to what physicist call bound states. The pure point spectrum corresponds to "localization" in a vague sense. In fact one means by localization also that the eigenfunctions are exponentially decaying, or sometimes dynamical properties.

In order to have a relation to the physics it might be nice to have some reference to the RAGE theorem! This theorem tells us how the Schrödinger flow evolves depending on the spectral type of the operator.

And the claim that $L^{2}(\mathbb {R} )$ functions decay is just plain wrong. E.g. the function $f(x)={\begin{cases}n&x\in [n,n+{\frac {1}{n^{4}}}]\\0&else\end{cases}}$ is a counter example.

Can I just go ahead and edit this article?

ElMaison (talk) 04:39, 30 March 2008 (UTC)

Go for it. Or as we say here, Be Bold! People will complain if they do not like what you're doing, but if nobody edits the article, it will never improve. And obviously, if you change e.g. H_{cont} to H_{ac}, try to keep the whole article consistent. -- Jitse Niesen (talk) 10:16, 1 April 2008 (UTC)

The example of a non-decreasing L^2 function occurs in a confusing place and lacking context. When reading the article I got the impression that the function is supposed to be an example of an eigenfunction corresponding to a point in the point spectrum, but which is not localized. I had to go to the discussion to understand that it was really trying to show that L^2 does not imply decreasing. Actually it's not really successful on that count either, because each function in the sequence is compactly supported. Moreover, you would not normally call those functions 'increasing' because they are not monotone... — Preceding unsigned comment added by 81.224.188.19 (talk) 00:38, 14 November 2021 (UTC)

"Incorrect" discussion of invertibility?

I don't understand what is going on with the weird definition of the spectrum here. It is stated that T-lambda is invertible if it is bounded below and has dense range. But if T-lambda is bounded below, then it has closed range (consider a cauchy sequence in the range, the corresponding preimages will also be a cauchy sequence). So, the definition is just that T-lambda is bounded below and is onto, which thanks to the closed graph theorem just says that T-lambda is bijective. It is also said that bijectivity is stronger than being invertible, which is not true. --Roggles (talk) 13:43, 24 November 2008 (UTC)

Spectrum of Adjoint Operator

It is claimed that if T^* is the adjoint of T then sigma(T^*)=sigma(T). I have only seen sigma(T^*) subset sigma(T), not equality; either a proof should be supplied or a citation. —Preceding unsigned comment added by 193.188.46.249 (talk) 08:47, 24 March 2009 (UTC)

It is a general theorem of functional analysis that T is invertible if and only if T^* is invertible (Banach space adjoint): one direction yields the assertion you mention, the other direction is even easier, because if T is invertible, then the inverse of T' is simply the adjoint of T's inverse. Delio.mugnolo (talk) 20:24, 19 December 2011 (UTC)

Definition of spectrum should not be here

The article should focus on the topic (*decomposition* of the spectrum) for readers who are likely to come here; namely people who already know the definition of spectrum of a functional operator. There is a separate article for that definition; this article should link to that, keeping at most a discrete one-line refresher sentence. --Jorge Stolfi (talk) 18:03, 5 February 2013 (UTC)

Merge and split

There is a section in spectrum (functional analysis) that duplicates much of this article. Indeed it seems that there is little one can do with the spectrum of a functional operator without decomposing it first, so that material must be there. On the other hand this article is becoming hard to read and edit because it tries to cover "decomposition of the spectrum" in many different contexts.
Therefore, please consider merging the main part of the contents of this article back into spectrum (functional analysis), and moving the specialized parts to other articles, new or old, like spectrum of an unbounded operator, spectrum of a Banach algebra, etc.
All the best, --Jorge Stolfi (talk) 14:05, 13 February 2013 (UTC)

Nothing wrong with including the definitions as a reminder, IMHO. I do agree that spectral theory of various classes of operators need their own separate articles:

In this article, we happen to have the case of self-adjoint operators on Hilbert spaces. Even though it kind of mirrors the general Banach case, a case can be made that it should be split and the discussion extended to normal operators.

The spectral theory of compact operator on Hilbert space already has it own article.

On the level of Banach algebras, as far as I know, there is not much to say about the spectrum of an element. That one can restrict to maximal abelian algebras seems to be about it. Mct mht (talk) 17:22, 13 February 2013 (UTC)

Moved the spectrum of the adjoint operator to Spectrum (functional analysis) since it seems completely irrelevant to the decomposition of the spectrum. Added decomposition into essential and discrete spectrum which seems precisely within the scope of the page. Comech (talk) 16:00, 20 October 2019 (UTC)