Timeline for Interesting examples of spectral decompositions of BOUNDED operators with both continuous and discrete spectrum

Current License: CC BY-SA 4.0

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Dec 7, 2018 at 14:20	comment	added	Hugo Chapdelaine		Dear @Christian, I guess it depends what you want to do. For example, does the spectral theorem give you a bound on how many discrete eigenvalues are to be found in a given bounded interval ? I number theory we have the so-called Selberg Trace formula that provides such a bound and in order to get it one needs to get a good hold on these "generalized eigenvectors".
Dec 6, 2018 at 23:58	comment	added	Christian Remling		@HugoChapdelaine: Yes, such generalized eigenfunctions (if we want to call them that) can be introduced in many situations, but I don't think they make anything clearer. Quite on the contrary, they obscure what is going on, and the rigorous version of the spectral theorem gives a much clearer picture.
Dec 6, 2018 at 21:12	comment	added	Nik Weaver		Some keywords that might be helpful are "coherent states" and "reproducing kernel Hilbert space".
Dec 6, 2018 at 20:43	comment	added	Hugo Chapdelaine		Well I like when I can touch and compute things. For example Eisenstein series admit Fourier series expansion and this gives you a good hold on these objects. I read a little bit about the spectral theorem for unbounded self-adjoint operators and I did not get much insight regarding the associated spectral measure. At the end I would like to be able for any f in H to write it as a weighted sum of explicit eigenvectors for which I have a good hold.
Dec 6, 2018 at 20:07	comment	added	Nate Eldredge		On the other side of the Fourier transform, are you willing to consider the possibility of $g_t$ being a measure?
Dec 6, 2018 at 19:58	history	edited	Hugo Chapdelaine	CC BY-SA 4.0	added 4 characters in body
Dec 6, 2018 at 19:49	history	edited	Hugo Chapdelaine	CC BY-SA 4.0	added 503 characters in body
Dec 6, 2018 at 19:40	comment	added	Hugo Chapdelaine		Dear @ChristianRemling, I agree that my expression "physical incarnation" is vague. The best examples that come to my mind are the functions $x\mapsto e^{ix\xi}$ where $\xi\in \mathbb{R}$ for the usual Laplacian $-d/dx^2$ on $\mathbb{R}$. These function are not square integrable but they are still "small" since their absolute value is equal to 1 as $x$ varies over $\mathbb{R}$. Other simple examples are Eisenstein series when one replaces $\mathbb{R}$ by the Poincare upper half-plane quotiented by a Fuchsian group of the first kind.
Dec 6, 2018 at 18:40	comment	added	Nate Eldredge		@ChristianRemling: Yes, I was going to put that in later.
Dec 6, 2018 at 18:39	comment	added	Christian Remling		@NateEldredge: Your example doesn't have discrete spectrum, but of course that's easily fixed by adding a discrete part $\sum \delta_{x_j}$, with $x_j\notin [0,1]$, to $dx$.
Dec 6, 2018 at 18:01	comment	added	Nate Eldredge		Yeah, what's the definition of "physical incarnation"? I'm trying to think about the most basic example of a bounded operator with continuous spectrum: multiplication by $x$ on $L^2([0,1], dx)$. I can't figure out whether it satisfies your criteria or not.
Dec 6, 2018 at 17:59	comment	added	Christian Remling		Conditions (2), (3) are a bit vague and don't really make strict formal sense, but leaving that aside, discrete versions of the Laplacian such as $(Tu)_n=u_{n-1}+u_{n+1}$ on $\ell^2(\mathbb Z_+)$, with suitable modification at the endpoint $n=1$ (to get some eigenvalues), provide examples with these properties.
Dec 6, 2018 at 16:33	comment	added	Mateusz Wasilewski		If $T$ is a self-adjoint operator then $(i Id - T)^{-1}$ is a bounded normal operator, whose spectral decomposition can be deduced from the decomposition of $T$. So boundedness is not really an issue.
Dec 6, 2018 at 15:10	history	asked	Hugo Chapdelaine	CC BY-SA 4.0