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20 votes
8 answers
12k views

Can a self-adjoint operator have a continuous set of eigenvalues?

This should be a trivial question for mathematicians but not for typical physicists. I know that the spectrum of a linear operator on a Banach space splits into the so-called "point," "continuous" ...
Mahdiyar's user avatar
  • 355
8 votes
1 answer
844 views

A doubt about the parts of the spectrum of tensor products

Let $\mathcal{H}$ be any complex Hilbert space of infinite dimensional. By an operator $T$ I mean a linear bounded transformation from $\mathcal{H}$ into $\mathcal{H}$, i.e, $T:\mathcal{H}\rightarrow\...
portella's user avatar
8 votes
1 answer
393 views

A question about comparison of positive self-adjoint operators

I have the following question but have no idea on its proof (one direction is trivial): Let $A$ and $B$ be (bounded) positive self-adjoint operators on a complex Hilbert space $H$. Prove that $$\...
Lao-tzu's user avatar
  • 1,906
7 votes
3 answers
2k views

Essential spectrum of multiplication operator

Let $a\in \mathcal{L}(L^2([0, 1], \mathbb{R}))$ be a multiplication operator. I wonder whether there is any work on calculating its essential spectrum. Is there any way to explicitly compute its ...
potionowner's user avatar
6 votes
2 answers
539 views

Is there a reasonable notion of spectral theorem on a pre-Hilbert space?

I'm trying to understand how bad things could possibly get without Cauchy completeness as a criterion for Hilbert spaces in quantum mechanics. Obviously, doing calculus on a pre-Hilbert space would be ...
Sanchayan Dutta's user avatar
6 votes
1 answer
1k views

Is the sum of spectral projections a projection?

Let $T$ be a closed operator on a Hilbert space with discrete spectrum. Then for $\{\lambda_1,...\lambda_n\}\in\sigma(T)$ one can define the spectral projections $$P_{\{\lambda_1,...\lambda_n\}}=\frac{...
Frank's user avatar
  • 241
5 votes
1 answer
159 views

For self-adjoint $A$ and $B$, when is $(A+iB)^*$ the closure of $A-iB$?

Suppose that I have two self-adjoint operators $A$ and $B$ such that $\mathcal{D}(A)\cap\mathcal{D}(B)$ is dense and $B$ positive. Then $A\pm iB$ (with domains $\mathcal{D}(A)\cap\mathcal{D}(B)$) are ...
Mathmo's user avatar
  • 223
5 votes
1 answer
379 views

Hilbert representation of a bilinear form

Let $\sigma:\mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$ be a bilinear symmetric form which is non-degenerate in the sense that for every $0\neq u\in \mathbb{R}^n$ there is $v\in \mathbb{R}^n$ with $...
erz's user avatar
  • 5,529
5 votes
1 answer
1k views

Left and right eigenvectors are not orthogonal

Consider a compact operator $T$ on a Hilbert space with algebraically simple eigenvalue $\lambda$. Is it then true that left (the eigenvector of the adjoint with complex-conjugate eigenvalue) and ...
Guido Li's user avatar
5 votes
1 answer
230 views

Canonical multiplication representation of self-adjoint operator in quantum chemistry and coding theory research

In my applied math research group, we are studying and going over functional analysis results from papers and theses from our institution to generalize their results and apply them in our discrete ...
groupoid's user avatar
  • 620
4 votes
1 answer
301 views

Trying to recover a proof of the spectral mapping theorem from old thesis/paper with continuous functional calculus

In my research group in functional analysis and operator theory (where we do physics and computer science as well), we saw in an old Russian combination paper/PhD thesis in our library a nice claim ...
groupoid's user avatar
  • 620
4 votes
1 answer
249 views

Is the Sobolev space $H^1(\mathbb{R})$ contained in the domain of $(-\partial_x \alpha(x) \partial_x)^{1/2}$?

Let $\alpha(x) : \mathbb{R} \to (0,\infty)$ have bounded variation (BV) and suppose $\inf_{\mathbb{R}} \alpha > 0$. Consider the second order differential operator $$H : =-\partial_x (\alpha(x) \...
JZS's user avatar
  • 481
4 votes
0 answers
2k views

Eigenvalues and spectrum of the adjoint

In a finite-dimensional Hilbert space, the eigenvalues of the adjoint $A^*$ of an operator $A$ are the complex conjugates of the eigenvalues of $A$. But in infinite dimensions this need no longer be ...
Arnold Neumaier's user avatar
3 votes
1 answer
214 views

Non-point spectrum for diagonalisable self-adjoint unbounded operator

Given a (separable) Hilbert space H and an unbounded densely defined linear operator $T:{\cal D}(T) \to $H such that ${\cal D}$ is diagonalizable (it means $\exists$ an O.N.B. of H such that all basis ...
Dave Shulman's user avatar
3 votes
3 answers
3k views

Infinite hermitian matrix

Suppose we have a finite square n x n matrix of complex numbers H that is Hermitian and skew-symmetric: $H^\dagger = H$ and $H^T = -H$. (T denotes transpose, $\dagger$ denote conjugate transpose. I ...
StevenJ's user avatar
  • 195
3 votes
1 answer
335 views

Book on Hilbert spaces, including non-separable

I am looking for a book that develops the theory of Hilbert spaces, including the spectral theorems and unitary representations, but includes non-separable Hilbert spaces in the main exposition. Any ...
3 votes
1 answer
232 views

what is about the corresponding power series?

According to the papers The absolutely continuous spectrum of Jacobi matrices and these lecture notes: periodicity ~ potential well or lattice (order) lack of absolutely continued spectrum ~ Anderson ...
XL _At_Here_There's user avatar
3 votes
1 answer
209 views

Must solutions to the time-independent Schrodinger equation that have discrete or negative eigenvalues be square-integrable?

This is a very straightforward question (although the answer may not be!) about a ubiquitous textbook physics situation. I assume that the answer is probably well-known, but I asked it on both Math ...
tparker's user avatar
  • 1,311
3 votes
1 answer
497 views

Hilbert-Schmidt integral operator with missing eigenfunctions

I'm having some issues with the spectral decomposition of the integral operator \begin{equation} (Af)(x)=\int_0^1|x-y|f(y)dy,\text{ with $f\in L^2[0,1]$}. \end{equation} Since \begin{equation} ...
Evan Gorman's user avatar
3 votes
1 answer
588 views

orthonormal basis of eigenvectors for laplacian on a concave polygon

I am interested in the Laplace operator $\Delta$ on a concave polygon. When the polygon is convex, it is known that $\Delta: H^2(\Omega) \rightarrow L^2(\Omega)$ is boundedly invertible. In addition, ...
localizer's user avatar
3 votes
0 answers
214 views

Extended adjoint of Volterra operator

Let $V$ be a Volterra operator on $L^2 [0,1]$. Does there exist a nonzero operator $X $ satisfying the following system $VX=XV^∗$, where $V^∗$ is the adjoint of the Volterra operator? $$ V(f) (x) =\...
Anas Abbas H.'s user avatar
2 votes
1 answer
541 views

Does spectral theory assume separability

On an infinite dimensional space, the spectral theorem for compact normal operators says that the eigenvectors form an orthonormal basis which, from wikipedia, is equivalent to the space being ...
user108605's user avatar
2 votes
1 answer
237 views

On spectral calculus and commutation of operators

Let $\mathcal{H}$ be a Hilbert space, $B\in\mathcal{B}(\mathcal{H})$ be bounded and self-adjoint and $A:\mathcal{D}(A)\to\mathcal{H}$ closed (but not necessarily self-adjoint or bounded). The ...
B.Hueber's user avatar
  • 1,171
2 votes
2 answers
873 views

Spectral theory in non-separable Hilbert Spaces

I am wondering about what can be said about the spectral theorem for unbounded, self-adjoint operators in a non-separable Hilbert space. There is a comment in this sense to the question "Does spectral ...
sbisaf's user avatar
  • 21
2 votes
2 answers
1k views

Lebesgue integral with respect to vector measures?

Good evening, I'm reading some papers of Jim Agler and Nicholas Young, in which they prove a formula of integral representation with respect to a vector measure, but the integration is in the sense ...
Đức Anh's user avatar
2 votes
1 answer
92 views

On the dimension of the range of the resolution of the identity

I want to prove the following: Let $A,B$ be bounded self-adjoint operators in a complex-Hilbert space and $E_A(\lambda)$, $E_B(\lambda)$ its corresponding spectral resolutions, i.e., $$A=\int_{[m_A,...
Andrés Felipe's user avatar
2 votes
0 answers
69 views

Mathematical reason for scatter states being special?

In infinite spectral theory, we have the discrete and continuous spectrum, which are called "bound" and "scatter states" in physics. My understanding is, if $O \in B(H)$ is a self-...
pyroscepter's user avatar
2 votes
0 answers
145 views

How to prove that a finite rank perturbation on an infinite matrix does not change its continuous spectrum?

I have the discrete Laplace operator on an infinite Hilbert space with an orthonormal basis $\psi_x$ ($\forall x \in \mathbb Z$), given by $\Delta \psi_x=\psi_{x-1}+\psi_{x+1}$. If I introduce a ...
pyroscepter's user avatar
2 votes
0 answers
131 views

Does a spectral theorem exist for linear operator pencils?

I was wondering if a version of the spectral theorem (the projection valued measure case) holds for linear pencils of the form $$ A-\lambda B $$ where $A,B$ are self-adjoint on some Hilbert space $\...
Mathmo's user avatar
  • 223
2 votes
0 answers
238 views

Examples for Markov generators with pure point spectrum

I'm looking at symmetric diffusion Markov generators $L$ with pure point spectrum, i.e. infinitesimal generators of symmetric diffusion Markov semigroups, which are defined on $L^2(\mu)$ where $\mu$ ...
herrsimon's user avatar
  • 235
1 vote
1 answer
89 views

Show $A_0 + S$ is invertible, where $S$ is the Riesz projection of bounded $A_0$

Let $A_0$ be a bounded linear operator on a Hilbert space $H$. Suppose $0$ is an isolated point of the spectrum of $A_0$. Let $S$ be the corresponding Riesz projection, namely, $$S = -\frac{1}{2\pi i} ...
JZS's user avatar
  • 481
1 vote
1 answer
119 views

Relation between the solutions $v_t=Lv$ and $v_t=Av$ if $A$ is a relatively compact perturbation of the linear operator $L$

In a nutshell, here is my question. I read and know about the relation between the spectra of $L$ and $A$ if $A$ is a relatively compact perturbation of $L$. However, for my purpose, I am interested ...
Gateau au fromage's user avatar
1 vote
1 answer
977 views

Regarding essential spectrum of the unilateral shift operator

This is with context to Example 4.10 in Section 11 of the book : A course in functional Analysis by J.B Conway. Let $\sigma_{le}(S)$ and $\sigma_{re}(S)$ denote the left and right essential spectrum ...
user534666's user avatar
1 vote
1 answer
127 views

Integral representation of joint projection valued measures

Given two positive $\sigma$-finite measures $\mu_{1/2}$ on the spaces $X_{1/2}$ one can define the product measure $\mu_1\otimes\mu_2$ on the product space $X_1\times X_2$. It can be proved that the ...
moppio89's user avatar
  • 275
1 vote
1 answer
213 views

Weyl asymptotics vs. form perturbations

Consider Hilbert spaces $V$,$H$; a closed quadratic form $a$ with domain $V$; and its associated operator $A$ on $H$. (If necessary, the form can be assumed to be coercive.) For the sake of simplicity,...
Delio Mugnolo's user avatar
1 vote
0 answers
62 views

$L^2$ norm of a kernel with a variable width

Suppose you have a kernel operator on a torus, with a kernel of a spatially varying width $\epsilon(x)$, which might be zero at certain points. That is to say, for some approximate identity $\psi_h(x)$...
Caroline Wormell's user avatar
1 vote
1 answer
771 views

A bounded operator $T$ is compact if and only if $\sigma_{\mathrm{ess}}(T)=\{0\}$ [closed]

Theorem: Let $T$ be a bounded self-adjoint operator on a complex infinite-dimensional Hilbert space $H$. Then $T$ is compact if and only if $\sigma_{\mathrm{ess}}(T)=\{0\}$. Proof: If $T$ is compact ...
Andrés Felipe's user avatar
1 vote
0 answers
67 views

Spectral theorems for generalized Hermitian matrices

Let $k$ be a field, and let $\sigma$ be a nontrivial involutory automorphism of $k$. Let $A$ be a square matrix with entries in $k$, such that $(A^{\sigma})^T = A$; here $A^\sigma$ means the matrix $(...
THC's user avatar
  • 4,547
1 vote
0 answers
53 views

Spectrum of a $1$-parameter family of symmetric linear operators

I am working with certain submanifolds of symmetric spaces and, using a construction in Terng-Thorbergson, we ended up in the following Hilbert space problem: Let $H$ be a (real) Hilbert Space and $...
Renato Moreira's user avatar
0 votes
2 answers
465 views

Spectrum equals eigenvalues for unbounded operator

Let $D$ be an unbounded densely defined operator on a separable Hilbert space $H$. If $D$ is diagonalisable with all eigenvalues having finite multiplicity and growing towards infinity, does it follow ...
Bas Winkelman's user avatar
0 votes
1 answer
269 views

Determine if an integral expression is in $L^2(\mathbb{R})$

Note: This is a simplified version of the following question. I did not get a full response and realized can make it simpler to have my main interrogation answered. I decided to write it as a ...
Gateau au fromage's user avatar
0 votes
2 answers
2k views

Spectral decomposition of compact operators

Let T be a compact operator on an infinite dimensional hilbert space H. I am proving the theorem which says that $Tx=\sum_{n=1}^{\infty}{\lambda}_{n}\langle x,x_{n}\rangle y_{n}$ where ($x_{n}$) is an ...
user134724's user avatar
0 votes
1 answer
185 views

Spectrum of a product of a symmetric positive definite matrix and a positive definite operator

Let $\mathbf H$ be an infinite dimensional Hilbert space. I want to find an example of a $2\times 2$ real symmetric positive definite matrix $M$ and a positive definite bounded operator $A : \mathbf H ...
SAKLY's user avatar
  • 63
0 votes
0 answers
55 views

reference request: conditions for pointwise and operator-norm convergence of kernel projections

At a very high level, I’m interested in the following question. Suppose $X$ is a (separable) Hilbert space, and $T_n : X \rightarrow X$ is a sequence of finite rank self-adjoint maps that converges (...
Joe's user avatar
  • 101
0 votes
0 answers
45 views

Discrete and continuous representation in Hilbert space

I’m interested in using laplacian (−Δ) eigenfunction as a basis for H1(Rn) . I know that in H1(Ω) , Ω bounded this can be done so I was wandering about H1(Rn) . Now let eλ be an eigenfunction ...
Alucard-o Ming's user avatar
0 votes
0 answers
252 views

Self-adjoint operator with pure point spectrum

Suppose that A is a self-adjoint (possible unbounded) operator from a separable Hilbert space H to itself. I would like to know if the following statement is true: A has pure point spectrum (i.e., the ...
user3476591's user avatar
0 votes
0 answers
41 views

Uniqueness of solution to abstract wave equation with unsigned energy

Let $H$ be a self-adjoint operator on a Hilbert space $(\mathcal{H}, \langle \cdot, \cdot \rangle$). Suppose the spectrum of $H$ in $(-\infty, 0)$ consists of only finitely many eigenvalues $\mu^2_k &...
JZS's user avatar
  • 481
0 votes
0 answers
220 views

Eigenvalue multiplicity of tensor product of positive operator with itself

Let $H$ be a separable complex Hilbert space and let $A\in B(H)$ be positive with $||A||=1$ and have eigenvalue 1 with multiplicity 1. Suppose $A=T^*T$ for some $T\in B(H)$. Denote the spectrum of $A$ ...
Dasherman's user avatar
  • 203
0 votes
1 answer
102 views

Essential spectrum of constant invertible diagonal matrix acting on a product of Hilbert spaces [closed]

Let $M$ be a $3\times 3$ real invertible diagonal matrix and $H$ a Hilbert space of infinite dimension (for example, we can take $H$ as the space of square integrable functions over a bounded ...
SAKLY's user avatar
  • 63
0 votes
0 answers
57 views

Isolated eigenvalues of "bipartite" operators

Please note: This is a reformulation of a previous question of mine. The old question has been already answered to, so I prefer asking a new one. However, it looked like the old formulation did not ...
Maurizio Moreschi's user avatar