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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
1 vote
1 answer
155 views

Spectrum invariant under (generalised) transpose as operator on trace class operators

For matrices $A$ it is well known that the spectrum is invariant under transpose $\sigma(A^T) = \sigma(A)$. Furthermore, the spectrum of the adjoint matrix $\sigma(A^*) = \overline{ \sigma(A)}$ the ...
Frederik Ravn Klausen's user avatar
9 votes
0 answers
540 views

Why is spectral theory developed for $\mathbb C$

Spectral theory is a fundamental part of operator theory and the spectrum of many operators is investigated throughout the existing literature. And that is for a good reason: If $A$ is some closed ...
Yaddle's user avatar
  • 381
0 votes
0 answers
36 views

Regarding significance of spectral variation under algebraic operations

I have been reading the paper Determining elements in $C^∗$-algebras through spectral properties. The paper discusses about what would be the relation be between two elements $a$ and $b$ of a Banach ...
user332905's user avatar
6 votes
3 answers
3k views

Non-empty resolvent set, then operator closed?

On Hilbert spaces, the following is true: Let $T$ be a densely-defined linear operator with non-empty resolvent set, then $T$ is closed. The obvious proof I see to show this uses explicitly the ...
gipom's user avatar
  • 115
11 votes
1 answer
487 views

Is the spectrum of a "self adjoint" operator real on $\ell^p$?

There might be an obvious answer to the question, but it doesn't come to mind. Suppose we have an infinite matrix $A=(a_{ij})$, which defines a bounded linear operator on $\ell^p$, i.e. for all ...
an_ordinary_mathematician's user avatar
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
5 votes
1 answer
1k views

Reference request: The resolvent is analytic in the resolvent set

I am busy reading through Taylor's paper Spectral Theory of Closed Distributive Operators. On page 192, he defines the resolvent and spectrum of $T$: Later on in the paragraph, he then proceeds by ...
user860374's user avatar
12 votes
1 answer
191 views

Spectra on different spaces

This is a method request: I am looking for techniques that allow me to investigate problems like this: Let $T_1: \ell^1 \rightarrow \ell^1$ be a bounded operator with $\Re(\sigma(T_1)) \subset (-\...
Kinzlin's user avatar
  • 305
4 votes
0 answers
171 views

quasi-nilpotent part of a dual operator

Definitions and notation. Let $X$ be a complex Banach space and $T\in\mathcal{L}(X)$ a continuous linear operator on $X$. We define the quasi-nilpotent part of $T$ as \begin{equation*}H_0(T):=\left\{...
Ben W's user avatar
  • 1,591
9 votes
1 answer
893 views

Perturbations of an operator that disconnect the spectrum

The following question came to me while working on a technical matter about transversality in infinite dimension, and I'm really curious to know whether it has an affirmative answer at least under ...
Pietro Majer's user avatar
  • 60.5k