All Questions
11 questions
3
votes
2
answers
280
views
Question about the Bessel operator
For $\nu>-1$ denote by $\{\lambda_{k,\nu}\}_{k\in\mathbb{N}}$ the succesive positive zeros of the Bessel function of the first kind $J_{\nu}$. The Bessel operator is given by
\begin{equation*}
L_\...
- 421
2
votes
1
answer
137
views
Disturbance of self-adjoint operator
Assume that $ A $ is self-adjoint operator and $ B $ is a bounded self-adjoint operator. The definite domain of $ A,B $, denoted by $ D(A) $ and $ D(B) $ satisfies $ D(A)\subset D(B) $. Show that
\...
- 1,219
3
votes
1
answer
226
views
$\tau$-measurable operator
Problem: Let $M$ be a semifinite von Neumann algebra with a faithful semifinite normal trace $\tau$. Let $m$ be a positive element in $M$ and let $e_{(0,\infty)}(m)$ be the spectral projection of $m$ ...
- 85
-1
votes
1
answer
164
views
Closure of the point spectrum of an unbounded diagonalizable 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 ...
- 969
1
vote
0
answers
205
views
Convergence in the resolvent sense and spectral properties
Let $\{T_k\}_k, T$ be unbounded selfadjoint operators on a Hilbert space $H$. If $T_k\to T$ in the norm-resolvent sense, then for any $(a,b)\subset \mathbb R$ with $\{a,b\}\cap \sigma(T)=\emptyset$, ...
- 111
0
votes
0
answers
62
views
"Trade-off" between bound on the function and on the spectrum for functional calculus in spectral theory
Let $A$ be a self-adjoint (unbounded) operator on a separable Hilbert space $H$.
From the following form of spectral theorem, we may define a functional calculus by $f(A)=Q^{-1} M_{f\circ \alpha} Q$. (...
- 1,755
2
votes
1
answer
127
views
Spectral representation of closed operators with finite spectral bound
Assume $A$ is a closed linear operator on a Banach space $X$ and is densely defined. Assume the spectral bound $s(A) = \sup\{Re\lambda: \lambda\in \sigma(A)\}$ is finite. For example, if $A$ is the ...
- 903
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 ...
- 969
5
votes
1
answer
2k
views
Commuting with self-adjoint operator
Let $T$ be an (unbounded) self-adjoint operator. Assume that there is a bounded operator $S$ such that $TS=ST.$ For which kind of $f$ do we have that $f(T)S=Sf(T)?$
My thought was that using a ...
- 501
4
votes
1
answer
172
views
For self-adjoint $T$ on $L^2(\mathbb{R}^n)$, when does $(1 + |x|)^{-1} (T - i \varepsilon)^{-1}(1 + |x|)^{-1}$ have a limit as $\varepsilon \to 0$?
Let $T : L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ be a possibly unbounded self adjoint operator. Let $R_\varepsilon$ denote the resolvent $(T - i \varepsilon)^{-1}$, $\varepsilon > 0$. Suppose that,...
- 481
14
votes
2
answers
4k
views
What is a good reference that compact resolvent implies Fredholm operator?
Suppose $A \in \mathcal{L}(E_1, E_0)$ is a bounded linear operator between Banach spaces $E_1$ and $E_0$, and we also have that $E_1$ is densely, continuously embedded in $E_0$ (i.e. $A$ can be ...
- 191