19
votes
On commutator of bounded operators
If $H$ is finite dimensional there is a one-line solution: ${\rm tr}(JK - KJ) = 0$, so $JK - KJ$ cannot have positive spectrum.
But it is false in general! Let $V$ be a partial isometry from $H$ onto ...
- 42.8k
11
votes
Unbounded operators vs compact operators
Well, if $A$ is bounded and $B$ is compact then $AB$ is compact, so $AB$ cannot be the identity unless the Banach space is finite dimensional. Thus the left (or right) inverse of a compact operator on ...
- 42.8k
9
votes
Accepted
Diagonalizing selfadjoint operator on core domain
This sounds suspicious right away since the eigenvectors are what they are (nothing to choose here, unless you have degeneracies), but there is much choice for $D$ and we should be able to avoid ...
- 24.2k
8
votes
Accepted
Closable unbounded operators and Banach space adjoints
You can use essentially the same definition. If $T: E_1 \supseteq D(T)\rightarrow E_2$ is a linear map between Banach spaces, then we define $x^*\in D(T^*)$ with $T^*(x^*)=y^*$ to mean that $y^*(x) = ...
- 18.7k
7
votes
Accepted
The von Neumann algebra generated by a non-closable operator
The answer to Question 1. is positive. Namely, consider the polar decomposition of your operator $M=U|M|$ and define $X:= U f(|M|)$, where $f:[0,\infty) \to [0,\infty)$ is a bounded increasing ...
- 5,005
7
votes
Accepted
On the domains and extensions of unbounded operators
Yes, you've got it right. Given an unbounded self-adjoint operator $A$ with domain $D(A) \subset H$, using Zorn's lemma you can produce an everywhere defined operator $A'$ on $H$ which extends $A$. (...
- 29.7k
7
votes
Accepted
Unbounded Fredholms operators
The equality holds for closed unbounded operators, provided one interprets "cokernel" as "quotient by the closure of the image".
You can check it directly for closed self-adjoint operators by using ...
- 43.2k
7
votes
Accepted
Non-point spectrum for diagonalisable self-adjoint unbounded operator
Take $T$ to be the inverse of a bounded/continuous, self-adjoint operator with eigenvalues (an orthonormal basis) all rationals between $0$ and $1$. Then $T$ has an orthonormal basis of eigenvectors, ...
- 23k
7
votes
Accepted
If $A$ is a closed operator, is $A^k$ closed?
Here's a counterexample (subject perhaps to what you consider "natural").
Take a separable Hilbert space with orthonormal basis $\{u_n : n = 1, 2, \ldots\}$ and the operator $A$ defined by
$$...
- 54.2k
6
votes
Accepted
Unbounded version of continuous functional calculus
You find the precise abstract statment in the Internet Seminar lecture notes of Markus Haase, especially in Chapter 4.
See
https://www.math.uni-kiel.de/isem21/en/course/phase1
By the way, the answer ...
- 4,729
6
votes
On the domains and extensions of unbounded operators
Yes, of course. By definition of the adjoint operator, $\{[-A^* y, y]: y \in \mathscr D(A^*)\}$ is the orthogonal complement in $H \oplus H$ of the graph $\{[x, Ax]: x \in \mathscr D(A)\}$ of $A$. ...
- 54.2k
6
votes
Accepted
A detail in the proof of Schur's lemma: the closures of the $\mathcal{Ker}$ and $\mathcal{Im}$ of the intertwiner
Let $V$ and $W$ be Hilbert spaces with irreducible unitary $G$-actions and let $T:V \to W$ be a bounded intertwiner. Then the adjoint is an intertwiner as well and hence so are $T^\ast T$ and $TT^\ast$...
- 20.9k
6
votes
Accepted
Convergence criterion in the domain of an unbounded operator
If you have a uniform upper bound on $\|Ax_n\|$ then you can extract a weakly convergent subsequence $Ax_{n_k}$. Denote $\lim_n Ax_{n_k} = y_{\infty}$ to be the weak limit. Since $A$ is closed, it is ...
- 625
6
votes
Convergence criterion in the domain of an unbounded operator
The answer by Lars van der Laan gives a positive answer for the Hilbert space case (which was considered in the question), and it also works on reflexive Banach spaces.
It might be worthwhile to add ...
- 12.5k
5
votes
Notations for dual spaces and dual operators
Given your situation of having to juggle all these notational traditions at the same time, I would recommend for a space $X$ and an operator $A$:
$X^{\vee}$, 2. $X'$, 3. $A^{\rm T}$, 4. $A^{\rm T}$, ...
5
votes
Notations for dual spaces and dual operators
The standard notation don't overlap as much as you think if you are careful with types. For example $X'$ is perfectly fine for both 1. and 2. because vector spaces and topological vector spaces are ...
5
votes
Minimum eigenvalue of One-dimensional Schrodinger Operator
The Löwdin method to obtain lower bounds to energy eigenvalues of the Schrödinger equation has been reviewed in Lower Bounds to Energy Eigenvalues (1976). It has been applied to the quartic potential ...
- 188k
5
votes
Accepted
Are $\| \Delta u \|_{L^p(\Bbb R^d)} + \| u \|_{L^p(\Bbb R^d)}$ and $\| u \|_{W^{2,p}(\Bbb R^d)}$ equivalent norms?
First, the classical Calderón–Zygmund estimate gives
$$\left\|D^2 u\right\|_{L^p(\mathbb{R}^d)}\leq C(p,d) \left\|\Delta u\right\|_{L^p(\mathbb{R}^d)}.$$
By interpolation, $\left\|u\right\|_{W^{2,p}(...
- 166
4
votes
Accepted
Commuting with self-adjoint operator
Any bounded Borel function $f: \mathbb{R} \to \mathbb{R}$. If $TS = ST$ then (taking adjoint of both sides) $S^*T = TS^*$. Therefore both ${\rm Re}(S) = \frac{1}{2}(S + S^*)$ and ${\rm Im}(S) = \frac{...
- 42.8k
4
votes
Accepted
Symmetric diagonalizable operators and self-adjointness
Yes. If $x_n\in D(L)$ is an ONB of $H$ and $Lx_n=\lambda_n x_n$, then the operator $T$ acting in the obvious way on $D(T)=\{ \sum a_n x_n\in H : \sum \lambda_n^2|a_n|^2<\infty\}$ is self-adjoint. ...
- 24.2k
4
votes
Accepted
Spectral growth of One dimensional Schrodinger Operator
First of all, it is sufficient to consider only one parameter: making a change of the independent variable $z\mapsto kz$, with appropriate $k$ one can eliminate either $a$ or $b$. Let us eliminate $b$ ...
- 91.8k
4
votes
Accepted
Spectrum equals eigenvalues for unbounded operator
I agree with Andreas that the obvious straightforward interpretation of "the eigenvalues grow to infinity" is that the sequence of eigenvalues $(\lambda_n)$ increases to infinity. (And, counter to ...
- 42.8k
4
votes
Accepted
Spectral representation of closed operators with finite spectral bound
I've looked it up now. The formula in question does indeed hold in the following sense:
Theorem. Let $(e^{tA})_{t \in [0,\infty)}$ be a $C_0$-semigroup on a complex Banach space $X$. Let $\omega \in \...
- 12.5k
4
votes
Unbounded operators vs compact operators
The following ramblings are just night thoughts kicked off by your question (and so should really be a comment, but I am not entitled) which I am posting in the hope that they may be of interest to ...
- 131
4
votes
Accepted
$\tau$-measurable operator
I think that this is simply not true. Take $M = \ell^\infty(\mathbb{N})$ with the semifinite trace $\tau(F) = \sum_n F(n)$. When $p \in M$ is a nonzero projection, we have $\tau(p) \geq 1$. So the ...
- 4,351
4
votes
Accepted
Unbounded positive self-adjoint without $0$ in its spectrum: can we construct its inverse using functional calculus?
$\newcommand\si\sigma\newcommand\D{\mathscr D}$Yes, $P^-=P^{-1}$.
Indeed, the resolvent set of $P$ is open and $0$ is in this set. So, $\si(P)$ is bounded away from $0$, and hence the domain $\D(P^-)$ ...
- 128k
3
votes
Accepted
Commuting with an unbounded operator
Yes, your statement is true.
First observe that $$M := \{B \in \mathcal{B}(\mathcal{H}) : BA \subset AB \,\,\mbox{ and }\,\, BA^* \subset A^*B\}$$ is a von Neumann algebra (on the nose, without ...
- 398
3
votes
Accepted
Characterising closed range self-adjoint operators
This is a complete, but rather abstract characterisation.
$T$ has closed range if and only if there is $H_0\subseteq H$ closed and a bounded self-adjoint injective map $R:H_0\rightarrow H_0$ with $...
- 18.7k
3
votes
When the adjoint of an unbounded operator on a Hilbert space coincides with the formal adjoint on its natural domain?
Even after addressing the issues raised in the comments, the matrix coefficients $A_{ij}$ don't give you enough information to find $D(A^*)$. For example, consider $A_j=-d^2/dx^2$ on $L^2(0,1)$, $j=1,...
- 24.2k
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