All Questions
31 questions
3
votes
0
answers
95
views
Commutator of $A\otimes I$ and $I \otimes B$ vanishes?
Consider two Hilbert spaces $H_1$ and $H_2$, and $A$, $B$ unbounded operators on $H_1$, $H_2$ respectively. $(A \otimes I)$ is classically defined as the closure of the operator defined on the set of ...
- 31
5
votes
2
answers
148
views
Showing an operator is (or not) closed on $L^2(\mathbb{R})$
I am linearizing nonlinear waves and get operators of the form below. Everything is considered in $L^2(\mathbb{R})$.
Consider the operator $L_1=\frac{d}{dx}$. The domain is $H^1(\mathbb{R})$ and it is ...
2
votes
0
answers
91
views
How to show $ |(Bx,x)|\leq (Ax,x) $ for any $ x\in D(A) $ here?
On the Hilbert space $ H $, $ A $ is a non-negative self-adjoint operator and $ B $ is a symmetric operator. Let $ D(B)\supset D(A) $, where $ D(A) $ and $ D(B) $ are definite domain for $ A $ and $ B ...
- 1,219
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
4
votes
1
answer
228
views
Diagonalizing selfadjoint operator on core domain
Let $A$ be a densely-defined, positive, self-adjoint operator with compact resolvent on a Hilbert space $H$. Then, $\text{Range}(1+A)=H$ and there is a basis for $H$ consisting of eigenvectors of $1+A$...
- 143
1
vote
0
answers
153
views
Maximal domain of an unbounded linear operator in a weighted Hilbert-space
Let's consider the following (unbounded) linear operator. (So called transport operator in some context.)
$$ \mathrm{T}: \mathcal{H} \supset \mathcal{D}(\mathrm{T}) \to \mathcal{H} , f \mapsto \mathrm{...
- 185
0
votes
1
answer
138
views
Antilinear unbounded operator has closed graph
Let $H$ and $K$ be Hilbert spaces and $D(T)$ a vector subspace of $H$. Let $T: D(T) \to K$ be a densely defined antilinear operator. Its adjoint $T^*: D(T^*)\to K$ is defined by the relation
$$\langle ...
- 175
3
votes
3
answers
1k
views
Unbounded operators vs compact operators
The operator $L:\operatorname{dom}(L)\subset C[0,1]\to C[0,1]$ given by $Lx=x'$
a) is closed, unbounded and densely defined
b) also has a compact right inverse, namely $K:C[0,1] \to C[0,1]$ given by $...
- 31
1
vote
1
answer
335
views
A consequence of the Min-Max Principle for self-adjoint operators
Let $H=(H, (\cdot, \cdot))$ be a Hilbert space. Let $T_1,T_2:D \subset H \longrightarrow H$ be a self-adjoint operators (not necessarily bounded). It's well-know that the spectrum $\sigma(T_i)$ of $...
- 205
6
votes
2
answers
514
views
Convergence criterion in the domain of an unbounded operator
Cross-post from math.sx.
My question is somewhat close to this one, but the counterexamples given there do not apply here.
Setup. Given a Hilbert space $\mathcal H$, a closed operator $A$ and a ...
- 245
2
votes
0
answers
654
views
Convergence of operator in norm resolvent sense and their eigenvectors
Let $\{T_n\}_{n=1}^\infty$ and $T$ be (unbounded) self-adjoint operators and $T_n\to T$ in norm resolvent sense, that is, for some $z\in \mathbb{C} \setminus \mathbb{R}$, $\|(zI- T_n)^{-1}- (zI- T)^{-...
- 21
4
votes
0
answers
230
views
Spectral theorem for unbounded operators
Part of the Spectral theorem for unbounded operators states that if $A$ is a self adjoint unbounded operator and $B$ is a bounded operator such that $BA$ is contained in $AB$, then $B$ commutes with ...
- 771
2
votes
0
answers
144
views
Extensions of symmetric unbounded operators
I saw it claimed that every symmetric operator on a Hilbert space $H$ can be extended to a self-adjoint operator on some potentially larger space K. But I seem to be able to prove from this that every ...
- 771
2
votes
0
answers
395
views
On strong resolvent convergence of unbounded operators
Given a sequence of unbounded self-adjoint operators $\Delta_N$ defined on some fixed domain of the Hilbert space $L^2(\mathbb{R})$ converging to i*identity in the sense that: for all $\phi\in\...
- 63
6
votes
2
answers
665
views
Unbounded Fredholms operators
Motivated by the situation of bounded Fredholm operators, I have the following question about "unbounded Fredholm operators".
Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be two Hilbert spaces, and
$$
D: ...
- 969
1
vote
1
answer
253
views
Power of an infinitesimal generator of a $C_0$-semigroup in a Banach space
Let $A$ be the infinitesimal generator of a $C_0$-semigroup of linear operators in a Banach space. Let $n$ be a positive integer, $n\geq2$. Is $A^n$ closed?
Here (setting $A^1$ $:=$ $A$, and ...
- 31
3
votes
1
answer
126
views
The imaginary exponential of a tangent field on a manifold
If $M$ is a compact Riemannian manifold and $X$ is a tangent field, I am seeking to define the object $\exp {\mathrm i t X}$ for $t \in \mathbb R$, and I do not know how to do it.
One option was to ...
- 5,407
2
votes
1
answer
486
views
Common eigenvector of commuting unbounded operators
Let $T$, $S$ be two self-adjoint linear operators on a Hilbert space $\mathcal{H}$ with pure point spectrum.
Then $T$ and $S$ commute if and only if they have a complete set of common eigenvectors.
...
- 21
2
votes
0
answers
40
views
Invariance of simple functions
Let $(A(s),D(A(s)))_{s\in\mathbb{R}}$ be a family of unbounded operators on a Banach space $X$ and $g:\mathbb{R}\rightarrow X$ be a simple function, i.e.,
\begin{align*}
g=\sum_{i=1}^n{x_i\textbf{1}_{...
4
votes
1
answer
878
views
Commuting with an unbounded operator
Let $H$ be a Hilbert space. Let $A$ be a closed unbounded operator, and let $B\in B(H)$ be a bounded operator.
Definition:
$A$ and $B$ strong-commute if the partial isometry in the polar ...
- 43.2k
8
votes
1
answer
302
views
Does every integer map generate a von Neumann algebra of type I?
Consider a map $m: \mathbb{N} \to \mathbb{N}$ (we call it an integer map). Let $E_r$ be the set $m^{-1}(\{r\})$.
Let $H$ be the Hilbert space $\ell^2(\mathbb{N})$ and consider the densely defined ...
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
5
votes
2
answers
2k
views
On the domains and extensions of unbounded operators
I am not an expert in functional analysis but I was studying some, motivated from some mathematical physics considerations. I am not quite sure whether this is research-level, but let me state some ...
- 7,746
1
vote
1
answer
223
views
Stochastic integral is a continous or closed operator?
The Setup
Let $\xi_t$ be a process adapted to the filtration $\mathfrak{F_t}$ of the semi-martinagale $X_t$, such that both are square integrable. Then is the map
\begin{align}
F_T: L^2(\mathfrak{...
- 5,405
1
vote
1
answer
1k
views
Sum of two unbounded self-adjoint operators
Let $A$ and $B$ be two unbounded self-adjoint operators. From this mathoverflow post, for instance, we know that $A + B$ is self-adjoint on $\mathcal{D}(A) \cap \mathcal{D}(B)$ if $A$ and $B$ are ...
- 13
1
vote
1
answer
2k
views
Operator theory of the Hessian
How can I learn more about the operator theory of the Hessian? The Hessian of a function $u : \Omega \rightarrow \mathbb R$ over a domain $\Omega \subseteq \mathbb R^n$ is the matrix of second ...
- 5,327
1
vote
1
answer
394
views
Pullback via flow as operator group
Let $X$ be a vector field on a manifold $M$ that induces a complete flow $\Theta_t$. Then the operator family $\Theta_t^*$,
$$\Theta_t^*u(x) = u(\Theta_t(x))$$
is a strongly continuous semigroup of ...
- 9,853
1
vote
2
answers
398
views
A question on unbounded operators
Assume that $H$ is a separable Hilbert space.
Is there a polynomial $p(z)\in \mathbb{C}[x]$ with $deg(p)>1$ with the following property?:
Every densely defined operator $A:D(A)\to D(A),\;D(A)\...
- 356
0
votes
1
answer
559
views
Densely-defined unbounded operators with large support
Most densely-defined unbounded linear operators on Hilbert spaces have a very large domain. In fact, for a lot of natural operators the intersection of their domains are still dense.
Let us consider ...
- 5,327
11
votes
3
answers
445
views
Does the generator of a 1-parameter group of Banach space isometries know which elements are entire?
Let $X$ be a complex Banach space. Let $(\sigma_t)_{t \in \mathbb{R}}$ be a 1-parameter group of linear isometries of $X$ which is strongly continuous i.e. $t \mapsto \sigma_t(x)$ is continuous for ...
- 662
2
votes
0
answers
114
views
non-closed weak graph limit of symmetric operators
Hi Everyone,
I was recently reading Reed & Simon's functional analysis textbook (the first volume), and it mentions casually on page 294 that weak graph limits of a sequence of symmetric ...
- 21