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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 ...
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
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. ...
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 ...
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 \...
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 $1+A$...
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{...
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 ...
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 $...
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 $...
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 ...
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)^{-...
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 ...
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 ...
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\...
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: ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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
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 ...
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{...
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 ...
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 ...
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)\...
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 ...
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 ...