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This is motivated by a question of mine on math.stackexchange. I would like a counter-example to the following: Let $A,B$ be positive self-adjoint operators on a Hilbert space $H$, and suppose that $D(...
Matthew Daws's user avatar
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I have a quick question regarding the spectral calculus of unbounded operators. Let $(\mathcal{H},\langle\cdot,\cdot\rangle_{\mathcal{H}})$ be a Hilbert space and $A:\mathcal{D}(A)\to\mathcal{H}$ be ...
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In Existence of closed operators with arbitrary dense domain of a given Banach space, it is shown by some descriptive set theory results that if $T: D(T)\subset X\to X$ is a closed operator on a ...
Prox's user avatar
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Let $H$ be a Hilbert space. Let $A:D(A)\to H$ be a self-adjoint operator. When can we say $A^2:D(A^2) \to H$ is self-adjoint? Is it enough to define $D(A^2)=\bigl\{u\in D(A) | Au\in D(A)\bigr\}$? And ...
Alucard-o Ming's user avatar
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I would like to understand the concept of multiplier for vector valued functions and find appropriate references for the multiplier theorems out there. For instance say that we would like to express $\...
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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 ...
Hugo's user avatar
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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 ...
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Let $H$ be a Hilbert space and let $h$ be a fixed, densely defined, possibly unbounded, self-adjoint operator on $H$. Letting $B(H)$ denote the space of all bounded operators on $H$, it is well ...
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In this post, it is said that there are no constructive examples of everywhere-defined unbounded operators between Banach spaces; every example furnished must use the axiom of choice. This seems like ...
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Let $\{a_n\}_{n=1}^\infty$, $a_n\in \mathbb{R}$. Consider the following linear operator $A$ on functions $f:\mathbb{R}\to \mathbb{R}$: $$(Af)(x) = \sum_{n=1}^\infty a_n f(x+n)+ \sum_{n=1}^\infty a_n f(...
H A Helfgott's user avatar
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Let $\mathbb H$ be a Hilbert space and let $\mathcal B(\mathbb H)$ be the bounded operators on $\mathbb H$. Let $J,K\in \mathcal B(\mathbb H)$ such that $ J=J^*, K=-K^*. $ Then the commutator $[J,K]$ ...
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Let $H$ be a separable Hilbert space with the inner product $\langle, \rangle$ and $\{ T_n \}$ be a sequence of unbounded closed linear operators with a common dense domain $D \subset H$ such that $...
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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_\...
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Let $S,T$ be unbounded, closed operators in Hilbert spaces $H,K$. In that case, we can form the tensor product operator $S\otimes T$ on the Hilbert space $H\otimes K$ which is the closure of the ...
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Let $\rho, \delta, m$ be real parameters such that $0\le \delta\le \rho\le 1, \delta<1$. The set $S^m_{\rho, \delta}(\mathbb R^{2n})$ is defined as the set of smooth functions $a$ on $\mathbb R^n\...
Bazin's user avatar
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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 ...
Luis Yanka Annalisc's user avatar
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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 \...
Luis Yanka Annalisc's user avatar
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This is a question about the existence of a core $C$ for the generator $A$ of a regular Dirichlet form $\mathcal{E}$ having a carré du champ $\Gamma$, so that $C$ is an algebra with respect to ...
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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$...
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Consider the following fragments from Takesaki's second volume "Theory of operator algebras" (chapter VIII $\oint 3$, Modular Automorphism groups, p107-108: Why are the second and third line ...
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Let $A$ be a closed (densely defined) operator on a Hilbert space $H$. We define for a natural number $k$, the operator $A^k$ with its natural domain. Is $A^k$ closed?
Andromeda's user avatar
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I would appreciate it if a reference could be given for the following claim. Let $g$ be a Riemannian metric on $\mathbb R^n$, $n\geq 2$ that is equal to the Euclidean metric outside some compact set. ...
Ali's user avatar
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1 answer
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Let $(\Omega, \mathcal{F})$ be a measurable space. Let $E: \mathcal{F}\to B(H)$ be a regular resolution of the identity on the Hilbert space $H$, see e.g. Rudin's functional analysis book. Suppose ...
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Consider the following fragment from Takesaki's book "Theory of operator algebras II" (Lemma 1.13 on p8, in chapter VI "Left Hilbert algebras"): Here, we associate with an ...
Andromeda's user avatar
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Let $P$ be a positive, self-adjoint (unbounded) operator in a Hilbert space $H$ with $0\notin \sigma(P)$. Consider its spectral decomposition $$P = \int_{\sigma(P)} t dE(t).$$ Since $0 \notin \sigma(P)...
Andromeda's user avatar
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Problem: Let $M$ be a semifinite von Neumann algebra with a faithful semifinite normal trace $\tau$. Let $m$ be a positive element in $M$ and let $e_{(0,\infty)}(m)$ be the spectral projection of $m$ ...
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Let $ T $ be a closed, densely defined operator on a Hilbert space $ H $. Then there exists a positive self-adjoint operator $ A $, $ D(A)=D(T) $ and a isometric operator $ V:R(A)\to \overline{R(T)} $ ...
Luis Yanka Annalisc's user avatar
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Given a (separable) Hilbert space H and an unbounded densely defined linear operator $T:{\cal D}(T) \to $H such that ${\cal D}$ is diagonalizable (it means $\exists$ an O.N.B. of H such that all basis ...
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Let $H$ and $K$ be Hilbert spaces and $D(T)$ a vector subspace of $H$. Let $T: D(T) \to K$ be a densely defined antilinear operator. Its adjoint $T^*: D(T^*)\to K$ is defined by the relation $$\langle ...
Andromeda's user avatar
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Given a Borel function $f:\mathbb{R}\rightarrow\mathbb{R}\cup\{\infty\}$, functional calculus allows to calculate $F(x)$ for any unbounded selfadjoint operator $x$ on a Hilbert space $\mathcal{H}$, ...
rpk's user avatar
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I want an example of a linear operator $T:X\to Y$ such that graph of $T$ is not closed. My thoughts: $T$ must be unbounded. Again by closed graph theorem any unbounded linear map from a Banach space $...
Anupam's user avatar
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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 $\{a,b\}\cap \sigma(T)=\emptyset$, ...
Watanabe's user avatar
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3 answers
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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 $...
The Ju's user avatar
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1 answer
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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 $...
pre-Hilbert's user avatar
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2 answers
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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 ...
Benjamin's user avatar
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1 answer
259 views

Let us consider the fractional heat semigroup $\left(e^{-t(-\Delta)^\alpha}\right)_{t\ge 0}$ for $\alpha\in (0,1)$ (the fractional power is taken in whole $\mathbb R^d$). Is there any result about the ...
Migalobe's user avatar
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11 votes
1 answer
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$\renewcommand\Im{\operatorname{\mathcal{Im}}}\newcommand\Ker{\operatorname{\mathcal{Ker}}}$I was sure that this is a trivial question and placed it on Math Stackexchange https://math.stackexchange....
Michael_1812's user avatar
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0 answers
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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)^{-...
user270619's user avatar
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Suppose that $\mathcal H$ is a Hilbert space representing some physical system, $H$ is the Hamiltonian for the system, and $A$ is some observable for the system, that is, some unbounded self-adjoint ...
Isaac's user avatar
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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 ...
Isaac's user avatar
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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 ...
Isaac's user avatar
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Preamble Two (unbounded) self-adjoint operators $A, B$ on a Hilbert space $\mathcal{H}$ are said to (strongly) commute if the unitary groups they generate commute or equivilantely if all the ...
Photographer's user avatar
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1 answer
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This is almost a copy of https://math.stackexchange.com/questions/3931318/when-the-adjoint-of-an-unbounded-operator-on-a-hilbert-space-coincides-with-the I am trying to work with infinite matrices in ...
apyshkin's user avatar
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0 answers
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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 $f(A)=Q^{-1} M_{f\circ \alpha} Q$. (...
Ma Joad's user avatar
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2 votes
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440 views

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\...
Kris's user avatar
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For a linear system of ODEs in $\mathbb{R}^{n}$ (with the usual inner product), say $\dot{V}(t) = A(t) V(t)$, we know that if $\xi_{1},\ldots,\xi_{k} \in \mathbb{R}^{n}$ and $V_{j}(t)=V_{j}(t,\xi_{j})$...
demolishka's user avatar
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1 answer
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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 ...
Jacob Lu's user avatar
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1 answer
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Is there any simple way to characterize explicitly the domain of fractional powers for a given operator? For example, the domain of Dirichlet Laplacian on a bounded nice domain $\Omega \subset \mathbb{...
Migalobe's user avatar
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1 answer
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Let $T:\mathrm{dom}(T) \subseteq H \to H$ be a densely defined, self-adjoint operator on a Hilbert spaces $𝐻$. In general the range of $T$ is not guaranteed to be closed. What tools are available to ...
Dave Shulman's user avatar
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Let $\mathcal{H}_i$, for $i=1,2$, be Hilbert spaces, and $T:{\frak Dom}(T) \subseteq \mathcal{H}_1 \to \mathcal{H}_2$ a densely-defined closed operator. If the kernel of $T$, and the kernel of its ...
Max Schattman's user avatar
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