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Questions tagged [unbounded-operators]

Let $X$ and $Y$ be normed spaces and $T: D(T)\rightarrow Y$ a linear operator, where $D(T)\subset X$. The operator $T$ is said to be unbounded if there exists a sequence $\{x_n\}\subset D(T)$ s.t. $$\| Tx_n\| \geq n\| x_n\| $$

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Existence of a non bounded linear operator. [duplicate]

I want to prove using Hamel basis that for $X$, $Y$ normed spaces such that dim$(X)=\infty$ there exists a non bounded linear operator... but I do not know how to proceed. Any hint, please?
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How is this operator well defined? $\frac{D}{(1+D^2)^{1/2} }$.

Let $D_+ = \partial_x +x, D_-=-\partial_x+x$. $$D= \begin{pmatrix} 0 & D_- \\ D_+ & 0 \end{pmatrix} $$ which acts on a dense subspace $C_c(\Bbb R) \oplus C_c(\Bbb R)$ of $L^2(\Bbb R) \...
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41 views

Proof of strong limit $\displaystyle s-\lim_{\varepsilon \to 0} (I+i\varepsilon A)^{-1}=I$

Let $A$ be an unbounded self-adjoint operator on a Hilbert space $H$, I was wondering if there was an 'elementary proof', i.e. that doesn't use the full functional calculus, of the strong limit : $$\...
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40 views

Why adjoint operator is not a generalization of transpose matrix?

I know in finite Hilbert spaces, the adjoint operator is a sort of generalization of the transpose matrix, in the sense that if $M$ is symmetric, then $\langle Mx,y\rangle=\langle x,My\rangle$. So if $...
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Elliptic Differential Operators I, Higson's Book

Throughout we assume $D:C^\infty(M;S) \rightarrow C^\infty(M;S)$ to be an order $1$ differential operator. Def: Let $D$ be a differetial operator on a manifold $M$. We say that $M$ is complete for ...
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27 views

Hahn Banach Theorem implying existence of a nonzero linear functional taking 0 in a linear subspace

I am reading this paper. In the proof of theorem 1, it is stated By the Hahn-Banach theorem, there is a bounded linear functional on $C(I_n)$, call it $L$, with the property that $L\ne 0$ but $L(R)...
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operator core of a fourier multiplier

Let $\Lambda:\mathcal{D}(\Lambda)\to L^2(\mathbb{R})$ be a densely defined Fourier multiplier, i.e, for any $u\in\mathcal{D}(\Lambda)$, we have $\mathcal{F}(\Lambda u)=m_{\Lambda}\mathcal{F}(u)$, ...
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Example of unbounded operators defined on the entire complete space

In functional analysis I have come across unbounded operators defined in the following way: $$A:\mathcal D(A)\subseteq X\to Y,$$ where $X,Y$ are Hilbert spaces and $A$ is defined only for $x\in\...
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Adjoint of multiplication operator

To keep it simple, let $\phi : I\to\mathbb C$ be a measurable function on a finite interval $I\subset\mathbb R$. The multiplication operator $M_\phi$ is defined as $M_\phi f = \phi\cdot f$, $f\in\...
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Are “lower bound” and “upper bound” unambigous? What if one's in the negative axis?

Are "lower bound" and "upper bound" unambigous? What if one's in the negative axis? Consider e.g. $$\{-n: n \in \mathbb{N} \}$$ This has no lower bound, if one consider lower to mean towards $- \...
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39 views

If operator is closed and densely defined then $D(A^*)^\perp = \{0\}$

I'm a bit rusty in my Functional Analysis and couldn't solve this question: Let $X$ be a Banach space (over either $\mathbb{R}$ or $\mathbb{C}$) and $X^*$ its dual space. Show that, if $A:D(A) \...
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prove d/dx is bounded on L^2

Can someone take a look at my answer and tell me what I am missing, also, what do you think this problem is trying to teach me? Problem The functions $f_n=x^n$ are square integrable and continuous ...
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31 views

Show that if $\phi \circ T$ is bounded for every $\phi\in Y^*$ then $T$ is bounded

Show that if $T:X \to Y$ is a linear map between Banach spaces and $\phi \circ T$ is bounded for every $\phi\in Y^*$ then $T$ is bounded. The given hint is to prove the contrapositive. 'Progress': ...
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Convergence to the differential operator

This is not an accurate question. I am not so sure about unbounded operators on a Hilbert space. Let $D$ be the differential operator on $L^2(0,1)$. Well, to somewhat, we can extend $D$ to a normal ...
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1answer
11 views

Cesaro average semigroup

Let $(P_t)_{t\geq 0}$ be a symmetric strongly continuous semigroup on $L^2(X,\mu)$ with generator $(L,\mathcal{D}(L))$. Given $f\in\mathcal{D}(L^*)$, define the Cesaro average $$ A_tf=\frac{1}{t}\...
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Every nonempty closed subset $M\subset \mathbb C$ is the spectrum of a closed operator

Several questions on this cite concern themselves with the operator $T:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$ where $T(x_n)_{n=1}^\infty=(r_nx_n)_{n=1}^\infty$. Here the sequence $\{r_n:n\in\mathbb N\}...
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$\frac{d}{dx}:C^1([0,1])\to C([0,1])$ as a closed, unbounded operator

First recall that a (potentially unbounded) operator $T:D(T)\subseteq X\to Y$ is closed whenever $(x_n)_{n=1}^\infty\subset D(T)$ convergent to $x\in X_0$ and $(Tx_n)_{n=1}^\infty$ convergent in $X_1$ ...
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“bounding” an unbounded operator

I was wondering if, given a certain unbounded operator on a Hilbert space, it can (naively speaking) be "cutted" (or "bounded") by certain projections. So, thinking about this in a more sensible way, ...
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55 views

Weyl sequence for oscillator harmonic

Consider the operator $T=-\Delta_{\Bbb{R}^2}-(x^2+y^2)$. How can I construct a sequence $ (u_n)\in D(T)=\{u\in L^2(R^2); Tu\in L^2(R^2)\}$ such that $\|u_n\|=\sqrt{\int_{R^2}|u(x,y)|^2}=1,$ $u_n\to ...
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About functional calculus of unbounded operator

I am learning spectrum theory of unbounded operator. I get confused on lots of places. One of them is the functional calculus for unbounded operator. I don't know what exact answer I am looking for. ...
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Norm of “resolvent” for nonlinear eigenvalue problems

I have been looking at extensions of the usual Linear eigenvalue problems for (possibly unbounded but closed) operators. These are defined as follows. Let $T$ be an analytic function of $\lambda \in \...
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Bounded operator but not Compact

Let $T : (C([-1,1]),||.||_{\infty}) \rightarrow (C([-1,1]),||.||_{\infty}) $ Such as : $(Tf)(x)=\frac{1}{2}(f(x)+f(-x))$ . For all $f\in C([-1,1])$ Why $T$ isn't Compact ? I tried to use the ...
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70 views

Unbounded self-adjoint operator with compact resolvent: Expansion

Given an unbounded self-adjoint operator $T\colon H \supset \mathrm{dom}(T) \to H$ on a Hilbert space $H$ with compact resolvent, i.e. $T$ has a purely discrete spectrum, the eigenfunctions $e_n$ form ...
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Convergence of (unbounded) self-adjoint operators

I'm learning about the dynamical convergence (i.e, convergence of the unitary group associated with each operator) and resolvent convergence of (unbounded) self-adjoint densely defined operators. I ...
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2answers
63 views

Continuous map but not bounded on Banach space

Let $X$ be a Banach space. I am looking for an example of a function $f\in C(X)$ but $f\notin B(X)$, i.e., there exist a bounded set $S\subset X$ such that $f(S)$ is not bounded in $X$. Note that ...
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Adjoint of operators with dense range

Let $X$ and $Y$ be Hilbert spaces and $A:X\to Y$ be a linear bounded operator. Assume further that range of $A$ is dense in $Y$. Is the operator $A^*$ bounded?
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equations of unbounded operator

$T$ is densely defined closed operator on $\mathcal{H}$ which is a Hilbert space. Prove that $\forall a,b\in H$,the equations \begin{equation} \label{eq:1} \left\{ \begin{aligned} -Tx+y &...
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34 views

Continuous linear functional on $c_{00}$

Let $X$ be the space of real sequences having finitely many nonzero terms with $||\ ||_p , 1\leq p \leq \infty$ . Define $f: X\to\mathbb{R}$ by $f(x)=\sum_{j=1}^{\infty} x_j $ for $x=(x_j) \in X$. ...
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1answer
97 views

Unbounded inverse of a bounded operator

Suppose that $T\in B(H)$. As we know $T^{-1}$ is bounded if and only if $T$ is bijective. Also, $T^{-1}T=TT^{-1}=I$. In the other word, $T^{-1}:\mathcal{H}\to \mathcal{H}$ such that $T^{-1}T\xi=\xi$ ...
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1answer
57 views

A question about a theorem in *Quantum dynamical semigroups generated by noncommutative unbounded elliptic operators*

I'm studying the paper Quantum dynamical semigroups generated by noncommutative unbounded elliptic operators by Changsoo Bahn and Chul Ki Ko and Yong Moon Park(https://arxiv.org/abs/math-ph/0505026v1)....
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product of a self-adjoint operator and a bounded operator, reference request

Let $A$ be an unbounded self-adjoint operator on a Hilbert space $H$ and $B$ be a bounded self-adjoint operator from $B(H)$. Let $A=\int \lambda dE_\lambda$ be the spectral resolution. If $B$ ...
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1answer
38 views

Non bounded linear operator

If $ T : X \rightarrow \mathbb{R}$ is a non bounded linear operator then $\forall \alpha \in \mathbb{R^{+}} \ \exists\{x_n\}\subseteq X$ such that $x_n \rightarrow 0$ implies $Tx_n \rightarrow \alpha$ ...
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Stability of the dimension of spectral subspaces of a self-adjoint operator under a relatively bounded perturbation

Consider a self-adjoint operator $T$ on a Hilbert space $\mathcal{H}$, with domain $\mathcal{D}(T)$ and suppose that $A$ is a symmetric operator satisfying: $\mathcal{D}(T) \subset \mathcal{D}(A)$ and ...
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1answer
72 views

Boundedness of an operator on a Bochner spaces

Given two Banach spaces $(X,\Vert \cdot\Vert_X)$ and $(Y,\Vert \cdot\Vert_Y)$ such that $X\subset Y$ with continuous embedding (i.e. there exits a constant $c>0$ such that $\Vert x\Vert_Y \leq c\...
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203 views

Spectrum of two operators

Let $A$ be a self-adjoint operator on a Hilbert space $H=\big(L^2(\Bbb{R}^2),||.||\big)$, assumed to be unbounded. We have the following characterization of its spectrum $\sigma(A)$: $\lambda\in\...
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2answers
107 views

Unbounded differential operator in $C^1.$

Why is the operator $\dfrac{d}{dx}$ unbounded? Let us take $f\in C^1[a,b]$ and the operator norm $$\left\|\frac{d}{dx}\right\|=\sup\left(\left\|\frac{d}{dx}f(x)\right\|:\|f(x)\|\leq 1\right).$$ I ...
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1answer
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Question about Hilbert Space

An operator $A: D(A)\to H$ of an Hilbert space $H$ is closed if his graph is closed (in the product topology). An operator is closable if there exists an extension such that his graph is the closure ...
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2answers
134 views

Domain of sum of self-adjoint operators $A \otimes 1 + 1 \otimes B$?

Thinking about some quantum mechanics issues, I stumbled across the following functional analysis problem which confuses me a lot. Let $A$, $B$ be self-adjoint, unbounded and positive operators with ...
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Example of Range-Kernel relation on unbounded operators.

I am reading the book "Methods of Modern Mathematical Physics, Vol 1", and going through the chapter VIII (Unbounded operators) I just wondered if the relation $(\ker A)^{ \perp}=\overline{A^{*}(D(A^{*...
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108 views

Show that a compact operator is bounded

Definition: A linear operator $T: V \to W$ is compact if and only if the image of the unit ball in $V$ is precompact (= every sequence has a cauchy subsequence $\iff $ totally bounded). Prove: ...
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1answer
68 views

Cayley transform for unbounded operator M(f)(t)=t(f)(t)

Let $M : C_c(\mathbb{R})\to L^2(\mathbb{R})$ be the unbounded, densely defined and symmetric operator defined by $M(f)(t)=t\cdot f(t)$. I want to determine the Cayley transform $$U=(M-iI)(M+iI)^{-1}$$...
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27 views

$A$ is bounded if $\displaystyle\lim_{h\to 0}\|e^{ihA}-{\rm id}_\mathcal{H}\|=0$

I am looking for the solution to the problem: Let $A$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$. Show that $A$ is bounded if and only if $$\lim_{h\to 0}\|e^{ihA}-{\rm id}_\mathcal{...
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1answer
132 views

Bloch's theorem in one dimension, confusion about proof

I was looking at the derivation of Bloch's theorem in Griffith's QM: If $V(x+a)=V(x)$ for any $x$ and some $a$, and $\psi$ solves $$ H\psi =\lambda \psi $$ for $H=-\frac{\hbar^2}{2\pi}\frac{d^2}{dx^2}+...
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Understanding the Core of an Operator or Map.

I first saw a reference to the core of an unbounded linear operator in the context of Hilbert space and subsequently in the more general context of Banach space. The definition here https://math....
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25 views

Why can we assume that the range of the generator of a $C_0$ semigroup is already the full space?

I am reading this script: http://www.hairer.org/notes/SPDEs.pdf about stochastic partial differential equations. I have a problem with the footnote on page 46. $L$ is the generator of some $C_0$ ...
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$A$ is self-adjoint and $X$ is bounded commuting with $A$, why $AX$ is densely defined?

Let $A$ be a self-adjoint linear operator on a Hilbert space and $X$ is a positive bounded linear operator on this Hilbert space. Assume that $X$ strongly commute with $A$ (commute with all the ...
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44 views

Positive symmetric densely defined operator with dense range: essentially self-adjoint?

Suppose $A$ is densely defined, symmetric ($\langle Ax,y \rangle = \langle x,Ay \rangle$ for $x,y \in \textrm{dom}(A)$), and semi-strictly positive ($\langle Ax, x \rangle > 0$ for all $0 \neq x \...
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1answer
25 views

Approximations to unbounded operators

In what sense, if any, can we construct an unbounded operator as a limit of an unbounded sequence of bounded operators? A matrix representation of an algebra of raising and lowering operators, for ...
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1answer
25 views

von Neumann-Liouville expansion in Quantume Field Theory

Here are two questions related to the following form $$U(t,t_0)=1+(\frac{-i}{\hbar})\int_{t_0}^tdt'\,H^I_1(t')U(t',t_0) \tag{1}$$ This is known as the iterative form of solution to the differential ...
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82 views

For $B=\int \lambda d E_\lambda $ and $X$ commutes with every $E_\lambda $, why $BX$ is positive and self-adjoint?

Let $B$ be an unbounded closed operator on a Hilbert space $H$. If $B=\int \lambda d E_\lambda $ is positive self-adjoint and a positive bounded operator $X$ commutes with every $E_\lambda $, then why ...