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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1answer
21 views

Is a dense subset in the domain of a closed, densely defined linear operator a core?

Let $X_0,X_1$ be Banach spaces. Let $A:D(A)\subseteq X_0\to X_1$ be a closable linear operator. Recall the definition of a core for such an operator: A set $\mathcal D\subseteq D(A)$ is called a core ...
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1answer
52 views

$A > B$ implies $1 > A^{-1} B$ for operators on Hilbert Spaces

As the title suggests, I have two operators on Hilbert spaces. They are both unbounded but I have bounded the inverse of $A$ (by showing it is strictly positive.) I seek to bound the composition $A^{-...
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57 views

Limit with unbounded operator and spectral measure

Let $T$ be a unbounded positive self adjoint linear operator ($T: \mathcal{D} (T) \to \mathcal{H}, \mathcal{D} (T) \subset \mathcal{H}$, where $\mathcal{H}$ is a Hilbert space) such that $\mathcal{N} (...
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1answer
47 views

Why the statement of one of the spectral theorems involves an integral rather than a sum?

My question is about this theorem regarding projection-valued measures. Suppose $H$ is a separable Hilbert space (in many cases we are just concerned about the separable case). Suppose $A$ is an ...
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1answer
31 views

Example of discontinuity of position operator $X\colon f(x) \mapsto xf(x)$ for $f(x) \in L^2(\mathbb{R})$

It is well known that continuous linear operators are bounded and vice versa. It is also well known that the position operator (which I shall call $X$) causes many headaches in quantum mechanics due ...
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1answer
79 views

On the resolvent set of an unbounded operator

Suppose $A$ is the infinitesimal generator of a $C_0$ semigroup $S(t)$ on an Hilbert space $X$. If $$\langle Ax, x\rangle \leq \omega \|x \|^2 \ \ \ \forall x \in \mathfrak{D}(A)$$ then $$\|S(t)\...
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1answer
54 views

The unbounded antipode for Woronowicz's quantum group $\operatorname{SU}_q(2)$

For non-zero $q\in [-1,1]$, Woronowicz's quantum group $\operatorname{SU}_q(2)$ is given as the universal unital $\mathrm{C}^*$-algebra generated by elements $a,c\in C(\operatorname{SU}_q(2))$ subject ...
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1answer
32 views

Couterexample to Littlewood-Paley theorem

Let $d\geq 2$ and let $P_j$ be the Fourier multiplier defined on $L^2(\mathbb{R}^d)$ by $\widehat{P_kf}(\xi)=\mathbf{1}_{2^k<|\xi|\leq2^{k+1}} \hat f(\xi)$, for any $k \ge 0$. It has been proven by ...
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89 views

Is $Tf(x)=\frac{1}{x}\int_{0}^{x}f(y)dy$ bounded as operator on $L^2((0,1);\mathbb{R} )$?

Given the operator $T:L^2((0,1);\mathbb{R} )\rightarrow L^2((0,1);\mathbb{R} )$ defined by $Tf(x)=\dfrac{1}{x}\displaystyle\int_{0}^{x}f(y)\,\mathrm dy$, say if it is well defined and discuss its ...
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34 views

Show that A is a bounded operator in ℓ2 and ∥A∥2≤∑∞j=1∑∞i=1|aij|2

I want to prove this: Let $(a_{ij})_{i,j=1}^{\infty}$ be an infinite matrix such that $\sum_{j=1}^{\infty }\sum_{i=1}^{\infty} \left | a_{ij} \right |<\infty $ and $A: \ell^2 \rightarrow \ell^2$ ...
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28 views

Reference request for unbounded linear operators

The following question is not as well formulated as I would like, but here goes. How should one think about unbounded linear operators on a Hilbert space $H$? Even though I have read a little on the ...
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45 views

Is this counterexample for $T$ closed, symmetric $\iff T$ self-adjoint valid?

Lemma Let $T: \text{dom}(T) \to \mathscr{H}$, where $\mathscr{H}$ is a Hilbert space, be densely defined. Then $T$ is closed and symmetric if and only if $T = T^{**} \subset T^*$. To better get a ...
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42 views

Are all bounded linear operators including the ones that are Banach, also isometries?

Are all bounded linear operators including Banach bounded linear operators, also isometries? An isometry is a homeomorphism that preserves distance, i.e. only reorders the points. May an unbounded ...
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16 views

Proving Boundness of Two Linear Operators

I have that $K:C[0,1] \rightarrow C[0,1] $ and $K_N:C[0,1] \rightarrow C[0,1]$ where: $$K \phi (x) = \int_0^1 k(x,t) \phi (t) dt $$ $$K_N \phi (x) = \int_0^1 k_N(x,t) \phi (t) dt $$ Where $k(x,t):= ...
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40 views

Unbounded operators with dense domain in Banach space.

Let $X$ and $Y$ two Banach spaces, and let $T: X \rightarrow Y$ be a closed operator. we known that if $T$ with dense domain in hilbert space $X$, then the domain of $T^{*}$ is also dense in hilbert ...
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33 views

Description of the adjoint of this closed operator

Consider the Hilbert space $L^2(\mathbb R)$ and the multiplication operator $A \colon D(A) \to L^2(\mathbb R)$ defined by $$(Af)(x) := (1+x^2)f(x)$$ with domain $$D(A) := \left \{f \in L^2(\mathbb R): ...
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27 views

Essentially self-adjoint operator A

A is essentially self-adjoint if and only if the equation $A^ * x = -x$ has no non-trivial solutions. Where in addition, A is a densely defined, symmetric and positive operator. I considered a ...
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28 views

Closure of a Hilbert subset with respect to a norm not defined on the entire Hilbert space

I have some problems to understand the definition of the closure of an operator. Let's $H$ be a Hilbert space with the norm $||\cdot||_H$, and let $A$ be a linear and symmetric operator defined on ...
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1answer
48 views

Domain of $T^\star T$ in Von Neumman's Theorem for closed operators

If $H$ is a separable Hilbert complex or real space and $T: D(T) \subseteq H \to H$ is a closed densely defined operator then $T^\star T$ is densely defined and self adjoint. Does $D(T^\star T) = D(...
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1answer
25 views

What are the main properties of eigenvalues of normal unbounded operators?

I am interested in the properties of the eigenvalues of unbounded normal operators. For compact linear operators we have that for every $t >0$, the set of distinct eigenvalues $\lambda$ such that $|...
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21 views

Which entrywise $p,q$-norms are comparable to the $\ell^2$ induced operator norm?

Consider an infinite-dimensional matrix $[A_{ij}\mid i,j\in\mathbb N]$, which corresponds to a linear operator $A:c_{00}\to\mathbb R^\mathbb N$ from finite sequences to arbitrary sequences: $$x\...
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Unbounded operator such that $P^2=P$

Does there exist an Unbounded operator $P$ on some Banach space $X$ such that $Dom(P)=X$ and $P^2=P$? If we don’t require $Dom(P)=X$, we can easily construct a Unbounded operator on $L^2[0,2π]$ by ...
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1answer
35 views

Resolvent convergence with only zero intersection of domains?

Let $ X $ be a Hilbert space. Can we find unbounded self-adjoint operators $ A, A_n $ on $ X $ such that $ A_n \stackrel{s.r.s}{\longrightarrow} A, $ that is, \begin{equation*} \ R_\lambda(A_n) \...
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27 views

$L^p$-Boundedness of Riesz Transform

The Problem From E.M. Stein's "Singular Integrals and Differentiability Properties of Functions". We wish to prove the bound: $||\frac{\partial^2 f}{\partial x_j \partial x_k}||_p \leq A_p ||\...
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2answers
257 views

Determine, by its action on an orthonormal basis, whether a linear operator can be continuous

We have a set of scalars $(A_{ij}\mid i,j\in\mathbb N)$, which are supposed to be the coefficients of a continuous linear operator $A$ on a real Hilbert space, with respect to an orthonormal basis $(...
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1answer
37 views

If a bilinear operator $E\times F\to G$ is bounded, is the corresponding linear operator $E\otimes F\to G$ bounded?

Consider a bilinear operator between Hilbert spaces $E\times F\to G$, denoted by $(x,y)\mapsto x\odot y$. There is a corresponding linear operator $\odot':E\otimes F\to G$, given by $\odot'(x\otimes y)...
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1answer
53 views

Showing that $Tx(n)=nx(n)$ in inner-product space $l^{2}$ satisfies $(Tx,y)=(y,Tx) \ \forall x,y \in A$ yet T is unbounded

Define A as: $ A\equiv \left \{ x\in l^{2}:\sum_{1}^{\infty}n^{2}{\left |x\left (n\right )\right |}^{2}<{\infty} \right \} $ Define a linear operator $T: A\rightarrow l^{2}$ (where $l^{2}$ is ...
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1answer
36 views

Why is $B_n$ in the domain of $H$ on a Hilbert Space?

Let $H:D(H)\subset \mathcal{H}\to\mathcal{H}$ be an unbounded self-adjoint operator. Let $B_n:= in(H+in)^{-1}$. For $\phi\in\mathcal{H}$ it is given in my notes that $B_n\phi\in D(H)$, where $D(H)$ ...
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28 views

Unbounded operator can not be extended

Densely Defined Operator In this wiki link, it gives the definition of densely defined operator. After the definition, it gives an example of a densely defined unbounded operator, differentiation ...
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1answer
30 views

Compactness and being finite-rank operators

Suppose $T$ is a selfadjoint unbounded operator with compact resolvent on a given separable Hilbert space $H$, and $P$ a projection onto $\ker T$. Define $T_1$ to be $T+P$. Is it true that $|T_1|^{-1}$...
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56 views

Two commuting self adjoint operators

Let $H$ be a Hilbert space and $A \in \mathcal{L}(H)$ be a bounded self adjoint operator. Let $B : \mathcal{D}(B) \subset H \to H$ be (possibly unbounded) self-adjoint operator such that \begin{...
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34 views

Implications of relative operator bounds.

Setting: Let $(\mathcal{H}, \langle\cdot,\cdot\rangle)$ be a complex Hilbert space and consider two (possibly unbounded) self-adjoint operators $A = A^\ast\colon \mathcal{D}_A \mapsto \mathcal{H}$ and ...
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34 views

Isomorphism between a quotient space and a subspace

When $L$ is an unbounded operator on a Hibert space $H$, and $K = Ker L$, $H/ K \simeq K^\perp$? I'd appreciate it if you'd give me any help! Thank you. **I've added the source below.
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79 views

Invariant subspace for unbounded operator

For a linear bounded operator $T$ on a Hilbert space $H$ that admits an invariant subspace $V$ (meaning: $TV\subset V$), the fact that $T\overline{V}\subset\overline{V}$, where $\overline{V}$ denotes ...
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How is the quotient space H/K is isomorphic to the orthogonal complement of K?

I am trying to construct an isomorphic map between $H/ K$ and $K^\perp$. Specifically, $L$ is an unbounded operator on a Hibert space $H$, with kernel $K$. How could I show $H/ K \simeq K^\perp$? ...
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1answer
42 views

Is it true that $U^{-1}\overline{A}U=\overline{U^{-1}AU}?$

Let $X_0$ and $X_1$ be Banach spaces and suppose that $A:D(A)\subseteq X_0\times X_1$ is a closable linear operator. If $U: X_0\to X_1$ is a unitary operator, is it always true that $$U^{-1}\overline{...
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0answers
59 views

Contradiction with Uniform Boundedness Principle using Hamel basis

Let $\{e_{\alpha}\}_{\alpha \in \mathcal{A}}$ be a Hamel basis of an infinite dimensional Banach space $(X, \| \cdot \|_X)$. Without loss of generality, we assume $\|e_{\alpha}\|_X = 1$. It is known ...
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1answer
92 views

Definition of Inverse of Unbounded Operator

I am currently studying unbounded operators. However, in the text I am using, the definition for the inverse of an unbounded operator is given. I also went to several other books I own and the ...
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1answer
39 views

Product of $0$ operator with another (unbounded) operator is nonzero?

I am a physicist who only knows a bit of functional analysis. In a Quantum Field Theory class, we seem to have an operator $A$ with the property $$A = 0$$ $$\langle\psi,AB\phi\rangle\neq0$$ where $...
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30 views

Can we cycle operators when we calculate operator norms?

Let $A,B$ be unbounded self-adjoint operators on a Hilbert space, that don't commute. Let $f,g : \mathbb{R} \to \mathbb{R}$ be functions such that $f(A)$ and $g(A)$ are well-defined by the spectral ...
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92 views

Proof of multiplication operator form of spectral theorem for unbounded normal operators

I'm looking for a proof that Let $H$ be a separable Hilbert space and $N$ an unbounded normal operator on $H$. Then there is a finite measure space $(X,\scr{M},\mu)$, a unitary operator $U:H\to L^2(...
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23 views

Proof that any commutative family of unbounded self-adjoint operators has a “generating” operator

In Quantum Mechanics for Mathematicians by Leon A. Takhtajan, the author remarks on page 73: According to von Neumann's theorem on a generating operator, for every commutative family $\mathbf{A}$ ...
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40 views

Domain of unbounded selfadjoint operator

Let $A:D(A)\subset H \to H$ be a selfadjoint unbounded operator on a complex Hilbertspace $H$. Let $\lambda_0\in I$ for some open interval $I$. Define $g(\lambda):=(\lambda_0-\lambda)^{-1}(1-\chi_{I}(...
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1answer
44 views

Functional calculus for diagonalisable unbounded operators

Let $H$ be a separable Hilbert space, and $T:{\frak Dom}(T) \to H$ a densely defined self-adjoint operator. As is well-known, for $\mathrm{spec}(T)$ the spectrum of $T$, and $f:\mathrm{spec}(T) \to \...
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71 views

Integral with respect to spectral measure

Let $A:D(A)\subset H \to H$ be a self-adjoint unbounded operator on complex Hilbertspace $H$ with corresponding spectral measure $E:\mathcal{B}(\mathbb{R})\to\mathcal{L}(H)$. I want to show that an ...
5
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1answer
110 views

Bounded superposition operator in the space of bounded variation functions

For $n\in\mathbb N$ put $A_n:=\{k/2^n\ |\ k\in\{1,\ldots,2^n-1\}\}$. Consider $g:[0,1]\times\mathbb R\to\mathbb R$, defined by \begin{align*} g(t,u)= \begin{cases} 2^{-n}\chi_{A_n}(t) & ...
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2answers
60 views

Is it a mistake if I delete the condition in Banach - Steinhaus Theorem?

I'm studying the Banach-Steinhaus theorem recently. There is a question to me, I'm sure I have made a mistake, but I can't find out where the mistake is. The Banach - Steinhaus theorem says: Let X be ...
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1answer
183 views

If $T$ is self-adjoint then $T^2$ is also self-adjoint,

Let $T:D(T)\subset H \to H$ be selfadjoint unbounded linear operator on a complex Hilbertspace $H$. Show that $T^2$ is self-adjoint. Since $T$ is selfadjoint it's spectrum must be real, so $T\pm i Id$...
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0answers
38 views

Spectral measure of an eigenvector

Let $T$ be an unbounded selfadjoint operator and let $P_T$ denote it's spectral measure such that $T= \int_\mathbb{R}\lambda dP_T (\lambda)$. Suppose $\psi$ is an eigenvector of $T$ such that $T\psi=\...
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3answers
71 views

Show $\{f \in L^2[0,1] : f \text{ is absolutely continuous},\ f\ '\in L^2[0,1],\ f(0)=f(1)=0\}$ is dense in $L^2[0,1]$.

Define $D(T)=\{f \in L^2[0,1] : f \text{ is absolutely continuous},\ f\ '\in L^2[0,1],\ f(0)=f(1)=0\}$. I want to show that $D(T)$ is dense in $L^2[0,1]$. My argument is as follows: $\{0,1\} $ ...

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