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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Is there a name for this class of operators, $T(x) = \sum_{j\geq 1} \lambda_j \langle x, u_j \rangle u_j$?

Let $\lambda_j$ be an increasing sequence of positive numbers such that $\lambda_j \to \infty$ as $j \to \infty$. Let us consider the map which takes $x \in \ell^2(\mathbb{N})$ and sends it to $$ Tx = ...
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Finding domain the unbounded operator $\exp(-H)$ where $H=id/dx$ on $L^2(\mathbb{R})$

let, $H=id/dx$ be the operator define on its maximal domain $D(H)=\{f\in L^2(\mathbb{R}):f'\in L^2(\mathbb{R})\}\subseteq L^2(\mathbb{R})$. Now, define $\exp(-H)=\sum_{n=0}^\infty(-1)^n\frac{H^n}{n!}$....
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Maximal domain of unbounded linear differential operator

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{...
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What does strong resolvent convergence tell about spectrum of the limit?

I have a sequence of operators $A_n$ on a separable Hilbert space (not necessarily with the same domain). These operators are unbounded, self-adjoint, and converge in the strong resolvent sense to ...
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3 votes
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On the domain of unbounded self-adjoint linear operators on Hilbert spaces

Suppose I have a linear unbounded operator $A:\operatorname{dom}(A) \to \mathscr{H}$, with $\operatorname{dom}(A)$ carefully chosen to make $A$ self-adjoint ($A^\dagger=A$). Could $A^n$ be ...
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Show that the set is not bounded - formal proof - $D = \{(x,y)\in R^2|y=\frac 1 x\}$

is there any way to prove with epsilon \ other way that the set is unbounded? $$D=\{(x,y)\in R^2\mid xy=1\} = \{(x,y)\in R^2\mid y=\frac 1 x\}$$ Here is a proof my exerciser at class did ( which is ...
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Is $e^{ax}$ a bounded operator in $\mathcal{L}^2_{[0,1]}$?

A similar question was asked a couple years ago: Is multiplication with $e^{ax}$ a bounded operator in $L^2(\mathbb{R})$ But I'd like to prove it in the closed interval $[0,1]$. Suppose the operator $\...
2 votes
2 answers
117 views

Uniqueness spectral decomposition unbounded self-adjoint operator

Let $A$ be a self-adjoint (unbounded) operator in a Hilbert space $H$. In Rudin's book "Functional analysis", theorem 13.30, the following is proven: There exists a unique spectral ...
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Why unbounded operators require a restricted domain?

Usually, unbounded operators on a Hilbert space $\mathcal{H}$ are defined as linear operators from a subset of $\mathcal{H}$ to $\mathcal{H}$. My question is why is it necessary to restrict the domain....
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Precise assumption in spectral theorem of unbounded operators

The most general version of the spectral theorem I am aware of is the spectral theorem for unbounded normal operators (firstly proven by von Neumann in 1932, I think). An operator $T:\mathcal{D}(T)\to\...
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Factorization of an unbounded operator as composition of a bounded operator composed with one whose inverse is bounded

Does every densely-defined unbounded operator $P$ have a ``factorization'' of the form $$P = A B^{-1},$$ with $A$ and $B$ bounded? Given two bounded operators $A,B$ such that $B^{-1}$ is a densely-...
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1 answer
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Construct an Injective and onto unbounded operator.

i was study functional analysis and i found a interesting problem. Let $X$ an infinite dimensional normed space. Construct an operator $T: X \rightarrow X$ such that $T$ is injective and onto. Also, ...
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Is this linear map continuous?

Let $A:L^{2}\left( (0,1)\right) \rightarrow L^{2}\left( (0,1)\right) $ be defined as $Af\left( x\right) =-\ln \left( x\right) f\left( \frac{x}{2}% \right) $ for every $f\in L^{2}\left( (0,1)\right) $ ...
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Adjoint of a Densely Defined Unbounded Operator is Unique

Let $\mathcal H$ be a Hilbert space and $A: \mathcal D_A \to \mathcal H$ be an unbounded linear operator. Suppose also that $\mathcal D_A$ is dense in $\mathcal H$. We define the adjoint of $A$, $A^*$ ...
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Calculating the commutator with functions of a certain self-adjoint unbounded operator

Let $A$ be an unbounded operator in an infinite-dimensional Hilbert space $\mathcal{H}$ such that the commutator with its adjoint $A^*$ is given by $[A, A^*] = A A^* - A^* A = \mathbb{I}$, where $\...
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Why these operators are not bounded

In the space $L^2(\mathbb{R})$ I'm studying the wave equation $$\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}.$$ By taking the second derivative operator $Au=u''$, we can define ...
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1 answer
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Prove the graph of this operator is closed

Consider the subspace $\mathcal L_0$ of $L^2=L^2([0,1])$ consisting of functions of integral zero. Consider the map $T:L^2\to L^2$ given by $T(f)(x)=\int_0^x{\rm d}t \, f(t)$. Prove that it is one-to-...
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Is $T$ self-adjoint if it is closed, symmetric and $\ker T=\ker T^*$?

Let $T$ be a densely defined, closed, unbounded, and symmetric linear operator (i.e., $T\subset T^*$) defined in a Hilbert space, with domain $D(T)$. Is it true that if we further suppose $\ker T=\ker ...
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The domain of the cube operator.

Let $E$ be a Banach space, and $A$ an unbounded operator with domain $D(A)$, we assume that $D(A)$ is dense in $E$ and $A$ is closed (i.e its graph is closed) and $\rho(A)$ (the resolvent set of $A$) ...
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Venn diagram for basic types of operators

This Venn diagram is an attempt to visually classify densely-defined linear operators between Banach spaces (self-adjoint operators are an exception, defined between Hilbert spaces). Operators are ...
1 vote
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Prove this Fourier series converges to a continuous function

Problem Consider the space $\mathcal D$ of continuously differentiable functions on the unit circle and the operator on $L^2$ (of the unit circle) with domain $\mathcal D$ given for $f\in\mathcal D$ ...
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Unbounded closed operator

Let $E,F$ be two Banach spaces and $A$ an unbouded operator from $D(A) \subset E$ to $F$. We want to prove the following lemma : Let $A$ be a closed operator. The following are equivalent : $A$ is ...
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What are simple conditions for the adjoint of a positive, unbounded, densely defined operator on a Hilbert space to be positive?

I'm reasking this deleted question because I believe I've made a some progress towards an answer, which I'm also interested in knowing. Here's the restatement: Suppose $\ A\ $ is a densely defined, [...
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Bounded Operator with unbounded Inverse

For an exercise I need to investigate the boundedness for an operator and its inverse. Concretely, the exercise states : Consider the linear mapping $A : l^2 \rightarrow l^2$ defined by $A : \{x_1, ...
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Natural Powers of Unbounded Operators densely defined

Let $A$ be a closed densely defined operator on some Hilbert space $H$ with non-empty resolvent set $$ \rho(A)=\{ \lambda \in \mathbb{C} ; (\lambda - A)^{-1} \in L(H) \}. $$ Then the operators $A^n$ ...
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Why is it that a closed unbounded operator can never have a closed domain?

I have seen a similar question here that tries to answer this. My question is why in practice closed unbounded operators cannot have a closed domain. Is it because a closed domain would imply the ...
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What is the norm of $T$ : $L^1$ (($-1$,$1$) , $\mathbb{R}$) $\rightarrow$ $\mathbb{R}$

I have the following question What is the norm of $T$ : $L^1(-1,1)\to \mathbb{R}$, where $T(f)= \int_{-1}^{1} t f(t) dt$? We know that $$\|T\|= \sup_{f \ne 0} \dfrac{|T(f)|}{\|f\|_{1}}$$ I was able ...
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Example 4.8 ; functional analysis book by S.kesavan

Let $V=L^2(\mathbb{R})$ and let ${s(t)}$ be the semigroup defined by $(s(t)f)(s)=f(s +t)$. Find its infinitesimal generator $A$.
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Does $(u \delta u^{*})^{it} = u \delta^{it} u^{*}$ hold, where $\delta$ is a (positive) self-adjoint unbounded operator and $u$ unitary?

Let $\delta$ be a positive self-adjoint unbounded operator and let $u$ be a unitary operator on an infinite-dimensional Hilbert space. Is it true that $(u \delta u)^{it} = u \delta^{it} u^{*}$ for ...
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1 vote
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Prove the operator defined by $Lf=x^2f(x)$ is not bounded.

Let the operator $L: dom(L) \subset L^2(\mathbb{R} ) \to L^2 (\mathbb{R})$ defined by $Lf(x)=x^2f(x)$ where $dom(L)=\{ f \in L^2(\mathbb{R} ) : x^2 f(x) \in L^2 (\mathbb{R}) \}$. Prove that $L$ is ...
1 vote
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Sum of adjoints of two operators is a subset of adjoint of sums.

Let $T : \mathcal{D} (T) \subset H \to H$ and $A: \mathcal{D} (A) \subset H \to H$ two densely defined operators and $H$ a complex Hilbert space. Prove $T^*+A^* \subseteq (T+A)^*$. My solution is as ...
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1 answer
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Multiplication operator is densely defined.

I saw the question Multiplication Operator on $L^2$ is densely defined Disintegration by parts answers using the following argument. If $f \perp \mathcal{D} (L)$ then $\frac{1}{m^2+1} f \in \mathcal{D}...
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2 answers
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What does it mean that a function is unbounded below in every neighborhood?

In this paper Strong Convexity Does Not Imply Radial Unboundedness In [3], Tapia gives this result showing that a strongly convex functional is either radially unbounded (and so minima-existence ...
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Bounded inverse for a closed range closed operator

I am dealing with an unbounded operator $T$ on an Hilbert space $H$. I am interested in proving that it has a bounded inverse $T^{-1}$. I managed to prove that the operator is closed and self-adjoint. ...
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Does this example of symmetric unbounded operators need $\mu$ to be $\sigma$-finite?

From Chapter X. of John B. Conway's textbook A Course in Functional Analysis: 1.10 Example Let $(X,\Omega,\mu)$ be a $\sigma$-finite measure space and let $\phi:X\to\mathbb C$ be a $\Omega$-...
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When is the composition of (unbounded) closed operators closed/closeable?

Let $H_{i}$ be Hilbert spaces for $i=1,2,3$. Let $T_{21}:H_{1} \rightarrow H_{2}$ and $T_{32}: H_{2} \rightarrow H_{3}$ be closed, densely defined, unbounded operators. What are appropriate conditions ...
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Is $TT^*$ (or $T^*T$) densely defined if $T$ is densely defined and symmetric?

Is $TT^*$ (or $T^*T$) densely defined if $T$ is a densely defined and symmetric linear operator? I feel this is untrue, but do you have a counterexample? Thanks
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Compact linear operator definition

It is well known what we mean by a compact linear operator $A:X\to Y$ where $X,Y$ are Banach spaces (see https://en.wikipedia.org/wiki/Compact_operator#Compact_operator_on_Hilbert_spaces). I wonder ...
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Prove the discrete spectrum of $A$ equals to the set of those complex $\lambda$ such that $\lambda I-A$ is Fredholm.

There is considerable divergence in the literature concerning the definition of the essential spectrum of a densely defined closed operator $A$ on a Banach space $X$. I want to find a direct ...
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Why is the maximal domain $D_\text{max}$ of a Sturm-Liouville operator defined the way it is?

Given a Sturm-Liouville type operator which acts on functions on the interval $(a,b)$ $$T:= \frac{1}{w(x)}\left(-\frac{d}{dx}\left[p(x) \frac{d}{dx}\right]+q(x)\right)$$ where $w$, $p^{-1}$ and $q \in ...
1 vote
1 answer
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Action of an operator-valued function of the momentum operator $\hat{p}$ and its unboundedness

I am currently dealing with an operator-valued function $f(\hat{T})$ of the following kind: $$f(\hat{T}) =\sqrt{1 + b\hat{T}^2} $$ where $b$ $\in \mathbb{R}$ and $\hat{T}$ is a linear operator acting ...
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Must Linear operators on entire Banach space $l^1$ be bounded?

Assume that it is possible to construct an unbounded linear functional $f$ defined on entire Banach space $l^1$. Then let us consider in unit ball the standard linearly independent sequence $\{e_i\}_{...
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Bounded kernel operator

I have to verify that this operator in $L^2(\Bbb R^3): (Gf)(x)=\int dy f(y) \frac{e^{-|x-y|}}{|x-y|} $ is bounded, using the following rule for the integral kernel: $ sup_x\int dy|a(x,y)|< \infty$ ...
2 votes
1 answer
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If $E,F$ are reflexive Banach spaces, then the graphs of unbounded linear operators $A$ and $A''$ are isometrically isomorphic

I'm reading Theorem 3.24 in Brezis's book of Functional Analysis. The statement of the theorem is: Let $E$ and $F$ be two reflexive Banach spaces. Let $ A:D(A) \subseteq E \rightarrow F$ be an ...
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How to prove that the graphs of unbounded linear operators $A$ and $A''$ are isomorphic?

I'm reading Theorem 3.24 in Brezis's book of Functional Analysis. Let $E$ and $F$ be two reflexive Banach spaces. Let $A: D(A) \subseteq E \rightarrow$ $F$ be an unbounded linear operator that is ...
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Proof of involvement

Let $A$ and $B$ be two bounded operators on a Hilbert space $H$. $1)$ $\exists M>0,$ $\forall x\in H$ : $||A^{\ast}x|| \leq M||B^{\ast}x||,$ $2)$ $\exists C$ bounded operator on $H$ : $A=BC.$ We ...
3 votes
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How to show that the Hilbert-adjoint operator $T^*$ of a (not necessarily bounded) linear operator $T$ on a complex Hilbert space $H$ is bounded

I want to show that if a (not necessarily bounded) linear operator $T$ is defined everywhere on a complex Hilbert space $H$, then its Hilbert-adjoint operator $T^*$ is bounded. Now given that $T$ may ...
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1 answer
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About bounded and closable operators.

Let $T:H\to \mathbb{C}^m$ a densely defined operator with $H$ Hilbert space. Is it true that if $T$ is closable, then $T$ is bounded? For example the differential operator $ T:C^1[0,1]\subset C[0,1]\...
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1 vote
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How to prove T is a closable operator

Let $T:H\to H$ a densely defined operator, with $H$ a Hilbert space such that: $$Re(x,Tx)\geq 0, \forall x\in Dom(T) $$ I want to prove that $T$ is a closable operator, that means... that there exists ...
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If $A^\star$ is surjective, then there is $c>0$ such that $|u| \leq c |A u|$ for all $u \in D(A)$

I'm trying to prove Theorem 2.21 in Brezis' book of Functional Analysis. The author leaves the proof as an exercise. Could you have a check on my attempt? Let $E, F$ be Banach spaces. Let $A: D(A) \...
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