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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38 views

Prove that the formula correctly defines an operator on a Banach space

Let $A$ be a bounded operator on a Banach space. Prove that the formula $e^A = \sum_{k=0}^\infty \frac{A^k}{k!}$ correctly defines operator $e^A$. Prove also that $\|e^A\| \le e^{\|A\|}$. I suppose ...
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63 views

Convergence Criterion in the Domain of an Unbounded Operator

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 convergent sequence $(x_n)_{n\...
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33 views

Boundary conditions for self-adjoint extensions of Laplacian

Suppose $T: C^\infty_c(J) \to L^2(J)$ is the kinetic operator $T\psi = -\frac{\hbar^2}{2m} \partial_x^2\psi$ on the open half-line $J=(0,\infty)$. This operator is symmetric and has infinitely many ...
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1answer
22 views

On properties of symmetric operators on Hilbert spaces

Let ${T: D(T) \subset H \to H}$ be a linear operator. $T$ is symmetric if: $(\phi, T\psi)= (T\phi, \psi)$ holds $\forall \phi, \psi \in D(T) $ and $D(T)$ is dense in $H$. The point 2. is necessary ...
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1answer
27 views

Linear functional and bounded norm

Let be given the function (x,y)-->f(x,y) differentiable at (a,b) from R^2 We define the function \begin{array}{l}T:R^2\rightarrow R\end{array} \begin{array}{l}T\left(h,l\right)=\frac{\partial f\...
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25 views

For an unbounded operator A, when is $D(A^{-1}) = D(A)'$? Is there a canonical isomorphism?

I am reading a text at the moment and I am not sure what the authors mean by $D(A^{-1})$. In this situation $A: D(A) \subset H \longrightarrow H$ is an unbounded operator on a hilbert space $H$ which ...
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1answer
34 views

Isometric operators and boundedness on Hilbert spaces

Let $ T: D(T) \rightarrow \scr H $ be a densely defined isometric operator, i.e. $$ \langle T \phi, T \psi \rangle = \langle \phi, \psi \rangle \quad \forall \ \phi,\psi \in D(T) $$ ...
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1answer
82 views

Show that the Cayley transform of a bounded self-adjoint operator on a Hilbert space has the number 1 in its resolvent.

This question is from section 10.6 in Kreyszig's Introductory Functional Analysis text. Let $H$ be a Hilbert space and let $D$ be a dense subspace of $H$. Let $T:D\rightarrow H$ be a self-adjoint ...
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1answer
177 views

A detail in the proof of Schur's lemma: the closures of the $\ker$ and $\operatorname{img}$ of the intertwiner.

Consider two irreducibles of a topological group $G$, acting in respective Hilbert spaces $\,\mathbb V\,$ and $\,{\mathbb{V}}^{\,\prime}\,$. Schur's lemma says: An intertwiner $\,M\,:\; {\mathbb{V}}\...
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51 views

Bound in the Hille-Yosida Theorem

I saw the following version of the Hille-Yosida theorem in a book: However, from the proof that I have seen, they only proof the bound $$\left|\left|\frac{\partial Y}{\partial s}(s)\right|\right|=\...
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1answer
30 views

Is the inverse of a restricted compact operator unbounded?

Suppose we have two separable Hilbert spaces $\mathbb{H}_{1},\mathbb{H}_{2}$ and the compact operator $\mathscr{T}:\mathbb{H}_{1}\to\mathbb{H}_{2}$. We know that since $\mathscr{T}$ is compact, its ...
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11 views

Bounded and bounded inverse of an operator. Multipliying summation of their spectral decompositions

Let $X$ be a random variable in an infinite-dimensional Hilbert space $(H,\langle·,·\rangle)$. Moreover, let $T$ be a self-adjoint, Hermitian positive definite operator in $H$ but,... with unbounded ...
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1answer
54 views

the boundedness of the derivative operator

I need to show that the linear operator $ \frac{d}{dx}: \mathcal{D} \rightarrow L^2(-\infty,\infty)$ is bounded with respect to the inner product $(f,g) = \int_{-\infty}^{\infty} {f(t) \overline{g(t)}+...
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38 views

Can $e^{X+Y}=e^XO$ be solved for $O$, when the operators $X$ and $Y$ don't commute?

What is the exact solution $O$ of $e^{X+Y}=e^XO$, when $X$ and $Y$ are non-commuting operators on a Hilbert space? All I found is the Baker–Campbell–Hausdorff formula, but it gives the solution $Z$ of ...
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1answer
27 views

Cauchy Equation and Continuity

Suppose that $f$ is a real-valued function defined on all of $R$ and satisfying the identity $$ f(x+y)=f(x) f(y)$$ for all $x, y$ in $\mathbb{R}$. Prove that, if $f$ is continuous at $x=0$, then $f$ ...
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Linear functional over a subspace is bounded or not

I have a problem trying to prove that a linear functional is bounded (in the sense of bounded linear operators in functional analysis). Proposition: Consider $\mathbb{K} := \mathbb{R}$ or $\mathbb{C}$ ...
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18 views

Bounds on norms of smallest-norm inverses for unbounded operators in banach spaces

Let $X, Y$ be two Banach spaces, and let $F: X \rightarrow Y$ be a (possibly unbounded) linear operator with dense image $R \subseteq Y$. Define the (possibly non-linear) function $F^+: R \rightarrow ...
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1answer
53 views

Most direct way to prove the domain of $A^2$ is dense.

Let $A\colon \operatorname D(A)\to \mathcal H$ be a (generally unbounded but densely defined) self-adjoint operator in a Hilbert space. $$\operatorname D(A^2):=\{\psi \in \operatorname D(A) \text{ s.t....
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1answer
34 views

On positive self adjoint unbounded operator

This is a theorem from Rudin Functional Analysis. $T$ is a self adjoint operator in $H$ (a Hilbert space). $T$ is not necessarily bounded and $\mathscr D(T)$ denotes the domain of $T$. We are to show ...
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1answer
23 views

$T$ unbounded operator not closed $\implies$ resolvent of T is empty?

I am studying the subject of unbounded operators and I'm wondering why if an operator is not closed than his resolvent is empty. Thanks !
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1answer
69 views

Symmetric and self-adjoint unbounded operator

I'm using the definitions here: Distinguishing between symmetric, Hermitian and self-adjoint operators. Let $\mathcal{D}(D) = \{ (x_n)_n \in \ell^2 : \sum_{n} n^2|x_n|^2 < \infty \}$. Let $D: \...
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1answer
38 views

On the adjoint of a densely defined unbounded operator

Let $H,K$ be Hilbert spaces, $D(T)\subset H$ be a dense vector subspace. Suppose $T: D(T) \rightarrow K$ and $D(T^*)=K$. What can you say about $T$? By definition, we have $D(T^*)= \{k \in K: \exists ...
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41 views

Green's functions in weak formulations: unbounded linear form

Consider a weak formulation of some PDE: $$\text{Find } u\in H^1_0(\Omega)\text{ s.t.:}\qquad\qquad\\ B(u,w) = L(w)\quad \forall\, w\in H^1_0(\Omega),$$ with $H^1_0(\Omega)$ the typical first order ...
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0answers
32 views

Unbounded operator spectrum theorem

$\textbf{Thm}$Let $A$ be an unbounded self-adjoint operator on a separable Hilbert space $H$ with domain $D(A)$. There there exists a measure sapce $(M,\mu)$ with $\mu$ a finite measure, a unitary ...
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36 views

Is the composition of closed unbounded operators closed?

Let $H$ be a complex Hilbert space with linear subspaces $U,V$. Then a (not necessarily bounded) linear operator $T:U\to V$ is said to be closed if $$\text{Graph}(T)\equiv\lbrace(u,Tu):u\in U\rbrace$$ ...
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1answer
48 views

condition for postive symmetric operator to be essentially self-adjoint

Suppose $A$ is a densely defined symmetric operator on Hilbert space which is positive. (a) prove $||(A+I)\phi||^2\geq ||\phi||^2+||A\phi||^2$ (b) Show $Ran(A+I)$ is closed if $A$ is a closed operator ...
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40 views

adjoint of unbounded differential operator

Consider $T=-\frac{d^2}{dx^2}$ as an operator on $L^2(\mathbb{R})$ with domain $C_0^\infty(\mathbb{R})$. What's the adjoint of T? This is an exercise from Reed&Simon's book. I have managed to ...
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37 views

Fractional powers of an operator associated to a sesquilinear form

Let $U,H$ be complex Hilbert spaces such that $U$ is densely embedded into $H$, $\mathfrak a$ be a bounded coercive symmetric sesquilinear form on $U$, $$\mathcal D(A):=\left\{v\in U\mid\exists a\in H:...
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1answer
31 views

Show that $\|B\varphi\| \le C(\|A \varphi\| + \|\varphi\|)$ extends from a core to entire domain of $A$

Let $A$, $B$ be two densely defined operators on a Hilbert space $H$. Suppose the domains obey $D(B) \supseteq D(A)$, and suppose $A$ is a closed operator with $\mathcal{D} \subseteq D(A)$ a core for $...
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20 views

Understanding the spectral measure of unbounded self-adjoint operator

Let $\mathcal{H}$ be a Hilbert space and $$\Delta: D(\Delta) \rightarrow \mathcal{H}$$ a (unbounded) self-adjoint operator, densely defined on $D(\Delta) \subset \mathcal{H}$. Denote by $\mu^{\Delta}$ ...
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1answer
42 views

Extending the Spectral Theorem of Unbounded Self-Adjoint Operators on Infinite-Dimensional Hilbert Spaces

I'm a physics student trying to do the maths of the Hilbert space in quantum mechanics with a bit more rigour than I'm accustomed to. I am trying to find ways to extend the spectral theorem for ...
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32 views

Unitary equivalent operators and spectrum

Suppose $T:\mathcal{D}(T)\to \mathcal{H}$ is a densly-defined and not-necessarily bounded operator on a Hilbert space $\mathcal{H}$. Furhtermore, let $U:\mathcal{H}\to\mathcal{H}$ be unitary. Then I ...
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1answer
37 views

Multplication operator self-adjoint

I have to show that the multiplication operator $M_{f}:D\to L^{2}(\mathbb{R})$ with $M_{f}g:=f\cdot g$ is self-adjoint if and only if $f:\mathbb{R}\to\mathbb{C}$ is real-valued, where $D$ denotes the ...
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83 views

Symmetry of Laplace operator

Good evening, I have a question regarding the proof of the symmetry property of Laplace operator. It's a well known fact that $-\Delta$ is an unbounded operator on $L^2(\mathbb{R}^n)$ and it is ...
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3answers
133 views

Showing derivative operator is self-adjoint

Consider the Hilbert space $L^2(\mathbb{R})$, and unbounded operator $Au:=iu’$ with domain $$D(A)= \{u \in L^2(\mathbb{R}) | u \text{ is absolutely continuous and } u’ \in L^2(\mathbb{R})\} $$ I’m ...
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0answers
80 views

Relationship between the spectrum of $AA^*$ and $A^*A$

I'm currently reading something on unbounded operators and their spectra and the author notes that the nonzero spectra of $AA^*$ and $A^*A$ are equal. $A$ here is unbounded and densely defined (on $L^...
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1answer
52 views

Prove $i\frac{d}{dx}$ is not essentially self-adjoint.

Consider $T=i\frac{d}{dx}$ on $C_c^{\infty}(0,\infty)\subset L^2(0,\infty)$, the infinitely differentiable functions with compact support away from the origin. Prove that $T$ is not essentially self-...
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29 views

When the adjoint of an unbounded operator on a Hilbert space coincides with the formal adjoint on its maximal domain?

Suppose we have a closable and densely defined operator $A$ with a domain $dom(A)$ which is a subspace of a Hilbert space $\mathcal{H}$. Let $\mathcal{H}$ have an orthonormal basis $\{e_n\}_{n=1}^\...
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2answers
236 views

Example of a basis which is not a Riesz basis?

I'm looking for an example of a countable "basis" $B=(\phi_i)_{i\in I}$ in a real Hilbert space $\mathcal{X}$ which is not a Riesz basis. So, we only require that the closure of the span of $...
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1answer
29 views

Essential self-adjointness of restriction of symmetric operator

Suppose $T$ is a (unbounded) symmetric operator on Hilbert space $\mathscr{H}$, defined on $D$. Let $D_1 \subseteq D$ is a linear subset dense in $D$, $T|_{D_1}$ be the restriction of $T$ on $D_1$. ...
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54 views

Self-adjoint extension of closed symmetric operator

Question: Let A be a closed symmetric operator that is semi-bounded from below, which means there is a $\alpha\in\mathbb R$, such that $$(Ax,x)\geq \alpha (x,x)\,,\quad\forall \,x\in D(A)\,.$$ Suppose ...
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1answer
26 views

What am I missing in this proof with uniform boundedness principle

I have the following exercise: Suppose $y=\left(y_{j}\right)_{j=1}^{\infty}$ is a sequence of complex numbers. Prove that if the series $\sum_{j=1}^{\infty} y_{j} \bar{x}_{j}$ is convergent for any ...
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1answer
37 views

Extension of unbounded symmetric operator ranges equal implies trivial extension

This is a problem I have encountered in my work and studies in operator theory and functional analysis. We take a Hilbert space $H$. We take a symmetric (possibly unbounded) operator $C$ which ...
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0answers
19 views

Unitarily equivalent unbounded operators

I know that for bounded operators $S,T$ on Hilbert spaces $H1,H2$, respectively, if they are unitarily equivalent, meaning there is a unitary $U$ such that $USU^*=T$, then $S$ and $T$ have the same ...
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0answers
47 views

Show that $u\in\text{dom}(\overline{\partial_t-\partial_x})$

Let $f\in L^2(\mathbb R)$ and suppose that $u(t,x)=f(x+t)$. Consider the equation $$\partial_t u=\partial_x u\iff\left(\partial_t-\partial_x\right)u=0.$$ I see that since we have assumed $f\in L^2(\...
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0answers
50 views

Unbounded spectra and operators

In a paper I have encountered the following non-autonomous infinite dimensional dynamical system: $$ \frac{d}{dt} \begin{pmatrix} f \\ g \end{pmatrix} = \underbrace{\begin{pmatrix} ...
2
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2answers
149 views

Computing adjoint of differentiation operator

This is a problem I have encountered in my studies and work in unbounded differential operators Let us define the operator $T = -\frac{d^2}{dx^2}$ as an operator on $L^2(\mathbb{R})$ with domain $C_0 ...
3
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1answer
51 views

When the following functional is bounded?

Let $p \geq 1$. I would like to know for what values $p$ the following functional $F$ is bounded in $L_p[0,1]$ $$ F(f) = \int_{0}^{1}f(x^2)dx $$ My attempt. I can show directly that for $p\in [1,2)$ ...
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1answer
25 views

Possibly unbounded operator which is closed and has a lower bound has closed range

This is an exercise I have encountered in my functional analysis class where we assume the operator may be unbounded and we assume nothing about the domain except that it is a vector subspace of $H$. ...
2
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2answers
69 views

How is multiplicative form of the spectral theorem interpreted as a spectral theorem?

Consider the following theorem taken from Reed & Simon's book. Theorem [Spectral Theorem - Multiplication Operator Form] Let $A$ be a (unbounded) self-adjoint operator on a separable Hilbert space ...

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