All Questions
Tagged with operator-theory unbounded-operators
158
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Commutators of unbounded operators on Hilbert spaces
Commutation seems to be a tricky business when it comes to unbounded operators, because of the domain questions. I have some trouble understanding the usual material about commutators of unbounded ...
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40
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Can an unbounded operator and its adjoint both have full domains?
Let $H$ be a complex Hilbert space. This post shows that there exist unbounded (which I will use to mean “not bounded”) operators on $H$ whose domain is all of $H$, i.e., $\mathcal D(T) = H$ (although ...
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On self-adjoint extensions and multiplicity of eigenvalues
I hope you can help me with the following question.
Let $B$ be a densely defined closed symmetric operator on a infinite-dimensional separable complex Hilbert space $\mathcal{H}$ with deficiency ...
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$T$, $S$ are closable implies that $\bar{T}\circ \bar{S}$ is closable
Let $T$ and $S$ be two unbounded closable densely defined operators on $H$. Assume additionally that $T\circ S$ is densely defined (and, if needed, closable and also that $\text{im}S \subset T$). Is ...
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Computing the adjoint of $-\Delta$
In B. Helffer's Spectral Theory and its Application, Remark 2.7 p. 16 the author is considering the two following operators
$T_0=-\Delta $ with $D(T_0)=C^{\infty}_{c}(\mathbb{R}^N)$
$T_1=-\Delta $ ...
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Application of Stone's theorem to regular representation
Consider the left regular representation
$$\lambda: \mathbb{R}\to U(L^2(\mathbb{R})), \quad (\lambda_x f)(y) = f(y-x), \quad x,y \in \mathbb{R}.$$
By Stone's theorem, there is a positive, self-adjoint,...
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Multiplication of two unbounded operators and functional calculus
Let $A$ be a positive, self-adjoint unbounded operator defined on a Hilbert space $H$.
Let $f,g: [0, \infty]\to \mathbb{R}$ be Borel measurable functions that are bounded on compact subsets. We can ...
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When is $\rho := e^{-\beta H}$ trace-class?
Suppose $H$ is a self-adjoint operator acting on a separable Hilbert space and $H$ has a discrete spectrum with eigenvalues converging to $+\infty$. I want to investigate under what condition the ...
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Proof Position Operator Is Dense
This is an exercise from my last homework sheet, proofing that $P$ is unbounded and self-adjoint was clear, however I'm having trouble proofing that $P$ is densely defined.
How my Instructor solved ...
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Bounded extension of an operator with limited numerical range
In an excercise I'm asked to prove that a densely defined operator, whose numerical range:
$$\nu(T)=\{ (\psi,T\psi) \space | \space \psi \in D(T) \wedge ||\psi||=1 \}$$
is a limited subset of $\...
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Bijective Linear Map Construction for continuous and bounded maps: Linear Operator $A$ and $A^{-1}$
Can any one construct a bijective linear map, with $C(E)\simeq C_0(E)$?
That is, consider a linear operator $A$, such that $A\colon C(E)\rightarrow C_0(E)$, which can make sense for both $A$ and its ...
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Showing that a graph which is a subspace of $\bigoplus \mathcal{H}_i \oplus \bigoplus \mathcal{H}_i $ is closed.
Let $\mathcal{H}_i$ be a collection of Hilbert spaces, and $T_i \in \mathbf{B}(\mathcal{H}_i)$. Assume that $\sup ||T_i|| = \infty$. Consider $\bigoplus \mathcal{H}_i$(the completetion). Given an ...
3
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182
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Reed and Simon, Fourier Analysis and Self-Adjointness, second corollary to Theorem X.$25$: how to show that $D(A^2)$ is dense in $D(A)$ for its norm?
This question arose while trying to figure out the proof of the second corollary to Theorem X.$25$ in Reed and Simon's Fourier Analysis, Self-Adjointness, stated as follows:
Theorem X.$25$: Let $A : ...
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1
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129
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The Spectrum of the derivative operator in a specific Banach space
Consider the Banach space $X=\left\{u\in C^1([0,1]):\, u(0)=0\right\}$ and the subspace $D=\{u\in C^2([0,1]):\, u(0)=u(1)=u'(0)=0\}$, and the operator $A:D\longrightarrow X$ defined by $Au=u'$. I have ...
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Exponential of operators and commutation
Consider an unbounded self-adjoint and strictly positive operator $A\:\mathcal{D}(A)\to\mathcal{H}$. With strictly positive, I mean $\sigma(A)\subset [a,\infty)$ for some $a>0$. Now, with the ...