All Questions
13 questions
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Operator-form correspondence without lower semiboundedness
When dealing with a normal unbounded operator $A$, it is often useful to frame questions about the operator in terms of questions about the associated form $\omega,$ which has domain $D(|A|^{1/2})$ ...
- 109
3
votes
1
answer
332
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Takesaki II Lemma 1.13: stuck in proof
Consider the following fragment from Takesaki's book "Theory of operator algebras II" (Lemma 1.13 on p8, in chapter VI "Left Hilbert algebras"):
Here, we associate with an ...
- 175
-1
votes
1
answer
164
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Closure of the point spectrum of an unbounded diagonalizable operator
Given a (separable) Hilbert space H and an unbounded densely defined linear operator $T:{\cal D}(T) \to $H such that ${\cal D}$ is diagonalizable (it means $\exists$ an O.N.B. of H such that all basis ...
- 969
0
votes
1
answer
138
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Antilinear unbounded operator has closed graph
Let $H$ and $K$ be Hilbert spaces and $D(T)$ a vector subspace of $H$. Let $T: D(T) \to K$ be a densely defined antilinear operator. Its adjoint $T^*: D(T^*)\to K$ is defined by the relation
$$\langle ...
- 175
2
votes
1
answer
335
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When the adjoint of an unbounded operator on a Hilbert space coincides with the formal adjoint on its natural domain?
This is almost a copy of https://math.stackexchange.com/questions/3931318/when-the-adjoint-of-an-unbounded-operator-on-a-hilbert-space-coincides-with-the
I am trying to work with infinite matrices in ...
- 21
2
votes
1
answer
916
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Characterising closed range self-adjoint operators
Let $T:\mathrm{dom}(T) \subseteq H \to H$ be a densely defined, self-adjoint operator on a Hilbert spaces $𝐻$. In general the range of $T$ is not guaranteed to be closed. What tools are available to ...
- 969
0
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1
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113
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An adjoint characterization of (unbounded) Fredholm operators
Let $\mathcal{H}_i$, for $i=1,2$, be Hilbert spaces, and $T:{\frak Dom}(T) \subseteq \mathcal{H}_1 \to \mathcal{H}_2$ a densely-defined closed operator. If the kernel of $T$, and the kernel of its ...
- 993
1
vote
0
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277
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Adjoint for a non-densely defined unbounded operator on a Hilbert space
Let $\mathbf{H}$ be a Hilbert space, and $D$ an unbounded densely-defined operator on $\mathbf{H}$. As is well-known, every such operator admits an adjoint, with domain possibly different from that ...
- 993
0
votes
2
answers
465
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Spectrum equals eigenvalues for unbounded operator
Let $D$ be an unbounded densely defined operator on a separable Hilbert space $H$. If $D$ is diagonalisable with all eigenvalues having finite multiplicity and growing towards infinity, does it follow ...
- 219
3
votes
1
answer
214
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Non-point spectrum for diagonalisable self-adjoint unbounded operator
Given a (separable) Hilbert space H and an unbounded densely defined linear operator $T:{\cal D}(T) \to $H such that ${\cal D}$ is diagonalizable (it means $\exists$ an O.N.B. of H such that all basis ...
- 969
6
votes
1
answer
1k
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Unbounded version of continuous functional calculus
For a normal operator $T$ on a Hilbert space ${\cal H}$, it is well known that for any continuous complex valued function $f$ on the spectrum of $T$, we have a well-defined operator $f(T) \in B({\cal ...
- 969
3
votes
1
answer
229
views
Symmetric diagonalizable operators and self-adjointness
Given a densely defined symmetric operator $L$ on a Hilbert space $H$, which is also assumed to be diagonalizable, will there always exist a unique extension of $L$ to a self-adjoint operator?
2
votes
1
answer
926
views
Eigenvalues and Compact Resolvent
For $A$ an unbounded (densely defined) operator on a separable Hilbert space, what conditions on its eigenvalues will show that, for $\lambda \notin $spec$(A)$, we have that $(A-\lambda)^{-1}$ is a ...
- 465