All Questions
Tagged with sp.spectral-theory functional-calculus
7 questions
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A representation of positive matrix
Let $\mathcal H$ be a Hilbert space. Let $-\frac{1}{2}<r<0.$ Denote $c_p:=\int_{0}^\infty\frac{t^r}{1+t}dt.$ Suppose $A$ be a positive invertible operator in $B(\mathcal H).$ Is it true that $A^...
- 1,879
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Relations between two Schwartz kernels in dimensions $n$ and $n+1$
Let $(M,g)$ be an $n$-dimensional pseudo-Riemannian manifold and $\Box_g$ be the Laplace-Beltrami operator on $M$. Consider $z \in \mathbb{C}$ such that $\mathrm{ Im}(z)>0$, and we define $P_0 := \...
- 319
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Generalization of Borel functional calculus
[Repost from https://math.stackexchange.com/questions/4802593/generalization-of-borel-functional-calculus]
Let $A$ be a normal operator on a Hilbert space $V$. The continuous functional calculus gives ...
- 95
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A gap in the proof of uniqueness of functional calculus based on a spectral theorem
This question considers the proof of a fundamental theorem of functional calculus, given in the book Spectral Theory -
Basic Concepts and Applications by David Borthwick (Theorem 5.9).
Firstly we have ...
- 1,755
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"Trade-off" between bound on the function and on the spectrum for functional calculus in spectral theory
Let $A$ be a self-adjoint (unbounded) operator on a separable Hilbert space $H$.
From the following form of spectral theorem, we may define a functional calculus by $f(A)=Q^{-1} M_{f\circ \alpha} Q$. (...
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2
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306
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Interesting examples of spectral decompositions of BOUNDED operators with both continuous and discrete spectrum
I would like to have a few basic examples of bounded self-adjoint operators $T$ (more generally bounded normal would be fine) on a Hilbert space $(H,\langle,\rangle)$ for which the following criteria ...
- 7,609
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Integrating the resolvent of a self-adjoint operator across a continuous part of the spectrum
Let $A$ be a closed self-adjoint operator on a Hilbert space $H$, possibly unbounded and hence defined on a dense domain $D(A) \subset H$. It is well known that integrating the resolvent $R_z = (z I - ...
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