All Questions
Tagged with hilbert-spaces harmonic-analysis
8 questions
2
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139
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Multidimensional weighted Paley-Wiener spaces are Hilbert spaces?
How to rigorously demonstrate that multidimensional weighted Paley-Wiener spaces are Hilbert spaces?
I am utilizing the exponential type definition established by Elias Stein in the book 'Fourier ...
- 21
1
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210
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Is this a well known space? Perhaps homogeneous Sobolev-like space?
The homogeneous Sobolev space $\dot H^s(\mathbb{R}^n) $ is often defined as the closure of $\mathcal{S}(\mathbb{R}^n)$ under the norm
$$ || |\omega|^s \widehat{f} ||_{L^2(\mathbb{R}^d)} =\int_{\...
- 197
2
votes
1
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Orthonormal bases in RKHSs via interpolating sequences
Definitions and setting
Let $\mathcal{H}$ be a separable, infinite-dimensional, reproducing kernel Hilbert space on a nonempty set $X$. As usual, denote the reproducing kernel on $\mathcal{H}$ by $K$ ...
- 5,405
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Norm of the Linear Operator
Let $G$ be a compact group, and $\pi : G \rightarrow \mathcal{U}(H)$ be a continuous unitary representation. Let $f \in L^{1}(G)$ be arbitrary.
By Riesz Representation Theorem we can find a bounded ...
- 581
20
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333
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Existence of orthonormal basis for $L^2(G)$ in $C_c(G)$
Remark: I cross-posted this question on MSE and added a bounty to it.
Suppose that $G$ is a locally compact (Hausdorff) group endowed with the Haar measure. It is well-known that the compactly ...
- 321
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46
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How to projectivize an infinite-dimensional $L^{2}$ space and do fourier analysis on said projectivization
I'm doing work with repelling (that is, non-contracting) linear operators on hilbert spaces, and I wondered if it might be worth my while to study my operators on a projectivization of my hilbert ...
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5
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207
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Is unitary group paracompact?
In this paper Martin Schottenloher notices that the unitary group $U(H)$ of a separable Hilbert space $H$ is metrizable in the strong operator topology. As a corollary (see R.Engelking, 5.1.3), it is ...
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A specific Schwartz function $f$ on $\mathbb C^2$
Choose a Schwartz function on $\mathbb C$ of the form $f(z)=f(r e^{i\theta})= f_0(r) e^{in\theta}$. Then $$(*) \quad f(e^{i\alpha} z)= e^{in\alpha} f(z), \quad \forall z\in \mathbb C.$$
Now, let $f$ ...
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