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Does weak $L^2$ approximation implies $L^2$ approximation under a condition similar to convexity?

(Cross posted from Math StackExchange: Does weak $L^2$ approximation implies $L^2$ approximation under a condition similar to convexity?) Assume $(\Omega, \mu)$ is a probability space. Consider a ...
David Gao's user avatar
  • 2,830
1 vote
1 answer
89 views

Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace - Part II

This is a follow-up to this previous question, but under stronger assumptions. Let $(X, \mu)$ be a (say, $\sigma$-finite) measure space, let $g \in L^2$ (say, over the real scalar field). Let $\tilde ...
Jochen Glueck's user avatar
7 votes
2 answers
464 views

Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace

Let $(X, \mu)$ be your favourite measure space (finite or $\sigma$-finite if you like), let $g \in L^2$ (say, the scalar field of $L^2$ is $\mathbb{R}$, though this probably doesn't matter). Let $\...
Jochen Glueck's user avatar
1 vote
0 answers
133 views

Subspace of RKHS generated by kernel mean embeddings

Suppose $\mathcal{H_k}$ is a reproducing kernel Hilbert space (RKHS) with reproducing kernel $k: \mathcal{X} \times \mathcal{X} \rightarrow \mathbb{R}$. I am looking for results characterising the ...
Athere's user avatar
  • 93
1 vote
0 answers
191 views

Dual of union of Reproducing Kernel Hilbert Spaces

I have a union of Reproducing Kernel Hilbert Spaces $\mathcal{B}$. I am interested in finding the dual of $\mathcal{B}$. Knowing what the dual is might help to write an alternate formulation for the ...
ChocolateRain's user avatar
2 votes
0 answers
131 views

Does a spectral theorem exist for linear operator pencils?

I was wondering if a version of the spectral theorem (the projection valued measure case) holds for linear pencils of the form $$ A-\lambda B $$ where $A,B$ are self-adjoint on some Hilbert space $\...
Mathmo's user avatar
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0 votes
0 answers
107 views

Norm equivalences for Gaussian random functions (Cameron-Martin space)

Preliminaries Consider the Hilbert space $H :={L^2_{\text{per}}(\mathbb{R})}$ of Gaussian random functions, $2\pi$-periodic in $\mathbb{R}$. These random functions are drawn from a Gaussian measure $\...
ares's user avatar
  • 101
1 vote
0 answers
83 views

Embedding random variables in infinite-dimensional spaces

Let $H$ be a reproducing kernel Hilbert space of functions $f:E\to F$ with kernel $k$. A point in $E$ may be embedded into $H$ via the canonical embedding $x\mapsto k(x,\cdot)$. Similarly, a random ...
JohnA's user avatar
  • 710
3 votes
0 answers
225 views

Defining a trace-class operator with a Bochner integral

I had asked this question previously on Math.StacheExchange but did not get an answer there in several months. This isn't strictly speaking research level mathematics but I hope it is sufficiently ...
Adomas Baliuka's user avatar
2 votes
1 answer
651 views

Some integrals with respect to a Gaussian measure on a Hilbert space

Assume that $(H,\langle\cdot,\cdot\rangle)$ is a real separable Hilbert space equipped with a Gaussian measure $\mu$ with a mean $m$ and a covariance operator $C$. Let $x\in H$ be a fixed vector. What ...
Pirx's user avatar
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8 votes
1 answer
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Definitions of Hilbert Bundles

I have some doubts regarding definitions and conventions on Hilbert Bundles. Some authors like Peter Kuchment (Floquet Theory for Partial Differential Equations) and Serge Lang (Differential and ...
Thangachelli Debopritama's user avatar
5 votes
2 answers
310 views

Error estimate in the spectral theorem of compact operators on a Hilbert space

Given a compact self-adjoint operator $K$ mapping $L^2(\mathbb{R}^d) \rightarrow L^2(\mathbb{R}^d)$ as $f \rightarrow \int K(x,y) f(y) d\mu(y)$, let us define its eigenvalues $\lambda_i$ and eigen-...
gradstudent's user avatar
  • 2,246
4 votes
0 answers
185 views

A strongly open set which is not measurable in the weak operator topology

Let $H$ be a non-separable Hilbert space and $\{e_i\}_{i\in I}$ be an orthonormal basis for $H$. Let $J$ be a uncountable proper subset in $I$. Let us put $$E=\{x\in B(H): \lVert xe_j\rVert <1: \...
ABB's user avatar
  • 4,058
2 votes
1 answer
183 views

Visualizing ANOVA Decomposition [closed]

Let $f \in L^2[0,1]^d$ be a measurable function where $d \in \mathbb{N}$. For a given subset $u \subseteq D := \{1,2,\ldots,d\}$ consider the projections $P_u : L^2[0,1]^d \to L^2[0,1]^{|u|}$ given ...
bcf's user avatar
  • 121
0 votes
1 answer
517 views

Injective inclusion map from RKHS function space to $L_p(\mu)$

Let $X$ be a measurable space, $\mu$ be a $\sigma$-finite measure on $X$, and $H$ be a separable reproducing kernel Hilbert space over $X$ with a measurable kernel $k$. At a certain part in a proof I ...
Eric's user avatar
  • 103
3 votes
1 answer
1k views

If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$?

Let $H$ be an infinite-dimensional, separable Hilbert space, and let $\gamma$ be a Radon probability measure on $H$ with mean zero and covariance operator the identity $I$. Let $H^*$ denote the space ...
Tom LaGatta's user avatar
  • 8,512
7 votes
1 answer
1k views

If $H$ is a separable Hilbert space, is $L^2(H)$ separable?

Let $H$ be a separable Hilbert space, and let $\gamma$ be a Radon probability measure on $H$ with mean zero and covariance operator the identity $I$. Is the Hilbert space $L^2(H,\gamma)$ separable?
Tom LaGatta's user avatar
  • 8,512
0 votes
1 answer
319 views

Hilbert space automorphisms realized as induced by transformations of some base-spaces

Following question may be soft. Fix abstract hilbert space H and consider any automorphism A in banach-spaces sence (i.e. no conditions on metric). Call A is realizable if exist measure space $(X,\mu)$...
Bad English's user avatar
13 votes
4 answers
5k views

What is known about the Gaussian measure of the unit ball in a Hilbert Space?

Let $X$ be an infinite dimensional separable Hilbert Space with norm $||\cdot||$ and let $\mu$ be a Gaussian measure on $X$ such that $\mu(X) = 1$. What do we know about $\mu(B(0,1))$, where $B(0,1)$ ...
RadonNikodym's user avatar