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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.

1 vote
0 answers
83 views

Topologies on spaces of linear sections

Let $X$ and $Y$ be topological linear spaces which are complete & Hausdorff, and admit dual spaces which separate points. Suppose the topologies are non-separable and non-metrizable. Let $f : X \to Y …
Tom LaGatta's user avatar
  • 8,532
1 vote
1 answer
359 views

Convergence of operators to the identity on Banach spaces

Let $U_\infty$ be a compact space, and let $U_r$ be an increasing family of compact subspaces whose closure is all of $U_\infty$. That is, $U_r \subseteq U_{r'}$ if $r \le r'$ and $U_\infty = \overli …
Tom LaGatta's user avatar
  • 8,532
-1 votes

Unbounded operator bounded in a dense subset

No. Bounded means continuous, for linear operators. If a linear operator is bounded (continuous) on a dense subset, then you can extend it continuously to the whole space, which means it is bounded. …
Tom LaGatta's user avatar
  • 8,532
5 votes
1 answer
399 views

Example: a locally convex TVS which is not compactly generated

Is there an example of a locally convex topological vector space which is not compactly generated? (any such example must be non-Fréchet, since all Fréchet spaces are compactly generated) (note: I a …
Tom LaGatta's user avatar
  • 8,532
9 votes
1 answer
780 views

Topological Generalization of Whitney's Extension Theorem

From Wikipedia: In mathematics, in particular in mathematical analysis, the Whitney extension theorem is a partial converse to Taylor's theorem. Roughly speaking, the theorem asserts that if $A$ i …
Tom LaGatta's user avatar
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5 votes
1 answer
2k views

Dual of the space of continuous functions

Let $T \subseteq \mathbb R$ be a closed set of real numbers. Let $X := C(T, \mathbb R)$ denote the Fréchet space of continuous real-valued functions on $T$. The topology on $X$ is generated by seminor …
Tom LaGatta's user avatar
  • 8,532
7 votes
1 answer
524 views

Are dual spaces barreled?

Let $X$ denote a topological affine space (with no additional assumptions). Let $X^*$ denote its dual space of continuous affine functionals, equipped with the weak-$*$ topology. It is easy to see tha …
Tom LaGatta's user avatar
  • 8,532
6 votes

Corollaries of the Yoneda Lemma in Analysis?

Thanks William for reaching out (and thanks David Roberts for the hat tip to my talk). Let me give an intentionally fuzzy, high-level answer. Generally speaking, the Yoneda Lemma allows you to make a …
Tom LaGatta's user avatar
  • 8,532
4 votes
1 answer
472 views

Isomorphisms between topological vector spaces [closed]

Let $f : X \to Y$ be a continuous map of complete topological vector spaces. Suppose that $A \subseteq X$ and $B \subseteq Y$ are proper, dense linear subspaces, and that the restriction map $f : A \t …
Tom LaGatta's user avatar
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2 votes

If Gaussian measures on a Hilbert space converge weakly to 0, how do their covariance operat...

Let $X$ be a separable Banach space, and let $\theta \mapsto \mathbb P_\theta$ be a weakly-continuous family of Radon probability measures. Suppose that each $\mathbb P_\theta$ admits a mean vector $\ …
Tom LaGatta's user avatar
  • 8,532
3 votes
0 answers
176 views

Extending a Hilbert space isometrically

Let $H$ be a Hilbert space, and let $X$ be a topological vector space. Under what conditions on the topologies of $X$ and $H$ does there exist an injective, continuous linear map $f : H \to X$? Su …
Tom LaGatta's user avatar
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1 vote

Conditional probabilities are measurable functions - when are they continuous?

Here's an answer in the case that $X$ and $Y$ are Gaussians. There's such a rigid algebraic structure there that I don't know whether I'm exploiting that, or a more general property of the distributi …
Tom LaGatta's user avatar
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2 votes
1 answer
381 views

Properties of the weak-$*$ topology

Let $X$ be a topological affine space over a complete base field $\mathbb S := \mathbb C$, $\mathbb R$ or $\mathbb Q_p$. Let $X^*$ be the dual space of continuous affine functionals equipped with the …
3 votes
Accepted

Conditional probabilities are measurable functions - when are they continuous?

Since a troll bumped this question to the front page, I might as well answer it. The technology which provides the solution is called regular conditional probability or disintegration.
Tom LaGatta's user avatar
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8 votes

Is there a natural measures on the space of measurable functions?

Sorry for the necromancy. Here's an attempt at constructing a $\sigma$-algebra using the tensor product of $\sigma$-algebras. This should likely not result in a Borel structure (i.e., a $\sigma$-algeb …

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