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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 …
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 …
-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. …
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 …
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 …
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 …
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 …
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 …
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 …
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 $\ …
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 …
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 …
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.
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 …