Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 61785

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

12 votes
Accepted

Is $X\times X$ homeomorphic to $X$ for a space of probability measures?

The answer is yes, following from Klee's extension of Keller's theorem on homeomorphisms of infinite dimensional compact convex sets: Klee, V. L., Some topological properties of convex sets, Trans. Am …
Robert Furber's user avatar
1 vote

Topology of ${\mathcal D}(\Omega)$ (space of test functions)

There's a general principle for proving that a topology on a vector space $E$ is not a weak topology (in the general sense, a topology of the form $\sigma(E,F)$ for some $F \subseteq E^*$). For $\sigm …
Robert Furber's user avatar
10 votes
Accepted

Banach space with uncountable basis

I will pull together the comments into a community wiki answer with some of my own remarks so that the question isn't left on the unanswered questions list. If you're willing to accept that it is cons …
4 votes
Accepted

Baire category theorem for uncountable unions

The hyperstonean case can be dealt with using a result from Fremlin's Measure Theory. For every hyperstonean space $X$, we can find a semi-finite measure $\mu$ defined on the sets with the Baire prope …
Robert Furber's user avatar
7 votes
Accepted

Equivalence of σ-convex hull and closed convex hull

Wlod AA gave a good counterexample for the case when $K$ is not required to be compact, here I give a counterexample $K$ compact, first in a locally convex space, and then for a(n infinite-dimensional …
Robert Furber's user avatar
14 votes

A topological vector space $X$ is separable if its dual space $X^*$ is separable?

YCor has given a counterexample for topological vector spaces. The statement is still false for locally convex spaces. Consider the space $X$ defined to be a locally convex coproduct of $\newcommand{\ …
Robert Furber's user avatar
5 votes

Set of w*-continuous operators closed for the weak* topology or not?

The answer is no. I know that for some people here, saying "It's false for $X = \ell^1$" would be a good enough hint, but I also know that this question originated on Math StackExchange, so I've inclu …
Robert Furber's user avatar
3 votes
Accepted

Convergence in $\sigma(\mathcal{E}',\mathcal{E})$ versus $\beta(\mathcal{E}',\mathcal{E})$

The answer is yes. First, since $\newcommand{E}{\mathcal{E}}\E$ is a Fréchet space, it is barrelled, and so any $\sigma(\E',\E)$-bounded subset of $\E'$ is equicontinuous, and therefore bounded in any …
Robert Furber's user avatar
3 votes

Is the space of Radon measures a Polish space or at least separable?

The other answers very adequately explain why the norm topology is not Polish except for trivial cases, so this answer is about the weak-* topology. Also, most results in the literature are about the …
Robert Furber's user avatar
3 votes

Complete dual of bornological space

Jochen is quite right. I have another example, just using any irreflexive Banach space $A$. The space $E = (A^*,\mu(A^*,A))$ is Mackey, by definition. The bounded sets in $E$ are the same as the norm- …
Robert Furber's user avatar
12 votes
Accepted

Unconditionally convergent series in some functional spaces

A good resource for these things is Section IV.10 of Schaefer's Topological Vector Spaces, so you should look there for the proofs of the following statements. For $E$ a locally convex space, let $\el …
Robert Furber's user avatar
6 votes
Accepted

Examples of non-isomorphic $C^\ast$ algebras with isomorphic quasi-state spaces

For any C$^*$-algebra $A$, we can define its opposite algebra $A^{\mathrm{op}}$, which is the algebra where $ab$ is defined to be $ba$, as calculated in $A$. Let's restrict to unital algebras for simp …
Robert Furber's user avatar
5 votes
Accepted

Is the compact-open topology on the dual of a separable Frechet space sequential?

Yes. In the next paragraph I will show that if $X$ is a Fréchet space (without requiring separability) then $X'_c$ with the compact-open topology is a $k$-space. As you note, this implies sequentialit …
Robert Furber's user avatar
3 votes

Non-separable metric probability space

Iosif Pinelis has given an answer to question 1 and partial answers to 2 and 3. Since he advised me to turn my comments into an answer, here it is. I will deal with the case where the axiom of choice …
Robert Furber's user avatar
9 votes

A characterization of $L_1(\mu)$ in $L_\infty(\mu)^*$

The criterion suggested in the question works fine for $\sigma$-finite spaces, and Michael Greinecker's answer is correct under this assumption. However, the suggested criterion is not (provably) suf …
Robert Furber's user avatar

15 30 50 per page