Timeline for If $S\subseteq A^*$ is separating, does $S$ also separate $M(A)$?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 6, 2022 at 16:11 | vote | accept | Andromeda | ||
| Jul 6, 2022 at 16:10 | comment | added | Andromeda | Yes, I can show this using a standard approximation argument. Just wanted to make sure that I actually had to check this! Thanks for all the help. Everything is clear now! | |
| Jul 6, 2022 at 16:05 | comment | added | Nik Weaver | You need to check this yourself. | |
| Jul 6, 2022 at 16:03 | comment | added | Andromeda | Yes, I see that. My question is, if $\omega \in c_0^* = \ell^1$, can we say how the strict extension $\omega\in (\ell^\infty)^*$ looks like? I would suspect that if $\omega$ is given by $f \in \ell^1$, then $\omega(g) = \sum_n f(n)g(n)$ for $f\in \ell^\infty$. This is definitely a well-defined extension, but shouldn't we check this is a strictly continuous? | |
| Jul 6, 2022 at 15:59 | comment | added | Nik Weaver | I am quoting your own post. | |
| Jul 6, 2022 at 15:59 | comment | added | Nik Weaver | "Every element $\omega \in A^∗$ extends uniquely to a strictly continuous functional $\omega \in M(A)^*$" | |
| Jul 6, 2022 at 15:32 | comment | added | Andromeda | Alright, it becomes a little clearer to me: we have functionals in $\ell^1 = c_0^*$, so indeed we start with functionals on $c_0$. We can also view elements in $\ell^1$ as functionals on $\ell^\infty$, and I guess my question is: why are these strictly continuous (on bounded subsets). I think I can show this using a standard estimate. Would you agree that showing this strict continuity is necessary? Or does it somehow follow trivially? (sorry if I am overcomplicating this, really appreciate the help). | |
| Jul 6, 2022 at 14:16 | comment | added | Nik Weaver | It seems that you're having difficulty with abstraction. Can you look at the specific example $S = \{\vec{b} \in l^1: \sum b_n = 0\}$ and verify directly that it separates $c_0$ but not $l^\infty$? | |
| Jul 6, 2022 at 14:12 | comment | added | Nik Weaver | "You start with functionals on $M(A)$" --- look for the place in my answer where I said this, and realize that I didn't. | |
| Jul 6, 2022 at 12:08 | comment | added | Andromeda | You start with functionals on $M(A)$ (where $A= c_0$), so how do we know that these come from strictly continuous functionals on $A=c_0$? It can happen that an element $\omega \in M(A)^*$ does not come from a strictly continuous functional on $A$ right? Do I miss something? | |
| Jul 6, 2022 at 11:29 | comment | added | Nik Weaver | "Every element $\omega \in A^*$ extends uniquely to a strictly continuous functional $\omega \in M(A)^*$" | |
| Jul 6, 2022 at 7:29 | comment | added | Andromeda | Thanks for the reply. One more thing: How are we sure that the elements $\ell^1 \subseteq (\ell^\infty)^*$ are continuous w.r.t. the strict topology on $\ell^\infty = M(c_0)$? Maybe the strict topology on $\ell^\infty$ admits a nice description relating it with some weak$^*$-topology (on bounded subsets)? | |
| Jul 5, 2022 at 23:09 | comment | added | Nik Weaver | $l^1$ as the dual of $c_0$. A codimension 1 subspace of $l^1$ is weak* closed iff it is the kernel of the map evaluating on some $\vec{a} \in c_0$; since our $\vec{a}\not\in c_0$ its kernel cannot be weak* closed. | |
| Jul 5, 2022 at 21:26 | comment | added | Andromeda | Thanks for your answer. Maybe some short follow-up questions: What do you mean with the weak$^*$-topology on $l^1$? Does it come from the natural embedding $l^1 \hookrightarrow (l^\infty)^*$? Why is $S$ weak$^*$-dense in $l^1$? | |
| Jul 4, 2022 at 22:02 | history | answered | Nik Weaver | CC BY-SA 4.0 |