Timeline for Under what conditions on $\mu^{\beta}$ we have $L_1(\beta X,\mu^{\beta})$ isometrically isomorphic to $L_1(X,\mu)$?
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6 events
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| Mar 31, 2017 at 14:04 | comment | added | Nate Eldredge | Well, I believe the same thing will happen when $X$ is $\mathbb{R}$ or any non-compact Polish space. | |
| Mar 31, 2017 at 11:21 | comment | added | Leandro | Hi @NateEldredge thanks for the comment as I was suspecting the answer in general should be no. Although I forgot to mention I was more interested in cases where $X$ is uncountable. | |
| Mar 31, 2017 at 5:08 | comment | added | Nate Eldredge | For a cruder idea, if $X = \mathbb{N}$ then $L^1(\mathbb{N},\mu)$ is separable for any $\mu$, but $L^1(\beta \mathbb{N}, \mu^\beta)$ is typically not separable. For instance, suppose $\mu^\beta$ comes from a non-principal ultrafilter on $\mathbb{N}$. The equivalence relation on $2^{\mathbb{N}}$ induced by the ultrafilter has uncountably many equivalence classes. If $\{A_i\}$ are pairwise inequivalent subsets of $\mathbb{N}$, let $f_i$ be the continuous extension of $1_{A_i}$ to $\beta \mathbb{N}$. These functions are at pairwise distance 1 in $L^1(\mu^\beta)$. | |
| Mar 31, 2017 at 5:03 | comment | added | Nate Eldredge | Well, a dumb example where this works is when $\mu^\beta$ is atomic, since then you can just arbitrarily choose a countable subset of $X$ and give it the same weights. But it makes it hard to see why the question is interesting because $\mu$ and $\mu^\beta$ need not be related in any canonical way. | |
| Mar 31, 2017 at 0:42 | comment | added | Christian Remling | I don't think this can work, as you suspected: if $X=\mathbb N$, then $L^1(X)$ is a weighted $\ell^1$ space whose unit ball has many extreme points, and it must be possible (easily, one would think) to come up with a measure on $\beta\mathbb N$ for which the unit ball of $L^1(\beta\mathbb N)$ has no extreme points. | |
| Mar 31, 2017 at 0:28 | history | asked | Leandro | CC BY-SA 3.0 |