Timeline for Is there an inclusion of $L_\infty(G)$ into $C_0(G)^{**}$?
Current License: CC BY-SA 3.0
15 events
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| Jan 12, 2012 at 9:39 | comment | added | Yulia Kuznetsova | Indeed, it is {\it norm-preserving}... Thanks a lot. | |
| Jan 12, 2012 at 3:37 | comment | added | Bill Johnson | The lifting gives a mapping from (indicator functions of) measurable sets modulo null sets to (indicator functions of) measurable sets; extend by linearity to a well defined mapping from the elements of $L_\infty$ that are simple modulo null sets to simple functions--this mapping is well defined, linear, and norm preserving, hence extends by continuity to a linear isometry from $L_\infty$ into $B(G)$. | |
| Jan 11, 2012 at 20:01 | comment | added | Yulia Kuznetsova | @Bill: Your 1st comment explained to me what was the problem that bothered me, that's why I prefer this as an answer. Also I am interested in the general case. But: as I remember from Tulcea's book, there is a lifting on every l.c.group. It is a selector of measurable sets; would you please indicate how to transform this into a selector of $L_\infty$-functions? | |
| Jan 11, 2012 at 19:19 | vote | accept | Yulia Kuznetsova | ||
| Jan 11, 2012 at 19:17 | vote | accept | Yulia Kuznetsova | ||
| Jan 11, 2012 at 19:18 | |||||
| Jan 11, 2012 at 18:08 | answer | added | Bill Johnson | timeline score: 6 | |
| Jan 11, 2012 at 18:08 | comment | added | Bill Johnson | Right, but note that my second comment (if it is indeed correct) allows you to interpret pointwise evaluation of $L_\infty$ functions in many cases (e.g., when the group is metrizable). | |
| Jan 11, 2012 at 16:45 | comment | added | Yulia Kuznetsova | @Bill: thank you, I would accept the comment on the direct sum if it were posted as an answer. As Matth points out, this $f(t)$ is 0, so nothing is contre-intuitive! | |
| Jan 11, 2012 at 16:40 | comment | added | Matthew Daws | Ah, so if you follow Bill's first comment... then as each point mass $\delta_t$ lives in $S(G)$ (assuming $G$ has no discrete parts) then actually your way of "associating" a value at $t$ just always gives $0$ (and hence nothing to worry about, but also not very useful...) | |
| Jan 11, 2012 at 15:15 | comment | added | Bill Johnson |
In some cases, you get a better embedding of $L_\infty(G)$ into $M(G)^*$ by first embedding $B(G)$, the bounded measurable functions under the sup norm, into $M(G)^*$ in the obvious way and using the Tulcea lifting theorem (which applies e.g. if $G$ is sigma compact) to embed $L_\infty(G)$ into $B(G)$. This is better in that it extends the canonical embedding of $C_0(G)$ into its second dual. Well, I think this is right; I am going from memory rather than thinking through the argument.
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| Jan 11, 2012 at 14:36 | comment | added | Bill Johnson |
Since you are in the commutative setting, you can present the construction more simply. $M(G)= L_1(G)\oplus_1 S(G)$, where $S(G)$ is the space of complex measures that are singular with respect to Haar measure, and hence $M(G)^*= L_\infty(G)\oplus_1 S(G)^*$.
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| Jan 11, 2012 at 14:00 | comment | added | Yulia Kuznetsova | I was lucky to see it. I didn't insist that this inclusion is a homomorphism, just that it is continuous. Is your example continuous? Maybe you would post it again, at least as a comment? | |
| Jan 11, 2012 at 13:51 | comment | added | Matthew Daws | I deleted a tentative answer-- your argument gives a *-homomorphism of $L^\infty(G)$ into $C_0(G)^{**}$, while I gave some more ``concrete'' arguments as to why this was plausible, they didn't seem to address why the "inclusion" would be an algebra homomorphism. | |
| Jan 11, 2012 at 13:28 | answer | added | Matthew Daws | timeline score: 1 | |
| Jan 11, 2012 at 13:17 | history | asked | Yulia Kuznetsova | CC BY-SA 3.0 |