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S Jun 17, 2021 at 15:06 history bounty ended CommunityBot
S Jun 17, 2021 at 15:06 history notice removed CommunityBot
Jun 9, 2021 at 16:52 comment added Jochen Wengenroth If $X$ is compact, the dual of $L^1(X,\mu)$ is $L^\infty(X,\mu)$. In the hemi-compact case, the dual is the projective limit of $L^\infty(X_n,\mu)$. What else do you want? Somewhat artificially, you can identify $L^\infty(X,\mu)$ with a space of measures, assigning to each bounded function $f$ the measure with $\mu$-density $f$.
S Jun 9, 2021 at 13:31 history bounty started Catologist_who_flies_on_Monday
S Jun 9, 2021 at 13:31 history notice added Catologist_who_flies_on_Monday Authoritative reference needed
Jun 8, 2021 at 12:06 comment added Catologist_who_flies_on_Monday @JochenWengenroth I was especially wondering if it had an interpretation as some sort of (Radon?) measures... do you think so? Also yes to your first point.
Jun 8, 2021 at 12:05 history edited Catologist_who_flies_on_Monday CC BY-SA 4.0
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Jun 4, 2021 at 16:14 comment added Jochen Wengenroth Roughly, the dual of the LB-space is the projective limit of the duals. My guess is thus, that $L_c(X)'$ is the space of measurable function whose restrictions to all $X_n$ are $\mu$-a.e. bounded.
Jun 4, 2021 at 16:11 comment added Jochen Wengenroth Do you really mean the category of locally convex spaces and continuous maps or rather the continuous and linear maps? Moreover, $X_n$ is probably a compact exhaustion of $X$, right?
Jun 4, 2021 at 11:43 history asked Catologist_who_flies_on_Monday CC BY-SA 4.0