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Let $L^1_{loc}$ denote the set of all functions from $\mathbb{R}$ to itself which are locally integrable. For every infinite compact subset $K\subseteq \mathbb{R}$, let $L^1_{m_K}$ denote the space of Lebesgue measurable functions supported on $K$.

Clearly the collection $\mathcal{K}$ of all such compact subsets of $\mathbb{R}$ form a poset wrt inclusion $i^{K_1}_{K_2}:K_1\hookrightarrow K_2$ if and only if $K_1\subseteq K_2$, for $K_i \in \mathcal{K}$. Therefore, we may define the colimit $$ \operatorname{colim}_{\mathcal{K}} L^1_{m_K}, $$ in Top.

How are $\operatorname{colim}_{\mathcal{K}} L^1_{m_K}$ and $L^1_{loc}$ related?

Note/Edit: Top is the category of topological spaces and continuous maps and LCS is the category of locally convex spaces and continuous linear maps.

Related: $L^1_{\mu}$ as limit

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  • $\begingroup$ Using the standard definition of L^1, a function in L^1_{m_K} can take arbitrary values outside of K. Thus L^1_{m_K1} is not a subset of L^1_{m_K2} for K1⊂K2. $\endgroup$
    – Dmitri Pavlov
    Commented Nov 30, 2019 at 23:48
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    $\begingroup$ If you change the definition of L^1_{m_K} to say that it is a subspace of L^1 consisting of functions supported on K, the resulting colimit is isomorphic to L^1_loc, pretty much by definition of L^1_loc. $\endgroup$
    – Dmitri Pavlov
    Commented Dec 1, 2019 at 0:13
  • $\begingroup$ @DmitriPavlov If $L^1_{m_K}$ is the space of $L^1$-functions with support in $K$ then we have an inductive spectrum of spaces and hence an inductive limit (which is the same as a colimit). The corresponding topology in TOP (and also in the category LCS of locally convex spaces) is much finer than the Frechet topology of $L^1_{loc}$ (which is the reverse or projective limit in LCS with respect to the restriction mappings). $\endgroup$
    – Jochen Wengenroth
    Commented Dec 1, 2019 at 10:30
  • $\begingroup$ @JochenWengenroth I would expect $\operatorname{colim} L^1_{m_K}$ (essentially definition almost) to be at-least as fine as $L^1_{loc}$, but why ("how much") finer? From your argument I see that it should be finer, but is there a concrete example of [functions] converging in $L^1_{loc}$ but not in $\operatorname{colim}L^1_{m_K}$, for example? $\endgroup$
    – ABIM
    Commented Dec 1, 2019 at 11:57
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    $\begingroup$ @JochenWengenroth: The original question (before it was edited) clearly talked about a set L^1_loc, so the colimit was assumed to be in the category of sets. Now that the question was retroactively edited, my answer no longer makes sense. $\endgroup$
    – Dmitri Pavlov
    Commented Dec 1, 2019 at 15:39

1 Answer 1

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Interpreting the question as Dmitri Pavlov I assume that $L^1_K$ is the space of $L^1$-functions with support in $K$ so that we have an inductive spectrum of Banach spaces (identifying almost everywhere equal functions). The colimit (=inductive limit) in the category of locally convex spaces (i.e., the union endowed with the finest locally convex topology making the inclusions from all $L^1_K$ continuous) is then a complete locally convex space (by a classical result of Dieudonne and Schwartz about strict countable inductive limits -- instead of all compact sets it is of course enough to consider a countable exhaustion). On the other hand it is dense in $L^1_{loc}$ (the projective limit of all $L^1_K$ with respect to the restrictions). Therefore, the inductive limit topology is strictly finer. (There are several other ways to see this. For example, a countable inductive limit of normed spaces $X_n$ such that $X_n\neq X_{n+1}$ is never metrizable.)

The limit topology in the category TOP is even finer than that in LCS.

If you consider uncountable colimits in LCS the situation is slightly different: As a Frechet spaces $L^1_{loc}$ is ultrabornological and hence the inductive limit of Banach spaces, namely of all Banach spaces generated by absolutely convex closed bounded sets (generated means that you take the linear span endowed with the Minkowski functional). You can describe these spaces as weighted $L^1$-spaces $\{f\in L^1: \int |f|wd\mu<\infty\}$ with suitable weights.

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  • $\begingroup$ This answer is fantastic, however it uses some concepts I'm not fully familier with. Would it be possible to add some references (which you feel are good). I can give a small bounty for that in a couple days. Also, is there a classical interpretation/application of the colimit in LCS? $\endgroup$
    – ABIM
    Commented Dec 1, 2019 at 17:21
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    $\begingroup$ "Older" books on Functional Analysis usually treated inductive limits. E.g., Köthe's Topological Vector Spaces I, Jarchow's Locally Convex Spaces or Bonet & Perez-Carreras' Barrelled Locally Convex Spaces. $\endgroup$
    – Jochen Wengenroth
    Commented Dec 3, 2019 at 7:18
  • $\begingroup$ Yes, the countable inductive limit of separable spaces is separable (the union of countable dense subsets of the steps is dense in the inductive limit). $\endgroup$
    – Jochen Wengenroth
    Commented Dec 3, 2019 at 11:05
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    $\begingroup$ By definition, the colimit topology in TOP is the finest topology making the inclusions continuous hence it is finer than the colimit topology in LCS. I believe that the TOP-colimit is even strictly finer but I don't have an argument for this. $\endgroup$
    – Jochen Wengenroth
    Commented Dec 3, 2019 at 14:37
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    $\begingroup$ I did not know that paper. Such weighted spaces of integrable functions had been investigated long time ago e.g. by Konrad Reiher ''Weighted inductive and projective limits of normed Köthe function spaces'', Results in Math.13, 147–161(1988). $\endgroup$
    – Jochen Wengenroth
    Commented May 3, 2020 at 11:30

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