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Setup:

Let $\mu$ be a measure on a measurable space $(X,\Sigma)$, such that for every $p ,q\in [1,\infty)$, $L^p_{\mu}(\Sigma)\subseteq L^q_{\mu}(\Sigma)$ if $p\geq q$. Furthermore, the inclusions are continuous. (A precise characterization of when this happens can be found in Rudin's book, for example.)

Background:

Now, let $X^p$ denote the closed unit ball in each $L^p_{\mu}(\Sigma)$; by construction we have that there are continuous inclusions maps $i^{p,q}:X^p\hookrightarrow X^q$. From classical topology, we known that the Top is (co)complete and from the Semadeni-Zidenberg Theorem we also know that Ban$_1$ (the category of unit Balls in Banach spaces and of short linear maps) is (co)complete. Therefore, the limit $\varprojlim X^p$ is well-defined in both Top and in Ban$_1$.

Question:

What is $\varprojlim X^p$ equal to when the limit is taken in Top and in Ban$_1$; and are they the same?

Intuition My intuition is that in $Ban_1$, $\varprojlim_n X^n$ is equal to $$ \left\{ f \in L^{\infty}_{\mu}(\Sigma):\, \|f\|_{\infty}\leq 1 \right\} , $$ with topology given by the norm $\|\cdot\|_1+\|\cdot\|_{\infty}$; but I cannot find a rigorous justification of my hypothesis (I have not shown minimiality in $Ban_1$ of the corresponding cone).

Furthermore, I expect that $\varprojlim X^p$ in Top has finer topology than $\varprojlim X^p$ in Ban$_1$. Am I correct in this direction?

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    $\begingroup$ Two comments. The underlying set for the inductive limit in the category of topological spaces is the union and this can never be a Banach space (except in trivial cases) by the Baire category theorem. Secondly, a typical situation where your conditions are satisfied is the discrete case, i.e., $\ell^p$. Here the Banach space limit is $c_0$, not $\ell^\infty$. $\endgroup$
    – user131781
    Commented Nov 29, 2019 at 9:24
  • $\begingroup$ Your setup is that $L^p$ goes into $L^q$ when $p\leq q$, e.g. when the measure space is atomic but not if it is $[0,1]$ with Borel-sigma algebra and Lebesgue measure. Therefore surely the intuition would be that the limit (inverse limit, projective limit) is the ball of $L^1$. In fact, with the definitions you have chosen, there is no arrow going into the ball of $L^1$, all the arrows go out of it, so why isn't this object already the limit of the cone? $\endgroup$
    – Yemon Choi
    Commented Nov 29, 2019 at 15:55
  • $\begingroup$ Actually I am now even more confused because you talk at the end about a cocone but you are consistently using the notation for limit (=projective limit) rather than colimit (=inductive limit). Which object is your question about? $\endgroup$
    – Yemon Choi
    Commented Nov 29, 2019 at 15:58
  • $\begingroup$ OK, this is starting to get ridiculous. Instead of engaging with other people's comments, you have now changed the $p\leq q$ condition to $p\geq q$ which changes everything completely. I notice that your question has been edited several times in non-trivial ways. Can you please leave some comments or add some remarks to your original question to make it clear, hopefully once and for all, what your firm asssumptions are and what your definite question is? $\endgroup$
    – Yemon Choi
    Commented Nov 30, 2019 at 3:09
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    $\begingroup$ IIRC the limit in Ban1 is nothing but the subset of the algebraic limit consisting of the elements whose all the component norms are at most 1. In your case, that means all the Lp-norm is at most 1, and according to your (implicit) assumption, Lp-norm increases as p increases. (otherwise, i^p,q makes no sense in Ban1.) Hence, the result is the 1-ball of Linf since the measure is finite (again according to your implicit assumption.) $\endgroup$
    – Junekey Jeon
    Commented Dec 4, 2019 at 20:38

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