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Let $(X_n,\pi_n^{m})$ be a countable projective system in the category Ban$_1$ of Banach spaces and short linear maps (is (continuous) linear constructions). Then (co)-completeness of this category implies that $\projlim X_n$ is a well-defined Banach space.

Suppose further that, a strict variant of the Mittag-Leffer condition holds true: for all $n \in \mathbb{N}$, there exists $m\geq n$ such that $$ \overline{\pi_m^n(X_n)} \subseteq X_k \qquad \forall k \geq n, $$ and that $U\subseteq \projlim X_n$. What reasonable conditions would imply that $U$ is dense in $\projlim X_n$

What I propose (so far...):

Condition:

$$ \bigcap_{k\leq n; \, k,n \in \mathbb{N}} \pi^k_n(X_n) \subseteq U $$

Proof:

Then, by Bourbaki's Mittag-Leffer theorem, the first condition implies that $\cap_{k\leq n; \, k,n \in \mathbb{N}} \pi^k_n(X_n)$ is dense in $\projlim X_n$. The second condition then implies that $U$ is dense in $\cap_{k\leq n; \, k,n \in \mathbb{N}} \pi^k_n(X_n)$. The transitivity of density implies that $U$ is dense in $\projlim X_n$.

Comment: I think the assumption is much too restrictive to be interesting.

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    $\begingroup$ I‘n not sure what you mean but consider the case of $\ell^\infty$ and its non-dense subspace $c_0$. The former is the projective limit of the finite dimensional spaces $\ell^\infty_n$ and this would seem to provide a counterexample. $\endgroup$
    – user131781
    Commented Jan 13, 2020 at 13:51
  • $\begingroup$ Oh, then how could one (non-trivially) modify the hypothesis so that $U$ is dense? $\endgroup$
    – ABIM
    Commented Jan 13, 2020 at 13:54
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    $\begingroup$ I think this example shows that you can‘t, at least not in any simple way. The classical abstract Mittag-Leffler theorem (which, to my knowledge, is due to Bourbaki and works in the context of metric spaces) applies to the „normal“ projective limit, i.e., with no restrictions on the Lipschitz constants of the morphisms. You can save the density property but only at the cost of abandoning the category of Banach spces, e.g., by using strict or mixed topologies but that is another story entirely. $\endgroup$
    – user131781
    Commented Jan 13, 2020 at 14:12
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    $\begingroup$ That is the whole point of them. Let me give you an example (for the case of continuous functions—this is more transparent). If you have a locally compact space $X$—let‘s say $\sigma$– compact to simplify things, then the family ${C(K)}$ of continuous functions on its compacta forms a projective spectrum and you can take its projective limit in three senses: as lcs‘s, as Banach spaces and in the sense of mixed topologies. This gives you successively the Fréchet space $C(X)$, the Banach space $C^b(X)$ and the strict topology on the latter (R.C. Buck) .... $\endgroup$
    – user131781
    Commented Jan 13, 2020 at 14:49
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    $\begingroup$ The third space has better density and duality properties than the other two ( which is why Buck introduced them). Similar situations arise for spaces of measurable or holomorphic functions, resp. for of opererators, say on Hilbert spaces and they have all been provided with appropriate strict topologies to deal with such problems. $\endgroup$
    – user131781
    Commented Jan 13, 2020 at 14:54

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