I learned in J. Castillo's Hitchhiker guide to categorical Banach space theory that, by a theorem of Semadeni and Zidenberg, limits and colimits exist in the category $\text{Ban}_1$ of Banach spaces and contractive maps. I am hoping now that the Banach space dual of a projective limit is the inductive limit of the corresponding duals; is this true? I wasn't able to get a hold on Semadeni and Zidenberg's paper, so I don't know if this question is addressed by them.
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$\begingroup$ Your title asks about the dual of a colimit. You question asks about the dual of a limit. Which one did you mean? $\endgroup$– Yemon ChoiCommented May 20, 2016 at 16:16
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$\begingroup$ Ops, yes, I meant limit. But in order not to outdate to your answer I guess it better stays like that. $\endgroup$– Rodrigo VargasCommented May 20, 2016 at 16:27
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$\begingroup$ @RodrigoVargas Have you ever found an answer to this question $\endgroup$– ABIMCommented Sep 6, 2019 at 11:00
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$\begingroup$ Yes, it is below! $\endgroup$– Rodrigo VargasCommented Sep 7, 2019 at 11:37
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1 Answer
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If I recall correctly, the duality functor $D: {\sf Ban}_1 \to {\sf Ban}_1^{\rm op}$ is a left adjoint -- the analogous statement for ${\sf Vect}$ and ${\sf Vect}^{\rm op}$ is in Mac Lane's CFTWM somewhere -- and so $D$ should preserve colimits. That is, if we think of it as a contravariant functor on ${\sf Ban_1}$, it should take colimits to limits.
However, you are asking about duals of limits (in the category-theoretic sense) and so one would not expect the dual of a limit to be the colimit of the duals. For instance, the product in ${\sf Ban}_1$ of countably many copies of the ground field is $\ell_\infty$, whose dual is (assuming Axiom of Choice in some form) not the same as $\ell_1$ (and $\ell_1$ is the coproduct of countably many copies of the ground field.
Actually, I think for your particular question we have a counterexample which doesn't use AC/Hahn-Banach. Consider an "inverse system" $\dots E_n \to E_{n-1} \to \dots E_0$ where $E_n=\ell_1^n$ and the linking maps are given by "omit the last entry of the vector". Then the inverse/projective limit should be $\ell_1$, whose dual is $\ell_\infty$. But the dual of the inverse system would be $E_0^* \to E_1^*\to\dots E_n^* \to \dots$ and here $E_n^*=\ell_n^\infty$ with linking maps given by obvious inclusion. The direct/inductive limit of this system will be $c_0$, not $\ell_\infty$.
answered May 20, 2016 at 16:15
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$\begingroup$ Although your examples make me skeptical, I will still ask: can one hope then for some simple characterization of the dual of a limit in $\text{Ban}_1$? $\endgroup$ Commented May 20, 2016 at 16:31
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$\begingroup$ In what I would call the "standard" setting, the dual of $\ell_\infty$ is a fairly nasty Banach space that is much bigger than $\ell_1$. So I think one would need to be looking at very special limits. I admit I haven't thought about pullbacks $\endgroup$ Commented May 20, 2016 at 16:35
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$\begingroup$ But on a second thought, that might not be a problem. From looking at your examples, one might have $(\lim E_n)^* = (\text{colim } E_n^*)^{**}$... This should be the case at least for reflexive spaces: $(\lim E_n)^* = (\lim E_n^{**})^* = (\text{colim } E_n^*)^{**}$. $\endgroup$ Commented May 20, 2016 at 17:19