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. 



Dual of colimit in $\text{Ban}_1$
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 $\ell\_\infty$.


