Finite-dimensional inverse limits of double-dual spaces Let $k$ be a field and $\{V_i\}_{i \in I}$ a filtered projective system of $k$-spaces with transition maps $f_{ji}: V_j \rightarrow V_i$ for $i \leq j$ (for my purposes we may assume the index set is $\mathbb{N}$).  For each $i$, let $V_i^{**}$ be the double dual of $V_i$: here this is simply the algebraic dual, $Hom_k(Hom_k(V_i, k), k)$.  The $V_i^{**}$ also form an projective system.

If $\varprojlim V_i^{**}$ is finite-dimensional over $k$, do we have $\varprojlim V_i \simeq \varprojlim V_i^{**}$? (Here there is no assumption that the $V_i$ themselves are finite-dimensional.)

Since $V_i \hookrightarrow V_i^{**}$ canonically for each $i$ and projective limit is left-exact, it follows that $\varprojlim V_i \hookrightarrow \varprojlim V_i^{**}$.  Therefore the hypothesis implies that $\varprojlim V_i$ is also finite-dimensional, and isomorphic to its own double dual $(\varprojlim V_i)^{**}$.  It would suffice to show that there is an injection $\varinjlim V_i^* \hookrightarrow (\varprojlim V_i)^*$, as this would dualize to a surjection $(\varprojlim V_i)^{**} \twoheadrightarrow (\varinjlim V_i^*)^* = \varprojlim V_i^{**}$ and the proof would be complete by comparing dimensions.  There is a natural map $\varinjlim V_i^* \rightarrow (\varprojlim V_i)^*$ , induced by the duals of the structural maps $\pi_i: \varprojlim V_i \rightarrow V_i$, but even with the assumption that both source and target are finite-dimensional I can't seem to show that this map is injective (nor can I think of a counterexample).
 A: I think this is true. I'll try to prove the statement
$$
\varinjlim V_i^* \hookrightarrow (\varprojlim V_i)^*
$$
if the source is finite dimensional and $\varprojlim V_i$ is finite dimensional. I'll assume the indexing set is $\mathbb N$ but it should work for any cofiltered system.
First a reduction. If we let $V_i' = \operatorname{im}(\varprojlim V_i → V_i)$ then we can choose complements $V_i = V_i' \oplus V_i''$ such that $\{V_i''\}_i$ is an inverse system itself (using that $k$ is a field). Then the tower $V'$ is levelwise finite dimensional with limit $\varprojlim V_i$, and $\varprojlim V_i'' = 0$.
Now we have
$$
\varinjlim V_i^* = \varinjlim (V_i')^* \oplus \varinjlim (V_i'')^* → (\varprojlim V_i')^* \oplus (\varprojlim V_i'')^* = (\varprojlim V_i)^*,
$$
and on the left summands, this map is an isomorphism. So in the first statement we want to show, we can assume that $(\varprojlim V_i)^* = 0$, and hence $\varprojlim V_i = 0$.
If $\{V_i\}$ is pro-trivial (this means for every $i$ there's a $j>i$ such that $V_j → V_i$ is the zero map) then $\varinjlim V_i^* = 0$ as well, so let's assume it's not. Possibly by passing to a cofinal subtower, we can choose nonzero elements $α_i\colon V_i → k$ such that $α_i|_{\operatorname{im}V_{i+1}}=0$ and $α_i$ maps nontrivially to $V_{j}$ for $j<i$. All $α_i$ map trivially to the colimit $\varinjlim V_i^*$, but by their construction, we can take infinite sums $α_I = \sum_{i ∈ I} α_i$ for any $I \subseteq \mathbb N$ and those are nontrivial precisely if $I$ is infinite. Moreover, $α_I$ and $α_J$ are linearly independent in the colimit iff $I$ and $J$ differ at infinitely many places, and similarly for any finite set of $a_I$. Thus $\varinjlim V_i^*$ is infinite-dimensional.
