Is the dual of a complete topological vector space always complete? Let $X$ be a complete topological vector space (over $\mathbb{C}$ say), and $X'$ its dual with the weak*-topology. Then is $X'$ always complete? You may assume $X$ is locally convex if you like.
 A: In the weak* topology, the dual is a (topological) subspace of
$$\mathbb{C}^X = \prod_{x\in X} \mathbb{C}.$$
A product of complete spaces is always complete, so $\mathbb{C}^X$ is complete ($\mathbb{C}$ is complete), and thus $X'$ is complete in the weak* topology if and only if it is a closed subspace of $\mathbb{C}^X$. The algebraic dual $X^\ast$ (the space of all linear functionals, continuous or not) is a closed subspace of $\mathbb{C}^X$, hence complete.
If the topological and the algebraic dual coincide, $X'$ is hence complete in the weak* topology.
If $X$ is locally convex, then $X'$ is dense in $X^\ast$ in the topology of pointwise convergence, hence in that case $X'$ is complete if and only if $X' = X^\ast$.
To see that $X'$ is dense in $X^\ast$ for locally convex $X$, take any linear functional $\lambda\colon X\to\mathbb{C}$. A neighbourhood of $\lambda$ contains a set of the form
$$V(\lambda;\varepsilon,x_1,\dotsc,x_n) = \left\{ \mu \in X^\ast : \lvert \mu(x_i) - \lambda(x_i)\rvert < \varepsilon, 1 \leqslant i \leqslant n \right\}$$
for some $\varepsilon > 0$ and finitely many $x_i \in X$. We need to see that any such set contains a continuous functional. But $Y = \operatorname{span} \{x_1,\,\dotsc,\,x_n\}$ is finite-dimensional, hence $\lambda\lvert_Y$ is continuous. By Hahn-Banach, there is a $\varphi \in X'$ with $\varphi\lvert_Y = \lambda\lvert_Y$, and thus
$$\varphi \in V(\lambda;\varepsilon,x_1,\dotsc,x_n) \cap X',$$
even for all $\varepsilon > 0$, where only the $x_i$ are kept fixed.
