If $X$ is a Banach space, and $X^{(n+1)} = (X^{(n)})^*$ where $X^{(0)} = X$, then the sequence $X^{(2n)}$ stabilizes iff $X$ is reflexive.
Consider $\hat{X} = \bigcup_{n=0}^\infty X^{(2n)}$ where $X^{(2n)}$ is identified as a subspace of $X^{(2n+2)}$ in the norm topology using the standard isometry into the double dual (more concretely, this could be defined as $\hat{X} = X\cup \bigcup_{n=0}^\infty \left(X^{(2n+2)}\setminus X^{(2n)}\right)$).
Clearly, $\hat{X}$ is a normed space with $||x|| := ||x||_{2n}$ where $||x||_{2n}$ is the norm of $x\in X^{(2n)}$.
Is $\hat{X}$ a Banach space? Does it have a name?
Is the completion of $\hat{X}$ reflexive?
I came up with the question myself while wondering about Banach spaces and their duals. Motivation is pure curiosity.
Note: I am working in ZFC, which means I include the axiom of choice.
As for 2), note that $X$ is a closed subspace of $\overline{\hat{X}}\subseteq(\hat{X})^{**}$, meaning that if the completion of $\hat{X}$ is reflexive, then $X$ is reflexive as its closed subspace. Conversely, if $X$ is reflexive then $\hat{X} = X$ is reflexive.
As for 1), if $X$ is not reflexive then $\hat{X}$ is not a Banach space since it's a countable union of closed subspaces with empty interior, so it's not a Baire space.