One version of James's theorem in functional analysis states the following.
A Banach space is reflexive if, and only if, every bounded linear functional attains its norm on the unit sphere.
For example if $X$ is a reflexive Banach space, then by Banach Alaoglu the unit sphere is weakly compact, so every bounded linear functional attains its supremum. If $X$ is irreflexive, then James's theorem states there exists a bounded linear functional that does not attain its norm on the unit sphere, which is significantly harder to show.
My question is whether this theorem still holds when we drop the completeness requirement. So is a general normed space reflexive if, and only if, every bounded linear functional attains its norm on the unit sphere? Since incomplete spaces are never reflexive, the question is really whether every incomplete normed space admits a bounded linear functional that does not attain its norm on the unit sphere.
My first thought is to look at the completion $\overline{X}$ of an incomplete normed space $X$. If the completion is irreflexive then James's theorem already suffices, so we may assume $\overline{X}$ to be reflexive. Then for any vector $x$ on the unit sphere of $\overline{X}$ that is not in $X$ the Hahn Banach separation theorem gives a bounded linear functional $\phi$ which attains its norm at $x$. If $\overline{X}$ is strictly convex, then $\phi$ can not attain its norm anywhere else, so it does not attain its norm on the unit sphere of $X$.
However I am lost at what to do if $\overline{X}$ is reflexive but not strictly convex. I have tried to use Krein Milman to ensure $x$ is an extreme point of the unit sphere, but then I can not ensure that $x$ is not in $X$ and it does not even really solve the strict convexity problem, because an extreme point can still be an endpoint of a line segment in the unit sphere.
Side note: If $X$ is strictly convex, does it follow that $\overline{X}$ is also strictly convex?
