A (real) normed space $(V, \lVert \cdot \rVert\_V)$ is called strictly convex if for all $x, y \in V \setminus \{ 0 \}$ we have
\begin{equation}
\lVert x + y \rVert\_V = \lVert x \rVert\_V + \lVert y \rVert\_V \implies \exists c >0 \enspace x = cy. 
\end{equation}
A completion of a normed space $(V, \lVert \cdot \rVert\_V)$ is a Banach space $( W, \lVert \cdot \rVert\_W)$ such that there exists an isometric embedding $i \colon V \to W$ such that $i[V]$ is dense in $W$. It is unique up to a (surjective) isometry.




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Question
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If $(V, \lVert \cdot \rVert\_V)$ is a strictly convex (real) normed space, does it follow that its completion is, again, strictly convex?




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I'm not sure whether it (regardless of whether the answer is positive or negative) is a well-known fact, but my initial tries to find a counter-example have failed. It possibly is due to the fact that I tried some modifications of rather well-behaved spaces, like $\ell^p$ for $p \in (1,\infty)$, so they might have some other properties, which are preserved by completion.



Is a completion of strictly convex normed space strictly convex?
No, the completion of a strictly convex normed space can fail to be strictly convex.


To put it differently, there are non strictly convex Banach spaces with a dense strictly convex subspace.


Here is a possible construction. To make things easier, it is tempting to start with a space where there is a good control on the non strictly convex part. For example, let $E,F$ be you favorite separable infinite dimensional strictly convex Banach spaces (for example $\ell\_2$), and consider $E\oplus\_{\ell\_1} F$, that is the direct sum of $E$ and $F$ for the norm $\|(v\_1,v\_2)\| = \|v\_1\|\_E + \|v\_2\|\_F$. In that way, for $v=(v\_1,v\_2)$ and $w=(w\_1,w\_2)$, $\|v+w\| = \|v\|+\|w\|$ if and only if for each $i \in \{1,2\}$, $v\_i$ and $w\_i$ are positively proportional. But if $v$ and $w$ are not proportional, this implies that there is a nonzero linear combination of $v$ and $w$ that is in $0\oplus F$ or $E\oplus 0$. In other words, any subspace $V \subset E\oplus\_{\ell\_1} F$ which interesects trivially $E \oplus 0$ and $0 \oplus F$ is strictly convex.


But it is a small exercise that, given a finite number of infinite codimensional subspaces $E\_i$ of a separable Banach space, there is a dense subspace which does intersects them all trivially.


(solution: if $(x\_n)$ is a dense sequence, construct by induction a sequence $(y\_n)$ such that $\|x\_n-y\_n\| \leq \frac 1 n$ and $\operatorname{span}(y\_1,\dots,y\_n) \cap E\_i = \{0\}$ for all $i$; the space spanned does the job).


