The following popped up when I was contemplating the completion of normed linear spaces.
Question: Let $E$ be an NLS (normed linear space) and $X$ a topological space. Let $E$ sit inside $X$ as a dense subset. Under what conditions can one ensure that there exists at most one NLS structure on $X$ that extends that on $E$?
Thoughts
If I demand that $X$ is a complete metric space with $$ d(x, y) = \| x - y\|\tag{1} $$ for $x, y\in E$, then the above does hold with many more consequences: There exists a unique Banach space structure on $X$ that extends the NLS structure of $E$ . However, I am wondering about weakening this seemingly stringent condition (1).
Say $X$ does admit an NLS structure that extends that on $E$. Then $X$ gets another topology from this extended norm. If this new topology coincides with $X$'s original topology, then $E$ is dense in the NLS $X$ as well. Now, the fact that continuous functions on a Hausdorff space are determined by their values on a dense subset of the domain, we can conclude that addition, scalar multiplication and norm (all continuous) are determined uniquely, as desired.
Note that this also implies that $X$'s original topology came from a metric satisfying (1).
Edit: I have managed to show the following:
Result. Let $E$ be an NLS that is contained in a complete metric space $X$ as a dense subset with its norm being uniformly equivalent to the metric inherited from $X$. Then there's a unique Banach space structure on $X$, topologically equivalent to the original metric on $X$, that extends the NLS structure on $E$. Further, the extended norm is uniformly related to the original metric on $X$ in the same way as the original norm was to the restricted metric on $E$.