TITLE: extending a norm on an $\mathbb{R}$-vector space
QUESTION [2 upvotes]: Let $(V,\lVert\cdot \rVert_0)$ be a normed vector space and let $(\overline{V}, \overline{d_0}) $ be a completion of $(V,d_0)$ where $d_0(x,y) = \lVert x - y \rVert_0.$ Show that $\overline{V}$ is an $\mathbb{R}$-vector space and the norm $\lVert\cdot \rVert_0$ extends to a norm $\lVert\cdot \rVert$ on $\overline{V}.$

To show that $\overline{V}$ is an $\mathbb{R}$-vector space, one should show that it satisfies the properties of a vector space (e.g. associativity, distributivity, commutativity of addition, etc.) and I believe that is relatively routine, though shouldn't this use the property that $\overline{V}$ is complete and contains a dense vector space that's isomorphic to $V$ ?
Also, how should I extend the norm? Should I use the completion $\overline{d_0}$ and define $\lVert x\rVert := \overline{d_0}(x,0)$ (I'm quite sure all normed vector spaces contain zero)?

Or is it necessary to consider the functions $A: V\times V\to V, A(v,w) = v+w$ and $S_\alpha : V\to V, S_\alpha(v) = \alpha v, \alpha\in \mathbb{R}$?

REPLY [1 votes]: The metric completion is done in such a way that $V$ is dense in $\bar V$. You extend the norm to elements $x\in \bar V\setminus V$ by setting $\|x\|=\bar d_0(x,0)$ as you suggested. The fact that this is a norm follows from the continuity of the extended metric $d_0$ and the fact that this metric, restricted to $V$, is associated to a norm (you can take limits in all the properties: triangle inequality, homogeneity, etc.).
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