# Showing normed space is $\mathbb{R}$-vector space

Let $(V,\|\cdot\|_0)$ be a normed vector space with $\overline{V}$ being its completion. Show that $\overline{V}$ is an $\mathbb{R}$-vector space and that the norm $\|\cdot\|_0$ extends to a norm $\|\cdot\|$ on $\overline{V}$.

I don't quite understand this problem. What does it mean for a normed space to be an $\mathbb{R}$-vector space?

Also, does a norm on $V$ have extension on the completion of $V$ because it has to operate on the additional (w.r.t. $V$) elements in the completion of $V$? But then, how can a norm not have an extension on a completion, even if it is the same as the norm on $V$?

• $V$ is a normed $\mathbb{R}$-vector space. $x \in \overline{V}$ iff there exists a sequence $(x_n) \in V$ such that $\lim_{n \to \infty} \|x_n-x\| = 0$. By definition if $(y_n)$ is a Cauchy sequence in $V$ then $y = \lim_{n \to \infty} y_n$ is in $\overline{V}$. And $\overline{V}$ is a normed $\mathbb{R}$-vector space by setting $\|x-y\| = \lim_{n \to \infty} \|x_n-y_n\|$ and checking the axioms. The context is just to formalize how something like $L^1([0,1])$ is obtained by completion of the continuous functions on $[0,1]$ with the $L^1$ norm (which leads to the Lebesgue integral) – reuns Oct 26 '17 at 2:45
• @reuns Can you please clarify: is any normed vector space necessarily an $\mathbb R$-valued vector space? Then, since $V$ is an $\mathbb{R}$-valued vector space, for any sequence $(x_k)_k\subset V$ and any $k\in \mathbb{N}$, $\|x_k\|_0\in \mathbb{R}$, so that $\lim\limits_{k\to\infty} \|x_k\|_0 \in \mathbb{R}$, since $\mathbb{R}$ is complete. Thus $x\in \overline{V}$. Now define the completion norm as $\|x\| = \| \lim\limits_{k\to\infty} x_k \|_0$ and check that norm axioms hold. Do you think this is the right approach? – sequence Oct 26 '17 at 3:29
• This is the only approach. In functional analysis we look at $\mathbb{R}$ (or $\mathbb{C}$) vector spaces, ie. spaces $X$ where $au+bv$ is well-defined for $u,v \in X, a,b \in \mathbb{R}$ (or $\mathbb{C}$)... Example the space of real sequences with the $\sup$ norm. – reuns Oct 26 '17 at 3:34
• As $\mathbb(R)$ is a subfield of $\mathbb(C)$, any real or complex vector space can be regarded as a real vector space (that is, an $\mathbb(R)-$vector space). The fact that the norm extends to the completion in a unique way is quite immediate once you know what the completion is. Probably you introduced the completion of $V$ as the space of all Cauchy sequences of elements of $V$, quotiented on a certain equivalence relation. Thus $V$ is embedded in $\bar(V)$ upon an identification of certain elements of the latter with the elements of the former;the question is not completely trivial! – SiD Oct 26 '17 at 9:24