I have given the following theorem
Let $X$ be a metric space, with metric completion $\bar{X}$, and $Y$ be a complete metric space. Then a uniformly continuous map $f: X \to Y$ extends uniquely to $\bar{X}$.
which I want to use to show that
the metric-completion of a normed space is canonically a Banach space.
Obviously $\bar{X}$ is complete and normed, but I do not find a way to proof that $\bar{X}$ is a normed vector space, i.e. that $(\bar{+}): \bar{X} \times \bar{X} \to \bar{X}$ is continuous in that norm.
The way to go should be to proof that $(+): X \times X \to \bar{X}$ is uniformly continuous and then use the theorem stated above, but the uniform continuity of $(+)$ is where I stuck.
Best Answer
You can't really use that result as is, simply because the domain of $+$ is not $X$, but $X \times X$. You would really need to show that $\overline{X} \times \overline{X}$ is really the closure of $X \times X$. Of course, to do that, you'd need to nominate an appropriate norm (or a metric, at least) on $X \times X$.
I'm not convinced that it's worth doing. Instead, recall (or prove) that the sum of Cauchy sequences is Cauchy. Note that, if $(x_n), (y_n), (z_n)$ are Cauchy sequences with $(x_n)$ and $(y_n)$ equivalent, i.e. $\|x_n - y_n\| \to 0$, then $(x_n + z_n)$ and $(y_n + z_n)$ are equivalent (pretty obviously). This gives us a well-defined $+$ operation on the completion: $[x_n] + [y_n] := [x_n + y_n]$. You can do something similar with scalar multiplication.
Next comes the legwork. You have to show that $\overline{X}$ is a vector space with respect to the two operations. That means verifying all the axioms of a vector space, bar none. It's not a subspace of a known vector space, so you can't just show closure and non-emptiness.
You can use the theorem to extend the norm of $X$ to $\overline{X}$ (note that $\|\cdot\|$ is uniformly continuous, as $|\|x\| - \|y\|| \le \|x - y\|$). You then need to prove that this is a norm, using the axioms.
To conclude completeness, simply note that the metric on $\overline{X}$ agrees with the norm metric. You know that the completion is complete, hence $\overline{X}$ will therefore be a Banach space under the given norm.
I've brushed over a lot of detail! It's too much for me to write out, but none of it is tricky. In fact, most of it follows directly from $X$ being a vector space.