# completion of $C^{\infty}\left(S\right)$ is $L^2(S)$?

I have the space of infinitely derivable functions, i.e. $C^{\infty}\left(S\right)$ with the following inner product $$\left\langle f,g\right\rangle =\intop_{0}^{2\pi}\intop_{0}^{2\pi}\left(\overline{f\left(\theta,\varphi\right)}g\left(\theta,\varphi\right)\sin\theta\right)d\theta d\varphi,$$ I would like to understand what does it mean that the completion of this space is $L^2(S)$?

Edit: If it's possible I would also like to understand why $L^2(S)$ is the completion of $C^{\infty}\left(S\right)$ with this inner product. I found that this might be a quite general statement so I'd like to know if there's a theorem of some kind abou it. Thanks

A) Have a look at https://en.wikipedia.org/wiki/Complete_metric_space ex: $\mathbb{R}$ is the completion of $\mathbb{Q}$ for the natural distance.
B) Here, your scalar product defines a norm, so a distance. For this distance, the completion of $\mathcal{C}^{\infty}(S)$ is $L^2(S)$.
To be more precise, any Cauchy sequence of $\mathcal{C}^{\infty}(S)$ cannot converge to an element of $\mathcal{C}^\infty(S)$ but it does converge to an element of $L^2(S)$ if you considered it as a sequence of $L^2(S)$ (here $\mathcal{C}^{\infty}(S)\subset L^2(S)$ because $S=[0,2\pi]\times[0,2\pi]$ is compact)
• ok thank, you but do you have any idea why is this? I mean why this particular metric space has his completion in $L^2(S)$ – Dac0 Apr 20 '16 at 8:58
• 1) $L^p$ spaces are Banach spaces when $p>1$ – MJ73550 Apr 20 '16 at 9:38
• I thought answer was quite clear mathworld.wolfram.com/L2-Space.html when $X$ is a compact metric space $C^\infty(X)\subset L^2(X)$ why $C^\infty(S)$ is not complete is because you can construct a counter example (take any approximation of the identity en.wikipedia.org/wiki/…) – MJ73550 Apr 20 '16 at 9:50