# What is $\mathcal{C}(S^{1})$? (Where $S^1$ denotes unit circle)

What is $\mathcal{C}(S^{1})$ (Continuous function on a unit circle)? (Where $S^1$ denotes unit circle)

I saw this in a proof of showing Fourier Basis $S:=\{1,\sqrt{2}\cos{nx},\sqrt{2}\sin{nx}\}$ is an orthonormal basis of $L^{2}[-\pi ,\pi]$

The proof says $\mathcal{C}(S^{1})$ is equivalent to

$C^{*}[-\pi,\pi]=\{f\in C[-\pi,\pi]:f(-\pi)=f(\pi)\}$

then used the facts

S is dense in $\mathcal{C}(S^{1})$

$C^{*}[-\pi,\pi]$ is dense in $C[-\pi,\pi]$

and $C[-\pi,\pi]$ is dense in $L^{2}[-\pi ,\pi]$

• What do you mean? It is the space of continuous functions on the unit circle. It doesn't really have a simpler description than that. (When you say "dense" you should specify what topology you're referring to.) What step of the proof don't you understand? Jun 9 '12 at 21:55
• @QiaochuYuan Now I got it. It's just continuous functions on a unit circle in $R^{2}$. I thought it was talking about something related to complex plane or isomorphic transformation, then I wondered why it was equivalent to $C^{*}[-\pi,\pi]$. Jun 9 '12 at 22:01
• @Polymorpher I think you should post your comment as an answer to the question. Otherwise the users will be opening this "unanswered" question only to find there is nothing to do here.
– user31373
Jun 9 '12 at 23:56

Here $\mathcal{C}(S^{1})$ is merely space of functions defined on unit circle in $\mathbb{R}^{2}$ sense. It make sense to say this is equivalent to space of functions defined on $[-\pi,\pi]$ that have the same value on $\pi$ and $-\pi$.