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 $L^{2}[-\pi ,\pi]$

  • 2
    $\begingroup$ 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? $\endgroup$ Jun 9 '12 at 21:55
  • $\begingroup$ @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 $C^{*}[-\pi,\pi]$. $\endgroup$
    – aaronqli
    Jun 9 '12 at 22:01
  • $\begingroup$ @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. $\endgroup$
    – user31373
    Jun 9 '12 at 23:56

This question is already answered. Thanks to @Qiaochu Yuan.

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 $-\pi$.

  • $\begingroup$ so then please also "accept" your "answer", otherwise Community will keep on bumping your question from time to time, see here. In addition, you can turn this answer to the CW (community wiki) mod: Up/down votes won't affect your reputation. And I don't think, you'll up votes for that... $\endgroup$
    – draks ...
    Jun 10 '12 at 11:41
  • $\begingroup$ You might want to put the content of the answer too, not just to point out the comment. $\endgroup$
    – Asaf Karagila
    Jun 10 '12 at 21:02
  • $\begingroup$ Sorry. Now I have fixed it. $\endgroup$
    – aaronqli
    Jun 11 '12 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.