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Let $\mathbb{T}=\{ z\in \mathbb{C}\; |\; |z|=1\}$ denote the unit circle. Define $L^2 (\mathbb{T})=\{ f:\mathbb{T}\to \mathbb{C}\; |\; f\; \text{is measurable and}\; \int_{\mathbb{T}}|f(z)|^2 dz<\infty\}$. My question is that why $(z^n)_{n\in \mathbb{Z}}$ is a basis for $L^2 (\mathbb{T})$?

$L^2 (\mathbb{T})$ is a Hilbert space with $$\langle f,g\rangle =\int_{\mathbb{T}}f(z)\overline{g(z)}dz,\quad f,g\in L^2 (\mathbb{T}).$$

I could porve that $(z^n)_{n\in \mathbb{Z}}$ is orthonormal but I don't have any idea about how to show it is a basis.

M.Ramana
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  • Where are you getting stuck in showing that it's a basis? – jgon Oct 11 '22 at 18:00
  • I assume you're talking about a Hilbert basis? Since otherwise you'd be limited to just polynomials. – Alan Oct 11 '22 at 18:05
  • The complex version of the Stone-Weierstrass – Mittens Oct 11 '22 at 18:11
  • @jgon I could prove that $(z^n)_{n\in \mathbb{Z}}$ is orthonormal but I don't have any idea about how to prove it is a basis. – M.Ramana Oct 11 '22 at 18:11
  • @Alan Yes, $L^2 (\mathbb{T})$ is a Hilbert space. – M.Ramana Oct 11 '22 at 18:12
  • There probably is a better duplicate target somewhere but that was the best I could find. As indicated in the answer there, you can first show that the pan of the $z^n$ is dense in $C(\mathbb{T})$ by Stone-Weierstrass, and then it follows that it is also dense in $L^2(\mathbb{T})$. – Eric Wofsey Oct 11 '22 at 19:57
  • @EricWofsey Yes, Understood. Thank you so much for your help and the link. I appreciate it. – M.Ramana Oct 17 '22 at 03:53

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