I have some doubt about the proof of the following theorem using SWT (Rudin 8.15):

If $f$ is continuous (with period $2\pi$) and if $\epsilon >0$, then there is a trigonometric polynomial $P$ such that $\forall\ x \in \Bbb{R}$, $|P(x)-f(x)|< \epsilon$.

The proof is as follows:

If we identify $x$ and $x +2\pi$ (typo by Rudin?), we may regard the $2\pi$-periodic functions on $\Bbb{R}$ as functions on the unit circle $T$, by means of the mapping $x \mapsto e^{ix}$. The trigonometric polynomials form a self adjoint subalgebra $\mathscr{A}$, which separates points on $T$ and vanishes at no point of $T$. Since $T$ is compact, SWT tells us that $\mathscr{A}$ is dense in $C(T)$. This is exactly what the theorem asserts.

I have already verified that $\mathscr{A}$ is an subalgebra of $C(T)$ which is self adjoint, separates points, and vanishes at no point. However, I think the theorem asserts that $\mathscr{A}$ is dense in the set of (complex valued) continuous 2$\pi$ periodic functions $C_{per}$, not exactly $C(T)$. The proof looks intuitive to me but how can we regard the periodic functions on $\Bbb{R}$ as functions on $T$, rigorously?

Shall I attempt to show that the mapping $\phi: C_{per} \to C(T)$ defined by $\phi(f) = f \circ \arg$ is a homeomorphism so that the density can be taken to $C_{per}$?

  • 1
    $\begingroup$ Rigorously, the map $f\mapsto f\circ e^{it}$ is from $C(T)\to C_{per}$ and yes, you can rather handily prove that said map is a surjective isometry if both spaces have the $\lVert \bullet\rVert_\infty$ norm. $\endgroup$ – user228113 Jul 11 '17 at 9:17
  • $\begingroup$ Yes. I fixed the typo. $\endgroup$ – Li Chun Min Jul 11 '17 at 9:18
  • $\begingroup$ By the way, Terence Tao's book Analysis II has a good explanation of this theorem. $\endgroup$ – littleO Jul 11 '17 at 9:22
  • $\begingroup$ That would a lot of things to prove though...(1) well defineness of $\phi$ ($\phi(f)$ is continuous on $T$), (2) $\phi$ is bijective, (3) $\phi: (C_{per}, \| \cdot \|_\infty) \to (C(T), \| \cdot \|_\infty) $ is continuous, (4) its inverse is continuous.... $\endgroup$ – Li Chun Min Jul 11 '17 at 9:25
  • $\begingroup$ @LiChunMin Four easy (and mandatory) things: if your purpose is rigorous explicitation of all the details, would you not prove that you can actually define such a map? By the way, surjective isometries are automatically (4),(2) and (3). $\endgroup$ – user228113 Jul 11 '17 at 9:29

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