Let $H$ be the inner product space = $\{f: \mathbb{R} \to \mathbb{C} \mid f \text{ is continuous and has period }2 \pi\}$ where the inner product is: $$\langle f,g \rangle = \int_{0}^{2\pi}f(t)\overline{g(t)} dt $$ How do I prove that for $n \in \mathbb{Z}$ and $e_n(t)=\dfrac{1}{\sqrt{2\cdot\pi}} e^{i n t}$, $(e_n)_{n \in N}$ is a basis of $H$ (that it is orthonormal is easy).

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    $\begingroup$ One way is to show that the linear span of $\{e_n\}$ satisfies the hypothesis of the Stone-Weierstrass Theorem so that $\mbox{span }\{e_n\}$ is dense in $C(0,2\pi)$ with respect to the supremum norm and hence also the $L^2$-norm induced by the inner product. $\endgroup$ – user38355 Feb 27 '13 at 21:39
  • $\begingroup$ Hmm, i've thought of that but i can only use the Stone-Weierstrass Theorem for algebric polynomials. And i haven't quite figured out how to translate density in algebric polynomials to trigonometric ones. $\endgroup$ – benfstokes Feb 28 '13 at 10:59
  • $\begingroup$ Ah, it seems there a nomenclature confusion here. There's also a version of the theorem that applies to any non-vanishing algebra of functions that separates points. In my book (Rudin PMA) this is called the Stone-Weierstrass Theorem whereas the special case of polynomials is called just Weierstrass' Theorem. $\endgroup$ – user38355 Feb 28 '13 at 11:06
  • $\begingroup$ ...and the one about more general algebras applies trivially to the problem. But of course if you don't want to use the theorem there are other ways to go. $\endgroup$ – user38355 Feb 28 '13 at 11:08
  • $\begingroup$ Yeah, it would be easy that way. The answer probably lies in the generalization of the Weierstrass theorem to the Stone-Weierstrass theorem $\endgroup$ – benfstokes Feb 28 '13 at 11:10

This is probably not the most elegant way, but this method generalizes to showing the unitarity of the Fourier transform. The essential part of this proof is showing that the map $L^2(\mathbb{R}/2\pi \mathbb{Z}) \rightarrow \ell^2(\mathbb{Z})$ given by taking the Fourier coefficients from the discrete Fourier transform is an isometry, since the discrete Fourier transform is essentially the orthogonal projection onto the closure of the span of the functions $e^{inx}$. Then, by basic Hilbert space theory, it would follow that $\{e^{inx}\}$ is a maximal orthogonal set.

To start, let $\{a_n\} \subset \ell^2(\mathbb{Z})$ be the sequence of Fourier coefficients, a.k.a. $a_n = \langle f, e^{inx}/2\pi \rangle$. Then \begin{align*} \sum_{-n}^n |a_n|^2 & = \sum_{-n}^n \frac{1}{2 \pi} \int_0^{2\pi} \int_0^{2\pi} f(x) \overline{f(y)} e^{in(x-y)} \, dx \, dy \\ & = \frac{1}{2 \pi} \int_0^{2\pi} \int_0^{2\pi} f(x) \overline{f(y)} D_n(x-y) \, dx \, dy, \end{align*} where $D_n$ is the Dirichlet kernel given by $D_n(t) = \sum_{-n}^n e^{int}$. Now, we can prove several basic facts about this Dirichlet kernel:

  1. It integrates to $2\pi$.
  2. When $t \neq 0$, $D_n(t) \rightarrow 0$ as $n \rightarrow \infty$.

Therefore the Dirichlet kernel behaves as an approximate identity times $2\pi$, so $D_n(x-y)$ behaves like a Dirac mass about $x = y$. Substituting this in (one of course needs to use an appropriate theorem to justify integrating a limit), we have precisely that $$ \sum_{-n}^n |a_n|^2 \rightarrow \|f\|_{L^2}^2, \quad n \rightarrow \infty$$

  • $\begingroup$ Thanks! That works for $L^2(\mathbb{R}/2\pi \mathbb{Z}) $. I should've been clearer in my question though. I'm not working in $L^2$ but just in C($\\mathbb{R}/2\pi \mathbb{Z} \to \mathbb{C}$} which is not a Hilbert space with the $L^2$ norm. And so the all of the Hilbert space machinery is not available to me $\endgroup$ – benfstokes Feb 28 '13 at 10:52
  • $\begingroup$ Ah, ok. However, since that space is a subspace of $L^2$, then the result still holds by the same argument. $\endgroup$ – Christopher A. Wong Feb 28 '13 at 20:41
  • $\begingroup$ Makes sense. I'll try to prove that rigorously $\endgroup$ – benfstokes Feb 28 '13 at 20:46
  • $\begingroup$ Is this either Carleson's theorem or Riesz-Fischer theorem? I confuse them up :-) $\endgroup$ – Ciro Santilli TRUMP BAN IS BAD Jun 21 '20 at 20:15
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    $\begingroup$ @CiroSantilli郝海东冠状病六四事件法轮功 It's Riesz-Fischer. Carleson's theorem is a much more complicated result; pointwise convergence requires much more modern harmonic analysis methods. $\endgroup$ – Christopher A. Wong Jun 22 '20 at 23:29

This argument is from Rudin's Real and Complex Analysis (section 4.24 Completeness of the Trigonometric System). It considers $C(-\pi,\pi)$ instead of $C(0,2\pi)$, but this makes no difference of course.

Suppose we had trigonometric polynomials $Q_k \geq 0$ such that

  • $$\frac{1}{2\pi}\int_{-\pi}^{\pi} \! Q_k(t) \, dt = 1$$
  • $Q_k \to 0$ uniformly on $[-\pi, -\delta] \cup [\delta, \pi]$ for every $\delta >0$

(i.e. the $Q_k$ are a summation kernel). Associate to each continuous $f$ the function $$P_k(t) = f \ast Q_k(t) = \frac{1}{2\pi}\int_{-\pi}^\pi \! f(t-s)Q_k(s) \, ds.$$ Substitution shows that also $$P_k(t) = \frac{1}{2\pi}\int_{-\pi}^\pi \! f(s)Q_k(t-s) \, ds$$ so that each $P_k$ is a trigonometric polynomial. Now let $\epsilon > 0$. Since $f$ is uniformly continuous, there exists $\delta > 0$ such that $|f(t) - f(s)| < \epsilon$ whenever $|t-s| < \delta$. Since each $Q_k$ has average equal to one, we see that $$P_k(t) - f(t) = \frac{1}{2\pi}\int_{-\pi}^\pi \! (f(t-s) - f(s)) Q_k(s) \, ds$$ and the positivity of $Q_k$ implies $$|P_k(t) - f(t)| \le \frac{1}{2\pi}\int_{-\pi}^\pi \! |f(t-s) - f(s)|Q_k(s) \, ds$$ $$= \frac{1}{2\pi}\int_{|s| \le \delta} \! |f(t-s) - f(s)|Q_k(s) \, ds + \frac{1}{2\pi}\int_{|s|\geq \delta} \! |f(t-s) - f(s)|Q_k(s) \, ds$$ In the first term, the uniform continuity of $f$ shows that the integrand is smaller than $\epsilon Q_k(s)$ and the assumption on the average of $Q_k$ makes this term small. For the second the integrand converges uniformly to $0$ as $k \to \infty$ (since $f$ is bounded). Since the estimates are independent of $t$ we have shown that $$\|P_k - f\|_\infty < \epsilon$$ for $k$ large enough and thus the trigonometric polynomials are dense. We need to now construct the $Q_k$ with the desired properties. To do this we let $$Q_k(t) = c_k \left(\frac{1 + \cos t}{2}\right)^k$$ where the $c_k$ is chosen in such a way as to satisfy the assumption on the averages. Since the $Q_k$ are clearly positive, we only need to show that $Q_k \to 0$ uniformly away from $0$. But $Q_k$ is even (on $(-\pi,\pi)$) and decreasing on $[0,\pi]$. Thus for any $\delta > 0$ we have $$|Q_k(t)| \le Q_k(\delta) \le \frac{\pi(k+1)}{2}\left(\frac{1 + \cos \delta}{2} \right)^k \to 0$$ independently of $t$ as $k \to \infty$ since $$\frac{1 + \cos \delta}{2} < 1.$$

  • $\begingroup$ Hmm, i see now. Thanks for all the help! $\endgroup$ – benfstokes Feb 28 '13 at 20:40
  • $\begingroup$ @benfstokes Of course; I'm glad I could be of help. $\endgroup$ – user38355 Mar 3 '13 at 17:56
  • $\begingroup$ Umm, the proof all makes sense, except for one part: when estimating the distance between Pk and f, seems to me like the integral should have f(t) and not f(s). It doesn't break the proof. $\endgroup$ – Roee Shenberg Aug 15 '19 at 14:52

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