If we take the inner product $\langle f, g \rangle = \displaystyle\int_{-\pi}^{\pi} f(t) \overline{g(t)} dt$ on $L^2 ((-\pi, \pi))$, which allows functions to $\mathbb{C}$, then it's not hard to check that $\Big(\frac{e^{int}}{\sqrt{2\pi}}\Big)_{n=1}^{\infty}$ is an orthonormal sequence in the space, but I'm having trouble showing it to be an orthonormal basis -- i.e. a complete orthonormal sequence. I tried using the characterisation of complete orthonormal sequences telling us that for any $f \in L^2 ((-\pi, \pi))$, $$\displaystyle\sum_{n=1}^{\infty}|\langle f, e_n \rangle|^2 = ||f||^2,$$ where we define $e_n =\frac{e^{int}}{\sqrt{2\pi}},$ but the products on the left were difficult because they're not as easily transformable into ordinary complex line integrals as $\langle e_m, e_n \rangle$ is . Trying to use the definition -- where $\langle f, e_n \rangle = 0$ $\forall n \in \mathbb{N}$ $\implies f=0$ -- but that presents the same difficulty.

Is there a simple way of proving the completeness of this sequence?

  • $\begingroup$ You only need to prove it for $f$ in a dense subspace of $L^2$. For example, step functions, or continuous functions. $\endgroup$ – Lord Shark the Unknown Apr 3 '18 at 11:21
  • $\begingroup$ OK, that sounds quite difficult too. But if that's the simplest approach, I guess there's just not one of those very neat solutions. $\endgroup$ – SPS Apr 3 '18 at 11:27
  • $\begingroup$ Have you had a course in Complex Variables? $\endgroup$ – DisintegratingByParts Apr 4 '18 at 0:41
  • $\begingroup$ Is $L^2((-\pi, \pi))$ the same as $L^2([-\pi, \pi])$? The answers seem all assume they're same. $\endgroup$ – namasikanam Sep 16 at 12:40

It isn't a basis - you need to let $n$ range over $\mathbb Z$. Specifcally, if $e_n(t)=\frac{e^{int}}{\sqrt{2\pi}}$, then $\{e_n\}_{n\in\mathbb Z}$ is a basis of $L^2(-\pi,\pi)$. The easiest way to prove this is to use Stone-Weierstrass.

Let $X:=\{f:[-\pi,\pi]\to\mathbb R: f\text{ is continuous, }f(-\pi)=f(\pi)\}$. Observe that we can identify $X$ with the set of continuous functions on the torus $\mathbb T$, which is a compact metric space. Now if we define $\mathcal A=\operatorname{span}\{e_n:n\in\mathbb Z\}$, $\mathcal A$ is a sub-algebra of $X$; it is clearly a subspace and it is closed under multiplication since $e_ne_m=\frac1{\sqrt{2\pi}}e_{n+m}$. We apply the Stone-Weierstrass theorem to deduce that $\mathcal A$ is dense in $X$ (in the uniform topology). There are three conditions we need to check:

  1. $\mathcal A$ is unital. This is clear since $\sqrt{2\pi}e_0=1$.
  2. $\mathcal A$ is closed under conjugation. Observe that $\overline{e_n(t)}=e_{-n}(t)$, so this follows too.
  3. $\mathcal A$ separates points. If $s,t\in(-\pi,\pi]$ with $s\neq t$ then either $\cos s\neq\cos t$ or $\sin s\neq \sin t$, so in particular $e_1(s)\neq e_1(t)$.

Hence the Stone-Weierstrass theorem applies: for every $f\in X$, there exists a sequence $\{f_k\}$ in $\mathcal A$ suchb that $f_k\to f$ uniformly. This implies $f_k\to f$ in $L^2$, and hence $\mathcal A$ is $L^2$-dense in $X$. Since $X$ is $L^2$-dense in $L^2(-\pi,\pi)$, the proof is complete.

  • $\begingroup$ Thanks for the answer. I've never heard of the Stone-Weierstrass theorem -- I was wondering more if an elementary proof existed, but I get the feeling it doesn't. $\endgroup$ – SPS Apr 3 '18 at 20:29
  • 1
    $\begingroup$ Stone-Weierstrass theorem (complelx version) in Wikipedia uses $C(X, \mathbb{C})$(complex locally continued functions) as function space rather than $L^2$ functions. Stone-Weierstrass theorem can also apply on $L^2$ functions? $\endgroup$ – namasikanam Sep 17 at 1:04
  • $\begingroup$ Uniform convergence implies $L^2$-convergence, as I mention in the last paragraph. $\endgroup$ – Jason Sep 17 at 18:53

One of the more interesting proofs of completeness uses complex analysis. Consider the expression $$ F(\lambda) = \frac{i}{1-e^{-2\pi i\lambda}}\int_{0}^{2\pi}e^{-i\lambda t}g(t)dt. $$ The function $F(\lambda)$ is holomorphic in $\mathbb{C}\setminus\mathbb{Z}$, with possible first order poles at $\lambda=0,\pm 1,\pm 2,\cdots$. The residue at $\lambda = n\in\mathbb{Z}$ is the Fourier coefficient, $$ \lim_{\lambda\rightarrow n}(\lambda-n)F(\lambda)=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-int}g(t)dt. $$ Suppose $g\in L^2[0,2\pi]$ is orthogonal to $e^{int}$ for all $n=0,\pm 1,\pm 2,\cdots$. Then $F(\lambda)$ extends to an entire function of $\lambda$. It can be shown that $F$ is uniformly bounded on squares centered at the origin of width $1,3,5,7,\cdots$. Revisiting the proof of Liouville's Theorem, it is not hard to conclude that $F$ is uniformly bounded on $\mathbb{C}$ and, hence, must be constant (and the constant must be $0$.) Hence, $F\equiv 0$ on $\mathbb{C}$. From that one obtains $\int_{0}^{2\pi}e^{-i\lambda t}g(t)dt=0$ for all $\lambda\in\mathbb{C}$, which leads to the conclusion that all derivatives are $0$ and, hence, $\int_{0}^{2\pi}t^n g(t)dt=0$ for $n=0,1,2,3,\cdots$. Because the polynomials are dense in $L^2[0,2\pi]$, it follows that $g=0$ a.e.. Therefore, the orthogonal complement of the span of $\{ e^{int} \}_{n=-\infty}^{\infty}$ in $L^2[0,2\pi]$ is $\{ 0 \}$, which proves that $\{ e^{int}/\sqrt{2\pi} \}_{n=-\infty}^{\infty}$ is a complete orthonormal basis of $L^2[-\pi,\pi]$.

  • $\begingroup$ Thanks for this. I'll look at it carefully and see if I understand properly before accepting. $\endgroup$ – SPS Apr 5 '18 at 13:50

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