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I have problems showing an orthonormal system is a basis. The question is as follows:

Show that $\left\{\frac{e^{int}}{\sqrt{2\pi}}e^{i\varphi(t)} \right\}_{n\in\mathbb{Z}}$ is indeed an orthonormal basis of $L^2[-\pi,\pi]$; with $\varphi:[-\pi,\pi]\rightarrow\mathbb{R}.$

Proving that is orthonormal is rather easy, because is showing $$\left< e_n,\overline{e_m} \right>=\begin{cases}1, & if\ n=m \\ 0, & if\ n\neq m\end{cases},$$ with $e_n:=\frac{e^{int}}{\sqrt{2\pi}}e^{i\varphi(t)}$.

But I have problems showing it is complete. I have tried showing $\left< f,e_n\right>=0, \forall f\in L^2[-\pi,\pi]$.

  1. How do I integrate $\int_{-\pi}^\pi f(t)\frac{e^{-int}}{\sqrt{2\pi}}e^{-i\varphi(t)}dt$? By parts?
  2. Do I have to prove it for a particular $\varphi(t)$? Like $\varphi(t)=t$ or $\varphi(t)=t^2$
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    You've misstated what you need to prove. To show it is a basis you should show that if $f \in L^2([-\pi,\pi])$ and $\langle f,e_n \rangle = 0$ for all $n$, then $f = 0$. – 2017-02-28
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    But how do I show that? – 2017-02-28

2 Answers 2

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I am assuming that you know that the basis $e_n(t) = {1 \over \sqrt{2 \pi}} e^{i n t}$ is complete. Let $b_n$ be the basis in the question.

The basis is complete iff $\langle b_n, f \rangle = 0$ for all $n$ implies $f = 0$.

Suppose $\langle b_n, f \rangle = 0$ for all $n$. Then $\langle e_n, t \mapsto e^{-i \psi(t)} f(t) \rangle = 0$ for all $n$ and so $t \mapsto e^{-i \psi(t)} f(t)$ is the zero function form which we get $f=0$.

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    I don't get, in the third line, when supposing $$ you arrive at $=0$ for all $n$. – 2017-02-28
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    You want to show that $\langle b_n, f \rangle = 0$ for all $n$ implies that $f=0$. If $\langle b_n, f \rangle = 0$, then if you write out the definition of the $L^2$ inner product, you will get $\int \overline{b_n} f = \int \overline{e_n(t) } e^{-i \psi(t)} f(t) dt = 0$. The latter can be written as $\langle e_n, g \rangle$ where $g(t) = e^{-i \psi(t)} f(t)$. – 2017-02-28
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    Ok, I get that. Thanks! One last thing, in the inner product should't it be $\int b_n\overline{f}$, instead of $\int \overline{b_n}f$? – 2017-02-28
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    It can be defined either way as long as the usage is consistent. I prefer linearity in the second parameter, but that is a personal preference. In any event, the above analysis is essentially the same. – 2017-02-28
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My favourite method is:

(1) Prove that your exponential functions are dense in $C[S^1]$ (i.e. the space of periodic continuous functions on $[-\pi, \pi]$ endowed with the $\sup $ norm). Since your exponential functions span a subspace that is closed under algebra operations and complex conjugation and includes the constants, this follows from the Stone-Weierstrass theorem.

(2) Use the fact that continuous functions are dense in $L^2[-\pi, \pi]$.

Note that the natural norm on $L^2[-\pi, \pi]$ is different from the sup norm on $C[-\pi,\pi]$, but this shouldn't trip you up.