Prove that complex exponentials $(\phi_n(x)=\frac{1}{\sqrt{a}}e^{i\frac{2\pi nx}{a}})_{n=0, \pm 1, \pm2,...}$ are complete and orthonormal on $[0,a]$

Question

I want to prove that the complex exponentials $(\phi_n(x)=\frac{1}{\sqrt{a}}e^{i\frac{2\pi nx}{a}})_{n=0, \pm 1, \pm2,...}$ form a complete orthonormal system for the space of square-integrable functions on the interval $[0,a]$. I have shown orthonormality, which wasn't too hard. Now I want to show completeness. I believe the easiest way to show completeness is to show the completeness relation: $\sum_{n=-\infty}^{\infty}\phi_n^*(x) \phi_n(x')= \delta(x-x'). $ (see https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/lecture-notes/MIT8_05F13_Chap_01.pdf for an explantion of this relation). The last time I wanted to show that some function was equal to the dirac delta function, I integrated it's product with an appropriate test function $f(x)$, and proved that it acted just like the dirac delta. In this case, this would look like so:$\int \sum_{-\infty}^{\infty}\phi_n^*(x) \phi_n(x') \cdot f(x)dx= \int \delta(x-x')\cdot f(x) dx = f(x'). $ So I want to show (assuming my approach works) $\int \sum_{n=-\infty}^{\infty}\phi_n^*(x) \phi_n(x') \cdot f(x)dx= \int\sum_{n=-\infty}^{\infty}\frac{1}{\sqrt{a}}e^{-i\frac{2\pi nx}{a}} \frac{1}{\sqrt{a}}e^{i\frac{2\pi nx'}{a}}f(x) dx = f(x') $. First of all is this a good approach? And second, how would you proceed from here?

Answer 1

I'll try to reduce it to standard Fourier theory. We begin by substituting $y= x/a$ to obtain

\begin{align}&\int_{0}^{a}\sum_{n=-\infty}^{\infty}\frac{1}{\sqrt{a}}e^{-i\frac{2\pi nx}{a}} \frac{1}{\sqrt{a}}e^{i\frac{2\pi nx'}{a}}f(x) dx = \sum_{n=0}^{\infty}e^{i\frac{2\pi nx'}{a}}\int_{0}^{1} e^{-i2\pi ny}f(ay)~\mathrm dy \\ &= \sum_{n=0}^{\infty} e^{i\frac{2\pi n x'}{a}} \hat{f_a}(n). \end{align} According to standard fourier theory, this is the Fourier series of the funtion $f_a(x) := f(ax)$ evaluated at the point $x'/a$. Accordingly, we find $$\sum_{n=0}^{\infty} e^{i\frac{2\pi n x'}{a}} \hat{f_a}(n) = f(x').$$