Showing that the functions $\sqrt{2x}\exp(2\pi inx^2)$ form a complete orthonormal system for $L^2([0,1])$

Question

How do I show that the functions

$$g_n(x) :=\sqrt{2x}\exp(2\pi inx^2)$$

where $n$ is a integer, are a complete orthonormal set in $L^2([0, 1])$?

I am relatively new to this and need some help getting started.

Answer 1

Hint:

$$\langle g_n, g_n\rangle =\int_{0}^{1}2x\exp(2\pi inx^2-2\pi i n x^2)dx$$

by definition of the scalar product $\langle,\rangle$ on $L^2([0,1])$. Then

$$\langle g_n, g_n\rangle = \int_{0}^{1}2xdx=2\frac{1}{2}x^2|^{1}_0=1,$$

for all $n$.

Now we consider $$\langle g_n, g_m\rangle =\int_{0}^{1}2x\exp(2\pi inx^2-2\pi i mx^2)dx= \int_{0}^{1}2x\exp(2\pi i(n-m)x^2)dx$$

with $m\neq n$. Using

$$\frac{d}{dx}\exp(2\pi i(n-m)x^2)=4\pi i(n-m)x\exp(2\pi i(n-m)x^2)$$

we arrive at

$$\langle g_n, g_m\rangle =\frac{2}{4\pi i(n-m)}\int_{0}^{1}\frac{d}{dx}\exp(2\pi i(n-m)x^2)dx=\frac{1}{2\pi i(n-m)}\exp(2\pi i(n-m)x^2)|^{1}_{0}= \frac{1}{2\pi i(n-m)}[\exp(2\pi i(n-m))-1]=0,$$

because $q:=n-m\in\mathbb Z$ and $\exp(2\pi i q)=1$.