Completness of orthnormal system

Question

I have a complex orthnormal system. \begin{align*} \phi_{k}(t)= e^{-i a t^2}e^{i k t} \end{align*} where $k\in\mathbb{Z}$ and $ a $ is a constant. How can I prove it is complete in $L^2[-\pi,\pi]$ with respect to inner product \begin{align*} \langle\phi_{n}(t),\phi_{m}(t)\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\phi_{n}(t)\overline{\phi_{m}(t)}~dt. \end{align*} I read one proof in "An Introduction to Non Harmonic Fourier Series" by R. M. Young. He proved that for orthonormal system $e^{i k t}$, $f\in L^2[-\pi,\pi]$ \begin{align*} \int_{-\pi}^{\pi}f(t)e^{i k t}~dt=0 \end{align*} then $ f=0$ a.e. Suppose \begin{align*} g(t)=\int_{-\pi}^{t}f(u)~du \end{align*} for $t\in [-\pi,\pi]$. Integration by parts shows that \begin{align*} \int_{-\pi}^{\pi}(g(t)-c)e^{i k t}~dt=0 \end{align*} for every constant $ c$ and $n\in \mathbb{Z}$. The problem is how to integrate with my system? It involves $e^{-i at^2}$.

Answer 1

Lemma: Suppose that $\{f_k\}_{k=1}^{\infty}$ is a complete orthonormal system in $L^2[-\pi,\pi]$, with respect to the usual inner product. Let $g\in L^2[-\pi,\pi]$ be a fixed element such that $g(t)\neq 0$ for all $t\in[-\pi,\pi]$. Then the system $\{gf_k\}_{k=1}^{\infty}$ is complete in $L^2[-\pi,\pi]$, that is, the only element of $L^2[-\pi,\pi]$ orthogonal to $gf_k$ for all $k$ is the zero element.

Proof. Suppose $\phi\in L^2[-\pi,\pi]$ is orthogonal to $gf_k$ for every $k$. Then $\phi\cdot g$ is orthogonal to $f_k$ for every $k$, and since $\{f_k\}$ is complete, we must have $\phi\cdot g=0$ in $L^2[-\pi,\pi]$. Therefore, $\phi\cdot g=0$ almost everywhere. Now if $\phi$ is not zero almost everywhere, then there is some subset $E\subset[-\pi,\pi]$ with positive measure such that $\phi$ is not zero on $E$, and so $\phi\cdot g$ is also non-zero on $E$, because $g$ is never zero. Subsequently, $\phi$ must be zero almost everywhere.

Now apply the lemma to $g=e^{-iat^2}$, and the result follows.