Proving the Fourier Inversion Formula for the Discrete Fourier Transform

analysisfourier analysisfourier seriesfourier transform

On pages 98-99 of Numerical Partial Differential Equations: Finite Difference Methods by J.W. Thomas, the author defines the discrete Fourier transform of a square-summable function as follows.

Definition 3.1.1. The discrete Fourier transform of $\mathbf{u} \in \ell^2$ is the function $\hat{u} \in L^2([-\pi,\pi])$ defined by
$$ \hat{u}(\xi) = \frac{1}{\sqrt{2\pi}} \sum_{m=-\infty}^{\infty} e^{-im \xi} \, u_m, \quad \text{for } \xi \in [-\pi,\pi]. $$

Then the author states the following Fourier Inversion Formula.

Proposition 3.1.2: If $\mathbf{u} \in \ell^2$ and $\hat{u}$ is the discrete Fourier transform of $\mathbf{u}$, then
$$ \hspace{2cm} u_m = \frac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^{im \xi} \, \hat{u}(\xi)\,d\xi \qquad \textit{for all } m \in \mathbb{Z}. $$

The author gives a "formal" proof that glosses over issues of convergence (specifically, the interchange of the sum and integral). I would like to prove the proposition rigorously and be able to justify each step. Here is my attempt so far:

Fix $m \in \mathbb{Z}$. Then by the definition of $\hat{u}$ we have
\begin{align*}
\frac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^{im \xi} \,\hat{u}(\xi) \,d\xi &= \frac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^{im \xi} \, \frac{1}{\sqrt{2\pi}} \sum_{n=-\infty}^{\infty} e^{-in \xi} \,u_n \,d\xi \\[5pt]
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \sum_{n=-\infty}^{\infty} e^{i(m-n)\xi} \,u_n \,d\xi.
\end{align*}

Now if we can switch the order of the sum and integral then the result will easily follow since
\begin{align*}
\frac{1}{2\pi} \sum_{n=-\infty}^{\infty} \int_{-\pi}^{\pi} e^{i(m-n)\xi} u_n \,d\xi &=
\frac{1}{2\pi} \sum_{n=-\infty}^{\infty} u_n \int_{-\pi}^{\pi} e^{i(m-n)\xi} \,d\xi \\[5pt]
&= \frac{1}{2\pi} \sum_{n=-\infty}^{\infty} u_n \cdot 2\pi \delta_{nm} \\[5pt]
&= u_m.
\end{align*}

So my question is: Why are we allowed to switch the order of the sum and integral?

I consulted Folland and according to Theorem 2.25 (p.55), if $(f_n)_{n=1}^{\infty}$ is a sequence in $L^1(\mathbb{R}^d)$ and $\sum_{n=1}^{\infty} \int |f_n| < \infty$, then $\int \sum_{n=1}^{\infty} f_n = \sum_{n=1}^{\infty} \int f_n$. However, I don't think we can apply this theorem if we are only assuming that $\mathbb{u} \in \ell^2$. If we assume that $\mathbb{u} \in \ell^1$, then we would have
\begin{align*}
\sum_{m=-\infty}^{\infty} \int \left| e^{-im \xi} \,u_m \right| \,d\xi = \sum_{m=-\infty}^{\infty} |u_m| < \infty
\end{align*}

and so the sum-integral swap would be permissible by the aforementioned theorem. But without assuming that $\mathbb{u} \in \ell^1$, can the sum and integral swap be justified?

Best Answer

Indeed, we cannot switch the order of the sum and the integral. For instance, when $\xi=0$ then $\sum_{n=-\infty}^{\infty} e^{i(m-n)\xi} \,u_n=\sum_{n=-\infty}^{\infty} u_n$. This series can diverge when ${\bf u}\in\ell^2\setminus\ell^1$, for instance, when $u_n=1/|n|$ for each nonzero integer $n$.

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