Advanced Analysis

The Fourier Inversion Formula

Theorem (Fourier Inversion Formula)
Suppose \(f\) and \(\hat{f}\) are both continuous and absolutely improperly Riemann integrable on \({\mathbb R}^n\) (i.e., the improper integrals of \(|f|\) and \(|\hat{f}|\) exist.) Then for each \(x \in {\mathbb R}^n\),
\[ f(x) = \int_{-\infty}^{\infty} e^{2 \pi i x \cdot \xi} \hat{f}(\xi) d \xi. \]
Corollary
The Fourier inversion formula holds for all Schwartz functions \(f\).
The proof of the Fourier inversion formula can be understood in terms of three major steps:
Step 1 (Approximating the Inversion Integral)
We will show that for each \(x \in {\mathbb R}^n\),
\[{}\int e^{2 \pi i x \cdot \xi} \hat{f}(\xi) d\xi{}\]
\[{}= \lim_{\eta \rightarrow 0^+} \int e^{2 \pi i x \cdot \xi} e^{-\eta \pi ||\xi||^2} \hat{f}(\xi) d \xi.{}\]
Here \(||\xi||\) denotes the Euclidean norm of \(\xi\).
Proof
Meta (Main Idea)
For each \(R > 0\), let \(B_R := [-R,R]^n\). For any \(\epsilon > 0\), let \(R > 0\) be chosen so that
\[ \int_{B_R^c} |\hat{f}(\xi)| d \xi < \frac{\epsilon}{3}. \]
Then let \(\eta\) be chosen sufficiently small that
\[{}|1 - e^{-\eta \pi ||\xi||^2}|{}\]
\[{}< \frac{\epsilon}{3} (1 + (2R)^n \sup_{\xi \in B_R} |\hat{f}(\xi)|)^{-1}{}\]
for all \(\xi \in B_R\). Then show that, for all such small \(\eta\),
\[{}\Bigg| \int e^{2 \pi i \xi \cdot x} \hat{f}(\xi) d \xi{}\]
\[{}- \int e^{2 \pi i x \cdot \xi} e^{-\eta \pi ||\xi||^2} \hat{f}(\xi) d \xi \Bigg| < \epsilon{}\]
by splitting both integrals into parts \(\xi \in B_R\) and \(\xi \in B_R^c\). Combine the integrals on \(B_R\) into a single integral on that interval and use the triangle inequality. The integrand will be less than \(\epsilon(2R)^{-n}/3\) in magnitude at each point.
Step 2 (Fubination)
For each \(\eta > 0\),
\[{}\int e^{2 \pi i x \cdot \xi} e^{-\eta \pi ||\xi||^2} \hat{f}(\xi) d \xi{}\]
\[{}= \int f(y) \eta^{-\frac{n}{2}} e^{-\eta^{-1} \pi ||x-y||^2} dy.{}\]
Proof
Meta (Main Idea)
Establish the identity
\[{}\int e^{-2 \pi i \xi \cdot y} e^{2 \pi i \xi \cdot x} e^{-\eta \pi ||\xi||^2} d\xi{}\]
\[{}= \eta^{-\frac{n}{2}} e^{-\eta^{-1} \pi ||x-y||^2}{}\]
by making the substitution \(\xi \mapsto \eta^{-1/2} \xi\) and reducing to the fact that \(e^{-\pi ||\xi||^2}\) is its own Fourier transform. Then use the Multiplication Formula
\[ \int g(\xi) \hat{f}(\xi) d \xi = \int \hat{g}(y) f(y) dy, \]
which is proved using Fubini's Theorem and is valid when both \(g\) and \(f\) are improperly Riemann integrable on \({\mathbb R}\).
Step 3 (Approximate Identity)
For each \(x \in {\mathbb R}^n\),
\[ \lim_{\eta \rightarrow 0^+} \int f(y) \eta^{-\frac{n}{2}} e^{-\eta^{-1} \pi ||x-y||^2} dy = f(x). \]
Proof
Meta (Main Idea)
We use the fact that the integral of \(\eta^{-n/2} e^{-\pi \eta^{-1} ||x - \cdot||^2}\) is one to give that
\[{}\int f(y) \eta^{-\frac{n}{2}} e^{-\eta^{-1} \pi ||x-y||^2} dy - f(x){}\]
\[{}= \int f(y) \eta^{-\frac{n}{2}} e^{\eta^{-1} \pi ||x-y||^2} dy{}\]
\[{}- \int f(x) \eta^{-\frac{n}{2}} e^{-\eta^{-1} \pi ||x-y||^2} dy{}\]
\[{}= \int (f(y)-f(x)) \eta^{-\frac{n}{2}} e^{-\eta^{-1} \pi ||x-y||^2} dy.{}\]
Next, use the triangle inequality. For a given \(\epsilon\), chose some \(\delta\) such that \(|f(y) - f(x)| < \epsilon/2\) when \(||x-y|| < \delta\). Then choose \(\eta\) sufficiently small that both
\[ \int_{||y-x|| \geq \delta} |f(y)| \eta^{-\frac{n}{2}} e^{-\eta^{-1} \pi ||x-y||^2} dy \]
and
\[ |f(x)| \int_{||y-x|| \geq \delta} \eta^{-\frac{n}{2}} e^{-\eta^{-1} \pi ||x-y||^2} dy \]
are less than \(\epsilon/3\) (in the former case, it will be because \(\eta^{-n/2} e^{-\pi \eta^{-1} \delta^2}\) is very small, and in the latter, it will be because the integral over the region \(||y-x|| \geq \delta\) is very small).