Advanced Analysis

The Fourier Transform

1. Definition and Basic Properties

Definition
For any improperly Riemann integrable function \(f\) on \({\mathbb R}^n\), the Fourier transform \(\hat{f}\) is the function on \({\mathbb R}^n\) given by
\[ \hat{f}(\xi) := \int e^{-2 \pi i x \cdot \xi} f(x) dx. \]
Note
If the function \(f\) is improperly Riemann integrable or is a Schwartz function and so is \(f(x) e^{-2\pi i x \cdot \xi}\) for each \(\xi \in {\mathbb R}^d\). Thus the integral defining the Fourier transform always exists in the improper Riemann integral sense.
Lemma (Riemann-Lebesgue)
If \(f\) is improperly Riemann integrable on \({\mathbb R}^n\), then \(\hat{f}\) is continuous and tends to \(0\) at infinity. Moreover,
\[{}|\hat{f}(\xi)| \leq \int |f| dx {}\]
for all \(\xi \in {\mathbb R}^n\).
Proof
Meta (Main Ideas)
The quantitative bound on the size of \(\hat{f}\) can be achieved using the triangle inequality.

To show that \(\hat{f}\) is continuous, choose some ball \(B_R\) such that \(\int_{B_R^c} |f| < \epsilon/4\) and then argue that
\[{}\Bigg| \int_{B_R} e^{2 \pi i x \cdot \xi_1} f(x) dx{}\]
\[{}- \int_{B_R} e^{2 \pi i x \cdot \xi_2} f(x) dx \Bigg|{}\]
\[{}< \epsilon/2{}\]
when \(||\xi_1 - \xi_2||\) is sufficiently small.

To show that \(|\hat{f}(\xi)| \rightarrow 0\) as \(||\xi|| \rightarrow \infty\), you can follow the proof of Riemann-Lebesgue for Fourier series and approximate \(f\) by simple functions on some large box, where in this case simple functions would be finite linear combinations of indicator functions of boxes.
Proposition (Multiplication Formula)
If \(f\) and \(g\) are improperly Riemann integrable on \({\mathbb R}^n\), then
\[ \int \hat{f} g = \int f \hat{g}. \]
Proof
Meta (Main Ideas)
There are several reasonable approaches here. One idea is to prove the formula when both \(f\) and \(g\) are indicator functions of boxes (which will boil down to applying the definition of the Fourier transform and using Fubini's Theorem). Then use linearity to deduce that the formula remains true when \(f\) and \(g\) are finite linear combinations of indicator functions.
If \(f_a\) and \(g_a\) are simple functions approximating \(f\) and \(g\), then use the observation
\[{}\int \hat{f} g - \int \hat{f}_a g_a{}\]
\[{}= \int \widehat{(f - f_a)} g + \int \hat{f}_a (g-g_a){}\]
combined with the Riemann-Lebesgue lemma to deduce that \(\int \hat{f} g\) and \(\int \hat{f}_a g_a\) must be close to one another when the approximations are very close to \(f\) and \(g\) in the integral sense.

2. Fourier Transform Symmetries

Proposition (Symmetries Involving Fourier Transforms and Differentiation)
Suppose \(f\) is a Schwartz function on \({\mathbb R}^d\).
  • If \(e_j\) is the unit vector pointing in the \(j\)-th coordinate direction,
    \[{}\lim_{h \rightarrow 0} \frac{\hat{f}(\xi + h e_j) - \hat{f}(\xi)}{h}{}\]
    \[{}= \int e^{- 2 \pi i x \cdot \xi} (-2 \pi i x \cdot e_j) f(x) dx{}\]
  • Similarly,
    \[ (2 \pi i \xi \cdot e_j) \hat{f}(\xi) = \int e^{-2 \pi i x \cdot \xi} \frac{\partial f}{\partial x_j}(x) dx. \]
Proof
Meta (Main Ideas)
For the first property, fix a \(\xi \in {\mathbb R}^d\). Simplifying exponentials gives that
\[{}\frac{\hat{f}(\xi + h e_j) - \hat{f}(\xi)}{h}{}\]
\[{}= \int e^{-2 \pi i x \cdot \xi} \left( \frac{e^{- 2 \pi i x \cdot h e_j}-1}{h} \right) f(x) dx.{}\]
The function \(\varphi(z) := | z + 2 \pi i x \cdot e_j|\) is a convex function of \(z \in {\mathbb C}\), so by the Mean Value Theorem,
\[{}\left| \frac{e^{- 2 \pi i x \cdot h e_j}-1}{h} + 2 \pi i x \cdot e_j \right|{}\]
\[{}\leq \left| - 2 \pi i (x \cdot e_j) e^{- 2 \pi i x \cdot e_j \tilde h}+ 2 \pi i x \cdot e_j \right|{}\]
for some \(\tilde h\) between \(0\) and \(h\). A second application of the Mean Value Theorem gives that
\[{}|-e^{-2 \pi i x \cdot e_j \tilde h} + 1|{}\]
\[{}\leq |(2 \pi i x \cdot e_j) \tilde h e^{-2 \pi i x \cdot e_j \tilde h}|{}\]
\[{}\leq |2 \pi i x \cdot e_j| |h|.{}\]
Therefore if \(g(x) := (-2 \pi i x \cdot e_j) f(x)\),
\[{}\left| \frac{\hat{f}(\xi + h e_j) - \hat{f}(\xi)}{h} - \hat{g}(\xi) \right|{}\]
\[{}\leq |h| \int |2 \pi x \cdot e_j|^2 |f(x)| dx.{}\]
The integral on the right-hand side is finite because \(f\) decays rapidly, so the right-hand side tends to zero as \(h \rightarrow 0\).

For the second property, use the integration by parts formula for Schwartz functions on the integral from the right-hand side.
Corollary (The Fourier Transform Maps Schwartz Space to Itself)
The Fourier transform \(\hat{f}\) is also a Schwartz function.
Proof
Meta (Main Idea)
Show that \(\xi^\beta \partial_\xi^\alpha \hat f(\xi)\) is the Fourier transform of \(\partial^\alpha_x ( x^\beta f(x)) (-2 \pi i)^{|\beta|} (2 \pi i)^{-|\alpha|}\). Then establish that this latter function is a Schwartz function and use Minkowski's inequality to show that its Fourier transform is a bounded function.
Example (The Gaussian)
The funciton \(e^{- \pi ||x||^2}\) (where \(||\cdot||\) is the Euclidean norm on \({\mathbb R}^d\)) is its own Fourier transform.
Proof
Meta (Main Ideas)
By Fubini's Theorem applied to a cube centered at the origin combined with a limiting argument as the side length tends to infinity, it suffices to prove the result in one dimension. Let \(f(\xi)\) be the Fourier transform of \(e^{-\pi x^2}\). By the computations above, \(f'(\xi)\) is the Fourier transform of \(-2 \pi i x e^{-\pi x^2}\). But \(- 2 \pi i x e^{-\pi x^2}\) is itself the derivative of \(i e^{- \pi x^2}\), so using the integration by parts formula gives that \(f'(\xi)\) is \((2 \pi i \xi)i\), or merely \(-2 \pi \xi\), times \(f\) itself. In other words \(f'(\xi) = - 2 \pi \xi f(\xi)\). By the product rule, the derivative of \(e^{\pi \xi^2} f(\xi)\) is zero, which means it is constant. Thus \(f(\xi) = C e^{- \pi \xi^2}\) for some constant \(C\).

The constant \(C\) can be computed by a trick: by definition of the Fourier transform, \(C e^{-\pi 0^2}\) must equal the integral of \(e^{-\pi x^2}\). The integral can be computed by squaring it, using Fubini, and converting to polar coordinates (after a suitable limiting argument as the outer radius tends to infinity), i.e.,
\[ C^2 = \int_0^\infty \int_0^{2 \pi} e^{-\pi r^2} r d \theta dr = 1. \]

3. Illustrations of the Action of the Fourier Transform

The plots below illustrate the behavior of the Fourier transform in a number of important situations. The first few plots demonstrate the effect of dilations, translations, and modulations:
Figure. Functions \(f\) and Fourier transforms \(\hat{f}\) illustrating translation, dilation, and modulation symmetries

The first plot is simply \(e^{-\pi x^2}\), which is its Fourier transform. The Fourier transform exhibits dilation symmetry: if \(f_a (x) := f(a^{-1} x)\) for some constant \(a > 0\), then \(\widehat{f_a}(\xi) = a \widehat{f} (a \xi)\). In other words, dilations on the “physical side” correspond to inverse dilations (and an overall multiplicative factor) on the “frequency” side. The second plot above is an example: If we dilate \(f\) to become \(e^{-\pi (3 x)^2}\), the Fourier transform gets an inverse dilation to become \(\frac{1}{3} e^{-\pi (\xi/3)^2}\). Other important symmetries are translation and modulation: if \(\tau_h f(x) := f(x-h)\) and \(m_a f(x) = e^{2 \pi i a x} f(x)\), then \(\widehat{\tau_h f} = m_{-h} (\xi) \widehat{f}(\xi)\) and \(\widehat{m_a f} = \tau_a \widehat{f}\). We see above that multiplying \(e^{-\pi x^2}\) by \(\cos(4 \pi x)\) makes the Fourier transform split into two lumps at frequencies \(\xi = \pm 2\). Similarly, multiplying by \(\sin 4 \pi x\) also makes two lumps on the Fourier side at frequencies \(\pm 2\), but one shows up with a minus sign (and there's an overall factor of \(i\)). Multiplying by higher frequencies would move the lumps further away from the origin (note that we consider \(e^{4 \pi i x}\) to be a “pure” frequency, which is why \(\cos (4 \pi x) = (e^{4 \pi i x} + e^{-4 \pi i x})/2\) looks like a sum of two pure frequencies).

Below we have more examples, this time illustrating the relationship between regularity and decay. Roughly speaking, smoother functions have Fourier transforms which decay more rapidly and vice-versa. This can be understood rigorously via the identities above for differentiation and multiplication by monomials.
Figure. Functions \(f\) and Fourier transforms \(\hat{f}\) illustrating the relationship between regularity and decay
In the first example, the function \(f\) is the indicator function of \([-1/2,1/2]\); its Fourier transform is \(\frac{\sin \pi \xi}{\pi \xi}\), which decays like \(O(|\xi|^{-1})\). The next function is a tent, which happens to be the convolution of the indicator function with itself. On the Fourier side, convolutions become products, so the Fourier transform is \(\left(\frac{\sin \pi \xi}{\pi \xi}\right)^2\), which has decay \(O(|\xi|^{-2})\) as \(\xi \rightarrow \pm \infty\). In the latter cases, we have another function with a discontinuity; its Fourier transform also decays like \(O(|\xi|^{-1})\). The last function is \(e^{-\pi |x|}\) which is \(C^0\) but not \(C^1\). Its Fourier transform is \(\frac{2}{\pi (1 + 4 \xi^2)}\).