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Fourier Inversion

Having defined the Fourier transform of an integrable function f : \mathbb R \to \mathbb C by \mathcal F\{f\} := \hat f : \mathbb R \to \mathbb C,

\displaystyle \hat f(\xi) := \frac 1{2\pi}\int_{-\infty}^{\infty} f(x) e^{-i \xi x}\, \mathrm dx,

can we take its inverse? That is, is there a reasonable definition of \mathcal F^{-1} such that \mathcal F^{-1} \circ \mathcal F = \mathcal F \circ \mathcal F^{-1} = \mathrm{id}?

We also need to consider a suitable collection \mathcal S of functions \mathbb R \to \mathbb C so that, in more technical parlance, \mathcal F : \mathcal S \to \mathcal S is bijective with inverse \mathcal F^{-1} : \mathcal S \to \mathcal S, but we will consider that later on.

Let f : \mathbb R \to \mathbb C be integrable with Fourier transform \hat f. Denote the set of such functions by L^1(\mathbb R). Furthermore, denote the set of functions \mathbb R \to \mathbb C continuous on \mathbb R by C(\mathbb R).

Lemma 1. For any f \in L^1(\mathbb R), \hat f \in C(\mathbb R).

Proof. For any \xi_n \to \xi, the sequence of functions \{ f(x) e^{-i \xi_n x} \} is dominated by the function |f(x)| and converges pointwise to f(x) e^{-i \xi x}. By the dominated convergence theorem,

\begin{aligned} \lim_{n \to \infty} \hat f(\xi_n) &= \lim_{n \to \infty} \frac 1{2\pi}\int_{-\infty}^{\infty} f(x) e^{-i \xi_n x}\, \mathrm dx \\ &= \lim_{n \to \infty} \frac 1{2\pi}\int_{-\infty}^{\infty} f(x) e^{-i \xi_n x}\, \mathrm dx \\ &= \frac 1{2\pi}\int_{-\infty}^{\infty} \lim_{n \to \infty} f(x) e^{-i \xi_n x}\, \mathrm dx \\ &= \frac 1{2\pi}\int_{-\infty}^{\infty} f(x) e^{-i \xi x}\, \mathrm dx = \hat f(\xi). \end{aligned}

Therefore, \hat f \in C(\mathbb R).

Following this article, we will prove the Fourier inversion formula.

Theorem 1 (Fourier Inversion Formula). For any f \in L^1(\mathbb R) \cap C(\mathbb R) such that \hat f \in L^1(\mathbb R),

\displaystyle f(x) := \int_{-\infty}^{\infty} \hat f(\xi) e^{i \xi x}\, \mathrm d\xi.

Proof. We first use the dominated convergence theorem to create an epsilon of room, and then use Fubini’s theorem to interchange the integrals:

\begin{aligned} \int_{-\infty}^{\infty} \hat f(\xi) e^{i \xi x}\, \mathrm d\xi &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} \underbrace{ \phantom{.} e^{-\epsilon^2 \xi^2/2 + i \xi x} \phantom{.}}_{g_x(\xi)}\cdot \hat f(\xi) \, \mathrm d\xi \\ &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} g_x(\xi) \cdot \int_{-\infty}^{\infty} f(y) e^{-i \xi y}\, \mathrm dy \, \mathrm d\xi \\ &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} f(y) \int_{-\infty}^{\infty} g_x(\xi) e^{-i \xi y}\, \mathrm d\xi\, \mathrm dy \\ &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} f(y) \cdot \hat g_x(y) \, \mathrm dy .\end{aligned}

By the Fourier transform of the Gaussian function h(\xi) = e^{-\epsilon^2 \xi^2/2 } so that \hat h(y) = \frac{1}{\epsilon \sqrt{2 \pi}} e^{-y^2/2\epsilon^2} =: \phi(y/\epsilon)/\epsilon, where \phi denotes the probability density function of \mathcal N(0, 1),

\begin{aligned} \hat g_x (y) &= \int_{-\infty}^{\infty} e^{-\epsilon^2 \xi^2/2 + i \xi x} e^{-i\xi y}\, \mathrm d\xi \\ &= \int_{-\infty}^{\infty} e^{-\epsilon^2 \xi^2/2 } e^{-i\xi (y-x)}\, \mathrm d\xi \\ &= \hat h(y-x) = \frac{1}{\epsilon } \cdot \phi \left( \frac{x-y}{\epsilon} \right). \end{aligned}

Now denote \phi_{\epsilon}(\cdot) := \hat h = \phi(\cdot/\epsilon)/\epsilon = \hat h so that \hat g_x(y) = \phi_{\epsilon}(x-y), and we have

\begin{aligned} \int_{-\infty}^{\infty} \hat f(\xi) e^{i \xi x}\, \mathrm d\xi &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} f(y) \cdot \phi_{\epsilon}(x-y) \, \mathrm dy.\end{aligned}

All that remains is to prove that the limit approaches f(x) as \epsilon \to 0. Define

\begin{aligned}\psi(\epsilon) := \int_{-\infty}^{\infty} f(y) \cdot \phi_{\epsilon}(x-y) \, \mathrm dy - f(x).\end{aligned}

Since \phi_{\epsilon}(x-\cdot) is the probability density function of some normal distribution, it has a total integral of 1, so that

\begin{aligned}\psi(\epsilon) = \int_{-\infty}^{\infty} (f(y) - f(x)) \cdot \phi_{\epsilon}(x-y) \, \mathrm dy .\end{aligned}

Fix \eta > 0. Since f is continuous at x, for any k > 0, there exists \delta > 0 such that

\displaystyle |x-y| < \delta \quad \Rightarrow \quad |f(x) - f(y)| < k \cdot \eta.

Therefore,

\begin{aligned} | \psi(\epsilon)| &\leq \left( \int_{|x-y| < \delta} + \int_{|x-y| \geq \delta} \right) |f(y) - f(x)| \cdot |\phi_{\epsilon}(x-y)| \, \mathrm dy \\ &\leq k \cdot \eta \cdot 2\delta + \phi_\epsilon(\delta) \cdot \int_{\mathbb R} |f(y) | \, \mathrm dy + 2|f(x)| \cdot \Phi_\epsilon(-\delta), \end{aligned}

where \Phi_\epsilon denotes the cumulative distribution function of \mathcal N(0, \epsilon). For sufficiently small \delta > 0, all three summands can be bounded above by \eta/3, so that |\psi(\epsilon)| < \eta. In particular, setting \eta = \epsilon, \phi(\epsilon) \to 0 as \epsilon \to 0^+, as required:

\begin{aligned} \int_{-\infty}^{\infty} \hat f(\xi) e^{i \xi x}\, \mathrm d\xi &= \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} f(y) \cdot \phi_{\epsilon}(x-y) \, \mathrm dy = f(x).\end{aligned}

This theorem is arguably one of the most important ones in Fourier analysis, though not the most powerful, since \hat f may not be Lebesgue-integrable even if f is (e.g. f = \mathbb I_{[-1,1]} implies \hat f(\xi) = \sin(\xi)/\pi \xi). Using Theorem 1 to define

\mathcal G : \mathcal F(L^1(\mathbb R) \cap C(\mathbb R)) \cap L^1(\mathbb R) \to L^1(\mathbb R) \cap C(\mathbb R)

by \mathcal G\{\hat f\} = f, we have proven that (\mathcal G \circ \mathcal F)\{f\} = f for suitable f.

In a limited sense then, \mathcal G = \mathcal F^{-1}, although a more proper proof requires bijectivity, which we are far from a complete proof, but thankfully do not need at the moment. What’s perhaps somewhat more surprising is the connection we recover with inverse Laplace transforms.

Theorem 2. For continuous Lebesgue-integrable f,g : [0,\infty) \to \mathbb C, if \mathcal L\{f\} = \mathcal L\{g\} is defined on \gamma + i \mathbb R, then f = g. Therefore, for any F = \mathcal L\{f\} defined on i \mathbb R, f = \mathcal L^{-1}\{F\} is well-defined.

Proof. Suppose F:=\mathcal L\{f\} = \mathcal L\{g\} =: G. We first prove the \gamma = 0 case. Given \xi \in \mathbb R such that F(i \xi) is well-defined,

\displaystyle \hat f(\xi) = \frac{1}{2\pi} \cdot \mathcal L \{ f\}(i\xi) = \frac 1{2\pi} \cdot F(i \xi).

Similarly, \hat g(\xi) = \frac 1{2\pi} \cdot G(i \xi). By Fourier inversion, \hat f = \hat g implies f = g.

For the \gamma \neq 0 case, we note that F(\gamma + \cdot) = G(\gamma + \cdot) is defined on i \mathbb R. By the previous result and the shift theorems,

\displaystyle e^{-\gamma t} \cdot f(t) = e^{-\gamma t}\cdot g(t) \quad \Rightarrow \quad f = g.

What if f is not integrable, say the simplest case f = \mathbb I_{[0, \infty)}? As much as I wanted to end complex analysis here, it would do the theory of inverse Laplace transforms injustice. We plug in this hole in our final post, next time.

—Joel Kindiak, 27 Aug 25, 1519H

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