Some thoughts on proving Fourier Inversion(Complex analysis method)

March 2025
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In complex analysis, we have the following well-known theorem:

Theorem If $f \in \mathfrak{F}$ , then the Fourier inversion holds, namely, given

$\widehat{f}(\xi) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i x \xi} \, dx \quad \text{for all } \xi \in \mathbb{R}.$

We have

$f(x) = \int_{-\infty}^{\infty} \widehat{f}(\xi) e^{2\pi i x \xi} \, d\xi \quad \text{for all } x \in \mathbb{R}.$

The proof is not easy for a beginner who just touched this theorem. As for myself, I spent a tons of time on managing the proof. So I would like to share the logical thoughts on proving this theorem.

It is intuitive to plug in the first equation into the second one and see if the relation holds. It will lead us to prove

$f(x) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x) e^{-2\pi i x \xi} \, dx e^{2\pi i x \xi} \, d\xi$

After some change of order, we have

$f(x) = \int_{-\infty}^{\infty} f(x)\int_{-\infty}^{\infty}e^{-2\pi i (x \xi-x \xi)} \, d\xi \, d x$

Problem arises. The integral $\int_{-\infty}^{\infty}e^{-2\pi i (x \xi-x \xi)} \, d\xi=\int_{-\infty}^{\infty}1 \, d\xi$ is unbounded. How could we solve this issue?

Well, if the exponential isn’t $e^{0}$ but instead $e^{-(a+bi)\xi}$ , where $a>0$ and $b\in \mathbb{R}\neq 0$ then the integratal from $0$ to infinity would exist, as the integral of $|e^-{(a+bi)\xi}|=e^{-a\xi}$ converges. We could easily compute the integral, which is $\int_{0}^{\infty}e^{-(a+bi)\xi}\, d\xi=\frac{1}{ai+b}$ .

This gives us a hint. One common technique we can do is integrating over a toy contour, often a rectangle or a circle. Here, the condition $f \in \mathfrak{F}$ , which gives moderate decay of $f$ , indicates the choice of rectangle. This is intuitive, as by our previous idea, we want to shift $x$ a bit so that the power of $e$ would be negative. A rectangle could achieve it by setting to sides, one on the real line while another on $\pm bi$ . When integrating along this curve, the two vertical sides would vanish by the condition given by the moderate decay. Hence, we have

$\widehat{f}(\xi) = \int_{-\infty}^\infty f(u - ib) e^{-2\pi i (u - ib) \xi} \, du,$

Plugging back into the integral, we can successfully integrate the exponential part (Here we need to separate the integral into positive and negative parts). We could get

$\begin{align*}\int_{0}^{\infty} f(\xi) e^{2\pi i x \xi} \, d\xi &= \int_{0}^{\infty} \int_{-\infty}^{\infty} f(u - ib) e^{-2\pi i (u - ib) \xi} e^{2\pi i x \xi} \, du \, d\xi\\&= \int_{-\infty}^{\infty} f(u - ib) \int_{0}^{\infty} e^{-2\pi i (u - ib - x) \xi} \, d\xi \, du\\&= \int_{-\infty}^{\infty} f(u - ib) \frac{1}{2\pi b + 2\pi i (u - x)} \, du\\&= \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{f(u - ib)}{u - ib - x} \, du\\&= \frac{1}{2\pi i} \int_{L_1} \frac{f(\xi)}{\xi - x} \, d\xi\end{align*}$

where $L_1$ is just the line $\text{Im}(z)=-b$ .

This hints us the residue formula,

$f(x)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi)}{\xi-x}\, d \xi$

Shifting the two integral back gives us the desired result!

Some thoughts on proving Fourier Inversion(Complex analysis method)

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