📂Complex Anaylsis

Proof of Jordan's Lemma

Theorem ¹

When the semicircle $\Gamma$ is represented as $z(\theta) = R e^{i \theta} , 0 \le \theta \le \pi$, if the function $f: \mathbb{C} \to \mathbb{C}$ is continuous on $\Gamma$ and $\displaystyle \lim_{z \to \infty} f(z) = 0$, then for any positive $m \in \mathbb{R}^{+}$, $$ \lim_{R \to \infty} \int_{\Gamma} e^{m i z } f(z) dz = 0 $$

Explanation

The pronunciation [Jordan] is not Konglish but comes from French. It’s a lemma, so it’s difficult to grasp its meaning immediately, and it’s enough to know that it’s used in various integration techniques. The proof may seem tedious, but it’s surprisingly straightforward, so it’s not a bad idea to study it properly at least once.

Fourier Transform

Fourier Transform: $$\mathcal{F}f(\xi):=\int _{-\infty} ^{\infty}f(x)e^{-i \xi x}dx$$

If $-\xi = m > 0$ and the continuous function $f : \mathbb{R} \to \mathbb{R}$ satisfies $\lim_{x \to \infty} f(x) = 0$, then formally, the integrand in Jordan’s lemma resembles the form appearing in the Fourier transform.

Proof

First, $$ \begin{align*} \left| \int_{\Gamma} e^{m i z} dz \right| =& \left| \int_{0}^{\pi} e^{m i z} dz \right| \\ \le & \int_{0}^{\pi} \left| e^{m i R \cos{\theta}} \right| \left| e^{- m R \sin{\theta}} \right| \left| i R e^{i \theta} \right| d \theta \\ =& R \int_{0}^{\pi} e^{-m R \sin \theta} d \theta \\ =& 2 R \int_{0}^{\pi/2} e^{-m R \sin \theta} d \theta \end{align*} $$ will be shown to be bounded. Since $0 = \sin 0$ and $$ {{ 2 } \over { \pi }} = \left. {{ d } \over { d \theta }} {{ 2 \theta } \over { \pi }} \right|_{\theta = 0} < \left. {{ d } \over { d \theta }} \sin \theta \right|_{\theta = 0} = 1 $$ therefore, in $\displaystyle \theta \in \left[ 0 , {{\pi} \over {2}} \right]$, $\displaystyle {{2} \over {\pi}} \theta \le \sin{\theta}$ is. If it goes up to an exponential function, then $\displaystyle e^{-m R \sin \theta} \le e^{-m R {{2} \over {\pi}} \theta }$, therefore, $$ \begin{align*} \left| \int_{\Gamma} e^{m i z} dz \right| \le & 2 R \int_{0}^{\pi/2} e^{-m R {{2} \over {\pi}} \theta} d \theta \\ =& 2R \left[ - {{ \pi } \over { mR 2 }} e^{-m R {{2} \over {\pi}} \theta} \right]_{0}^{ \pi / 2} \\ =& {{\pi} \over {m}} (1 - e^{-mR} ) \\ <& {{\pi} \over {m}} \end{align*} $$ namely, it has been shown to be bounded. From the assumption, since $f$ is continuous on $\Gamma$ and $\displaystyle \lim_{z \to \infty} f(z) = 0$, for any $\varepsilon >0$, $$ \left| {{1} \over {z}} \right| < \delta \implies |f(z)| < \varepsilon $$ a $\delta > 0$ that satisfies this will exist. Since $|z|=R$ on $\Gamma$, summarizing for $R$, $$ {{1} \over {R}} < \delta \implies |f(z)| < \varepsilon $$ namely, for any $\varepsilon >0$, $$ {{1} \over {R}} < \delta \implies \left| \int_{\Gamma} e^{m i z} f(z) dz \right| < \varepsilon \left| \int_{\Gamma} e^{m i z} dz \right| < {{ \varepsilon \pi} \over {m}} $$ a $\delta > 0$ that satisfies this exists, and the following is obtained: $$ \lim_{R \to \infty} \int_{\Gamma} e^{m i z } f(z) dz = 0 $$

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