📂Complex Anaylsis

Proof of Jordan's Lemma

Theorem ¹

When the semicircular arc $\Gamma$ 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 a positive number $m \in \mathbb{R}^{+}$, $$ \lim_{R \to \infty} \int_{\Gamma} e^{m i z } f(z) dz = 0 $$

Explanation

The pronunciation “Jordan” is not Konglish but comes from French. Being a lemma, its meaning is hard to grasp right away, and it is enough to just know that it is used in various integration techniques. The proof looks tedious, but it is surprisingly not much of a big deal, so it is 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 of Jordan’s lemma is similar to the form that appears in the Fourier transform.

Proof

First, we will show that $$ \begin{align*} \left| \int_{\Gamma} e^{m i z} dz \right| =& \left| \int_{0}^{\pi} e^{m i z} dz \right| \\ \le & \int_{0}^{\pi} \left| e^{m i R \cos{\theta}} \right| \left| e^{- m R \sin{\theta}} \right| \left| i R e^{i \theta} \right| d \theta \\ =& R \int_{0}^{\pi} e^{-m R \sin \theta} d \theta \\ =& 2 R \int_{0}^{\pi/2} e^{-m R \sin \theta} d \theta \end{align*} $$ is bounded. Since $0 = \sin 0$ and $$ {{ 2 } \over { \pi }} = \left. {{ d } \over { d \theta }} {{ 2 \theta } \over { \pi }} \right|_{\theta = 0} < \left. {{ d } \over { d \theta }} \sin \theta \right|_{\theta = 0} = 1 $$ we have $\displaystyle {{2} \over {\pi}} \theta \le \sin{\theta}$ for $\displaystyle \theta \in \left[ 0 , {{\pi} \over {2}} \right]$. Raising this to the exponential function, since $\displaystyle e^{-m R \sin \theta} \le e^{-m R {{2} \over {\pi}} \theta }$, $$ \begin{align*} \left| \int_{\Gamma} e^{m i z} dz \right| \le & 2 R \int_{0}^{\pi/2} e^{-m R {{2} \over {\pi}} \theta} d \theta \\ =& 2R \left[ - {{ \pi } \over { mR 2 }} e^{-m R {{2} \over {\pi}} \theta} \right]_{0}^{ \pi / 2} \\ =& {{\pi} \over {m}} (1 - e^{-mR} ) \\ <& {{\pi} \over {m}} \end{align*} $$ that is, we have shown that it is bounded. By assumption, $f$ is continuous on $\Gamma$ and $\displaystyle \lim_{z \to \infty} f(z) = 0$, so for any $\varepsilon >0$ there exists a $\delta > 0$ satisfying $$ \left| {{1} \over {z}} \right| < \delta \implies |f(z)| < \varepsilon $$ Since $|z|=R$ on $\Gamma$, rearranging with respect to $R$ gives $$ {{1} \over {R}} < \delta \implies |f(z)| < \varepsilon $$ that is, for any $\varepsilon >0$ there exists a $\delta > 0$ satisfying $$ {{1} \over {R}} < \delta \implies \left| \int_{\Gamma} e^{m i z} f(z) dz \right| < \varepsilon \left| \int_{\Gamma} e^{m i z} dz \right| < {{ \varepsilon \pi} \over {m}} $$ and we obtain the following. $$ \lim_{R \to \infty} \int_{\Gamma} e^{m i z } f(z) dz = 0 $$

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