Complex Analysis MT23, Jordan's lemma
Flashcards
Can you state Jordan’s lemma, which lets you discard the semicircular part of a contour integral for a particular class of functions?
Suppose:
- $f : \mathbb H \to \mathbb C$ is a meromorphic function on upper half-plane
- $f(z) \to 0$ as $z \to \infty$ on $\mathbb H$
- $\gamma _ R(t) = Re^{it}$, $t \in [0, \pi]$
- $\alpha \in \mathbb R$, $\alpha > 0$.
Then
\(\int _ {\gamma _ R} f(z)e^{i a z} \text dz \to 0\) as $R \to \infty$.
Quickly prove Jordan’s lemma, i.e. that if
- $f : \mathbb H \to \mathbb C$ is a meromorphic function on upper half-plane
- $f(z) \to 0$ as $z \to \infty$ on $\mathbb H$
- $\gamma _ R(t) = Re^{it}$, $t \in [0, \pi]$
- $\alpha \in \mathbb R$, $\alpha > 0$.
then
\(\int _ {\gamma _ R} f(z)e^{i a z} \text dz \to 0\)
as $R \to \infty$, and actually establish in the process the more general result that even if $f(z) \not\to 0$ as $z \to \infty$ on $\mathbb H$, we have
\(\left \vert \int _
{\gamma _
R} f(z) e^{iaz} \text d z \right \vert \le \frac{\pi}{a} \cdot \max _
{\theta \in [0, \pi]} \big \vert f(Re^{i\theta})\big \vert\)
\(\begin{aligned} \left \vert \int _ {\gamma _ R} f(z) e^{iaz} \text d z \right \vert &= \left \vert \int^\pi _ 0 f(Re^{i\theta}) e^{iaR(\cos \theta + i \sin \theta)} iRe^{i\theta} \text d \theta\right \vert \\\\ &= R \left \vert \int^\pi _ 0 f(Re^{i\theta}) e^{aR(i\cos \theta - \sin \theta)} \text d \theta\right \vert \\\\ &\le R \int^\pi _ 0 \left \vert f(Re^{i\theta}) e^{aR(i\cos \theta - \sin \theta)} \right \vert \text d \theta \\\\ &= R \int^\pi _ 0 \vert f(Re^{i\theta}) \vert e^{-aR\sin \theta} \text d \theta \\\\ &\le R \cdot \left(\max _ {\theta \in [0, \pi]} \left \vert f(Re^{i \theta})\right \vert \right) \cdot \int^\pi _ a e^{-a R\sin \theta} \text d \theta \\\\ &= 2R \cdot \left(\max _ {\theta \in [0, \pi]} \left \vert f(Re^{i \theta})\right \vert \right) \cdot \int^{\pi/2} _ 0 e^{-a R \sin \theta} \text d \theta \quad (\star1) \\\\ &\le 2R \cdot \left(\max _ {\theta \in [0, \pi]} \left \vert f(Re^{i \theta})\right \vert \right) \cdot \int^{\pi/2} _ {0} e^{-2aR\theta/\pi} \text d\theta \quad (\star2) \\\\ &= \frac \pi a\cdot \left(\max _ {\theta \in [0, \pi]} \left \vert f(Re^{i \theta})\right \vert \right) \cdot (1 - e^{-aR}) \\\\ &\le \frac \pi a \left(\max _ {\theta \in [0, \pi]} \left \vert f(Re^{i \theta})\right \vert \right) \end{aligned}\) There are 2 non-obvious steps above.
- $(1\star)$: This follows from the identity that $\sin \theta = \sin (\pi - \theta)$, so
- $(2\star)$: $\sin(\theta)$ is concave on $[0, \pi / 2]$, so will be bounded below by the straight line from the two endpoints, i.e. $\frac{2\theta}{\pi} \le \sin \theta$. Then negating gives $-\sin \theta \le -\frac{2\theta}{\pi}$.
The final result, about tending to zero, follows from the assumption that $f(z) \to 0$ for sufficiently large $z$.
Suppose you are considering Jordan’s lemma, i.e. that if
- $f : \mathbb H \to \mathbb C$ is a meromorphic function on upper half-plane
- $f(z) \to 0$ as $z \to \infty$ on $\mathbb H$
- $\gamma _ R(t) = Re^{it}$, $t \in [0, \pi]$
- $\alpha \in \mathbb R$, $\alpha > 0$.
then $\int _ {\gamma _ R} f(z)e^{i a z} \text dz \to 0$ as $R \to \infty$. However, you have some $\alpha < 0$. What “version” of Jordan’s lemma works instead?
Considering $f$ as a meromorphic function on the lower half-plane, and using a semicircular contour on the lower half-plane.