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Spring 2012 - Problem 9

Jordan's lemma

Prove Jordan's lemma: If f ⁣:C→C\func{f}{\C}{\C} is meromorphic, R>0R > 0, and k>0k > 0, then

∣∫Γf(z)eikz dz∣≤100ksup⁡z∈Γ ∣f(z)∣\abs{\int_\Gamma f\p{z} e^{ikz} \,\diff{z}} \leq \frac{100}{k} \sup_{z \in \Gamma}\,\abs{f\p{z}}

where Γ\Gamma is the quarter-circle z=Reiθz = Re^{i\theta} with 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2}. (It is possible to replace 100100 here by π2\frac{\pi}{2}, but you are not required to prove that.)
Solution.

By parametrizing Γ\Gamma, we get

∣∫Γf(z)eikz dz∣=∣∫0π/2f(Reiθ)eikReiθiReiθ dθ∣≤∫0π/2sup⁡z∈Γ ∣f(z)∣Re−kRsin⁡θ dθ.\begin{aligned} \abs{\int_\Gamma f\p{z} e^{ikz} \,\diff{z}} &= \abs{\int_0^{\pi/2} f\p{Re^{i\theta}} e^{ikRe^{i\theta}} iRe^{i\theta} \,\diff\theta} \\ &\leq \int_0^{\pi/2} \sup_{z \in \Gamma}\,\abs{f\p{z}} Re^{-kR\sin\theta} \,\diff\theta. \end{aligned}

By concavity of sin⁡θ\sin\theta, we have sin⁡θ≥2θπ\sin\theta \geq \frac{2\theta}{\pi} on [0,π2]\br{0, \frac{\pi}{2}} and so

∫0π/2sup⁡z∈Γ ∣f(z)∣Re−kRsin⁡θ dθ≤sup⁡z∈Γ ∣f(z)∣∫0π/2Re−kR2θπ dθ=sup⁡z∈Γ ∣f(z)∣π2k(1−e−kR)≤π2ksup⁡z∈Γ ∣f(z)∣,\begin{aligned} \int_0^{\pi/2} \sup_{z \in \Gamma}\,\abs{f\p{z}} Re^{-kR\sin\theta} \,\diff\theta &\leq \sup_{z \in \Gamma}\,\abs{f\p{z}} \int_0^{\pi/2} Re^{-kR\frac{2\theta}{\pi}} \,\diff\theta \\ &= \sup_{z \in \Gamma}\,\abs{f\p{z}} \frac{\pi}{2k}\p{1 - e^{-kR}} \\ &\leq \frac{\pi}{2k} \sup_{z \in \Gamma}\,\abs{f\p{z}}, \end{aligned}

which completes the proof.