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I came across the following (unproven) claim in the lecture notes of a course I'm reading. For the purpose of this claim given $a,R,R_{0}\in\mathbb{R}$ such that $R,R_{0}>0$ we mark $$E=\left\{ z\in\mathbb{C}\:|\:\left|Z\right|\geq R_{0}\;\wedge\;\mathfrak{I}\left(z\right)\geq a\right\}$$ $$S_{R}=\left\{ z\in E\,|\;\left|z\right|=R\right\}$$ Let $f:E\to\mathbb{C}$ be a continuous function such that $${\displaystyle \lim_{z\in E\: z\to\infty}}f\left(z\right)=0$$Prove that for all $k>0$: $$\lim_{R\to\infty}\oint\limits _{s_{R}}e^{i\cdot k\cdot z}f\left(z\right)dz=0$$

I must admit I have no idea how one would really begin proving this so I would greatly appreciate a detailed proof if possible.

p.s: I had a problem coming up with an informative title for the question so if anyone feels an edit is in order please go ahead.

Serpahimz
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1 Answers1

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Define

$$\mu(r) = \sup \{ \lvert f(z)\rvert : \lvert z\rvert = r, z\in E\}.$$

For $R \geqslant R_0$, we have

$$\left\lvert\int_{S_R} e^{ikz}f(z)\,dz \right\rvert \leqslant \int_{S_R} \lvert e^{ikz}\rvert\cdot \mu(R)\, \lvert dz\rvert \leqslant \mu(R)\cdot \left(c(R,a) + R\int_0^\pi e^{-kR\sin\varphi}\,d\varphi\right),$$

where $c(R,a)$ is the integral over the part of $S_R$ in the lower half plane, thus $c(R,a) = 0$ if $a \geqslant 0$, and

$$c(R,a) \leqslant \pi \lvert a\rvert e^{k\lvert a\rvert}$$

in all cases, so that part is bounded independent of $R$. To see that the other summand in the parenthesis is also bounded, note that by symmetry

$$\int_0^\pi e^{-kR\sin\varphi}\,d\varphi = 2\int_0^{\pi/2} e^{-kR\sin\varphi}\,d\varphi.$$

For $0 \leqslant \varphi \leqslant \pi/2$, we have $\sin\varphi \geqslant \frac{2}{\pi}\varphi$, hence $e^{-kR\sin\varphi} \leqslant e^{-2kR/\pi\cdot\varphi}$, and

$$\int_0^{\pi/2} e^{-kR\sin\varphi}\,d\varphi \leqslant \int_0^{\pi/2} e^{-2kR/\pi\cdot\varphi}\,d\varphi \leqslant \int_0^\infty e^{-2kR/\pi\cdot\varphi}\,d\varphi = \frac{\pi}{2kR}.$$

Wrapping up,

$$\left\lvert\int_{S_R} e^{ikz}f(z)\,dz \right\rvert \leqslant \mu(R)\left(\pi \lvert a\rvert e^{k\lvert a\rvert} + \frac{\pi}{k}\right),$$

and $\lim\limits_{\substack{z\in E\\z\to\infty}} f(z) = 0$ means $\lim\limits_{R\to\infty} \mu(R) = 0$.

Daniel Fischer
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