Appendix
Proof of Lemma 1
Proof
For simplicity in notation we focus on a single neutral fraction and drop the superscript i notation. By assumption, \(x^2v_0(x)\mathrm{e}^{sx} \in L^{1}({\mathbb {R}})\cap L^{\infty }({\mathbb {R}})\). Thus, we have
$$\begin{aligned} x^2v_0(x)\mathrm{e}^{sx} \le (1+x^2)v_0(x)\mathrm{e}^{sx} \le C \end{aligned}$$
(93)
for all \(x \in {\mathbb {R}}\) where C is a positive constant. Rearranging the previous inequality,
$$\begin{aligned} v_0(x) \le \frac{C\mathrm{e}^{-sx}}{1+x^2} \end{aligned}$$
(94)
for all \(x \in {\mathbb {R}}\). Thus, there exists a positive constant C such that the function \(w_0(x)\) defined by
$$\begin{aligned} w_0(x) := \frac{C\mathrm{e}^{-sx}}{1+x^2} \end{aligned}$$
(95)
satisfies \(v_0(x) \le w_0(x)\) for all \(x \in {\mathbb {R}}\). It is easy to see that \(w_0(x)\mathrm{e}^{sx}\in L^{1}({\mathbb {R}})\cap L^{\infty }({\mathbb {R}})\). Hence, the Fourier transform of \(w_0(x)\mathrm{e}^{sx} \in L^1({\mathbb {R}})\). To calculate the Fourier Transform of \(w_0(x)\mathrm{e}^{sx}\), note that
$$\begin{aligned} {\mathcal {F}}\left[ \mathrm{e}^{-|x|}\right] (\omega )&=\int _{-\infty }^{\infty }\mathrm{e}^{-|x|}\mathrm{e}^{-i\omega x}\,\mathrm{d}x \end{aligned}$$
(96)
$$\begin{aligned}&= \int _{-\infty }^0 \mathrm{e}^{(1-i\omega )x}\,\mathrm{d}x+ \int _{0}^\infty \mathrm{e}^{-(1+i\omega )x}\,\mathrm{d}x \end{aligned}$$
(97)
$$\begin{aligned}&= \lim _{b\rightarrow \infty } \left[ \frac{\mathrm{e}^{(1-i\omega )x}}{(1-i\omega )}\bigg |_{-b}^0 - \frac{ \mathrm{e}^{-(1+i\omega )x}}{(1+i\omega )}\bigg |_0^b\right] \end{aligned}$$
(98)
$$\begin{aligned}&= \lim _{b\rightarrow \infty } \left[ \frac{1}{(1-i\omega )}-\frac{\mathrm{e}^{-(1-i\omega )b}}{(1-i\omega )} - \frac{ \mathrm{e}^{-(1+i\omega )b}}{(1+i\omega )}+\frac{1}{(1+i\omega )}\right] \end{aligned}$$
(99)
$$\begin{aligned}&=\left[ \frac{1}{(1-i\omega )} +\frac{1}{(1+i\omega )}\right] \end{aligned}$$
(100)
$$\begin{aligned}&= \frac{2}{1+\omega ^2}. \end{aligned}$$
(101)
From the inverse Fourier transform,
$$\begin{aligned} \pi \mathrm{e}^{-|x|}&= \frac{\pi }{2\pi } \int _{-\infty }^\infty \frac{2}{1+\omega ^2}\mathrm{e}^{i\omega x}\,\mathrm{d}\omega \end{aligned}$$
(102)
$$\begin{aligned}&= \int _{-\infty }^\infty \frac{1}{1+\omega ^2}\mathrm{e}^{i\omega x}\,\mathrm{d}\omega . \end{aligned}$$
(103)
Using the above result,
$$\begin{aligned} {\mathcal {F}}\left[ \frac{C}{1+x^2}\right] (\omega )&={\mathcal {F}}\left[ \frac{C}{1+(-x)^2}\right] (\omega ) \end{aligned}$$
(104)
$$\begin{aligned}&=C \int _{-\infty }^{\infty }\frac{1}{1+(-x)^2}\mathrm{e}^{-i\omega (-x)}\,\mathrm{d}x \end{aligned}$$
(105)
$$\begin{aligned}&=C \int _{-\infty }^{\infty }\frac{1}{1+x^2}\mathrm{e}^{i\omega x}\,\mathrm{d}x \end{aligned}$$
(106)
$$\begin{aligned}&= C\pi \mathrm{e}^{-|\omega |}. \end{aligned}$$
(107)
The proof of the lemma is complete. \(\square \)