Why is taking the inverse Laplace transform valid in this case? Assume $F \in L^2([0,\infty))$, so that the Laplace-transform $L[F]$ is well-defined. Assume furthermore, that
$$
y \mapsto \frac{L[F](iy)}{1+L[F](iy)}
$$
is in $L^2(\mathbb{R})\cap L^1(\mathbb{R})$, so that we obtain a well-defined function $R \in L^2(\mathbb{R})$ by the $L^2$-limit
$$
R(t) = \frac{1}{2\pi}\lim_{T\to\infty} \int_{-T}^{T} \frac{L[F](iy)}{1+L[F](iy)} e^{iyt}dy.
$$
Is it true that then $R(t)=0$ for $t < 0$?
This argument has been implicitly used in the paper http://www.tandfonline.com/doi/abs/10.1080/00411459408203873 by Glassey and Schaeffer. But I doubt that this argument is valid in general. By the classical theory of Laplace transforms one would make sure that the function $z \mapsto 1+L[F](z)$ has no poles in the complex half-plane $\Re(z) > 0$. Otherwise, if there are poles in the right half-plane, the property $R(t) =0$ for $t<0$ cannot be assumed to hold true. Furthermore, don't you have to assume in addition that 
$$
z \mapsto \frac{L[F](z)}{1+L[F](z)}
$$
tends to $0$ as $\Re(z) \to \infty$? The author of the recent article http://www.tandfonline.com/doi/abs/10.1080/23324309.2015.1075556 makes exactly the same mistake.
 A: Fourier transforms $f(z)=\int_0^{\infty} F(x)e^{-ixz}\, dx$ of functions $F\in L^2(0,\infty)$ are one way of obtaining the Hardy space $H^2$ on the lower half plane (normally I would prefer the upper half plane here, but I'll stick to your choice of sign in the exponent). So you are asking if $f/(1+f)\in H^2$.
This is not the case in general, for the simple reason that (as you already suspected) nothing prevents $f$ from taking the value $-1$ somewhere.
The Paley-Wiener Theorem says that $g\in H^2$ precisely if $g$ is holomorphic on $\mathbb C^-$ and $\sup_{y<0}\int_{\mathbb R} |g(x+iy)|^2\, dx<\infty$. So
$$
f(z)= 6i\left(\frac{1}{z-i}- \frac{1}{z-2i}\right) \in H^2 ,
$$
and this function satisfies your additional condition: Observe that the solutions of
$f(z)=-1$ are $z=4i, -i$, so
$f/(1+f)\in L^1\cap L^2$ is perfectly well behaved on the real line. However, $f/(1+f)\notin H^2$; this function isn't even defined everywhere on $\mathbb C^-$ since $f(-i)=-1$.
A: In both papers cited above, $L[F]$ happens to be the Hilbert transform of an absolutely integrable function. It's never stated in those papers, but
$$
L[F](z) = \frac{1}{\xi^2}\int_{-\infty}^{\infty}
        \frac{f(u)}{i z/\xi - u}
    \mathrm{d} u.
$$
for some function $f$ and some $\xi > 0$. Hence, $L[F]$ is analytic in the right half plane and continuous up to the boundary (since $f$ is also Lipschitz continuous).
By the argument principle, it therefore suffices to show that $y \mapsto L[F](iy)$ never crosses the real axis left of $-1$. In fact, the authors prove that either $\Re(L[F](iy)) > -1/2$ or $|\Im(L[F](iy))|$ is positive for all $y$.
This proves that $\frac{L[F]}{1+L[F]}$ is analytic in the right half plane and therefore has an inverse laplace transform in the required sense.
