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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.

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  • $\begingroup$ "Laplace transforms" are mentioned for confusion only. You are taking Fourier transforms here, which is certainly safe for $L^2$ functions. $\endgroup$ Oct 3, 2015 at 17:30
  • $\begingroup$ Unfortunately, it's not that easy. The convolutions are taken only from $0$ to $t$. That's an important property of Laplace transform. If you take the inverse Fourier transform without additional constraints, you might end up with a function that is non-zero for $t<0$ and thus the convolutions will be classical convolutions... $\endgroup$
    – thomas
    Oct 3, 2015 at 18:15
  • $\begingroup$ You can't simply "extend" the function $R$ to $\mathbb{R}$. It's already defined everywhere. If you change the definition of $R$ on the negative real axis, it's Laplace transform won't be anymore what one naturally expects. I think, I have to change the question in order to prevent this confusion. $\endgroup$
    – thomas
    Oct 3, 2015 at 18:37
  • $\begingroup$ I'm not extending $R$, I'm viewing $F$, $\rho$ as functions on $\mathbb R$, and then $R$ also when it gets introduced. $R$ will of course not be zero on the negative half line, so the final formula is incorrect, the integral needs to be over $(-\infty, t)$. $\endgroup$ Oct 3, 2015 at 18:40
  • $\begingroup$ In the new version, you answered your own question already: you are equivalently asking if $f\in H^2(\mathbb C^+)$ implies that $f/(1+f)\in H^2$, and, as you pointed out, the answer is trivially "no" because $f$ can take the value $-1$. $\endgroup$ Oct 3, 2015 at 18:51

2 Answers 2

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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 $f(-i)=-1$.

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  • $\begingroup$ Thanks for that example. I'm still hoping for some clarification regarding the quoted papers. Schaeffer and Glassey are two established mathematicians and I can't believe that the essence of their paper is plain wrong. I suspect that they have some further assumptions that they don't mention explicitely. $\endgroup$
    – thomas
    Oct 4, 2015 at 19:03
  • $\begingroup$ Your answer is still valid, but I posted another answer to clarify the confusion about the cited papers. I think, this makes clear that the authors did actually make some additional assumptions, but they never explicitly mentioned the role of these assumptions. $\endgroup$
    – thomas
    Oct 25, 2015 at 14:40
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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.

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