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Given $h:\mathbb{R}\to \mathbb{R}^+$, I want to know whether $f(t)\to 0$ as $t\to \infty$ where $f(t)$ is defined in terms of its Laplace transform $F(s)$: $$F(s)=\frac{\langle h, z_s\rangle}{1-\langle h^2, z_s\rangle}$$

with

$$z_s(i)=\frac{1}{s-2h(i)}$$

Empirically it appears that $\int \mathrm{d}i\ h(i)\le 1$ implies $\lim_{t\to\infty}f(t)=0$, can this be shown more rigorously?

Motivation: $F(s)$ describes evolution of continuous mean-field model solved here

Yaroslav Bulatov
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    There is a typo in the first line. – geetha290krm Jun 01 '23 at 04:53
  • Did you check this: Riemann–Lebesgue lemma ? – Joako Jun 04 '23 at 03:56
  • If $F(s)\in \mathbb{R}$ don't have complex terms so you could treat the variable $s$ is were a real-valued variable, then if you could prove that $\int_{\mathbb{R}} |F(s)| ds<\infty$ then by the Riemann-Lebesgue lemma I mentioned earlier you should prove that $f(t)$ vanishes at infinity (seeing $f(t)$ as the Fourier Transform of $F(s)$). – Joako Jun 04 '23 at 23:41
  • Could you explain the meaning of the used notation? $\langle h, z_s\rangle$ ? $h$ ? – Cesareo Jun 07 '23 at 18:15
  • Using natural inner product definition for $L2$ integrable functions $\langle f, g\rangle=\int\mathrm{d}x\ h(x)g(x)$. $h$ is a function from real numbers to positive real numbers – Yaroslav Bulatov Jun 07 '23 at 18:27

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