Let $f\in L^{\infty}(\mathbb{R}^+)$ with support in an open interval $(a,b)\subset \mathbb{R}$ ($a,b$ finite) and $g\in H^{s}(\mathbb{R}^+)$ for some $s\in \mathbb{R}^{+}$, with $g$ a positive function (i.e. $g\geq 0$). In other words,
$$ \int_{0}^{\infty} (1+x^2)^{s}\left|\int_{0}^{\infty}e^{-ixt}g(t )dt\right|^2dx < \infty $$
I am trying to understand why $fg\notin H^{s}(\mathbb{R}^+)$ in general. Indeed, isn't it true that
$$ \int_{0}^{\infty} (1+x^2)^{s}\left|\int_{0}^{\infty}e^{-ixt}f(t)g(t )dt\right|^2dx < |f|_{L^{\infty}} |g|_{H^{s}}^2 < \infty $$
I am sure there is something I am missing, because if $fg$ is discontinuous then it will not belong to $H^{s}$ for $s\geq 1/2$. But I don't see which part of the previous inequality is incorrect.