At least, I think it might deserve to be called a Tauberian theorem, inasmuch as it would generalize the Tauberian theorem mentioned by Liviu Nicolaescu in his reply to my question Using a quadratic kernel instead of a linear kernel in the Laplace transform.
Is the following conjecture true?
${\bf Suppose}$ $f$ and $g$ are non-negative decreasing continuous functions in $L^1(\mathbb{R}_{>0})$ with $f(0)=g(0)=1$. (Some of these conditions might turn out to be unneeded, but since I am hoping for an affirmative answer to the question, I thought I would be conservative and include any hypotheses that might make the conjecture more likely to hold true.) Suppose that $a(t)$ is in $L_{loc}^1(\mathbb{R}_{>0})$ (integrable on compacts), and to be conservative let's assume that $a(t)$ is also continuous. Define $F(x) = \int_0^\infty f(xt) \: a(t) \: dt$ and $G(x) = \int_0^\infty g(xt) \: a(t) \: dt$, and suppose furthermore that $\sup_{x>0} |F(x)|$ and $\sup_{x>0} |G(x)|$ exist. ${\bf Then}$ we can conclude that $\liminf_{x \searrow 0} F(x) = \liminf_{x \searrow 0} G(x)$.
Note that my question from a year ago (referenced above) concerns the special case where $f(u) = e^{-u}$ and $g(u) = e^{-u^2}$. But my earlier question was restricted to the case in which $F(x)$ and $G(x)$ approach limits as $x$ decreases to 0; in the new version, I want to know about the lim inf even if the limits don't exist (as well as replace $e^{-u}$ and $e^{-u^2}$ by more general functions $f(u)$ and $g(u)$).
If the answer is "no", I would like to know why, and I would like to know if there are additional hypotheses satisfied by the functions $e^{-u}$ and $e^{-u^2}$ whose imposition for more general $f$ and $g$ would suffice to make a restricted version of the conjecture true.