I am trying to show that
$$\int^{\infty}_{0}\vert f(x)\vert^2e^x dx < \infty \implies \int^{\infty}_{0}\vert f(x)\vert dx < \infty,$$
for a Borel measurable function $f:[0,\infty) \rightarrow \mathbb{R}$.
So far I could find a few simple cases such as $f(x) = e^{-x}$, or could do a simple step like $\int^{\infty}_{0}\vert f(x)\vert^2e^x dx < \infty \implies \int^{\infty}_{0}\vert f(x)\vert^2 dx < \infty,$ but I was not able to prove the general case. Could anybody give me some hints?
$\int_0^{\infty} |f(x)| \, dx =\int_0^{\infty} |f(x)| e^{x/2} e^{-x/2} \, dx \leq (\int_0^{\infty} |f(x)|^{2} e^{x}\, dx)^{1/2} (\int_0^{\infty} e^{-x} \, dx)^{1/2} < \infty$.