Suppose $g(x)$ is bounded, and $$ f(x) := e^{x^2/2} \int_{-\infty}^x g(t) e^{-t^2/2} \,dt.$$ I have to show that $$ f \mbox{ bounded } \,\,\, \iff \,\,\, \int_{-\infty}^\infty g(t) e^{-t^2/2} \,dt=0.$$
The direction $\Rightarrow$ is clear. If the integral on the RHS equals $a\neq0$, then as $x \to \infty$, $|f|$ also tends to infinity.
How do I show $\Leftarrow$ ?
Using the fact that the total integral is $0$ you can rewrite $f$ as: $$f(x)=-e^\frac{x^2}{2} \int_x^{+\infty} g(t) e^\frac{-t^2}{2} dt$$ with $M$ a bound on $g$: $$|f(x)| \leq M e^\frac{x^2}{2} \int_x^{+\infty} e^\frac{-t^2}{2} dt$$ and since: $$\int_x^{+\infty} e^\frac{-t^2}{2} dt=\int_x^{+\infty} \frac{t}{t} e^\frac{-t^2}{2} dt \leq \frac{1}{x} \int_x^{+\infty} t e^\frac{-t^2}{2} dt=\frac{e^\frac{-x^2}{2}}{x}$$ you obtain: $$|f(x)| \leq \frac{M}{x}$$