TITLE: How to show the function $ f(x) := e^{x^2/2} \int_{-\infty}^x g(t) e^{-t^2/2} \,dt$ is bounded?
QUESTION [5 upvotes]: 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 $\Leftarrow$ ?

REPLY [7 votes]: 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:
$$|f(x)| \leq \frac{M}{x}$$