If $f(x) \neq 0$ on $[0,\infty)$, does there exists some $\alpha \in [0,\infty)$ such that $\int_0^\infty f(x)e^{-\alpha x^2}dx \neq 0$?

Question

Ok,I have refined my question.

Suppose that $f : [0,\infty) \to \mathbb{R}$ is measurable and $\lvert f(x) \rvert \leq x$. Further suppose that $f(x)$ is not zero almost everywhere.

Then, I wonder if there exists some $\alpha \in [0, \infty)$ such that \begin{equation} \int_0^\infty e^{-\alpha x^2} f(x) dx \neq 0. \end{equation}

Due to Hermite expansion, I believe this is the case, but cannot rigorously prove myself.

Could anyone hlep me?

Answer 1

Hint: Suppose there is no such $\alpha$. Make the change of variable $x=\sqrt t$. Check that the Laplace transform of the integrable function $e^{-t}\frac {f(\sqrt t)}{ \sqrt t}$ is $0$. Hence, $e^{-t}\frac {f(\sqrt t)}{ \sqrt t}=0$ a.e.