TITLE: Solutions to Spring 2020 UCLA Analysis Qual Problem 1? Or: an identity implies function is odd
QUESTION [6 upvotes]: I would like to see an elegant/simple solution  to the following problem from the Spring 2020 UCLA Analysis Qual.
Suppose that $f\in C_c^{\infty}(\mathbb R)$ satisfies
\begin{equation}
\int_{\mathbb R}e^{-tx^2}f(x)\,dx=0\qquad\text{for any }t\geq0.
\end{equation}
Show that $f$ is odd; that is, show that $f(x)=-f(-x)$ for each $x\in\mathbb R$.
The proof that I have in mind is quite long. Any help is appreciated!

REPLY [4 votes]: You can rewrite the integral, substituting $x^2 =s$
$$0=\int_{\mathbb{R}} e^{-tx^2}f(x)dx = \int_0^\infty e^{-tx^2} (f(x)+f(-x))dx = \int_0^\infty e^{-ts}(f(\sqrt{s})+f(-\sqrt{s}))\frac{1}{2\sqrt{s}}ds. $$
This then is the Laplace transform of the even part of $g:s\mapsto \frac{f(\sqrt{s})}{\sqrt{s}}$, so the even part is zero almost everywhere. So $g$ has to be odd almost everywhere, which then implies the same for $f$ and since $f$ is continuous you can drop the almost everywhere.