# Solutions to Spring 2020 UCLA Analysis Qual Problem 1? Or: an identity implies function is odd

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 $$$$\int_{\mathbb R}e^{-tx^2}f(x)\,dx=0\qquad\text{for any }t\geq0.$$$$ Show that $$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!

Say the support of $$f$$ is contained in $$[-M,M]$$. Let $$g(x) = f(x) + f(-x)$$ and note that it is enough to show that $$g=0$$ on the interval $$[0,M]$$. It's clear that $$\int_0^M g(x) e^{-tx^2} \,dx \ = \ \int_\mathbb{R} f(x) e^{-tx^2} \,dx \ = \ 0$$ for all $$t \geq 0$$. Let $$\mathcal{F} = \operatorname{span} \{ x \mapsto e^{-tx^2} : t \geq 0 \} \subseteq C([0,M],\mathbb{R})$$. We see that $$\mathcal{F}$$ is closed under linear combinations by definition, and closed under multiplication because $$e^{-tx^2} e^{-sx^2} = e^{-(t+s)x^2}$$. We also see that $$\mathcal{F}$$ contains a nonzero constant by taking $$t=0$$. Finally, it's clear that $$\mathcal{F}$$ separates points because each $$x \mapsto e^{-tx^2}$$ is strictly decreasing on $$[0,M]$$. Therefore the Stone-Weierstrass theorem implies that $$\mathcal{F}$$ is dense in $$C([0,M],\mathbb{R})$$, so by the equation above and a standard argument it follows that $$\int_0^M g(x) \phi(x) \,dx = 0$$ for all continuous $$\phi$$ and therefore $$g = 0$$ on $$[0,M]$$ as desired.
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.