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Spring 2020 - Problem 1

Stone-Weierstrass

Assume $f \in C_{\mathrm{c}}^\infty\p{\R}$ satisfies

\int_\R e^{-tx^2} f\p{x} \,\diff{x} = 0 \quad\text{for any } t \geq 0.

Show that $f\p{x} = -f\p{-x}$ for any $x \in \R$ .

Solution.

Since $f$ is compactly supported, it is supported in some $\br{-R, R}$ with $R > 0$ . Thus, because the power series for $e^x$ converges uniformly on compact sets and $f$ is bounded, we have

\int_\R e^{-tx^2} f\p{x} \,\diff{x} = \int_{-R}^R \sum_{n=0}^\infty \frac{\p{-tx^2}^n}{n!} f\p{x} \,\diff{x} = \sum_{n=0}^\infty \frac{\p{-t}^n}{n!} \int_{-R}^R x^{2n} f\p{x} \,\diff{x},

which is identically $0$ for $t \geq 0$ by assumption. Thus, by uniqueness of power series, it follows that

\int_{-R}^R x^{2n} f\p{x} \,\diff{x} = 0

for all $n \geq 0$ . Via the change of variables $x \mapsto -x$ , we also get

\int_{-R}^R x^{2n} f\p{-x} \,\diff{x} = 0 \implies \int_{-R}^R x^{2n}\p{f\p{x} + f\p{-x}} \,\diff{x} = 0.

If we set $g\p{x} = f\p{x} + f\p{-x}$ , then clearly $g$ is even, and so $x^{2n+1} g\p{x}$ is an odd function. Hence,

\int_{-R}^R x^{2n+1} g\p{x} \,\diff{x} = 0

by symmetry, i.e., $\int_{-R}^R x^n g\p{x} \,\diff{x}$ for all $n \geq 0$ . Since $g$ is continuous and $\br{-R, R}$ is compact, Stone-Weierstrass gives us a sequence of polynomials $\set{p_n}_n$ such that $p_n \to g$ uniformly. Then

0 = \int_{-R}^R p_n\p{x} g\p{x} \,\diff{x} \xrightarrow{n\to\infty} \int_{-R}^R \abs{g\p{x}}^2 \,\diff{x},

so because $g$ is continuous, we have $g\p{x} = 0$ everywhere, i.e., $f\p{x} = -f\p{-x}$ for every $x \in \R$ .