It is known that a random variable $X$ which is symmetric about $0$ (i.e $X$ and $-X$ have the same distribution) must have all its odd moments (when they exist!) equal to zero. The converse is a strong temptation which has been around for perhaps a hundred years.

>**Question.** *Suppose $\mathbb E[X^n] = 0$ for all odd $n$. Is it true that $X$ is symmetric ?*

This question was solved [Churchill (1946)](https://doi.org/10.1214/aoms/1177730987). In fact, he proved something much stronger

>**Theorem**. *Let $X$ be a random variable and let $(a_m)_{m \in \mathbb N}$ be a sequence of real numbers. Then for every $\epsilon > 0$, the exists a random variable $Y$ such that (1) $\mathbb E[Y^{2m+1}] = a_m$ for all $m \in \mathbb N$, and (2) The Kolmogorov distance between $X$ and $Y$ is at most $\epsilon$.*

Of course this theorem immediately implies a negative answer to the above question.

The proof given in the paper is constructive, but somewhat mysterious.

>**Question.** *Is there simple / modern way to prove the above theorem using functional-analytic tools ?*

After all, the theorem simply says (roughly) that the set of random variables of with odd-moments given by the sequence $(a_m)$ is dense in the space of random variables equipped with Komolgorov distance.

**Note.** The expected advantage of a general functional-analytic solution is that it would perhaps extend to constraints which are more general than those implied by prescribed odd moments.

  [1]: https://projecteuclid.org/download/pdf_1/euclid.aoms/1177730987