A Simple but Surprising Answer

Recently I was astonished by the following problem.

Question.Let $f$ be a continuous function such that \begin{equation}\label{eqn:cond} \int_{-\infty}^{\infty} x^n f(x) \; dx = 0 \end{equation} for all $n = 0, 1, 2, \cdots$. Then is $f = 0$ everywhere?

Answer. It might seem at the first glance that this is true. Surprisingly, however, this is not true. Let $$ g(x) = \exp\left(-x^2-\frac{1}{x^2}\right). $$ Clearly $g(x)$ is of Schwartz class. Since the Fourier transform is a linear isomorphism on Schwartz space, its inverse Fourier transform $$ f(x) := \mathcal{F}^{-1}g(x) $$ is also of Schwartz class. In particular, the LHS of \eqref{eqn:cond} is defined for every $n$. Therefore, we have $$ \int_{-\infty}^{\infty} x^n f(x) \; dx = \left. \frac{1}{(-2\pi i)^n} \frac{d^n}{dw^n} \int_{-\infty}^{\infty} f(x) e^{-2\pi i w x} \; dx \right|_{w=0} = \frac{1}{(-2\pi i)^n} g^{(n)}(0) = 0 $$ and the equation \eqref{eqn:cond} is satisfied. Since $f \neq 0$, we have found a non-trivial counter-example.

In fact, after a success on a class of functions with much stronger condition

$$ \forall s, \ f(x)e^{s|x|} \in L^1(\mathbb{R}), $$

I tried for a generalized case, which of course included the case of Schwartz space. But now it turned out that this was just too-good-to-be-true.

References

1. Does $\int_{\mathbb R} f(x)x^n dx = 0$ for $n=0,1,2,\ldots$ imply $f=0$ a.e.? in "Math StackExchange"