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A Simple but Surprising Answer
sos440
2012. 8. 24. 21:55
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.