Consider trying to find a function $f \in L^2(0,1)$ satisfying
$$a_n = \int_0^1 f(x)x^n dx$$
Where $n$ is a nonnegative integer. Is there any method to go about doing this in general for any reasonably behaved sequence of real numbers $\{ a_n \}$?
I've been looking that this for a while and I have at least one interesting example:
When $f(x) = \log(x)$, let $a_n = \frac{1}{n+1}$, and it satisfies the above. Not only this, it satisfies something a little bit stronger:
$$(a_n)^s = \frac{(-1)^{s-1}}{(s-1)!} \int_0^1 f(x)^{s-1} x^n dx$$
Where $s \in \mathbb{N}$. I did not derive this analytically, I found it as a result of messing around with a somewhat nasty integral, which is what made me think of the more general case. Has anyone dealt with something like this before? I realize that if we use the $L^2$ inner product, this is equivalent to saying $$\langle f, x^n \rangle = a_n$$
Indeed this is awfully reminiscent of working with orthogonal polynomials: If $f$ had some sort of expansion $$\sum_{k=0}^\infty c_k \phi_k(x)$$ such that $\langle \phi_k, x^n \rangle = 1$ when $k=n$ and $0$ otherwise, then this could be easier, but again this comes down to solving the very first equation when $a_n$ is just a simpler sequence.
Any help is appreciated! Thanks.
This is an instance of the moment problem. By the Cauchy-Schwarz inequality, $$ a_n^2 \leq \frac{1}{2n+1}\int_{0}^{1}f(x)^2\,dx,\qquad a_n=O\left(\frac{1}{\sqrt{n}}\right)\tag{1}$$ and by considering $$ \varphi(\xi) = \int_{0}^{1}f(x) e^{2\pi i\xi x}\,dx \tag{2} $$ we have: $$ \varphi^{(n)}(0) = (2\pi i)^n a_n \tag{3}$$ and: $$ \varphi(\xi) = \sum_{n\geq 0}\frac{(2\pi i)^n a_n}{n!}\,\xi^n. \tag{4}$$ By $(2)$, it follows that $f$ is essentially given by the inverse Fourier transform of $(4)$.