Let $f:[0,1]\to\mathbb{R}$ be continuous. Consider for every $n\in\mathbb{N}$ :
$$M_n(f)=\int_0^1t^n\,f(t)\,dt$$
It is easy to see that the sequence $\left(M_n(f)\right)_{n\in\mathbb{N}}$ converges to $0$.
But is it possible to choose $f$ in such a way that :
$$\forall n\in\mathbb{N},\,M_n(f)=e^{-\lambda n^2}$$
where $\lambda$ is some positive constant ?
It can be seen that such a function could not be positive and would necessarily verify $f(1)=0$, but I wasn't able to get much more than that ...
Any hint would be appreciated :)
This is not a complete answer.
Note that the set of polynomials is dense in $L^2([0,1])$. Hence there exists a complete orthogonal system in $L^2([0,1])$ $$p_0(t),p_1(t),...,p_n(t),...$$ consisting of polynomials. Note that $p_n(t)$ is a polynomial of degree $n$. Suppose that $$p_n(t) = c(n)_n t^n + ... + c(n)_1t + c(n)_0$$ where $c(n)_i\in \mathbb{R}$ and $c(n)_n\neq 0$.
Fix $f\in L^2([0,1])$. We have $$A_n(f) = \int^1_0p_n(t)f(t)dt = c(n)_n\cdot M_n(f) + ... + c(n)_1\cdot M_1(f) + c(n)_0\cdot M_0(f)$$ for $n\in \mathbb{N}$. On the other hand we know from the theory of Hilbert spaces that the only constraint on the sequence of numbers $\{A_n(f)\}_{n\in \mathbb{N}}$ is
$$\sum_{n\in \mathbb{N}}A_n(f)^2 < +\infty$$
Thus if you calculate (recursively, by means of Gram-Schmidt orthogonalization) the sequence $\{p_n(t)\}_{n\in \mathbb{N}}$, then you may check if there exists $f\in L^2([0,1])$ such that $$M_n(f) = e^{-\lambda n^2}$$ simply by calculating $\{A_n(f)\}_{n\in \mathbb{N}}$ and checking that its $l_2$-norm is finite. I don't know, if this is tractable.