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I'm stuck on the following problem :

Does there exist a real borel measure $\mu$ on $[0,1]$ such that

$\int_{0}^{1}x^n d\mu = e^{-n^2}$ for all $n \geq 1$ ?

Does anyone have any hint?

Thank you

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    Theo's link only seems to deal with the positive case. If you're interested in necessary and sufficient conditions in the general case, a good reference is chapter 3 of Widder's *The Laplace transform*. (Note that real finite measures correspond to functions of bounded variation, and Widder's exposition uses Stieltjes integrals.)2011-01-20

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If $\mu$ is supposed to be positive, then $\langle f,g\rangle=\int fgd\mu$ defines a pre-inner product on the real vector space of polynomial functions on $[0,1]$. Gramian matrices with respect to this inner product must be positive semidefinite. For example, $\begin{bmatrix} \langle x,x\rangle & \langle x,x^2\rangle \\ \langle x^2,x\rangle & \langle x^2,x^2\rangle \end{bmatrix}$ should be positive semidefinite.

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    in fact you were right, one cannot conclude from the Gram matrix based on $1$ and $x$ because the OP does not specify the total mass of $\mu$. Sorry about the noise.2011-02-19
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By the simple argument given by Jonas and explained in the Wiki page Theo referred to, there can exist no such positive measure $\mu$ since, using the notation below, $M_2M_4<(M_3)^2$.

As regards the existence of a signed measure $\mu$ of bounded variation on the interval $[0,1]$ with a given sequence of moments $ M_n=\displaystyle\int_0^1x^n\mathrm{d}\mu(x), $ it might be worthwhile, following Jonas's hint, to recall the result stated in chapter 3 section 2 of Widder's The Laplace transform: such a measure $\mu$ exists iff the sequence $(S_n)$ is bounded, where for every nonnegative $n$, $ S_n=\sum_{k=0}^n{n\choose k}\left|\sum_{i=0}^k(-1)^i{k\choose i}M_{n-i}\right|. $ Hence to determine if $(S_n)$ is bounded or not when $M_n=\mathrm{e}^{-n^2}$ for every $n\ge1$ would give the answer.