Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Is $\mu$ supposed to be positive? – Jonas Meyer Jan 20 '11 at 3:10
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.) – Jonas Meyer Jan 20 '11 at 4:42

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.

share|cite|improve this answer
yes. And already the Gram matrix based on $(1,x)$ (instead of your $(x,x^2)$) is not. – Did Feb 19 '11 at 11:18
@Didier: I don't know; I used $(x,x^2)$ because the problem says $n\geq 1$. But yes, my point was that it is not possible in the positive case. I only wish the OP had responded with clarification. – Jonas Meyer Feb 19 '11 at 17:12
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. – Did Feb 19 '11 at 21:41

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.