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Let $\phi(t)$ be some a contionous infinitly differentialbe function such that $\phi(0)=1$ and $\phi(t)$ is symmmetric.
Let \begin{align} m_{2n} =i^{-2n} \phi^{n}(0) \end{align}

Suppose, that \begin{align} \sum_{n=0}^\infty m_{2n}^{-\frac{1}{2n}}=\infty \end{align}

That is $m_n$'s satisfy the Carleman's condition.

Does this imply that $\phi(t)$ is a characteristic function of some distribution?

Thanks.

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    Of course not. For instance, the inequality $m_4\ge m_2^2$ does not follow from Carleman's condition. – 2017-01-17
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    @zhoraster Of course. Thanks. So, in Humberger problem, it is assumed that the sequence $m_{2k}$ is already a valid sequence of moments? – 2017-01-17

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I don't know neither Humburger, nor Humberger problem. If you're speaking about Hamburger problem, then no, this is not assumed. Carleman's condition supplies that there is at most one solution to the moment problem. The existence of solution is provided by another condition (in fact, a criterion) that the matrix $(m_{i+j})$ is positive definite. Returning to your particular question, even if $(m_n)$ is a valid sequence of moments, $\phi$ does not need to be a characteristic function: you impose conditions in the point $0$ only, but elsewhere $\phi$ can misbehave arbitrarily. There can be some positive answers. Say, if $\phi$ is analytic, I think it can be verified (and I welcome you to) from the criterion I cite that $\phi$ is a characteristic function.

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    Going back to the Hamburger problem. You saying, that in the Hamburger problem we don't have to assume apriori that $(m_n)$ is a valid sequence of moments, right? But if Carleman's condition is verified then it implies a positive definite condition and $(m_n)$ is a valid sequence of moments, is this correct? – 2017-01-18
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    @Boby, it doesn't imply positive definiteness. – 2017-01-18
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    I think showing that $\phi$ is a valid characteristic function under the assumption of analyticity is not that difficult the proof would essentially go as follows: 1) On the one hand, If $\phi$ is analytic on some neighborhood, then it has power series expansion where the coefficients are given by $\phi^n(0)/n!=\frac{(i)^n m_n}{n!}$; 2) on the other hand, since $(m_i)$ is a valid sequence of moments it corresponds to some distribution $F$ with a characteristic function $g(t)$; 3) $g(t)$ can be written as power series where coefficient are give by $\frac{(i)^n m_n}{n!}$; – 2017-01-19
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    4) by uniqueness of power series $g(t)=f(t)$ and therefore $f(t)$ is a characteristic function. – 2017-01-19
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    I have another question. Is there a sufficient condition (simpler than positive definite one) that guarantees that sequence of numbers is a valid sequence of moments? – 2017-01-19