Carleman's condition

Remarks

The proof goes through quasi-analytic functions. The condition puts the Fourier–Stieltjes transform of the difference of two solutions in a quasi-analytic class. That transform has all its derivatives zero at $0$, so it vanishes identically. Example: for the standard normal distribution, $m_{2n} = (2n - 1)!! \le (2n)^n$, so $m_{2n}^{-1/(2n)} \ge (2n)^{-1/2}$ and the series diverges; the normal law is determined by its moments. Example: if $\mu$ has bounded support in $[-R, R]$, then $m_{2n} \le m_0 R^{2n}$, so every term is at least $m_0^{-1/(2n)}/R$, and the series diverges. The condition is sufficient but not necessary for determinacy. The hypothesis $m_{2n} > 0$ only excludes the measures concentrated at $0$, which are determinate anyway.

Going deeper

For a student

It is the test used in practice to check that a law is determined by its moments, for instance before applying the method of moments to prove convergence in distribution. A simpler sufficient condition is that the moment generating function is finite near $0$, and that condition implies Carleman's. In the other direction, Krein's condition gives indeterminacy for densities with very heavy tails, such as the log-normal density.