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Let $\lambda_n$ be an increasing and unbounded sequence of positive real numbers and $a_n$ be a sequence of real numbers such that

$$\sum_{n=1}^\infty a_n \lambda_n^k=0 \ \ \text{ for all }\ \ k\geq 0.$$

Is $a_n=0$ for all $n$?

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  • $\begingroup$ Should the summation index be $n$? Also, do you assume integer or real $k$? $\endgroup$ – Mikhail Tikhomirov Sep 23 '17 at 0:56
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    $\begingroup$ Also, the answer is trivially no if we take $\lambda_n = 1$ and $a_n$ to be elements of any series converging to 0. Is a condition missing? $\endgroup$ – Mikhail Tikhomirov Sep 23 '17 at 0:59
  • $\begingroup$ Yes, I forgot to mention that $\lambda_n$ is increasing and unbounded. They are indeed the eigenvalues of an elliptic differential equation. $\endgroup$ – User4966 Sep 23 '17 at 1:03
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With $d\mu = \sum a_n\delta_{\lambda_n}$, your question can be rephrased as: Does $\int t^k\, d\mu(t) = 0$ imply that $\mu=0$? Since there are indeterminate moment problems (that is, collections of moments that do not come from one unique measure), it's now clear that the answer is no.

To make this more concrete, you can start out with any limit circle Jacobi matrix; any two of its spectral measures $\nu_1,\nu_2$ will have the same moments, and you can take $\mu=\nu_1-\nu_2$. These measures will automatically have most of the extra properties you require here (pure point, with discrete disjoint spectra) if you take two spectral measures corresponding to two distinct boundary conditions at infinity. What is not automatic is that these measures will be supported by $[0,\infty)$, but you can get that, too, since there are indeterminate Stieltjes moment problems.

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