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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asked Sep 23, 2017 at 0:52
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$\begingroup$ Should the summation index be $n$? Also, do you assume integer or real $k$? $\endgroup$– Mikhail TikhomirovCommented Sep 23, 2017 at 0:56
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1$\begingroup$ 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 TikhomirovCommented Sep 23, 2017 at 0:59
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$\begingroup$ Yes, I forgot to mention that $\lambda_n$ is increasing and unbounded. They are indeed the eigenvalues of an elliptic differential equation. $\endgroup$– A random mathematicianCommented Sep 23, 2017 at 1:03
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1 Answer
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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.
answered Sep 23, 2017 at 1:31