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Christian Remling
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In terms of the finite and compactly supported measure $\mu=\sum x_j^2 \delta_{x_j^2}$, we are given the moments $\int t^k\, d\mu(t)$, $k=0,1,2 \ldots$.

Moment problems with compactly supported solutions are determinate, that is, they have unique solutions.

In fact, if this is combined with the Muntz-Szasz theorem, we see that it would be enough to know $\sum x_j^{a+bN_n}$ to recover the $x_j$ as long as $\sum 1/N_n=\infty$. (In this case, define $\mu=\sum x_j^a \delta_{x_j^b}$.)

In terms of the finite and compactly supported measure $\mu=\sum x_j^2 \delta_{x_j^2}$, we are given the moments $\int t^k\, d\mu(t)$, $k=0,1,2 \ldots$.

Moment problems with compactly supported solutions are determinate, that is, they have unique solutions.

In terms of the finite and compactly supported measure $\mu=\sum x_j^2 \delta_{x_j^2}$, we are given the moments $\int t^k\, d\mu(t)$, $k=0,1,2 \ldots$.

Moment problems with compactly supported solutions are determinate, that is, they have unique solutions.

In fact, if this is combined with the Muntz-Szasz theorem, we see that it would be enough to know $\sum x_j^{a+bN_n}$ to recover the $x_j$ as long as $\sum 1/N_n=\infty$. (In this case, define $\mu=\sum x_j^a \delta_{x_j^b}$.)

Source Link
Christian Remling
  • 24.2k
  • 2
  • 48
  • 83

In terms of the finite and compactly supported measure $\mu=\sum x_j^2 \delta_{x_j^2}$, we are given the moments $\int t^k\, d\mu(t)$, $k=0,1,2 \ldots$.

Moment problems with compactly supported solutions are determinate, that is, they have unique solutions.