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We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

 
  1. the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.

    the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.

  2. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

  1. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Since $m^+[0,1]=m^-[0,1]$, we can rescale these measures in order to get probability measures. Then with the same notations as in the OP, $m^+_n=m^-_n$ for all $n\in \{0\}\cup S$, but $m^+\neq m^-$.

We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

 
  1. the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.
  1. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Since $m^+[0,1]=m^-[0,1]$, we can rescale these measures in order to get probability measures. Then with the same notations as in the OP, $m^+_n=m^-_n$ for all $n\in \{0\}\cup S$, but $m^+\neq m^-$.

We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

  1. the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.

  2. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Since $m^+[0,1]=m^-[0,1]$, we can rescale these measures in order to get probability measures. Then with the same notations as in the OP, $m^+_n=m^-_n$ for all $n\in \{0\}\cup S$, but $m^+\neq m^-$.

correct the argument.
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Davide Giraudo
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We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

  1. the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.
  1. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Then the moments of $1/2(\mu+m^+)$ and $1/2(\nu+m^-)$ coincide onSince $S$$m^+[0,1]=m^-[0,1]$, and we cannot have both equalities $\mu=\nu$ andcan rescale these measures in order to get probability measures. Then with the same notations as in the OP, $\mu+m^+=\nu+m^-$$m^+_n=m^-_n$ for all (notice that$n\in \{0\}\cup S$, but $\mu^+[0,1]=\mu^-[0,1]$)$m^+\neq m^-$.

We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

  1. the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.
  1. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Then the moments of $1/2(\mu+m^+)$ and $1/2(\nu+m^-)$ coincide on $S$, and we cannot have both equalities $\mu=\nu$ and $\mu+m^+=\nu+m^-$ (notice that $\mu^+[0,1]=\mu^-[0,1]$).

We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

  1. the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.
  1. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Since $m^+[0,1]=m^-[0,1]$, we can rescale these measures in order to get probability measures. Then with the same notations as in the OP, $m^+_n=m^-_n$ for all $n\in \{0\}\cup S$, but $m^+\neq m^-$.

added 51 characters in body
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Davide Giraudo
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We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

  1. the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.
  1. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Then the moments of $\mu-m^+$$1/2(\mu+m^+)$ and $\nu+m^-$$1/2(\nu+m^-)$ coincide on $S$, and we cannot have both equalities $\mu=\nu$ and $\mu-m^+=\nu+m^-$$\mu+m^+=\nu+m^-$ (notice that $\mu^+[0,1]=\mu^-[0,1]$).

We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

  1. the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.
  1. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Then the moments of $\mu-m^+$ and $\nu+m^-$ coincide on $S$, and we cannot have both equalities $\mu=\nu$ and $\mu-m^+=\nu+m^-$.

We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

  1. the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.
  1. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Then the moments of $1/2(\mu+m^+)$ and $1/2(\nu+m^-)$ coincide on $S$, and we cannot have both equalities $\mu=\nu$ and $\mu+m^+=\nu+m^-$ (notice that $\mu^+[0,1]=\mu^-[0,1]$).

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Davide Giraudo
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Davide Giraudo
  • 4k
  • 2
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  • 43
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