Consider the discrete distribution $\mu = \sum\_{i = 0}^{n - 1} \delta(x - x\_i)$ with all $a \leq x\_0 < \ldots < x\_{n - 1} \leq b$ and $[a, b] \in \mathbb{R}$. Suppose that $u\_0(x), \ldots, u\_n(x)$ are continuous real valued functions defined on $[a, b]$, and that the $u\_0(x), \ldots, u\_n(x)$ form a [Chebyshev system](https://www.semanticscholar.org/paper/Tchebycheff-Systems%3A-With-Applications-in-Analysis-Karlin-Studden/d448484d3808c1c9b6ff9efd2249ff5d093dfdf3). That is, any nontrivial linear combination $u(x) = \sum\_{i = 0}^{n}a\_i u\_i(x)$ has at most $n$ distinct zeros, and is uniquely defined by the values of $u(x)$ at any $n + 1$ distinct points.


Suppose that the $n + 1$ generalized moments $c\_i = \sum\_{j = 0}^{n} u\_i(x\_j) = \int\_a^b u\_i(x) d\mu$ are known. Is this sufficient to uniquely determine the $x\_i$ for $0 \leq i \leq n - 1$? It is clear that this is possible when $u\_i(x) = x^i$, since then the $c\_i$ are precisely the power sum symmetric polynomials. The values of these are related to the elementary symmetric polynomials by [Newton's identities](https://en.wikipedia.org/wiki/Newton%27s_identities), allowing one to find an order-$n$ polynomial in a single variable with zeros at the $x\_i$ (this is addressed in [other](https://mathoverflow.net/questions/201719/moment-problem-for-discrete-distributions) [posts](https://math.stackexchange.com/questions/1785297/solving-a-multivariate-polynomial-system-involving-the-power-sums)). But what about other Chebyshev systems?


For example, the spherical Bessel functions of the first kind of order $\nu$ $j\_\nu(xu\_{\nu k})$ where $u\_{\nu k}$ is the $(k + 1)$th nonzero root of $j\_\nu(r)$ satisfy the orthogonality condition
\begin{equation}
\int\_{0}^{1}x^2j\_\nu(xu\_{\nu k})j\_\nu(xu\_{\nu k'})\mathrm d x=\frac{\delta \_{kk'}}{2}\left [j\_{\nu+1}(u\_{\nu k}) \right ]^2
\end{equation}
and therefore not only form a Chebyshev system in $k$ but are smooth orthogonal functions. They are not linear combinations of a finite number of monomials though. Would it be possible to find the $x\_i$ from the generalized moments in this case?


Edit: A geometric way of thinking about this uses moment curves. Let $u\_i(x)$ be the $i$th coordinate of a curve in $\mathbb{R}^{n + 1}$ parameterized by $x \in [a, b]$. When $u\_i(x) = x^i$, the fact that the $x\_i$ can be recovered amounts to saying that every nondegenerate $(n - 1)$-simplex with vertices on the curve has a unique barycenter. Is there a reason why this should be true for $u\_i(x) = x^i$ and not other Chebyshev systems?



Generalized moment problem for discrete distributions
Note that $1\cdot\begin{bmatrix}1\\0\\0\end{bmatrix}+3\cdot\begin{bmatrix}1\\2\\4\end{bmatrix}=3\cdot\begin{bmatrix}1\\1\\1\end{bmatrix}+1\cdot\begin{bmatrix}1\\3\\9\end{bmatrix}=\begin{bmatrix}4\\6\\12\end{bmatrix}$


Now take your favorite positive continuous function $f$ on the real line with $f(0)=f(3)=1, f(1)=f(2)=3$ and consider the Chebyshev system $u\_j(x)=f(x)x^j$ ($j=0,1,2$) and the pairs of points $(0,2)$ and $(1,3)$.


So, the Chebyshev property is certainly too weak to imply what you want in general. What one really needs remains a mystery to me.


