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. 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, allowing one to find an order-$n$ polynomial in a single variable with zeros at the $x_i$ (this is addressed in other posts). 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?
