# Generalized moment problem for discrete distributions

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 $$c_i = sum_{j = 0}^{n} u_i(x_j) = int_a^b u_i(x) dmu$$ 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?

MathOverflow Asked by harharkh on February 8, 2021

Note that $$1cdotbegin{bmatrix}1\0\0end{bmatrix}+3cdotbegin{bmatrix}1\2\4end{bmatrix}=3cdotbegin{bmatrix}1\1\1end{bmatrix}+1cdotbegin{bmatrix}1\3\9end{bmatrix}=begin{bmatrix}4\6\12end{bmatrix}$$

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.

Answered by fedja on February 8, 2021

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