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

1 AnswersNote 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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