English

Inequalities for exponential polynomials with applications to moment sequences

Classical Analysis and ODEs 2026-04-07 v1 Functional Analysis

Abstract

Let ΦΛn\Phi_{\Lambda_{n}} be the unique solution of the differential operator L=j=0n(ddxλj)L=\prod_{j=0}^{n}\left( \frac{d}{dx}-\lambda_{j}\right) such that ΦΛn(j)(0)=0\Phi_{\Lambda_{n}}^{\left( j\right) }\left( 0\right) =0 for j=0,...,n1,j=0,...,n-1, and ΦΛn(n)(0)=1.\Phi_{\Lambda_{n}}^{\left( n\right) }\left( 0\right) =1. Assume that ΦΛn\Phi_{\Lambda_{n}} is real-valued and ΦΛn(n+1)(x)0\Phi_{\Lambda_{n} }^{\left( n+1\right) }\left( x\right) \geq0 for all x[0,B].x\in\left[ 0,B\right] . Then, if a polynomial R(x)=k=0nakxkR\left( x\right) = {\displaystyle\sum_{k=0}^{n}} a_{k}x^{k} is non-negative on the interval [0,B],\left[ 0,B\right] , the inequality k=0nakk!ΦΛn(nk)(x)R(x) {\displaystyle\sum_{k=0}^{n}} a_{k}k!\Phi_{\Lambda_{n}}^{\left( n-k\right) }\left( x\right) \geq R\left( x\right) holds for x[0,B]x\in\left[ 0,B\right] . From this we derive several interesting inequalities for exponential polynomials. An important consequence is that for a non-negative measure μ\mu over the interval [a,b]\left[ a,b\right] with ba<Bb-a<B the sequence defined by sk:=abk!ΦΛn(nk)(xa)dμ(x) s_{k}:=\int_{a}^{b}k!\Phi_{\Lambda_{n}}^{\left( n-k\right) }\left( x-a\right) d\mu\left( x\right) for k=0,...,nk=0,...,n is a moment sequence, i.e. there exists a non-negative measure ν\nu with support in [a,b]\left[ a,b\right] such that sk=ab(ta)kdν(t)s_{k}=\int_{a} ^{b}\left( t-a\right) ^{k}d\nu\left( t\right) for k=0,....,n.k=0,....,n.

Keywords

Cite

@article{arxiv.2409.18136,
  title  = {Inequalities for exponential polynomials with applications to moment sequences},
  author = {Ognyan Kounchev and Hermann Render and Tsvetomir Tsachev},
  journal= {arXiv preprint arXiv:2409.18136},
  year   = {2026}
}

Comments

14 pages

R2 v1 2026-06-28T18:58:36.458Z