Bernstein operators for exponential polynomials

J. M. Aldaz, O. Kounchev, H. Render

Published 2008-05-12Version 1

Let $L$ be a linear differential operator with constant coefficients of order $n$ and complex eigenvalues $\lambda_{0},...,\lambda_{n}$. Assume that the set $U_{n}$ of all solutions of the equation $Lf=0$ is closed under complex conjugation. If the length of the interval $[ a,b] $ is smaller than $\pi /M_{n}$, where $M_{n}:=\max \left\{| \text{Im}% \lambda_{j}| :j=0,...,n\right\} $, then there exists a basis $p_{n,k}$%, $k=0,...n$, of the space $U_{n}$ with the property that each $p_{n,k}$ has a zero of order $k$ at $a$ and a zero of order $n-k$ at $b,$ and each $% p_{n,k}$ is positive on the open interval $(a,b) .$ Under the additional assumption that $\lambda_{0}$ and $\lambda_{1}$ are real and distinct, our first main result states that there exist points $% a=t_{0}<t_{1}<...<t_{n}=b$ and positive numbers $\alpha_{0},..,\alpha_{n}$%, such that the operator \begin{equation*} B_{n}f:=\sum_{k=0}^{n}\alpha_{k}f(t_{k}) p_{n,k}(x) \end{equation*} satisfies $B_{n}e^{\lambda_{j}x}=e^{\lambda_{j}x}$, for $j=0,1.$ The second main result gives a sufficient condition guaranteeing the uniform convergence of $B_{n}f$ to $f$ for each $f\in C[ a,b] $.