Abstract

Necessary conditions are given for the Hermite–Fejér interpolation polynomials based at the zeros of orthogonal polynomials to converge in weighted Lp spaces at the Jackson rate. These conditions are known to be sufficient in the case of the generalized Jacobi polynomials.

Introduction

The first detailed study of weighted mean convergence of Hermite–Fejér interpolation based at the zeros of orthogonal polynomials was accomplished in [13] and [14], where it was shown that some of the most delicate problems associated with mean convergence of Hermite–Fejér interpolation can be approached through the general theory of orthogonal polynomials; in particular, a distinguished role is played by Christoffel functions. As opposed to Lagrange interpolation operators, Hermite–Fejér interpolation operators are not projectors, and thus in general the rate of convergence cannot be expected to equal the rate of the best approximation. Nevertheless, Jackson rates can be obtained.

Unaware of the general theory in [13] and [14] and of a variety of technical tools developed in [6], [9] and [10] (see [11] for a survey), A. K. Varma & J. Prasad in [22] investigated mean convergence of Hermite–Fejér interpolation in a particular case, namely in the case of interpolation based at the zeros of the Chebyshev polynomials. Subsequently, P. Vértesi & Y. Xu [23] wrote a paper dealing with the case of generalized Jacobi polynomials. However, their results left a significant gap between the necessary and the sufficient conditions.

Abstract

Extending a theorem of Alon, we prove a conjecture of Katchalski that every graph of order n and minimal degree at least n/k > 1 contains a cycle of length at least n/(k - 1). The result is best possible for all values of n and k (2 ≤ k < n).

Introduction

A well-known result of Erdös and Gallai [4] states that, for n ≥ k ≥ 3, a graph of order n and size greater than ½(k - 1)(n - 1) has circumference at least k, that is, it contains a cycle of length at least k. According to Dirac's [3] classical theorem, every graph of order n ≥ 3 and minimal degree at least ½n is Hamiltonian. What can one say about the circumference of a graph of order n and minimal degree at least d ≥ 2? Recently Alon [1] came close to giving a complete answer to this question when he proved that for 2 ≤ k < n every graph of order n and minimal degree at least n/k has circumference at least [n/(k - 1)]. Our aim here is to improve on this slightly, namely to show that the assertion holds without the integer sign, as conjectured by Katchalski. Although this seems to need a surprising amount of work, we feel it is worth it since the new result is best possible for all values of n and k (2 ≤ k < n) and implies a complete answer to the question above concerning the minimal circumference of a graph of order n and minimal degree d ≥ 2.

This volume is dedicated to Paul Erdős, who has profoundly influenced mathematics this century. He has worked in number theory, complex analysis, probability theory, geometry, interpolation theory, algebra, set theory and, perhaps above all, in combinatorics. His theorems and conjectures have had a decisive impact. In particular, he, more than anybody else, is the founder of modern combinatorics, he pioneered probabilistic number theory, he is the master of random methods in analysis and combinatorics, and he has created the fields of Ramsey theory and the partition calculus of set theory.

Paul Erdős is the consummate problem solver: his hallmark is the succinct and clever argument, often leading to a solution from ‘the book’. He loves areas of mathematics which do not require an excessive amount of technical knowledge but give scope for ingenuity and surprise. The mathematics of Paul Erdos is the mathematics of beauty and insight.

One of the most attractive ways in which Paul Erdős has influenced mathematics is through a host of stimulating problems and conjectures, to many of which he has attached money prizes, in accordance with their notoriety. He often says that he could not pay up if all his problems were solved at once, but neither could the strongest bank if all its customers withdrew their money at the same time. And the latter is far more likely.