Markov Inequalities for Polynomials with Restricted Coefficients

Abstract

Essentially sharp Markov-type inequalities are known for various classes of polynomials with constraints including constraints of the coefficients of the polynomials. For and we introduce the class as the collection of all polynomials of the form , , , . In this paper, we prove essentially sharp Markov-type inequalities for polynomials from the classes on . Our main result shows that the Markov factor valid for all polynomials of degree at most on improves to for polynomials in the classes on .

1. Introduction

In this paper, always denotes a nonnegative integer; and always denote absolute positive constants. In this paper will always denote a positive constant depending only on the value of which may vary from place to place. We use the usual notation to denote the Banach space of functions defined on with the norms

(1.1)

We introduce the following classes of polynomials. Let

(1.2)

denote the set of all algebraic polynomials of degree at most with real coefficients. Let

(1.3)

denote the set of all algebraic polynomials of degree at most with complex coefficients. For we introduce the class as the collection of all polynomials of the form

(1.4)

So obviously

(1.5)

The following so-called Markov inequality is an important tool to prove inverse theorems in approximation theory. See, for example, Duffin and Schaeffer [1], Devore and Lorentz [2], and Borwein and Erdelyi [3].

Markov inequality. The inequality

(1.6)

holds for every

It is well known that there have been some improvements of Markov-type inequality when the coefficients of polynomial are restricted; see, for example, [3–7]. In [5], Borwein and Erdélyi restricted the coefficients of polynomials and improved the Markov inequality as in following form.

Theorem 1.1.

There is an absolute constant such that

(1.7)

for every .

We notice that the coefficients of polynomials in only take three integers: and . So, it is natural to raise the question: can we take the coefficients of polynomials as more general integers, and the conclusion of the theorem still holds? This question was not posed by Borwein and Erdélyi in [5, 6]. Also, we have not found the study for the question by now. This paper addresses the question. We shall give an affirmative answer. Indeed, we will prove the following results.

Theorem 1.2.

There are an absolute constant and a positive constant depending only on such that

(1.8)

Our proof follows [6] closely.

Remark 1.3.

Theorem 1.2 does not contradict [6, Theorem 2.4] since the coefficients of polynomials in are assumed to be integers, in which case there is a room for improvement.

2. The Proof of Theorem

In order to prove our main results, we need the following lemmas.

Lemma 2.1.

Let and . Suppose , is analytical inside and on the ellipse , which has focal points and , and major axis

(2.1)

Let be the ellipse with focal points and , and major axis

(2.2)

Then there is an absolute constant such that

(2.3)

Proof.

The proof of Lemma 2.1 is mainly based on the famous Hadamard's Three Circles Theorem and the proof [6, Corollary 3.2]. In fact, if one uses it with replaced by and replaced by , Lemma 2.1 follows immediately from [6, Corollary 3.2].

Lemma 2.2.

Let with , . Suppose and . Then there is a constant such that

(2.4)

Proof.

By Chebyshev's inequality, there is an such that

(2.5)

for every with . Therefore, Because of the assumption on , we can write

(2.6)

Recalling the facts that

(2.7)

, and we obtain

(2.8)

Now by Lemma 2.1 we have

(2.9)

Let , then there is an absolute constant such that

(2.10)

By Cauchy's integral formula and the above inequality, we obtain

(2.11)

The proof of Lemma 2.2 is complete.

Proof of Theorem 1.2.

Noting and the fact

(2.12)

proved by [6], we only need to prove the upper bound. To obtain

(2.13)

we distinguish four cases.

Case 1.

. Let be an arbitrary number in , then

(2.14)

Case 2.

and , where and denotes the number of zeros of at 1. Let be a positive integer. If satisfies the assumptions, then , and . Therefore, Markov inequality implies

(2.15)

Hence

(2.16)

So, the last inequality and imply

(2.17)

Now using Taylor's Theorem, Lemma 2.2 with , the above inequality, and the fact , we obtain

(2.18)

Case 3.

and . Let . We have , where and are the major axis and minor axis of , respectively, and . Let , we see

(2.19)

Denote

(2.20)

The solution of equation is

(2.21)

It is obvious that

(2.22)

So, and the assumption of Lemma 2.2 imply

(2.23)

And from (2.17) and Cauchy's integral formula, it follows that for every ,

(2.24)

and there holds

(2.25)

Case 4.

. Applying Lemma 2.1 with and , we obtain that there is constant such that

(2.26)

Indeed, noting that

(2.27)

we get the result want to be proved by a simple modification of the proof of Lemma 2.2. We omit the details. The proof of Theorem 1.2 is complete.