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        Erd?s Type Inequality for Lorentz Polynomials

        2019-01-08 00:58:16LaiyiZhuDapengZhouandZhiyongHuang
        Analysis in Theory and Applications 2018年3期

        Laiyi Zhu,Dapeng Zhouand Zhiyong Huang

        School of Mathematics,Renmin University of China,Beijing 100872,China

        Abstract.An elementary,but very useful tool for proving inequalities for polynomials with restricted zeros is the Bernstein or Lorentz representation of polynomials.In the present paper,we give two classes of Lorentz polynomials,for which the Erd?s-type inequality holds.

        Key Words:Lorentz representation of polynomials,constrained polynomials,Morkov-type inequalities,Erd?s-type inequality.

        1 Introduction and main results

        Let πndenote the set of all polynomials of degree at most n with real coefficients,and πcndenote the set of all polynomials of degree at most n with complex coefficients,where n is a nonnegative integer.

        Let

        denote the supremum norm of a function f defined on a set A.The Markov inequality is that

        holds for all pn∈.And the Bernstein inequality is that

        holds for all pn∈and for all x∈(-1,1).For proofs of these see[2]or[3].Polynomial inequalities are very basic in several disciplines and they play a foundational role in approximation theory.There are a lot of papers devoted to them;see,e.g.,[2]and[11].In[7],Erd?s gave a class of restricted polynomials for which the Markov factor n2in(1.1)improves to cn.He proved the following theorem.

        Theorem 1.1.The inequality

        holds for all pn∈πnhaving all their zeros in R(-1,1).

        And in[7],Erd?s claimed the following theorem but did not give a hint to prove it.Recently,Erdélyi proved it in[5].Using different approach,We gave another proof of Theorem E2in[14].

        Theorem 1.2.The inequality

        holds for all pn∈πnhaving all their zeros in R(-1,1)and

        These results motivated several people to study Markov- and Bernstein-type inequalitiesfor polynomials with restricted zeros and under some other constraints. These peopleinclude Lorentz[8],Scheick[12],Szabados[13],Máté[10],P.Borwein[1],Erdélyi[4,5]and others.

        An elementary,but very useful tool for proving inequalities for polynomials with restricted zeros is the Bernstein or Lorentz representation of polynomials;see,e.g.,[2]and[6].

        Let

        And if pn(x)∈Bn(-1,1)or-pn(x)∈Bn(-1,1),we call it Lorentz polynomial.As Lorentz observed that,if pn(x)∈πnhaving all their zeros outside the open unit disk,then pn(x)is a Lorentz polynomial.

        In this paper,we give two classes of Lorentz polynomials,for which the Erd?s-type inequality(1.4)holds.Our main results are the following.

        Theorem 1.3.Let n≥1 be an integer and pn(x)∈πnπn-1be a Lorentz polynomial.If,then

        Equality holds only for pn(x)=σc(1+σx)nor pn(x)=σc[2n-(1-σx)n]withaconstant0≠c∈R and σ∈{-1,1}.

        Theorem 1.4.Let n≥3 be an integer and pn(x)∈πnπn-1be a Lorentz polynomial with Lorentz polynomial representation

        then

        Moreover,if

        then equality holds only for pn(x)=c1(1+x)n-2[(1+x)2+c2(1-x)2]with constants c1>0 and 0≤c2≤1.

        2 Some lemmas

        In order to prove our main results,we need the following lemmas.The first one is the well known Lucas'theorem(to see[9]or[2]).

        The next lemma is a result ascribed to Lorentz,to see[2]or[5].

        Lemma 2.2.Suppose p∈πnhas its zeros outside the open unit disk,then pn(z)is a Lorentz polynomial.

        To look at the conditions for a Lorentz polynomial pn(x)∈πn,we have the following two lemmas.

        Lemma 2.3.Let n≥1 be an integer and pn(x)∈πn.For a Lorentz polynomial

        or

        Proof.This follows from the simple calculations

        So,we complete the proof.

        From(2.1)and(2.2),we obtain immediately the following lemma.

        Lemma 2.4.Let n≥1 be an integer and pn(x)∈πn.For a Lorentz polynomial

        moreover,if a0=an,thenholds if and only if

        3 Proofs of the main results

        Proof of Theorem 1.3.Let pn∈πnπn-1be a Lorentz polynomial.Without loss of generality we may assume that

        Thus,(2.3)and(2.4)imply that for all x∈[-1,1],

        So that

        and for all x∈[-1,1],

        Hence

        For the case that equality holds,without loss of generality,we may assume that there exists an x0∈[-1,1]such that

        Then(3.2),yields x0=1 or x0=-1.If x0=1,then it is from(2.3)that na0-a1=na0,which implies

        hence

        If x0=-1,then it is from(2.3)that an-1-nan=na0.From(2.1),we can deduce that an=0 and

        So

        from which it follows that

        and the Theorem is proved.

        Remark 3.1.(a)If pn∈πnhas all its zeros in R(-1,1)and(x)≠0,x∈(-1,1),then it is from Lemma 2.2 that pn∈πnπn-1is a Lorentz polynomial and pn(x)∈Bn(-1,1)or-pn(x)∈Bn(-1,1).The following example shows that Theorem 1.3 extends essentially Theorem 1.2.

        Example 3.1.Let m≥1 be an integer.Then for

        with all its zeros at x=1±i and x=m.It follows from Lemma 2.1 that p(k)(x),k=1,2,···,n-1,have all their zeros in the triangle formed by x=1±i and x=m which are certainly outside the open unit disk.Hence from Lemma 2.2,pn(x)∈πnπn-1is a Lorentz polynomial and-(x)∈ Bn-1(-1,1).By simple calculations,we get

        and

        Remark 3.2.(b)There is an alternative proof of(1.6).In fact,suppose pn(x)∈Bn(-1,1)andThen it follows from Lemma 2.3 and Lemma 2.4(or even by one's geometric intuition)that

        Applying Theorem 2.3 in[5]with Qn(x),we obtain

        which gives the alternative proof of(1.6).

        We emphasize here that the methods we used to deduce(1.6)is innovative and quiet different from T.Erdélyi's.We provide a direct and concise elementary proof to Theorem 1.3 using only undergraduate calculus.

        Proof of Theorem 1.4.Let n≥3be an integerand pn(x)∈πnπn-1with Lorentz polynomialrepresentation

        It follows from(1.7)and(2.3)that for x∈[-1,1]

        Moreover,if

        Then(3.7)implies x0=1 or x0=-1.If x0=1,then it is from(2.3)that na0-a1=na0.So(1.8)deduce that

        and

        If x0=-1,then it is from(2.3)that an-1-nan=na0.However,(1.8)implies an-1≤a0,so nan+na0≤a0,a contradiction,which finishes the proof. ?

        Remark 3.3.(a).By the transformation t=-x,we obtain the following result:Let n≥3 be an integer and pn(x)∈πnπn-1be a Lorentz polynomial with Lorentz polynomial representation

        then

        Moreover,if

        then equality holds only for pn(x)=c1(1-x)n-2[(1-x)2+c2(1+x)2]with constants c1>0 and 0≤c2≤1.

        (b).The following example satisfies the conditions of Theorem 1.4,which also shows that the assumption ”(x)∈ Bn-1(-1,1)or-(x)∈ Bn-1(-1,1)” is not necessary for Lorentz polynomial pn(x)to satisfy the Erd?s inequality,and the necessary conditions for a Lorentz polynomial pn(x)to satisfy the Erd?s inequality remain to be solved.

        Example 3.2.Let n≥3 be an integer.Consider Lorentz polynomial

        Then

        then

        Acknowledgements

        The research has been supported by the National Natural Science Foundation of China under Grant No.11571362.The authors would like to thank the referee for useful comments and suggestions which improved the presentation of this paper.

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