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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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