We point out certain flaws in two papers published in Ann. Univ. Mariae Curie-Skłodowska Sect. A, one in 2009 and the other in 2011. We discuss in detail the validity of the results in the two papers in question

Qazi, M. A.

English

University of Maria Curie-Skłodowska (UMCS): Scientific e-Journals / Uniwersytet Marii Curie-Skłodowskiej: e-czasopisma naukowe

doi: 10.2478/v10062-012-0031-zANNALESU N I V E R S I T A T I S M A R I A E C U R I E - S K Ł O D O W S K AL U B L I N – P O L O N I AVOL. LXVII, NO. 2, 2013 SECTIO A 59–64M. A. QAZIRemarks on some recent results about polynomialswith restricted zerosAbstract. We point out certain flaws in two papers published in Ann. Univ.Mariae Curie-Skłodowska Sect. A, one in 2009 and the other in 2011. Wediscuss in detail the validity of the results in the two papers in question.1. Introduction. The following result was proved by Govil [3].Theorem A. Let P (z) be a polynomial of degree n having all its zeros inthe disk |z| ≤ k for some k ≥ 1. Then(1) max|z|=1|P ′(z)| ≥ n1 + knmax|z|=1|P (z)| .The result is best possible and equality holds for P (z) = zn + kn.The next result is also due to Govil [4, p. 184, Theorem D].Theorem B. Let P (z) =∑nk=0 akzk be a polynomial of degree n havingall its zeros on |z| = k for some k ≤ 1. Then(2) max|z|=1|p′(z)| ≤ nkn + kn−1max|z|=1|p(z)| .In [2] the authors state and I quote: “In this paper, we consider a classof polynomials P (z) = c0 +∑nν=µ cνzν , 1 ≤ µ ≤ n and generalize as well as2010 Mathematics Subject Classification. 30A10, 30C10, 30C15.Key words and phrases. Polynomials, extremal properties, Bernstein’s inequality.60 M. A. Qaziimprove upon Theorem A and also generalize Theorem B by proving the fol-lowing results”. They state their “so-called generalizations” of Theorem A,etc. as follows.Theorem 1. If P (z) = c0 +∑nν=µ cνzν , 1 ≤ µ < n is a polynomial ofdegree n, having all its zeros in the disk |z| ≤ k, k ≥ 1, then(3) max|z|=1|P ′(z)| ≥ n1 + kn−µ+1max|z|=1|P (z)| .The result is best possible and equality holds forP (z) = (zn−µ+1 + kn−µ+1)nn−µ+1 .Theorem 2. If P (z) = cnzn +∑nν=µ cn−νzn−ν , 1 ≤ µ < n is a polynomialof degree n, having all its zeros on |z| = k, k ≤ 1, thenmax|z|=1|P ′(z)|≤ nkn−µ+1(n|cn|k2µ + µ|cn−µ|kµ−1µ|cn−µ|(1 + kµ−1) + n|cn|kµ−1(1 + kµ+1))max|z|=1|P (z)| .Theorem 3. If P (z) = c0 +∑nν=µ cνzν , 1 ≤ µ < n is a polynomial ofdegree n, having all its zeros in the disk |z| ≤ k, k ≥ 1, then(4) max|z|=1|P ′(z)| ≥ n1 + kn−µ+1{max|z|=1|P (z)|+ min|z|=k|P (z)|}.The result is best possible and equality holds forP (z) = (zn−µ+1 + kn−µ+1)nn−µ+1 .2. Some comments on Theorems 1, 2 and 3. Unfortunately, Theo-rems 1 and 3 are false. As regards Theorem 2, its proof is based on a lemmathat is erroneous.To see that Theorem 1 is false, let us consider the example P (z) := zn+kn.This is a polynomial which does have the form c0 +∑nν=µ cνzν , 1 ≤ µ < nwithc0 = kn, cν = 0 for ν = µ, . . . , n− 1 , and cn = 1 ,where µ can be taken to be any integer in {1, 2, . . . , n−1}. Besides, it has allits zeros on |z| = k. Clearly, max|z|=1 |P (z)| = 1+kn and max|z|=1 |P ′(z)| =n. Thus, if (3) was true, then we would haven ≥ n1 + kn−µ+1(1 + kn)for any µ ∈ {1, . . . , n − 1}, which amounts to saying that kn−µ+1 ≥ kn forany µ ∈ {1, . . . , n−1}. For k > 1, this is obviously false except when µ = 1.Even if kn−µ+1 = kn when µ = 1 or k = 1, it is of no significance sincewhen µ = 1 or k = 1, the so-called Theorem 1 says nothing more than whatTheorem A does.Remarks on some recent results about polynomials... 61In the face of this counter-example, the authors of [2] might claim thatin zn + kn, which is our counter-example, the coefficients c1, . . . , cn−1 areall zero whereas in Theorem 1, cµ is supposed to be different from 0. So,we shall give a counter-example in which cµ 6= 0.Take any a > 1 and consider the polynomial P (z) := zn+δzµ+an, whereδ is supposed to be positive and small. Since the zeros of P are continuousfunctions [5, p. 9] of δ and those of zn+ an all lie on |z| = a the polynomialP has all its zeros in |z| ≤ k, where |k − a| → 0 as δ → 0. Now, note thatmax|z|=1|P (z)| = 1 + δ + an and max|z|=1|P ′(z)| = n+ δµ .Then, according to Theorem 1, we would have(5)(µ− n1 + kn−µ+1)δ ≥(1 + an1 + kn−µ+1− 1)n =(an − kn−µ+11 + kn−µ+1)n .As δ → 0, (an − kn−µ+11 + kn−µ+1)n→(an − an−µ+11 + an−µ+1)n ,which is strictly positive if 1 < µ < n − 1. Hence, for any such µ, thereexists a positive number δ0 such that(an − kn−µ+11 + kn−µ+1)n >12(an − an−µ+11 + an−µ+1)n for 0 < δ < δ0 .Now, from (5) it follows that(µ− n1 + kn−µ+1)δ >12(an − an−µ+11 + an−µ+1)nfor 0 < δ < δ0. This cannot be true since the expression on the left-hand side of the inequality tends to 0 as δ → 0 whereas the expressionon the right-hand side is a positive constant. The second sentence in thestatement of Theorem 1 is: “The result is best possible and equality holdsfor P (z) = (zn−µ+1 + kn−µ+1)nn−µ+1 ”. This statement implicitly presumesthat (zn−µ+1 + kn−µ+1)n/(n−µ+1) is a polynomial. However, for (zn−µ+1 +kn−µ+1)n/(n−µ+1) to be a polynomial, n must be divisible by n − µ + 1.Surprisingly, the authors do not seem to realize this. This remark alsoapplies to the second sentence in the statement of Theorem 3.Since Theorem 1 is false, as we have shown above, Theorem 3 cannot betrue either because it clearly says more than what Theorem 1 does.The above comments clearly debunk Theorems 1 and 3 of Dewan andHans.62 M. A. Qazi2.1. The principal error in the proofs of Theorems 1 and 3. SinceTheorems 1 and 3 are invalid, there must be something wrong with theirproofs. This had to be looked into, which we did. We found a seriousmistake in the proof of Lemma 1 of their paper [2]. It is applied to obtainLemma 2, which the authors use to prove Theorem 1. Here is what Lemma 1of Dewan and Hans says.Lemma 1. If P (z) = c0+∑nν=µ cνzν , 1 ≤ µ < n is a polynomial of degreen, having all its zeros in the disk |z| ≤ k, k ≥ 1, then for |z| = 1(6) kn+µ−3|Q′(z)| ≤ |P ′(k2z)| ,where Q(z) = znP (1/z).The polynomial P (z) := zn+kn satisfies the conditions of Lemma 1 withany µ such that 1 ≤ µ < n. For this polynomial, (6) reduces to kµ−1 ≤ 1,which clearly does not hold for any µ > 1 if k > 1. This shows that Lemma 1is false for 2 ≤ µ ≤ n− 1 and k > 1.The authors use the faulty Lemma 1 to prove Lemma 2, stated as follows.Lemma 2. If P (z) = c0 +∑nν=µ cνzν , 1 ≤ µ < n is a polynomial of degreen, having all its zeros in the disk |z| ≤ k, k ≥ 1, thenmax|z|=1|Q′(z)| ≤ kn−µ+1 max|z|=1|P ′(z)| ,where Q(z) = znP (1/z).The example zn+kn shows that this lemma is also false for 2 ≤ µ ≤ n−1and k > 1.We note that the proof of Theorem 1, as given by Dewan and Hans, usesLemma 2. Since Lemma 2 is deduced from Lemma 1, it is desirable toidentify the error in the proof of Lemma 1 as presented by the authors onpages 57–58 of [2]. So, we shall do that.Using a standard argument, the authors conclude that ([2, p. 58], see(2.3))kn−1|Q′(z/k)| ≤ k|P ′(kz)| for |z| ≥ 1 .This is fine. Since c1 = · · · = cµ−1 = 0, this can be written askn−1|Q′(z/k)| ≤ k∣∣∣∣∣(kz)µ−1n∑ν=µνcν(kz)ν−µ∣∣∣∣∣ for |z| ≥ 1 .In particular, the authors say (see inequality (2.4) of their paper) that(7) kn−1|Q′(z/k)| ≤ kµ∣∣∣∣∣n∑ν=µνcν(kz)ν−µ∣∣∣∣∣for |z| = 1. We agree with this. Next, they say that ∑nν=µ νcν(kz)ν−µ 6= 0in |z| > 1 and we agree once again. Then they make the bizarre assertionRemarks on some recent results about polynomials... 63that by maximum modulus principle it (by which they mean (7)) also holdsfor |z| > 1. They overlook that for this to be truekn−1|Q′(z/k)|kµ∣∣∣∑nν=µ νcν(kz)ν−µ∣∣∣must tend to a finite limit as z →∞. Except in the case where c0 = 0 theabove mentioned quotient tends to infinity as z → ∞. Thus, the proof ofLemma 1 is based on a false application of the maximum modulus principle.We are sorry to add that the authors apply Lemma 2 to prove anotherlemma which they state as follows.Lemma 3. If P (z) = cnzn +∑nν=µ cn−νzn−ν , 1 ≤ µ < n is a polynomialof degree n, having no zeros in the disk |z| ≤ k, k ≤ 1, thenkn−µ+1 max|z|=1|P ′(z)| ≤ max|z|=1|Q′(z)| ,where Q(z) = znP (1/z).Once again, the example zn + kn shows that this lemma is also invalidfor 2 ≤ µ ≤ n− 1 and k < 1.3. Another related paper. The authors have gone on to use their faultyLemmas 1, 2 and 3 in another paper, namely [1] published in Ann. Univ.Mariae Curie-Skłodowska Sect. A in the year 2011. As we shall explain,Theorems 1 and 2 of [1] are not true. The results in [1] involve the notionof polar derivative. The polar derivative of a polynomial P (z) with respectto a point α, denoted by DαP (z), is defined byDαP (z) = nP (z) + (α− z)P ′(z) .Theorem 1 of [1] can be stated as follows. Because of its obvious relation-ship with Theorem 2 of [2], stated above as Theorem 2, we shall name itTheorem 2b.Theorem 2b. If P (z) = cnzn+∑nj=µ cn−jzn−j , 1 ≤ µ < n, is a polynomialof degree n having all its zeros on |z| = k, k ≤ 1, then for every real orcomplex number α with |α| ≥ k, we have(8)max|z|=1|DαP (z)|≤ n(|α|+ kµ)kn−µ+1n|cn|k2µ + µ|cn−µ|kµ−1µ|cn−µ|(1 + kµ−1) + n|cn|kµ−1(1 + kµ+1) max|z|=1 |P (z)| .For P (z) := zn+kn, k ≤ 1, which is a polynomial satisfying the conditionsof Theorem 2b, inequality (8) says that(9) kn + |α| ≤ kµ + |α|kn−µ+1k2µkµ−1(1 + kµ+1)64 M. A. Qaziand so a fortiori kn−2µ(kn + |α|) ≤ kµ + |α|, that isk2n−2µ + kn−2µ|α| ≤ kµ + |α| .If k < 1, then k2n−2µ > kµ and kn−2µ > 1 for any µ > 2n/3. Thus, (9)and so also (8) cannot hold for any µ > 2n/3. As indicated by the authors(see [1, pp. 6–7, §3]) the proof of Theorem 2b uses Lemma 2 of [1], which isthe same as Lemma 3 of [2], cited above. Since Lemma 3 of [2] is false, aswe have already indicated, their proof of Theorem 2b is invalid and there isreally no need to look for counter-examples to (8) for µ ≤ 2n/3.4. The polynomials considered by Dewan and Hans. It seems thatDewan and Hans overlooked the fact that in inequality (1) of Govil, equalityholds for P (z) := zn + kn, which is a polynomial of the form P (z) :=c0+∑nν=µ cνzν . To think that they could improve upon (1), by consideringpolynomials which are of the form P (z) := c0 +∑nν=µ cνzν , was not apromising idea to start with. They could obtain a stronger conclusion thanthat of Theorem A only if they considered a class of polynomials which didnot contain the polynomial zn + kn. In fact, there is no raison d’ e`tre forTheorems 1 and 3. Not only are their proofs not correct, their statementsare false. The problem with Theorem 2 is of a different nature; its proofuses Lemma 3, which is faulty.References[1] Ahuja, A., Dewan, K. K., Hans, S., Inequalities concerning polar derivative of polyno-mials, Ann. Univ. Mariae Curie-Skłodowska Sect. A 65 (2011), 1–9.[2] Dewan, K. K., Hans, S., On maximum modulus for the derivative of a polynomial ,Ann. Univ. Mariae Curie-Skłodowska Sect. A 63 (2009), 55–62.[3] Govil, N. K., On the derivative of a polynomial , Proc. Amer. Math. Soc. 41 (1973),543–546.[4] Govil, N. K., On a theorem of S. Bernstein, J. Math. Phys. Sci. 14 (1980), 183–187.[5] Rahman, Q. I., Schmeisser, G., Analytic Theory of Polynomials, Clarendon Press,Oxford, 2002.M. A. QaziDepartment of MathematicsTuskegee UniversityTuskegee, Alabama 36088U.S.A.e-mail: qazima@aol.comReceived May 12, 2013

Remarks on some recent results about polynomials with restricted zeros

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Remarks on some recent results about polynomials with restricted zeros

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