Some Sharpening and Generalizations of a result of T. J. Rivlin

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by  N. K. Govil

ANALYSISINTHEORYANDAPPLICATIONS
Anal.TheoryAppl.,Vol.x,No.x(201x),pp.1-9
SomeSharpeningandGeneralizationsofaresultofT.J.Rivlin∗
7 N. K. Govil1,∗andEze R.Nwaeze2
1
0 1DepartmentofMathematics,AuburnUniversity,Auburn,AL,36849,USA
2 2DepartmentofMathematics,TuskegeeUniversity,Tuskegee,AL,36088,USA

n
a
J

6 Abstract. Letp(z)=a +a z+a z2+a z3+···+a znbeapolynomialofdegreen. Rivlin
0 1 2 3 n
1
r+1 n
[12]provedthatifp(z)6=0intheunitdisk,thenfor0<r≤1, max|p(z)|≥ max|p(z)|.
] |z|=r 2 |z|=1
V In this paper, we prove a sharpening and generalization of this result(cid:16), and(cid:17)show by
C meansofexamplesthatforsomepolynomialsourresultcansignificantlyimprovethe
. boundobtainedbytheRivlin’sTheorem.
h
at KeyWords: Inequalities,Polynomials,Zeros.
m
AMSSubjectClassifications:15A18;30C10;30C15;30A10
[

ChineseLibraryClassifications:O175.27
1
v
1
6
1
4 1 Introduction
0
. n
1 Let p(z)=∑a zj beapolynomialofdegreen, M(p,r):=max|p(z)|, r>0,
0 j
7 j=0 |z|=r
1 ||p||:=max|p(z)|, and D(0,K):={z:|z|<K}, K>0.Thenitiswellknownthat
: |z|=1
v
i M(p′,1)≤n||p||, (1.1)
X
r and
a
M(p,R)≤Rn||p||, R≥1. (1.2)
The above inequalities are known as Bernstein inequalities, and have been the starting
point of a considerable literature in approximation theory. Several papers and research
monographs have been written on this subject (see, for example Govil and Mohapa-
tra [3], Milovanovic´, Mitrinovic´ and Rassias [6], Rahman [9], Nwaeze [7], and Rahman
∗Thisisapreprintofapaperwhosefinalanddefiniteformispublishedin”AnalysisinTheoryandAppli-
cations.”
∗Correspondingauthor. Emailaddresses: [email protected] (N.K.Govil),[email protected]
(EzeR.Nwaeze)
http://www.global-sci.org/ata/ 1 (cid:13)c201xGlobal-SciencePress
2 N.K.GovilandEzeR.Nwaeze/Anal.TheoryAppl.,x(201x),pp.1-9
andSchmeisser[10,11]).
Forpolynomialsofdegreennotvanishingintheinterioroftheunitcircle, theabove
inequalitieshavebeenreplacedby:
n
M(p′,1)≤ ||p||,
2
and
Rn+1
M(p,R)≤ ||p||, R≥1.
2
(cid:16) (cid:17)
If one applies Inequality (1.2) to the polynomial P(z)=znp(1/z) and use maximum
modulusprinciple,oneeasilygets
n
Theorem1.1. Let p(z)=∑a zj beapolynomialofdegree n.Thenfor0<r≤1,
j
j=0
M(p,r)≥rn||p||. (1.3)
Equalityholdsfor p(z)=αzn,αbeingacomplexnumber.
TheaboveresultisduetoVarga[13]whoattributesittoE.H.Zarantonello.
ItwasshownbyGovil,QaziandRahman[4]thattheinequalities(1.1),(1.2)and(1.3)
areallequivalentinthesensethatanyoftheseinequalitiescanbederivedfromtheother.
TheanalogueofInequality(1.3)forpolynomialsnotvanishingintheinteriorofaunit
circlewasprovedin1960byRivlin[12],whoinfactproved
n
Theorem1.2(Rivlin[12]). Let p(z)=∑a zj6=0in D(0,1).Thenfor0<r≤1,
j
j=0
r+1 n
M(p,r)≥ ||p||.
2
(cid:16) (cid:17)
α+βz n
Theinequalityisbestpossibleandequalityholdsfor p(z)= ,where|α|=|β|=1.
2
(cid:16) (cid:17)
Govil[1]generalizedTheorem1.2byproving
n
Theorem1.3. Let p(z)=∑a zj6=0in D(0,1).Thenfor0<r≤ρ≤1,
j
j=0
1+r n
M(p,r)≥ M(p,ρ). (1.4)
1+ρ
(cid:16) (cid:17)
1+z n
Theresultisbestpossible andequalityholdsforthepolynomial p(z)= .
1+ρ
(cid:16) (cid:17)
N.K.GovilandEzeR.Nwaeze/Anal.TheoryAppl.,x(201x),pp.1-9 3
TherearemanyextensionsofInequality(1.4)(See,forexampleGovil, QaziandRah-
man [4], Govil and Qazi [5], and Qazi [8]). Also, for somemore results in this direction,
seeZirehetal.[14–16].
In this paper, we present some further extensions and sharpening of Rivlin’s result,
Theorem1.2.
2 Main Results
Our first result is the following which, besides generalizing and sharpening several re-
sultsinthisdirection,generalizesandsharpensTheorem1.2duetoRivlin[12].
n
Theorem2.1. Let p(z)=a +∑a zj, 1≤µ<n.If p(z)6=0in|z|<1, thenfor0<r<1,
0 j
j=µ
(1+r)n/µ 2
M(p,r)≥ M(p,1)+nmin|p(z)|ln .
(1+rµ)n/µ+µ2n/µ−µ(1+r)n/µ |z|=1 1+r
h (cid:16) (cid:17)i
Theaboveinequalitybecomesequalityforthepolynomial p(z)=(1+z)n.
If p(z) is a polynomial of degree n having no zeros in |z|<K, K>0, then the poly-
nomial P(z)=p(Kz)6=0 for |z|<1. Further, if 0<r<K, then 0<r/K<1, and applying
Theorem2.1to P(z),weget
(1+r/K)n/µ 2
M(P,r/K)≥ M(P,1)+nmin|P(z)|ln ,
(1+(r/K)µ)n/µ+µ2n/µ−µ(1+r/K)n/µ |z|=1 1+r/K
h (cid:16) (cid:17)i
whichyields
K−n/µ(r+K)n/µ 2K
M(p,r)≥ M(p,K)+nmln ,
K−n(rµ+Kµ)n/µ+µ2n/µ−µK−n/µ(r+K)n/µ r+K
h (cid:16) (cid:17)i
wherem=min|p(z)|.
|z|=K
This,infact,leadstothefollowingmoregeneralresult.
n
Theorem2.2. Let p(z)=a +∑a zj, 1≤µ<n.If p(z)6=0in|z|<K,K>0,thenfor0<r<K,
0 j
j=µ
K−n/µ(r+K)n/µ 2K
M(p,r)≥ M(p,K)+nmln ,
K−n(rµ+Kµ)n/µ+µ2n/µ−µK−n/µ(r+K)n/µ r+K
h (cid:16) (cid:17)i
wherem=min|p(z)|.Again,theequalityholdsforthepolynomial p(z)=(1+z)n.
|z|=K
AsageneralizationandsharpeningofTheorem1.3,wewillbeproving
4 N.K.GovilandEzeR.Nwaeze/Anal.TheoryAppl.,x(201x),pp.1-9
n
Theorem2.3. Let p(z)=∑a zj. If p(z)6=0in|z|<K, K≥1,thenfor0<r<R≤1,
j
j=0
(1+r)n R+K
M(p,r)≥ M(p,R)+nmln ,
(1+r)n+(R+K)n−(r+K)n r+K
h (cid:16) (cid:17)i
wherem=min|p(z)|.
|z|=K
OntakingK=1,theabovetheoremreducesto
n
Corollary2.1. Let p(z)=∑a zj. If p(z)6=0in|z|<1, thenfor0<r<R≤1,
j
j=0
1+r n 1+R
M(p,r)≥ M(p,R)+nmln ,
1+R 1+r
(cid:18) (cid:19)
h (cid:16) (cid:17)i
wherem=min|p(z)|.
|z|=1
Clearly,theabovecorollarysharpensTheorem1.3duetoGovil[1].
Ifwetake R=1,inTheorem2.3,weget
n
Corollary2.2. Let p(z)=∑a zj. If p(z)6=0in|z|<K, K≥1,thenfor0<r<1,
j
j=0
(1+r)n 1+K
M(p,r)≥ M(p,1)+nmin|p(z)|ln .
(1+r)n+(1+K)n−(r+K)n |z|=K r+K
h (cid:16) (cid:17)i
SettingK=1inCorollary2.2gives
n
Corollary2.3. Let p(z)=∑a zj. If p(z)6=0in|z|<1, thenfor0<r<1,
j
j=0
1+r n 2
M(p,r)≥ M(p,1)+nmin|p(z)|ln .
2 |z|=1 1+r
(cid:16) (cid:17) h (cid:16) (cid:17)i
The above corollary clearly sharpens Theorem 1.2 due to Rivlin [12], and excepting
thecasewhenmin|p(z)|=0,theCorollary2.3alwaysgivesaboundthatissharperthan
|z|=1
theboundobtainablefromTheorem1.2.
N.K.GovilandEzeR.Nwaeze/Anal.TheoryAppl.,x(201x),pp.1-9 5
3 Lemmas
FortheproofsofTheorems2.1and2.3,wewillneedthefollowinglemmas.
Inthisdirection,ourfirstlemmaisaresultduetoGovil[2,Corollary1].
Lemma3.1. Let p(z) beapolynomialofdegree nhavingnozerosin|z|<K, K≥1,then
n
max|p′(z)|≤ max|p(z)|−min|p(z)| .
|z|=1 1+K |z|=1 |z|=K
h i
n
Lemma 3.2 (Qazi [8]). Let p(z)=a +∑a zj, 1≤µ<n. If p(z)6=0 for |z|<1, then for
0 j
j=µ
0<r<R≤1,
1+rµ n/µ
M(p,r)≥ M(p,R);
1+Rµ
(cid:16) (cid:17)
moreprecisely,
R tµ+(µ/n)|a /a |tµ−1
µ 0
M(p,r)≥exp −n dt M(p,R).
Zr tµ+1+(µ/n)|aµ/a0|(tµ+t)+1 !
4 Proofs
ProofofTheorem2.1. Let0<r<1,andθ∈[0,2π). Thenwehave:
1
p(eiθ)−p(reiθ) = eiθp′(teiθ)dt ,
(cid:12)Zr (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
whichimplies (cid:12) (cid:12)
(cid:12) (cid:12)
1
p(eiθ) ≤ p(reiθ) + eiθp′(teiθ)dt . (4.1)
(cid:12)Zr (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
If p(z)6=0 in |z|<1, th(cid:12)en p(t(cid:12)z)6=(cid:12) 0 in |z(cid:12)|<(cid:12)(cid:12)1/t. Further, if(cid:12)(cid:12)0<t≤1, then 1/t≥1 and
(cid:12) (cid:12)
hencebyLemma3.1weget
nt
t|p′(tz)|≤ M(p,t)−min|p(z)|
1+t |z|=1
h i
whichisequivalentto
n
|p′(tz)|≤ M(p,t)−min|p(z)| . (4.2)
1+t |z|=1
h i
Combining(4.1)and(4.2)yield
1 n 1 1
p(eiθ) ≤ p(reiθ) + M(p,t)dt−nmin|p(z)| dt.
Zr 1+t |z|=1 Zr 1+t
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
6 N.K.GovilandEzeR.Nwaeze/Anal.TheoryAppl.,x(201x),pp.1-9
whichclearlygives,
1 n 1 1
M(p,1)≤M(p,r)+ M(p,t)dt−nmin|p(z)| dt.
Zr 1+t |z|=1 Zr 1+t
< < <
OnapplyingLemma3.2andnotingthat0 r t 1, weobtain
1 n 1+tµ n/µ 1 1
M(p,1)≤M(p,r)+ M(p,r)dt−nmin|p(z)| dt
Zr 1+t 1+rµ |z|=1 Zr 1+t
(cid:16) (cid:17)
1 n 1+t n/µ 1 1
≤M(p,r)+ M(p,r)dt−nmin|p(z)| dt
Zr 1+t 1+rµ |z|=1 Zr 1+t
(cid:16) (cid:17)
nM(p,r) 1(1+t)n/µ 1 1
=M(p,r)+ dt−nmin|p(z)| dt
(1+rµ)n/µZr 1+t |z|=1 Zr 1+t
nM(p,r) µ 1 1
=M(p,r)+ 2n/µ−(1+r)n/µ −nmin|p(z)| dt
(1+rµ)n/µ n |z|=1 Zr 1+t
h i
µM(p,r) 2
=M(p,r)+ 2n/µ−(1+r)n/µ −nmin|p(z)|ln .
(1+rµ)n/µ |z|=1 1+r
h i (cid:16) (cid:17)
Thusweget
µ2n/µ µ(1+r)n/µ 2
M(p,r) 1+ − ≥M(p,1)+nmin|p(z)|ln
" (1+rµ)n/µ (1+rµ)n/µ# |z|=1 1+r
(cid:16) (cid:17)
whichimplies
(1+rµ)n/µ+µ2n/µ−µ(1+r)n/µ 2
M(p,r) ≥M(p,1)+nmin|p(z)|ln .
" (1+rµ)n/µ # |z|=1 1+r
(cid:16) (cid:17)
Theaboveisclearlyequivalentto
(1+rµ)n/µ 2
M(p,r)≥ M(p,1)+nmin|p(z)|ln ,
(1+rµ)n/µ+µ2n/µ−µ(1+r)n/µ" |z|=1 1+r #
(cid:16) (cid:17)
andthiscompletestheproofofthetheorem.
ProofofTheorem2.3. AsintheproofofTheorem2.1,weobtainsimilarly that
R
p(Reiθ) ≤ p(reiθ) + eiθp′(teiθ)dt . (4.3)
(cid:12)Zr (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Nowif p(z)6=0in|z|<K,K≥1,then p(tz)6=0in(cid:12)(cid:12) |z|<K/t.Furth(cid:12)(cid:12)er,if0<t≤1,then1/t≥1
andK/t≥1.
N.K.GovilandEzeR.Nwaeze/Anal.TheoryAppl.,x(201x),pp.1-9 7
ByLemma3.1,weget
n
|p′(tz)|≤ M(p,t)−min|p(z)| . (4.4)
K+t |z|=K
h i
Usingrelations(4.3)and(4.4),weget
R n R 1
p(Reiθ) ≤ p(reiθ) + M(p,t)dt−nmin|p(z)| dt,
Zr K+t |z|=K Zr K+t
(cid:12) (cid:12) (cid:12) (cid:12)
whichim(cid:12)plies (cid:12) (cid:12) (cid:12)
R n R 1
M(p,R)≤M(p,r)+ M(p,t)dt−nmin|p(z)| dt.
Zr K+t |z|=K Zr K+t
NowusingLemma3.2,weobtain
R n 1+t n R 1
M(p,R)≤M(p,r)+ M(p,r)dt−nmin|p(z)| dt
Zr K+t 1+r |z|=K Zr K+t
(cid:16) (cid:17)
nM(p,r) R(1+t)n R 1
=M(p,r)+ dt−nmin|p(z)| dt
(1+r)n Zr K+t |z|=K Zr K+t
nM(p,r) R(K+t)n R 1
≤M(p,r)+ dt−nmin|p(z)| dt
(1+r)n Zr K+t |z|=K Zr K+t
nM(p,r) 1 K+R
=M(p,r)+ (K+R)n−(K+r)n −nmin|p(z)|ln .
(1+r)n n |z|=K K+r
h i (cid:16) (cid:17)
Therefore,weget
(1+r)n+(K+R)n−(K+r)n K+R
M(p,r) ≥M(p,R)+nmin|p(z)|ln
" (1+r)n # |z|=K K+r
(cid:16) (cid:17)
whichisequivalentto
(1+r)n K+R
M(p,r)≥ M(p,R)+nmin|p(z)|ln ,
(1+r)n+(K+R)n−(K+r)n" |z|=K K+r #
(cid:16) (cid:17)
andtheproofofthetheoremisnowcomplete.
5 Examples
Although, in general, for any polynomial having no zeros on |z|=1, our Theorem 2.3
alwaysgivesaboundsharperthanobtainablebytheknownresults,howeverinthissec-
tionwepresentanexampleofapolynomialtoshowthatinsomecasestheimprovement
canbeconsiderablysignificant,andwedothisbyusingMATLAB.
8 N.K.GovilandEzeR.Nwaeze/Anal.TheoryAppl.,x(201x),pp.1-9
Example
(a). Letp(z)=z3+64,apolynomialofdegreen=3. ThenbyusingMATLAB,onecansee
thatthezerosofthispolynomialare: −4, 2+3.4641i,and2−3.4641i,hence p(z)6=0
in|z|<1. IfweuseTheorem1.3with R=0.5and r=0.1,weget
M(p,r)≥(0.3943704)M(p,R).
Notethat,forthispolynomialm=63,soonusingCorollary2.1ofourTheorem2.3,
weeasilyget
M(p,r)≥(0.3943704)M(p,R)+23.117715,
animprovementofmorethan23overtheboundobtainedbyTheorem1.3.
(b). IfintheaboveexamplewetakeR=1andr=0.1,asin(a),thenwecanapplyRivlin’s
Theorem1.2,andget
M(p,r)≥(0.166375)M(p,R),
whileCorollary2.3ofourTheorem2.3gives
M(p,r)≥(0.166375)M(p,R)+18.79891,
which is an improvementofabout 18.8 overthebound obtainedby Rivlin’s Theo-
rem1.2.
Remark. It may be remarked that in fact one can always construct a polynomial for
whichthisimprovementisgreaterthananygivenpositivenumber.
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