L On the Markov inequality in the 2-norm with the Gegenbauer weight 7 G.Nikolov, A.Shadrin 1 0 2 n Abstract a J Letwλ(t):=(1−t2)λ−1/2,whereλ>−21,betheGegenbauerweightfunction,letk·kwλ be 6 theassociatedL2-norm, 1 1/2 2 2 kfk = |f(x)| w (x)dx , wλ (cid:26)Z−1 λ (cid:27) ] A anddenotebyPnthespaceofalgebraicpolynomialsofdegree≤n.Westudythebestconstant C cn(λ)intheMarkovinequalityinthisnorm . kp′k ≤c (λ)kp k , p ∈P , h n wλ n n wλ n n t a namelytheconstant m kp′k c (λ):= sup n wλ . [ n pn∈Pn kpnkwλ 1 WederiveexplicitlowerandupperboundsfortheMarkovconstantcn(λ),whicharevalidfor v allnandλ. 2 MSC2010:41A17 8 6 Keywordsandphrases:Markovtypeinequalities,Gegenbauerpolynomials,matrixnorms 7 0 . 1 Introduction 1 0 7 Letw (t) := (1 t2)λ 1/2,whereλ > 1,betheGegenbauerweightfunction, let bethe 1 λ − − −2 k·kwλ associatedL -norm, : 2 v 1 1/2 i f = f(x)2w (x)dx , X k kwλ | | λ (cid:26)Z−1 (cid:27) r a anddenoteby n thespaceofalgebraicpolynomialsofdegree n. Inthispaper,westudythe P ≤ bestconstantc (λ)intheMarkovinequalityinthisnorm n kp′nkwλ ≤cn(λ)kpnkwλ, pn ∈Pn, (1.1) namelytheconstant p c (λ):= sup k ′nkwλ . n p pn∈Pn k nkwλ Our goal is to derive good and explicit lower and upper bounds for the Markov constant c (λ) n whicharevalidforallnandλ,i.e.,tofindconstantsc(n,λ)andc(n,λ)suchthat c(n,λ) c (λ) c(n,λ), n ≤ ≤ withasmallratio c(n,λ). c(n,λ) 1 Itisknownthat,forafixedλ,c (λ)growslike (n2),andthattheasymptoticvalue n O c (λ) n c (λ):= lim ∗ n n2 →∞ isequalto1/(2j2λ−3),withjν beingthefirstpositivezerooftheBesselfunctionJν,see[2,Thms. 4 1.1–1.3],wherebyitcanbeshownthatc (λ)behaveslike (λ 1). Thereisalsoanumberofmore − ∗ O preciseresults. Forλ= 12 (theconstantweightw21 ≡1),itfollowsfromtheSchmidtresult[4]that 1 3 1 1 (n+ )2 c ( ) (n+2)2. π 2 ≤ n 2 ≤ π Forλ=0,1(theChebyshevweightsw (x)= 1 andw (x)=√1 x2,respectively),Nikolov 0 √1 x2 1 − [3]provedthat − 0.472135n2 c (0) 0.478849(n+2)2, n ≤ ≤ (1.2) 5 0.248549n2 c (1) 0.256861(n+ )2. n 2 ≤ ≤ In[1],weobtainedanupperboundvalidforallnandλ, (n+1)(n+2λ+1) c (λ) , (1.3) n ≤ 2√2λ+1 however,thealreadymentionedasymptoticsc (λ)= (λ 1)showsthatthisresultisnotoptimal. − ∗ O The main result of this paper is lower and upper bounds for c (λ) which are uniform with n respecttonandλ. Theyshow,inparticular,that 1 [c (λ)]2 n(n+2λ)3. n ≍ λ2 Forn=1,2theexactvaluesoftheMarkovconstantareeasilycomputable: 4(2+λ)(2+2λ) [c (λ)]2 =2(1+λ), [c (λ)]2 = . (1.4) 1 2 2λ+1 Therefore,weconsiderbelowthecasen 3. Ourmainresultis ≥ Theorem1.1 Forallλ> 1 andn 3,thebestconstantc (λ)intheMarkovinequality −2 ≥ n kp′nkwλ ≤cn(λ)kpnkwλ, pn ∈Pn, admitstheestimates 1 n2(n+λ)2 n(n+2λ+2)3 < [c (λ)]2 < , λ 2; (1.5) n 4(λ+1)(λ+2) (λ+2)(λ+3) ≥ (n+λ)2(n+2λ′)2 < [c (λ)]2 < (n+λ+λ′′+2)4, λ> 1, (1.6) n 2 (2λ+1)(2λ+5) 2(2λ+1)√2λ+5 − whereλ =min 0,λ , λ =max 0,λ . ′ ′′ { } { } Asaconsequence,wecanspecifythefollowingboundsfortheasymptoticvaluec (λ): ∗ Corollary1.2 For any λ > 1, theasymptotic Markov constant c (λ) = lim n 2c (λ) satisfies the −2 ∗ n − n inequalities →∞ 1 , 1 <λ λ , 1 2(2λ+1)√2λ+5 −2 ≤ ∗ <[c (λ)]2 < (2λ+1)(2λ+5) ∗  1 , λ>λ∗, (λ+2)(λ+3) whereλ∗ 25.  ≈ 2 Thelowerboundin(1.5)followsfromthatin(1.6)andislessaccurate,weputitinthisform tomakethecomparisonbetweenthetwoboundsin(1.5)moreobvious. Theupperboundin(1.6)doesnothavetherightorder (n4/λ2)in λ (forλfixed),however O thisboundservesnotonlyforthecase 1 < λ < 2,butforafixedλ [2,λ ]andn n (λ)itis −2 ∈ ∗ ≥ 0 alsobetterthantheonein(1.5). Inthenextcorollary,wesetλ = 0,1intheupperestimate(1.6),andthatimprovestheupper estimatesin(1.2)forthe Chebyshevweights. Whencoupledwiththelower estimatefrom(1.2), thisgivesrathertightbounds. Corollary1.3 FortheChebyshevweightsw (x)= 1 andw (x)=√1 x2,wehave 0 √1 x2 1 − − 0.472135n2 c (0) 0.472871(n+2)2, n ≤ ≤ 0.248549n2 c (1) 0.250987(n+4)2. n ≤ ≤ Thelower andupperestimatesin(1.5)havedifferentorderswith respecttoλ. Howeverwe cangetaperfectmatchwithslightlylessaccurateconstants. Theorem1.4 Forallλ 7andn 3,thebestconstantc (λ)intheMarkovinequalitysatisfies n ≥ ≥ 1 n(n+2λ)3 n(n+2λ)3 [c (λ)]2 . (1.7) 16 λ2 ≤ n ≤ λ2 Corollary1.5 FortheMarkovconstantc (λ)wehavethefollowingasymptoticestimates: n c (λ) i) √n lim n √3n; ≤λ √2λ ≤ →∞ 1 3 ii) (n )(n 1) lim c (λ) 2√2λ+1 (n+ )2. 2 n 2 − − ≤λ 1 · ≤ →−2 Partii)followsfrom(1.6).Thoughparti)doesnotformallyfollowfromTheorem1.4,itfollows fromapartofitsproof. Letusdescribebrieflyhowtheseresultsareobtained. Itiswell-knownthatthesquaredbestconstantintheMarkovinequalityintheL -normwith 2 arbitrary(andpossiblydifferent)weightsforpandp isequaltothelargesteigenvalueofacertain ′ positivedefinitematrix,inourcasewehave [c (λ)]2 =µ (B ), (1.8) n max n wherethematrixB isspecifiedinSect.2. Weobtainthenlowerandupperboundsforµ (B ) n max n usingthreevaluesassociatedwiththematrixB anditseigenvalues(µ )(notethatµ >0): n i i a)thetrace tr(B ):= b = µ ; n ii i b)themax-norm X X B =max b ; n ij k k∞ i | | j X c)theFrobeniusnorm B 2 := b 2 =tr(B BT)= µ2. k nkF | ij| n n i i,j X X Clearly,wehave i) µ tr(B ), ii) µ B , iii) µ B , (1.9) max n max n max n F ≤ ≤k k∞ ≤k k 3 andgenerallyµ B ,where isanymatrixnorm. Theupperestimate(1.3)citedfrom max n [1]isexactlythefirs≤tinkequka∗lityµ k·kt∗r(B ),andaswenoted,thisestimateisnotoptimal. The max n ≤ betterupperbounds(1.5)-(1.6)inTheorem1.1areobtainedfrom(1.9.ii)and(1.9.iii),respectively. Forthelowerboundsweusetheinequalities µ2 B 2 i′) µmax ≥ Pµii = tkr(BnknF), ii′) µmax(Bn)≥miaxbii. (1.10) Inequality (i’) givesthe lower ePstimates in (1.5)-(1.6),and combination of (i’) and (ii’) yields the lowerboundin(1.7). The paper is organised as follows. In Sect.2, following our previous studies [1], we give an explicit form of the matrix B appearingin (1.8). Sects.2-4 contain some auxiliary inequalities. n In Sect.5, we find anupper bound for the max-norm B , and in Sect.6 we give both lower n and upper estimates for the Frobinuis norm B . Fkinalkl∞y, in Sect.7 we prove the upper and n F k k thelowerestimatesinTheorems1.1-1.4usinginequalities(1.9)-(1.10)andrelation(1.8). Herewe haveusedtheexpressionfortr(B )andfordiagonalelementsb foundin[1]. n ii The formulas for the trace, the max-norm and the Frobenius norm of a matrix are straight- forwardonce thematrixelementsareknown, sothemaintechnicalissuesare,firstly, infinding reasonableupperandlowerboundsfortheentriesofthematrixB = (b )whichareexpressed n ij initiallyintermsoftheGammafunctionΓ,and,secondly,infindingreasonableestimatesfortheir sums. ThefirstissueisdealtwithinSect.3,whereweshowthat b fσ(j) , f (x)=xα1(x+ λ)α2(x+λ)α3, jk α 2 ≍ f (k) τ and the second one in Sect.4, where we give elementarybut effectiveupper andlower bounds fortheintegralsofthetype x f(t)dt, f(x)=(x+γ )α1(x+γ )α2 (x+γ )αr. 1 2 r ··· Zx0 2 Preliminaries Inthissection,wequotearesultobtainedearlierin[1],whichequatetheMarkovconstantc (λ) n withthelargesteigenvalueofaspecificmatrixB . n Definition2.1 Forn N,setm:= n+1 anddefinesymmetricpositivedefinitematricesA ,A Rm m withentriesa∈anda give⌊n2by⌋ m m ∈ × kj kj e min(k,j) min(k,j) e a := α2 β β , a := α2 β β , (2.1) kj i k j kj i k j (cid:16) Xi=1 (cid:17) (cid:16) Xi=1 (cid:17) e e e e sothat α2β2 α2β β α2β β α2β β 1 1 1 1 2 1 1 3 ··· 1 1 m α2β β 2 α2 β2 2 α2 β β 2 α2 β β  1 1 2 i=1 i 2 i=1 i 2 3 ··· i=1 i 2 m Am :=α21β1β3 (cid:16)P2i=1α2i (cid:17)β2β3 (cid:16)P 3i=1α(cid:17)2i β32 ··· (cid:16)P3i=1α2i(cid:17)β3βm , (2.2)  ... (cid:16)P ... (cid:17) (cid:16)P ... (cid:17) ... (cid:16)P ... (cid:17)      α2β β 2 α2 β β 3 α3 β β m α2 β2   1 1 m i=1 i 2 m i=1 i 3 m ··· i=1 i m   (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)  P P P 4 withthesameoutlookforA . Thenumbersα ,β andα ,β aregivenby m k k k k 1 eαk :=(2k−1+λ)h2k−1, βek :=eh2k ; (2.3) 1 α :=(2k 2+λ)h , β := , (2.4) k 2k 2 k h − − 2k−1 where e e Γ(i+2λ) h2 :=h2 := . (2.5) i i,λ (i+λ)Γ(i+1) Notethat αk =αk 1 , βk =βk 1 . (2.6) −2 −2 Definition2.2 Forn∈N,set e e 4A , n=2m; B := m (2.7) n  4A , n=2m 1.  m − Theorem2.3([1],Theorem3.2) Letcn(λ)beethebestconstantintheMarkovinequality(1.1).Then [c (λ)]2 =µ (B ), n max n whereµ (B )isthelargesteigenvalueofthematrixB . max n n Remark2.4 AppearanceoftwomatricesA andA reflectsthefactthattheextremepolynomial m m p fortheMarkovinequalitywithanevenweightfunctionw(x) = w( x)iseitheroddoreven. n − Thelatterisarelativelysimpleconclusion,whatisenotobviousthoughiswhetherp isofdegree n ebxactlynandnotn 1. In[1],weprovedthatfortheGegenbauerweightswλ, − b µ (A )<µ (A )<µ (A ) max m max m max m+1 andthisimpliesthatdegpn =n,heence[cn(λ)]2 isthelargesteeigenvalueofAm orAm forn=2m orn=2m 1,respectively. − b e We finish this section by simplifying the expressions for a and thus for the matrix A as kj m follows. From(2.1),wederive min(k,j) βk β2 j α2 , j <k, a := α2 β β = βj j i=1 i kj (cid:16) Xi=1 i(cid:17) k j  ββkj (cid:0)βk2Pki=1α2i(cid:1), j >k, (cid:0) P (cid:1) sothat  j βk a , j <k, a =β2 α2, a = βj jj (2.8) jj j i kj  βj a , j >k. Xi=1  βk kk Respectively,  a β2a β3a βma 11 β1 11 β1 11 ··· β1 11 β2a a β3a βma  β1 11 22 β2 22 ··· β2 22 A =β3a β3a a βma  . m β1 11 β2 22 33 ··· β3 33  ... ... ... ... ...    ββm1 a11 ββm2 a22 ββm3 a33 ··· amm  Note thatA and A areembedded. An analogous representationand embedding hold for m m+1 A . m e 5 a β 3 Estimates for and k kk β j WewillneedupperandlowerestimatesfortheelementsofmatricesA andA ,namely m m k βk a , j <k, e a =β2 α2, a = βj jj kk k i kj  βj a , j >k. Xi=1  βk kk Wefoundexpressionfora anda in[1,Lemmas2.1(ii)and2.2(ii)],thosearequotedinPropo- kk kk sition3.1,andinthissectionweobtaininequalitiesfortheratios βk. βj e Proposition3.1([1]) Thefollowingidentitieshold: k (i) a :=β2 α2 = c f (k), (3.1) kk k i 0 0 i=1 X k 1 (ii) a :=β2 α2 = c f (k ), (3.2) kk k i 0 0 − 2 i=1 X e e e where 4 λ c := , f (x):=x(x+ )(x+λ). 0 0 2 2λ+1 Proposition3.2 Letj,k N,j <k. Thenthecoefficientsβ in(2.3)satisfythefollowingrelations: k ∈ (i) If 1 <λ 0 or λ 1,then −2 ≤ ≥ j 2λ 2 β2 j+λ 2λ 2 − k − . (3.3) k ≤ β2 ≤ k+λ j (cid:16) (cid:17) (cid:16) (cid:17) (ii) If 0<λ 1,then ≤ j 2λ 2 β2 j+λ 2λ 2 − k − . (3.4) k ≥ β2 ≥ k+λ j (cid:16) (cid:17) (cid:16) (cid:17) Proof. Denotetheleft-hand,themiddleandtheright-handsidetermsin(3.3)-(3.4)byℓ(λ),m(λ) andr(λ),respectively.Fromdefinitions(2.3)and(2.5)wehave β2 Γ(2j+2λ) Γ(2k+2λ) 1 m(λ):= k = − , (3.5) β2 (2j+λ)Γ(2j+1) (2k+λ)Γ(2k+1) j (cid:16) (cid:17) andusingthefunctionalequationΓ(t+1)=tΓ(t)weseethat k 2, λ=0, m(λ)= j ℓ(λ)=m(λ)=r(λ), λ=0,1. (3.6)  ⇒ (cid:0) 1(cid:1), λ=1,  Weshallproveinequalities(3.3)-(3.4)forthelogarithmsofthevaluesinvolved. 1) Let us start with the proof of the left-hand side inequalities in (3.3)-(3.4). Consider the differenceofthelogarithmsofthemiddleandtheleft-handsideterms, j g(λ):=logm(λ) logℓ(λ)=logm(λ) (2λ 2)log − − − k Weneedtoprovethatg(λ) 0forλ [0,1]andthatg(λ) > 0otherwise. Sinceg(0) = g(1) = 0 ≤ ∈ by(3.6),itsufficestoshowthatg (λ)>0forallλ> 1,i.e.,that[logm(λ)] >0. ′′ −2 ′′ 6 From(3.5),wehave 2j+λ Γ(2j+1) logm(λ)=logΓ(2j+2λ) logΓ(2k+2λ) log log , − − 2k+λ − Γ(2k+1) therefore,usingthedigammafunctionψ(t):=Γ(t)/Γ(t),weobtain ′ 1 1 [logm(λ)] =2 ψ(2j+2λ) ψ(2k+2λ) . ′ − − 2j+λ − 2k+λ (cid:2) (cid:3) h i FromtheequationΓ(t+1)=tΓ(t)itfollowsthatψ(t+1)=ψ(t)+1/t,andthelatterimplies 2k 1 − 1 1 1 [logm(λ)]′ = 2 , (3.7) − i+2λ − 2j+λ − 2k+λ iX=2j h i whence 2k 1 − 1 1 1 [logm(λ)] =4 + >0, ′′ (i+2λ)2 (2j+λ)2 − (2k+λ)2 iX=2j h i andthatprovestheleft-handinequalitiesin(3.3)-(3.4). 2)Weapproachinthesamewaytotheproofoftheright-hand inequalitiesin(3.3)and(3.4), bytakingthedifferenceofthelogarithmsofthemiddleandtheright-handterms, j+λ h(λ):=logm(λ) logr(λ)=logm(λ) (2λ 2)log . (3.8) − − − k+λ Weneedtoshowthath(λ) 0forλ [0,1]andthath(λ) < 0otherwise. Sinceh(0) = h(1) = 0 ≥ ∈ by(3.6),itsufficestoshowthath(λ)<0forλ>1andthath (λ)<0forλ ( 1,1]. ′ ′′ ∈ −2 2a)Letusshowthath(λ) 0forλ 1. From(3.8)using(3.7),weobtain ′ ≤ ≥ 2k 1 − 1 1 1 j+λ 1 1 h′(λ)= 2 2log (2λ 2) . (3.9) − i+2λ − 2j+λ − 2k+λ − k+λ − − j+λ − k+λ iX=2j h i h i Forthesum,sincethefunctionf(x)=(x+2λ) 1isdecreasing,wehave − 2k−1 1 2k 1 j+λ 2 < 2 dx=2 log , − i+2λ − x+2λ k+λ i=2j Z2j X hence 1 1 1 1 h′(λ)< (2λ 2) , (3.10) − 2j+λ − 2k+λ − − j+λ − k+λ h i h i andforλ>1andj <k,theright-handsideisnegative. Thus,h(λ)<0forλ>1. ′ 2b)Next,weprovethatifλ ( 1,1],thenh (λ)<0. From(3.9),wederive ∈ −2 ′′ 2k 1 − 1 1 1 h′′(λ) = 4 + (3.11) (i+2λ)2 (2j+λ)2 − (2k+λ)2 iX=2j h i 1 1 1 1 4 +(2λ 2) . (3.12) − j+λ − k+λ − (j+λ)2 − (k+λ)2 h i h i Thefirsttermintheright-handsideisestimatedasfollows 2k 1 2k − 1 1 1 1 4 = 4 + (i+2λ)2 (i+2λ)2 (j+λ)2 − (k+λ)2 iX=2j i=X2j+1 h i 1 1 1 1 < 2 + , j+λ − k+λ (j+λ)2 − (k+λ)2 h i h i 7 whereforthesumwehaveusedtheinequality 2k (i+2λ) 2 < 2k(x+2λ) 2dx. i=2j+1 − 2j − 1 Next,forλ∈(−21,1]andx≥ 2 thefunctionfP(x)=(2x+λ)−2−(xR+λ)−2isincreasing,hence forthesecondtermin(3.11)wehave 1 1 1 1 < . (2j+λ)2 − (2k+λ)2 (j+λ)2 − (k+λ)2 h i h i Substitutingtheaboveupperboundsintheexpression(3.11)-(3.12)forh (λ),weobtain ′′ 1 1 1 1 2(k j)(kj λ2) h′′(λ)< 2 +2λ = − − <0, (3.13) − j+λ − k+λ (j+λ)2 − (k+λ)2 − (j+λ)2(k+λ)2 h i h i since1 j <kandλ ( 1,1]. (cid:3) ≤ ∈ −2 Proposition3.3 Letj,k N,j <k. Thenthecoefficientsβ in(2.4)satisfythefollowingrelations. k ∈ (i) If 1 <λ 0 or λ 1,then −2 ≤ ≥ e j 1 2λ 2 β2 j 1 +λ 2λ 2 − 2 − k − 2 − . (3.14) k 1 ≤ β2 ≤ k 1 +λ (cid:16) − 2(cid:17) ej (cid:16) − 2 (cid:17) (ii) If 0<λ 1,then e ≤ j 1 2λ 2 β2 j 1 +λ 2λ 2 − 2 − k − 2 − . (3.15) k 1 ≥ β2 ≥ k 1 +λ (cid:16) − 2(cid:17) ej (cid:16) − 2 (cid:17) e Proof. Byequality(2.6),wehave β =β , β =β . j j 1 k k 1 −2 −2 Thenalltherelationsthroughout(3e.5)-(3.13)remainevalidwiththesubstitution 1 1 j j , k k . 2 2 → − → − Theonlyexceptionisinequality(3.13)whichfailsforj =1,k =2,andλ [√3,1],sincethefactor ∈ 2 (k 1)(j 1) λ2 isnotpositivethen. − 2 − 2 − Let us prove that h(λ) 0 in this case as well. Since h(1) = 0, it is sufficient to prove that (cid:2) (cid:3) ≥ h(λ)<0forλ [√3,1]andj =1,k =2. Wehave ′ ∈ 2 e e e h′(λ) =h′(λ) j,k j 1,k 1 (cid:12) (cid:12) −2 −2 e (cid:12) (cid:12) sosubstitutingj = 1,k = 3 into(3.10),w(cid:12)efindthat(cid:12)forλ [3,1] [√3,1] 2 2 ∈ 4 ⊃ 2 1 1 1 1 h(λ) =h(λ) < (2λ 2) ′ 1,2 ′ 1,3 − 1+λ − 3+λ − − 1 +λ − 3 +λ (cid:12) (cid:12)2 2 h i h2 2 i e (cid:12) (cid:12) 1 1 1 1 1 (cid:12) (cid:12) + ≤ − 1+λ − 3+λ 2 1 +λ − 3 +λ h i h2 2 i 2 2 = + <0. −(1+λ)(3+λ) (1+2λ)(3+2λ) 8 4 Three lemmas In the next two sections, we deal with lower and upper estimates for the sums ℓ f(j), in j=1 particular for f = F , where F ,F are given in (4.1) below. For that purpose, we need the ν 1 2 P followingthreelemmas. Weusethefollowingnotation: ℓ ℓ 1 ′′ 1 − 1 f(i)= f(1)+ f(i)+ f(ℓ). 2 2 i=1 i=2 X X Lemma4.1 Foraconvexintegrandf,wehave ℓ ℓ+1 ℓ ℓ 2 ′′ f(i) f(x)dx, f(i) f(x)dx. Xi=1 ≤Z21 Xi=1 ≥Z1 Proof. Theinequalitiesrevealwell-knownpropertiesofthemidpointandthetrapezoidalquadra- tureformulasrelativetothecorrespondingintegrals. (cid:3) Lemma4.2 Forλ> 1,thefunctions −2 λ λ F (x)=x2λ(x+ )2(x+λ)2, F (x)=x2(x+ )2(x+λ)2λ (4.1) 1 2 2 2 areconvexon[1, )andincreasingon[1, ). 2 ∞ ∞ Proof. 1)Forλ 1,allthefactorsofF ,F in(4.1)areconvex,positiveandincreasingon[0, ), 1 2 ≥ ∞ hencethestatement. 2)Forλ [0,1]thefunctions ∈ u (x):=xλ(x+λ), u (x):=x(x+λ)λ 1 2 are non-negative and increasing on [0, ). Further, u is convex on [0, ), because it can be 2 ∞ ∞ writtenintheform u (x)=(x+λ)λ+1 λ(x+λ)λ, 2 − wherebothtermsareconvexforλ [0,1],whereasu isconvexon[1, ]because ∈ 1 2 ∞ 1 1 u′1′(x)=[xλ+1+λxλ]′′ =λxλ−2 (λ+1)x+λ(λ−1) >λxλ−2 x− 4 ≥0, x≥ 2. h i h i Therefore,bothF (x) = [u (x)]2(x+ λ)2 andF (x) = [u (x)]2(x+ λ)2 areconvexon[1, )and 1 1 2 2 2 2 2 ∞ increasingon(0, ). ∞ 3)Letλ ( 1,0]. Then ∈ −2 u′1(x)=xλ−1 (λ+1)x+λ2 >0, x>0, h i and u (x)=(x+λ)λ 1 (λ+1)(x+λ) λ2 (x+λ)λ 1 (λ+1)2 λ2 >0, x 1, ′2 − − ≥ − − ≥ h i h i henceF andF areincreasingon[1, ). Further,thefunction 1 2 ∞ λ 3λ λ2 v (x):=xλ(x+ )(x+λ)=xλ+2+ xλ 1+ xλ 1 2 2 − 2 9 is convex for x > 0 because all the terms are convex for λ ( 1,0], hence F (x) = [v (x)]2 is ∈ −2 1 1 convexwheneverv isnonnegative,i.e.,forx> λ,thusforx 1. Finally,for 1 − ≥ 2 λ 3λ λ2 v (x):=x(x+ )(x+λ)λ =yλ+2 yλ+1+ yλ, y =x+λ, 2 2 2 2 − weobtain 3 1 v2′′(x)=yλ−2 (λ+2)(λ+1)y2− 2λ2(λ+1)y+ 2λ3(λ−1) =:yλ−2p2(y), h i anditiseasytocheckthat,forλ ( 1,0],thequadraticpolynomialp hasnorealzeros. Hence, v isconvexandsoisF (x)=[v ∈(x)−]22forx 1. 2 (cid:3) 2 2 2 ≥ 2 Lemma4.3 Let α >0,γ γ γ ,1 i r,andlet i min i max ≤ ≤ ≤ ≤ r f(x):=(x+γ )α1(x+γ )α2 (x+γ )αr, s:= α . 1 2 r i ··· i=1 X Then,forany x>x ,wherex +γ 0,wehave 0 0 min ≥ x 1 x 1 (t+γ )f(t) < f(t)dt< (x+γ )f(x). (4.2) min max s+1h ix0 xZ0 s+1 Proof. Set 1 1 G(x):= (x+γ )f(x), F(x):= (x+γ )f(x). min max s+1 s+1 ItsufficestoshowthatG(t)<f(t)<F (t)forx t x. Wehave ′ ′ 0 ≤ ≤ r r 1 t+γ 1 min G(t)= 1+ α f(t) 1+ α f(t)=f(t), ′ i i s+1 t+γ ≤ s+1 i h Xi=1 i h Xi=1 i andsimilarly s s 1 t+γ 1 F (t)= f(t) 1+ α max 1+ α f(t)=f(t). (cid:3) ′ i i s+1 t+γ ≥ s+1 i h Xi=1 i h Xi=1 i Remark4.4 Wecanrefinetheupperestimateasfollows: x 1 s+1 f(t)dt< [f(x)] s . s+1 Zx0 Indeed, with F(x) := s+11[f(x)]s+s1, it suffices to show that F′(t) ≥ f(t) for every t > x0. We havetheequivalentrelations 1 1 1 s F (t)= f(t) s f (t) f(t) f(t) s , ′ s ′ ≥ ⇔ ≥ f′(t) f(t) (cid:2) (cid:3) (cid:2) (cid:3) andthelatterissimplytheinequalitybetweenthegeometricandharmonicmeans 1 α (x+γ )αi Pαi i . i ≥ αi (cid:16)Y (cid:17) Px+γi P 10