On the Markov inequality in the $L_2$-norm with the Gegenbauer weight

0.2 MB·

by  Geno Nikolov

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