J.ofInequal. &Appl., 1997, Vol. 1, pp. 301-310 Reprintsavailabledirectlyfrom thepublisher Photocopying permittedbylicenseonly
(C)1997 OPA(OverseasPublishersAssociation) AmsterdamB.V.Publishedin TheNetherlands under licensebyGordonand Breach Science Publishers Printed inMalaysia
Discrete Inequalities of Wirtinger’s
Type for Higher Differences
GRADIMIR V. MILOVANOVI
andIGOR :. MILOVANOVI_,
Faculty ofElectronicEngineering,
Department
ofMathematics, P. O. Box73, 18001Ni, Yugoslaviae-mail:grade @gauss.elfak.ni.ac,yu (Received28July1996)
Discrete versionofWirtinger’s type inequalityforhigherdifferences,
2< (Amxk)2 nn,m
An,m x x
k= k=Im k=
wherelm [m/2],Um n [m/2]and
mmxk
(--1)
Xk+m_ii=0
isconsidered. Undersomeconditions, thebestconstantsAn,mandBn,medeteined.
Keywords: Discreteinequality;differenceof higher order; eigenvalue; eigenvector.
AMS1991 Subjectclassifications:Primary 26D 15;Secondary41 A44.
1
INTRODUCTION AND PRELIMINARIES
In
1(see
also[2])wepresentedageneralmethod forfindingthe bestpossible constantsAn
andBn
ininequalitiesofthe formpx
<_ rlc(xl- <Bn px, (1.1)
An
k=l k=0 k=l
ThisworkwassupportedinpartbytheSerbianScientificFoundation, grant number 04M03.
301
where p (pk)and r
(rk)
aregiven weightsequencesande(x)
isan arbitrary sequenceofthe real numbers. The basic discreteinequalitiesofthe form(1.1)
for pr
1 weregivenby K.Fan,
O. Taussky,andJ.
Todd [3].Here,
we mention some references in this direction[4-8].
The first results for the second difference were
proved
byFan,
Taussky and Todd[3]:
THEOREM1.1
If XO(-"
0), XI,X2 Xn, Xn+l("- O)
are given real numbers, thenn-1 n
7 2
(Xk 2Xk+l - Xk+2)
2 >__ 16sin42(n + 1) x:, (1.2)
k=0 k=l
withequalityin
(1.2) if
andonlyif x A
sin kTr k 1 2, n,whereA
isanarbitraryconstant.n+l
THEOREI 1.2
If
XO,Xl Xn,Xn+l are given real numbers such that XO Xl,Xn+ Xn andn
x
0,(1.3)
k=l
then
(Xk
2Xk+l +
Xk+2 >_ 16 sin4nn x:.
k=0 k=l
Theequalityin
(1.4)
isattainedif
andonlyif (2k
xk
A
cos whereA
is anarbitraryconstant.2nk-- 1,2 n,
(1.4)
A
converseinequality of(1.2)
wasproved by Lunter [9], Yin [10] and Chen[11] (seealsoAgarwal[8]).
/f X0(=
0), Xl,X2 Xn, xn+l(=
0) are given realn-1 n
zr
1
2(1.5)
(x 2x+ + x+e)
e _< 16cos42(n + 1) xc’
k=0
withequalityin
(1.5) if
andonlyifx: A(-1)
sinn-I-
kri"
k 1 2 n,where
A
is anarbitraryconstant.Chen
[11
alsoprovedthefollowingresult:DISCRETEINEQUALITIES OF WIRTINGER’S TYPE 303
THEOREM1.4
/f
X0,Xl Xn,Xn/ are given real numbers such thatxo
x andXn+ Xn,thenn-1 n
(x 2X+l -F x+2)
2 <_ 16cos4n x,
k=O k=l
withequality holding
if
andonlyif
x A(-1)
sin(2k- 1)re
k 1,2,... ,n,where
A
isanarbitraryconstant.In
thiscase, then n symmetricmatrixcorrespondingtothe quadratic formn-1
F2 __,(Xk 2Xk+l + Xk+2)
2(Hn,2,
,)k=O
2 -3 1
-3 6 -4 1
1 -4 6 -4 1
"’. "’. "’. "’. "’.
1 -4 6 -4 1
1 -4 6 -3
1 -3 2_
This matrixisthesquareof then n matrix 1 -1
1 2 -1
-1 2 -1
-1 2 -1
-1 1
(1.6)
Theeigenvaluesof
Hn
are.v v(Hn)
4cos2(n
v2n+ 1)zr
v=l n,andtherefore,thelargest eigenvalueof
Hn
is)n(Hn)
4cos2nn
> ,kn-1(Hn).
Thecorresponding eigenvectorisan Xln X2n Xnn where Xn
(-1)
sin(2v
12n v--- 1,2 n.
Thus,thelargest eigenvalueof
Hn,2
isn(Hn,2)-- 16cos4 n
n >,n-l(Hn,2),
and the associatedeigenvectorisxn
Notice that the minimal eigenvalue of the matrix
Hn
(and alsoHn,2)
isX 0.Therefore,the condition
(1.3)
mustbe included in Theorem 1.2(see Agarwal [8, Ch.11])
and the best constant is the square of the relevant eigenvalue(n- 1)re
re2
4COS2 4 sin22n 2n
For
any n-dimensional vector a [xl x2 Xn]r,
Pfeffer [12]introduced a periodically extended n-vector by setting Xi+rn xi for 1,2 n and r 6
N,
andused the rnth difference of a given by(m) Amxl Amx2 Amxn ]T,
wherem
()
Amxi (--1)
m-r Xi-[m/2]+r, r---O rl<i<n, inorder toprovethefollowingresult:
THEOREM 1.5
If
a is aperiodicallyextendedn-vectorand(1.3)
holds,then(X(rn) x(rn))
>(4sin
2withequalitycase
if
andonlyif
a isthe periodicextensionof
a vectorof
theform Clu + Czv,
whereU Ul U2 Un T and v Vl 132 Vn
]T
have the following components
2krc 2krc
uk cos
,
vk sin,
k 1 n,g/ n
and
C1
andC2
arearbitraryrealconstants.DISCRETE INEQUALITIES OF WIRTINGER’S TYPE 305
2
MAIN RESULTS
In
thispaperwe considerinequalitiesof the form2 2 2
(2.1)
An,m
xk< (Amxk) < Bn,m
xk,k= k--Im k=
where
Im
1[m/2],
Um n[m/2]
andAmXk (--1)
Xk+m-i.i=0
/m
Thequadraticform
Fm (AmXk)
2form 1reduces ton-1 n-I
2
2xkxk+l
Vl x
-Jr-2x +
xnk=2 k=l
withcorresponding tridiagonal symmetricmatrix
Itn Hn,
givenby(1.6).
We
considerinequalities(2.1)
under conditionsXs Xl-s, Xn+l-s Xn+s (lm < S <
0) (2.2)
and defineAJxl-[j/2]
x(j)
AJ
x2-[j/2]AJXn-[j/2]
Thequadraticform
Fm
canbeexpressedthen inthefollowingformUm
Fm_
Fro(x,)_(Amxk)
2((m), x,(m)),
where
Xl
(0)
x2Xn
At
thebeginingweprovethreeauxiliaryresults:(2.3)
(2.4)
LEMMA
2.1If
j is an eveninteger, under conditions(2.2),
wehavethatAJ+lx-[j/2]
0 andzxJ+lxn-[j/2]
O.(2.5)
Proof Let
q 0or q n.Puttingj 2pwehaveAJ+
2p+l
(2p +
1Xq-[j/2]
A2p+lxq-p E (-1)i \
i=0
Xq+p+l-i
2p+l
t
,(_i
x//_i+ , (-
x//_i=0 i=p+l
P
(2p+l)
E(--1)
Xq-p+ii--0
Xq+p+l-i
E(--1)
2p+
1i=0
( )
E(_I)i
2p+
1i=0
(Xq+p+
l_i Xq-p+i 0 becauseofthe conditions(2.2).
LEMMA
2.2If
j is an eveninteger, under conditions(2.2),
wehavethatHnx(j) _x(j+2),
where thematrix
Hn
isgivenby(1.6).
Proof We
haveUn
x(J)AJ
xI_[j/2]A
jx2-[j/2]--AJ+2xI_[j/2
AJ+2xn
2-[j/2]A J Xn-1-[j/2] "1- AJXn-[j/2]
Since and
AJ+2x_[j/2] AJ+lxl_[j/2] AJ+lx_[j/2]
AJ+2xn_l_[j/2] AJ+lxn_[j/2] AJ+lxn_l_[j/2],
(2.6)
DISCRETEINEQUALITIESOFWIRTINGER’S TYPE 307
because ofLemma 2.1,weconclude that
AJ+2x_[j/2]- AJ+IxI_[j/2]
andAJ+2xn_l_[j/2] -"--AJ+lxn_l_[j/2],
respectively. Therefore,
AJ+lxl _AJ
AJ
xI-[j/2]AJ
x2-[j/2] [j/2]+2x-[j/2]
and
--AJ
Xn-l-[j/2]+
Ajxn-[j/2]AJ+I
Then
(2.6)
becomesAJ+2x_[j/2 AJ+2xl_[j/2]
Hnx(J)
A]+2xn-2-[j/2]
AJ+2Xn-l-[j/2]
Xn-l-[j/2]
AJ+2xn
-1-[j/2].AJ+2xI-[(j+2)/2]
AJ+2x2-[(j+2)/2]
A
j+2x
n_1-[(j+2)/21AJ+2xn-[(j+2)/2]
_x(J+2).
LEMMA2.3
If
j isaneveninteger, under conditions(2.2),wehave that(x(j), x(j+2)) _((j+l), (j+l)).
Proof Let
j is an even integer.Using(2.3)
wehaven
E AJ+2xk
(X
(j) X(j+2))
Ajxk-[j/2]
-1-[j/21 k=lE AJxk-[J/z](AJxk-l-[j/2] 2AJxk-[j/2] + AJXk+l-[j/2])
k=l n
E AJxk-[j/2](AJxk+l-[j/2] AJxk-[j/2])
k=l
}2
k=l
AJXk-[j/2]AJ+lxk-[j/2] E
AjXk-[j/2]AJ+l
Xk-l-[j/2]k=l k=l
n n-1
E Ajxk-[j/2lAj+lxk-[j/2l E AJxk+l-[j/2lAJ+lxk-[j/2]"
k=l k=O
Becauseof
(2.5)
we can write(X (j), x(J+2))-- -AJxk-[j/2]AJ+" 1Xk_[j/2] AJxk+I_[j/2]A
j+lk=l k=l
n 2
1Xk-[j
k=l
Xk-[j/2]
Sincej is an evenintegerwehave that
(x(j) x(j+2)) (Aj
+lxk-[(j+l)/2] )2
k--1
_(x(J+I), x(J+l)).
Now,
wegivethe main result:THEOREM2.4
If
Xl,X2 Xnaregiven real numbers and conditions(2.2)
are
satisfied,
thenbl n
(AmXk)
2 <_ 4mCOS2m Xk’k=lm k=l
(2.7)
wherelm
1[m/2]
andUm n[m/2].
Theequalityin(2.7)
isattainedif
andonlyif
Xk
A(-1)
sin (2k1)re
k-- 1,2 n,
where
A
is anarbitraryconstant.Proof We
willprovethat thecorrespondingmatrix ofthequadraticform(2.4)
is exactly the mthpower of the matrixHn Hn,1
so that the bestconstantin theright inequality
(2.1),
i.e., (2.7),isgivenby4m
7gBn,m cs2m
2n Evidently,
An,m
O.Let
mbe an even integer.Then,usingLemma2.2, we find Fm(x,(m),
x,(m)) (Un3(m-2), Unx,(m-2)),
DISCRETEINEQUALITIESOFWIRTINGER’S TYPE 309
em <>, Hnm/’ <>)
Similarly,for an odd m,using
Lemmas
2.3and2.4,we obtainFrn (x (m), x(m)) _((m--1), (m+l)) ((m-1), nn(m-1)).
Now,
usingLemma
2.2 again,we findFrn (nn(m-1)/2x (0), nn(m+l)/2 (0)) (anm, ).
By
restriction(1.3),we canobtainthefollowingresult:TI-IEORM2.5
If
xl, X2 Xnaregiven real numbers and conditions(2.2)
and(1.3)
aresatisfied,
then7z" 2 2
4m sinZm
2--
<(Amxl)
k=l k--Im
(2.8)
wherelm
1[m/2]
andUm n [m/2].
Theequalityin(2.8)
isattainedif
andonlyif
(2k 1)Tr
xk
Acos
k 1,2 n,2n where
A
is anarbitraryconstant.For
othergeneralizationsof discreteWirtinger’s inequalities see 13-15].There are alsogeneralizations for multidimensionalsequences and partial differences(see 16]and
17]).
Acknowledgements
Thiswork wassupportedinpartbytheSerbianScientific Foundation,grant number04M03.
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