Volume 2007, Article ID 94325,15pages doi:10.1155/2007/94325
Research Article
Relations between Limit-Point and Dirichlet Properties of
Second-Order Difference Operators
A. Delil
Received 24 July 2006; Revised 6 March 2007; Accepted 11 April 2007
Dedicated to Professor W. D. Evans on the occasion of his 65th birthday
Recommended by Martin J. Bohner
We consider second-order difference expressions, with complex coefficients, of the form
w−1
n [−Δ(pn−1Δxn−1) +qnxn] acting on infinite sequences. The discrete analog of some known relationships in the theory of differential operators such asDirichlet,conditional Dirichlet,weak Dirichlet, andstrong limit-pointis considered. Also, connections and some relationships between these properties have been established.
Copyright © 2007 A. Delil. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and repro-duction in any medium, provided the original work is properly cited.
1. Introduction
In this paper, we will deal with the second-order formally symmetric difference expres-sionMacting on complex valued sequencesx= {xn}∞−1defined by
Mxn:= ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
1
wn
−Δpn−1Δxn−1+qnxn , n≥0,
−wp−1
−1Δxn, n= −1,
(1.1)
them are crucial. The work we present here is the discrete analogue of the work by Race [9] for differential expressions.
2. Preliminaries
We use the following notation throughout:RandCdenote the real and complex number fields, andNis the set of nonnegative integers.zdenotes the complex conjugate ofz∈C.
(·) and(·) represent the imaginary and real part of a complex number.1is the space of all absolutely summable complex sequences.2and2
ware the Hilbert spaces
2= ditionally convergent. We associate a maximal operator,T(M), in2
wwith the linear dif-ference expression
Note that definingMby (2.3) makes the difference equation
gives rise to the equalities
m
n=0
xnMynwn= m
n=0
qnynxn+ m
n=0
pnΔynΔxn−pmΔymxm+1+p−1Δy−1x0 (2.9)
and, for allx,y∈DT(M),
∞
n=0
pnΔynΔxn+qnynxn=
∞
n=0
xnT(M)ynwn+ limm→∞pmΔymxm+1−p−1Δy−1x0.
(2.10)
The left-hand side of (2.10) is called theDirichlet sum, and (2.10) is called theDirichlet formula. The following also holds for allx,y∈DT(M):
∞
n=0
xnT(M)yn−ynT(M)xnwn=mlim→∞pmΔxmym+1−Δymxm+1−p−1Δx−1y0−Δy−1x0.
(2.11)
Following (2.10) we have, forx∈DT(M),
∞
n=0
pnΔxn2+qnxn2=
∞
n=0
xnT(M)xnwn+ limm→∞pmΔxmxm+1−p−1Δx−1x0.
(2.12)
An immediate consequence of (2.11) together with (2.7) is that
lim m→∞pm
Δxmym+1−Δymxm+1
exists and is finite∀x,y∈DT(M). (2.13)
Moreover, the expression in (2.13) is a constant for allm∈Nwhenx,yare the solutions of (2.5), which is easy to prove. We also have the followingvariation of parameters formula: letφ= {φn}∞−1andψ= {ψn}∞−1be linearly independent solutions of (2.5) and suppose that [φ,ψ]n:=pn[(Δφn)ψn+1−(Δψn)φn+1]=1 for alln. Then,Φ= {Φn}∞−1defined by
Φn= n
m=0
−ψmφn+φmψnwmfm (n∈N),
Φ−1=0
(2.14)
satisfies
MΦn=λΦn+fn, n∈N,λ∈C, (2.15a)
Φ−1=Φ0=0. (2.15b)
Any solution of (2.15a) is of the form
Ψ=Φ+Aφ+Bψ (2.16)
Definition 2.1. If there is precisely one2
w solution (up to constant multiples) of (2.5) for(λ)=0, then the expressionMis said to be in thelimit-point (LP) case; otherwise all solutions of (2.5) are in2
w for allλ∈CandM is said to be in thelimit-circle(LC) case, see Atkinson [11] and Hinton and Lewis [6]. Note that in the limit-circle (LC) case, the defect numbers are equal and the limit-point case does not hold. An alternative but equivalent characterization ofMbeingLPis that
lim m→∞pm
Δxmym+1−Δymxm+1
=0 (2.17)
or
lim m→∞pm
ymxm+1−ym+1xm
=0 (∗1)
for allx,y∈DT(M), see Hinton and Lewis [6, page 425]. It may also be observed that this condition is equivalent to saying that
lim m→∞pm
Δxmxm+1−Δxmxm+1
=0 (2.18)
or
lim m→∞pm
xmxm+1−xm+1xm
=0 (∗2)
for allx∈DT(M). To see that, takex=yin (∗1) to get the implication in one direction.
For the implication on the other side, takexto be the linear combination ofzandy, that is,x=z+αyin (∗2), and then choose the complex numberαasα=1 andα=ito get
(∗1).
Definition 2.2. Mis said to bestrong limit-point(SLP) onDT(M)if
lim
m→∞pmΔymxm+1=0 ∀x,y∈DT(M). (2.19)
Definition 2.3. Mis said to be (i)Dirichlet(D) onDT(M)if
pn1/2
Δxn∞−1, qn 1/2
xn∞−1∈2 ∀x∈DT(M); (2.20)
(ii)conditional Dirichlet(CD) onDT(M)if
pn1/2
Δxn∞−1∈2, ∞
n=0
qnxn2is convergent∀x∈DT(M), (2.21)
(iii)weak Dirichlet(WD) onDT(M)if
∞
n=0
Observe that (2.19) is equivalent to
lim
m→∞pmΔxmxm+1=0 or mlim→∞pmΔxmxm+1=0 ∀x∈DT(M). (2.23)
Also, byDirichletformula (2.10), it is seen that theWDproperty, (2.22), is equivalent to
lim
m→∞pmΔymxm+1 exists and is finite ∀x,y∈DT(M), (2.24)
and this is equivalent to
lim
m→∞pmΔxmxm+1 exists and is finite ∀x∈DT(M). (2.25)
Note also that in (iii), for allx,y∈DT(M), pn1/2
Δxn∞−1∈2 ⇐⇒
pnΔxn2∞−1∈1 ⇐⇒
pnΔxnΔyn∞−1∈1. (2.26)
Following the above definitions and subsequent comments, we have the following.
Corollary2.4. The following implications hold for allx,y∈DT(M): (a)D⇒CD⇒WD;
(b)SLP⇒WD; (c)SLP⇒LP.
3. Statement of results
In this section, we would like to obtain some implications additional toCorollary 2.4by imposing conditions onp,q, andwwhich are as weak as possible. The motivation of the problem and parts (a) and (b) of the following theorem was previously presented at the 17th National Symposium of Mathematics, Bolu, Turkey[12]. It is presented here for the sake of completeness.
Theorem3.1. Let pandqbe complex-valued.
(a)If1/p∈l1, thenCD⇒SLPonD
T(M). (b)If1/p∈l1but∞
n=0qnis not convergent, thenCD⇒SLPonDT(M). (c)Ifw,1/p,q∈l1, thenMis bothDandLC.
Proof. (a) We assume that 1/p∈1 and M isCDon D
T(M). Let x,y∈DT(M) then, by (2.10),
α:= lim
m→∞pmΔymxm+1<∞. (3.1)
We need to prove thatα=0 under the conditions in the hypothesis. Suppose the contrary thatα=0, then for somem0∈N,
pmΔymxm+1≥|α|
2 ∀m≥m0, (3.2)
which implies that
pmΔymΔxm≥|α|
2 Δxm
xm+1
However,M isCDand this implies that, summing overm, the left-hand side of (3.3)
xn is convergent,
lim
However, summing overm, the left-hand side of (3.11) belongs to1by the hypothesis thatM isCD. Hence, so does the right-hand side of (3.11) which is a contradiction to saying that 1/p∈1. Henceα=0, provingMisSLP.
(b) Assume thatp−1∈1but∞
n=0qnis not convergent andM isCD. Letx∈DT(M) and, as in (a) above, suppose that
α= lim
Asm→ ∞, we see that the right-hand side of (3.20) tends to a finite limit since∞n=0qnxn is convergent and limn→∞xn=β=0, which contradicts the hypothesis that∞n=0qn is divergent. This provesα=0 which guarantees thatMisSLP.
(c) If 1/p,w,q∈1, thenMisLCandD. For the proof, we need the matrix represen-tation of (2.5); forn≥0, we have the recurrence relation
pnxn+1−xn=−λwn+qnxn+pn−1xn−xn−1, (3.21)
which is equivalent to (2.5). So, taking
Xn=
xn yn
, An= ⎛ ⎜ ⎜ ⎜ ⎜ ⎝
0 1
pn−1
−λwn+qn −λwpn+qn n−1
⎞ ⎟ ⎟ ⎟ ⎟
⎠, (3.22)
we get
Xn=I+AnXn−1, n=0, 1, 2,..., (3.23)
whereIis the identity matrix and
xn=xn−1+ yn−1
pn−1
yn=
xn−1+pyn−1
n−1
−λwn+qn+yn−1.
(3.24)
We are going to give the proof for theLCandDcases separately.
(i) The LC case. We prove that, for some λ, say λ=0, for all solutions of (3.21), ∞
n=−1|xn|2wn<∞holds. Moreover, since ∞n=−1wn<∞, it is sufficient to prove that all solutions of (3.21), withλ=0, are bounded. For this purpose, we make use of the following theorem due to Atkinson [11, page 447].
Theorem3.2 (Atkinson). Let the sequence ofk-by-kmatrices,
An, n=0, 1, 2, 3,...; An=anrs, r,s=1, 2, 3,...,k, (3.25)
satisfy
∞
n=0
An<∞, An:=k r=1
k
s=1
anrs. (3.26)
Then, the solutions of the recurrence relation
Xn−Xn−1=An−1Xn−1, n=0, 1, 2,..., (3.27)
whereXnis ak-vector, converge asn→ ∞. If in addition the matricesI+Anare all
So, applying this theorem to our case,{Xn}∞0 is convergent, that is, the entries ofXn,
Xn1∞0 =xn∞0,
Xn2∞0 =yn∞0 =
pnΔxn∞0, (3.28)
are convergent, so they are bounded and hence (i) of condition (c) is proved.
(ii)TheDcase. We will state the proof forλ=0 only, but the proof also applies to all
λ∈C. Letx∈DT(M)and define f = {fn}∞−1by
fn=Mxn. (3.29)
Then∞n=−1|fn|2wn<∞. Also, by the variation of parameters formula, ifϕ= {ϕn}∞−1and
ψ= {ψn}∞−1are linearly independent solutions of (2.5) with
[ϕ,ψ]n:=pn−1ϕnΔψn−1−ψnΔϕn−1=1 ∀n∈N, (3.30)
then any solution of
Mxn=λxn+fn (3.31)
is of the form
xn=Φn+Aϕn+Bψn (3.32)
in whichAandBare constants, and
Φn= n
m=0
ψmϕn−ϕmψnwmfm, n∈N,Φ−1=0. (3.33)
Since {ϕ}∞−1 and {ψ}∞−1 are bounded by case (i) of condition (c), using also Cauchy-Schwarz inequality in2, it follows that
Φn≤Cn m=0
wmfm, (3.34)
whereCis a positive constant. Hence,Φis bounded. This implies that{xn}∞−1is bounded from the fact that{Aϕn+Bψn}∞−1and{Φn}∞−1are bounded in (3.32). So, sinceq∈1and following the above result,
∞
n=0
qnxn2
<∞. (3.35)
We also need to prove that∞n=0|pn||Δxn|2<∞. For, from (3.32),
pnΔxn=pnΔΦn+pnΔAϕn+Bψn,
pnΔΦn= n
m=0
and since{pnΔϕn}∞−1,{pnΔψn}∞−1,{ϕn}−1∞, and{ψn}∞−1are bounded by the theorem of Atkinson,{pnΔΦn}∞−1is also bounded, and so is{pnΔxn}∞−1. By the hypothesis thatp−1∈
1, we obtain
∞
n=0
pnΔxn2
=∞
n=0
pnΔxn2
pn <∞. (3.37)
Hence,MisDand the proof ofTheorem 3.1is complete.
Corollary3.3. (1) Following the Dirichlet formula, (2.23), andTheorem 3.1(a)-(b), it may be deduced that if eitherp−1∈1or p−1∈1but∞
n=0qnis not convergent, thenCD
implies that the sum∞n=0(pn|Δxn|2+qn|xn|2)is convergent for allx∈DT(M). (2) Under the conditions ofTheorem 3.1(a)-(b),D⇒CD⇒SLP⇒LPonDT(M).
Remarks 3.4. (1) Whenw,p−1,q∈1, it is proved by Atkinson [11, page 134] thatM is
LC. We have additionally proved thatM is alsoD. (2) The condition imposed onq in Theorem 3.1(a) is in general weaker thanq∈1. Indeed, inExample 3.5, we prove that
q∈1is not sufficient to ensure thatCD⇒SLP.
Example 3.5. In this example, we want to establish an expressionM of the form (2.3) such that∞n=0qn is conditionally convergent andw, 1/p∈1 whileM isCD andLC, hence notSLP, at the same time. This proves thatq∈1is not sufficient to ensure that the implicationCD⇒SLP. This example is a direct analogue of the example given in Kwong [7, page 332]. Let∞n=0rnbe a conditionally convergent real series. Choose a constantC1 so that the sequence
Rn∞0 = n
k=0
rk ∞
0
+C1 (3.38)
be positive, that is,Rn>0 for all,n=0, 1, 2,.... Then{Rn}∞0 is bounded, forpn>0n∈N and given thatC2>0, the sequence
xn∞0 = n
k=0
Rk−1
pk−1 ∞
0
+C2, R−1=0, pn−1>0∀n∈N,x−1≥x0 (3.39)
is also positive. Note that{xn}∞−1is monotonic increasing, that is,xn+1≥xnfor alln, from the fact thatxnare the sum of positive numbers. Now,
X=lim
n→∞xnexists (3.40)
since{Rn}∞−1 is bounded and p−1= {pn−1}−∞1∈1. Moreover,x∈w2 sincew∈1 and
{xn}∞−1is bounded. We see that if{qn}∞−1is given by
qn=xrn
then{xn}∞−1is a solution of (2.5) withλ=0. Note that, in
qn=rn xn ≥
rn
X ∀n, (3.42)
summing overn, we have{qn}∞−1∈1 from the fact that ∞
0 rnis conditionally conver-gent. Now, summation-by-parts formula gives, for allN∈N,
N
n=0
qn= N
n=0
rn xn=
RN xN −
N−1
n=−1
Rn xn+1+
N−1
n=−1
Rn
xn. (3.43)
For the first expression on the right-hand side, the limits limn→∞Rnand limn→∞xn ex-ist andX=limn→∞xn>0. For the sums on the right, since∞n=0Rnis convergent and
{1/xn}∞−1is positive and decreasing, bothNn=−1(Rn/xn+1) andNn=−1(Rn/xn) are conver-gent, and therefore∞n=0qnis convergent. Now, let{yn}∞−1be another solution of (2.5) together with (3.41) complementary to{xn}∞−1, that is, such that [x,y]n:=pn−1(ynxn−1−
yn−1xn) is constant, or equivalently, [x,y]n=1. Then,
Δyn−1
xn−1
= p 1
n−1xnxn−1 =⇒yn=xn
n
k=0 1
pk−1xkxk−1. (3.44)
So, since{yn}∞−1is bounded and increasing,
lim
n→∞ynexists. (3.45)
We note that∞k=0(1/pk−1xkxk−1) is absolutely convergent since{xn}∞−1 is bounded and
p−1∈1. So,y∈2
wsincew∈1. We also see thatMyn=0. Hence, we have shown that MisLC, and hence notSLPsincex,y∈2
wandx,yare linearly independent solutions of Mxn=λxn,λ∈C. We now show thatM isCD. Since, from the identity (2.12), theCD property is equivalent to
(a){pn|Δzn|2}∞−1∈1,
(b) limn→∞pnΔznzn+1exists∀z∈DT(M),
and we will show both (a) and (b) above. So, letz∈DT(M). Then,
T(M)zn∞−1=
Mzn∞−1=
fn∞−1∈2w, w∈1. (3.46)
The method of variation of parameters gives
zn=Axn+Byn+ n
m=0
xnym−ynxmfmwm z−1=0,n∈N, (3.47)
whereAandBare constants. Note that limn→∞nm=0(xnym−ynxm)fmwm<∞, (3.40) and (3.45) together imply that
lim
We see that{p1/2
n Δxn}∞−1,{pn1/2Δyn}∞−1∈2 since{Rn}∞0 is bounded and{p−n1}∞−1∈1. Also, using the Cauchy-Schwarz inequality in2,n, we see that, for alln∈N,
n
whereCis a constant. Hence,
It is a consequence of (3.50) and (3.52) thatMisCD. This completes the desired example.
and hence
This follows by using the summation-by-parts formula. Asm→ ∞, we see that the con-dition inTheorem 3.6is equivalent to the condition that
For example, ifm <∞andw=1, this condition becomes
∞
Proof. SinceSLPalways impliesWDbyCorollary 2.4, we only need to prove thatWD⇒
SLPunder the conditions in the hypothesis. So, suppose thatMsatisfies theWDproperty, that is,β=limm→∞pnΔxnxn+1exists and is finite for allx∈DT(M), butMis notSLP, that is,β=0. We show thatβ=0 leads to a contradiction under the hypothesis, and henceM isSLP. So, suppose that
β= lim
m→∞pmΔxmxm+1=0 ∀x∈DT(M). (3.63)
Now, multiplying both sides of the following byβandwm, and summing overm:
xm+1Δxm=x2m+1−xmxm+1, (3.64)
we have ∞
m=0
βpmΔxmxm+1
wmpm−1
=β
∞
m=0
wm+1xm2+1 w
m wm+1
−∞
m=0
wmwm+1 1/2
xmxm+1 w
m wm+1
1/2
.
(3.65)
Under the conditions of the hypothesis, the left-hand side of this equality is∞while the right-hand side is finite. This contradiction leads us to say thatβ=0 andM isSLPon
DT(M). Hence the theorem is proved. Remark 3.9. As a final remark, Theorem 3.1(c) demonstrates that whenw,p−1,q∈1
WD does not implySLP or even LP. Thus, for the equivalency of WD andSLP, the hypothesis ofTheorem 3.8is needed. For example, whenw=1, the requirements for the resultSLP ⇐⇒ WDbecome∞n=−1p−1n = ∞.
Acknowledgment
The author is grateful to the referee for a careful scrutiny of the manuscript and for point-ing out a number of ambiguities.
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