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fference EquationsVolume 2007, Article ID 94325, 15 pages doi:10.1155/2007/94325 Research Article Relations between Limit-Point and Dirichlet Properties of Second-Order Difference Operato

fference Equations

Volume 2007, Article ID 94325, 15 pages

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(p n −1Δx n −1) +q n x n] acting on infinite sequences The discrete analog of some known relationships in the theory of differential operators such as Dirichlet, conditional

Dirichlet, weak Dirichlet, and strong limit-point is 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-sionM acting on complex valued sequences x = { x n } ∞

−1defined by

Mx n:=

⎧

⎪

⎨

⎪

⎩

1

w n



−Δp n −1Δx n −1



+q n x n

, n ≥0,

− p −1

(1.1)

with complex coefficients p= { p n } ∞

−1,q = { q n } ∞

−1and weightw = { w n } ∞

−1 In differential operators case, when the coefficients p and q are real-valued, the terms limit-point (LP),

strong limit-point ( SLP), Dirichlet (D), conditional Dirichlet (CD), and weak Dirichlet

(WD) at the regular endpoint are often used to describe certain properties associated

with the differential expression under consideration, see [1–10] Here, we introduce the discrete analogue of these properties and some relations between them In studying in-equalities involving expression (1.1), such as HELP (after Hardy, Everitt, Littlewood and Polya) and Kolmogorov-type inequalities, these properties and the relationships between

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.z denotes the complex conjugate of z ∈ 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= x =x n∞

−1:

∞

n =−1

x n 2

< ∞



,

2

w = x =x n∞

−1:

∞

n =−1

x n 2

w n < ∞

withw n > 0 for all n and the inner products

(x, y) =

∞

n =−1

x n y n, (x, y) =

∞

n =−1

x n y n w n, (2.2) respectively If{ x n } ∞

−1∈ 1but∞

n =−1x n < ∞, then we say that the sum∞

n =−1x nis con-ditionally convergent We associate a maximal operator,T(M), in 2

wwith the linear dif-ference expression

Mx n:=

⎧

⎪

⎨

⎪

⎩

1

w n



−Δp n −1Δx n −1



+q n x n

, n ≥0,

− p −1

(2.3)

whereΔx n = x n+1 − x n, the forward difference, and the coefficients{ p n } ∞

−1and{ q n } ∞

−1are complex valued with

p n =0, q −1=0, w n > 0, ∀ n = −1, 0, 1, (2.4) Note that definingM by (2.3) makes the difference equation

Mx n = λx n, n =0, 1, 2, (λ ∈ C), (2.5)

a three-term recurrence relation The operatorT(M) is defined on D T(M)into2

was



T(M)x n = T(M)x n:= Mx n, n = −1, 0, 1, , (2.6)

D T(M):= x =x n∞

−1∈ 2

w:

∞

n =−1

T(M)x n 2

w n < ∞



The summation-by-parts formula

m

n = k

x n Δy n = x m+1 y m+1 − x k y k −

m

n = k

y n+1 Δx n, k ≤ m, k,m ∈ N, (2.8)

gives rise to the equalities

m

n =0

x n My n w n =

m

n =0

q n y n x n+

m

n =0



p n Δy n

Δx n − p m Δy m x m+1+p −1Δy −1x0 (2.9) and, for allx, y ∈ D T(M),

∞

n =0



p n Δy n Δx n+q n y n x n

=

∞

n =0



x n T(M)y n

w n+ lim

m →∞ p m Δy m x m+1 − p −1Δy −1x0.

(2.10) The left-hand side of (2.10) is called the Dirichlet sum, and (2.10) is called the Dirichlet formula The following also holds for all x, y ∈ D T(M):

∞

n =0



x n T(M)y n − y n T(M)x n

w n =lim

m →∞ p m

Δx m y m+1 − Δy m x m+1

− p −1 

Δx −1y0− Δy −1x0 

.

(2.11) Following (2.10) we have, forx ∈ D T(M),

∞

n =0



p nΔx n 2

+q nx n 2 

=

∞

n =0



x n T(M)x n

w n+ lim

m →∞ p m Δx m x m+1 − p −1Δx −1x0.

(2.12)

An immediate consequence of (2.11) together with (2.7) is that

lim

m →∞ p m

Δx m y m+1 − Δy m x m+1

exists and is finite∀ x, y ∈ D T(M) (2.13) Moreover, the expression in (2.13) is a constant for allm ∈ Nwhenx, y are the solutions

of (2.5), which is easy to prove We also have the following variation of parameters formula:

letφ = { φ n } ∞

−1andψ = { ψ n } ∞

−1be linearly independent solutions of (2.5) and suppose that [φ,ψ] n:= p n[(Δφ n)ψ n+1 −(Δψ n)φ n+1]=1 for alln Then, Φ = {Φn } ∞

−1defined by

Φn =

n

m =0



− ψ m φ n+φ m ψ n

w m f m (n ∈ N),

Φ−1=0

(2.14)

satisfies

MΦ n = λΦ n+f n, n ∈ N,λ ∈ C, (2.15a)

Φ−1=Φ0=0. (2.15b) Any solution of (2.15a) is of the form

for some constantsA,B ∈ C

Definition 2.1 If there is precisely one 2

w solution (up to constant multiples) of (2.5) for(λ) =0, then the expressionM is said to be in the limit-point (LP) case; otherwise

all solutions of (2.5) are in2

w for allλ ∈ CandM is said to be in the limit-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 ofM being LP is that

lim

m →∞ p m

Δx m y m+1 − Δy m x m+1

or

lim

m →∞ p m

y m x m+1 − y m+1 x m

for allx, y ∈ D T(M), see Hinton and Lewis [6, page 425] It may also be observed that this condition is equivalent to saying that

lim

m →∞ p m

Δx m x m+1 − Δx m x m+1

or

lim

m →∞ p m

x m x m+1 − x m+1 x m

for allx ∈ D T(M) To see that, takex = y in ( ∗1) to get the implication in one direction For the implication on the other side, takex to be the linear combination of z and y, that

is,x = z + αy in ( ∗2), and then choose the complex numberα as α =1 andα = i to get

(∗1)

Definition 2.2 M is said to be strong limit-point (SLP) on D T(M)if

lim

m →∞ p m Δy m x m+1 =0 ∀ x, y ∈ D T(M) (2.19)

Definition 2.3 M is said to be

(i) Dirichlet ( D) on D T(M)if

p n 1/2 Δx n∞

−1, q n 1/2 x n∞

−1∈ 2 ∀ x ∈ D T(M); (2.20)

(ii) conditional Dirichlet ( CD) on D T(M)if

p n 1/2 Δx n∞

−1∈ 2,

∞

n =0

q nx n 2

is convergent∀ x ∈ D T(M), (2.21)

(iii) weak Dirichlet ( WD) on D T(M)if

∞

n =0



p n Δx n Δy n+q n x n y n

is convergent∀ x, y ∈ D T(M) (2.22)

Observe that (2.19) is equivalent to

lim

m →∞ p m Δx m x m+1 =0 or lim

m →∞ p m Δx m x m+1 =0 ∀ x ∈ D T(M) (2.23)

Also, by Dirichlet formula (2.10), it is seen that theWD property, (2.22), is equivalent to

lim

m →∞ p m Δy m x m+1 exists and is finite ∀ x, y ∈ D T(M), (2.24) and this is equivalent to

lim

m →∞ p m Δx m x m+1 exists and is finite ∀ x ∈ D T(M) (2.25) Note also that in (iii), for allx, y ∈ D T(M),

p n 1/2 Δx n∞

−1∈ 2 ⇐⇒ p n

Δx n 2 ∞

−1∈ 1 ⇐⇒ p n Δx n Δy n∞

−1∈ 1. (2.26) Following the above definitions and subsequent comments, we have the following

Corollary 2.4 The following implications hold for all x, y ∈ D T(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, and w which 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

Theorem 3.1 Let p and q be complex-valued.

(a) If 1 /p ∈ l1, then CD ⇒ SLP on D T(M)

(b) If 1 /p ∈ l1but∞

n =0q n is not convergent, then CD ⇒ SLP on D T(M) (c) If w, 1/p, q ∈ l1, then M is both D and LC.

Proof (a) We assume that 1 /p ∈ 1 and M is CD on D T(M) Let x, y ∈ D T(M) then, by (2.10),

α : = lim

m →∞ p m Δy m x m+1 < ∞ (3.1)

We need to prove thatα =0 under the conditions in the hypothesis Suppose the contrary thatα =0, then for somem0∈ N,

p m Δy m x m+1  ≥ | α |

which implies that

p m Δy m Δx m  ≥ | α |

2



Δx m

x m+1



 ∀ m ≥ m0,∀ x, y ∈ D T(M) (3.3)

However,M is CD and this implies that, summing over m, the left-hand side of (3.3) belongs to1 Thus,

∞

n =−1



Δx n

x n+1



and hence in particular| Δx n /x n+1 | →0 asn → ∞ So, asn → ∞,



logx n+1

x n



 =−log



1− Δx n

x n+1



∼

Δx n

x n+1



since

lim

t →0

log (1− t)

Hence,

∞

n =−1



logx n+1

x n



< ∞ =⇒

∞

n =−1

logx n+1

x n is convergent, lim

N →∞

N

n = m0

logx n+1

x n exists form0∈ N

(3.7)

This implies that

lim

N →∞

N

n = m0

Δlogx n

= lim

N →∞



logx N+1 −logx m0



exists. (3.8) So,

β : = lim

Thus, sinceα : =limm →∞ p m Δy m x m+1 < ∞,

lim

m →∞ p m Δy m = αβ −1, (3.10) and, for somem0∈ N,

p m

Δy m 2 ≥1

4αβ −1  2 p −1

m  ∀ m ≥ m0. (3.11) However, summing overm, the left-hand side of (3.11) belongs to1by the hypothesis thatM is CD Hence, so does the right-hand side of (3.11) which is a contradiction to saying that 1/p ∈ 1 Henceα =0, provingM is SLP.

(b) Assume thatp −1∈ 1but∞

n =0q nis not convergent andM is CD Let x ∈ D T(M)

and, as in (a) above, suppose that

α = lim

m →∞ p m x m+1 Δx m =0. (3.12)

Then, limm →∞ x m = β =0 exists and it follows that

lim

m →∞ p m Δx m = αβ −1=0=⇒ lim

m →∞ Δx m = lim

m →∞ αβ −1p −1

m (3.13)

So, sincep −1∈ 1, we have

∞

m =−1

Δx m< ∞, that is,

Δx n∞

−1∈ 1 

x ∈ D T(M)

Now, sincex ∈ D T(M), using Cauchy-Schwarz inequality in2, we have

∞

n =−1

x n w1/2

n 

−Δp n −1Δx n −1



+q n x n

w −1/2

n 

≤

n =−1

x n w1/2

n  2

 1/2 ∞

n =−1

−Δp n −1Δx n −1



+q n x n

w −1/2

n  2

 1/2 (3.15)

which gives

∞

n =−1

x n

−Δp n −1Δx n −1



+q n x n < ∞ . (3.16)

Also, since limm →∞ x m = β =0, we have that

∞

n =−1

−Δp n −1Δx n −1



+q n x n < ∞ . (3.17)

Now,

∞

n =0



−Δp n −1Δx n −1



+q n x n

= −lim

m →∞ p m Δx m+p −1Δx −1+

∞

n =0

q n x n (3.18) implies that

∞

n =0

q n x n = lim

m →∞ p m Δx m − p −1Δx −1+

∞

n =0



−Δp n −1Δx n −1



+q n x n

, (3.19) which proves the convergence of the sum∞

n =0q n x n Sinceβ =limm →∞ x m =0, thenx m =

0 for all largem ∈ N On the other hand, using summation-by-parts formula and sup-posingk ∈ Nis such thatx n =0 for alln ≥ k, we have

m

n = k

q n = m

n = k

1

x n



q n x n

= 1

x m+1

m

s = k −1

q s x s − 1

x k

k −1

s = k −1

q s x s − m

n = k

s = k −1

q s x s



Δx1

n



=

m

n = k −1q n x n

x m+1 − q k −1x k −1

m

n = k

s = k −1

q s x s

n

x n+1 x n



.

(3.20)

Asm → ∞, we see that the right-hand side of (3.20) tends to a finite limit since∞

n =0q n x n

is convergent and limn →∞ x n = β =0, which contradicts the hypothesis that∞

n =0q n is divergent This provesα =0 which guarantees thatM is SLP.

(c) If 1/p, w, q ∈ 1, thenM is LC and D For the proof, we need the matrix

represen-tation of (2.5); forn ≥0, we have the recurrence relation

p n

x n+1 − x n

=− λw n+q n

x n+p n −1 

x n − x n −1 

which is equivalent to (2.5) So, taking

X n =



x n

y n



, A n =

⎛

⎜

⎜

⎝

p n −1



− λw n+q n  − λw n+q n

p n −1

⎞

⎟

⎟

⎠, (3.22)

we get

X n =I + A n

X n −1, n =0, 1, 2, , (3.23) whereI is the identity matrix and

x n = x n −1+ y n −1

p n −1

y n =



x n −1+ y n −1

p n −1



− λw n+q n

+y n −1. (3.24)

We are going to give the proof for theLC and D cases separately.

(i) The LC case We prove that, for some λ, say λ =0, for all solutions of (3.21),

∞

n =−1| x n |2w n < ∞holds Moreover, since ∞

n =−1w n < ∞, 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]

Theorem 3.2 (Atkinson) Let the sequence of k-by-k matrices,

A n, n =0, 1, 2, 3, ; A n =a nrs

, r,s =1, 2, 3, ,k, (3.25)

satisfy

∞

n =0

A n< ∞, A n:= k

r =1

k

s =1

a nrs. (3.26)

Then, the solutions of the recurrence relation

X n − X n −1= A n −1X n −1, n =0, 1, 2, , (3.27)

where X n is a k-vector, converge as n → ∞ If in addition the matrices I + A n are all nonsin-gular, then lim n →∞ X n = 0, unless all the X n are zero vectors.

So, applying this theorem to our case,{ X n } ∞

0 is convergent, that is, the entries ofX n,

X n1∞

0 =x n∞

X n2∞

0 =y n∞

0 =p n Δx n∞

0, (3.28) are convergent, so they are bounded and hence (i) of condition (c) is proved

(ii) The D case We will state the proof for λ =0 only, but the proof also applies to all

λ ∈ C Letx ∈ D T(M)and define f = { f n } ∞

−1by

Then∞

n =−1| f n |2w n < ∞ Also, by the variation of parameters formula, ifϕ = { ϕ n } ∞

−1and

ψ = { ψ n } ∞

−1are linearly independent solutions of (2.5) with

[ϕ,ψ] n:= p n −1 

ϕ n Δψ n −1− ψ n Δϕ n −1 

=1 ∀ n ∈ N, (3.30) then any solution of

is of the form

x n =Φn+Aϕ n+Bψ n (3.32)

in whichA and B are constants, and

Φn =

n

m =0



ψ m ϕ n − ϕ m ψ n

w m f m, 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  ≤ C n

m =0

w mf m, (3.34)

whereC is a positive constant Hence, Φ is bounded This implies that { x n } ∞

−1is bounded from the fact that{ Aϕ n+Bψ n } ∞

−1and{Φn } ∞

−1are bounded in (3.32) So, sinceq ∈ 1and following the above result,

∞

n =0

q nx n 2

We also need to prove that∞

n =0| p n || Δx n |2< ∞ For, from (3.32),

p n Δx n = p nΔΦn+p nΔAϕ n+Bψ n

,

p nΔΦn =

n

m =0



ψ m

p n Δϕ n

− ϕ m

p n Δψ n

w m f m; (3.36)

and since{ p n Δϕ n } ∞

−1,{ p n Δψ n } ∞

−1,{ ϕ n } ∞

−1, and{ ψ n } ∞

−1are bounded by the theorem of Atkinson,{ p nΔΦn } ∞

−1is also bounded, and so is{ p n Δx n } ∞

−1 By the hypothesis thatp −1∈

1, we obtain

∞

n =0

p nΔx n 2

= ∞

n =0

p nΔx n 2

p n < ∞ (3.37) Hence,M is D and the proof ofTheorem 3.1is complete 

Corollary 3.3 (1) Following the Dirichlet formula, ( 2.23), and Theorem 3.1(a)-(b), it may be deduced that if either p −1∈ 1or p −1∈ 1but∞

n =0q n is not convergent, then CD implies that the sum∞

n =0(p n | Δx n |2+q n | x n |2) is convergent for all x ∈ D T(M) (2) Under the conditions of Theorem 3.1(a)-(b), D ⇒ CD ⇒ SLP ⇒ LP on D T(M)

Remarks 3.4 (1) When w, p −1,q ∈ 1, it is proved by Atkinson [11, page 134] thatM is

LC We have additionally proved that M is also D (2) The condition imposed on q in

Theorem 3.1(a) is in general weaker thanq ∈ 1 Indeed, inExample 3.5, we prove that

q ∈ 1is not sufficient to ensure that CD⇒ SLP.

Example 3.5 In this example, we want to establish an expression M of the form (2.3) such that∞

n =0q n is conditionally convergent andw,1/p ∈ 1 whileM is CD and LC,

hence notSLP, at the same time This proves that q ∈ 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 =0r nbe a conditionally convergent real series Choose a constantC1

so that the sequence

R n∞

0 =

n

k =0

r k

∞

0

be positive, that is,R n > 0 for all, n =0, 1, 2, Then { R n } ∞

0 is bounded, forp n > 0 n ∈ N

and given thatC2> 0, the sequence

x n∞

0 =

n

k =0

R k −1

p k −1

∞

0 +C2, R −1=0, p n −1> 0 ∀ n ∈ N,x −1≥ x0 (3.39)

is also positive Note that{ x n } ∞

−1is monotonic increasing, that is,x n+1 ≥ x nfor alln, from

the fact thatx nare the sum of positive numbers Now,

X =lim

since{ R n } ∞

−1 is bounded and p −1= { p −1

n } ∞

−1∈ 1 Moreover,x ∈ 2

w sincew ∈ 1 and

{ x n } ∞

−1is bounded We see that if{ q n } ∞

−1is given by

q n = r n

x n, n ≥0, q −1=0, (3.41)

However,M is CD and this implies that, summing over m, the left-hand side of (3.3) belongs... solution of (2.15a) is of the form

for some constantsA,B ∈ C

Definition...

(3.20)