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Iterated remainder operator, tests for multiple convergence of series, and solutions of difference equations

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Abstract

We establish some properties of iterations of the remainder operator which assigns to any convergent series the sequence of its remainders. Moreover, we introduce the spaces of multiple absolute summable sequences. We also present some tests for multiple absolute convergence of series. These tests extend the well-known classical tests for absolute convergence of series. For example we generalize the Raabe, Gauss, and Bertrand tests. Next we present some applications of our results to the study of asymptotic properties of solutions of difference equations. We use the spaces of multiple absolute summable sequences as the measure of approximation.

MSC:39A10.

1 Introduction

Let r denote the operator which assigns to any convergent series the sequence of its remainders. The purpose of this paper is to study the basic properties of iterations r m of r and apply these results to the study of asymptotic properties of solutions of difference equations. We also study the spaces of multiple absolute summable sequences. Moreover, we obtain extensions of some classical tests for absolute convergence of series.

The paper is organized as follows. In Section 2, we introduce our notation and terminology. In Section 3, we define the iterations of the remainder operator and the spaces S(m) of m-times summable sequences. Moreover, we establish some basic properties of r m . For example, we show that

Δ m r m x= ( 1 ) m xand r m Δ m z= ( 1 ) m z
(1)

for any sequence xS(m) and any sequence z convergent to zero. This means that the operator ( 1 ) m r m is inverse to the restriction Δ m |Z where Z denotes the space of all convergent to zero sequences.

In Section 4, we introduce the spaces A(m) of absolutely m-times summable sequences and establish the relationships between A(m) and the spaces O( n s ). We also extend some classical tests for absolute convergence of series. For example, the Raabe test states that

if lim infn ( | a n | | a n + 1 | 1 ) >1,then  n = 1 | a n |<.

In Lemma 4.5 we show that

if lim infn ( | a n | | a n + 1 | 1 ) >2,then  n = 1 k = n | a k |<

and, more generally,

if lim infn ( | a n | | a n + 1 | 1 ) >m,then  i 1 = 1 i 2 = i 1 i m = i m 1 | a i m |<.

On the other hand,

if n ( | a n | | a n + 1 | 1 ) mfor large n,then  i 1 = 1 i 2 = i 1 i m = i m 1 | a i m |=.

Similarly in Lemma 4.4 we show that

lim sup log | a n | log n <mimplies i 1 = 1 i 2 = i 1 i m = i m 1 | a i m |<

and

log | a n | log n mfor large nimplies i 1 = 1 i 2 = i 1 i m = i m 1 | a i m |=.

For xS(m), by definition, we have

r m (x)(n)= i 1 = n i 2 = i 1 i m = i m 1 x i m .
(2)

If |x|S(m), i.e., xA(m), then xS(m) and

r m (x)(n)= k = 0 ( m 1 + k m 1 ) x n + k = i = n ( m + i n 1 m 1 ) x i .
(3)

We use (3) to obtain basic properties of the operator r m . At the end of Section 4, using all possible Δ p A(m) and r p A(m), we obtain an infinite stratification of the space of all sequences convergent to zero. This stratification induces a substratification of A(m) for any m.

In Section 5, we present some applications of our results to the study of asymptotic properties of solutions of difference equations. We use the spaces A(p) as the measure of approximation. For example, we apply our results and fixed point theorems to the study of solutions with prescribed asymptotic behavior. More precisely, using the Schauder fixed point theorem and the Knaster-Tarski fixed point theorem, we establish conditions under which for a given sequence b and a solution y of the equation Δ m y=b there exists a solution x of the equation

Δ m x n = a n f( x σ ( n ) )+ b n
(4)

such that

xy+A(p).

We also show that if a,bA(m+p) and x is a solution of (4) such that the sequence (f( x n )) is bounded, then

xKer Δ m +A(p).

Hence x is asymptotically polynomial. The equality

Δ m ( ( 1 ) m r m x ) =x

for xS(m), which is a consequence of (1), plays a crucial role in the application of fixed point theorems to the study of solutions of difference equations. The value r m x is used, mainly implicitly, in the study of solutions with prescribed asymptotic behavior. In some papers the multiple sums (2) appear explicitly; see for example [14] or [5]. This paper is a continuation of the papers [4, 6] and [7]. Our studies were inspired by the papers [812] and the papers [1317], and [18].

Some applications of our results to the study of asymptotic properties of solutions of nonautonomous difference equations are presented in [19].

2 Notation and terminology

Let , , denote the set of positive integers, the set of all integers and the set of real numbers, respectively. The space of all sequences x:NR we denote by SQ.

If p,kZ, pk, then N(p), N(p,k) denote the sets defined by

N(p)={p,p+1,},N(p,k)={p,p+1,,k}.

If x, y in SQ, then xy denotes the sequence defined by pointwise multiplication

xy(n)= x n y n .

Moreover, |x| denotes the sequence defined by |x|(n)=| x n | for every n.

We use the symbols ‘big O’ and ‘small o’ in the usual sense but for aSQ we also regard o(a) and O(a) as subspaces of SQ. More precisely, let

o(1)={xSQ:x is convergent to zero},O(1)={xSQ:x is bounded}

and for aSQ let

o(a)=ao(1)= { a x : x o ( 1 ) } ,O(a)=aO(1)= { a x : x O ( 1 ) } .

For mN(0) we define

Pol(m1)=Ker Δ m = { x SQ : Δ m x = 0 } .

Then Pol(m1) is the space of all polynomial sequences of degree less than m. For mN(0), kN we define numbers s k m by

s k m = ( m + k 1 m ) = k ( k + 1 ) ( k + m 1 ) m ! .

For pN we define

Fin(p)={xSQ: x n =0 for np}.

We say that a subset U of a metric space X is a uniform neighborhood of a subset Y of X if there exists a positive number ε such that

y Y B(y,ε)U,

where B(y,ε) denotes an open ball of radius ε about y.

A sequence xSQ is called nonoscillatory if x n x n + 1 0 for large n.

Assume f:RR. We say that a sequence xSQ is f-bounded if the sequence fx is bounded. Note that xσ is f-bounded for any f-bounded sequence x and any sequence σ:NN.

Example 2.1 If f(t)= e t , then f-boundedness of a sequence x is equivalent to the boundedness above of x. Assume f(t)= t 1 for t0, f(0)=3 and let L(x) denote the set of limit points of a given sequence x. Then f-boundedness of x is equivalent to the condition 0L(x).

3 Iterated remainder operator

In this section we introduce the remainder operator r, the spaces S(m) of m-times summable sequences and iterations r m :S(m)S(0). Next in Lemma 3.1 we establish some basic properties of r m and the relationships between r m and Δ m . Let

S ( 0 ) = o ( 1 ) = { x SQ : lim x n = 0 } , S ( 1 ) = { x SQ : the series  x n  is convergent } .

For xS(1), we define the sequence r(x) by the formula

r(x)(n)= j = n x j .

Then r(x)S(0) and we obtain the remainder operator

r:S(1)S(0).

For mN we define, by induction, the linear space S(m+1) and the linear operator r m + 1 :S(m+1)S(0) by

S(m+1)= { x S ( m ) : r m ( x ) S ( 1 ) } , r m + 1 (x)=r ( r m ( x ) ) .

The value r m (x)(n) we denote also by r n m (x) or simply r n m x.

Lemma 3.1 Assume x,ySQ, m,kN, pN(0) and qN(0,m1). Then

  1. (01)

    if |x|S(m), then xS(m) and | r m x| r m |x|,

  2. (02)

    |x|S(m) if and only if n = 1 s n m 1 | x n |<,

  3. (03)

    |x|S(m) if and only if n = 1 n m 1 | x n |<,

  4. (04)

    |x|S(m) if and only if O(x)S(m),

  5. (05)

    if |x|S(m), then r n m x= s 1 m 1 x n + s 2 m 1 x n + 1 + s 3 m 1 x n + 2 + ,

  6. (06)

    if |x|S(m), then r k m |x| n = k n m 1 | x n |,

  7. (07)

    if xS(m), then Δ m r m x= ( 1 ) m x,

  8. (08)

    if x=o(1), then Δ m xS(m) and r m Δ m x= ( 1 ) m x,

  9. (09)

    Δ m S(0)=S(m), r m S(m)=S(0),

  10. (10)

    Δ p S(m)=S(m+p), r p S(m+p)=S(m),

  11. (11)

    if x,yS(m) and x n y n for np, then

    r n m x r n m yfor np,
  12. (12)

    if xS(m) and y n = x n for np, then yS(m) and

    r n m y= r n m xfor np,
  13. (13)

    if yS(m) and 0xy, then xS(m),

  14. (14)

    if |x|S(m) and y is bounded, then yxS(m) and

    | r m (yx)||y| r m |x|,
  15. (15)

    if |yx|S(m), |y| is nondecreasing and positive, then |x|S(m),

    |y| r m |x| r m |yx|and Δ q r m x=o ( n q y 1 ) ,
  16. (16)

    if t(,0] and n = 1 n m t 1 | x n |<, then Δ q r m x=o( n t q ).

Proof Assertion (01) is proved in Lemma 1 of [6]. Assertion (02) is proved in Lemma 2 of [6]. (03) is proved in Lemma 3 of [6]. Since |x|O(x) we see that (04) is a consequence of (03). (05) is proved in Lemma 2 of [6]. Assertion (06) follows from (05). Assertion (07) is proved in Lemma 5 of [6]. Assertion (08) follows from Lemma 6 of [6]. Assertion (09) is an easy consequence of (08). Assertion (10) is a consequence of (09) and (08). Assertion (11) is obvious for m=1. For m>1 it can easily be proved by induction. Assertion (12) is an easy consequence of (11). Assertion (13) is well known for m=1. Assume it is true for certain m1. Let 0xy and yS(m+1). Then yS(m) and, by assumption, xS(m). Moreover, using (11) we have

0 r m x r m y

and, by (10), r m yS(1). Hence r m xS(1) and we obtain xS(m+1). Assertion (14) follows from (03) and Lemma 4 of [6]. Assume |y| is nondecreasing and positive and |yx|S(m). Then the sequence y 1 is bounded and using (04) we have |x|= | y | 1 |yx|S(m). Moreover, using (05), we have

| y n | r n m | x | = | y n | s 1 m 1 | x n | + | y n | s 2 m 1 | x n + 1 | + | y n | s 3 m 1 | x n + 2 | + s 1 m 1 | y n x n | + s 2 m 1 | y n + 1 x n + 1 | + s 3 m 1 | y n + 2 x n + 2 | + = r n m | y x |

for any n. Hence |y| r m |x| r m |yx|. Using (01), we obtain

|y| | r m x | |y| r m |x| r m |yx|=o(1)and r m x=o ( y 1 ) .
(5)

By (03), |y n q x|S(mq). Hence, replacing y by n q y and m by mq in (5), we obtain r m q x=o( n q y 1 ). Therefore

Δ q r m x= Δ q r q r m q x= ( 1 ) q r m q x=o ( n q y 1 )

and we obtain (15). Using (03) and taking y n = n t in (15) we obtain (16). □

Remark 3.1 For xS(m) and nN, by definition of r m , we have

r n m x= i 1 = n i 2 = i 1 i m = i m 1 x i m .

Moreover, if |x|S(m), then, by Lemma 3.1(05), we have

r n m x= k = 0 ( m 1 + k m 1 ) x n + k .

Remark 3.2 By Lemma 3.1(07) and (08) the restriction Δ m |S(0):S(0)S(m) is a bijection with inverse ( 1 ) m r m .

Remark 3.3 If X is a linear subspace of S(0) and Y is a linear subspace of S(m), then, using Lemma 3.1(08) and (07), we have

r m Δ m X=X, Δ m r m Y=Y.

For any mN we have S(m)S(0)=o(1)=o( n 0 ). In the following example we show that for any mN and any ε>0 there exists a sequence x such that

xS(m)O ( n ε )

and, moreover, lim n n ε | x n |=.

Example 3.1 Let mN and ε>0. Choose δ(0,ε). Let y n = ( 1 ) n n δ and x= Δ m y. Then yS(0), xS(m) and

| x n | = | Δ m y n | = | i = 0 m ( 1 ) m + i ( m i ) y n + i | = | i = 0 m ( 1 ) m + i ( m i ) ( 1 ) n + i ( n + i ) δ | = | ( 1 ) m + n i = 0 m ( m i ) ( n + i ) δ | n δ .

Hence n ε | x n | n ε δ and xS(m)O( n ε ).

4 Absolute summable sequences

In this section we introduce the spaces A(m) of absolutely m-times summable sequences. We establish the relationships between A(m) and the spaces o( n s ) and O( n s ). There exist many tests for the absolute convergence of series. Most of them may be extended to the case A(m). For example we present five of them in Lemmas 4.3-4.7. At the end of the section, using all possible Δ p A(m) and r p A(m) we obtain an infinite stratification of the set S(0)A().

For mN(0) we define the set A(m) by

A(m)= { a SQ : | a | S ( m ) } .

Moreover, let

A()= k = 0 A(k),o ( n ) = s R o ( n s ) ,Fin= p = 1 Fin(p).

Remark 4.1 Note that A(0)=S(0) and the condition aA(1) is equivalent to the absolute convergence of the series n = 1 a n . Moreover, A(m) is a linear subspace of S(m) for any mN(0). Note also that if pN and λ(1,1), then

Fin(p)Fino ( λ n ) A().

Lemma 4.1 Assume aSQ, mN(0) and sR. Then

  1. (a)

    aA(m+1)( n m a n )A(1),

  2. (b)

    aA(m+1)(n a n )A(m),

  3. (c)

    ( n s )A(m)s<m.

Proof Assertion (a) is a consequence of Lemma 3.1(03), (b) is a consequence of (a). Assertion (c) follows from (b) and from the fact that the condition ( n s )A(1) is equivalent to the condition s<1. □

Example 4.1 Let mN(0). If we define a sequence x by

x n = 1 n m log n ,

then, using Lemma 4.1(a), we obtain xo( n m )A(m).

Example 4.2 Assume mN and ε>0. Choose pN such that p 1 <ε. Let

A= { k 2 p : k N } ,B=NA, a n ={ n p 1 for  n A , n 2 for  n B .

Then

n A a n = k = 1 ( k 2 p ) p 1 = k = 1 k 2 <, n B a n n = 1 n 2 <,

and n ε a n = n ε p 1 for nA. Hence lim sup n n ε a n = and we obtain aA(1)O( n ε ). Let b n = n 1 m a n . Then, by Lemma 4.1(a),

bA(m)O ( n m + 1 ε ) .

Remark 4.2 Note that for real s, t the conditions o( n s )o( n t ) and O( n s )O( n t ) are equivalent to the condition st.

Lemma 4.2 Assume mN, ε>0 and sR. Then

  1. (a)

    O( n m ε )A(m)o( n m + 1 ),

  2. (b)

    o( n s )A(m)s<m,

  3. (c)

    A(m)O( n s )sm+1,

  4. (d)

    A()=o( n ).

Proof Assertion (a) follows from Lemma 4.1, (b) is a consequence of (a) and Example 4.1, (c) is a consequence of (a) and Example 4.2. Let mN(0). If s<m, then, by (b), o( n s )A(m). Hence o( n )A(m) for any mN(0). Therefore

o ( n ) A().

Let sR. Choose mN(0) such that m>1s. If aA(m), then, by the convergence of the series n = 1 n m 1 | a n |, we have n m 1 a n =o(1) and so ao( n 1 m )o( n s ). Hence A(m)o( n s ). Therefore A()o( n s ) for any sR and we obtain

A()o ( n ) .

 □

Lemma 4.3 (Comparison test)

Assume a,bSQ and mN. Then

  1. (a)

    if | a n || b n | for large n and bA(m), then aA(m),

  2. (b)

    if | a n || b n | for large n and bA(m), then aA(m),

  3. (c)

    if lim sup| a n / b n |< and bA(m), then aA(m),

  4. (d)

    if lim inf| a n / b n |>0 and aA(m), then bA(m),

  5. (e)

    if | a n + 1 / a n || b n + 1 / b n | for large n and bA(m), then aA(m).

Proof Assertion (a) follows from Lemma 3.1(13). Assertions (b), (c), (d), and (e) are consequences of (a). □

Lemma 4.4 (Generalized logarithmic test)

Assume aSQ, mN and

u n = log | a n | log n .

Then

  1. (a)

    if lim sup u n <m, then aA(m),

  2. (b)

    if u n m for large n, then aA(m),

  3. (c)

    if lim inf u n >m, then aA(m),

  4. (d)

    if lim u n =, then aA().

Proof If lim sup u n <m, then there exists a number s<m such that u n <s for large n. Then | a n | n s for large n. Hence, using Lemma 4.1 and Lemma 4.3, we obtain (a). If u n m for large n, then | a n | n m for large n and (b) follows from Lemma 4.3 and the fact that ( n m )A(m). Assertion (c) follows immediately from (b), and (d) is a consequence of (a). □

Lemma 4.5 (Generalized Raabe test)

Assume aSQ, mN and

u n =n ( | a n | | a n + 1 | 1 ) .

Then

  1. (a)

    if lim inf u n >m, then aA(m),

  2. (b)

    if u n m for large n, then aA(m),

  3. (c)

    if lim sup u n <m, then aA(m),

  4. (d)

    if lim u n =, then aA().

Proof For m=1 assertion (a) follows from the usual Raabe test. Assume it is true for certain m1 and lim inf u n >m+1. Let

b n =n a n , w n =n ( | b n | | b n + 1 | 1 ) .

Then

w n =n ( n | a n | ( n + 1 ) | a n + 1 | n + 1 n + 1 ) = n n + 1 ( n | a n | | a n + 1 | n 1 ) = n n + 1 ( u n 1).

Hence lim inf w n =lim inf u n 1>m and, by inductive hypothesis, bA(m). Hence, by Lemma 4.1(b), aA(m+1) and we obtain (a). Similarly we may obtain (b), by taking b n =n a n , using the usual Raabe test and Lemma 4.1(b). Assertion (c) follows from (b), and (d) is a consequence of (a). □

Lemma 4.6 (Generalized Gauss test)

Let aSQ, mN, α,βR, s<1 and

| a n | | a n + 1 | =α+ β n +O ( n s ) .

Then

  1. (a)

    if α>λ>1, then ao( λ n ),

  2. (b)

    if α<1, then ao(1),

  3. (c)

    if α=1 and β>m, then aA(m),

  4. (d)

    if α=1 and βm, then aA(m).

Proof Note that

lim n | a n | | a n + 1 | =α.

Hence (a) and (b) follow from the d’Alembert ratio test. For m=1 assertions (c) and (d) follow from the usual Gauss test. Assume they are true for certain m1. Let b n =n a n , u n =1/(n+1). Then

| b n | | b n + 1 | = n n + 1 | a n | | a n + 1 | = ( 1 u n ) ( 1 + β n + O ( n s ) ) = 1 + β n + O ( n s ) 1 n + 1 β n ( n + 1 ) 1 n + 1 O ( n s ) = 1 + β 1 n + ( 1 n 1 n + 1 ) + O ( n s ) + O ( n 2 ) + O ( n s 1 ) = 1 + β 1 n + O ( n s )

for certain s <1. If β>m+1, then β1>m and, by inductive hypothesis, bA(m). Similarly, if βm+1, then bA(m). Now, assertions (c) and (d) follow from Lemma 4.1(b). □

Lemma 4.7 (Generalized Bertrand test)

Assume aSQ, mN and

| a n | | a n + 1 | =1+ m n + λ n n log n .
(6)

Then

  1. (a)

    if lim inf λ n >1, then aA(m),

  2. (b)

    if λ n 1 for large n, then aA(m),

  3. (c)

    if lim sup λ n <1, then aA(m).

Proof Let b n =n a n . Then

| b n | | b n + 1 | = n n + 1 ( 1 + m n + λ n n ln n ) = n n + 1 + m n + 1 + λ n ( n + 1 ) log n = 1 + m 1 n + 1 + ( n n + 1 ) λ n n log n = 1 + m 1 n c n + z n λ n n log n ,

where

z n = n n + 1 , c n = m 1 n m 1 n + 1 = m 1 n ( n + 1 ) .

Let

w n = c n nlogn= ( m 1 ) log n n + 1 , τ n = z n λ n w n .

Then

lim w n =0,lim z n =1,lim inf τ n =lim inf λ n
(7)

and

| b n | | b n + 1 | =1+ m 1 n + τ n n log n .
(8)

Moreover,

w n 0,0< z n <1, τ n λ n .
(9)

For m=1 assertion (a) follows from the classical Bertrand test. Assume it is true for certain m11. Then using the inductive hypothesis, (8), and (7), we have bA(m1). Hence, by Lemma 4.1, aA(m) and we obtain (a). Analogously, using (9), we obtain (b). Assertion (c) is a consequence of (b). The proof is complete. □

Remark 4.3 Computing λ n from (6) we have

λ n = ( n ( | a n | | a n + 1 | 1 ) m ) logn.
(10)

Replacing (6) by (10) in Lemma 4.7 one can obtain another form of the Bertrand test.

Lemma 4.8 Assume mN. Then

  1. (a)

    ΔA(m)A(m),

  2. (b)

    A(m)rA(m),

  3. (c)

    rA(m)A(m1),

  4. (d)

    A(m)ΔA(m1).

Proof For ySQ let EySQ be defined by

Ey(n)=y(n+1).

Let xS(m). By Lemma 3.1(09) there exists yS(0) such that x= Δ m y. Obviously EΔy=ΔEy and, by induction, E Δ m y= Δ m Ey. Moreover, EyS(0). Hence

Ex=E Δ m y= Δ m EyS(m).

Now assume xA(m). Then |x|S(m) and |Ex|=E|x|S(m). Hence

0|Δx||Ex|+|x|S(m).

By Lemma 3.1(13) we have ΔxA(m). Hence

ΔA(m)A(m)andA(m)=rΔA(m)rA(m).

From |x|S(m) we have r|x|S(m1). Moreover, 0|rx|r|x|. Hence |rx|S(m1). Therefore rxA(m1) and we obtain

rA(m)A(m1),A(m)=ΔrA(m)ΔA(m1).

The proof is complete. □

Example 4.3 Let mN, x n = n m , y=Δx. By Theorem 2.2 in [7], we have yO( n m 1 ). Hence, by Lemma 4.2, yA(m). On the other hand xA(m) and we obtain y=ΔxΔA(m). Therefore yA(m)ΔA(m).

Example 4.4 Let mN, s(m,m+1], x n = ( 1 ) n n s . Then, by Lemma 4.1, xA(m)A(m+1). Assume xrA(m+1). Choose yA(m+1) such that x=ry. Then, using Lemma 3.1(07), we obtain Δx=Δry=yA(m+1). Since x is alternating we have |Δx||x|. Hence, by Lemma 4.3, y=ΔxA(m+1). This contradiction shows that xA(m)rA(m+1).

Remark 4.4 Using Lemma 4.8 we obtain the following infinite diagram, where arrows denote inclusions:

Using Remark 3.2 and Example 4.3 we can see that any vertical arrow represents a proper inclusion. Analogously, using Remark 3.2 and Example 4.4 we can see that any horizontal arrow represents a proper inclusion.

Remark 4.5 Introducing new notation

S k k + p = Δ p A ( k ) , S k k m = r m A ( k ) , S k = n = 0 S k n , S n = k = 0 S k n , S = k = 0 S k k

for k,pN(0) and mN(0,k) we can extend the diagram from Remark 4.4 in the following way:

Note that if k,n,pN(0) and mN(0,n), then

Δ p S k n = S k n + p , r m S k n = S k n m , Δ p S n = S n + p , r m S n = S n m , Δ m S k = S k = r m S k , Δ m S = S = r m S .

Moreover,

S =A()=o ( n ) .

5 Approximative solutions of difference equations

In this section we present some applications of our previous results in the study of asymptotic properties of solutions of difference equations. We use the spaces A(p) to measure the ‘degree of approximation’.

Assume mN, a,bSQ, f:RR, σ:NN and lim n σ(n)=. We consider the equation

Δ m x n = a n f( x σ ( n ) )+ b n .
(E)

By a solution of (E) we mean a sequence x:NR satisfying (E) for all large n. Note that the assumption σ:NN does not exclude the case of equations with delayed argument. For example we may define σ(n)=n5 for n>5 and σ(n)=1 for n5.

In Theorem 5.1, using fixed point theorems and the iterated remainder operator, we establish conditions under which there exist solutions of (E) with prescribed asymptotic behavior. In Theorem 5.2 we show that in many cases some assumptions of Theorem 5.1 are necessary. In Theorem 5.3 we establish conditions under which all f-bounded solutions of (E) are asymptotically polynomial.

In the proof of our first theorem we will use the Schauder fixed point theorem and the following version of the Knaster-Tarski fixed point theorem.

Lemma 5.1 If X is a complete partially ordered set and a map T:XX is nondecreasing then there exists x 0 X such that T( x 0 )= x 0 .

A simple proof of this result can be found in [20] (or in [19]).

Theorem 5.1 Assume pN(0), aA(m+p), ySQ, Δ m y=b, f is bounded on some uniform neighborhood U of the set y(N) and one of the following conditions is satisfied:

  1. (a)

    f is nondecreasing on U and ( 1 ) m a n 0 for large n,

  2. (b)

    f is nonincreasing on U and ( 1 ) m a n 0 for large n,

  3. (c)

    f is continuous on U.

Then there exists a solution x of (E) such that

xy+A(p).

Proof For xSQ let x ¯ =fxσ. Choose c>0 and M>0 such that

n = 1 ( y n c, y n +c)U,|f(t)|Mfor tU.
(11)

Let

ρ=M r m |a|.

Choose numbers k 0 and k such that ρ n <c for n k 0 and σ(n) k 0 for nk. Let

S= { x SQ : | x y | ρ  and  x n = y n  for  n < k } .

It is easy to see that S, with natural order defined by xz if x n z n for every nN, is a complete partially ordered set. If xS and nk, then | x σ ( n ) y σ ( n ) |<c. Hence

x σ ( n ) U
(12)

for any xS and nk. If xS, then, by (11) and (12), the sequence x ¯ is bounded and so a x ¯ A(m+p). We define an operator H:SSQ by

H(x)(n):={ y n for  n < k , y n + ( 1 ) m r n m ( a x ¯ ) for  n k .

If nk, then, using Lemma 3.1(01) and Lemma 3.1(14), we obtain

|H(x)(n) y n |=| r n m (a x ¯ )| r n m |a x ¯ |M r n m |a|= ρ n .

Hence H(S)S.

Assume now that the condition (a) is satisfied. We can assume that ( 1 ) m a n 0 for nk. Let x,zS and xz. Since f is nondecreasing on U we have f( x σ ( n ) )f( z σ ( n ) ) for nk. Hence

( 1 ) m a n f( x σ ( n ) ) ( 1 ) m a n f( z σ ( n ) )

for nk. By Lemma 3.1(11), we have H(x)(n)H(z)(n) for nk. Moreover, H(x)(n)= y n =H(z)(n) for n<k. Hence H(x)H(z). Now, using the Knaster-Tarski fixed point theorem we obtain xS such that H(x)=x. Analogously if condition (b) is satisfied then H(x)=x for certain xS.

Assume (c) and let d be a metric on S defined by

d(x,z)=xz= sup n N | x n z n |.

Let BS denote the Banach space of all bounded sequences xSQ with the norm x= sup n N | x n | and let

T= { x BS : | x | ρ  and  x n = 0  for  n < k } .

It is easy to see that T is a convex and closed subset of BS. Choose an ε>0. Then there exists qN such that ρ n <ε for nq. For n=1,,q let G n denote a finite ε-net for the interval [ ρ n , ρ n ] and let

G={xT: x n G n  for nq and  x n =0 for n>q}.

Then G is a finite ε-net for T. Hence T is a complete and totally bounded metric space and so, T is compact. Hence T is a convex and compact subset of the Banach space BS and, by the Schauder fixed point theorem, any continuous map TT has a fixed point. Let F:TS be a map given by F(x)(n)= x n + y n . Then F is an isometry of T onto S. Assume U:SS is a continuous map and let W= F 1 UF. Then W:TT is continuous and there exists a point zT such that Wz=z. Let x=Fz. Then

x=Fz=FWz=F F 1 UFz=Ux.

Hence any continuous map U:SS has a fixed point. Let ε>0. Choose qN(k) and α>0 such that

M m = q n m 1 | a n |<εandα n = k q n m 1 | a n |<ε.

Let

W= n = k q [ y n ρ n , y n + ρ n ].

Then WU. By compactness of W, f is uniformly continuous on W. Hence there exists δ>0 such that if s,tW and |st|<δ, then |f(s)f(t)|<α. Assume x,zS, d(x,z)<δ. Let u= x ¯ z ¯ . Then

d ( H x , H z ) = sup n k | r n m ( a x ¯ ) r n m ( a z ¯ ) | = sup n k | r n m ( a u ) | sup n k r n m | a u | = r p m | a u | n = k q n m 1 | a n u n | + n = q n m 1 | a n u n | α n = k q n m 1 | a n | + 2 M n = q n m 1 | a n | < 3 ε .

Hence H is continuous and there exists xS such that H(x)=x. Then

x n =H(x)(n)= y n + ( 1 ) m r n m (a x ¯ )

for nk. Hence

x ( y + ( 1 ) m r m ( a x ¯ ) ) Fin(k).

Since the sequence x ¯ is bounded, we have a x ¯ O(a). Hence

xy+ r m O(a)+Fin(k)y+A(p).

Moreover, for nk, by Lemma 3.1(07), we have

Δ m x n = Δ m y n + ( 1 ) m Δ m r n m (a x ¯ )= b n + a n f( x σ ( n ) ).

Hence x is a solution of (E). The proof is complete. □

Remark 5.1 The conclusion xy+A(p) of Theorem 5.1 may be written in the form

yx+A(p).

We can say that y is an approximative solution of (E) with ‘degree of approximation’ A(p).

Remark 5.2 Using Lemma 4.5 we can replace the assumption aA(m+p) of Theorem 5.1 by

lim infn ( | a n | | a n + 1 | 1 ) >m+p.

Analogously, the conclusion of this theorem may be replaced: there exist a solution x of (E) and a sequence zSQ such that x=y+z and

lim supn ( | z n | | z n + 1 | 1 ) p.

Similarly, using other tests, we can obtain many formulations of this theorem.

Theorem 5.2 Assume qN, pN(0), ε>0, s(,p), a,b,ySQ, Δ m y=b, a is nonoscillatory, there exists a uniform neighborhood U of the set y(N(q)) such that f|Uε or f|Uε, and one of the following conditions is satisfied:

  1. (a)

    there exists a solution x of (E) such that x=y+o( n s ),

  2. (b)

    there exists a solution x of (E) such that xy+S(p).

Then aA(m+p).

Proof By Lemma 4.2 we have o( n s )A(p)S(p). Hence, there exists a solution x of (E) and a sequence zS(p) such that x=y+z. Let x ¯ =fxσ. For large n we have

a n x ¯ n + b n = a n f( x σ ( n ) )+ b n = Δ m x n = Δ m y n + Δ m z n = b n + Δ m z n .

By Lemma 3.1(10), Δ m zS(m+p). Hence, by Lemma 3.1(12), we have a x ¯ S(m+p). Since z n =o(1), we have x σ ( n ) U for large n. Hence the sequence a x ¯ is nonoscillatory and we obtain |a x ¯ |S(m+p), which means a x ¯ A(m+p). Moreover,

| a n |=| a n x ¯ n | | x ¯ n | 1 | a n x ¯ n | ε 1

for large n. Hence, by Lemma 3.1(04), aA(m+p). The proof is complete. □

Remark 5.3 Theorem 5.2 extends Theorem 1 of [8] and parts of Theorems 6 and 7 of [6].

Lemma 5.2 Assume mN, pN(0), xSQ and Δ m xA(m+p). Then

xPol(m1)+A(p).

Proof Using Lemma 4.8(c) we obtain

r m A(m+p)A(p).
(13)

Let z= ( 1 ) m r m Δ m x. Then Δ m z= Δ m ( 1 ) m r m Δ m x= Δ m x. Hence

xzKer Δ m =Pol(m1).

By (13) we have zA(p). Hence

x=(xz)+zPol(m1)+A(p).

 □

Theorem 5.3 Assume pN(0), a,bA(m+p) and x is an f-bounded solution of (E). Then

xPol(m1)+A(p).

Proof The sequence x ¯ =fxσ is bounded. Hence a x ¯ O(a) and we obtain Δ m xO(|a|+|b|). Therefore Δ m xA(m+p) and, by Lemma 5.2, we have

xPol(m1)+A(p).

 □

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Keywords

  • remainder operator
  • Raabe’s test
  • Gauss’s test
  • Bertrand’s test
  • difference equation
  • prescribed asymptotic behavior
  • asymptotically polynomial solution