R E S E A R C H
Open Access
Iterated remainder operator, tests for multiple
convergence of series, and solutions of
difference equations
Janusz Migda
**Correspondence:
[email protected] Faculty of Mathematics and Computer Science, A. Mickiewicz University, Umultowska 87, Pozna ´n, 61-614, Poland
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
Keywords: remainder operator; Raabe’s test; Gauss’s test; Bertrand’s test; difference equation; prescribed asymptotic behavior; asymptotically polynomial solution
1 Introduction
Letrdenote the operator which assigns to any convergent series the sequence of its re-mainders. The purpose of this paper is to study the basic properties of iterationsrm of
rand 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 , we introduce our notation and terminol-ogy. In Section , we define the iterations of the remainder operator and the spaces S(m) ofm-times summable sequences. Moreover, we establish some basic properties ofrm. For
example, we show that
mrmx= (–)mx and rmmz= (–)mz ()
for any sequencex∈S(m) and any sequencezconvergent to zero. This means that the operator (–)mrmis inverse to the restrictionm|ZwhereZdenotes the space of all
con-vergent to zero sequences.
In Section , we introduce the spaces A(m) of absolutelym-times summable sequences and establish the relationships between A(m) and the spaces O(ns). We also extend some
classical tests for absolute convergence of series. For example, the Raabe test states that
Similarly in Lemma . we show that
lim suplog|an|
We use () to obtain basic properties of the operatorrm. At the end of Section , using
all possiblepA(m) andrpA(m), we obtain an infinite stratification of the space of all sequences convergent to zero. This stratification induces a substratification of A(m) for anym.
solutions with prescribed asymptotic behavior. More precisely, using the Schauder fixed point theorem and the Knaster-Tarski fixed point theorem, we establish conditions un-der which for a given sequenceband a solutionyof the equationmy=bthere exists a solutionxof the equation
mxn=anf(xσ(n)) +bn ()
such that
x∈y+ A(p).
We also show that ifa,b∈A(m+p) andxis a solution of () such that the sequence (f(xn)) is bounded, then
x∈Kerm+ A(p).
Hencexis asymptotically polynomial. The equality
m(–)mrmx=x
forx∈S(m), which is a consequence of (), plays a crucial role in the application of fixed point theorems to the study of solutions of difference equations. The valuermxis used,
mainly implicitly, in the study of solutions with prescribed asymptotic behavior. In some papers the multiple sums () appear explicitly; see for example [–] or []. This paper is a continuation of the papers [, ] and []. Our studies were inspired by the papers [–] and the papers [–], and [].
Some applications of our results to the study of asymptotic properties of solutions of nonautonomous difference equations are presented in [].
2 Notation and terminology
LetN,Z,Rdenote the set of positive integers, the set of all integers and the set of real numbers, respectively. The space of all sequencesx:N→Rwe denote by SQ.
Ifp,k∈Z,p≤k, thenN(p),N(p,k) denote the sets defined by
N(p) ={p,p+ , . . .}, N(p,k) ={p,p+ , . . . ,k}.
Ifx,yin SQ, thenxydenotes the sequence defined by pointwise multiplication
xy(n) =xnyn.
Moreover,|x|denotes the sequence defined by|x|(n) =|xn|for everyn.
We use the symbols ‘big O’ and ‘small o’ in the usual sense but fora∈SQ we also regard o(a) and O(a) as subspaces of SQ. More precisely, let
and fora∈SQ let
o(a) =ao() =ax:x∈o(), O(a) =aO() =ax:x∈O().
Form∈N() we define
Pol(m– ) =Kerm=x∈SQ :mx= .
ThenPol(m– ) is the space of all polynomial sequences of degree less thanm. Form∈
N(),k∈Nwe define numberssm k by
smk =
m+k–
m
=k(k+ )· · ·(k+m– )
m! .
Forp∈Nwe define
Fin(p) ={x∈SQ :xn= forn≥p}.
We say that a subsetUof a metric spaceXis a uniform neighborhood of a subsetY ofX
if there exists a positive numberεsuch that
y∈Y
B(y,ε)⊂U,
where B(y,ε) denotes an open ball of radiusεabouty.
A sequencex∈SQ is called nonoscillatory ifxnxn+≥ for largen.
Assumef :R→R. We say that a sequencex∈SQ isf-bounded if the sequencef◦xis bounded. Note thatx◦σ isf-bounded for anyf-bounded sequencexand any sequence
σ:N→N.
Example . Iff(t) =et, thenf-boundedness of a sequencexis equivalent to the
bound-edness above ofx. Assumef(t) =t–fort= ,f() = and letL(x) denote the set of limit points of a given sequence x. Then f-boundedness ofx is equivalent to the condition /∈L(x).
3 Iterated remainder operator
In this section we introduce the remainder operator r, the spaces S(m) of m-times summable sequences and iterationsrm: S(m)→S(). Next in Lemma . we establish some basic properties ofrmand the relationships betweenrmandm. Let
S() = o() ={x∈SQ :limxn= },
S() =
x∈SQ : the seriesxnis convergent
.
Forx∈S(), we define the sequencer(x) by the formula
r(x)(n) = ∞
Thenr(x)∈S() and we obtain the remainder operator
r: S()→S().
Form∈Nwe define, by induction, the linear space S(m+ ) and the linear operatorrm+:
S(m+ )→S() by
S(m+ ) =x∈S(m) :rm(x)∈S(), rm+(x) =rrm(x).
The valuerm(x)(n) we denote also byrm
n(x) or simplyrnmx.
Lemma . Assume x,y∈SQ,m,k∈N,p∈N()and q∈N(,m– ).Then
() if|x| ∈S(m),thenx∈S(m)and|rmx| ≤rm|x|,
() |x| ∈S(m)if and only if∞n=sm–n |xn|<∞, () |x| ∈S(m)if and only if∞n=nm–|xn|<∞,
() |x| ∈S(m)if and only ifO(x)⊂S(m), () if|x| ∈S(m),thenrm
nx=sm– xn+sm–xn++sm– xn++· · ·,
() if|x| ∈S(m),thenrm k|x| ≤
∞
n=knm–|xn|,
() ifx∈S(m),thenmrmx= (–)mx,
() ifx= o(),thenmx∈S(m)andrmmx= (–)mx,
() mS() = S(m),rmS(m) = S(),
() pS(m) = S(m+p),rpS(m+p) = S(m), () ifx,y∈S(m)andxn≤ynforn≥p,then
rnmx≤rmny forn≥p,
() ifx∈S(m)andyn=xnforn≥p,theny∈S(m)and
rnmy=rmnx forn≥p,
() ify∈S(m)and≤x≤y,thenx∈S(m),
() if|x| ∈S(m)andyis bounded,thenyx∈S(m)and
rm(yx)≤ |y|rm|x|,
() if|yx| ∈S(m),|y|is nondecreasing and positive,then|x| ∈S(m),
|y|rm|x| ≤rm|yx| and qrmx= on–qy–,
() ift∈(–∞, ]and∞n=nm–t–|xn|<∞,thenqrmx= o(nt–q).
an easy consequence of (). Assertion () is well known form= . Assume it is true for certainm≥. Let ≤x≤yandy∈S(m+ ). Theny∈S(m) and, by assumption,x∈S(m). Moreover, using () we have
≤rmx≤rmy
and, by (),rmy∈S(). Hencermx∈S() and we obtainx∈S(m+ ). Assertion () follows
from () and Lemma of []. Assume|y|is nondecreasing and positive and|yx| ∈S(m). Then the sequencey–is bounded and using () we have|x|=|y|–|yx| ∈S(m). Moreover,
using (), we have
|yn|rnm|x|=|yn|sm– |xn|+|yn|sm– |xn+|+|yn|sm– |xn+|+· · · ≤sm– |ynxn|+sm– |yn+xn+|+sm– |yn+xn+|+· · ·=rnm|yx|
for anyn. Hence|y|rm|x| ≤rm|yx|. Using (), we obtain
|y|rmx≤ |y|rm|x| ≤rm|yx|= o() and rmx= oy–. ()
By (),|ynqx| ∈S(m–q). Hence, replacingyby nqyandmby m–qin (), we obtain
rm–qx= o(n–qy–). Therefore
qrmx=qrqrm–qx= (–)qrm–qx= on–qy–
and we obtain (). Using () and takingyn=n–tin () we obtain ().
Remark . Forx∈S(m) andn∈N, by definition ofrm, we have
rmnx= ∞
i=n ∞
i=i · · ·
∞
im=im–
xim.
Moreover, if|x| ∈S(m), then, by Lemma .(), we have
rmnx= ∞
k=
m– +k m–
xn+k.
Remark . By Lemma .() and () the restrictionm|S() : S()→S(m) is a bijec-tion with inverse (–)mrm.
Remark . IfXis a linear subspace of S() andYis a linear subspace of S(m), then, using Lemma .() and (), we have
rmmX=X, mrmY=Y.
For anym∈Nwe have S(m)⊂S() = o() = o(n). In the following example we show that for anym∈Nand anyε> there exists a sequencexsuch that
x∈S(m)\On–ε
Example . Letm∈Nandε> . Chooseδ∈(,ε). Letyn= (–)nn–δandx=my. Then
In this section we introduce the spaces A(m) of absolutelym-times summable sequences. We establish the relationships between A(m) and the spaces o(ns) and O(ns). 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 .-.. At the end of the section, using all possiblepA(m) andrpA(m) we obtain an infinite stratification of the
set S()\A(∞).
Remark . Note that A() = S() and the conditiona∈A() is equivalent to the absolute convergence of the series∞n=an. Moreover, A(m) is a linear subspace of S(m) for any
Proof Assertion (a) is a consequence of Lemma .(), (b) is a consequence of (a). Asser-tion (c) follows from (b) and from the fact that the condiAsser-tion (ns)∈A() is equivalent to
the conditions< –.
Example . Letm∈N(). If we define a sequencexby
xn=
nmlogn,
Example . Assumem∈Nandε> . Choosep∈Nsuch thatp–<ε. Let
A=kp:k∈N, B=N\A, an= ⎧ ⎨ ⎩
n–p– forn∈A,
n– forn∈B.
Then
n∈A an=
∞
k=
kp–p–= ∞
k=
k–<∞,
n∈B an≤
∞
n=
n–<∞,
andnεan=nε–p–
forn∈A. Hencelim supn→∞nεan=∞and we obtaina∈A()\O(n–ε). Letbn=n–man. Then, by Lemma .(a),
b∈A(m)\On–m+–ε.
Remark . Note that for reals, tthe conditions o(ns)⊂o(nt) and O(ns)⊂O(nt) are equivalent to the conditions≤t.
Lemma . Assume m∈N,ε> and s∈R.Then
(a) O(n–m–ε)⊂A(m)⊂o(n–m+),
(b) o(ns)⊂A(m)⇔s< –m,
(c) A(m)⊂O(ns)⇔s≥–m+ , (d) A(∞) = o(n–∞).
Proof Assertion (a) follows from Lemma ., (b) is a consequence of (a) and Example ., (c) is a consequence of (a) and Example .. Letm∈N(). Ifs< –m, then, by (b), o(ns)⊂ A(m). Hence o(n–∞)⊂A(m) for anym∈N(). Therefore
on–∞⊂A(∞).
Lets∈R. Choosem∈N() such thatm> –s. Ifa∈A(m), then, by the convergence of the series∞n=nm–|an|, we havenm–an= o() and soa∈o(n–m)⊂o(ns). Hence A(m)⊂
o(ns). Therefore A(∞)⊂o(ns) for anys∈Rand we obtain
A(∞)⊂on–∞.
Lemma .(Comparison test) Assume a,b∈SQand m∈N.Then
(a) if|an| ≤ |bn|for largenandb∈A(m),thena∈A(m), (b) if|an| ≥ |bn|for largenandb∈/A(m),thena∈/A(m), (c) iflim sup|an/bn|<∞andb∈A(m),thena∈A(m), (d) iflim inf|an/bn|> anda∈A(m),thenb∈A(m),
(e) if|an+/an| ≤ |bn+/bn|for largenandb∈A(m),thena∈A(m).
Proof Assertion (a) follows from Lemma .(). Assertions (b), (c), (d), and (e) are
Lemma .(Generalized logarithmic test) Assume a∈SQ,m∈Nand
Lemma .(Generalized Raabe test) Assume a∈SQ,m∈Nand
un=n
Proof Form= assertion (a) follows from the usual Raabe test. Assume it is true for cer-tainm≥ andlim infun>m+ . Let
bn=nan, using the usual Raabe test and Lemma .(b). Assertion (c) follows from (b), and
(b) ifα< ,thena∈/o(),
Hence (a) and (b) follow from the d’Alembert ratio test. Form= assertions (c) and (d) follow from the usual Gauss test. Assume they are true for certainm≥. Letbn=nan,
Lemma .(Generalized Bertrand test) Assume a∈SQ,m∈Nand
Let
wn=cnnlogn=
(m– )logn
n+ , τn=znλn–wn.
Then
limwn= , limzn= , lim infτn=lim infλn ()
and
|bn|
|bn+| = + m–
n +
τn
nlogn. ()
Moreover,
wn≥, <zn< , τn≤λn. ()
Form= assertion (a) follows from the classical Bertrand test. Assume it is true for certain
m–≥. Then using the inductive hypothesis, (), and (), we haveb∈A(m–). Hence, by Lemma .,a∈A(m) and we obtain (a). Analogously, using (), we obtain (b). Assertion (c)
is a consequence of (b). The proof is complete.
Remark . Computingλnfrom () we have
λn=
n |
an|
|an+|
–
–m
logn. ()
Replacing () by () in Lemma . one can obtain another form of the Bertrand test.
Lemma . Assume m∈N.Then
(a) A(m)⊂A(m), (b) A(m)⊂rA(m),
(c) rA(m)⊂A(m– ), (d) A(m)⊂A(m– ).
Proof Fory∈SQ let Ey∈SQ be defined by
Ey(n) =y(n+ ).
Letx∈S(m). By Lemma .() there existsy∈S() such thatx=my. Obviously Ey=
Eyand, by induction, Emy=mEy. Moreover, Ey∈S(). Hence
Ex= Emy=mEy∈S(m).
Now assumex∈A(m). Then|x| ∈S(m) and|Ex|= E|x| ∈S(m). Hence
By Lemma .() we havex∈A(m). Hence
A(m)⊂A(m) and A(m) =rA(m)⊂rA(m).
From|x| ∈S(m) we haver|x| ∈S(m– ). Moreover, ≤ |rx| ≤r|x|. Hence|rx| ∈S(m– ). Thereforerx∈A(m– ) and we obtain
rA(m)⊂A(m– ), A(m) =rA(m)⊂A(m– ).
The proof is complete.
Example . Letm∈N,xn=n–m,y=x. By Theorem . in [], we havey∈O(n–m–).
Hence, by Lemma .,y∈A(m). On the other handx∈/ A(m) and we obtainy=x∈/
A(m). Thereforey∈A(m)\A(m).
Example . Letm∈N,s∈(m,m+ ],xn= (–)nn–s. Then, by Lemma .,x∈A(m)\
A(m+ ). Assume x∈rA(m+ ). Choosey∈A(m+ ) such that x=ry. Then, using Lemma .(), we obtain x=ry= –y∈A(m+ ). Since x is alternating we have |x| ≥ |x|. Hence, by Lemma ., y=x∈/ A(m+ ). This contradiction shows that
x∈A(m)\rA(m+ ).
Remark . Using Lemma . we obtain the following infinite diagram, where arrows denote inclusions:
· · · ––––––→ rA() ––––––→ rA() ––––––→ rA() ––––––→ S() ⏐
⏐ ⏐⏐ ⏐⏐ ⏐⏐
· · · ––––––→ rA() ––––––→ rA() ––––––→ A() ––––––→ S() ⏐
⏐ ⏐⏐ ⏐⏐ ⏐⏐
· · · ––––––→ rA() ––––––→ A() ––––––→ A() ––––––→ S()
⏐
⏐ ⏐⏐ ⏐⏐ ⏐⏐
· · · ––––––→ A() ––––––→ A() ––––––→ A() ––––––→ S()
⏐
⏐ ⏐⏐ ⏐⏐ ⏐⏐
..
. ... ... ...
Using Remark . and Example . we can see that any vertical arrow represents a proper inclusion. Analogously, using Remark . and Example . we can see that any horizontal arrow represents a proper inclusion.
Remark . Introducing new notation
Sk+pk =pA(k), Sk–mk =rmA(k),
S∞k = ∞
n=
Snk, Sn∞= ∞
k=
Snk, S∞∞= ∞
k=
fork,p∈N() andm∈N(,k) we can extend the diagram from Remark . in the follow-ing way:
S
∞ ––––––→ · · · ––––––→ S ––––––→ S ––––––→ S ––––––→ S ⏐
⏐ ⏐⏐ ⏐⏐ ⏐⏐ ⏐⏐
S
∞ ––––––→ · · · ––––––→ S ––––––→ S ––––––→ S ––––––→ S ⏐
⏐ ⏐⏐ ⏐⏐ ⏐⏐ ⏐⏐
S
∞ ––––––→ · · · ––––––→ S ––––––→ S ––––––→ S ––––––→ S ⏐
⏐ ⏐⏐ ⏐⏐ ⏐⏐ ⏐⏐
S
∞ ––––––→ · · · ––––––→ S ––––––→ S ––––––→ S ––––––→ S ⏐
⏐ ⏐⏐ ⏐⏐ ⏐⏐ ⏐⏐
..
. ... ... ... ...
⏐
⏐ ⏐⏐ ⏐⏐ ⏐⏐ ⏐⏐
S∞∞ ––––––→ · · · ––––––→ S∞ ––––––→ S∞ ––––––→ S∞ ––––––→ S∞
Note that ifk,n,p∈N() andm∈N(,n), then
pSnk= Sn+pk , rmSnk= Sn–mk , pSn∞= Sn+p∞ , rmSn∞= Sn–m∞ ,
mS∞k = S∞k =rmS∞k , mS∞∞= S∞∞=rmS∞∞.
Moreover,
S∞∞= A(∞) = on–∞.
5 Approximative solutions of difference equations
In this section we present some applications of our previous results in the study of asymp-totic properties of solutions of difference equations. We use the spaces A(p) to measure the ‘degree of approximation’.
Assumem∈N,a,b∈SQ,f :R→R,σ:N→Nandlimn→∞σ(n) =∞. We consider the equation
mxn=anf(xσ(n)) +bn. (E)
By a solution of (E) we mean a sequencex:N→Rsatisfying (E) for all largen. Note that the assumptionσ:N→Ndoes not exclude the case of equations with delayed argument. For example we may defineσ(n) =n– forn> andσ(n) = forn≤.
In Theorem ., 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 . we show that in many cases some assumptions of Theorem . are necessary. In Theorem . we establish conditions under which allf-bounded solu-tions of (E) are asymptotically polynomial.
Lemma . If X is a complete partially ordered set and a map T:X→X is nondecreasing then there exists x∈X such that T(x) =x.
A simple proof of this result can be found in [] (or in []).
Theorem . Assume p∈N(),a∈A(m+p), y∈SQ,my=b, f is bounded on some uniform neighborhood U of the set y(N)and one of the following conditions is satisfied:
(a) f is nondecreasing onUand(–)man≥for largen, (b) f is nonincreasing onUand(–)ma
n≤for largen,
(c) f is continuous onU.
Then there exists a solution x of (E)such that
x∈y+ A(p).
Proof Forx∈SQ letx¯=f ◦x◦σ. Choosec> andM> such that
∞
n=
(yn–c,yn+c)⊂U, f(t)≤M fort∈U. ()
Let
ρ=Mrm|a|.
Choose numberskandksuch thatρn<cforn≥kandσ(n)≥kforn≥k. Let
S=x∈SQ :|x–y| ≤ρandxn=ynforn<k.
It is easy to see thatS, with natural order defined byx≤zifxn≤znfor everyn∈N, is a
complete partially ordered set. Ifx∈Sandn≥k, then|xσ(n)–yσ(n)|<c. Hence
xσ(n)∈U ()
for anyx∈Sandn≥k. Ifx∈S, then, by () and (), the sequencex¯is bounded and so
ax¯∈A(m+p). We define an operatorH:S→SQ by
H(x)(n) :=
⎧ ⎨ ⎩
yn forn<k,
yn+ (–)mrm
n(ax¯) forn≥k.
Ifn≥k, then, using Lemma .() and Lemma .(), we obtain
H(x)(n) –yn=rnm(ax¯)≤rmn|ax¯| ≤Mrmn|a|=ρn.
HenceH(S)⊂S.
Assume now that the condition (a) is satisfied. We can assume that (–)man≥ for n≥k. Letx,z∈Sandx≤z. Sincef is nondecreasing onUwe havef(xσ(n))≤f(zσ(n)) for n≥k. Hence
forn≥k. By Lemma .(), we haveH(x)(n)≤H(z)(n) forn≥k. Moreover,H(x)(n) =yn=
H(z)(n) forn<k. HenceH(x)≤H(z). Now, using the Knaster-Tarski fixed point theorem we obtainx∈S such thatH(x) =x. Analogously if condition (b) is satisfied thenH(x) =x
for certainx∈S.
Assume (c) and letdbe a metric onSdefined by
d(x,z) =x–z=sup
n∈N|xn–zn|.
Let BS denote the Banach space of all bounded sequencesx∈SQ with the normx=
supn∈N|xn|and let
T=x∈BS :|x| ≤ρandxn= forn<k.
It is easy to see thatT is a convex and closed subset of BS. Choose anε> . Then there existsq∈Nsuch thatρn<εforn≥q. Forn= , . . . ,qletGndenote a finiteε-net for the
interval [–ρn,ρn] and let
G={x∈T:xn∈Gnforn≤qandxn= forn>q}.
ThenGis a finiteε-net forT. HenceTis a complete and totally bounded metric space and so,T is compact. HenceT is a convex and compact subset of the Banach space BS and, by the Schauder fixed point theorem, any continuous mapT→Thas a fixed point. Let
F:T→Sbe a map given byF(x)(n) =xn+yn. ThenFis an isometry ofTontoS. Assume
U:S→Sis a continuous map and letW=F–◦U◦F. ThenW:T→T is continuous
and there exists a pointz∈Tsuch thatWz=z. Letx=Fz. Then
x=Fz=FWz=FF–UFz=Ux.
Hence any continuous mapU:S→Shas a fixed point. Letε> . Chooseq∈N(k) and
α> such that
M
∞
m=q
nm–|an|<ε and α q
n=k
nm–|an|<ε.
Let
W=
q
n=k
[yn–ρn,yn+ρn].
ThenW⊂U. By compactness ofW,f is uniformly continuous onW. Hence there exists
δ> such that ifs,t∈W and|s–t|<δ, then|f(s) –f(t)|<α. Assumex,z∈S,d(x,z) <δ. Letu=x¯–z¯. Then
d(Hx,Hz) =sup n≥k
rm
n(ax¯) –rmn(az¯)
=sup n≥k
rmn(au)≤sup n≥k
≤
q
n=k
nm–|anun|+ ∞
n=q
nm–|anun|
≤α
q
n=k
nm–|an|+ M
∞
n=q
nm–|an|< ε.
HenceHis continuous and there existsx∈S such thatH(x) =x. Then
xn=H(x)(n) =yn+ (–)mrmn(ax¯)
forn≥k. Hence
x–y+ (–)mrm(ax¯)∈Fin(k).
Since the sequencex¯is bounded, we haveax¯∈O(a). Hence
x∈y+rmO(a) +Fin(k)⊂y+ A(p).
Moreover, forn≥k, by Lemma .(), we have
mxn=myn+ (–)mmrm
n(ax¯) =bn+anf(xσ(n)).
Hencexis a solution of (E). The proof is complete.
Remark . The conclusionx∈y+ A(p) of Theorem . may be written in the form
y∈x+ A(p).
We can say thatyis an approximative solution of (E) with ‘degree of approximation’ A(p).
Remark . Using Lemma . we can replace the assumptiona∈A(m+p) of Theorem . by
lim infn |
an|
|an+|–
>m+p.
Analogously, the conclusion of this theorem may be replaced: there exist a solutionxof (E) and a sequencez∈SQ such thatx=y+zand
lim supn
|zn|
|zn+|–
≥p.
Similarly, using other tests, we can obtain many formulations of this theorem.
Theorem . Assume q∈N,p∈N(), ε> ,s∈(–∞, –p),a,b,y∈SQ,my=b,a is nonoscillatory,there exists a uniform neighborhood U of the set y(N(q))such that f|U≥ε or f|U≤–ε,and one of the following conditions is satisfied:
(b) there exists a solutionxof (E)such thatx∈y+ S(p).
Then a∈A(m+p).
Proof By Lemma . we have o(ns)⊂A(p)⊂S(p). Hence, there exists a solutionxof (E)
and a sequencez∈S(p) such thatx=y+z. Letx¯=f◦x◦σ. For largenwe have
anxn¯ +bn=anf(xσ(n)) +bn=mxn=myn+mzn=bn+mzn.
By Lemma .(),mz∈S(m+p). Hence, by Lemma .(), we haveax¯∈S(m+p). Since
zn= o(), we havexσ(n)∈Ufor largen. Hence the sequenceax¯is nonoscillatory and we
obtain|ax¯| ∈S(m+p), which meansax¯∈A(m+p). Moreover,
|an|=|anxn¯ ||¯xn|–≤ |anxn¯ |ε–
for largen. Hence, by Lemma .(),a∈A(m+p). The proof is complete.
Remark . Theorem . extends Theorem of [] and parts of Theorems and of [].
Lemma . Assume m∈N,p∈N(),x∈SQandmx∈A(m+p).Then
x∈Pol(m– ) + A(p).
Proof Using Lemma .(c) we obtain
rmA(m+p)⊂A(p). ()
Letz= (–)mrmmx. Thenmz=m(–)mrmmx=mx. Hence
x–z∈Kerm=Pol(m– ).
By () we havez∈A(p). Hence
x= (x–z) +z∈Pol(m– ) + A(p).
Theorem . Assume p∈N(),a,b∈A(m+p)and x is an f -bounded solution of (E).
Then
x∈Pol(m– ) + A(p).
Proof The sequencex¯=f ◦x◦σ is bounded. Henceax¯∈O(a) and we obtainmx∈
O(|a|+|b|). Thereforemx∈A(m+p) and, by Lemma ., we have
x∈Pol(m– ) + A(p).
Competing interests
The author declares that he has no competing interests.
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