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Aequat. Math. 87 (2014), 379–389 c Springer Basel 2013 0001-9054/14/030379-11 published online May 4, 2013 DOI 10.1007/s00010-013-0194-x Aequationes Mathematicae Behaviour at infinity of solutions of some linear functional equations in normed spaces Janusz Brzd ˛ ek and Stevo Stevi´ c Abstract. Let K ∈{R, C},I =(d, ): I I be unbounded continuous and increasing, X be a normed space over K, F := {f X I : limt→∞ f (t) exists in X}, ˆ a K, Aa) := {α K I : limt→∞ α(t)=ˆ a}, and X := {x X I : lim sup t→∞ x(t) < ∞}. We prove that the limit limt→∞ x(t) exists for every f ∈F ∈Aa) and every solution x ∈X of the functional equation x(φ(t)) = α(t)x(t)+ f (t) if and only if |ˆ a|= 1. Using this result we study the behaviour of bounded at infinity solutions of the functional equation x(φ [k] (t)) = k1 j=0 α j (t)x(φ [j] (t)) + f (t), under some conditions posed on functions α j (t),j =0, 1,...,k 1and f . Mathematics Subject Classification (2010). 39B22; 39B52. Keywords. Linear functional equation, existence of a limit, bounded solution, strictly increasing function. 1. Introduction The study of various types of linear equations or equations closely related to them, has re-attracted some attention recently, among others, because of numerous applications (see, for example, [16, 1224] and the related refer- ences therein). For some classical results on linear functional equations see, e.g. [911]. Motivated by our papers [13, 14, 17, 22] where we studied the relationship between the boundedness of a solution of a difference equation with discrete or continuous argument and the behaviour of the difference operator at infin- ity, in [25] we studied a related problem for some linear functional equations

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Page 1: Behaviour at infinity of solutions of some linear functional equations in normed spaces

Aequat. Math. 87 (2014), 379–389c© Springer Basel 20130001-9054/14/030379-11published online May 4, 2013DOI 10.1007/s00010-013-0194-x Aequationes Mathematicae

Behaviour at infinity of solutions of some linear functionalequations in normed spaces

Janusz Brzdek and Stevo Stevic

Abstract. Let K ∈ {R, C}, I = (d, ∞), φ : I → I be unbounded continuous and increasing,X be a normed space over K, F := {f ∈ XI : limt→∞ f(t) exists in X}, a ∈ K, A(a) :={α ∈ K

I : limt→∞ α(t) = a}, and X := {x ∈ XI : lim supt→∞ ‖x(t)‖ < ∞}. We prove thatthe limit limt→∞ x(t) exists for every f ∈ F , α ∈ A(a) and every solution x ∈ X of thefunctional equation

x(φ(t)) = α(t)x(t) + f(t)

if and only if |a| �= 1. Using this result we study the behaviour of bounded at infinitysolutions of the functional equation

x(φ[k](t)) =

k−1∑

j=0

αj(t)x(φ[j](t)) + f(t),

under some conditions posed on functions αj(t), j = 0, 1, . . . , k − 1, φ and f .

Mathematics Subject Classification (2010). 39B22; 39B52.

Keywords. Linear functional equation, existence of a limit, bounded solution,

strictly increasing function.

1. Introduction

The study of various types of linear equations or equations closely relatedto them, has re-attracted some attention recently, among others, because ofnumerous applications (see, for example, [1–6,12–24] and the related refer-ences therein). For some classical results on linear functional equations see,e.g. [9–11].

Motivated by our papers [13,14,17,22] where we studied the relationshipbetween the boundedness of a solution of a difference equation with discreteor continuous argument and the behaviour of the difference operator at infin-ity, in [25] we studied a related problem for some linear functional equations

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380 J. Brzdek and S. Stevic AEM

with real argument and complex-valued solutions. Here we continue this lineof research by studying a related problem for some linear functional equationsin normed spaces.

If φ : I → I, I ⊆ R, is a function, then we define inductively the operatorφ[n] by φ[n] = φ ◦ φ[n−1], n ∈ N, where φ[0](t) = t (the identity function on I).If φ is injective, then by φ−1 = φ[−1] we denote the inverse of function φ andφ[−n] := (φ[n])−1 for n ∈ N.

Let d ∈ R, I := (d,∞), and X be a normed space over the field K ∈ {R,C}with norm ‖ · ‖. Further, let

F :={f ∈ XI : lim

t→∞ f(t) exists in X},

A(a) :={α ∈ K

I : limt→∞α(t) = a

}, a ∈ K,

and

X :={x ∈ XI : lim sup

t→∞‖x(t)‖ < ∞

}.

2. Auxiliary result

In this section we give an auxiliary result which is used in the proof of themain result in the paper.

Lemma 1. Let a ∈ K, I = (d,∞), and φ : I → I be an unbounded continuousand strictly increasing function on the interval I. Then the limit limt→∞ x(t)exists for every f ∈ F , α ∈ A(a) and every solution x ∈ X of the functionalequation

x(φ(t)) = α(t)x(t) + f(t) (1)

if and only if |a| �= 1.

Proof. First assume that |a| �= 1. Let f ∈ F , α ∈ A(a), x ∈ X be a solutionof Eq. (1) and f0 := limt→∞ f(t). Set

αn(t) =n∏

i=0

α(φ[i](t)), t ∈ I, n ∈ N0.

It is easy to show by induction on n that

x(φ[n](t)) = αn−1(t)x(t) +n−1∑

j=0

αn−1(t)αj(t)

f(φ[j](t)), (2)

for every t ∈ I and n ∈ N, such that αn−1(t) �= 0.Since φ is continuous and increasing, the following three situations are pos-

sible.(a) There is a strictly increasing sequence {tn}n∈N of fixed points of φ with

limn→∞ tn = ∞.

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Vol. 87 (2014) Linear functional equations in normed spaces 381

(b) There is t0 ∈ I such that φ(t) > t for t > t0.(c) There is t0 ∈ I such that φ(t) < t for t > t0.

Clearly, if (a) holds, then

φ([tn, tn+1]) = [tn, tn+1], n ∈ N. (3)

Moreover, in view of the definition of X , there exists t ∈ I such that x isbounded on the interval [ t,∞).

First consider the situation where either (a) or (b) holds. We start with thecase |a| > 1. Note that (2) yields

x(t) +f0

a− 1=

1αn−1(t)

x(φ[n](t)) +(

1a− 1

−n−1∑

j=0

1αj(t)

)f0

−n−1∑

j=0

1αj(t)

(f(φ[j](t)) − f0), (4)

for every t ∈ I and n ∈ N, such that αn−1(t) �= 0.Next, for each ε > 0 there exist t(ε) > t and ξ > 0 such that

|α(t)| > 1 + ξ, t > t(ε), (5)‖f(t) − f0‖ ≤ ε, t > t(ε). (6)

Clearly, in case (a), we can take t(ε) = tn for some fixed n ∈ N, while in case(b) we assume that t(ε) > t0. Then

φ(t) > t(ε), t ∈ (t(ε),∞), (7)

because in the case of (a) we have (3).Observe that (5)–(7) mean that

|αj(t)| > (1 + ξ)j+1, t > t(ε), j ∈ N0, (8)

‖f(φ[j](t)) − f0‖ ≤ ε, t > t(ε), j ∈ N0. (9)

Consequently, by (4) and (7)–(9), for every ε > 0 we have∥∥∥∥x(t) +

f0α− 1

∥∥∥∥ ≤ 1|αn−1(t)| ‖x(φ

[n](t))‖

+

∣∣∣∣∣∣1

a− 1−

n−1∑

j=0

1αj(t)

∣∣∣∣∣∣‖f0‖ +

n−1∑

j=0

‖f(φ[j](t)) − f0‖|αj(t)|

≤ 1(1 + ξ)n

‖x(φ[n](t))‖ +

∣∣∣∣∣∣1

a− 1−

n−1∑

j=0

1αj(t)

∣∣∣∣∣∣‖f0‖

+n−1∑

j=1

ε

|αj(t)| , t > t(ε). (10)

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382 J. Brzdek and S. Stevic AEM

Letting n → ∞ in (10) and using the fact that 1 + ξ > 1, we get∥∥∥∥x(t) +

f0a− 1

∥∥∥∥ ≤∣∣∣∣∣∣

1a− 1

−∞∑

j=0

1αj(t)

∣∣∣∣∣∣‖f0‖ + ε

∞∑

j=0

1|αj(t)| , (11)

for every t > t(ε).Estimate (8) guarantees that the series

∑∞j=0

1αj(t)

is uniformly convergenton the interval (t(ε),∞). Hence by a known theorem we have that

limt→∞

∞∑

j=0

1αj(t)

=∞∑

j=1

1aj

=1

a− 1, lim

t→∞

∞∑

j=0

1|αj(t)| =

1|a| − 1

,

which along with inequality (11) and since ε is an arbitrary positive numberimplies that

limt→∞x(t) =

f01 − a

.

Next assume that 0 < |a| < 1 (the result in the case a = 0 trivially followsfrom the assumptions of the lemma) and (a) or (b) holds. Then (2) implies that

x(t) +f0

a− 1= αn−1(φ[−n](t))x(φ[−n](t)) +

(1

a− 1+

n−1∑

j=0

αn−1(φ[−n](t))αj(φ[−n](t))

)f0

+n−1∑

j=0

αn−1(φ[−n](t))αj(φ[−n](t))

(f(φ[j−n](t)) − f0), (12)

for n ∈ N and t > inf φ[n](I), such that αn−1(t) �= 0.Note that there exist t > t and ξ ∈ (0, 1) such that

0 < |α(t)| < 1 − ξ, t > t.

Further, for each ε > 0 there exist t(ε) > t with

‖f(t) − f0‖ ≤ ε, t > t(ε).

As in case (a), we can take t(ε) = tn for some n ∈ N and, in case (b), wemay assume that t(ε) > t0. This means that

|α(φ[−j](t))| < 1 − ξ, m ∈ N, t > φ[m](t(ε)), j = 1, . . . ,m,

‖f(φ[−j](t)) − f0‖ ≤ ε, m ∈ N, t > φ[m](t(ε)), j = 1, . . . ,m.

Next, (12) implies that∥∥∥∥x(t)+

f0

a − 1

∥∥∥∥ ≤ |αn−1(φ[−n](t))| ‖x(φ[−n](t))‖+

∣∣∣∣∣1

a − 1+

n−1∑

j=0

αn−1(φ[−n](t))

αj(φ[−n](t))

∣∣∣∣∣ ‖f0‖

+

n−1∑

j=0

|αn−1(φ[−n](t))|

|αj(φ[−n](t))| ‖f(φ[j−n](t)) − f0‖

for n ∈ N and t > φ[n](t(ε)).

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Vol. 87 (2014) Linear functional equations in normed spaces 383

Consequently, for every ε > 0, n ∈ N, and t > φ[n](t(ε)), we have that∥∥∥∥x(t) +

f0a− 1

∥∥∥∥ ≤ (1 − ξ)n ‖x(φ[−n](t))‖ +∣∣∣∣

1a− 1

+n−1∑

j=0

αn−1(φ[−n](t))αj(φ[−n](t))

∣∣∣∣ ‖f0‖

+n−1∑

j=0

(1 − ξ)jε. (13)

Take arbitrary k ∈ N and ε > 0. Then there is n0 ∈ N and wk > φ[n0](t(ε))such that

(1 − ξ)n0 ‖x(φ[−n0](t))‖ ≤ 1k, t > wk, (14)

∣∣∣∣∣∣1

a− 1+

n0−1∑

j=0

aj

∣∣∣∣∣∣‖f0‖ ≤ 1

k(15)

and∣∣∣∣∣∣

n0−1∑

j=0

αn0−1(φ[−n0](t))αj(φ[−n0](t))

−n0−1∑

j=0

aj

∣∣∣∣∣∣‖f0‖ ≤ 1

k, t > wk, (16)

because

limt→∞

n0−1∑

j=0

αn0−1(φ[−n0](t))αj(φ[−n0](t))

=n0−1∑

j=0

aj .

Clearly, inequalities (13)–(16) yield∥∥∥∥x(t) +

f0a− 1

∥∥∥∥ ≤ 3k

+∞∑

j=0

(1 − ξ)jε =3k

ξ, t > wk.

Since k ∈ N and ε > 0 have been chosen arbitrary, we have that

limt→∞x(t) =

f01 − a

.

Now assume that (c) holds. Then there is d0 ≥ d such that the functionψ := φ

∣∣J

is bijective, where J := (d0,∞). Clearly, ψ−1(t) > t for t ∈ J and(1) implies that

x(ψ−1(t)) =1

α(ψ−1(t))x(t) − 1

α(ψ−1(t))f(ψ−1(t)), t ∈ J.

By the previous part of the proof, it follows that there is the limitlimt→∞ x(t).

Now suppose that |a| = 1, and that (a) holds. Then, for every n ∈ N, φn :[tn, tn+1] → [tn, tn+1], given by

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384 J. Brzdek and S. Stevic AEM

φn(t) := φ(t), for t ∈ [tn, tn+1],

is a bijection.Fix x0 ∈ X with ‖x0‖ = 1. For each n ∈ N, choose sn ∈ (tn, tn+1), and

define x ∈ XI by

x(φn[j](sn)) = ajx0, j ∈ Z,

and

x(t) = 0, for t ∈ I \⋃

n∈N

{φn[j](sn) : j ∈ Z}.

It is easily seen that x ∈ X is a solution to equation (1) with f(t) ≡ 0 andα(t) ≡ a. Moreover,

limn→∞ ‖x(sn)‖ = 1, lim

n→∞x(tn) = 0,

so the limit limt→∞ x(t) does not exist.If (b) or (c) is valid, then again we take x0 ∈ X with ‖x0‖ = 1 and s0 > t0,

write

Z0 := N ∪ {0} ∪ {−j : j ∈ N and there is s ∈ I with φ[j](s) = s0},and define x ∈ XI by

x(φ[j](s0)) = ajx0, j ∈ Z0,

and

x(t) = 0, t ∈ I \ {φ[j](s0) : j ∈ Z0}.It is easy to see that x ∈ X is a solution to (1) with f(t) ≡ 0 and α(t) ≡ a.Next, in the case of (b),

limn→∞ ‖x(φ[n](s0))‖ = 1, lim

n→∞φ[n](s0) = ∞,

and, in the case of (c),

limn→∞ ‖x(φ[−n](s0))‖ = 1, lim

n→∞φ[−n](s0) = ∞.

Since

x(I \ {φ[i](s0) : i ∈ Z0}

)= {0},

this means that limt→∞ x(t) does not exist. �

Let k ∈ N, a0, . . . , ak−1 ∈ K and r1, . . . , rk denote the complex roots of theequation

zk =k−1∑

j=0

ajzj . (17)

Page 7: Behaviour at infinity of solutions of some linear functional equations in normed spaces

Vol. 87 (2014) Linear functional equations in normed spaces 385

Now, we investigate the behaviour at infinity of the solutions x ∈ XI ofthe linear functional equation

x(φ[k](t)) =k−1∑

j=0

αj(t)x(φ[j](t)) + f(t) (18)

with given functions αj ∈ A(aj) for j = 0, . . . , k− 1. We show that it dependson r1, . . . , rk.

The following well known result will be useful for this (see, e.g., [7]).

Lemma 2. Assume that the monic complex polynomial

f(z) = zn + c1zn−1 + · · · + cn ∈ C[z]

is factored as

f(z) = (z − a1) · · · (z − an),

where the roots aj ∈ C, j = 1, . . . , n, need not be distinct. Then for every ε > 0,there is δ > 0 such that every polynomial g(z) = zn +d1z

n−1 + · · ·+dn ∈ C[z],with |dj − cj | < δ for j = 1, . . . , n, can be written in the form g(z) = (z −b1) · · · (z − bn) with some bj ∈ C such that |bj − aj | < ε for j = 1, . . . , n.

3. Main result

Now we formulate and prove the main result in this paper.

Theorem 3.1. Let I = (d,∞), φ : I → I be an unbounded continuous andstrictly increasing function on the interval I, k ∈ N and a0, . . . , ak−1 ∈ K. Thenthe limit limt→∞ x(t) exists for every f ∈ F , every αj ∈ A(aj), j = 0, . . . , k−1,and every solution x ∈ X of Eq. (18) if and only if the roots rj , j = 1, . . . , k,of polynomial (17) satisfy the condition

|rj | �= 1, j = 1, . . . , k. (19)

Proof. First we assume that K = C, and that (19) holds. We give a proof byinduction on k.

Clearly, case k = 1 follows at once from Lemma 1. So fix m ∈ N and assumethat the statement is valid for k = m. We are to show that this is the case fork = m+ 1.

To this end observe that there exist functions rm+1, b0, . . . , bm−1 : I → C

such that

zm+1 −m∑

j=0

αj(t)zj = (z − rm+1(t))

⎝zm −m−1∑

j=0

bj(t)zj

⎠ t ∈ I, z ∈ C.

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386 J. Brzdek and S. Stevic AEM

In view of Lemma 2, without loss of generality we can assume that

rm+1 := limt→∞ rm+1(t).

Next, it is easily seen that, for every t ∈ I,

αj(t) =

⎧⎪⎨

⎪⎩

rm+1(t) + bm−1(t), if j = m;−rm+1(t)bj(t) + bj−1(t), if j = 1, . . . ,m− 1;−rm+1(t)b0(t), if j = 0.

(20)

Consequently the limit

bj := limt→∞ bj(t)

exists for j = 0, 1 . . . ,m− 1. Moreover, r1, . . . , rm are roots of the equation

zm =m−1∑

j=0

bjzj .

Fix a solution x ∈ XI of Eq. (18) (with k = m+ 1) and write

y(t) := x(φ(t)) − rm+1(t)x(t) t ∈ I.

Then, in view of (20), it is easy to check that

y(φ[m](t)) =m−1∑

j=0

bj(t)y(φ[j](t)) + f(t),

where b0, . . . , bm−1 ∈ KI are uniquely determined by the formula

αj(t) =

⎧⎪⎨

⎪⎩

rm+1(φ[m](t)) + bm−1(t), if j = m;−rm+1(φ[j](t))bj(t) + bj−1(t), if j = 1, . . . ,m− 1;−rm+1(t)b0(t), if j = 0.

In view of (20), it is easily seen that

limt→∞ bj(t) = lim

t→∞ bj(t) = bj , j = 0, 1 . . . ,m− 1.

So, by the inductive hypothesis, the limit

y = limt→∞ y(t)

exists.Further,

x(φ(t)) = rm+1(t)x(t) + y(t),

whence, by Lemma 1, the limit

x := limt→∞x(t) (21)

exists, too.

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Vol. 87 (2014) Linear functional equations in normed spaces 387

Next, suppose that (19) does not hold. Again, the case k = 1 follows atonce from Lemma 1. So assume that k > 1. Without loss of generality wecan assume that |rk| = 1. According to the proof of Lemma 1 there exists asolution x ∈ XI of the equation

x(φ(t)) = rkx(t)

for which the limit (21) does not exist. Next, the function y ∈ XI , y(t) ≡ 0, isa solution of the equation

y(φ[k−1](t)) =k−2∑

j=0

bjy(φ[j](t)), (22)

where b0, . . . , bk−2 are uniquely determined by the formula

aj =

⎧⎪⎨

⎪⎩

rk + bk−2, if j = k − 1;−rkbj + bj−1, if j = 1, . . . , k − 2;−rkb0, if j = 0.

Since

y(t) = 0 = x(φ(t)) − rkx(t), t ∈ I,

(22) implies that x is a solution of Eq. (18) with f(t) ≡ 0 and αj(t) ≡ aj forj = 0, . . . , k − 1.

To complete the proof it remains to consider the case when K = R. Then(see, e.g., [8, 1.9.6, p. 66]), X2 is a complex normed space with the linearstructure defined by the operations

(x, y) + (z, w) := (x+ z, y + w), x, y, z, w ∈ X,

(α+ iβ)(x, y) := (αx− βy, βx+ αy), x, y ∈ X,α, β ∈ R

and the Taylor norm ‖ · ‖T given by the formula

‖(x, y)‖T := sup0≤θ≤2π

‖(cos θ)x+ (sin θ)y‖, x, y ∈ X.

Clearly,

max {‖x‖, ‖y‖} ≤ ‖(x, y)‖T ≤ ‖x‖ + ‖y‖, x, y ∈ X. (23)

Define f : I → X2 by

f(t) = (f(t), f(t)), t ∈ I.

Then, in view of (23), the limit f0 = limt→∞ f(t) exists if and only if thereexists the limit

f0 = limt→∞ f(t).

Analogously, for x ∈ XI we write

x(t) = (x(t), x(t)), t ∈ I.

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388 J. Brzdek and S. Stevic AEM

Then x ∈ XI satisfies (18) if and only if x satisfies the equation

x(φ[k](t)) =k−1∑

j=0

αj(t)x(φ[j](t)) + f(t)

and, by (23), the limit (21) exists if and only if there exists the limit

x0 = limt→∞ x(t).

Further, x ∈ X if and only if

lim supt→∞

x(t) < ∞.

Consequently the statement for K = R follows from the first part of the proof.�

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[13] Stevic, S.: A generalization of the Copson’s theorem concerning sequences which satisfya linear inequality. Indian J. Math. 43(3), 277–282 (2001)

[14] Stevic, S.: Asymptotic behavior of solutions of a nonlinear difference equation with acontinuous argument. Ukr. Math. J. 56(8), 1300–1307 (2004)

[15] Stevic, S.: More on a rational recurrence relation. Appl. Math. E-Notes 4, 80–85 (2004)[16] Stevic, S.: A short proof of the Cushing-Henson conjecture. Discrete Dyn. Nat. Soc.

2006, 5 (Article ID 37264) (2006)

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Vol. 87 (2014) Linear functional equations in normed spaces 389

[17] Stevic, S.: Bounded solutions of a class of difference equations in Banach spaces pro-ducing controlled chaos. Chaos Solitons Fractals 35(2), 238–245 (2008)

[18] Stevic, S.: On a system of difference equations. Appl. Math. Comput. 218, 3372–3378 (2011)

[19] Stevic, S.: On a system of difference equations with period two coefficients. Appl. Math.Comput. 218, 4317–4324 (2011)

[20] Stevic, S.: On the difference equation xn = xn−2/(bn + cnxn−1xn−2). Appl. Math.Comput. 218, 4507–4513 (2011)

[21] Stevic, S.: On a third-order system of difference equations. Appl. Math. Com-put. 218, 7649–7654 (2012)

[22] Stevic, S.: On some linear difference equations and inequalities with continuous argu-ment. Appl. Math. Comput. 218, 9831–9838 (2012)

[23] Stevic, S.: On some solvable systems of difference equations. Appl. Math. Com-put. 218, 5010–5018 (2012)

[24] Stevic, S.: On the difference equation xn = xn−k/(b + cxn−1 · · · xn−k). Appl. Math.Comput. 218, 6291–6296 (2012)

[25] Stevic, S.: Behaviour of solutions of some linear functional equations at infinity. Appl.Math. Comput. 219, 6134–6141 (2013)

Janusz BrzdekDepartment of MathematicsPedagogical UniversityPodchorazych 230-084 Krakow, Polande-mail: [email protected]

Stevo StevicMathematical Institute of the Serbian Academy of SciencesKnez Mihailova 36/III11000 Beograd, Serbiae-mail: [email protected]

King Abdulaziz UniversityDepartment of MathematicsJeddah 21859, Saudi Arabia

Received: December 29, 2012

Revised: February 26, 2013