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

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 = ∞.

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)

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)).

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

+ε

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

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)

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.

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.

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.

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.�

References

[1] Brzdek, J., Popa, D., Xu, B.: Hyers-Ulam stability for linear equations of higherorder. Acta Math. Hungarica. 120, 1–8 (2008)

[2] Brzdek, J., Popa, D., Xu, B.: Remarks on stability of linear recurrence of higherorder. Appl. Math. Lett. 231, 459–1463 (2010)

[3] Brzdek, J., Popa, D., Xu, B.: On nonstability of the linear recurrence of order one. J.Math. Anal. Appl. 367, 146–153 (2010)

[4] Brzdek, J., Popa, D., Xu, B.: Note on nonstability of the linear functional equation ofhigher order. Comput. Math. Appl. 62, 2648–2657 (2011)

[5] Brzdek, J., Popa, D., Xu, B.: Remarks on stability of the linear functional equation insingle variable. In: Pardalos, P., Srivastava, H.M., Georgiev, P. (eds.) Nonlinear analy-sis: stability, approximation, and inequalities, vol. 68, pp. 91–119 (Springer Opt. Appl.).Springer, New York (2012)

[6] Brzdek, J., Popa, D., Xu, B.: A note on stability of the linear functional equation ofhigher order and fixed points of an operator. Fixed Point Theory 13(2), 347–356 (2012)

[7] Cucker, F., Corbalan, A.G.: An alternate proof of the continuity of the roots of a poly-nomial. Amer. Math. Monthly 96, 342–345 (1989)

[8] Kadison, R.V., Ringrose, J.R.: Fundamentals of the theory of operator algebras, v. I:elementary theory. Am. Math. Soc., Providence, RI (reprint of the 1983 original) (1997)

[9] Kuczma, M.: Functional equations in a single variable. Panstwowe Wydawnictwo Nau-kowe, Warszawa (1968)

[10] Kuczma, M.: An introduction to the theory of functional equations and inequalities.Cauchy’s equation and Jensen’s inequality. Birkhauser Verlag, Basel (2009)

[11] Kuczma, M., Choczewski, B., Ger, R.: Iterative functional equations. Cambridge Uni-versity Press, Cambridge (1990)

[12] Papaschinopoulos, G., Stefanidou, G.: Asymptotic behavior of the solutions of a classof rational difference equations. Inter. J. Differ. Equ. 5(2), 233–249 (2010)

[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)

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: jbrzdek@up.krakow.pl

Stevo StevicMathematical Institute of the Serbian Academy of SciencesKnez Mihailova 36/III11000 Beograd, Serbiae-mail: sstevic@ptt.rs

King Abdulaziz UniversityDepartment of MathematicsJeddah 21859, Saudi Arabia

Received: December 29, 2012

Revised: February 26, 2013