Existence of Ψ− Bounded Solutions for Linear Difference Equations on Z

In this paper 1 , we give a necessary and sufficient condition for the existence of Ψ− bounded solutions for the nonhomogeneous linear difference equation x(n + 1) = A(n)x(n) + f(n) on Z. In addition, we give a result in connection with the asymptotic behavior of the Ψ− bounded solutions of this equation.


Introduction
The problem of boundedness of the solutions for the system of ordinary differential equations x = A(t)x + f(t) was studied by Coppel in [2].In [3], [4], [5], the author proposes a novel concept, Ψ− boundedness of solutions (Ψ being a matrix function), which is interesting and useful in some practical cases and presents the existence condition for such solutions.Also, in [1], the author associates this problem with the concept of Ψ− dichotomy on R of the system x = A(t)x.
Naturally, one wonders whether there are any similar concepts and results on the solutions of difference equations, which can be seen as the discrete version of differential equations.
In [7], the authors extend the concept of Ψ− boundedness to the solutions of difference equation (via Ψ− bounded sequence) and establish a necessary and sufficient condition for existence of Ψ− bounded solutions for the nonhomogeneous linear difference equation (1) in case f is a Ψ− summable sequence on N.
In [6], the author proved a necessary and sufficient condition for the existence of Ψ− bounded solutions of (1) in case f is a Ψ− bounded sequence on N.
Similarly, we can consider solutions of (1) which are bounded not only N but on the Z.
In this case, the conditions for the existence of at least one Ψ−bounded solution are rather more complicated, as we will see below.
In this paper, we give a necessary and sufficient condition so that the nonhomogeneous linear difference equation ( 1) have at least one Ψ−bounded solution on Z for every Ψ−summable function f on Z Here, Ψ is a matrix function.The introduction of the matrix function Ψ permits to obtain a mixed asymptotic behavior of the components of the solutions.

Preliminaries
Let R d be the Euclidean d-space.For

d and let the matrix function
Consider the nonautonomous difference linear equation where the d × d real matrix A(n) is invertible at n ∈ Z.Let Y be the fundamental matrix of (2) with Y(0 . the solution of (2) with the initial condition y(0) = y 0 is Let the vector space R d represented as a direct sum of three subspaces X − , X 0 , X + such that a solution y of (2) is Ψ− bounded on Z if and only if y(0) ∈ X 0 and Ψ− bounded on Z + = {0,1,2,• • • } if and only if y(0) ∈ X − ⊕ X 0 .Also let P − , P 0 , P + denote the corresponding projection of R d onto X − , X 0 , X + respectively.EJQTDE, 2008 No. 26, p. 2

Main result
The main result of this paper is the following.
Theorem 1.The equation ( 1) has at least one Ψ− bounded solution on Z for every Ψ− summable function f on Z if and only if there is a positive constant K such that Proof.First, we prove the "only if" part.We define the sets: ∈ B} Obviously, B Ψ , B and D are vector spaces over R and the functionals x Step 1.It is a simple exercise that (B Ψ , • B Ψ ) and (B, • B ) are Banach spaces.
Step 2. (D, • D ) is a Banach space.Let (x p ) p∈N be a fundamental sequence in D.Then, (x p ) p∈N is a fundamental sequence in B Ψ .Therefore, there exists a Ψ− bounded function x : On the other hand, the sequence By the above relations, we have that Thus, (D, • D ) is a Banach space.
Step 3.There exists a positive constant K such that, for every f ∈ B and for corresponding solution x ∈ D of (1), we have x Clearly, T is linear and bounded, with T ≤ 1.Let Tx = 0 be.Then, x ∈ D and x(n + 1) = A(n)x(n).This shows that x is a Ψ− bounded solution of (2) with x(0) ∈ X − ⊕ X + .From the Definition of X 0 , we have x(0) ∈ X 0 .Thus, x(0) ∈ X 0 ∩ (X − ⊕ X + ) = {0}.It follows that x = 0.This means that the operator T is one-to-one.Now, for f ∈ B, let x be a Ψ− bounded solution of the equation ( 1).Let z be the solution of the Cauchy problem z(n + 1) = A(n)z(n) + f(n), z(0) = (P − + P + )x(0).Then, the function u = x − z is a solution of the equation (2) with u(0) = x(0) − z(0) = P 0 x(0) ∈ X 0 .It follows that the function u is Ψ− bounded on Z. Thus, the function z is Ψ− bounded on Z.It follows that z ∈ D and Tz = f.Consequently, T is onto.EJQTDE, 2008 No. 26, p. 4 From a fundamental result of Banach "If T is a bounded one-to-one linear operator from a Banach space onto another, then the inverse operator T −1 is also bounded", we have that Thus, we have (4), where K = T −1 − 1.
Step 4. The end of the proof.For a fixed but arbitrary k ∈ Z, ξ ∈ R d , we consider the function f : Obviously, f ∈ B and f B = ξ .The corresponding solution x ∈ D of ( 1) is Indeed, we prove this in more cases: Finally, we have From the Definitions of X − , X 0 and X + , it follows that the function x is Ψ− bounded on Z − and N. Thus, x is the solution of (1) in D. Now, we have, Step 5.The function u is well-defined.For p, q ∈ Z, q < 0 < p, we have EJQTDE, 2008 No. 26, p. 6 and then, and then, and then, k) is an absolutely convergent series for n > 0.
Thus, the function u is well defined for n ≥ 0.
Similarly, the function u is well defined for n < 0.
Step 6.The function u is a solution of the equation ( 1).Indeed, using the expresion of the function u, we obtain: These relations show that the function u is a solution of the equation ( 1).
Step 7. The function u is Ψ− bounded on Z. Indeed, for n > 0 we have For n < 0, we have Thus, the solution u of the equation ( 1) is Ψ− bounded on Z.
The proof is now complete.
Corollary 1.If the homogeneous equation ( 2) has no nontrivial Ψ− bounded solution on Z, then, the equation ( 1) has a unique Ψ− bounded solution on Z for every Ψ− summable function f on Z if and only if there exists a positive constant K such that, for k, n ∈ Z, EJQTDE, 2008 No. 26, p. 9 (5) Proof.Indeed, in this case, P 0 = 0. Now, the Corollary follows from the above Theorem.
Finally, we give a result in which we will see that the asymptotic behavior of Ψ− bounded solutions of ( 1) is determined completely by the asymptotic behavior of the fundamental matrix Y of (2).
Theorem 2. Suppose that: 1 • .the fundamental matrix Y of (2) satisfies the conditions (3) for some K > 0 and the conditions i). lim Then, every Ψ− bounded solution x of (1) satisfies the condition Proof.Let x be a Ψ− bounded solution of ( 1).Let u be the Ψ− bounded solution of (1) from the proof of Theorem 1 ("if" part).
Let the function y It is easy to see that y is a Ψ− bounded solution of (2) and then y(0) ∈ X 0 .
On the other hand, y(0) = (I − P For n > 0, we have The proof is now complete.