5.1. Proof of Theorem 3.1
To prove Theorem 3.1 we need the next result from [20].
Theorem A.
Let us consider the initial value problem ( 3.1 ), ( 3.2 ). Suppose that there are
such that
Assume also that. Then, there is a nonnegative matrix, independent of and , such that the solutionof ( 3.1 ), ( 3.2 ) satisfies
Now, we prove some lemmas.
Lemma 5.1.
The hypotheses of Theorem 3.1 imply that the hypotheses of Theorem A are satisfied, and hence the solution of (3.1), (3.2) is bounded.
Proof.
Let be such that . This can be satisfied because . Then, there is an for which
and hence for an , we have
Thus,
therefore,
But the matrices are nonnegative in the above inequality, thus
and this shows (5.1). Since condition holds, we get
therefore,
thus (5.2) is satisfied.
In the next lemma we give an equivalent form of .
Lemma 5.2.
Let be a sequence of real by matrices which satisfies . Then, there exists a real by matrix such that
if and only if
is finite. In both cases
If satisfies too, and (5.11) holds, then .
Proof.
First we show that
is finite if and only if
is finite, and in both cases
These come from , since
Suppose is a real by matrix. Then, by for every
and hence
Now, suppose that (5.11) holds. Then by (5.19), either
or
Both of the previous cases implies that
which shows that
is finite and
As we have seen, this is equivalent with (5.12). If (5.12) is true or equivalently
is finite, then by (5.19)
satisfies (5.11). follows from . The proof is now complete.
Lemma 5.3.
The hypotheses of Theorem 3.1 imply that
is the only vector satisfying the equation
Proof.
Since the matrix is invertible, which shows the uniqueness part of the lemma. On the other hand, by Lemma 5.1 we have that is a bounded sequence, and hence is finite. Thus, is well defined and satisfies (5.28). The proof is complete.
Lemma 5.4.
The vector defined by (5.27) satisfies the relation
for any , where the sequence , , satisfies
Proof.
Let be arbitrarily fixed and
But under the hypotheses of Theorem 3.1 we find
Now, by Lemma 5.2
and hence (5.30) holds. On the other hand, it can be easily seen that by the above definition of the relation (5.29) also holds. The proof is complete.
Now, we prove Theorem 3.1.
Proof.
Let be arbitrarily fixed. Then, (3.1) can be written in the form
Subtracting (5.29) from the above equation, we get
On the other hand, by Lemma 5.1, is bounded and hence
is finite. Let be arbitrarily fixed and . Then there is an such that
Thus, (5.35) yields
From this it follows:
Thus,
and hence Lemma 5.4 implies that
Since is a nonnegative matrix with , we have that . Thus,
and hence the proof of Theorem 3.1 is complete.
5.2. Proof of Theorem 3.3
Theorem 3.3 will be proved after some preparatory lemmas.
In the next lemma, we show that (3.7) can be transformed into an equation of the form (3.1) by using the transformation
Lemma 5.5.
Under the conditions of Theorem 3.3, the sequence defined by (5.43) satisfies (3.1), where the sequences and are defined by
Proof.
Let be defined by (5.43). Then,
Thus,
On the other hand, from (3.7) it follows:
Thus,
and hence
But
Therefore,
where is defined in (5.45). By interchanging the order of the summation we get
This and (5.52) yield
By using the definition , we have
and hence
But by using the definition of in (5.44) the proof of the lemma is complete.
In the next lemma, we collect some properties of the function defined in (3.13).
Lemma 5.6.
Let be a sequence such that for some and
Then, the function defined in (3.13) has the following properties.

(a)
The series of functions
is convergent on and it is divergent on .

(b)

(c)

(d)
is strictly decreasing.

(e)
If , then the equation has a unique solution.
Proof.
(a) The root test can be applied. (b) The series of functions (5.58) is uniformly convergent on for every , and this, together with , implies the result. (c) If is finite, then the series of functions (5.58) is uniformly convergent on , hence is continuous on . Suppose now that and . Let be fixed and let such that
Since
there is a such that
whenever , and this shows . Finally, we consider the case . Then, follows from the condition . (d) The series of functions (5.58) can be differentiated termbyterm within , and therefore , . Together with (c) this gives the claim. (e) We have only to apply (d), (c), and (b). The proof is complete.
We are now in a position to prove Theorem 3.3.
Proof.
(a) Let for all . Then, it is easy to see that there is a such that (3.10) holds if and only if (3.15) is true, and in this case (3.11) is also satisfied. Suppose for some . Let be finite. By the root test, the series
are convergent for all . Moreover, it can be easily verified that the series
are absolutely convergent, whenever is finite. Define the functions by
where if is finite and , otherwise. The series of functions in (5.64) are uniformly convergent on for every , hence and are continuous. Further,
Let if , and let if . It now follows from the previous inequalities and Lemma 5.6(d) that
and hence is strictly increasing on . It is immediate that . If (3.10) is hold for some , then the convergence of the series implies . Suppose . It is simple to see that there is a satisfying (3.10) if and only if either (in case ) or (in case ). Moreover, the existence of a satisfying (3.11) is equivalent to either (in case ) or (in case ). Now, the result follows from the properties of the functions . The proof of (a) is complete. (b) In virtue of Lemma 5.5 the proof of the theorem will be complete if we show that the sequences and satisfy the conditions (H_{2})–(H_{5}) in Section 3. Since the series is convergent,
and hence . Thus holds. Now, let and consider . In fact
Thus,
Thus,
is finite and . In a similar way, one can easily prove that
is also finite. It is also clear that
is finite. Thus, by Theorem 3.1 we get that the limit
is finite and satisfies the required relation (3.22). The proof is now complete.
Lemma 5.7.
Let be a sequence such that for some . Then, there is at most one satisfying (3.10) and (3.11).
Proof.
Suppose on the contrary that there exist two different numbers from , denoted by and , such that (3.10) and (3.11) hold. Then,
It follows from (5.74), the mean value inequality and (5.75) that
and this is a contradiction.