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 solution
of ( 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 term-by-term 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 (H2)–(H5) 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.