Consider now the proof of Theorem 1.4.
Proof.
-
(a)
If
such that
, then (1.9) follows directly from (1.1).
Now, fix a point
with
. To estimate the second term on the right of (1.1), we can apply Jensen's inequality (see [13]), by (
):
and therefore
This inequality, together with Definition 1.3(a), implies that the expression
satisfies the inequality
By Lemma 2.3 (b) and Definition 1.3 (b),
, and hence (1.9) can be deduced from
The properties defining
are trivial if
with
. So assume
such that
.
First, we show that the function
is
-almost measurable on
. It is an easy consequence of (
) and Lemma 2.4 that the functions
are
-almost measurable on
. This means that there exists a measurable subset
of
such that
and (3.6), (3.7) are measurable on
. Further, since
is
-almost measurable on
, it can be supposed that
is measurable on
. Thus we need to show that the function
is measurable on
. To prove this let
The measurability of (3.6) on
implies that
. Since
it is enough to show that (3.8) is measurable on
. It follows from the definitions of
and
that the function
is measurable. Hence, by Lemma 2.3 (d), (3.8) is measurable on
.
It is now clear that
is
-almost measurable on
.
Next, we prove that
is
-integrable over
.
To prove this, it is enough to show that the function
is bounded on
. Since
we need to verify that (3.12) is bounded on
. By (
), we can find a
such that
It therefore follows from
that
for all
, and hence
Thus
and therefore another application of (3.14) gives
and from this the claim follows. Consequently,
is
-integrable over
, as required.
The result is completely proved.
Now we are in a position to prove Theorem 1.5.
Proof.
We begin with the proof of (c) and (d).
-
(c)
To prove that
we use induction on
. Clearly
belongs to
. Let
such that the assertion holds. Then Lemma 2.6 yields that
. We show now that the sequence
is increasing. By our hypotheses on
, it follows that
, and we again complete the proof by induction. Suppose
such that
. Then, by Lemma 2.1 and the induction hypothesis
Since
Theorem 1.4(a) implies that
Because of (3.22) and the fact that
is increasing, there exists a function
such that
converges pointwise on
to
. Then
is a consequence of
together with (3.22), Lemma 2.1(a) and, Theorem 1.4(b). If
with
, then according to the continuity of
on
,
converges to
. Let
with
. As we have seen
is increasing, and therefore
for all
. Consequently
Now (1.11) and the monotone convergence theorem give that
is a solution of (1.2).
-
(d)
We can see exactly as in the proof of (c) that
,
and the sequence
is decreasing. It follows from an easy induction argument that
is nonnegative for all
. Linking up with the foregoing, there exists a function
such that
converges pointwise on
to
.
can be shown as in the proof of (c).
The continuity of
on
implies that
converges to
for every
.
Assume now that
. If
such that
for every large enough
, then
If
such that
for every
, then
which leads to
. In both cases
converges to
for every
.
According to (1.11) and the monotone convergence theorem
is a solution of (1.2).
(
) For convergence of the successive approximations
to a solution
of (1.2) it suffices, in view of (c), to show that
is a solution of (1.1), which is evident. It remains to prove that if
is a solution of (1.6), then
for every
. To this end, it is enough to show that
This is true for
, since the functions
and
are nonnegative. Let
for which the result holds. Then, because of the nonnegativity of
and the fact that
is increasing,
and the proof of the induction step is complete.
(
) Let
Choose
. The set of the upper bounds from
for the solutions of (1.1) on
is denoted by
. By Theorem 1.4,
is not empty. Let
Then
is a minorized subset of
(the elements of
are nonnegative). Since this space is order complete (see [12]),
exists. Let
, and let
be a solution of (1.1) such that
.
,
gives that
It follows that
because
is increasing. Since
the proof of Lemma 2.5 shows that
, and hence by (
), the function
defined by
belongs to
, and therefore
. It now comes from (3.33) and the definition of
that
Since
is increasing,
and thus
We can see that
If we set
then (3.38) shows that
is a solution of (1.2). (3.33) gives that
for every solutions
of (1.1) for which
.
Repeat the preceding construction for every
. We thus obtain a set of functions
, each a solution of (1.2) on its domain. Moreover, if
is a solution of (1.1), then for every
we have
Introduce the next functions: for every
with
let the function
be defined on
by
Obviously, these functions are also solutions of (1.2) on their domains.
Now let
,
such that
and
. Using (3.41), it is easy to verify that
and therefore Lemma 2.10 is applicable to the solutions
. This gives a unique solution
of (1.2) for which
It remains to prove that
is maximal. Let
be a solution of (1.1), and let
. If
, then
while if
, then by (3.41)
so that Lemma 2.1 implies that
We first show that every solution
of (1.2) with
has an extension
that is a solution of (1.2) such that
is a proper subset of
. This follows from (c) as soon as it is realized that every solution
of (1.2) with
has an extension
that is a solution of (1.1) such that
is a proper subset of
. Really, in this case the successive approximations determined by
converge to a solution
of (1.2), which is obviously an extension of
.
That realization can be reached in finitely many steps.
Let
be a solution of (1.2) with
, and let
.
-
(i)
Suppose
.
If
, then
is an appropriate solution.
If
, then
is a solution of (1.1) that agrees with
on
.
-
(ii)
Suppose
and
. Let
If
, then the function
is a solution of (1.1) (
,
, thus
) that agrees with
on
.
If
, then we introduce the functions
,
,
which all lie in
, and they are all solutions of (1.1). By Theorem 1.4 (b),
is a majorized subset of
. Since this space is order complete (see [12]),
exists. Choose
. Then
-almost everywhere on
for every
, and therefore Lemma 2.1 yields that
Since
the proof of Lemma 2.5 shows that
, and hence by (A
), the function
belongs to
too. It now follows from
and (3.54) that
We set
By (3.56)
Fix
. From Lemma 2.9 (with
and
there being
and
resp.) we get that
-almost everywhere on
, and hence
Consequently,
,
. If
with
, then
. We can see that
By what we have already proved that
is a solution of (1.1) that agrees with
on
.
-
(iii)
Suppose
and
. By an argument entirely similar to that for the case (ii), we can get a solution
of (1.1) that agrees with
on
.
After these preparations, the proof can be concluded quickly. Let
be a solution of (1.2), and let
be the set of all solutions of (1.2) which agree with
on
. Since
,
is not empty. Partially order
by declaring
to mean that the restriction of
to the domain of
agrees with
. By Hausdorff's maximality theorem, there exists a maximal totally ordered subcollection
of
. Let
be the union of the domains of all members of
, and define
by
, where
occurs in
. It is easy to check that
is well defined, and it is a solution of (1.2). If
were a proper subset of
, then the first part of the proof would give a further extension of
, and this would contradict the maximality of
.
-
(e)
Let
be a solution of (1.2), and let
.
If
, then
, so that
is bounded on
.
Suppose
. Since
and
are bounded on
and
is finite, Theorem 1.4 (a) implies that it is enough to prove the boundedness of the function
on
. Moreover, we have only to observe that the function
is bounded on
. To prove this, let
If
and
,
, then
and
for every
, and therefore by Lemma 2.3(d), the function (3.63) is bounded on
. If
, then the definition of the function
gives that
. The claim about
is therewith confirmed.
The proof of the theorem is now complete.