Necessary Conditions for Optimal Control of Diffusions with Hard Terminal State Restrictions ()
1. Introduction
The purpose of the present paper is to prove necessary conditions for the optimal control of certain types of control problems involving diffusions where hard end constraints on solutions occur. The books [1] and [2] contain introductions to the topic of optimal control of diffusions. Various types of maximum principles have been proved for problems of control of diffusions in case of no or soft terminal state restrictions, see e.g. [3] [4] [5] and [6] . Maximum principles for problem with hard terminal restrictions are proved for certain types of continuous time controlled diffusion problems in [7] . In that paper only the drift term contained controls. In this paper controls are allowed to enter also the diffusion term. Singular controls are not discussed, below we merely consider problems where the controls appearing may be said to be absolutely continuous with respect to Lebesgue measure. The restriction to such controls makes it harder to operate with hard terminal state restrictions. In fact, the case discussed in detail in this paper is the case where hard terminal restrictions are only placed on states governed by differential equations not containing a Brownian motion. The necessary condititions are stated in terms of “weak” variations, not strong ones (needleshaped ones). Brownian motions will only appear in differential equations of states unconstrained at the terminal time. The constrained states are, however, influenced by other states directly influenced by Brownian motions, so, below, necessary conditions are stated and proved for such problems. Because the states are stochastic, the state space is infinite dimensional, so to obtain necessary conditions, one must impose a condition amounting to demanding sufficient variability of the first order variations in the problem.
2. The Control Problem and the Statement of the Necessary Condition
Let
and let
be a given point in the Euclidean space
, let
be a projection from
onto
,
, such that
and let U be a closed convex subset of a Euclidean space
. The symbol
is used for the Euclidean norm in any Euclidean space
, including
, and, applied to matrices, it is the linear operator norm. Furnish the interval
with the Lebesgue measure. Let
be a given filtered probability space, (i.e. for
, the
’s are sub-s-algebras of the given s-algebra
of subsets of
,
if
, P is a probability measure on
), and assume that
is complete, that
contains all the P-null sets in
and that
is right continuous. Let
,
, be the set of Lebesgue
-measurable functions
for which
and let
be the subset of essentially bounded functions in
. Related to
, let
be a vector the
components of which (denoted
) are independent one-dimensional Brownian motions all adapted to
, such that
is independent of
for all
,
, and is normally distributed with mean 0 and covariances
,
the identity matrix. In applications where the
’s are entities that can be observed, it is natural to take
as the natural filtration generated by the
’s. There are given functions
and
, from
into
. Let
be the set of functions
taking values in U, such that
, for each t, when restricted to
, is Borel
-measurable (i.e. progressively measurable).
Let
be the
-norm both on
and on
.
The following conditions are called the Basic Assumptions.
A1 The functions
and
have derivatives
and
with respect to
,
that are continuous in
.
A2 As a function of t, the functions f and
and their derivatives have one-sided limits with respect to t.
Write
for the
-matrix whose columns are
, let
(
) be the matrix with entries
(
), and write
. Also, write
for the indicator function of the set C.
The following assumptions are called the Global Assumptions.
B1
and
are uniformly continuous in
, uniformly in t.
B2 For some constant M, for some given
in
, for all
,
(1)
B3 A constant
exists such that for all
,
(2)
B4
(3)
Let
be the set of progressively measurable functions in
,
. From now on we require that all control functions
belong to
. The strong (unique) solution―continuous in t―of the equation
(4)
is denoted
.
Let
(a fixed,
) such that
, let
denote scalar product, and consider the problem
(5)
subject to
(6)
Below,
will be an optimal control in this problem, assumed to exist. Write
.
We have collected a few definitions that are going to be used in the sequel.
For
, let
(7)
(8)
(9)
(10)
(11)
where
. Let
(12)
Finally, let
(13)
(14)
In the subsequent necessary conditions, the following local linear controllability condition (15) is needed. There exist numbers
,
and
, and a progressively measurable function
such that for all
for which
,
(15)
A more “concrete” condition implying (15) is presented in Remark 3.
Theorem 1. Assume that
is optimal in problem (5),(6), that Assumptions A and B hold (the Basic and Global Assumptions) and that (15) is satisfied. Then the following necessary conditions hold: For some linear functional
on
, bounded on
, and some number
, for all
,
(16)
where
is the solution of
![]()
with
. Moreover1,
. ,
Remark 1. If (15) holds for
, then
. Moreover, when
, (21) below holds and
is (also) left continuous, then both
and
is a continuous linear functional on
((15) then holds for
). ,
Remark 2. Let
, and let
be the resolvent of the equation
(17)
, so
,
the identity map in
, and
. Let
,
. For
,
,
, is continuous in
-norm and hence can be represented
by an
-function
,
progressively measurable and right continuous in t, a.s. and in
(even continuous in this manner if
is continuous). Note that if
on
then (16) implies
.
Let
be the natural filtration generated by the
’s,
augmented by the P-null sets in
. The function
, now taken to be a row vector, then satisfies the following condition: On
, there exist
-valued, progressively measurable functions
,
,
, all row
vectors,
continuous in t, such that
,
for all
, and such that
(18)
and such that, for all
,
a.s.
In this case, we have that for all
, for a.e. t, a.s.
(19)
When
,
is left continuous, and (21) below holds, the following additional properties hold:
,
, the
-limit (and a.s. limit)
exists,
, and finally
(20)
Remark 3 The condition (15), with
, is implied by the following conditions:
and for some
, for all
,
(21)
The
-norm (or the norm
) used in the formulation of the condition (15) can have quite strong consequences. Assume for example that
, and, for some
, that
for all
. Then for some
,
for all
, if
and
(
and
balls in
-spaces,
), but if U is strictly smaller than
it is neither possible to obtain this inclusion nor (15). When
the situation may be different. Consider in particular the case where the original terminal constraint is, say
,
. Using auxiliary controls
and auxiliary states z and y, the system can be rewritten as a system where the end constraints are
,
, by adding new state equations
,
,
,
, and by changing the first
state equations to read
,
,
. Define
. We assume now
, so the control set in the rewritten system is now
. The control
is optimal in the rewritten system. Assume that
for some
for all
. Then, for some
, for all
,
, for
,
a ball in
. Then for all
,
for all
so for all
,
,
a ball in
. Hence, for all
,
, where
(Necessary conditions for
systems with inequality constraints are stated in Remark 5 below, the local linear controllability condition just obtained suffices for the controllability condition in Remark 5 to apply to the original system).
3. Proof of Theorem 1
The proof consists of three lemmas and six proof steps A.-F., and relies on an “abstract” maximum principle, Corollary I in Appendix.
Without loss of generality, from now on let
and let
. Note that
. For any
, let
be the solution of
(22)
By general existence results and (1) and (2),
, and
, see (4), do exist, both unique, strong solutions, continuous in t (both column vectors).
A) Growth properties.
When
, by (1),
and, similarly,
. By (2), (4), and
(57) in Appendix, Lemma A, (a consequence of Gronwall’s inequality), with
,
,
,
,
,
,
,
, we have, for some constant
independent of
(only dependent on
), that
(23)
Let
. Using (2), and (57) in Appendix, Lemma A, (with
,
,
,
,
,
,
,
, for some constant D independent of
and u (only dependent on
), we get
(24)
(the last inequality by (2)).
Let
. As explained below, for some
, we have
(25)
The inequality (25) follows from (57) in Appendix, Lemma A, together with (2), for
,
,
,
(
), and
![]()
![]()
Similarly2, for
,
(26)
In (25) and (26),
is independent of
(it only depends on
).
We need to prove that
(27)
This follows from (24) and (2), because, in a shorthand notation, for all t,
![]()
We also need:
(28)
This follows from (26) and (2), because, in a shorthand notation, for
,
, for all t,
![]()
Note finally that, when
,
, then
(29)
see Remark L in Appendix.
B) Local controllability of the linear perturbations.
Note that, by (2), in a shorthand notation,
(30)
when
,
. Then, from (15) and (29), when
,
(31)
where “co” can be omitted due to the concavity of U. To see this, apply Remark W in Appendix.
Next, let the number
satisfy
, (for
, see (25)), and let
. Let
, so
. When
, then
, hence
. Then, for all
, for all t
(32)
To see this, note that by (25), for all t,
![]()
![]()
Then, using again Remark W in Appendix and (31), (29), (32), for all
,
(33)
C) Relations between exact and linear perturbations.
Let
be given elements in
. Let
be arbitrarily given. Let us first prove that for
small enough, for any
, for
, for all t,
(34)
Write
,
. Define, in a shorthand notation,
![]()
![]()
![]()
![]()
For any
, there exists a
, such that
(35)
by Lemmas B and C in Appendix.
Consider now
![]()
By (56) in Appendix, Lemma A, for
, for some
only dependent on
, for all t
(36)
To see this, in Lemma A let
,
,
,
,
,
,
. To obtain (34), put
.
Next, given
and assume that
Let us prove that for any
, for
small enough, for any
, for
, we have that
(37)
Let
![]()
Now,
![]()
Because
and
, then Lemma J in Appendix gives that
when
. Next, let
,
. Using
and
, we get
, which
when
, by (36). Hence, (37) follows.
D) Continuity of
at
.
Let
. Define
. Let us first prove that
(38)
Now, in a shorthand notation,
![]()
![]()
![]()
Using the notation in Lemma A in Appendix, we write
where
,
,
![]()
![]()
(
,
,
).
Then, by (57),
![]()
where
![]()
for
![]()
and
![]()
for
![]()
Consider e.g the term
. Now,
,
, and
when
, Hence, by Remark T in Appendix,
when
. Similarly, the terms containing
converge to zero when
. So
uniformly in t when
.
Next, assume that
,
. We want to prove that then
satisfies
(39)
where, in a shorthand notation,
![]()
Now,
![]()
Let us show that the four terms in square brackets have
-norms
when
. Now,
when
. Hence, by Lemma J in Appendix, the second and fourth square bracket
in
when
(recall that
and
). Moreover the first and third square bracket have
-norms smaller than
and
, respectively. So the
-norms of both these terms
when
, recall that
when
, see (38). So (39) has been proved.
E) Dense subsets of
.
Let
(for this equality see Appendix L). Note that for each
, for some
,
. Let
. For each
, let
be the projection of
on the line containing 0 and
. Let
if
has the opposite direction of
or if
, let
if
is longer than
and has the same direction as
, and if
is smaller than
and has the same direction as
, let
. Evidently, for some measurable
,
, and
and
(
also depends on
, the notation hides this fact). Evidently,
and
is progressively measurable. Now, for any
, let
. We have that
is
-dense in
because if
, for any
, for some
,
(
, and hence,
, belong
to the linear space
). Then
, and so
for
(
is convex).
Note finally that if
, then
,
for
, so
, in fact
(40)
F) Final proof steps.
Define
(41)
Using Jensens inequality, for an arbitrary function
we get
. It follows that
(42)
For
, define
and
. Define
![]()
(43)
Furthermore, let
be the subset of
consisting of all element
such that
and such that
, (limit in
-norm,
). It is easily seen that elements of the type
,
,
, precisely make up the
set
. To see this, let
be such a function, and, using Jensen’s inequality three times (and for any
, that
), note that for any interval
,
(44)
so, in particular,
. This yields also, for
, that
(45)
so
, see (43). Moreover, by (44), similarly,
(46)
so
is a
-limit of
. Hence,
. Note also that
![]()
![]()
for all
(holding “the more” for
).
Finally, if
, then
, for
(47)
where
,
and where
,
progressively measurable. (So the linear map
satisfies
.)
Let
be the linear map from
(see (43)) into
defined by
and note that
,
. In (45) we have just proved that
has norm
for the norms
and
(or for
, as we shall express it).
Now, let
, let cl(2) be closure in
in
-norm, let
, let
, and let
be a
-ball in
.
Now, for
, we have
, where
(48)
Hence, by (33), using
and
, for
, (for
and
, used in a moment, see definitions subsequent to (31)), we get
![]()
where
(49)
(note that
is continuous in
, as shown above). Let
,
. Then
![]()
Using Remark W,
, and the fact that the first integrand has a
-norm
by (32) (hence the first integral has a
-norm
, we get from (49) that
(50)
Observe that, given
,
(
), by
for
and (37), for any given
, for
small enough, for any
,
(51)
And, by (39) and
, for
(
), then
(52)
To obtain the conclusion in Theorem 1, we will now apply Corollary I in Appendix, and for this end, we make the following identifications:
,
(for
see below),
,
,
, the norm
on
equal to
,
,
,
,
,
,
,
,
,
,
. By (34) and (51) it follows that the property (61) is satisfied for
for
(
by (40)), and (62) is trivially satisfied by concavity of U, both (61) and (62) in the manner required in Corollary I. By (50), for
,
for
, so for
,
,
), where
,
defined subsequent to (31). By
and (27) and (24),
and
are continuous. By (26), (28), and
,
is continuous in
, for any
, and by (38) and (52), for
,
is continuous at
. The required boundedness of
is satisfied because of (25). Evidently,
is complete, see Appendix, Remark M. Hence all conditions in Corollary I are satisfied. Thus, for some
, some
-continuous linear functional
on
,
, for all
,
.
By Hahn-Banach’s theorem,
has an
-continuous linear extension to all
, also denoted
.
Proof of Remark 1. Let
. Note that (50) and
gives
. If
, the necessary condition
(16) implies
and hence
, a contradiction. So
.
From (16), in a shorthand notation, we get, for
, and even for
, that
(53)
Fix any
, such that
By (21) and
, there exists a
, such that
,
. So
, where
,
, so
(
) and
.
By (25), for some constant K,
, so the absolute value of the scalar products on the right hand side of (53) are smaller than
(
,
) and
, respectively, the last number being small than
. Hence, by (53), for
,
. Replacing
by
, by the same argument, if
, we get
. In fact, we have
for all
,
.
Thus, on this subspace of
,
is
-continuous. Hence, by Hahn-Banach’s theorem (see [8] ) and a representation theorem, for some
in
,
, for all
(in fact on
).
Note that for
,
, which
means that
is
-continuous on
and has a unique
-continuous extension to the
-closure of
, which equals
(and includes
) in case of left continuity of
.
Proof of Remark 2.
Let
,
. Let
(
) and let
and
be the adjoints of
and of
.
It is easily proved (using Lemma A in Appendix and
for
) that for
,
for some constant D, see Remark P in Appendix. Hence, for any given
, by the
-continuity of
on
,
and hence of
, there exists a
-function
on
such that for any
-function
, we have
. The function
is progressively measurable and right continuous in
, both a.s. and in
, see arguments in Remark Q in Appendix.
Assume now
, (21) and left continuity of
, which implies
-continuity of
on all
(by extension). In this case the argument for
-continuity of
for
in Remark Q in Appendix extends to
. Let
be a representation of
and let
. Consider now
![]()
where
. In this case, we can evidently put
and we have
![]()
In case of
, (21), and left continuity of
, by
-continuity of
, the two definitions of
coincide, because for both definitions we have
for all
, t arbitrary in
.
Let now
be the natural filtration generated by the
’s , augmented by the P-null sets in
, and let
.
Then, consider the pair of equations
(54)
(55)
By Theorem 2.2, p. 349 in [6] , there is a unique progressively measurable collection
,
continuous in
, satisfying these equations (or more precisely, a.s. the integrated version of these equations),
,
, with
equal to
-a.s. for each t (see Remark O in Appendix). The uniqueness says that if two pairs
, (
) and
satisfy the pair of equations, then
for all
and
for a.e.
. We can let
when
(
) and obtain such functions
defined on each
. For
, by the fact that
a.s. and uniqueness, we have that
for a.e
a.s. Then, evidently, there exists a unique pair
on
satisfying (54), with
a.s. equal to
for any
,
,
for all
,
continuous.
Let us show (19). For any
,
if
on
, so in this case
(a consequence of (16)).
Define
and
,
, and
similarly. It turns out that
. From this and
it follows that for all
,
for a.e.
, a.s., see Remark S. In fact, this holds for a.e.
because
was arbitrary. An informal proof, using Ito’s formula, shows the last equality:
![]()
So
![]()
Remark 4. (Exact attainability). In Theorem 1, drop the assumption that
is optimal (and the optimization problem), so (
) is simply a pair satisfying
(4) and
a.s. Then, for each
,
, (cl2 and int = interior both corresponding to ![]()
and the space
), for all
, for some number
and some control
,
a.s. ,
Proof of Remark 4. Let
,
. Corollary G in Appendix will be applied. Let
,
,
, the norm
on Y equal to
,
,
,
,
,
,
,
,
,
,
. Recall that
,
when
. By (51) it follows that the property (61) is satisfied for
for
, and (62) is trivially satisfied. By (27) H is continuous, by (28),
is continuous in A for any
, and by (52), for
,
is continuous at
.
Assume that
and let
. For any
, it was shown earlier that there exists a
such that
,
(
,
,
), so
and
. Thus, if
, then
. Hence all conditions in Corollary G are satisfied and the conclusion in Remark 4 follows. ,
Remark 5 If it is required that
instead of
, where C is a
-closed convex set in
, and if (15) holds for
replaced by
,
(for
see (47)), then
(16) again holds, and with
. In particular, this holds if
a.s,
,
a.s.,
,
given numbers. In fact, because
automatically, the results here cover the case where the terminal constraints are
a.s,
,
a.s.,
. ,
Proof of Remark 5. A proof can be obtained (for
) by replacing
by
and
by
, with
replaced by
, and
replaced by
, keeping
and
as before and letting
,
. (Then
,
,
,
, gives
, i.e.
). The details are omitted.
Remark 6 (The case
for some or all j,
). Let
be closure in
as well as in
. For
, let
. When
, and
is the
natural filtration, augmented by null sets as before, the necessary condition of Theorem 1 can be obtained for
defined and continuous on
(with
,
), if, for some
, for some
, for some
,
a) for some
,
,
when
,
, where
![]()
If
, then
. The condition (A), with
, is implied by
b): For some
, for all
, for any
,
![]()
Letting
,
, and
, it is here sufficient to operate with the ![]()
-norm instead of
on
, with
. Note that
, so (A) holds for
replaced by
. The set arising by replacing
by
in
is
denoted
. Define
. The condition (A), so modified, implies that for some
,
,
close to
a ball in
, so
(
close to
). See Appendix, Remark V for the next to last inclusion and the implication (B) Þ (A).
Now, in the manner required in Corollary I, (34) Þ (61) (with
,
,
), (24) and (26) implies continuity of
and
, and (38) implies continuity of
at
.
4. Conclusion
In this paper, necessary conditions for optimal control of diffusions with hard end restrictions on the states have been obtained. The main case considered is the one where the states restricted at the terminal time correspond to differential equations not containing Brownian motions. Brownian motions only occur in differential equations for states unconstrained at the terminal time. A removal of this restriction is discussed in Remark 6.
Acknowledgements
The author is grateful to a referee for useful comments.
Appendix
The appendix contains, among other things, a number of wellknown results, included for the convenience of the reader. The first one concerns a result on comparison of solutions. Still
.
Lemma A. Assume that
(an
-vector) and
, (a
matrix, with columns
,
) are Lipschitz continuous in
with rank
and progressively measurable in
. Assume that six progressively measurable functions
and
exist (
,![]()
-matrices), satisfying
![]()
and
![]()
where
. We assume that the eight integrands belong to
-spaces. Then, for some constant
,
(56)
(applied to matrices the index j indicates columns), and for some constant
,
(57)
and
only dependent on
.
Proof of (57). We shall use a shorthand notation. Using the algebraic inequality
, then for some positive constant k,
![]()
![]()
The Burkholder-Davis-Gundy inequality yields, for a “universal” constant
, that
.
Similar inequalities hold for the other terms involving
. Hence (using also Jensen’s inequality) we get
![]()
![]()
![]()
![]()
Note that, by Gronwall’s inequality, for any functions
,
, if
, and
is increasing, then
. Hence, for
,
![]()
Using the fact that the square root of a sum of positive numbers is £ the sum of square roots of the numbers, we get
![]()
Note that
, and that
. Using this for the term containing
, and a similar argument for the term containing
, then (57) follows.
Proof of (56).
Using Ito’s isometry,
![]()
![]()
![]()
Then, again using
and Jensen’s inequality, for some positive constant k,
![]()
![]()
![]()
![]()
Thus, for
,
![]()
so (56) follows.
Simple results om Gâteaux derivatives appear in the next two lemmas.
Lemma B. Let
be continuously differentiable in z for each t, and have one-sided limits with respect to t, and assume
for all
. For each
, for each “direction”
, and for each t,
(
), has, in the norm
, a bounded linear Gâteaux derivative at
in direction
, which equals
. The derivative is uniform in t.
Proof. By Ito’s isometry,
![]()
When
, the term in curly brackets converges to zero for each
and is smaller than the
-function
. Hence, Lebesgue’s dominated convergence theorem gives that
when
. ,
Lemma C. Let
be continuously differentiable in z for each t, and have one-sided limits with respect to t, and assume
for all
. For each
, for each “direction”
, and for each t,
(
), has, in the norm
, a bounded linear Gâteaux derivative at
in direction
, which in equals
. The derivative is uniform in t.
Proof. Define
![]()
![]()
Jensen’s inequality yields the inequality
. The remaining arguments are as in the preceding proof, they yield
when
. ,
Below, on product spaces, maximum norms (= maximum of norms) and maximum metrics are used. In the sequel, the following entities are used:
Y is a normed space, A is a complete pseudometric space with pseudo-metric r, and a* is a given element in A. The function H(a) from A into Y is continuous. (58)
Theorem D. (Attainability) Let the entities in (58) be given. Let positive numbers
, and an element
in Y be given. Assume that the following properties hold for all
: For all
with
, for all
, a pair
exists, such that
and
. (59)
Then, for all
, there exists a pair
, such that
, where
.
Corollary E. Assume that
. Then, in (59), evidently
can be replaced by the stronger inequality
. ,
(On the other hand, when
, then
Þ
for
.)
Central ideas in the proof of Theorem D stem from the proof of the multifunction inverse function theorem Theorem 4, p. 431, in [9] .
Proof of Theorem D. The property (59) also holds for
in the set
. To see this, let
and let
. Then
for some
, some
such that
. Now, for all
, there exists a pair
,
, such that the inequalities in (59) hold. From these inequalities, for
, using
, it follows that
and
. Hence, (59) holds for
.
Below, write
. The following lemma is needed in the proof:
Lemma F. Let
. Assume that the pair
minimizes
![]()
in
. Then
.
Proof of Lemma F. By contradiction, assume
,
. The vector
satisfies
, so
belongs to
. Hence, by the extended property (59), there exist an
and a
,
, such that
(60)
Moreover,
, (use the first inequality for
), which implies
Define
. Then, using (60),
, and the definition of
, we get
![]()
Using
and
yields
![]()
a contradiction of the optimality of
.
Continued proof of the theorem. Let
,
let
be as in the conclusion of the theorem, and let
. Note that
. Let the distance between
and
be
in the complete space
. By Aubin and Ekeland (1984, Theorem 1, p. 255), (Ekeland’s variational principle), there exists a
such that
![]()
for all
and
![]()
which gives
,
. By Lemma F,
, so
, for
. ,
Below,
is a sort of Gâteaux derivative at a of
.
Corollary G. Let
, Y a normed space, A a complete pseudometric space with metric
,
a given element in A, and let
be continuous. For each
, let
be a set dense in A. Assume the existence of a function
, from
into Y and a positive constant
such that, for each
, for all
, all
, all
, there exists a pair
,
,
such that
(61)
Assume also that for all
,
(62)
Assume that
is continuous in A for any
, and that
is continuous at
for any
. Assume finally that b is an interior point in
, and that, for some
, some
,
for all
. Then, for some
and some
,
.
Proof. Write
, and let
for some
. Then, for some
,
. Define
. Evidently, b is an interior point in
if
. Then b is an interior point in
even if
is only an approximate equality, in fact there exist positive numbers
and
such that
for all
. Because
, by (62) there exists a
, such that
, and we can even assume
, by density of
and continuity of
. By the continuity of
in the corollary, for
small enough,
for
. We assume
,
. Evidently,
. Hence,
for all
. Thus, for
,
, because
and
is convex. Hence,
,
. It follows that if
,
, then, for any
, by (62), for some
,
,
(63)
We can even assume
by continuity of
. By (61), for
, for some
, and some arbitrarily small
,
(64)
Now, by (63),
. Then, by (64),
(65)
(
,
,
). In Theorem D, replace
by
, a by
,
by
and A by
, and let
,
,
,
,
,
. Then the conditions in Theorem D are satisfied when, in (59),
is replaced by
,
as just constructed,
, (
, for
,
, and
as
). Thus, we get that, for all
small enough,
for some
, or
.
Remark H. If
, then
can be taken to be arbitrary small (b can be replaced by
for any
). Hence, in this case,
holds for some arbitrarily small
.
Corollary I Let
be a normed space, let
, let
be a complete pseudometric space with metric
and let
be dense in
. Assume that
is a given element in
, let
be continuous. Let
,
. Assume that (61) and (62) are satisfies for
replaced by
and also that
is continuous at
for any
and that
is continuous for any
, with
for all
. Assume, for some given
, that
. Assume also, for some
, some
, that
for all
. Assume, finally, that
. Then, for some nonzero continuous linear functional
on
,
a number
, we have
for all
.
Proof. Define
, and
. Define
. Evidently, for
,
and
, so
. Assume by contradiction that
belongs to
for some
. Define
, and for
,
, let
,
and
, and let
,
. Then, for
,
(62) and (61) are evidently satisfied, (the latter for
when
). Obviously,
for each
where
(Þ
Þ
Þ
). Hence, by the preceding corollary, for some
,
for some
. Hence,
,
, or
, contradicting optimality. Thus the set
is disjoint from
, so the convex set L can be separated from the convex set int
by a nonzero continuous linear functional
such that
,
, which implies
.
Lemma J. Let
. Let
be continuous in z and measurable in t. Assume that
exists for all
and that
for all
. Let
be continuous in
at
, uniformly in
. Let
,
. When
, then
![]()
uniformly in
such that
. ,
Proof. Let an error function
be a nonnegative function on
such that
when
. By continuity of
, uniformly in
there exists an increasing error function
such that
for all
. Let
. Evidently,
, and for a.e. t,
a.s.3
Suppose by contradiction that some
exists, such that, for each
, there exist
such that, for ![]()
(66)
and
,
. Now for a.e. t,
, so
. Let
.
The set
has full Lebesgue measure
. Then there must exist
such that
(67)
Now,
![]()
There exists an increasing concave error function
, see Lemma K below. Evidently, by Jensen’s inequality,
.
So a subsequence
of
converges a.s. to zero, hence also
converges a.s. to zero (
is a.s. finite by the assumption on
in the Lemma). Moreover,
,
being a
-function by the assumption on
in the Lemma. By dominated convergence
when
, and a contradiction of (67) is obtained. ,
Lemma K. Let
be an increasing error function. There exists an increasing concave error function
such that
.
Proof. Let
and let
. The set
is compact. There is a line through
with minimal positive slope touching
“to the left of”
at some perhaps nonunique point
,
. Here, “touching” means that the line contains a point of
, but no point in
lies strictly above the line. We choose
as large as possible (perhaps
). All points
,
, ly below or on the segment
between
and
. If
, repeat the construction by replacing
by
. The segment
so obtained between
and some
,
,
chosen as large as possible, has a slope strictly larger than the slope of
, otherwise
would not be as large as possible. All points
,
, ly below or on the segment
. Continue this prosess. The points
obtained equals
for some increasing subsequence
of
Perhaps
for some
, or
for
in which case
. The union of the segments
forms the graph of an increasing concave error function
. We have that the points
on the segments have the property that they “dominate” the points
in the sense that
, and for
we have
. Then for all
,
, for
,
. (If
, the segment
contains
as well as
, so
is linear on
and
for
,
.)
Remark L (Proof of (29)) Let
and let
. Let
and let
. Then
a.s. when
as
is a.s. finite, because
. Then
a.s. and by Lebesgue’s dominated convergence theorem also in
. So
is
-dense in
(and then also
-dense). Evidently,
, moreover, as shown,
, so
.
If
, then
by (57) in Appendix. If
,
, then
![]()
so
. Because
, then
®
in
when (in
)
,
, hence
when
.
Now,
when
, by
, and then
in
when
in
,
,
, so
when
. Moreover, by (57) in Appendix,
when
, so
when
.
Hence
as well as
belong to
when
(
).
Remark M (Completeness of
)
Evidenly,
is complete, so a
-Cauchy-sequence
in
is
-convergent to some
. Then for a subsequence
, for a.e. t,
. From this we get that, for any
, if
when
are large, then for a.e. t even
, so
, which means that
is
-complete.
Remark N (
in
-norm and a.s. when
).
Let
, and let
. By weak compactness, a convex combination
,
,
, converges in
and also almost uniformly to some
-measurable limit4
, the latter convergence by considering subsequences if necessary. By
-convergence,
evidently satisfies
.
Let
and
be arbitrary, let M be a set on which
converges uniformly,
, and let j be the smallest j such that
and
. Let
. Then, for
,
,
, using Jensens inequality twice,
![]()
For
, this holds even for
, yielding
-convergence of
when
, using
yields almost uniform convergence.
Remark O. In this remark let “tr” mean tranpose. Note that
(
) and that, for
,
,
![]()
Taking now
to be a row vector, then
satisfies
. Letting
be the entities on pp. 348-350 in [6] , for
,
,
,
, we have that
, and then
by (2.20) in that book, so by (2.9) in that book, for some
,
satisfies (54),(55).
Remark P (Proof of continuity of
on
,
).
Let
,
,
and let k be the smallest k such that
. Note that if
, then
, for
, and that
. Hence
. On the other hand
. Hence, on
, the norms
and
are equivalent. (Thus, the spaces
,
are subspaces of
, in fact of
, because
-closure of
.) For
, define
(68)
Then an application of Appendix, Lemma A gives that
for some constant D independent of
. Then
![]()
Because
for
, then
(69)
Now,
, so
. For some constant
,
for
, so
,
.
Remark Q (Proof of
-continuity of
,
).
Let
, and fix
. By Remark P,
is
-continuous in
, so there exists an
-function
, such that
. Define
,
. Note that
, so
is continuous because
and
is continuous (compare Theorem 6.14 p. 47 in [6] , a similar theorem holds for random coeffficients). Because
by Lemma A in Appendix,
is
-continuous. Hence
is a.s. and
-continuous.
Evidently, for any
,
![]()
Then, for any
,
![]()
From this it follows that
for all
.
Let
be given. Note that if
and
is (also) left continuous, then
in
and a.s. , see Corollary A.9, Appendix C in [1] 5 and
in
and a.s., see Remark N in Appendix, or, for both these results, Theorem 6.23 in [10] . Using this for
and
and continuity of
yield that
is a.s. and
-continuous, as we shall see: Fix any
. When
is close to s,
. This is obvious for the
-norm, and also for a.s.-convergence, once we have shown that
almost uniformly. Note that by Egoroff’s theorem,
almost uniformly. Let
. For any
, for some set
, with
and some
,
when
. By almost uniform convergence again, we can find a set
, with
and a
such that
when
. Defining
and
, we have
and
and hence
when
. Thus,
![]()
when
. Then,
![]()
So we have
when
and
. Now,
, and
. Hence,
almost uniformly.
Remark R (Proof of (21)Þ (15), with
, when
)
Let
. For any
, there exists a
, with
,
(and with
), such that
and hence
![]()
Then for
, we have
![]()
note that
, so (15) holds for
,
.
Remark S. (Proof of (19)). Let u be any given element in U,
, and let
be an arbitrarily given Lebesgue point of
. Then, for any
,
,
(70)
when
. We have
, when
, for
. From this and (70) it follows that
![]()
Because
was arbitrary, (19) follows for
, i.e. for a.e. t.
Remark T. Let
be a finite measure space. If
(
Euclidean spaces) is continuous in x and
-measurable in
,
,
, and
in
-measure, (
and
-measurable), then
in
. This result, which is a special case of Krasnoselskii's theorem (see p. 20 in [9] ), can be proved as follows. By contradiction, assume for some
and for some subsequence
that
for all j. A subsequence
converges
-a.e. to
. Then, by continuity,
-a.e. and even in
, by Lebesgue's dominated convergence theorem. A contradiction has been obtained.
Remark U. (On
,
)
On
,
. To show this, for the component
of
, note that
![]()
![]()
so
![]()
Remark V. (Proof of (B) Þ (A) with
in Remark 6)
Let
. By Lemma A in Appendix, for some
,
when
. Let
![]()
let
satisfy
, let
and let
,
,
. By (B) in Remark 5, for any
, any
, for some progressively measurable
,
on
, for each
,
,
![]()
where
, so
. Hence, by slight abuse of notation, we have that
![]()
By the choice of c, for
,
![]()
By Remark W and the previous inclusion, then
, where
![]()
Now, by definition of
, for
small,
and
are
when
, so
and
, so
, where
![]()
Hence, (B) Þ (A) has been proved.
Let
be the natural filtration, augmented as before. Postulating now (A), by Ito’s representation theorem,
, so by the linear interior mapping theorem, for some
-ball
in
,
. But then
.
Remark W. Let I be a set and let X be a normed space. Let the maps v and w from I into X have the following properties. Given
,
,
, assume that
and that
for all i. Then
.
For a proof see p. 327 in [11] .
Remark X. (A nonzero continuous linear functional on
vanishing on all
,
)
Let
,
the natural filtration corresponding to some given
. For simplicity, assume
. Choose a
such that
. Then
, so
belongs to the
-boundary of the
-ball
in
. Then for some nonzero continuous linear functional
on
,
. Let
and let k be any given integer such that
. If
and
, then
for
and
for
, so
for all j (Þ
). Then the inequality involving
yields
, i.e.
vanishes on
,
. To show in detail that such a
exists, let
,
. Then
, so for
,
, and for
,
. Letting
, we get
, so
and
.
Note that
belongs to the
-boundary of
and the f's belong to
, so the arguments above actually yield a nonzero
-continuous linear functional on
, vanishing on all
,
.
Remark Y. (
is nonempty).
On
define
, where
are some arbitrarily chosen sets such that
,
. Then for all t,
, so
and
. Now
for all i, so
for all i.
Now
. Let us show that. Assume by contradiction that
for some
. Using the projection of
onto
as in E. in the proof of Theorem 1, we have for some progressively measurable
, that
,
,
.
For all t,
. Then
, or
. Then
where
, otherwise (using
)
. But then, for all i,
, contradicting
.
NOTES
1
*means restriction to
.
2Lemma A applies for
,
,
,
,
,
,
,
.
3Note that
, for all
, so
which implies that
for a.e. t and hence, for a.e. t, that
a.s.
4A.s. convergence of the component
to
,
, means
for any given
(a arbitrary), hence
.
5If
, this corollary ensures only
-convergence of
to
, but also of
in
(to
, which means that
is equicontionuous,
(the countable additivity is uniform). Then
in
, because
, so
is equicontionuous.