OJOp  Vol.7 No.1 , March 2018
Necessary Conditions for Optimal Control of Diffusions with Hard Terminal State Restrictions
Author(s) Atle Seierstad
ABSTRACT
Necessary conditions are proved for certain problems of optimal control of diffusions where hard end constraints occur. The main results apply to several dimensional problems, where some of the state equations involve Brownian motions, but not the equations corresponding to states being hard restricted at the terminal time. The necessary conditions are stated in terms of weak variations. Two versions of necessary conditions are given, one version involving solutions of variational equations, the other one involving first order adjoint equations.

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 ÞÞ). 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.

Cite this paper
Seierstad, A. (2018) Necessary Conditions for Optimal Control of Diffusions with Hard Terminal State Restrictions. Open Journal of Optimization, 7, 1-40. doi: 10.4236/ojop.2018.71001.
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