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Well posedness and porosity in the calculus of variations without convexity assumptions
Nonlinear Analysis 53 (2003) 1 – 22
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Well posedness and porosity in the calculus of variations without convexity assumptions Alexander J. Zaslavski Department of Mathematics, Technion-Israel Institute of Technology, Haifa 32000, Israel Received 18 December 2001; accepted 5 March 2002
Abstract The Tonelli existence theorem in the calculus of variations and its subsequent modi-cations were established for integrands f which satisfy convexity and growth conditions. In Zaslavski (Nonlinear Analysis 43 (200l) 339), a generic well-posedness result (with respect to variations of the integrand of the integral functional) without the convexity condition was established for a class of optimal control problems satisfying the Cesari growth condition. In Zaslavski (Communications in Applied Analysis, to appear) we extended this generic well-posedness result to a class of variational problems in which the values at the end points are also subject to variations. More precisely, we established a generic well-posedness result for a class of variational problems (without convexity assumptions) over functions with values in a Banach space E which is identi-ed with the corresponding complete metric space of pairs (f; (1 ; 2 )) (where f is an integrand satisfying the Cesari growth condition and 1 ; 2 ∈ E are the values at the end points) denoted by A. We showed that for a generic (f; (1 ; 2 )) ∈ A the corresponding variational problem is well posed. In this paper we study the set of all pairs (f; (1 ; 2 )) ∈ A for which the corresponding variational problem is well posed. We show that the complement of this set is not only of the -rst category but also a -porous set. ? 2003 Elsevier Science Ltd. All rights reserved. MSC: 49J99; 90C31 Keywords: Complete metric space; Generic property; Integrand; Optimal solution; Porous set; Variational problem
1. Introduction The Tonelli existence theorem in the calculus of variations [22] and its subsequent generalizations and extensions (e.g., [6,7,14,18,21]) were established for integrands f E-mail address: [email protected] (A.J. Zaslavski). 0362-546X/03/$ - see front matter ? 2003 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 0 2 ) 0 0 1 8 0 - 3
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which satisfy convexity and growth conditions. Moreover, certain convexity assumptions are also necessary for properties of lower semicontinuity of integral functionals which are crucial in most of the existence proofs, although there are some interesting theorems without convexity (see [6,4,16,17]). In [27], it was shown that the convexity condition is not needed generically, and not only for the existence but also for the uniqueness of a solution and even for well posedness of the problem (with respect to some natural topology in the space of integrands). Instead of considering the existence of a solution for a single integrand f, we investigated it for a space of integrands and showed that a unique solution exists for most of the integrands in the space. This approach has already been successfully applied in the theory of dynamical systems [8,19], approximation theory [9], as well as in the calculus of variations (see, e.g., [2,13,25]). Interesting generic existence results were obtained for particular cases of variational problems [1,5,15]. In [27], this approach allowed us to establish the generic existence of solutions for a large’class of optimal control problems without convexity assumptions. More precisely, in [27] we considered a class of optimal control problems (with the same system of diHerential equations, the same functional constraints and the same boundary conditions) which is identi-ed with the corresponding complete metric space of cost functions (integrands), say F. We did not impose any convexity assumptions. These integrands are only assumed to satisfy the Cesari growth condition. The main result in [27] establishes the existence of an everywhere dense G -set F ⊂ F such that for each integrand in F the corresponding optimal control problem has a unique solution. The next steps in this area of research were done in [12,26,29]. In [12], we introduced a general variational principle having its prototype in the variational principle of Deville et al. [10]. A generic existence result in the calculus of variations without convexity assumptions was then obtained as a realization of this variational principle. It was also shown in [12] that some other generic well-posedness results in optimization theory known in the literature and their modi-cations are obtained as a realization of this variational principle. Note that the generic existence result in [12] was established for variational problems but not for optimal control problems and that the topologies in the spaces of integrands in [27,12] are diHerent. In [26], we suggested a modi-cation of the variational principle in [12] and applied it to classes of optimal control problems with various topologies in the corresponding spaces of integrands. As a realization of this principle we established a generic existence result for a class of optimal control problems in which constraint maps are also subject to variations as well as the cost functions [26]. In [29], this variational principle was applied to obtain a generic existence result for a class of variational problems in which the values at the end points are also subject to variations as well as the integrands. More precisely, in [29] we established a generic existence result for a class of variational problems (without convexity assumptions) over functions with values in a Banach space E which is identi-ed with the corresponding complete metric space of pairs (f; (1 ; 2 )) (where f is an integrand satisfying the Cesari growth condition and 1 ; 2 ∈ E are the values at the end points) denoted by A and endowed with some natural topology. We
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
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showed that for a generic (f; (1 ; 2 )) ∈ A the corresponding variational problem has a unique solution. In this paper we extend the main result in [29]. We consider the set of all pairs (f; (1 ; 2 )) ∈ A for which the corresponding variational problem is well posed and show that the complement of this set is not only of the -rst category but also a -porous set. Note that this result is also an extension of the main result in [28] where we studied the spaces of integrands introduced in [26]. For each of these spaces we considered the set of all integrands for which the corresponding optimal control problem is well posed and showed that its complement is -porous. In the present paper, we obtain an extension of the main result in [28] to a class of variational problems in which the values at the end points are also subject to variations as well as the. integrands. We obtain our main result as a realization of a general variational principle which is established in Section 2. Before we continue we recall the concept of porosity [3,9,11,20,23]. Let (Y; d) be a complete metric space. For each y ∈ Y and each r ¿ 0 set Bd (y; r) = {z ∈ Y : d(z; y) 6 r}. A subset E ⊂ Y is called porous in (Y; d) if there exist ∈ (0; 1] and r0 ¿ 0 such that for each x ∈ Y and each r ∈ (0; r0 ] there exists y ∈ Y for which Bd (y; r) ⊂ Bd (x; r) \ E:
(1.1)
A subset of the space Y is called -porous in (Y; d) if it is a countable union of porous sets in (Y; d). It is known that in the above de-nition of porosity, the point x can be assumed to belong to E. Namely, the following result is true. Proposition 1.1. A subset E ⊂ Y is porous in (Y; d) if and only if the following properly holds: C(i) There exist ∈ (0; 1] and r0 ¿ 0 such that for each x ∈ E and each r ∈ (0; r0 ] there exists y ∈ Y for which (0.1) is valid. Other notions of porosity have been used in the literature [3,23]. We use the rather strong notion which appears in [9,11,20]. Since porous sets are nowhere dense, all -porous sets are of the -rst category. If Y is a -nite dimensional Euclidean space, then -porous sets are of Lebesgue measure 0. In fact, the class of -porous sets in such a space is much smaller than the. class of sets which have measure 0 and are of the -rst category. Also, every complete metric space without isolated points contains a closed nowhere dense set which is not -porous [24]. To point out the diHerence between porous and nowhere dense sets note that if E ⊂ Y is nowhere dense, y ∈ Y and r ¿ 0, then there is a point z ∈ Y and a number s ¿ 0 such that Bd (z; s) ⊂ Bd (y; r) \ E. If, however, E is also porous, then for small enough r we can choose s = r, where ∈ (0; 1) is a constant which depends only on E. Finally, we prove the following simple proposition. It shows that porosity is equivalent to another property which is easier to verify.
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Proposition 1.2. A subset E ⊂ Y is porous set in (Y; d) if and only if the following property holds: C(ii) There exist ∈ (0; 1] and r0 ¿ 0 such that for each x ∈ E and each r ∈ (0; r0 ] there exists y ∈ Y for which d(x; y) 6 r;
Bd (y; r) ∩ E = ∅:
Proof. Assume that C(ii) holds with ∈ (0; 1] and r0 ¿ 0. By Proposition 1.1; in order to prove the proposition it is suKcient to show that C(i) holds. Let x ∈ E and r ∈ (0; r0 ]. By C(ii) there exists y ∈ Y such that d(x; y) 6 r=4;
Bd (y; r=4) ∩ E = ∅:
Then Bd (y; r=4) ⊂ Bd (x; r) \ E. Therefore, property C(i) holds and Proposition 1.2 is proved.
2. Well posedness of optimization problems We use the convention that ∞=∞=1 and ∞−∞=0. For each function f : X → R1 ∪ {∞}, where X is nonempty, we set inf (f) = inf {f(x): x ∈ X }. We consider a metric space (X; ) and a complete metric space (A; d). Each of these spaces is equipped with the topology generated by its metric. We assume that with every a ∈ A a lower semicontinuous function fa on X is associated with values in R1 ∪ {∞} = (−∞; ∞] and that fa is not identically in-nity. The following property was introduced in [12]. Let ∈ A. We say that the minimization problem for fa on (X; ) is strongly well posed with respect to (A; d) if inf (fa ) is -nite and attained at a unique point xa ∈ X and the following assertion holds: For each ¿ 0 there exist a neighborhood V of a in (A; d) and ¿ 0 such that for each b ∈ V; inf (fb ) is -nite and if z ∈ X satis-es fb (z) 6 inf (fb )+, then (z; xa ) 6 and |fb (z) − fa (xa )| 6 . (In a slightly diHerent setting a similar property was introduced in [30].) In our study we use the following hypotheses about the functions. (H) If a ∈ A; inf (fa ) is -nite, {xi }∞ i=1 ⊂ X is a Cauchy sequence and the sequence is bounded, then the sequence {x n }∞ {fa (x n )}∞ n=1 n=1 converges in X . Clearly if (X; ) is complete, then (H) is true. Note that for classes of variational problems considered in this paper the space (X; ) is not complete. Fortunately, instead of the completeness assumption we can use (H) and this hypothesis holds for spaces of integrands which satisfy the Cesari growth condition. For each integer n ¿ 1 denote by A(n) the set of all a ∈ A which have the following property: (P1) There exist xN ∈ X; ¿ 0; ∈ R1 and a neighborhood U of a in (A; d) such that for each b ∈ U; inf (fb ) is -nite and if x ∈ X satis-es fb (x) 6 inf (fb ) + , then (x; x) N 6 1=n and |fb (x) − | 6 1=n.
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
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∞ Proposition 2.1. Let a ∈ n=1 A(n) . Then the minimization problem for fa on (X; ) is strongly well posed with respect to (A; d). Proof. By property (P1) for each integer n ¿ 1 there exist x n ∈ X; n ¿ 0; n ∈ R1 and a neighborhood Un of a in (A; d) such that the following property holds: (P2) For each b ∈ Un ; inf (fb ) is -nite and if x ∈ X satis-es fb (x) 6 inf (fb ) + n , then (x; x n ) 6 1=n and |fb (x) − n | 6 1=n. Assume that {zi }∞ i=1 ⊂ X and lim fa (zi ) = inf (fa ):
(2.1)
i→∞
Fix an integer n ¿ 1. It follows from property (P2) and (2.1) that for all suKciently large natural numbers i, (zi ; x n ) 6 1=n;
|fa (zi ) − n | 6 1=n:
(2.2)
Since n is an arbitrary natural number we conclude that {zi }∞ i=1 is a Cauchy sequence. By (H) there exists xa = limi→∞ zi . As fa is lower semicontinuous, we have fa (xa ) = inf (fa ). Clearly xa is a unique minimizer of fa for otherwise we would be able to construct a nonconvergent sequence {zi }∞ i=1 . (2.2) implies that for all natural numbers n, (xa ; x n ) 6 1=n;
|fa (xa ) − n | 6 1=n:
(2.3)
Let ¿ 0. Choose a natural number n ¿ 4=. Assume that b ∈ Un ; x ∈ X and fb (x) 6 inf (fb ) + n . By property (P2), (x; x n ) 6 1=n;
|fb (x) − n | 6 1=n:
Combined with (2.3) these inequalities imply that (x; xa ) 6 2=n 6 =2;
|fb (x) − fa (xa )| 6 2=n 6 =2:
This completes the proof of Proposition 2.1. 3. Variational principle We will obtain our main result as a realization of a variational principle which will be introduced in this section. We consider a metric space (X; ) and complete metric spaces (A1 ; d1 ) and (A2 ; d2 ). The set A = A1 × A2 is equipped with a metric d de-ned by d((a1 ; a2 ); (b1 ; b2 )) = d1 (a1 ; b1 ) + d2 (a2 ; b2 );
(a1 ; a2 ); (b1 ; b2 ) ∈ A:
(3.1)
Clearly (A; d) is a complete metric space. We assume that with every a ∈ A1 a lower semicontinuous function a on X is associated with values in R1 ∪ {∞} and that a is not identically in-nity. We also assume that with every a ∈ A2 a nonempty subset Sa ⊂ X is associated. For each a = (a1 ; a2 ) ∈ A = A1 × A2 de-ne fa (x) = a1 (x);
x ∈ S a2 ;
fa (x) = ∞;
x ∈ X \ S a2 :
(3.2)
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
Fix ∈ A2 . For each r ¿ 0 set A2(r) = {a ∈ A2 : d2 ( ; a) 6 r}:
(3.3)
In our study we use the following basic hypotheses about the functions. (A1) For each a ∈ A the function fa is lower semicontinuous on X and inf (fa ) is -nite. (A2). There exist ∈ (0; 1] and r0 ¿ 0 such that for each r ∈ (0; r0 ], each a; b ∈ A1 satisfying d1 (a; b) 6 r and each x ∈ X , the inequality | a (x) − b (x)| 6 r is valid. (A3) For each ¿ 0 there exist numbers ¿ 0 and r0 ¿ 0 such that the following property holds: For each a = (a1 ; a2 ) ∈ A = A1 × A2 and each r ∈ (0; r0 ] there exist aN = (aN1 ; a2 ) ∈ A and xN ∈ X such that d1 (a1 ; aN1 ) 6 r;
xN ∈ Sa2 ;
(3.4)
aN1 (x) N = faN(x) N 6 inf (faN) + r
(3.5)
and that for each x ∈ Sa2 which satis-es faN(x) 6 inf (faN) + r
(3.6)
the inequality (x; x) N 6 holds. ⊂ A1 and positive (A4) For each pair of integers i; j ¿ 1 there exist a set A(ij) 1 numbers rNi ; rNij ; ij such that ∞ A1 = A(ij) for all integers i ¿ 1 (3.7) 1 j=1
and for each pair of integers i; j ¿ 1 the following property holds: If a1 ∈ A1 ;
inf {d1 (a1 ; ): ∈ A(ij) 1 } ¡ rNi ;
a2 ∈ A2(i) ;
x ∈ S a2 ;
r ∈ (0; rNij ];
b2 ∈ A2 ;
f(a1 ;a2 ) (x) 6 inf (f(a1 ;a2 ) ) + 1; d2 (b2 ; a2 ) 6 r;
(3.8) (3.9) (3.10)
then there exists xN ∈ Sb2 such that (x; x) N 6r
ij ;
| a1 (x) N − a1 (x)| 6 r
ij :
(3.11)
Theorem 3.1 (Variational principle). Assume that (A1)–(A4) and (H) hold. Then there exists a set B ⊂ A such that A \ B is -porous in (A; d) and that for each a ∈ B the minimization problem for fa on (X; ) is strongly well posed with respect to (A; d). Proof. We recall that A(n) is the set of all a ∈ A which have the property (P1) ∞ (n = 1; 2; : : :). Set B = n=1 A(n) . By Proposition 2.1 in order to prove the theorem it is suKcient to show that the set A \ A(n) is -porous in (A; d) for all integers n ¿ 1.
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
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Let n ¿ 1 be an integer. We will show that A \ A(n) is -porous in (A; d). Since ∞ A \ A(n) = [(A \ A(n) ) ∩ (A1 × A2(i) )] i=1
we need only to show that the set (A \ A(n) ) ∩ (A1 × A2(i) ) is -porous in (A; d) for all integers i ¿ 1. For each pair of natural numbers i; j let a subset A(ij) 1 ⊂ A1 and positive numbers rNi ; rNij ; ij be as guaranteed by (A4). Let i ¿ 1 be an integer. By (A4), ∞ (i) (n) [(A \ A(n) ) ∩ (A1((i+1) j) × A2(i) )]: (A \ A ) ∩ (A1 × A2 ) = j=1
Therefore, in order to prove the theorem it is suKcient to show that for each integer j ¿ 1 the set Eij :=(A \ A(n) ) ∩ (A1((i+1) j) × A2(i) ) is porous in (A; d). Fix a natural number j ¿ 1. There exist ¿ 0 and r0 ¿ 0 such that (A3) holds with =(2n)−1 . There exist N ∈ (0; 1] and rN ∈ (0; 1] such that (A2) is valid with = ; N r0 = r. ˜ We choose positive numbers ∗ ; r∗ such that r∗ ¡ min{2−1 ; r0 ; 4−1 rNi+1 ; 4−1 rN(i+1) j ; (4 (8n
(i+1) j )
−1
; (8n)−1 (4
˜ =32; ˜ (24 ∗ ¡ min{a=2;
(i+1) j
(i+1) j )
(i+1) j )
−1
+ 2 + )−1 };
−1
}:
; 2−1 ˜r; ˜ (3.13) (3.14)
Let a = (a1 ; a2 ) ∈ Eij ;
r ∈ (0; r∗ ]:
(3.15)
By the de-nition of ; r0 , (A3), (3.13) and (3.15), there exist aN = (aN1 ; a2 ) ∈ A and xN ∈ X such that d1 (a1 ; aN1 ) 6 r=2;
xN ∈ Sa2 ;
aN1 (x) N = faN(x) N 6 inf (faN) + r=2
(3.16)
and the following property holds: N 6 (2n)−1 (P3) For each x ∈ Sa2 satisfying faN(x) 6 inf (faN)+r=2 the inequality (x; x) is valid. Clearly d(a; a) N = d1 (a1 ; aN1 ) 6 r=2: We will show that {b ∈ A: d(b; a) N 6 ∗ r} ⊂ A(n) ⊂ A \ Eij
(3.17)
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
(see (3.12)). Let b = (b1 ; b2 ) ∈ A and d1 (b1 ; aN1 ) + d2 (b2 ; a2 ) = d(b; a) N 6 2∗ r:
(3.18)
By (3.3), (3.12) – (3.15), and (3.18) a1 ∈ A1(i+1) j ; a2 ∈ A2(i) ; b2 ∈ A2(i+1) :
(3.19)
Now consider the pairs (b1 ; b2 ); (b1 ; a2 ) ∈ A. By (3.13) – (3.16), (3.18), and (3.19) inf {d1 (b1 ; h): h ∈ A1(i+1) j } 6 d1 (b1 ; a1 ) 6 d1 (b1 ; aN1 ) + d1 (aN1 ; a1 ) 6 2∗ r + r=2 ¡ 3r∗ ¡ rNi+1 :
(3.20)
By (3.13) – (3.15), and (3.18) d2 (a2 ; b2 ) 6 2∗ r 6 r 6 r∗ ¡ rN(i+1) j :
(3.21)
Assume now that h1 ; h2 ∈ A2 and either h1 = a2 ; h2 = b2
or
h1 = b2 ; h2 = a2 :
(3.22)
It follows from the de-nition of rNi+1 ; rN(i+1) j ; (i+1) j , (A4), (3.19) – (3.22) that the following property holds: (P4) If x ∈ X satis-es f(b1 ;h1 ) (x) 6 inf (f(b1 ;h1 ) ) + 1, then there exists x˜ ∈ Sh2 such that (x; x) ˜ 6 2∗ r
(i+1) j ;
| b1 (x) ˜ − b1 (x)| 6 2∗ r
(i+1) j :
(3.23)
Since inf (f(b1 ;hi ) ) = inf {f(b1 ;hi ) (x): x ∈ Shi ; f(b1 ;hi ) (x) 6 inf (f(b1 ;hi ) ) + 1};
i = 1; 2;
property (P4) implies that |inf (fb1 ;a2 ) − inf (fb1 ;b2 )| 6 2∗ r
(i+1) j :
(3.24)
Let x ∈ X and fb (x) 6 inf (fb ) + 2∗ r
(i+1) j :
(3.25)
Since inf (fb ) is -nite we have x ∈ Sb2 . By property (P4), (3.25), (3.13) – (3.15) there exists x˜ ∈ Sa2
(3.26)
for which (3.23) is valid. It follows from (3.23) – (3.25) that b1 (x) ˜ 6 b1 (x) + 2∗ r
(i+1) j
6 inf (f(b1 ;a2 ) ) + 6∗ r
6 inf (fb ) + 4∗ r (i+1) j :
(i+1) j
(3.27)
(3.13) – (3.15), and (3.18) imply that d1 (aN1 ; b1 ) 6 2∗ r 6 2∗ r∗ 6 a˜r: ˜
(3.28)
By (3.28), the de-nition of ; ˜ r˜ and (A2), | aN1 (y) − b1 (y)| 6 2∗ r() ˜ −1
for all y ∈ X:
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This inequality implies that | aN1 (x) ˜ − b1 (x)| ˜ 6 2∗ r() ˜ −1
(3.29)
|inf (f(aN1 ;a2 ) ) − inf (f(b1 ;a2 ) )| 6 2∗ r() ˜ −1 :
(3.30)
and It follows from (3.29), (3.30) and (3.27) that aN1 (x) ˜ 6 b1 (x) ˜ + 2∗ r(a) ˜ −1 6 inf (f(b1 ;a2 ) ) + 6∗ r
(i+1) j
+ 2∗ r(a) ˜ −1
6 inf (f(aN1 ;a2 ) ) + 4∗ r() ˜ −1 + 6∗ r
(i+1) j :
(3.31)
By (3.2), (3.26), (3.31) and (3.14), f(aN1 ;a2 ) (x) ˜ = aN1 (x) ˜ 6 inf (f(aN1 ;a2 ) ) + r=2: By this inequality, (3.26) and property (P3), (x; ˜ x) N 6 (2n)−1 . Combined with (3.23), (3.13) – (3.15) this inequality implies that (x; x) N 6 (x; x) ˜ + (x; ˜ x) N 6 2∗ r
(i+1) j
+ (2n)−1 ¡ n−1 :
(3.32)
It follows from (3.25), (3.24), (3.30) and (3.15) that |fb (x) − inf (faN)| 6 |fb (x) − inf (fb )| + |inf (f(b1 ;b2 ) ) − inf (f(b1 ;a2 ) )| + |inf (f(b1 ;a2 ) ) − inf (f(aN1 ;a2 ) )| 6 2∗ r
(i+1) j
+ 2∗ r
(i+1) j
+ 2∗ r() ˜ −1 6 (4n)−1 + (4n)−1
= (2n)−1 : Thus (x; x) N ¡ n−1 ;
|fb (x) − inf (faN)| ¡ n−1 :
(3.33)
We have shown that for each b ∈ A satisfying (3.18) and each x ∈ X satisfying (3.25), the inequality (3.33) hold. By the de-nition of A(n) and (3.12), {b ∈ A: d(b; a) N 6 ∗ r} ⊂ A(n) ⊂ A \ Eij :
(3.34)
Therefore, we have shown that for each a ∈ A and each number r satisfying (3.15) there exists aN ∈ A for which (3.17) and (3.34) are true. By Proposition 1.2 the set Eij is porous in (A; d). Theorem 3.1 is proved. 4. The main result For each function f : X → R1 , where (X; ) is a metric space, de-ne Lip(f) = sup{(f(x) − f(y))=(x; y): x; y ∈ X and x = y}: Let (E; · ) be a Banach space. We equip the space E with the metric dE (x; y) = x − y; x; y ∈ E.
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Let −∞ ¡ #1 ¡ #2 ¡ ∞. We say that a function x : [#1 ; #2 ] → E is diHerentiable if there exists a Bochner integrable function u : [#1 ; #2 ] → E such that t u(s) ds; t ∈ [#1 ; #2 ]: (4.1) x(t) = x(#1 ) + #1
It is known that if x is diHerentiable, then (4.1) de-nes a unique Bochner integrable function u which is called the derivative of x and is denoted by x . We denote by mes(&) the Lebesgue measure of a Lebesgue measurable set & ⊂ R1 . Let −∞ ¡ T1 ¡ T2 ¡ ∞. Denote by X the set of all diHerentiable functions x : [T1 ; T2 ] → E. For the set X we consider the metric de-ned by (x1 ; x2 ) = inf { ¿ 0: mes{t ∈ [T1 ; T2 ]: x1 (t) − x2 (t) + x1 (t) − x2 (t) ¿ } 6 };
x1 ; x2 ∈ X:
(4.2)
Let f : [T1 ; T2 ] × E × E → R1 . For each t ∈ [T1 ; T2 ] de-ne a function ft : E × E → R1 by ft (x; y) = f(t; x; y);
x; y ∈ E:
(4.3) 1
Denote by M the set of all functions f : [T1 ; T2 ] × E × E → R with the following properties: (i) f is measurable with respect to the -algebra generated by products of Lebesgue measurable subsets of [T1 ; T2 ] and Borel subsets of E × E; (ii) ft is lower semicontinuous for almost every t ∈ [T1 ; T2 ]; (iii) for each ¿ 0 there exists an integrable scalar function (t) 6 0; t ∈ [T1 ; T2 ] such that u 6
(t)
+ f(t; x; u)
for all (t; x; u) ∈ [T1 ; T2 ] × E × E;
(iv) for each M ¿ 0 there exists cM ¿ 0 such that for almost every t ∈ [T1 ; T2 ], sup{f(t; x; u): x; u ∈ E; x; u 6 M } 6 cM ; (v) for each M ¿ 0 there exist dM ¿ 0; rM ¿ 0; cM 1 ¿ 0 and cM 2 ¿ 0 such that for each t ∈ [T1 ; T2 ], each r ∈ (0; rM ], each x1 ; x2 ∈ E satisfying x1 ; x2 6 M;
x1 − x2 6 r
and each u ∈ E satisfying u ¿ dM , |f(t; x1 ; u) − f(t; x2 ; u)| 6 cM 1 r min{f(t; x1 ; u); f(t; x2 ; u)} + cM 2 r; (vi) for each M ¿ 0 there exists an integrable scalar function M (t) ¿ 0; t ∈ [T1 ; T2 ] such that M (t) ¿ Lip(gt ); t ∈ [T1 ; T2 ] where gt is the restriction of ft to the set {(x; y) ∈ E × E: x; y 6 M } for all t ∈ [T1 ; T2 ]. The growth condition in (iii) was proposed by Cesari [6] and its equivalents and modi-cations are rather common in the literature. It follows from property (i) that for any f ∈ M and any x ∈ X the function f(t; x(t); x (t)); t ∈ [T1 ; T2 ] is measurable. Clearly if f ∈ M and , ¿ 0, then ,f ∈ M and if f1 ; f2 ∈ M, then f1 + f2 ∈ M.
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
11
Remark 4.1. Assume that f ∈ M and g : [T1 ; T2 ] × E → R1 is a measurable function with respect to the -algebra generated by products of Lebesgue measurable subsets of [T1 ; T2 ] and Borel subsets of E such that 0 ¡ inf {g(t; u): (t; u) ∈ [T1 ; T2 ] × E} 6 sup{g(t; u): (t; u) ∈ [T1 ; T2 ] × E} ¡ ∞; for almost every t ∈ [T1 ; T2 ] the function (x; u) → f(t; x; u)g(t; u); (x; u) ∈ E ×E is lower semicontinuous; for each M ¿ 0 there is M ¿ 0 such that for each t ∈ [T1 ; T2 ] and each x1 ; x2 ; u1 ; u2 ∈ E satisfying xi ; ui 6 M; i = 1; 2 the following inequality holds: |g(t; u1 ) − g(t; u2 )| 6
M u1
− u2 :
We can show in a straightforward manner that the function (t; x; u) → f(t; x; u)g(t; u);
(t; x; u) ∈ [T1 ; T2 ] × E × E
belongs to M. Remark 4.2. Assume that a function f : [T1 ; T2 ] × E × E → R1 satis-es f(t; x; u) ¿ (u) − c0 ;
(t; x; u) ∈ [T1 ; T2 ] × E × E;
(4.4)
: [0; ∞) → [0; ∞) is an increasing function such that where c0 is a constant and limu→∞ (u)=u = ∞. Then we can verify in a straightforward manner that the property (iii) holds and the property (v) is equivalent to the following property: For each M ¿ 0 there exist dM ¿ 0; rM ¿ 0; cM ¿ 0 such that for each t ∈ [T1 ; T2 ], each r ∈ (0; rM ], each x1 ; x2 ∈ E which satisfy x1 ; x2 6 M and x1 − x2 6 r, and each u ∈ E satisfying u ¿ dM the following inequality holds: |f(t; x1 ; u) − f(t; x2 ; u)| 6 cM r min{f(t; x1 ; u); f(t; x2 ; u)}: Remark 4.3. Let f ∈ M satisfy (4.4) with a constant c0 and an increasing function : [0; ∞) → [0; ∞) such that limu→∞ (u)=u = ∞. Assume that g : [T1 ; T2 ] × E × E → R1 is a measurable function with respect to the -algebra generated by products of Lebesgue measurable subsets of [T1 ; T2 ] and Borel subsets of E × E with the following properties: for almost every t ∈ [T1 ; T2 ] the function (x; u) → f(t; x; u) + g(t; x; u); (x; u) ∈ E × E is lower semicontinuous; inf {g(t; x; u): t ∈ [T1 ; T2 ]; x; u ∈ E} ¿ 0; f is Lipschitzian on all bounded subsets of [T1 ; T2 ] × E × E; sup{Lip(ftu ): t ∈ [T1 ; T2 ]; u ∈ E} ¡ ∞ where ftu : E → R1 is de-ned by ftu (x) = f(t; x; u); x ∈ E for all (t; u) ∈ [T1 ; T2 ] × E. Using Remark 4.2 we can show in a straightforward manner that the function (t; x; u) → f(t; x; u) + g(t; x; u); (t; x; u) ∈ [T1 ; T2 ] × E × E also belongs to M.
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
It is an elementary exercise to show that a function f=f(t; x; u) ∈ C 1 ([T1 ; T2 ]×E×E) belongs to M if (iii) is true and the following conditions hold: for each M ¿ 0, sup{9f=9x(t; x; u) + 9f=9u(t; x; u): t ∈ [T1 ; T2 ]; x; u ∈ E and x; u 6 M } ¡ ∞; there exist increasing functions : [0; ∞) → [0; ∞) and that for each (t; x; u) ∈ [T1 ; T2 ] × E × E, max{9f=9x(t; x; u); 9f=9u(t; x; u)} 6
0 (x)
0
: [0; ∞) → [0; ∞) such
(u)
and (u) 6 f(t; x; u). For each f; g ∈ M de-ne d˜ 1 (f; g) = sup{|f(t; x; u) − g(t; x; u)|: (t; x; u) ∈ [T1 ; T2 ] × E × E} + sup{Lip(ft − gt ): t ∈ [T1 ; T2 ]}; d1 (f; g) = d˜ 1 (f; g)(d˜ 1 (f; g) + 1)−1 : Clearly (M; d1 ) is a complete metric space. Denote by Ml (respectively, Mc ) the set of all lower semicontinuous (respectively, continuous) functions f : [T1 ; T2 ] × E × E → R1 in M. Clearly Ml and Mc are closed subsets of the metric space (M; d1 ). We consider the complete metric spaces (Ml ; d1 ) and (Mc ; d1 ). For each f ∈ M we de-ne I f : X → R1 ∪ {∞} by T2 f f(t; x(t); x (t)) dt; x ∈ X: (4.5) I (x) = T1
We study the variational problem I f (x) → min;
x ∈ X;
x(Ti ) = i ;
i = 1; 2;
where f ∈ M and i ∈ E; i = 1; 2. Set A2 = E × E and de-ne d2 ((x1 ; y1 ); (x2 ; y2 )) = x1 − x2 + y1 − y2 ;
(xi ; yi ) ∈ E × E;
i = 1; 2:
We consider the complete metric space (A; d) where A = A1 × A2 ; A1 is either M or Ml or Mc and d((a1 ; a2 ); (b1 ; b2 )) = d1 (a1 ; b1 ) + d2 (a2 ; b2 );
(a1 ; a2 ); (b1 ; b2 ) ∈ A:
For each a = (f; (1 ; 2 )) ∈ A1 × A2 we de-ne Ja : X → R1 ∪ {∞} by Ja (x) = I f (x) if x(Ti ) = i ;
i = 1; 2;
otherwise Ja (x) = ∞:
(4.6)
By properties (iii) and (iv), inf (Ja ) is -nite for all a ∈ A. We will show that the functional Ja is lower semicontinuous for any a ∈ A (see Propositions 5.1 and 5.2). Now we are ready to state our main result. Theorem 4.1. There exists a set B ⊂ A such that A \ B is a -porous set in the space (A; d) and that for each a ∈ B the minimization problem for Ja on (X; ) is strongly well posed with respect to (A; d).
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
13
The proof of Theorem 4.1 consists in verifying that hypotheses (A1) – (A4) and (H) hold for the space (A; d). 5. Preliminary results for hypotheses (A1), (A3) and (H) For each f ∈ M and each 1 ; 2 ∈ E set 0(f; 1 ; 2 ) = inf (J(f; (1 ;2 )) ):
(5.1)
For the proof of the following proposition see [29, Proposition 3.1]. Proposition 5.1. Let f ∈ M; x ∈ X; {xi }∞ i=1 ⊂ X and let (xi ; x) → 0 as i → ∞. Then I f (x) 6 lim inf i→∞ I f (xi ). The following proposition is an auxiliary result for hypothesis (H). Its proof is analogous to the proof of Proposition 4.2 in [26]. Proposition 5.2. Assume that f ∈ M; {xi }∞ i=1 ⊂ X is a Cauchy sequence and that the sequence {I f (xi )}∞ i=1 is bounded. Then there is x∗ ∈ X such that (xi ; x∗ ) → 0 as i → ∞ and moreover; xi (t) → x∗ (t) as i → ∞ uniformly on [T1 ; T2 ]. Analogous to Lemma 5.1 of [28] we can prove the following result which implies (A3). Proposition 5.3. For each 1 ∈ (0; 1) there exists (1) ∈ (0; 1) such that for each f ∈ M; each 1 ; 2 ∈ E and each r ∈ (0; 1] there exist a continuous function h : [T1 ; T2 ] × E × E → R1 which satis6es 0 6 h(t; x; u) 6 r=2
for all (t; x; u) ∈ [T1 ; T2 ] × E × E;
Lip(ht ) 6 r=2 for all t ∈ [T1 ; T2 ] and xN ∈ X satisfying x(T N i ) = i ; i = 1; 2 such that for the function fN ∈ M de6ned by N x; u) = f(t; x; u) + h(t; x; u); (t; x; u) ∈ [T1 ; T2 ] × E × E f(t; the following properties hold: N N 1 ; 2 ) + (1)r; I f (x) N 6 0(f; for each y ∈ X satisfying y(Ti ) = i ;
i = 1; 2;
N ˜ 1 ; 2 ) + 2(1)r I f (y) 6 0(f;
the inequality (x; N y) 6 1 is valid. Moreover, h is the sum of two functions, one of them depending only on (t; x) while the other depending only on (t; u). For each 1 ; 2 ∈ E de-ne a function x1 2 : [T1 ; T2 ] → E by x1 2 (t) = 1 + (t − T1 )(T2 − T1 )−1 (2 − 1 );
t ∈ [T1 ; T2 ]:
(5.2)
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
Clearly for each 1 ; 2 ∈ E x1 2 (Ti ) = i ;
i = 1; 2;
x1 2 ∈ X
(5.3)
and x 1 2 (t) = (T2 − T1 )−1 (2 − 1 );
t ∈ [T1 ; T2 ]:
Lemma 5.1. Let f ∈ M and let cM ¿ 0 be as guaranteed by property (iv) for any M ¿ 0. Let M0 ¿ 0 and M1 =M0 (3+2(T2 −T1 )−1 ). Then for each 1 ; 2 ∈ E satisfying 1 ; 2 6 M0
(5.4)
the inequality I f (x1 2 ) 6 cM1 (T2 − T1 ) holds. Proof. Let 1 ; 2 ∈ E satisfy (5.4). Then by (5.2) – (5.4) for all t ∈ [T1 ; T2 ]; x1 2 (t) 6 3M0 6 M1 ;
x 1 2 (t) 6 2M0 (T2 − T1 )−1 6 M1 :
It follows from these inequalities and property (iv) that for almost every t ∈ [T1 ; T2 ]; f(t; x1 ;2 (t); x 1 ;2 (t)) 6 cM1 . This completes the proof of Lemma 5.1. Lemma 5.2. Let f ∈ M and M0 ¿ 0. Then there exists M ¿ 0 such that for each 1 ; 2 ∈ E satisfying; 1 ; 2 6 M0 ; the inequality |0(f; 1 ; 2 )| 6 M is valid. Proof. By Lemma 5.1 there exists M1 ¿ 0 such that for each 1 ; 2 ∈ E satisfying 1 ; 2 6 M0 the inequality 0(f; 1 ; 2 ) 6 M1 holds. By property (iii) there exists an integrable scalar function (t) ¿ 0; t ∈ [T1 ; T2 ] such that f(t; x; u) ¿ − (t)
for all (t; x; u) ∈ [T1 ; T2 ] × E × E:
Therefore; inf {0(f; 1 ; 2 ): 1 ; 2 ∈ E} = inf (I f ) ¿ −
T2
T1
(t) dt:
This completes the proof of Lemma 5.2. Lemma 5.3. Let f ∈ M and M0 ; M1 ¿ 0. Then there exists M2 ¿ 0 such that for each 1 ; 2 ∈ E satisfying i 6 M0 ;
i = 1; 2
(5.5)
and each x ∈ X which satis6es x(Ti ) = i ;
i = 1; 2;
and
I f (x) 6 0(f; 1 ; 2 ) + M1
(5.6)
the following inequality holds: x(t) 6 M2 ;
t ∈ [T1 ; T2 ]:
(5.7)
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
15
Proof. By Lemma 5.2 there exists c ¿ 0 such that for each 1 ; 2 ∈ E satisfying (5.5) the inequality |0(f; 1 ; 2 )| 6 c
(5.8)
is valid. By property (iii) (see the de-nition of M) there exists an integrable scalar function (t) ¿ 0; t ∈ [T1 ; T2 ] such that u 6 (t) + f(t; x; u) Set
for all (t; x; u) ∈ [T1 ; T2 ] × E × E:
M2 = c + M0 + M1 +
T2
T1
(5.9)
(t) dt:
(5.10)
Let 1 ; 2 ∈ E satisfy (5.5) and let x ∈ X satisfy (5.6). It follows from (5.5) and the de-nition of c; that (5.8) holds. (5.8) and (5.6) imply that I f (x) 6 M1 + c. Combined with (5.9) this inequality implies that T2 T2 T2 x (t) dt 6 (t) dt + f(t; x(t); x (t)) dt T1
T1
6 M1 + c +
T1
T2
T1
(t) dt:
By these inequalities; (5.5); (5.6) and (5.10) for each t ∈ [T1 ; T2 ]; T2 T2 x(t) 6 x(T1 ) + x (t) dt 6 M0 + M1 + c + (t) dt = M2 : T1
T1
Lemma 5.3 is proved. 6. Auxiliary result for hypothesis (A4) Let f ∈ M; M0 ; M1 ¿ 0 and let M2 ¿ M0 be as guaranteed by Lemma 5.3. Namely if 1 ; 2 ∈ E satisfy (5:5) and if x ∈ X satis-es (5:6); then (5:7) holds:
(6.1)
By Lemma 5.2 there exists M3 ¿ 0 such that |0(f; 1 ; 2 )| 6 M3
for all 1 ; 2 ∈ E satisfying i 6 M0 + 1;
i = 1; 2: (6.2)
By properly (iii) (see the de-nition of M) there exists an integrable function t ∈ [T1 ; T2 ] such that u 6
1 (t)
+ f(t; x; u)
for all (t; x; u) ∈ [T1 ; T2 ] × E × E:
1 (t) ¿ 0;
(6.3)
By property (v) (see the de-nition of M) there exist d0 ¿ M2 + 1;
r0 ∈ (0; 1]; cN1 ; cN2 ¿ 0
such that the following property holds:
(6.4)
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
For each t ∈ [T1 ; T2 ], each r ∈ (0; r0 ], each x1 ; x2 ∈ E which satisfy x1 − x2 6 r
x1 ; x2 6 M2 + 1;
(6.5)
and each u ∈ E satisfying u ¿ d0 , we have |f(t; x1 ; u) − f(t; x2 ; u)| 6 cN1 r min{f(t; x1 ; u); f(t; x2 ; u)} + cN2 r: Choose positive numbers d∗ ¿ 4(d0 + 1) 1 +
(6.6)
T2
1 (t) dt
T1
+ M3 + M2 (min{1; T2 − T1 })−1 ;
r∗ ∈ (0; 12−1 r0 min{1; T2 − T1 }):
(6.7) (6.8)
For each t ∈ [T1 ; T2 ] we denote by gt the restriction of ft to the set {(x; y) ∈ E × E: x; y 6 d∗ + 1}. By property (vi) (see the de-nition of M) there exists a scalar integrable function (t) ¿ 0; t ∈ [T1 ; T2 ] such that (t) ¿ Lip(gt );
t ∈ [T1 ; T2 ]:
(6.9)
Choose a number T2 (t) dt: ∗¿
(6.10)
T1
Lemma 6.1. Assume that 1 ; 2 ∈ E satisfy i 6 M0 ;
i = 1; 2;
(6.11)
x ∈ X satisfy x(Ti ) = i ;
i = 1; 2;
I f (x) 6 0(f; 1 ; 2 ) + M1 ;
(6.12)
r ∈ (0; r∗ ] and Ni ; N2 ∈ X satisfy i − Ni 6 r;
i = 1; 2:
(6.13)
Then there exists xN ∈ X such that x(T N i ) = Ni ;
i = 1; 2;
x(t) N − x(t) 6 3r;
(6.14) t ∈ [T1 ; T2 ];
xN (t) − x (t) 6 4r(T2 − T1 )−1
for almost every t ∈ [T1 ; T2 ]
(6.15)
and |I f (x) − I f (x)| N 6 r [3cN2 (T2 − T1 ) + 3cN1 (M1 + M3 ) T2 −1 + 3cN1 : (t) dt + (3 + 4(T − T )) 1 ∗ 2 1 T1
(6.16)
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
17
Proof. By (6.11); (6.12) and the de-nition of M2 (see (6.1)); x(t) 6 M2 ;
t ∈ [T1 ; T2 ]:
(6.17)
Set; &1 = {t ∈ [T1 ; T2 ]: x (t) ¿ d∗ };
&2 = [T1 ; T2 ] \ &1 :
(6.18)
We will show that mes(&1 ) 6 (T2 − T1 )=4. It follows from (6.18); (6.3) (6.12); (6.2) and (6.11) that T2 T2 d∗ mes(&1 ) 6 x (t) dt 6 x (t) dt 6 1 (t) dt T1
&1
+ 6
T1 T2
T1
6
T2
T2
T1
T1
f(t; x(t); x (t)) dt
1 (t) dt
+ 0(f; 1 ; 2 ) + M1
1 (t) dt
+ M 3 + M1 :
Combined with (6.7) these inequalities imply that T2 mes(&1 ) 6 d−1 (t) dt + M + M 1 3 1 6 (T2 − T1 )=4: ∗ T1
Thus mes(&1 ) 6 (T2 − T1 )=4;
mes(&2 ) ¿ 3(T2 − T1 )=4:
(6.19)
De-ne a function u : [T1 ; T2 ] → E by u(t) = x (t);
t ∈ &1 ;
u(t) = x (t) + (mes(&2 ))−1 [(N2 − N1 ) − (2 − 1 )];
(6.20) t ∈ &2 :
Clearly u is a Bochner integrable function. For t ∈ [T1 ; T2 ] set t x(t) N = N1 + u(t) dt: T1
(6.21)
(6.22)
Evidently xN ∈ X . It follows from (6.22) that x(T N i ) = Ni ; i = 1; 2. Thus (6.14) is valid. It follows from (6.19) – (6.22) and (6.13) that for almost every t ∈ [T1 ; T2 ] xN (t) = u(t)
(6.23)
and xN (t) − x (t) 6 (N2 − N1 ) − (2 − 1 )(mes(&2 ))−1 6 8r(3(T2 − T1 ))−1 :
(6.24)
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
(6.22); (6.23); (6.12); (6.13); (6.20) and (6.21) imply that for all t ∈ [T1 ; T2 ]; t t x(t) N − x(t) 6 N1 − 1 + xN (s) ds − x (s) ds 6r +
T1
t
T1
6r +
&2
T1
xN (s) − x (s) ds u(s) − x (s) ds 6 3r:
Combined with (6.24) these inequalities imply (6.15). Now we will show that (6.16) is valid. Let t ∈ &1 and (6.23) be valid. By (6.23), (6.20), (6.18) and (6.7), x (t) = xN (t);
x (t) ¿ d∗ ¿ d0 :
(6.25)
By (6.15), (6.8) and (6.4), x(t) N − x(t) 6 3r 6 3r∗ ¡ 4−1 r0 ¡ 4−1 :
(6.26)
(6.26) and (6.17) imply that x(t) 6 M2 ;
x(t) N 6 M2 + 1:
(6.27)
It follows from (6.25) – (6.27) and the de-nition of d0 ; r0 ; cN1 ; cN2 (see (6.4) – (6.6)) that |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| = |f(t; x(t); x (t)) − f(t; x(t); N x (t))| 6 3r cN1 min{f(t; x(t); x (t)); f(t; x(t); N x (t))} + 3r cN2 : Since these relations hold for almost every t ∈ &1 we obtain by (6.18), (6.12) and (6.3) that |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| dt &1
6 3r cN2 (T2 − T1 ) + 3r cN1
&1
f(t; x(t); x (t)) dt
f
= 3r cN2 (T2 − T1 ) + 3r cN1 I (x) −
&1
f(t; x(t); x (t)) dt
6 3r cN2 (T2 − T1 ) + 3r cN1 (0(f; 1 ; 2 ) + M1 ) − 3r cN1
&2
f(t; x(t); x (t)) dt
6 3r cN2 (T2 − T1 ) + 3r cN1 (0(f; 1 ; 2 ) + M1 ) − 3r cN1 x (t) dt − 1 (t) dt &2
&2
6 3r cN2 (T2 − T1 ) + 3r cN1 (0(f; 1 ; 2 ) + M1 ) + 3r cN1
T2
T1
1 (t) dt:
(6.28)
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
19
(6.28), (6.11) and (6.2) imply that |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| dt &1
6 3r cN2 (T2 − T1 ) + 3r cN1 M1 + M3 +
T2
1 (t) dt
T1
:
(6.29)
Now we will estimate &2 |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| dt. Let t ∈ &2 . By (6.23) and (6.15) we may assume that xN (t) = u(t);
xN (t) − x (t) 6 4r(T2 − T1 )−1 :
(6.30)
It follows from (6.30), (6.18), (6.8), (6.17), (6.15) and (6.7) that x (t) ¡ d∗ ;
xN (t) ¡ d∗ + 4r∗ (T2 − T1 )−1 6 d∗ + 1;
x(t) 6 M2 ¡ d∗ ;
x(t) N 6 x(t) + 3r 6 M2 + 3r∗ 6 M2 + 1 ¡ d∗ :
(6.31)
Recall that for each # ∈ [T1 ; T2 ]; g# is the restriction of f# to the set {(x; y) ∈ E × E: x; y 6 d∗ + 1}. It follows from (6.31), (6.15), (6.30) and (6.9) that |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| 6 Lip(gt )(x(t) − x(t) N + x (t) − xN (t)) 6 Lip(gt )(3r + 4r(T2 − T1 )−1 ) 6 r (t)(3 + 4(T2 − T1 )−1 ): Since these inequalities hold for almost every t ∈ &2 , (6.10) implies that −1 |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| dt 6 r(3 + 4(T2 − T1 ) ) (t) dt &2
&2
6 r(3 + 4(T2 − T1 )−1 ) 6r
∗ (3
T2
T1
(t) dt
+ 4(T2 − T1 )−1 ):
Combining these inequalities with (6.29) we obtain that T2 |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| dt T1
6 r 3cN2 (T2 − T1 ) + 3cN1 M1 + M3 +
T2
T1
(t) dt + 1
−1 (3 + 4(T − T ) ) : ∗ 2 1
Therefore (6.16) is valid. Lemma 6.1 is proved. 7. Proof of Theorem 4.1 By Theorem 3.1 we need to show that (A1) – (A4) and (H) hold. It follows from Propositions 5.1 and 5.2 that for each a=(a1 ; a2 ) ∈ A the functionals Ja : X → R1 ∪{∞}
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
and I a1 : X → R1 ∪ {∞} are lower semicontinuous. Therefore (A1) holds. Evidently (A2) is valid. It is easy to see that (A3) follows from Proposition 5.3 and (H) follows from Proposition 5.2. In order to prove Theorem 4.1 we need only to show that (A4) holds. Set = 0 ∈ E × E. Then for any r ¿ 0, A2(r) = {(x; y) ∈ E × E: x + y 6 r}
(7.1)
(see (3.3)). For each f ∈ M, each G ⊂ M and each r ¿ 0 we de-ne d1 (f; G) = inf {d1 (f; g): g ∈ G};
(7.2)
B1 (f; r) = {g ∈ M: d1 (f; g) 6 r}:
(7.3)
Let i ¿ 1 be an integer. For each integer s ¿ i + 2 denote by Ms(i) the set of all f ∈ M such that for each h ∈ B1 (f; 1=2) the following properties hold: (P5) There exists a scalar integrable function 1 (t) ¿ 0; t ∈ [T1 ; T2 ] such that u 6 T2 1 (t) dt 6 s; 1 (t) + h(t; x; u) for all (t; x; u) ∈ [T1 ; T2 ] × E × E and T1 (P6) |0(h; 1 ; 2 )| 6 s for all 1 ; 2 ∈ E satisfying 1 ; 2 6 i + 1; (P7) For each 1 ; 2 ∈ E satisfying 1 ; 2 6 i + 1 and each x ∈ X satisfying x(T1 ) = 1 ;
x(T2 ) = 2 ;
I h (x) 6 0(h; 1 ; 2 ) + 4
the inequality x(t) 6 s holds for all t ∈ [T1 ; T2 ]. Clearly (i) Ms(i) ⊂ Ms+1
for all integers s ¿ i + 2:
(7.2)
By the de-nition of d1 and d˜ 1 (see Section 4), Lemmas 5.2 and 5.3, and property (iii) (see the de-nition of M), M=
∞
Ms(i) :
(7.3)
s=i+2
Fix an integer s ¿ i + 2 and consider the set Ms(i) . For each pair of integers k1 ; k2 (i) the set of all f ∈ Ms(i) such that for each satisfying k2 ¿ k1 ¿ s + 1 denote by Msk 1 k2 h ∈ B1 (f; 1=2) the following property holds: (P8) For each t ∈ [T1 ; T2 ], each r ∈ (0; k1−1 ], each x1 ; x2 ∈ E which satisfy x1 ; x2 6 s + 1 and x1 − x2 6 r and each u ∈ E satisfying u ¿ k1 the inequality |h(t; x1 ; u) − h(t; x2 ; u)| 6 k1 r min{h(t; x1 ; u); h(t; x2 ; u)} + k2 r holds. By property (v) (see the de-nition of M) and the de-nition of d1 and d˜ 1 (see Section 4), (i) Ms(i) = {Msk : k1 and k2 are integers satisfying k2 ¿ k1 ¿ s + 1}: (7.4) 1 k2
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
21
(i) Fix a pair of integers k1 ; k2 satisfying k2 ¿ k1 ¿ s + 1 and consider the set Msk . 1 k2 De-ne
d(s; k1 ) = 4(k1 + 1)(1 + 3s) min{1; T2 − T1 }−1 + 2;
(7.5)
r(k1 ) = (16k1 )−1 min{1; T2 − T1 }:
(7.6)
(i) Msk 1 k2 q
(i) f ∈ Msk 1 k2
such that for each the set of all For each integer q ¿ 1 denote by h ∈ B1 (f; 1=2) the following property holds: T (P9) There exists an integrable scalar function (t) ¿ 0; t ∈ [T1 ; T2 ] such that T12 (t) dt ¡ q and (t) ¿ Lip(h˜t ); t ∈ [T1 ; T2 ] where h˜t is the restriction of ht to the set {(x; y) ∈ E × E: x; y 6 d(s; k1 )} for all t ∈ [T1 ; T2 ]. It follows from property (vi) (see the de-nition of M) and the de-nition of d1 and d˜ 1 (see Section 4) that ∞ (i) (i) Msk = Msk : (7.7) 1 k2 1 k2 q q=1
(i) . Lemma 6.1 and the properties Fix an integer q ¿ 1 and consider the set Msk 1 k2 q (P5) – (P9) imply the following lemma. Note that by (P5) – (P9), Lemma 6.1 can be applied to h (see the statement of Lemma 7.1) with M0 = i; M1 = 4; M2 = s (see (P7)); M3 = s (see (P6)); d0 = k1 ; r0 = k1−1 ; cN1 = k1 ; cN2 = k2 (see (P8)); d∗ = d(s; k1 ) − 1; (see (P5) and (7.5)); r∗ = r(k1 ) (see (7.6)); ∗ = q (see (P9)). (i) ; h ∈ B1 (f; 1=2); 1 ; 2 ∈ X satisfy 1 ; 2 6 i; Lemma 7.1. Assume that f ∈ Msk 1 k2 q x ∈ X satisfy x(Ti )=i ; i=1; 2 and I h (x) 6 0(h; 1 ; 2 )+4; r ∈ (0; r(k1 )] and N1 ; N2 ∈ X satisfy N1 − 1 ; N2 − 2 6 r. Then there exists xN ∈ X such that
x(T N j ) = Nj ;
j = 1; 2;
x(t) N − x(t) 6 3r;
xN (t) − x (t) 6 4r(T2 − T1 )−1
t ∈ [T1 ; T2 ];
for almost every t ∈ [T1 ; T2 ]
and |I h (x) − I h (x)| N 6 r[3(T2 − T1 )k2 + 3k1 (4 + 4s) + q(3 + 4(T2 − T1 )−1 )]: (i) Clearly the collection of all sets Msk where integers s; k1 ; k2 ; q satisfy s ¿ i + 1 k2 q 2; k2 ¿ k1 ¿ s + 1 and q ¿ 1, is countable. By (7.3), (7.4) and (7.7) the union of this collection is M. Then (A4) follows from Lemma 7.1. This completes the proof of Theorem 4.1.
References [1] E. Asplund, FrQechet diHerentiability of convex functions, Acta Math. 121 (1968) 31–47. [2] J.M. Ball, N.S. Nadirashvili, Universal singular sets for one-dimensional variational problems, Calc. Var. 1 (1993) 429–438. [3] Y. Benyamini, J. Lindenstrauss, Geometric Nonlinear Functional Analysis, American Mathematical Society, Providence, RI, 2000.
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[4] A. Cellina, G. Colombo, On a classical problem of the calculus of variations without convexity assumptions, Ann. Inst. H. Poincare, Anal. Non lineare 7 (1990) 97–106. [5] A. Cellina, C. Mariconda, The existence question in the calculus of variations: A density result Proc. Amer. Math. Soc. 120 (1994) 1145–1150. [6] L. Cesari, Optimization—Theory and Applications, Springer, New York, 1983. [7] F.H. Clarke, Optimization and Nonsmooth Analysis, Wiley Interscience, New York, 1983. [8] F.S. De Blasi, J. Myjak, Generic Tows generated by continuous vector -elds in Banach Spaces, Adv. in Math. 50 (1983) 266–280. [9] F.S. De Blasi, J. Myjak, On a generalized best approximation problem, J. Approx. Theory 94 (1998) 54–72. [10] R. Deville, R. Godefroy, V. Zizler, Smoothness and Renorming in Banach Spaces, Longman, New York, 1993. [11] R. Deville, J. Revalski, Porosity of ill-posed problems, Proc. Amer. Math. Soc. 128 (2000) 1117–1124. [12] A.D. IoHe, A.J. Zaslavski, Variational principles and well-posedness in optimization and calculus of variations, SIAM J. Control Optim. 38 (2000) 566–581. [13] M. Marcus, A.J. Zaslavski, The structure of extremals of a class of second order variational problems, Ann. Inst. H. Poincare, Anal. Non lineare 16 (1999) 593–629. [14] E.J. McShane, Existence theorem for the ordinary problem of the calculus of variations, Ann. Scoula Norm. Pisa 3 (1934) 181–211. [15] M.D.P. Monteiro Marques, A. Ornelas, Genericity and existence of a minimum for scalar integral functionals, J. Optim. Theory Appl. 86 (1995) 421–431. [16] B.S. Mordukhovich, Approximation Methods in Optimization and Control, Nauka, Moscow, 1988. [17] B.S. Mordukhovich, Existence theorems in nonconvex optimal control, Calculus of Variations and Optimal Control, CRC Press, Boca Raton, FL, 1999, pp. 175 –197. [18] Ch. Morrey, Multiple Integrals in the Calculus of Variations, Springer, Berlin, Heidelberg, New York, 1967. [19] S. Reich, A.J. Zaslavski, Convergence of generic in-nite products of nonexpansive and uniformly continuous operators, Nonlinear Anal.: Theory, Methods Appl. 36 (1999) 1049–1065. [20] S. Reich, A.J. Zaslavski, Asymptotic behavior of dynamical systems with a convex Lyapunov function, J. Nonlinear Convex Anal. 1 (2000) 107–113. [21] R.T. Rockafellar, Existence and duality theorems for convex problems of Bolza, Trans. Amer. Math. Soc. 159 (1971) 1–40. [22] L. Tonelli, Fondamenti di Calcolo delle Variazioni, Zanicelli, Bolonia, 1921–1923. [23] L. Zajicek, Porosity and -porosity, Real Anal. Exchange 13 (1987) 314–350. [24] L. Zajicek, Small non--porous sets in topologically complete metric spaces, Colloq. Math. 77 (1998) 293–304. [25] A.J. Zaslavski, Dynamic properties of optimal solutions of variational problems, Nonlinear Anal.: Theory, Methods Appl. 27 (1996) 895–932. [26] A.J. Zaslavski, Generic well-posedness of optimal control problems without convexity assumptions, S1AM J. Control Optim. 39 (2000) 250–280. [27] A.J. Zaslavski, Existence of solutions of optimal control problems without convexity assumptions, Nonlinear Anal.: Theory, Methods Appl. 43 (2001) 339–361. [28] A.J. Zaslavski, Well-posedness and porosity in optimal control without convexity assumptions, Calc. Var. 13 (2001) 265–293. [29] A.J. Zaslavski, Generic existence of solutions of variational problems without convexity assumptions, Comm. Appl. Anal., in press. [30] T. Zolezzi, Well-posedness criteria in optimization with application to the calculus of variations, Nonlinear Anal.: Theory, Methods Appl. 25 (1995) 437–453.
www.elsevier.com/locate/na
Well posedness and porosity in the calculus of variations without convexity assumptions Alexander J. Zaslavski Department of Mathematics, Technion-Israel Institute of Technology, Haifa 32000, Israel Received 18 December 2001; accepted 5 March 2002
Abstract The Tonelli existence theorem in the calculus of variations and its subsequent modi-cations were established for integrands f which satisfy convexity and growth conditions. In Zaslavski (Nonlinear Analysis 43 (200l) 339), a generic well-posedness result (with respect to variations of the integrand of the integral functional) without the convexity condition was established for a class of optimal control problems satisfying the Cesari growth condition. In Zaslavski (Communications in Applied Analysis, to appear) we extended this generic well-posedness result to a class of variational problems in which the values at the end points are also subject to variations. More precisely, we established a generic well-posedness result for a class of variational problems (without convexity assumptions) over functions with values in a Banach space E which is identi-ed with the corresponding complete metric space of pairs (f; (1 ; 2 )) (where f is an integrand satisfying the Cesari growth condition and 1 ; 2 ∈ E are the values at the end points) denoted by A. We showed that for a generic (f; (1 ; 2 )) ∈ A the corresponding variational problem is well posed. In this paper we study the set of all pairs (f; (1 ; 2 )) ∈ A for which the corresponding variational problem is well posed. We show that the complement of this set is not only of the -rst category but also a -porous set. ? 2003 Elsevier Science Ltd. All rights reserved. MSC: 49J99; 90C31 Keywords: Complete metric space; Generic property; Integrand; Optimal solution; Porous set; Variational problem
1. Introduction The Tonelli existence theorem in the calculus of variations [22] and its subsequent generalizations and extensions (e.g., [6,7,14,18,21]) were established for integrands f E-mail address: [email protected] (A.J. Zaslavski). 0362-546X/03/$ - see front matter ? 2003 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 0 2 ) 0 0 1 8 0 - 3
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
which satisfy convexity and growth conditions. Moreover, certain convexity assumptions are also necessary for properties of lower semicontinuity of integral functionals which are crucial in most of the existence proofs, although there are some interesting theorems without convexity (see [6,4,16,17]). In [27], it was shown that the convexity condition is not needed generically, and not only for the existence but also for the uniqueness of a solution and even for well posedness of the problem (with respect to some natural topology in the space of integrands). Instead of considering the existence of a solution for a single integrand f, we investigated it for a space of integrands and showed that a unique solution exists for most of the integrands in the space. This approach has already been successfully applied in the theory of dynamical systems [8,19], approximation theory [9], as well as in the calculus of variations (see, e.g., [2,13,25]). Interesting generic existence results were obtained for particular cases of variational problems [1,5,15]. In [27], this approach allowed us to establish the generic existence of solutions for a large’class of optimal control problems without convexity assumptions. More precisely, in [27] we considered a class of optimal control problems (with the same system of diHerential equations, the same functional constraints and the same boundary conditions) which is identi-ed with the corresponding complete metric space of cost functions (integrands), say F. We did not impose any convexity assumptions. These integrands are only assumed to satisfy the Cesari growth condition. The main result in [27] establishes the existence of an everywhere dense G -set F ⊂ F such that for each integrand in F the corresponding optimal control problem has a unique solution. The next steps in this area of research were done in [12,26,29]. In [12], we introduced a general variational principle having its prototype in the variational principle of Deville et al. [10]. A generic existence result in the calculus of variations without convexity assumptions was then obtained as a realization of this variational principle. It was also shown in [12] that some other generic well-posedness results in optimization theory known in the literature and their modi-cations are obtained as a realization of this variational principle. Note that the generic existence result in [12] was established for variational problems but not for optimal control problems and that the topologies in the spaces of integrands in [27,12] are diHerent. In [26], we suggested a modi-cation of the variational principle in [12] and applied it to classes of optimal control problems with various topologies in the corresponding spaces of integrands. As a realization of this principle we established a generic existence result for a class of optimal control problems in which constraint maps are also subject to variations as well as the cost functions [26]. In [29], this variational principle was applied to obtain a generic existence result for a class of variational problems in which the values at the end points are also subject to variations as well as the integrands. More precisely, in [29] we established a generic existence result for a class of variational problems (without convexity assumptions) over functions with values in a Banach space E which is identi-ed with the corresponding complete metric space of pairs (f; (1 ; 2 )) (where f is an integrand satisfying the Cesari growth condition and 1 ; 2 ∈ E are the values at the end points) denoted by A and endowed with some natural topology. We
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
3
showed that for a generic (f; (1 ; 2 )) ∈ A the corresponding variational problem has a unique solution. In this paper we extend the main result in [29]. We consider the set of all pairs (f; (1 ; 2 )) ∈ A for which the corresponding variational problem is well posed and show that the complement of this set is not only of the -rst category but also a -porous set. Note that this result is also an extension of the main result in [28] where we studied the spaces of integrands introduced in [26]. For each of these spaces we considered the set of all integrands for which the corresponding optimal control problem is well posed and showed that its complement is -porous. In the present paper, we obtain an extension of the main result in [28] to a class of variational problems in which the values at the end points are also subject to variations as well as the. integrands. We obtain our main result as a realization of a general variational principle which is established in Section 2. Before we continue we recall the concept of porosity [3,9,11,20,23]. Let (Y; d) be a complete metric space. For each y ∈ Y and each r ¿ 0 set Bd (y; r) = {z ∈ Y : d(z; y) 6 r}. A subset E ⊂ Y is called porous in (Y; d) if there exist ∈ (0; 1] and r0 ¿ 0 such that for each x ∈ Y and each r ∈ (0; r0 ] there exists y ∈ Y for which Bd (y; r) ⊂ Bd (x; r) \ E:
(1.1)
A subset of the space Y is called -porous in (Y; d) if it is a countable union of porous sets in (Y; d). It is known that in the above de-nition of porosity, the point x can be assumed to belong to E. Namely, the following result is true. Proposition 1.1. A subset E ⊂ Y is porous in (Y; d) if and only if the following properly holds: C(i) There exist ∈ (0; 1] and r0 ¿ 0 such that for each x ∈ E and each r ∈ (0; r0 ] there exists y ∈ Y for which (0.1) is valid. Other notions of porosity have been used in the literature [3,23]. We use the rather strong notion which appears in [9,11,20]. Since porous sets are nowhere dense, all -porous sets are of the -rst category. If Y is a -nite dimensional Euclidean space, then -porous sets are of Lebesgue measure 0. In fact, the class of -porous sets in such a space is much smaller than the. class of sets which have measure 0 and are of the -rst category. Also, every complete metric space without isolated points contains a closed nowhere dense set which is not -porous [24]. To point out the diHerence between porous and nowhere dense sets note that if E ⊂ Y is nowhere dense, y ∈ Y and r ¿ 0, then there is a point z ∈ Y and a number s ¿ 0 such that Bd (z; s) ⊂ Bd (y; r) \ E. If, however, E is also porous, then for small enough r we can choose s = r, where ∈ (0; 1) is a constant which depends only on E. Finally, we prove the following simple proposition. It shows that porosity is equivalent to another property which is easier to verify.
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
Proposition 1.2. A subset E ⊂ Y is porous set in (Y; d) if and only if the following property holds: C(ii) There exist ∈ (0; 1] and r0 ¿ 0 such that for each x ∈ E and each r ∈ (0; r0 ] there exists y ∈ Y for which d(x; y) 6 r;
Bd (y; r) ∩ E = ∅:
Proof. Assume that C(ii) holds with ∈ (0; 1] and r0 ¿ 0. By Proposition 1.1; in order to prove the proposition it is suKcient to show that C(i) holds. Let x ∈ E and r ∈ (0; r0 ]. By C(ii) there exists y ∈ Y such that d(x; y) 6 r=4;
Bd (y; r=4) ∩ E = ∅:
Then Bd (y; r=4) ⊂ Bd (x; r) \ E. Therefore, property C(i) holds and Proposition 1.2 is proved.
2. Well posedness of optimization problems We use the convention that ∞=∞=1 and ∞−∞=0. For each function f : X → R1 ∪ {∞}, where X is nonempty, we set inf (f) = inf {f(x): x ∈ X }. We consider a metric space (X; ) and a complete metric space (A; d). Each of these spaces is equipped with the topology generated by its metric. We assume that with every a ∈ A a lower semicontinuous function fa on X is associated with values in R1 ∪ {∞} = (−∞; ∞] and that fa is not identically in-nity. The following property was introduced in [12]. Let ∈ A. We say that the minimization problem for fa on (X; ) is strongly well posed with respect to (A; d) if inf (fa ) is -nite and attained at a unique point xa ∈ X and the following assertion holds: For each ¿ 0 there exist a neighborhood V of a in (A; d) and ¿ 0 such that for each b ∈ V; inf (fb ) is -nite and if z ∈ X satis-es fb (z) 6 inf (fb )+, then (z; xa ) 6 and |fb (z) − fa (xa )| 6 . (In a slightly diHerent setting a similar property was introduced in [30].) In our study we use the following hypotheses about the functions. (H) If a ∈ A; inf (fa ) is -nite, {xi }∞ i=1 ⊂ X is a Cauchy sequence and the sequence is bounded, then the sequence {x n }∞ {fa (x n )}∞ n=1 n=1 converges in X . Clearly if (X; ) is complete, then (H) is true. Note that for classes of variational problems considered in this paper the space (X; ) is not complete. Fortunately, instead of the completeness assumption we can use (H) and this hypothesis holds for spaces of integrands which satisfy the Cesari growth condition. For each integer n ¿ 1 denote by A(n) the set of all a ∈ A which have the following property: (P1) There exist xN ∈ X; ¿ 0; ∈ R1 and a neighborhood U of a in (A; d) such that for each b ∈ U; inf (fb ) is -nite and if x ∈ X satis-es fb (x) 6 inf (fb ) + , then (x; x) N 6 1=n and |fb (x) − | 6 1=n.
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
5
∞ Proposition 2.1. Let a ∈ n=1 A(n) . Then the minimization problem for fa on (X; ) is strongly well posed with respect to (A; d). Proof. By property (P1) for each integer n ¿ 1 there exist x n ∈ X; n ¿ 0; n ∈ R1 and a neighborhood Un of a in (A; d) such that the following property holds: (P2) For each b ∈ Un ; inf (fb ) is -nite and if x ∈ X satis-es fb (x) 6 inf (fb ) + n , then (x; x n ) 6 1=n and |fb (x) − n | 6 1=n. Assume that {zi }∞ i=1 ⊂ X and lim fa (zi ) = inf (fa ):
(2.1)
i→∞
Fix an integer n ¿ 1. It follows from property (P2) and (2.1) that for all suKciently large natural numbers i, (zi ; x n ) 6 1=n;
|fa (zi ) − n | 6 1=n:
(2.2)
Since n is an arbitrary natural number we conclude that {zi }∞ i=1 is a Cauchy sequence. By (H) there exists xa = limi→∞ zi . As fa is lower semicontinuous, we have fa (xa ) = inf (fa ). Clearly xa is a unique minimizer of fa for otherwise we would be able to construct a nonconvergent sequence {zi }∞ i=1 . (2.2) implies that for all natural numbers n, (xa ; x n ) 6 1=n;
|fa (xa ) − n | 6 1=n:
(2.3)
Let ¿ 0. Choose a natural number n ¿ 4=. Assume that b ∈ Un ; x ∈ X and fb (x) 6 inf (fb ) + n . By property (P2), (x; x n ) 6 1=n;
|fb (x) − n | 6 1=n:
Combined with (2.3) these inequalities imply that (x; xa ) 6 2=n 6 =2;
|fb (x) − fa (xa )| 6 2=n 6 =2:
This completes the proof of Proposition 2.1. 3. Variational principle We will obtain our main result as a realization of a variational principle which will be introduced in this section. We consider a metric space (X; ) and complete metric spaces (A1 ; d1 ) and (A2 ; d2 ). The set A = A1 × A2 is equipped with a metric d de-ned by d((a1 ; a2 ); (b1 ; b2 )) = d1 (a1 ; b1 ) + d2 (a2 ; b2 );
(a1 ; a2 ); (b1 ; b2 ) ∈ A:
(3.1)
Clearly (A; d) is a complete metric space. We assume that with every a ∈ A1 a lower semicontinuous function a on X is associated with values in R1 ∪ {∞} and that a is not identically in-nity. We also assume that with every a ∈ A2 a nonempty subset Sa ⊂ X is associated. For each a = (a1 ; a2 ) ∈ A = A1 × A2 de-ne fa (x) = a1 (x);
x ∈ S a2 ;
fa (x) = ∞;
x ∈ X \ S a2 :
(3.2)
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
Fix ∈ A2 . For each r ¿ 0 set A2(r) = {a ∈ A2 : d2 ( ; a) 6 r}:
(3.3)
In our study we use the following basic hypotheses about the functions. (A1) For each a ∈ A the function fa is lower semicontinuous on X and inf (fa ) is -nite. (A2). There exist ∈ (0; 1] and r0 ¿ 0 such that for each r ∈ (0; r0 ], each a; b ∈ A1 satisfying d1 (a; b) 6 r and each x ∈ X , the inequality | a (x) − b (x)| 6 r is valid. (A3) For each ¿ 0 there exist numbers ¿ 0 and r0 ¿ 0 such that the following property holds: For each a = (a1 ; a2 ) ∈ A = A1 × A2 and each r ∈ (0; r0 ] there exist aN = (aN1 ; a2 ) ∈ A and xN ∈ X such that d1 (a1 ; aN1 ) 6 r;
xN ∈ Sa2 ;
(3.4)
aN1 (x) N = faN(x) N 6 inf (faN) + r
(3.5)
and that for each x ∈ Sa2 which satis-es faN(x) 6 inf (faN) + r
(3.6)
the inequality (x; x) N 6 holds. ⊂ A1 and positive (A4) For each pair of integers i; j ¿ 1 there exist a set A(ij) 1 numbers rNi ; rNij ; ij such that ∞ A1 = A(ij) for all integers i ¿ 1 (3.7) 1 j=1
and for each pair of integers i; j ¿ 1 the following property holds: If a1 ∈ A1 ;
inf {d1 (a1 ; ): ∈ A(ij) 1 } ¡ rNi ;
a2 ∈ A2(i) ;
x ∈ S a2 ;
r ∈ (0; rNij ];
b2 ∈ A2 ;
f(a1 ;a2 ) (x) 6 inf (f(a1 ;a2 ) ) + 1; d2 (b2 ; a2 ) 6 r;
(3.8) (3.9) (3.10)
then there exists xN ∈ Sb2 such that (x; x) N 6r
ij ;
| a1 (x) N − a1 (x)| 6 r
ij :
(3.11)
Theorem 3.1 (Variational principle). Assume that (A1)–(A4) and (H) hold. Then there exists a set B ⊂ A such that A \ B is -porous in (A; d) and that for each a ∈ B the minimization problem for fa on (X; ) is strongly well posed with respect to (A; d). Proof. We recall that A(n) is the set of all a ∈ A which have the property (P1) ∞ (n = 1; 2; : : :). Set B = n=1 A(n) . By Proposition 2.1 in order to prove the theorem it is suKcient to show that the set A \ A(n) is -porous in (A; d) for all integers n ¿ 1.
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
7
Let n ¿ 1 be an integer. We will show that A \ A(n) is -porous in (A; d). Since ∞ A \ A(n) = [(A \ A(n) ) ∩ (A1 × A2(i) )] i=1
we need only to show that the set (A \ A(n) ) ∩ (A1 × A2(i) ) is -porous in (A; d) for all integers i ¿ 1. For each pair of natural numbers i; j let a subset A(ij) 1 ⊂ A1 and positive numbers rNi ; rNij ; ij be as guaranteed by (A4). Let i ¿ 1 be an integer. By (A4), ∞ (i) (n) [(A \ A(n) ) ∩ (A1((i+1) j) × A2(i) )]: (A \ A ) ∩ (A1 × A2 ) = j=1
Therefore, in order to prove the theorem it is suKcient to show that for each integer j ¿ 1 the set Eij :=(A \ A(n) ) ∩ (A1((i+1) j) × A2(i) ) is porous in (A; d). Fix a natural number j ¿ 1. There exist ¿ 0 and r0 ¿ 0 such that (A3) holds with =(2n)−1 . There exist N ∈ (0; 1] and rN ∈ (0; 1] such that (A2) is valid with = ; N r0 = r. ˜ We choose positive numbers ∗ ; r∗ such that r∗ ¡ min{2−1 ; r0 ; 4−1 rNi+1 ; 4−1 rN(i+1) j ; (4 (8n
(i+1) j )
−1
; (8n)−1 (4
˜ =32; ˜ (24 ∗ ¡ min{a=2;
(i+1) j
(i+1) j )
(i+1) j )
−1
+ 2 + )−1 };
−1
}:
; 2−1 ˜r; ˜ (3.13) (3.14)
Let a = (a1 ; a2 ) ∈ Eij ;
r ∈ (0; r∗ ]:
(3.15)
By the de-nition of ; r0 , (A3), (3.13) and (3.15), there exist aN = (aN1 ; a2 ) ∈ A and xN ∈ X such that d1 (a1 ; aN1 ) 6 r=2;
xN ∈ Sa2 ;
aN1 (x) N = faN(x) N 6 inf (faN) + r=2
(3.16)
and the following property holds: N 6 (2n)−1 (P3) For each x ∈ Sa2 satisfying faN(x) 6 inf (faN)+r=2 the inequality (x; x) is valid. Clearly d(a; a) N = d1 (a1 ; aN1 ) 6 r=2: We will show that {b ∈ A: d(b; a) N 6 ∗ r} ⊂ A(n) ⊂ A \ Eij
(3.17)
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(see (3.12)). Let b = (b1 ; b2 ) ∈ A and d1 (b1 ; aN1 ) + d2 (b2 ; a2 ) = d(b; a) N 6 2∗ r:
(3.18)
By (3.3), (3.12) – (3.15), and (3.18) a1 ∈ A1(i+1) j ; a2 ∈ A2(i) ; b2 ∈ A2(i+1) :
(3.19)
Now consider the pairs (b1 ; b2 ); (b1 ; a2 ) ∈ A. By (3.13) – (3.16), (3.18), and (3.19) inf {d1 (b1 ; h): h ∈ A1(i+1) j } 6 d1 (b1 ; a1 ) 6 d1 (b1 ; aN1 ) + d1 (aN1 ; a1 ) 6 2∗ r + r=2 ¡ 3r∗ ¡ rNi+1 :
(3.20)
By (3.13) – (3.15), and (3.18) d2 (a2 ; b2 ) 6 2∗ r 6 r 6 r∗ ¡ rN(i+1) j :
(3.21)
Assume now that h1 ; h2 ∈ A2 and either h1 = a2 ; h2 = b2
or
h1 = b2 ; h2 = a2 :
(3.22)
It follows from the de-nition of rNi+1 ; rN(i+1) j ; (i+1) j , (A4), (3.19) – (3.22) that the following property holds: (P4) If x ∈ X satis-es f(b1 ;h1 ) (x) 6 inf (f(b1 ;h1 ) ) + 1, then there exists x˜ ∈ Sh2 such that (x; x) ˜ 6 2∗ r
(i+1) j ;
| b1 (x) ˜ − b1 (x)| 6 2∗ r
(i+1) j :
(3.23)
Since inf (f(b1 ;hi ) ) = inf {f(b1 ;hi ) (x): x ∈ Shi ; f(b1 ;hi ) (x) 6 inf (f(b1 ;hi ) ) + 1};
i = 1; 2;
property (P4) implies that |inf (fb1 ;a2 ) − inf (fb1 ;b2 )| 6 2∗ r
(i+1) j :
(3.24)
Let x ∈ X and fb (x) 6 inf (fb ) + 2∗ r
(i+1) j :
(3.25)
Since inf (fb ) is -nite we have x ∈ Sb2 . By property (P4), (3.25), (3.13) – (3.15) there exists x˜ ∈ Sa2
(3.26)
for which (3.23) is valid. It follows from (3.23) – (3.25) that b1 (x) ˜ 6 b1 (x) + 2∗ r
(i+1) j
6 inf (f(b1 ;a2 ) ) + 6∗ r
6 inf (fb ) + 4∗ r (i+1) j :
(i+1) j
(3.27)
(3.13) – (3.15), and (3.18) imply that d1 (aN1 ; b1 ) 6 2∗ r 6 2∗ r∗ 6 a˜r: ˜
(3.28)
By (3.28), the de-nition of ; ˜ r˜ and (A2), | aN1 (y) − b1 (y)| 6 2∗ r() ˜ −1
for all y ∈ X:
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
9
This inequality implies that | aN1 (x) ˜ − b1 (x)| ˜ 6 2∗ r() ˜ −1
(3.29)
|inf (f(aN1 ;a2 ) ) − inf (f(b1 ;a2 ) )| 6 2∗ r() ˜ −1 :
(3.30)
and It follows from (3.29), (3.30) and (3.27) that aN1 (x) ˜ 6 b1 (x) ˜ + 2∗ r(a) ˜ −1 6 inf (f(b1 ;a2 ) ) + 6∗ r
(i+1) j
+ 2∗ r(a) ˜ −1
6 inf (f(aN1 ;a2 ) ) + 4∗ r() ˜ −1 + 6∗ r
(i+1) j :
(3.31)
By (3.2), (3.26), (3.31) and (3.14), f(aN1 ;a2 ) (x) ˜ = aN1 (x) ˜ 6 inf (f(aN1 ;a2 ) ) + r=2: By this inequality, (3.26) and property (P3), (x; ˜ x) N 6 (2n)−1 . Combined with (3.23), (3.13) – (3.15) this inequality implies that (x; x) N 6 (x; x) ˜ + (x; ˜ x) N 6 2∗ r
(i+1) j
+ (2n)−1 ¡ n−1 :
(3.32)
It follows from (3.25), (3.24), (3.30) and (3.15) that |fb (x) − inf (faN)| 6 |fb (x) − inf (fb )| + |inf (f(b1 ;b2 ) ) − inf (f(b1 ;a2 ) )| + |inf (f(b1 ;a2 ) ) − inf (f(aN1 ;a2 ) )| 6 2∗ r
(i+1) j
+ 2∗ r
(i+1) j
+ 2∗ r() ˜ −1 6 (4n)−1 + (4n)−1
= (2n)−1 : Thus (x; x) N ¡ n−1 ;
|fb (x) − inf (faN)| ¡ n−1 :
(3.33)
We have shown that for each b ∈ A satisfying (3.18) and each x ∈ X satisfying (3.25), the inequality (3.33) hold. By the de-nition of A(n) and (3.12), {b ∈ A: d(b; a) N 6 ∗ r} ⊂ A(n) ⊂ A \ Eij :
(3.34)
Therefore, we have shown that for each a ∈ A and each number r satisfying (3.15) there exists aN ∈ A for which (3.17) and (3.34) are true. By Proposition 1.2 the set Eij is porous in (A; d). Theorem 3.1 is proved. 4. The main result For each function f : X → R1 , where (X; ) is a metric space, de-ne Lip(f) = sup{(f(x) − f(y))=(x; y): x; y ∈ X and x = y}: Let (E; · ) be a Banach space. We equip the space E with the metric dE (x; y) = x − y; x; y ∈ E.
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
Let −∞ ¡ #1 ¡ #2 ¡ ∞. We say that a function x : [#1 ; #2 ] → E is diHerentiable if there exists a Bochner integrable function u : [#1 ; #2 ] → E such that t u(s) ds; t ∈ [#1 ; #2 ]: (4.1) x(t) = x(#1 ) + #1
It is known that if x is diHerentiable, then (4.1) de-nes a unique Bochner integrable function u which is called the derivative of x and is denoted by x . We denote by mes(&) the Lebesgue measure of a Lebesgue measurable set & ⊂ R1 . Let −∞ ¡ T1 ¡ T2 ¡ ∞. Denote by X the set of all diHerentiable functions x : [T1 ; T2 ] → E. For the set X we consider the metric de-ned by (x1 ; x2 ) = inf { ¿ 0: mes{t ∈ [T1 ; T2 ]: x1 (t) − x2 (t) + x1 (t) − x2 (t) ¿ } 6 };
x1 ; x2 ∈ X:
(4.2)
Let f : [T1 ; T2 ] × E × E → R1 . For each t ∈ [T1 ; T2 ] de-ne a function ft : E × E → R1 by ft (x; y) = f(t; x; y);
x; y ∈ E:
(4.3) 1
Denote by M the set of all functions f : [T1 ; T2 ] × E × E → R with the following properties: (i) f is measurable with respect to the -algebra generated by products of Lebesgue measurable subsets of [T1 ; T2 ] and Borel subsets of E × E; (ii) ft is lower semicontinuous for almost every t ∈ [T1 ; T2 ]; (iii) for each ¿ 0 there exists an integrable scalar function (t) 6 0; t ∈ [T1 ; T2 ] such that u 6
(t)
+ f(t; x; u)
for all (t; x; u) ∈ [T1 ; T2 ] × E × E;
(iv) for each M ¿ 0 there exists cM ¿ 0 such that for almost every t ∈ [T1 ; T2 ], sup{f(t; x; u): x; u ∈ E; x; u 6 M } 6 cM ; (v) for each M ¿ 0 there exist dM ¿ 0; rM ¿ 0; cM 1 ¿ 0 and cM 2 ¿ 0 such that for each t ∈ [T1 ; T2 ], each r ∈ (0; rM ], each x1 ; x2 ∈ E satisfying x1 ; x2 6 M;
x1 − x2 6 r
and each u ∈ E satisfying u ¿ dM , |f(t; x1 ; u) − f(t; x2 ; u)| 6 cM 1 r min{f(t; x1 ; u); f(t; x2 ; u)} + cM 2 r; (vi) for each M ¿ 0 there exists an integrable scalar function M (t) ¿ 0; t ∈ [T1 ; T2 ] such that M (t) ¿ Lip(gt ); t ∈ [T1 ; T2 ] where gt is the restriction of ft to the set {(x; y) ∈ E × E: x; y 6 M } for all t ∈ [T1 ; T2 ]. The growth condition in (iii) was proposed by Cesari [6] and its equivalents and modi-cations are rather common in the literature. It follows from property (i) that for any f ∈ M and any x ∈ X the function f(t; x(t); x (t)); t ∈ [T1 ; T2 ] is measurable. Clearly if f ∈ M and , ¿ 0, then ,f ∈ M and if f1 ; f2 ∈ M, then f1 + f2 ∈ M.
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
11
Remark 4.1. Assume that f ∈ M and g : [T1 ; T2 ] × E → R1 is a measurable function with respect to the -algebra generated by products of Lebesgue measurable subsets of [T1 ; T2 ] and Borel subsets of E such that 0 ¡ inf {g(t; u): (t; u) ∈ [T1 ; T2 ] × E} 6 sup{g(t; u): (t; u) ∈ [T1 ; T2 ] × E} ¡ ∞; for almost every t ∈ [T1 ; T2 ] the function (x; u) → f(t; x; u)g(t; u); (x; u) ∈ E ×E is lower semicontinuous; for each M ¿ 0 there is M ¿ 0 such that for each t ∈ [T1 ; T2 ] and each x1 ; x2 ; u1 ; u2 ∈ E satisfying xi ; ui 6 M; i = 1; 2 the following inequality holds: |g(t; u1 ) − g(t; u2 )| 6
M u1
− u2 :
We can show in a straightforward manner that the function (t; x; u) → f(t; x; u)g(t; u);
(t; x; u) ∈ [T1 ; T2 ] × E × E
belongs to M. Remark 4.2. Assume that a function f : [T1 ; T2 ] × E × E → R1 satis-es f(t; x; u) ¿ (u) − c0 ;
(t; x; u) ∈ [T1 ; T2 ] × E × E;
(4.4)
: [0; ∞) → [0; ∞) is an increasing function such that where c0 is a constant and limu→∞ (u)=u = ∞. Then we can verify in a straightforward manner that the property (iii) holds and the property (v) is equivalent to the following property: For each M ¿ 0 there exist dM ¿ 0; rM ¿ 0; cM ¿ 0 such that for each t ∈ [T1 ; T2 ], each r ∈ (0; rM ], each x1 ; x2 ∈ E which satisfy x1 ; x2 6 M and x1 − x2 6 r, and each u ∈ E satisfying u ¿ dM the following inequality holds: |f(t; x1 ; u) − f(t; x2 ; u)| 6 cM r min{f(t; x1 ; u); f(t; x2 ; u)}: Remark 4.3. Let f ∈ M satisfy (4.4) with a constant c0 and an increasing function : [0; ∞) → [0; ∞) such that limu→∞ (u)=u = ∞. Assume that g : [T1 ; T2 ] × E × E → R1 is a measurable function with respect to the -algebra generated by products of Lebesgue measurable subsets of [T1 ; T2 ] and Borel subsets of E × E with the following properties: for almost every t ∈ [T1 ; T2 ] the function (x; u) → f(t; x; u) + g(t; x; u); (x; u) ∈ E × E is lower semicontinuous; inf {g(t; x; u): t ∈ [T1 ; T2 ]; x; u ∈ E} ¿ 0; f is Lipschitzian on all bounded subsets of [T1 ; T2 ] × E × E; sup{Lip(ftu ): t ∈ [T1 ; T2 ]; u ∈ E} ¡ ∞ where ftu : E → R1 is de-ned by ftu (x) = f(t; x; u); x ∈ E for all (t; u) ∈ [T1 ; T2 ] × E. Using Remark 4.2 we can show in a straightforward manner that the function (t; x; u) → f(t; x; u) + g(t; x; u); (t; x; u) ∈ [T1 ; T2 ] × E × E also belongs to M.
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
It is an elementary exercise to show that a function f=f(t; x; u) ∈ C 1 ([T1 ; T2 ]×E×E) belongs to M if (iii) is true and the following conditions hold: for each M ¿ 0, sup{9f=9x(t; x; u) + 9f=9u(t; x; u): t ∈ [T1 ; T2 ]; x; u ∈ E and x; u 6 M } ¡ ∞; there exist increasing functions : [0; ∞) → [0; ∞) and that for each (t; x; u) ∈ [T1 ; T2 ] × E × E, max{9f=9x(t; x; u); 9f=9u(t; x; u)} 6
0 (x)
0
: [0; ∞) → [0; ∞) such
(u)
and (u) 6 f(t; x; u). For each f; g ∈ M de-ne d˜ 1 (f; g) = sup{|f(t; x; u) − g(t; x; u)|: (t; x; u) ∈ [T1 ; T2 ] × E × E} + sup{Lip(ft − gt ): t ∈ [T1 ; T2 ]}; d1 (f; g) = d˜ 1 (f; g)(d˜ 1 (f; g) + 1)−1 : Clearly (M; d1 ) is a complete metric space. Denote by Ml (respectively, Mc ) the set of all lower semicontinuous (respectively, continuous) functions f : [T1 ; T2 ] × E × E → R1 in M. Clearly Ml and Mc are closed subsets of the metric space (M; d1 ). We consider the complete metric spaces (Ml ; d1 ) and (Mc ; d1 ). For each f ∈ M we de-ne I f : X → R1 ∪ {∞} by T2 f f(t; x(t); x (t)) dt; x ∈ X: (4.5) I (x) = T1
We study the variational problem I f (x) → min;
x ∈ X;
x(Ti ) = i ;
i = 1; 2;
where f ∈ M and i ∈ E; i = 1; 2. Set A2 = E × E and de-ne d2 ((x1 ; y1 ); (x2 ; y2 )) = x1 − x2 + y1 − y2 ;
(xi ; yi ) ∈ E × E;
i = 1; 2:
We consider the complete metric space (A; d) where A = A1 × A2 ; A1 is either M or Ml or Mc and d((a1 ; a2 ); (b1 ; b2 )) = d1 (a1 ; b1 ) + d2 (a2 ; b2 );
(a1 ; a2 ); (b1 ; b2 ) ∈ A:
For each a = (f; (1 ; 2 )) ∈ A1 × A2 we de-ne Ja : X → R1 ∪ {∞} by Ja (x) = I f (x) if x(Ti ) = i ;
i = 1; 2;
otherwise Ja (x) = ∞:
(4.6)
By properties (iii) and (iv), inf (Ja ) is -nite for all a ∈ A. We will show that the functional Ja is lower semicontinuous for any a ∈ A (see Propositions 5.1 and 5.2). Now we are ready to state our main result. Theorem 4.1. There exists a set B ⊂ A such that A \ B is a -porous set in the space (A; d) and that for each a ∈ B the minimization problem for Ja on (X; ) is strongly well posed with respect to (A; d).
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
13
The proof of Theorem 4.1 consists in verifying that hypotheses (A1) – (A4) and (H) hold for the space (A; d). 5. Preliminary results for hypotheses (A1), (A3) and (H) For each f ∈ M and each 1 ; 2 ∈ E set 0(f; 1 ; 2 ) = inf (J(f; (1 ;2 )) ):
(5.1)
For the proof of the following proposition see [29, Proposition 3.1]. Proposition 5.1. Let f ∈ M; x ∈ X; {xi }∞ i=1 ⊂ X and let (xi ; x) → 0 as i → ∞. Then I f (x) 6 lim inf i→∞ I f (xi ). The following proposition is an auxiliary result for hypothesis (H). Its proof is analogous to the proof of Proposition 4.2 in [26]. Proposition 5.2. Assume that f ∈ M; {xi }∞ i=1 ⊂ X is a Cauchy sequence and that the sequence {I f (xi )}∞ i=1 is bounded. Then there is x∗ ∈ X such that (xi ; x∗ ) → 0 as i → ∞ and moreover; xi (t) → x∗ (t) as i → ∞ uniformly on [T1 ; T2 ]. Analogous to Lemma 5.1 of [28] we can prove the following result which implies (A3). Proposition 5.3. For each 1 ∈ (0; 1) there exists (1) ∈ (0; 1) such that for each f ∈ M; each 1 ; 2 ∈ E and each r ∈ (0; 1] there exist a continuous function h : [T1 ; T2 ] × E × E → R1 which satis6es 0 6 h(t; x; u) 6 r=2
for all (t; x; u) ∈ [T1 ; T2 ] × E × E;
Lip(ht ) 6 r=2 for all t ∈ [T1 ; T2 ] and xN ∈ X satisfying x(T N i ) = i ; i = 1; 2 such that for the function fN ∈ M de6ned by N x; u) = f(t; x; u) + h(t; x; u); (t; x; u) ∈ [T1 ; T2 ] × E × E f(t; the following properties hold: N N 1 ; 2 ) + (1)r; I f (x) N 6 0(f; for each y ∈ X satisfying y(Ti ) = i ;
i = 1; 2;
N ˜ 1 ; 2 ) + 2(1)r I f (y) 6 0(f;
the inequality (x; N y) 6 1 is valid. Moreover, h is the sum of two functions, one of them depending only on (t; x) while the other depending only on (t; u). For each 1 ; 2 ∈ E de-ne a function x1 2 : [T1 ; T2 ] → E by x1 2 (t) = 1 + (t − T1 )(T2 − T1 )−1 (2 − 1 );
t ∈ [T1 ; T2 ]:
(5.2)
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
Clearly for each 1 ; 2 ∈ E x1 2 (Ti ) = i ;
i = 1; 2;
x1 2 ∈ X
(5.3)
and x 1 2 (t) = (T2 − T1 )−1 (2 − 1 );
t ∈ [T1 ; T2 ]:
Lemma 5.1. Let f ∈ M and let cM ¿ 0 be as guaranteed by property (iv) for any M ¿ 0. Let M0 ¿ 0 and M1 =M0 (3+2(T2 −T1 )−1 ). Then for each 1 ; 2 ∈ E satisfying 1 ; 2 6 M0
(5.4)
the inequality I f (x1 2 ) 6 cM1 (T2 − T1 ) holds. Proof. Let 1 ; 2 ∈ E satisfy (5.4). Then by (5.2) – (5.4) for all t ∈ [T1 ; T2 ]; x1 2 (t) 6 3M0 6 M1 ;
x 1 2 (t) 6 2M0 (T2 − T1 )−1 6 M1 :
It follows from these inequalities and property (iv) that for almost every t ∈ [T1 ; T2 ]; f(t; x1 ;2 (t); x 1 ;2 (t)) 6 cM1 . This completes the proof of Lemma 5.1. Lemma 5.2. Let f ∈ M and M0 ¿ 0. Then there exists M ¿ 0 such that for each 1 ; 2 ∈ E satisfying; 1 ; 2 6 M0 ; the inequality |0(f; 1 ; 2 )| 6 M is valid. Proof. By Lemma 5.1 there exists M1 ¿ 0 such that for each 1 ; 2 ∈ E satisfying 1 ; 2 6 M0 the inequality 0(f; 1 ; 2 ) 6 M1 holds. By property (iii) there exists an integrable scalar function (t) ¿ 0; t ∈ [T1 ; T2 ] such that f(t; x; u) ¿ − (t)
for all (t; x; u) ∈ [T1 ; T2 ] × E × E:
Therefore; inf {0(f; 1 ; 2 ): 1 ; 2 ∈ E} = inf (I f ) ¿ −
T2
T1
(t) dt:
This completes the proof of Lemma 5.2. Lemma 5.3. Let f ∈ M and M0 ; M1 ¿ 0. Then there exists M2 ¿ 0 such that for each 1 ; 2 ∈ E satisfying i 6 M0 ;
i = 1; 2
(5.5)
and each x ∈ X which satis6es x(Ti ) = i ;
i = 1; 2;
and
I f (x) 6 0(f; 1 ; 2 ) + M1
(5.6)
the following inequality holds: x(t) 6 M2 ;
t ∈ [T1 ; T2 ]:
(5.7)
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
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Proof. By Lemma 5.2 there exists c ¿ 0 such that for each 1 ; 2 ∈ E satisfying (5.5) the inequality |0(f; 1 ; 2 )| 6 c
(5.8)
is valid. By property (iii) (see the de-nition of M) there exists an integrable scalar function (t) ¿ 0; t ∈ [T1 ; T2 ] such that u 6 (t) + f(t; x; u) Set
for all (t; x; u) ∈ [T1 ; T2 ] × E × E:
M2 = c + M0 + M1 +
T2
T1
(5.9)
(t) dt:
(5.10)
Let 1 ; 2 ∈ E satisfy (5.5) and let x ∈ X satisfy (5.6). It follows from (5.5) and the de-nition of c; that (5.8) holds. (5.8) and (5.6) imply that I f (x) 6 M1 + c. Combined with (5.9) this inequality implies that T2 T2 T2 x (t) dt 6 (t) dt + f(t; x(t); x (t)) dt T1
T1
6 M1 + c +
T1
T2
T1
(t) dt:
By these inequalities; (5.5); (5.6) and (5.10) for each t ∈ [T1 ; T2 ]; T2 T2 x(t) 6 x(T1 ) + x (t) dt 6 M0 + M1 + c + (t) dt = M2 : T1
T1
Lemma 5.3 is proved. 6. Auxiliary result for hypothesis (A4) Let f ∈ M; M0 ; M1 ¿ 0 and let M2 ¿ M0 be as guaranteed by Lemma 5.3. Namely if 1 ; 2 ∈ E satisfy (5:5) and if x ∈ X satis-es (5:6); then (5:7) holds:
(6.1)
By Lemma 5.2 there exists M3 ¿ 0 such that |0(f; 1 ; 2 )| 6 M3
for all 1 ; 2 ∈ E satisfying i 6 M0 + 1;
i = 1; 2: (6.2)
By properly (iii) (see the de-nition of M) there exists an integrable function t ∈ [T1 ; T2 ] such that u 6
1 (t)
+ f(t; x; u)
for all (t; x; u) ∈ [T1 ; T2 ] × E × E:
1 (t) ¿ 0;
(6.3)
By property (v) (see the de-nition of M) there exist d0 ¿ M2 + 1;
r0 ∈ (0; 1]; cN1 ; cN2 ¿ 0
such that the following property holds:
(6.4)
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
For each t ∈ [T1 ; T2 ], each r ∈ (0; r0 ], each x1 ; x2 ∈ E which satisfy x1 − x2 6 r
x1 ; x2 6 M2 + 1;
(6.5)
and each u ∈ E satisfying u ¿ d0 , we have |f(t; x1 ; u) − f(t; x2 ; u)| 6 cN1 r min{f(t; x1 ; u); f(t; x2 ; u)} + cN2 r: Choose positive numbers d∗ ¿ 4(d0 + 1) 1 +
(6.6)
T2
1 (t) dt
T1
+ M3 + M2 (min{1; T2 − T1 })−1 ;
r∗ ∈ (0; 12−1 r0 min{1; T2 − T1 }):
(6.7) (6.8)
For each t ∈ [T1 ; T2 ] we denote by gt the restriction of ft to the set {(x; y) ∈ E × E: x; y 6 d∗ + 1}. By property (vi) (see the de-nition of M) there exists a scalar integrable function (t) ¿ 0; t ∈ [T1 ; T2 ] such that (t) ¿ Lip(gt );
t ∈ [T1 ; T2 ]:
(6.9)
Choose a number T2 (t) dt: ∗¿
(6.10)
T1
Lemma 6.1. Assume that 1 ; 2 ∈ E satisfy i 6 M0 ;
i = 1; 2;
(6.11)
x ∈ X satisfy x(Ti ) = i ;
i = 1; 2;
I f (x) 6 0(f; 1 ; 2 ) + M1 ;
(6.12)
r ∈ (0; r∗ ] and Ni ; N2 ∈ X satisfy i − Ni 6 r;
i = 1; 2:
(6.13)
Then there exists xN ∈ X such that x(T N i ) = Ni ;
i = 1; 2;
x(t) N − x(t) 6 3r;
(6.14) t ∈ [T1 ; T2 ];
xN (t) − x (t) 6 4r(T2 − T1 )−1
for almost every t ∈ [T1 ; T2 ]
(6.15)
and |I f (x) − I f (x)| N 6 r [3cN2 (T2 − T1 ) + 3cN1 (M1 + M3 ) T2 −1 + 3cN1 : (t) dt + (3 + 4(T − T )) 1 ∗ 2 1 T1
(6.16)
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
17
Proof. By (6.11); (6.12) and the de-nition of M2 (see (6.1)); x(t) 6 M2 ;
t ∈ [T1 ; T2 ]:
(6.17)
Set; &1 = {t ∈ [T1 ; T2 ]: x (t) ¿ d∗ };
&2 = [T1 ; T2 ] \ &1 :
(6.18)
We will show that mes(&1 ) 6 (T2 − T1 )=4. It follows from (6.18); (6.3) (6.12); (6.2) and (6.11) that T2 T2 d∗ mes(&1 ) 6 x (t) dt 6 x (t) dt 6 1 (t) dt T1
&1
+ 6
T1 T2
T1
6
T2
T2
T1
T1
f(t; x(t); x (t)) dt
1 (t) dt
+ 0(f; 1 ; 2 ) + M1
1 (t) dt
+ M 3 + M1 :
Combined with (6.7) these inequalities imply that T2 mes(&1 ) 6 d−1 (t) dt + M + M 1 3 1 6 (T2 − T1 )=4: ∗ T1
Thus mes(&1 ) 6 (T2 − T1 )=4;
mes(&2 ) ¿ 3(T2 − T1 )=4:
(6.19)
De-ne a function u : [T1 ; T2 ] → E by u(t) = x (t);
t ∈ &1 ;
u(t) = x (t) + (mes(&2 ))−1 [(N2 − N1 ) − (2 − 1 )];
(6.20) t ∈ &2 :
Clearly u is a Bochner integrable function. For t ∈ [T1 ; T2 ] set t x(t) N = N1 + u(t) dt: T1
(6.21)
(6.22)
Evidently xN ∈ X . It follows from (6.22) that x(T N i ) = Ni ; i = 1; 2. Thus (6.14) is valid. It follows from (6.19) – (6.22) and (6.13) that for almost every t ∈ [T1 ; T2 ] xN (t) = u(t)
(6.23)
and xN (t) − x (t) 6 (N2 − N1 ) − (2 − 1 )(mes(&2 ))−1 6 8r(3(T2 − T1 ))−1 :
(6.24)
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
(6.22); (6.23); (6.12); (6.13); (6.20) and (6.21) imply that for all t ∈ [T1 ; T2 ]; t t x(t) N − x(t) 6 N1 − 1 + xN (s) ds − x (s) ds 6r +
T1
t
T1
6r +
&2
T1
xN (s) − x (s) ds u(s) − x (s) ds 6 3r:
Combined with (6.24) these inequalities imply (6.15). Now we will show that (6.16) is valid. Let t ∈ &1 and (6.23) be valid. By (6.23), (6.20), (6.18) and (6.7), x (t) = xN (t);
x (t) ¿ d∗ ¿ d0 :
(6.25)
By (6.15), (6.8) and (6.4), x(t) N − x(t) 6 3r 6 3r∗ ¡ 4−1 r0 ¡ 4−1 :
(6.26)
(6.26) and (6.17) imply that x(t) 6 M2 ;
x(t) N 6 M2 + 1:
(6.27)
It follows from (6.25) – (6.27) and the de-nition of d0 ; r0 ; cN1 ; cN2 (see (6.4) – (6.6)) that |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| = |f(t; x(t); x (t)) − f(t; x(t); N x (t))| 6 3r cN1 min{f(t; x(t); x (t)); f(t; x(t); N x (t))} + 3r cN2 : Since these relations hold for almost every t ∈ &1 we obtain by (6.18), (6.12) and (6.3) that |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| dt &1
6 3r cN2 (T2 − T1 ) + 3r cN1
&1
f(t; x(t); x (t)) dt
f
= 3r cN2 (T2 − T1 ) + 3r cN1 I (x) −
&1
f(t; x(t); x (t)) dt
6 3r cN2 (T2 − T1 ) + 3r cN1 (0(f; 1 ; 2 ) + M1 ) − 3r cN1
&2
f(t; x(t); x (t)) dt
6 3r cN2 (T2 − T1 ) + 3r cN1 (0(f; 1 ; 2 ) + M1 ) − 3r cN1 x (t) dt − 1 (t) dt &2
&2
6 3r cN2 (T2 − T1 ) + 3r cN1 (0(f; 1 ; 2 ) + M1 ) + 3r cN1
T2
T1
1 (t) dt:
(6.28)
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
19
(6.28), (6.11) and (6.2) imply that |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| dt &1
6 3r cN2 (T2 − T1 ) + 3r cN1 M1 + M3 +
T2
1 (t) dt
T1
:
(6.29)
Now we will estimate &2 |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| dt. Let t ∈ &2 . By (6.23) and (6.15) we may assume that xN (t) = u(t);
xN (t) − x (t) 6 4r(T2 − T1 )−1 :
(6.30)
It follows from (6.30), (6.18), (6.8), (6.17), (6.15) and (6.7) that x (t) ¡ d∗ ;
xN (t) ¡ d∗ + 4r∗ (T2 − T1 )−1 6 d∗ + 1;
x(t) 6 M2 ¡ d∗ ;
x(t) N 6 x(t) + 3r 6 M2 + 3r∗ 6 M2 + 1 ¡ d∗ :
(6.31)
Recall that for each # ∈ [T1 ; T2 ]; g# is the restriction of f# to the set {(x; y) ∈ E × E: x; y 6 d∗ + 1}. It follows from (6.31), (6.15), (6.30) and (6.9) that |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| 6 Lip(gt )(x(t) − x(t) N + x (t) − xN (t)) 6 Lip(gt )(3r + 4r(T2 − T1 )−1 ) 6 r (t)(3 + 4(T2 − T1 )−1 ): Since these inequalities hold for almost every t ∈ &2 , (6.10) implies that −1 |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| dt 6 r(3 + 4(T2 − T1 ) ) (t) dt &2
&2
6 r(3 + 4(T2 − T1 )−1 ) 6r
∗ (3
T2
T1
(t) dt
+ 4(T2 − T1 )−1 ):
Combining these inequalities with (6.29) we obtain that T2 |f(t; x(t); x (t)) − f(t; x(t); N xN (t))| dt T1
6 r 3cN2 (T2 − T1 ) + 3cN1 M1 + M3 +
T2
T1
(t) dt + 1
−1 (3 + 4(T − T ) ) : ∗ 2 1
Therefore (6.16) is valid. Lemma 6.1 is proved. 7. Proof of Theorem 4.1 By Theorem 3.1 we need to show that (A1) – (A4) and (H) hold. It follows from Propositions 5.1 and 5.2 that for each a=(a1 ; a2 ) ∈ A the functionals Ja : X → R1 ∪{∞}
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A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
and I a1 : X → R1 ∪ {∞} are lower semicontinuous. Therefore (A1) holds. Evidently (A2) is valid. It is easy to see that (A3) follows from Proposition 5.3 and (H) follows from Proposition 5.2. In order to prove Theorem 4.1 we need only to show that (A4) holds. Set = 0 ∈ E × E. Then for any r ¿ 0, A2(r) = {(x; y) ∈ E × E: x + y 6 r}
(7.1)
(see (3.3)). For each f ∈ M, each G ⊂ M and each r ¿ 0 we de-ne d1 (f; G) = inf {d1 (f; g): g ∈ G};
(7.2)
B1 (f; r) = {g ∈ M: d1 (f; g) 6 r}:
(7.3)
Let i ¿ 1 be an integer. For each integer s ¿ i + 2 denote by Ms(i) the set of all f ∈ M such that for each h ∈ B1 (f; 1=2) the following properties hold: (P5) There exists a scalar integrable function 1 (t) ¿ 0; t ∈ [T1 ; T2 ] such that u 6 T2 1 (t) dt 6 s; 1 (t) + h(t; x; u) for all (t; x; u) ∈ [T1 ; T2 ] × E × E and T1 (P6) |0(h; 1 ; 2 )| 6 s for all 1 ; 2 ∈ E satisfying 1 ; 2 6 i + 1; (P7) For each 1 ; 2 ∈ E satisfying 1 ; 2 6 i + 1 and each x ∈ X satisfying x(T1 ) = 1 ;
x(T2 ) = 2 ;
I h (x) 6 0(h; 1 ; 2 ) + 4
the inequality x(t) 6 s holds for all t ∈ [T1 ; T2 ]. Clearly (i) Ms(i) ⊂ Ms+1
for all integers s ¿ i + 2:
(7.2)
By the de-nition of d1 and d˜ 1 (see Section 4), Lemmas 5.2 and 5.3, and property (iii) (see the de-nition of M), M=
∞
Ms(i) :
(7.3)
s=i+2
Fix an integer s ¿ i + 2 and consider the set Ms(i) . For each pair of integers k1 ; k2 (i) the set of all f ∈ Ms(i) such that for each satisfying k2 ¿ k1 ¿ s + 1 denote by Msk 1 k2 h ∈ B1 (f; 1=2) the following property holds: (P8) For each t ∈ [T1 ; T2 ], each r ∈ (0; k1−1 ], each x1 ; x2 ∈ E which satisfy x1 ; x2 6 s + 1 and x1 − x2 6 r and each u ∈ E satisfying u ¿ k1 the inequality |h(t; x1 ; u) − h(t; x2 ; u)| 6 k1 r min{h(t; x1 ; u); h(t; x2 ; u)} + k2 r holds. By property (v) (see the de-nition of M) and the de-nition of d1 and d˜ 1 (see Section 4), (i) Ms(i) = {Msk : k1 and k2 are integers satisfying k2 ¿ k1 ¿ s + 1}: (7.4) 1 k2
A.J. Zaslavski / Nonlinear Analysis 53 (2003) 1 – 22
21
(i) Fix a pair of integers k1 ; k2 satisfying k2 ¿ k1 ¿ s + 1 and consider the set Msk . 1 k2 De-ne
d(s; k1 ) = 4(k1 + 1)(1 + 3s) min{1; T2 − T1 }−1 + 2;
(7.5)
r(k1 ) = (16k1 )−1 min{1; T2 − T1 }:
(7.6)
(i) Msk 1 k2 q
(i) f ∈ Msk 1 k2
such that for each the set of all For each integer q ¿ 1 denote by h ∈ B1 (f; 1=2) the following property holds: T (P9) There exists an integrable scalar function (t) ¿ 0; t ∈ [T1 ; T2 ] such that T12 (t) dt ¡ q and (t) ¿ Lip(h˜t ); t ∈ [T1 ; T2 ] where h˜t is the restriction of ht to the set {(x; y) ∈ E × E: x; y 6 d(s; k1 )} for all t ∈ [T1 ; T2 ]. It follows from property (vi) (see the de-nition of M) and the de-nition of d1 and d˜ 1 (see Section 4) that ∞ (i) (i) Msk = Msk : (7.7) 1 k2 1 k2 q q=1
(i) . Lemma 6.1 and the properties Fix an integer q ¿ 1 and consider the set Msk 1 k2 q (P5) – (P9) imply the following lemma. Note that by (P5) – (P9), Lemma 6.1 can be applied to h (see the statement of Lemma 7.1) with M0 = i; M1 = 4; M2 = s (see (P7)); M3 = s (see (P6)); d0 = k1 ; r0 = k1−1 ; cN1 = k1 ; cN2 = k2 (see (P8)); d∗ = d(s; k1 ) − 1; (see (P5) and (7.5)); r∗ = r(k1 ) (see (7.6)); ∗ = q (see (P9)). (i) ; h ∈ B1 (f; 1=2); 1 ; 2 ∈ X satisfy 1 ; 2 6 i; Lemma 7.1. Assume that f ∈ Msk 1 k2 q x ∈ X satisfy x(Ti )=i ; i=1; 2 and I h (x) 6 0(h; 1 ; 2 )+4; r ∈ (0; r(k1 )] and N1 ; N2 ∈ X satisfy N1 − 1 ; N2 − 2 6 r. Then there exists xN ∈ X such that
x(T N j ) = Nj ;
j = 1; 2;
x(t) N − x(t) 6 3r;
xN (t) − x (t) 6 4r(T2 − T1 )−1
t ∈ [T1 ; T2 ];
for almost every t ∈ [T1 ; T2 ]
and |I h (x) − I h (x)| N 6 r[3(T2 − T1 )k2 + 3k1 (4 + 4s) + q(3 + 4(T2 − T1 )−1 )]: (i) Clearly the collection of all sets Msk where integers s; k1 ; k2 ; q satisfy s ¿ i + 1 k2 q 2; k2 ¿ k1 ¿ s + 1 and q ¿ 1, is countable. By (7.3), (7.4) and (7.7) the union of this collection is M. Then (A4) follows from Lemma 7.1. This completes the proof of Theorem 4.1.
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