T HE UN IV ER SIT Y OF MI C I GA N COLLEGE OF LITERATURE, SCIENCE, AND THE ARTS Department of Mathematics Technical Report No. 13 LAGRANGE PROBLEMS OF OPTIMAL CONTROL AND CONVEX SETS NOT CONTAINING ANY STRAIGHT LINE Lamberto Cesari ORA Pro j ect: 0::6 Submitted for: UNITED STATES AIR FORCE AIR FORCE OFFICE OF SCIENTIFIC RESEARCH GRANT NO. AFOSR-69-l662 ARLINGTO, VRGINIA. administered through: OFFICE OF RESEARCH ADMIlNISTRATION ANN ARBOR July 1970

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LAGRANGE PROBLEMS OF OPTIMAL CONTROL AND CONVEX SETS NOT CONTAINING ANY STRAIGHT LINE* Lamberto Cesari 1. Introduction In a previous paper [1] we proved that the usual sets Q(x) of Lagrange problems of optimal control if bounded below have property (Q) i.f, and only if, they satisfy conditions similar to those of "weak seminormality" for free problems of the calculus of variations (see (4.iii) beloW). In the present paper we prove that the same sets of Q(x) if bounded below have property (Q) and contain no straight line if, and only if, they satisfy conditions similar to those for "seminormality" for free problems (see (4.vii) below). These results will be used in successive papers in proving theorems of lower closure and existence theorems for optimal solutionso In Section 2 we study real-valued functions F(u), u E Q C Em, which are convex on a given fixed convex subset Q of E. Under suitable conventions we m prove that these functions grow "'more than linearly" as u +- o with u c Q if, and only if, their graph does not contain any straight lineo This statement, 9 which had been proved by Tonelli and Turner for Q Em (see [1] for references) requires a partially new proof in the present situation where Q is any convex subset of EM, and F is only convex in Q, and hence may not be even continuous *This research was partially supported by AFOSR Research Project No. 69-1662 at the University of Michigan~ 1

on the boundary of Q. The proof is based on a series of lemmas. The boundary points may actually present exceptional behavior as examples show. In Section 3 we study real-valued functions f (xu), u E Q(z) C E, which are convex in u on the convex subset Q(x) of E, and where Q(x) may depend on x. No continuity requirement is made on f0, but only a suitable property of lower semicontinuity is assumed. Finally, in Section 4, we apply the results to the typical sets Q(x) of Lagrange problems, by consistent use of the related realvalued functions T(z;x), z E Q(x) c E, which has been introduced in [1], and m' for which we proved there the properties which we require for f in the present paper. 2. Real-Valued Convex Functions on a Convex Set Given a real-valued function F(u) on a subset Q of E, we shall often use the expression "limit of F(u) as ul -u * with u C Q." This expression is clear in itself when Q is unbounded. Whenever Q is bounded, the limit above is taken to be + o. In the present section we shall prove the following (2.i) Theorem. If F(u), u E Q. is a real-valued convex function on the convex subset Q of E, then the following properties are equivalent: 1. There is a linear function w(u) = r + b ~ u, u E E, r real, b = (bl,...,b ) real, with F(u) > w(u) for all u E Q, and F(u)w(u) + + as lul + o, u e Q. 2. For no two points uo Q, u1 E E, u1 j o, it occurs that u + ku1 E Q for all A real, and

F(u) = 2 [F(u + kXu) + F(u - kUl)] for all X > o. (1) For Q = E the function F is necessarily continuous, and then this theorem n reduces to a well known statement proved by Tonelli for smooth functions and by Turner under sole hypotheses of continuity and convexity (see [1] for references). Note that if Q is bounded, or Q is unbounded but contains no straight line, then 2. is trivial, but 1. is not. In this situation statement (2.i) yields the following corollary which is interesting and far from trivial. Corollary. If F(u), u E Q, is a real-valued convex function on a convex subset Q of E which contains no straight line, then 1. holds. m We shall prove first a series of lemmas, and then theorem (2.i). (2.ii) If a convex set Q E does not contain any straight line, then clQ does n not contain any straight line either. Proof. First, let us assume that Q has at least an interior point, and, if possible, that clQ contains a straight line. It is not restrictive to assume that this straight line is the ul-axiL. If.n = 1, then Q is contained on the same ul-axis, and we may well assume that o is the interior point. If Al, 1 2 are the infimum and supremum respectively, of the real numbers u1 with u1 C Q, then - co< 1 < o < X1 < + 0, and at least one of the two numbers Xl, x2 is finite. If, say, X1 is finite, hence -c < k1 < o < < + a, then the points of the half straight line (-o, Xl) are not in Q, are not points of accumulation of points of Q and therefore are not in clQ, a contradiction. 5

If n > 1, it is not restrictive to assume that some point p = (o,a,o,...,o) with a > o is interior to Q, and then all points in the direction of the u2-axis of a neighborhood of p, say all points u = (ul,...,U ) with a < u2 < a + 5, u1= U=... =o, are interior to Q. For any e, o<<2 -1(a + ),1 and t real, o, -1 the point (ul = e 5, u2 =... = u = o) is in clQ; hence, there are points M= (u1 + 1 u2 = 2'... u = ) which belong to Q with 1,.., as small in absolute value as we want. But then the points N = (ul = -(1-E) U = (1-) (a + 2 - u = -(1-)., n, are all in Q, since o < (1-) (a+2 -1) < a + 5, and provided we take M with ~11.,.-.,Inl sufficiently small. Now the point EM + (1-E)N is the point (t,a + 2-, o,...,o) and belongs to Q. This proves that all points of the straight line [u E E, u2 = a+2 5, u U = = o] are in Q, a contradiction. We have proved (2.ii) in the case Q has an interior point. If Q has no interior points, then let R denote the linear manifold of minimum dimension r, o < r < n, containing Q. If r = o then Q reduces to a single point and so does clQ. If r > 1, then Rint Q is not empty. It is not restrictive to assume that R is the linear (ul,...,u )-subspace of E, so that Q has interior points with respect to R = E, and Q c clQ c R. We are now in the same conditions above with n replaced by r. Statement (2.ii) is thereby proved. This statement will be used only in Section 4. Let Y denote the set of all vectors v c E with the property that Inf (u'v) > - co for all u E Q.

Obviously o E Y, and if v E Y and k > o, then kv also belongs to Y. Finally, if Vl, v2 E Y and a real, o < a < 1, then cvl + (1-a)v2 c Y. Thus, Y is a convex cone of vertex the origin. (2.iii) If a convex set Q c En does not contain any straight line, then the corresponding cone Y has interior points. Proof. First, let us assume that Q has at least an interior point. It is not restrictive to suppose that the origin is an interior point of Q. Then there is some sphere a of center the origin completely contained in Q. Suppose, if' possible that Y has no interior points. Then Y must be contained in an (n-l)-dimensional subspace E nl and it is not restrictive to assume that Enis the (u...u )-space. Let A be the linear set of all real numbers 5 such n-l n that (o,...,o,)) e Q. Then A is a. convex linear set, and ~= o is an interior n point of A. If x and A' denote the infimum and the supremurn of all numbers n n n A, then -o< < < o < X < a, but at least one of the numbers,' - n n n n must be finite (otherwise the entire u -axis would belong to Q). We may well assume that say, A is finite, hence -co < X < o <' < +oo. Then the half n n n straight line 1- of all points (o,...,o, k) with 5< A is outside Q, and n 1' there must be a plane A separating I from Q in E,: Clu +... + u +n(U - ) 0, 11 n-1 n-1 n n n with z(u) > o for all u E Q, z(u) < o for all u E I. We must have cy o since, otherwise we would have o c a, and A would separate some pairs of points n D7_ of 5 hichcertanly re boh inQ. Asoi,thnemutav >o

since X- contains points (o,...,o, ) with g < x as large in absolute value as we want. Thus, A can be written in the form u = x +'1u +... + P1 Un1 n- n-l' andn > + 1U +... + un_ Unl for all u e Q. In other words, vn (-l..- n_ll) c Y. A contradiction, since Y c En1l, that is, Y is made up only of vectors (u1,..., un1,O). We have proved (2.iii) when Q has interior points. If Q has no interior points, then let R denote the linear manifold of minimum dimension r, o < r < n, containing Q. If r = o then Q is reduced to a single point, and obviously Y = E. If r > 1, then Rint Q is not empty. It is not restrictive to assume that R is the linear (ul...,ur)-subspace E of E, 1 r r n so that Q has now interior points with respect to R = E. The set Y of all r 0 vectors v = (v,...,v2,o,...,o) with Inf (y-v) > - co, now has interior points 2 yEQ in E, and obviously Y = Y x E has interior points in E. Statement (2.iii) is thereby proved. When dealing with properties of convex sets which are invariant with respect to affine transformations in E, we shall freely change systems of coordinates. (2.iv) If a convex set Q c E does not contain a line, then there is a system n of coordinates vl,...,v in E such that Q c K = [vv. > o, i =,...,n]. 1'~vn n Proof. Since Y has interior points, then Y contains a system of n independent unit vectors ei = (eil'..',ein), with Inf (u'e) = e k > - i = 1,...,n. Then the n relations v = e u +...+ e u - K, i = 1,...,n, define a new system of coordinates v1,...,v in E, and v. > o for all v = (vl,...,vn) C Q. 6

Remark. Note that (2.ii), of which we had given a direct proof, is now a corollary of (2.iv). (2.v) If F(u), u E Q, is a real-valued convex function on a convex subset Q of E which contains no straight line, then there is a linear function n w(u) = r + b ~ u, u E En, r, b = (bl,...,bn) real, such that F(u) > w(u) for all n1 n u E Q, and f(u) -w(u) - + c as |u| -* + m with u c Q. An equivalent form of this statement is obtained by extending F to all of E by taking F = + c in E -Q, and then the contention of' (2.v) becomes F(u)-w(u) > o for all u c E E, and F(u)-w(u) + + o as Iul -* + + with u c E Statement (2.v) is trivial whenever Q is bounded. Note that (2.v) is precisely the corollary to (2.i) stated at the beginning. Proof of (2.v). First, let us choose in E a system of coordinates v1,...,v such that Q c K = [vjv. > o, i = l,...,n]. Let R denote the linear manifold of minimum dimensions r, o < r < n, containing Q. If r = o then Q reduces to a single point v, and we take w(v) = r = F(v), b = (o,...,o). If 1 < r < n, then RintQ is not empty, and if v c RintQ, then F possesses a supporting plane z(v) = F(v) + b ~ (v - v), b = (bl,...,b ), such that F(v) > z(v) for all v e Q. Finally, let ~ > o arbitrary, and let w(v) = z(v) - ~(v1 +...+ v ), or n w(v) = (F(v) - b'v) + (b-E)'v, where b-8 stands for the real vector (bl-E,..., b -E). Then n F(v) - w(v) F(v) - z(v) + ~(vl +...+ v ) > s(v +...+v ), where v c Q, and hence v. > o, i = l,...,n. To prove that F(v) - w(v) - + cc 1

as v - + oo, v ~ Q, we have only to take an arbitrary N > o, and note that, for IvI > ~ nN, v c Q, certainly one of the coordinates v. must be > e N. while the others are > o. Hence F(v)-w(v) > N for IvI > ~ nN, v E Q. (2.vi) If Q is a convex subset of E possessing interior points, and u int Q, u E E n u f o, and u + Xu E Q for all real A, then v + Au c Q for all v E int Q and all real X. Proof. For every ~ real, and a, o < ~ < 1, sufficiently small, the points M = u + ~ tu and N = (1-E) (v-Eu) certainly belong to Q. Hence aM + (1-E)N ~; Q. that is, the point aM + (1-a)N = a(u + a =u) + (-)(-) (v-au) = v + t belongs to Q. In other words, the entire straight line v + ku, X real, belongs to Q. Since v c int Q, all points of this straight line are interior points of Q. (2.vii) If Q is a. convex subset of E, and u c RintQ, u c E, and u + ku c Q n n for all real X, then v + Xu c RintQ for all v c RintQ and all real A. The proof is the same as above by considering the linear manifold R of minimum dimension r containing Q, o < r < n. (2.viii) If Q is an arbitrary convex subset of E which may contain straight lines, then there are integers s > o, o > o, s + o < n, and a decomposition E = E x E x E such that Q c E x E, and there is a convex subset Q n s oj n-s- c( s a o of E, open in E and not containing any straight line, such that 8

Q x E c Qc clQ x E (2) 0 C 0 a Note that Q ma.y be reduced to a single point, and in that case we take s = o. 0 Proof. As usual, let R be the linear manifold of minimum dimension containing Q, o < r < n. If r = o, then Q is reduced to a single point, and we take s = o, a = o, Q = Q. Let us assume now that r > 1, so that RintQ / 0. It is not restrictive to assume that R is the (u...,u )-space E. Let Z be the set r r of all vectors u such that u + \u E RintQ for all real A, where u c Rint Q. By (2.vii) we know that Z does not depend on the particular point u c Rint Q. Thus, Z is a linear space Z = E of E of some dimension a, o < C < r. Now let E Cv r s be the complementary space of E in E, so that E = E x E, s > o, s + a = r.,~ r r s s - The intersection Q n E is a. convex set, and we take Q = int(Q fl E ), so that Q c Q n E c cl Qo, RintQ = Q x E, and (2) follows. Proof of (2.i). (a) Let us prove that 1 implies 2. It is enough to prove tha~t property 1 and the negation of 2 together lead to a. contradiction. Thus, we assume that 1 holds, and that there exist points u c Q, u ~ E, u1 E o, a's 0 I n in property 2. Then, all points u + kul, A real, are points of Q; hence, u c RintQ. By (5.iv) of [!], F possesses a. supporting plane z(u) = r + b'u, u E, at u; thus r and b = (bl,...,b ) are real, F(u) > z(u) for all u c Q, and F(u ) = z(u ) = r + b'uo. Besides F(u + Xul) > r + b (uf + u ), F(u _-Ul) > o o o o -o 1 o 1r + b(-ku,) for all X > o. The proof proceeds now exactly as the proof of (5.x) in [i], and will not be repeated. (b) Let us prove that 2 implies i. First, let us assume that o ~ RintQ, and that F(u) > o for all u c Q with F(o) = o. Let E = E x E x E be - n s o n-s-u 9

the decomposition of E according to (2.viii) so that Q x E Q c cQ x E, where Q is a convex subset of E not containing any straight line and open in o S Es. Let us choose (vl...v )-coordinates in E so that E is the (vl...v )-space; E is the (v...v )-space, and E is the (v.v.v )>space in E, with ~a (s+l s PS+a n-s-a s+a+l n n s >o, a > o. Also, according to (2.iv), let us choose (vl...v )-coordinates in E so that Q c K = [(vl,....,vs)Iv > o, i =,...,s]. Let e1,.,e be the unit vectors of the v-space E defined by ei = (e, s = l,..,n), i. =,.,n, n 1 is with eii = 1, eis = o for s f i. Let T be the set of all real vectors b = (bl..o,b ) for which there is some real number r such that F(v) > r + b-v for all v E Q. Let us prove that T is a convex set. Indeed, if bl, b2 E T and rl, r2 are the corresponding numbers, then, for o < a < 1, we have F(v)- [l + (l-a)r - (ab + (-ca) b ) v] 2 1 2 = [F(v) - (rl+bl'v)] + (1-a) [F(v)-(r + b2 v)] > o for all v c Qo Hence, abl + (1-a) b2 E T. Moreover, T contains the origin since F(u) > o for all u E Q. Let us prove that T is not contained in any (n-l)-dimensional subspace of E. If it were, there would be a unit vector e = e1 e+ + t en such that e ~ b = o for all b E To Note that for b = ei, i = s + a + 1,..,n, and v = vle1 +... + v e, v E Q, we have v +... - v = o, and F(v) - bv > o - ei (ve +a.. +v e = o; hence, b = ei ~ T, and o = be = e (lel +. n+ ) ni 10

We have proved that oi = o for all i = s + a + l,...,n. For b = -ei, i = l,...,s, and v = vle +...+ v e, v E Q,we have v1 > o,..., i 11 n v > o, and S F(z) - b v > o + ei (v e *. n i) 0; hence b = -ei c T, and o = be = -ei (Ile~ +...+ Sne ) = -i. We have proved that fi = o for all i = 1,...,s. Thus, e = r e +...+ e, r+1 r+l r+a r+' or e E E. Then,\e belongs to Q for all X real, and F(ke) + F(-ke) > o for some X $ o by force of 2. Hence, either F(ke) > o,or F(-ke) / o, or both. It is not restrictive to assume F(ke) > o. Since ke E RintQ, then F has a supporting plane z(v) = F(Xe) + b'(v-ke) at the point ke, by force of (5.iii) of [1]. Then F(v) > z(v) for all v, so b E t, e'b = o, and z(ye) = F(ke) + + b'(y -ke) = F(ke) for all y real. Thus, in the directions + e, the function z(u) is constant and positive. But z(o) < F(o) = o, a contradiction. We have proved that T is n-dimensional. We know that a convex set in E contained in no (n-l)-dimensional manifold n has an interior point. Therefore, let b be an interior point of T, and let a > o be so chosen that |b-bj < a implies b E T. Let r be a constant such that F(v) > w(v) = r + b'v for all v E Q. Suppose that lim inf [F(v) - w(v)] f + c, where lim inf is taken as Ivl | + cc with v E Q. Then, there is a, constant a > o and a sequence [Vk] such that lvkl + + c, vk c Q, F(vk)-w(vk) < a for all k. Without loss of generality we can assume that Vk/lvkl converges to a unit 11

vector v as k + o. Then, b + E v c T, and there is a constant r1 such that z(v) = r1 + (b + ~ v)-v < F(v) for all v. Thus, F(V)-w(k)+ (b + v) v - r - b vk = r - r + Ev' vk rl - r + kl'(k/I vkl ) as k -+ oo, a contradiction. We have proved that F(v)-w(v) + oo as Ivl + + with v c Q. We have proved that 2 implies 1 for the case where o c Rint Q, F(u) > o and F(o) = o. In general, let R be as usual the linear manifold of minimum dimension r containing Q, o < r < n. If r = o, then Q is reduced to a single point u, we can take w(u) = F(u ), and 1. holds trivially in force of the chosen conventions. If r > 1, then Rint Q p o and we can take a point u E Rint Q. Then F(u) has a supporting plane at u by force of (5.iii) of [1], say z(u) = 0 F(u) + b1 (u-u o). Let G(u) = F(u)-z(u). Then G(u) > o for all u E Q and G(u ) = o. Finally, we can transfer u to the origin by a translation in E o o n We can now apply to G the results proved above, and translate them in terms of F. Theorem (2.i) is thereby proved. 3. Properties of Seminormality of Convex Functions Let A be a closed subset of the x-space E, and for every x E Q let Q(x) be a convex subset of the u-space E o Let M be the set of all (x,u) with x E A, u C Q(x), and let f (x,u) be a real-valued function on M. 12

(3.i) If for some x E A and u E Q(x) there are numbers r, b = (bl,...,bm) real, v > o such that f (x,u) > r + b u + vju-u| for all u E Q(x), 0 then for no u1 E, u1' o, it may occur that u + ul Q(x), (3) f (x,u) = 2 [f (x,u + Auf) + fo(x,u - Ul)] for all X real. Proof. Suppose that there is ul c E, u1 $ o such that both (3) and (4) hold for all X real. Then f (x,u) = 2- ff (x,u + f) + (X,u -u)] > 2 ([r + b'(u + u) + vXlull] + [r + b (u- ul) + vI-1u II) = r + b-u + 2vX! jull. This is impossible since X can be arbitrarily large. Statement (3.i) is thereby proved. Again, let A be a closed subset of the x-space E, and for every x E A let Q(x) be a convex subset of the u-space E. Let M be the set of all (x,u) with m x c A, u c Q(x), and let f (x,u) be a real-valued function defined on M. We shall assume that, for every x E A, f (x,u) is convex in u in the convex set Q(x). 13

For every x E A and 5 > o let N (x) be the set of all x e A with Ix-xl <. For every x E A let Q(x) be the set of all (z,u) C Ea+1 with z > f (x,u), u c Q(x). Then Q(x) is a convex subset of E whose projection on the u-space m+al E is Q(x). m For every x E A and 6 > o let Q(x;6) - U Q(x) c E, XEN (x) Q(x;b) U Q(x) c E xNs (x) m+l These sets may not be convex, but the sets co Q(x;6), co Q(x;6) certainly are convex subsets of E and Er+1 respectively, and the projection of m m+ co Q(x;b) on the u-space E is co Q(x;6). Let O(x,u,6) = Inf [z0 I(zO,U) co q(x;6)], u E co Q(x;6) (,.ii) Theorem. Suppose that for a given x c A and u C Rint Q.(x) it occurs that for no u1 E, u1 ~ o, we have u + kul c Q(x) and f (x,u) = 2 [f (xu + kXu) + f (x,u-Xul)] for all A real. Suppose that for any system of numbers r, b = (bl,...,bm) real such that f (x,u) > r + b'u for all u E Q(x), any compact subset K of E, and E > o, there is some 5 = 6 (E,K) > o such that u C K C] co Q(x;6 ) implies I(x,u,6) > r + b'u-E. Then,there are numbers r, b = (bl,...,b ) real and v > o, 6 > o such that f (x,u) < r + b'u + ~, and 14

f (x,u) > r + b * u + v |u-ul for all x E A, Ix-x I < 6, and all u c Q(x). Proof. Let v(u) = rl + b1 u, u c Em, be the supporting plane of f (x,u) at u = u. This supporting plane exists since u E Q(x) and by force of (5.iii) of [1]. Let w(u) = r + b ~ u, u c Em, be the linear function satisfying the requirements of (2.i), part 1, for f (x,u) thought of as a, convex function 0 of u alone on the convex set Q(x). Then, for o < a < 1 and all u c Q(x) we have f (x,u) - [aw(u) + (l-a)v(u)] 0 l[f o(,u) - w(u)] + (1-a) [f (x,u) - v(u)] > o Let a, o < a < 1, be so small that |Iw(u)-v(u)|< E/4, and let Z(U) = O w(U) + (l-c8)v(u) - e/4. Then f (x,u)-z(u) = ao[f (X,U)-w(u)] + (1-o)j[f (x,u)-v(u)] + E/4 > E/4; lim [f (x,u)-z(u)] = + as lul + +Oc with u c Q(x); (6) f (X,u)-z(u) = v(u)-z(u) = C [v(u)-w(u)] + E/4 < E/2. (7) From (6) we conclude that, for some m > o we have f (x,u) - z(u) > 2~ for all u c Q(x) with u-uJ = m. By force of the hypotheses in (3.ii) we conclude that there is some 6 > o such that 15

D(x,u,6)-z(u) > E/8 for all u E Q(x;8), |u-ul < m, (8) (x,u,5)-z(u) > 9~/8 for all u E Q(x;6), lu-u| = m. (9) If v = s/8m, then (8) implies o(x,u,6)-z(u)-v u-ul > (E/8)-vt u-ul > (8/8)-(E/8) o for all u c Q(x;6), |u-ut < m. For lu-ut > m, let a = m/ u-ut so o < a < 1, and let us take u = a(u-u) + u, or u = au + (l-a)u. Then u E co Q(x;O), It u -u= (m/lu-uI)lu-u| = m, and, by the convexity of (, also O(x,u,6)-z(uo) < [a4(x,u,3)-z(u)] + (1-c)[ o(x,u,6)-z(u)]. From here we deduce that ~(x,u,6)-z(u ) > O(x,u,6)-z(u) + +(1/ac) ([o(x,u,S)-z(u )] - [I(x,),y)-z(u)]) > (1/n) ((9~/8)-) = h/8a, Since ivlu-ul = ~/8 we have q(x,u,b)-z(u) - vlu-ul > (E/8a) - (E/8e) = o. We have proved that T(x,u,b) > z(u) + vtu-u' for all u E Q(x;6). This implies that f (x,u) > z(u) + v u-u| for all x c A, Ix-xt < 6, u c Q(x). Statement (3.ii) is thereby proved. 4. The Sets Q(x) and Their Seminormality Properties Let A be a given closed subset of the x-space E and for every x E A let U(x) be a given subset of the u-space E. Let M be the set of all (x,u) with 16

x E A, u E U(x), that we shall suppose closed. Let f (x,u), f(x,u) = (f],...,fn) be real-valued functions defined on M. For every x E A let Q(x), Q(x) be the sets Q(x) = [z E Ejz = f(x,u) for some u E U(x)] C E, Q(x) = [(z,z)lz >f (x,u), z = f(x,u) for some u E U(x)] c E Then, the projection of Q(x) on the z-space E is Q(x). We say that the sets Q(x) satisfy property (Q) at x E A provided Q(x) = U) clo Q(x;) = c! co u Q(x) xEN6(x) We say that the sets Q(x) satisfy property (Q) in A if they have the above mentioned property at every x c A. Sets satisfying property (Q) are necessarily convex and closed. For any fixed x E A let us consider the following scalar function defined on Q(x): T(z;x) = Inf [f (x,u)lz = f(x,u), u c U()] 0 - Inf [zo (z,z) c Q(x)], z c Q(x). Then,for x c A, we have < T(z;x) < + for all z c Q(x). We consider T(z;x) as a function of z in Q(x). Note that the convexity of Q(x) c E+1 implies the convexity of' Q(x) c E, but Q(x) may be not closed even if Q(x) is closed. As usual we denote by R = R(x) the linear manifold of minimum dimension r containing Q(x), so that Rint Q(x) = Q(x) R(x) c E 17

We say that condition (a) holds at a point x c A provided (a) If(z,z) e n c o Q(x,6), then z Q(x). We showed in ([1], no. 4) that (a) is a necessary condition for property (Q). We say that condition (X) holds at a point x c A provided (X) for every z E Q(x) there is at least one u E U(x) with z = f(x,u) such that the following holds: given e > o there are numbers b > o and r, b = (bl,...,b ) real such that (X') f (x,u) > r + Zj bjfj(x,u) for all x E N6(x) and u E U(x), (X") fo(x,i) < r + Zj bjfj(x,u) + E. (4.i) If conditions (ac) and (X) hold at the point x E A, then Q(x) is close>: and convex, and the sets Q(x) satisfy property (Q) at the point x. This statement was proved in ([1], (4.i)). (4.ii) If Q(x) is convex, then either T(z;x) = -a for all z E Rint Q(x), or T(z;x) > -co for all z E Q(x). In the latter case, T(z;x) is finite everywhere and a convex function on the convex set Q(x), T(z;x) is bounded below on every bounded subset of Q(x), and T(z;x) is continuous in the convex set Rint Q(x) open with respect to R(x). Finally, if Q(x) is convex and closed, and T(z;x) > -OC for all z E Q(x), then T(z;x) is (continuous on Rint Q(x) and) lower semicontinuous at every point z e Q(x) - Rint Q(x). This statement was proved in ([1], (8.i)).

(4.iii) Theorem. If T(z;x) > -x in Q(x), then the sets Q(x) have property (Q) at x if and only if properties (a) and (X) hold at the point x. This statement was proved in ([1], (9.i)). (4.iv) If Q(x) is convex, and T(z;x) > -o in Q(x), then the set Q(x) contains a straight line if, and only if, there are z E Q(x) and z1 E, z1 Z o, such that z + ez 1 Q(x) and o 1 T(z;x) = 2-1 [T(z + zl; x) + T(z - Xz;x)] (10) for all X real. Proof. Note that for any z E Q(x) the points (z,z) with z > T(z;x) belong to Rint Q(x). Note that Q(x) contains a straight line if and only if its closure cl Q(x) contains a straight line by force of (2.ii). Finally, the graph of T(z;x), z E Q(x), certainly belongs to cl Q(x). Thus, if (10) holds for some z; Z and all \ real, then certainly cl Q(x) and Q(x) contain straight linesa Conversely, if Q(x) contains a straight line Q, then Q cannot be vertical, and 0 0 - 0 thus = [z + XZl, z + XZl, real] for some z c E, zq o, z1 realo Then z > T(zix), and for every E > o the point (T(z;x) + E,z) belongs to Rint Q(x), 0 - hence the straight line [T(z;x) + s + XZl, z + XZl, X real] belongs certainly to Q(x), hence to cl Q(x), and finally T(z;x) + ~ + \Z1 > T(z + 2zl;x) for all A real. Since 8 > o is arbitrary, we have also 19

T(z;x) + Xz1 > T(z + Xzl;x) and by exchanging X with -X also T(z;x) - Xz1 > T(z -;x) By addition then T(z;x) > 21[T(z + Xzl;x) + T(Z - Xzl;x)], where T is convex in its first argument, and hence the opposite relation is also true. We conclude that T(z;x) = 2 [T(z + XZl;X) + T(z - Xzl;x)] for all X real. Statement (4.iv) is thereby proved. Note that for x c A, the set Q(x) convex, and T(z;x) > - o in Q(x), then certainly the set Q(x) is also convex, the real valued function T(z;x) is convex on the convex set Q(x), and hence T(z;x) possesses a supporting plant T: z = r + b'z, z c E,n) at every point z c Rint Q(x), by force of statement (5.iii) of [1]. That is, T(z;x) > r + b.z for all z E Q(x), T(z;x) = r + b'z, and o is actually the supporting plane of the convex set Q(x) at the point (ZO~z) z = T(z;x). At points z c Q(x)- Rint Q(x) (if any) the situation may be different. Indeed, T(z;x) may be not continuous at z, not even lower semicontinuous, T 20

may have no supporting plane at z, in the sense that the convex set Q(x) may have vertical supporting plane at (z,z), z = T(z;x). Actual examples of these occurrences have been given in ([1], Section 8). Even if Q(x) is closed, and hence T(z,x) is lower semicontinuous at the points z Q(x) - Rint Q(x), yet the supporting plane of Q(x) at (z,z) may be vertical. Nevertheless, the following statement holds: (4.v) If x E A, if T(z;x) > -cc in Q(x), if the sets Q(x) have property (Q) at x, then for every z E Q(x) and ~ > o there are numbers r, b = (bl,...,b ) real and 6 > o such that T(z;x) > r + b'z for all z C Q(x), T(z;x) < r + b-z + ~. For z E Rint Q(x) this statement is only a corollary of (4.ii) and of (50iii) of [1], as mentioned above. For z E Q(x) - Rint Q(x) statement (4.iv) has been actually proved in ([1], proof of (9.i)). (4.vi) If x E A, if T(z;x) > -c-in Q(x), if the sets Q(x) have property (Q) at x, if r, b = (bl,...,b ) are real numbers such that T(z;x) > r + b'z for all z E Q(x),then given any compact subset K of E and numbers E > o there is some other number 6 = 6 (~,K) > o such that T(z;x,b) > r + b-z - ~ for all z c co Q(x;6 ). Proof. If the statement were not true, then there would be some ~ > o and a sequence of numbers k and points zk with 6k > k -, Zk E K n co Q(x;6k) 21

SUCh that T(zk;z,6k) < r + b'zk - ~. Hence there would be also real numbers Zk with (z,zk) E CO Q(x6k), (11) r + b'zk - E < Zk < r + bzk - E/2, k 1,2... (12) Since the points zk belong to the compact set K, [Zk] is bounded, and, by force of (12), [Zk] also is bounded. Thus, there is a subsequence, say still [k], such that (zkzk) converges as k + o to some (z,z). Now, for every 6 > o and k sufficiently large so that 6k < 6,we deduce from (11) that (ZkZk) E co Q(x;o) and hence (z,z) E C1 CO Q(x;b). Since this holds for every 6 > o we have also (z,z) E n>0 cl co Q(x;) = Q(x) by force of property (Q) at x. On the other hand, from (12) we deduce z < r + b z - ~/2, a contradiction,since z > T(z;x) > r + b'z. Statement (4.v) is thereby proved. (4.vii) Theorem. If T(z;x) > - o in Q(x), if the set Q(x) contains no straight line, then the sets Q(x) have property (Q) at x if, and only if, the following 0 properties hold: (5) If (z,z) nbA cl co Q(x;6), then z c Q(x); (X*) for every z C Q(x) there is at least one point u c U(x) with z = f(x,u) such that given ~ > o, there are numbers 8 > o, v > o, and r, b = (bl,..,b ) real with (X'*) f (x,u) > r + j.b.f.(x,u) + vlf(x,u)l for all x E N6(x) and u e U(x), (x") f(X, r + E.bf(Jfjx) +. 22

Proof. Let us prove the sufficiency part. First conditions (a) and (X*) hold, hence condition (X) holds, and by force of (4.iii) the sets Q(x) have property (Q) at x. On the other hand, the set Q(x) contains no straight line by force of (3.i). Let us prove the necessity part. First, the sets Q(x) have property (Q) at x; hence conditions (a) and (X) hold by force of (4.iii). Also, the real-valued function T(z;x) is convex over the convex set Q(x), and statement (3.ii) holds. In addition, the set Q(x) contains no straight line by hypothesis, and hence the real function T(z;x) has the property that for no z c Q(x), z1 E, z1 i o it occurs that zo + Xz E Q(x) for all x real and T(z;x) = 21[T(z + zl1;x) + T(zo-\zl;x)] for all X real. Finally, by force of (4.ii), there are numbers r, b = (bl,...,bn) real and v > o, 6 > o, such that T(z;x) > r + b'z + vlz-zI for all x E A, fx-xl < 6, z E Q(x). Reference [1] Cesari, L., Seminormality and upper semincontinuity in optimal control. Journal of Optimization Theory and Applications,Vol. 6, 1970, pp. 114-137.

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