By Stuart E. Dreyfus

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I r r ( t ) ) i n Y ; + ' ( by I) Then 11' separates 11' and Y, contradicting 11' being a weak * limit of Y. Thus Y is closed, convex, and weak * compact. 2, T is LI E Y ) is a compact, convex weak * continuous; therefore T Y = {Ti#: subset of En. To complete the proof we must show the equality of TY and T V o . Clearly T Y oc T Y . If a E T Y , T - I ( u ) n 'I' is a weak * compact, convex 28 I . FUNCTIONAL ANALYSIS subset of Y and hence, by Lemma 8. I , has a n extreme point u. The proof will be complete if we can show II is in Yo.

Assume that u(t*) is an interior point of %(t*). 3 \z(t*) is an interior point of 3‘(t,) for some 0 < I, < t * . (t*) inside %‘(t,). @(t,)c % ( t ) for all t > t , , N is contained in % ( I )for all t > I , . The continuity of ~ i . (I,) for some t , < t*. As pointed at the beginning, this completes the proof. 5)] that if u* is an optimal control-and therefore for some q # 0 is of the form u*(t) = sgn[q’ Y ( t ) ] on [0, t*]-then the hyperplane n(t, q ) through y ( t ; u*) is a support plane to 9 ( t ) for each t E [0, t*].

I). 3 Let Y ( f ) be ail 17 x ( r + 1)-niatrix-valued f i ~ n c [ i ~with components in zI(I). Then (j, Y ( 7 ) 4 7 ) d ~ LI: measurable, I / ( ? ) E or for 7 E I } = { j I Y ( 7 ) ~ ( 7tlr: ) LI measurable, I I ( T ) E V(o'), r E / > and both of these sets are compact and convex Proof. To simplify notation, let 'tJ= Yo = {Id€Y~l(I):tI(f)€o ? €r I, ) {I( E9 r ; I(1): If([) E V(fJr), f € I) and define T:Y:f ' ( I ) --f E" by Tu = 1Y ( T )u ( r ) d7. -1 Clearly Y is convex and bounded i n the norm topology; hence.

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