In this case we have for n large enough ||xs(n)|| �� r2 and so zs(n) is bounded which ensures that ys(n) �� 0 and so y-=0. On the other hand, by continuity of �� and the convergence of xn to x-, we obtain ��(||x-||)=lim?n��(||xs(n)||)=lim?n??ys(n)=0 and so G(t-,x-)=0. Thus, we get y-��G(t-,x-)=0. Assume now that there exists AZD9291 FDA some �� > 0 and n0 such that ��(||xn||) > �� > 0 for all n �� n0. Then by continuity of �� we have zn=yn/��(||xn||)��z-:=y-/��(||x-||). Thus, by upper semicontinuity of F, we get z-��F(t-,x-); that is, y-�ʦ�(||x-||)F(t-,x-)=G(t-,x-). This completes the proof of the closedness of the graph of G and hence the proof is achieved.Now, we are ready to prove our main existence result under the nonlinear growth condition in Banach spaces.

Theorem 12 ��Let be a Banach space, D a nonempty closed set, and F : satisfying the following: F is u.s.c. on I �� D with F(t, x) being closed convex for all (t, x) I �� D;F(t, x) c(t)(||x|| + ||x||p) on I �� D, for some c C(I, +), and p with p �� 1, and for some convex compact set in ;F(t, x)��K(D; x) �� on I �� D.Then for every x0 D, there exists an absolutely continuous mapping x : I1 �� D such on??I1,(31)where I1 = [0,?a.e.??on??I1,x(0)=x0��D,x(t)��D,?thatx�B(t)��F(t,x(t)) T] when p (?��, 1) and I1 = [0, b1) for p (1, ��) and b1 is as in Lemma 7.Proof ��Let k �� 1 be such that k and assume that c-:=max?t��I?c(t)>1. Let k-=kc- and let M be as in Proposition 9 with k- instead of the function c. ThenM=exp?(k?b2)[||x0||1?p+1]?exp?(pk?b2),(32)where b2 is defined as in Proposition 9.

Set r:=k-(M+Mp)>M and let �� : [0, +��)��[0,1] be a continuous function such that ��(s) = 1 for s �� M and ��(s) = 0 for s �� r. Define now the on??I��D.(33)By?set-valued mapping G on I �� D as follows:G(t,x)=��(||x||)F(t,x) Proposition 11, the set-valued mapping G inherits the convexity and the upper semicontinuity of the set-valued mapping F with G(t, x) c(t)0, where 0 : = r 0. We have to check that G satisfies the tangential condition on D. Let t I and let x D��M. Then G(t, x) = F(t, x) and so the tangential condition is satisfied from (c). Assume now that t I and x D with ||x|| > M. Then by (c) there exists some z F(t, x) such that z K(D; x). Let y = ��(||x||)z. Clearly, y G(t, x) and y ��(||x||)K(D; x) K(D; x), since K(D; x) is a cone. So, G(t, x)��K(D; x) �� .

Therefore, the tangential condition is satisfied for G on all I �� D. Consequently, all the assumptions (a), (b), and (c) in Theorem 10 with the compact set 0 instead of are satisfied and hence for Cilengitide every x0 D there exists a solution x on I of ((DI)) associated with the set-valued mapping G defined above, that is, an absolutely continuous mapping x : I �� D such that x�B(t)��G(t,x(t)) a.e. on I, x(0) = x0, and x(t) D on I. Let us prove that x is the desired a.e.?on??I(34)and??��(||x(t)||)k?(||x(t)||+||x(t)||p)??=��(||x(t)||)F(t,x(t))?solution for (31).