Continuation Theorems for Nonexpansive Maps
Radu Precup · 2002
Proposition 3.1 Let (X, I . I ) be a uniformly convex Banach space, D C X nonempty closed bounded and convex. Assume T : D -t D is nonexpansive. Then T has at least one fixed point. For the proof we recommend [53] or [150]. Now we state and prove the corresponding Leray-Schauder type the- orem [104], [139]. Theorem 3.2 Let ( X , I . I ) be a uniformly convex Banach space, U C X open bounded convex with 0 E U and T : U + X nonexpansive. Assume For the proof we need the following two lemmas due to Browder [19]. Lemma 3.3 Let ( X , I. I ) be a uniformly convex Banach space, D C X bounded convex and T : D -t X nonexpansive. Then, for each E > 0 there exists S = S ( E ) > 0 such that lx - T ( x ) J E / (3d) . If 1 - X ~ / 3 . We have and similarly Let 30 = X-&s; 1x0 - ~ ~ 1 - l ( T ( x x ) - xo) Then it is easily seen that For S small enough, by the uniform convexity of X , we deduce that I yo - yl I 5 &/d. It follows that Lernrna 3.4 Let ( X , I . I ) be a uniformly convex Banach space, D C X closed bounded convex and T : D + X nonexpansive. If ( x k ) C D , xk -t xo weakly and xk - T ( x k ) -+ y~ strongly as k + m, then X 0 - T (xo) = YO. Proof. We may assume without loss of generality that y~ = 0. Let ~k = lxk - T ( x k ) I . Since E I , -+ 0 as k + m, we may suppose that 1 6 ( E , Z - ~ ) 5 ~ k - l for any k, where S ( E ) is given as in Lemma 3.3. Since xo E Dk : = m { x j ; j > k ) for every k, it follows that in order to prove xo - T ( x o ) = 0, it suffices to show X - T X 5 E for a11 X E D ~ . (3.2) Each element of Dk is the strong limit of some sequence of finite convex combinations of elements in { x j ; j > k) . Thus it is sufficient to prove inequality (3.2) only for the elements of D? : = conv { x j ; k 5 j 5 m} , m > k. We use induction with respect to k decreasing: For k = m, Ixm - T (xm)l = Em 5 and (3.2) holds. Let us now assume that (3.2) is true for all X E D?. Let y E Then y = (1 - X ) X ~ - ~ + X X for some X E D? and X E [0, 11 . We have From Lemma 3.3, we have ly - T ( y ) ( < q - 2 which proves (3.2) on or-"-1 Proof of Theorem 3.2. From Theorem 2.4, for each X E (0, l ) , there exists a unique xx E U with xx - AT ( x X ) = 0. Denote by xk the element xx for X = 1 - ilk. Taking if necessary a subsequence, we may suppose that xk + xo weakly as k + m. On the other hand, since it follows that xr, - T ( x k ) + Q strongly as k + m. Now Lemma 3.4 guarantees that xo - T ( x o ) = 0.