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# Continuation Theorems for Nonexpansive Maps

DOI link for Continuation Theorems for Nonexpansive Maps

Continuation Theorems for Nonexpansive Maps book

# Continuation Theorems for Nonexpansive Maps

DOI link for Continuation Theorems for Nonexpansive Maps

Continuation Theorems for Nonexpansive Maps book

## ABSTRACT

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 for all X = ( 1 - X ) xo + X X I , where X E ( 0 , l ) , xo, X I E D, 1x0 - T (xo)l < S and lxl - T ( X I ) l < 6. Proof. If 1x0 - x1 1 < ~ / 3 , then for X = xx := (1 - X ) xo + Xxl, we have

provided that S < ~ / 3 . Thus, we have only to deal with the case 1x0 - X I [ 2 ~ / 3 . If X < E / (3d) , where d = diam D, then Ixx - xol = X 1x0 - X I \ < ~ / 3 . Then, as above, we obtain Ixx - T (xx)I < E. Furthermore, we assume X > E / (3d) . If 1 - X < &/ (3d ) , then by a similar argument, we get Ixx - T ( x x ) / < E where S 5 e/3. Now we examine the case: E / (3d) < X < 1 - E / (3d) and 1x0 - X I 1 > ~ / 3 . We have

and similarly

Let 30 = X-' 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.