On the Convergence of Multistep Iteration for Uniformly Continuous Φ-Hemicontractive Mappings

and Applied Analysis 3 In 2005, C. E. Chidume and C. O. Chidume 3 proved the convergence theorems for fixed points of uniformly continuous generalized Φ-hemicontractive mappings and published in 3 . However, there exists a gap in the proof course of their theorems. The aim of this paper is to show the convergence of the multistep iteration with errors for fixed points of uniformly continuous Φ-hemicontractive mappings and revise the results of C. E. Chidume and C. O. Chidume 3 . For this, we need the following Lemmas. Lemma 1.1 see 4 . Let E be a real Banach space and let J : E → 2E be a normalized duality mapping. Then ∥∥x y∥∥2 ≤ ‖x‖ 2〈y, j(x y)〉, 1.7 for all x, y ∈ E and for all j x y ∈ J x y . Lemma 1.2 see 5 . Let {δn}n 0, {λn}n 0 and {γn}n 0 be three nonnegative real sequences and let Φ : 0, ∞ → 0, ∞ be a strictly increasing and continuous function with Φ 0 0 satisfying the following inequality: δ2 n 1 ≤ δ2 n − λnΦ δn 1 γn, n ≥ 0, 1.8 where λn ∈ 0, 1 with ∑∞ n 0 λn ∞, γn o λn . Then δn → 0 as n → ∞. 2. Main Results Theorem 2.1. Let E be an arbitrary real Banach space, D a nonempty closed convex subset of E, and T : D → D a uniformly continuous Φ-hemicontractive mapping with q ∈ F T / ∅. Let {an}, {bk n}, {cn}, {dk n} be real sequences in 0, 1 and satisfy the conditions: i an cn ≤ 1, b n d n ≤ 1, k 1, 2, . . . , p − 1; ii an, cn, b n, d k n → 0 as n → ∞, k 1, 2, . . . , p − 1; iii cn o an , ∑∞ n 0 an ∞. For some x0 ∈ D, let {un}, {w1 n}, {w2 n}, . . . , {wp−1 n } be any bounded sequences ofD, and let {xn} be the multistep iterative sequence with errors defined by 1.6 . Then 1.6 converges strongly to the fixed point q of T . Proof. Since T : D → D is Φ-hemicontractive mapping, then there exists a strictly increasing continuous function Φ : 0, ∞ → 0, ∞ with Φ 0 0 such that 〈 Tx − Tq, j(x − q)〉 ≤ ∥∥x − q∥∥2 −Φ(∥∥x − q∥∥), 2.1 for x ∈ D, q ∈ F T , that is 〈 Tx − x, j(x − q)〉 ≤ −Φ(∥∥x − q∥∥). @ Choose some x0 ∈ D and x0 / Tx0 such that ‖x0 − Tx0‖ · ‖x0 − q‖ ∈ R Φ and denote that r0 ‖x0 − Tx0‖ · ‖x0 − q‖, R Φ is the range of Φ. Indeed, if Φ r → ∞ as r → ∞, then 4 Abstract and Applied Analysis r0 ∈ R Φ ; if sup{Φ r : r ∈ 0, ∞ } r1 < ∞ with r1 < r0 here, we only give a example. If r0 2, Φ t t/ 1 t , then sup{Φ r : r ∈ 0, ∞ } r1 1 < 2 r0 , then for q ∈ D, there exists a sequence {wn} in D such that wn → q as n → ∞ with wn / q. Furthermore, we obtain that Twn → Tq as n → ∞. So {wn − Twn} is the bounded sequence. Hence, there exists natural number n0 such that ‖wn − Twn‖ · ‖wn − q‖ < r1/2 for n ≥ n0, then we redefine x0 wn0 and ‖x0 − Tx0‖ · ‖x0 − q‖ ∈ R Φ . This is to ensure that Φ−1 r0 is defined well. Step 1. We show that {xn} is a bounded sequence. Set R Φ−1 r0 , then from above formula @ , we obtain that ‖x0 − q‖ ≤ R. Denote B1 { x ∈ D : ∥∥x − q∥∥ ≤ R}, B2 { x ∈ D : ∥∥x − q∥∥ ≤ 2R}. 2.2 Since T is the uniformly continuous, so T is a bounded mapping. We let M sup x∈B2 {∥∥Tx − q∥∥ 1} max { sup n ∥∥w1 n − q ∥∥ } , sup n ∥∥w2 n − q ∥∥ } , . . . , sup n ∥∥wp−1 n − q ∥∥ } , sup n ∥un − q ∥∥} } . 2.3 Next, we want to prove that xn ∈ B1. If n 0, then x0 ∈ B1. Now, assume that it holds for some n, that is, xn ∈ B1. We prove that xn 1 ∈ B1. Suppose that it is not the case, then ‖xn 1 − q‖ > R > R/2. Since T is uniformly continuous, then for 0 Φ R/2 /8R, there exists δ > 0 such that ‖Tx − Ty‖ < 0 when ‖x − y‖ < δ. Denote τ0 min { 1, R M , Φ R/2 8R M 2R , δ 2M 4R } . 2.4 Since an, b n, cn, d k n → 0 as n → ∞ for k 1, 2, . . . , p−1. Without loss of generality, we assume that 0 ≤ an, b n, cn, d n ≤ τ0 for any n ≥ 0. Since cn o an , let cn < anτ0. Now, estimate ‖yk n − q‖ for k 1, 2, . . . , p − 1. By using 1.6 , we have ∥∥yp−1 n − q ∥∥ ≤ ( 1 − bp−1 n − dp−1 n ∥∥xn − q ∥∥ bp−1 n ∥Txn − q ∥∥ dp−1 n ∥∥wp−1 n − q ∥∥ ≤ R τ0M ≤ 2R, 2.5 then yp−1 n ∈ B2. Similarly, we have ∥∥yp−2 n − q ∥∥ ≤ ( 1 − bp−2 n − dp−2 n ∥∥xn − q ∥∥ bp−2 n ∥∥Typ−1 n − q ∥∥ dp−2 n ∥∥wp−2 n − q ∥∥ ≤ R τ0M ≤ 2R, 2.6 Abstract and Applied Analysis 5 then yp−2 n ∈ B2, . . .. We have ∥∥y1 n − q ∥∥ ≤ ( 1 − b1 n − d1 n ∥∥xn − q ∥∥ b1 n ∥∥Ty2 n − q ∥∥ d1 n ∥∥w1 n − q ∥∥ ≤ R τ0M ≤ 2R, 2.7and Applied Analysis 5 then yp−2 n ∈ B2, . . .. We have ∥∥y1 n − q ∥∥ ≤ ( 1 − b1 n − d1 n ∥∥xn − q ∥∥ b1 n ∥∥Ty2 n − q ∥∥ d1 n ∥∥w1 n − q ∥∥ ≤ R τ0M ≤ 2R, 2.7 then y1 n ∈ B2. Therefore, we get ∥xn 1 − q ∥∥ ≤ 1 − an − cn ∥xn − q ∥∥ an ∥∥Ty1 n − q ∥∥ cn ∥un − q ∥∥ ≤ R τ0M ≤ 2R. 2.8 And we have ‖xn 1 − xn‖ ≤ an ∥∥Ty1 n − xn ∥∥ cn‖un − xn‖ ≤ an ∥∥Ty1 n − q ∥∥ ∥xn − q ∥∥) cn ∥un − q ∥∥ ∥xn − q ∥∥) ≤ τ0 ∥∥Ty1 n − q ∥∥ ∥un − q ∥∥) 2∥xn − q ∥∥] ≤ τ0 M 2R ≤ Φ R/2 8R , ∥∥xn 1 − y1 n ∥∥ ≤ an ∥∥Ty1 n − xn ∥∥ cn‖un − xn‖ b1 n ∥∥Ty2 n − xn ∥∥ d1 n ∥∥w1 n − xn ∥∥ ≤ an ∥∥Ty1 n − q ∥∥ ∥xn − q ∥∥) cn ∥un − q ∥∥ ∥xn − q ∥∥) b1 n ∥∥Ty2 n − q ∥∥ ∥xn − q ∥∥) d1 n ∥∥w1 n − q ∥∥ ∥xn − q ∥∥) ≤ τ0 ∥∥Ty1 n − q ∥∥ ∥un − q ∥∥ 2∥xn − q ∥∥) ∥∥Ty2 n − q ∥∥ ∥∥w1 n − q ∥∥ 2∥xn − q ∥∥)] ≤ τ0 2M 4R ≤ δ. 2.9 Further, by using uniform continuity of T , we have ∥∥Txn 1 − Ty1 n ∥∥ < Φ R/2 8R . 2.10 6 Abstract and Applied Analysis In view of Lemma 1.1 and the above formulas, we obtain ∥xn 1 − q ∥∥2 ∥∥(xn − q ) an ( Ty1 n − xn ) cn un − xn ∥∥ 2 ≤ ∥xn − q ∥∥2 2an 〈 Ty1 n − xn, j ( xn 1 − q )〉 2cn 〈 un − xn, j ( xn 1 − q )〉 ≤ ∥xn − q ∥∥2 2an 〈 Txn 1 − xn 1 xn 1 − xn − Txn 1 Ty1 n, j ( xn 1 − q )〉 2cn‖un − xn‖ · ∥xn 1 − q ∥∥ ≤ ∥xn − q ∥∥2 − 2anΦ ∥xn 1 − q ∥∥) 2an‖xn 1 − xn‖ · ∥xn 1 − q ∥∥ 2an ∥∥Txn 1 − Ty1 n ∥∥ · ∥xn 1 − q ∥∥ 2cn ∥un − q ∥∥ ∥xn − q ∥∥xn 1 − q ∥∥ ≤ ∥xn − q ∥∥2 − 2anΦ ( R 2 ) 2an Φ R/2 8R · 2R 2an R/2 8R · 2R 2anτ0 R M 2R ≤ ∥xn − q ∥∥2 − anΦ ( R 2 ) 2an Φ R/2 8R M 2R R M 2R ≤ ∥xn − q ∥∥2 − an 2 Φ ( R 2 ) ≤ R2, 2.11 which is a contradiction. Hence, xn 1 ∈ B1, that is, {xn} is a bounded sequence; it leads to that {y1 n}, {y2 n}, . . . , {yp−1 n } are all bounded sequences as well. Step 2. We want to prove ‖xn − q‖ → 0 as n → ∞. Since an, b n, cn, d k n → 0 as n → ∞ and {xn}, {y1 n} are bounded. From 2.9 , we obtain lim n→∞ ‖xn 1 − xn‖ 0, lim n→∞ ∥∥xn 1 − y1 n ∥∥ 0, lim n→∞ ∥∥Txn 1 − Ty1 n ∥∥ 0. 2.12