APPROXIMATION OF FIXED POINTS OF STRONGLY PSEUDOCONTRACTIVE MAPPINGS IN UNIFORMLY SMOOTH BANACH SPACES

for all x ∈ E, where 〈·,·〉 denotes the generalized duality pairing. It is well known that if E is a uniformly smooth Banach space, then J is single valued such that J(−x) = −J(x), J(tx) = tJ(x) for all t ≥ 0, x ∈ E; and J is uniformly continuous on any bounded subset of E. In the sequel we will denote single-valued normalized duality mapping by j. In the following we give some concepts. Let T : D(T)→ E be a mapping with domain D(T) and range R(T). A mapping T is said to be pseudocontractive if for any x, y ∈ D(T) there exists j(x− y) ∈ J(x− y) such that 〈 Tx−Ty, j(x− y)≤ ‖x− y‖2. (1.2)


Introduction
Let E be an arbitrary real Banach space and E * the dual space on E. The normalized duality mapping J : E → 2 E * is defined by for all x ∈ E, where •, • denotes the generalized duality pairing.It is well known that if E is a uniformly smooth Banach space, then J is single valued such that J(−x) = −J(x), J(tx) = tJ(x) for all t ≥ 0, x ∈ E; and J is uniformly continuous on any bounded subset of E. In the sequel we will denote single-valued normalized duality mapping by j.In the following we give some concepts.Let T : D(T) → E be a mapping with domain D(T) and range R(T).A mapping T is said to be pseudocontractive if for any x, y ∈ D(T) there exists j(x − y) ∈ J(x − y) such that (1. 2) The mapping T is said to be strongly pseudocontractive if for any x, y ∈ D(T) there exists j(x − y) ∈ J(x − y) such that for some constant k ∈ (0,1).Recently, Zhou and Jia [5] proved the following result: let E be a real Banach space with a uniformly convex dual E * , and let K be a nonempty closed convex and bounded subset of E. Assume that T : K → K is a continuous and strong pseudocontraction, the Ishikawa iteration sequence {x n } ∞ n=1 generated by (IS) converges strongly to the unique fixed point of T. However, when T is continuous strongly pseudocontractive mapping, one question arises naturally: if T neither is Lipschitzian nor has the bounded range, whether or not the Ishikawa iterative sequence {x n } ∞ n=1 generated converges strongly to the unique fixed point of T. It is our purpose in this note to solve the above question by proving the following much more general result: E is a real uniformly smooth Banach space, and K is a nonempty closed convex subset of E. Assume that T : K → K is a continuous and strong pseudocontraction, and T neither is Lipschizian nor has the bounded range, then the Ishikawa iteration sequence converges strongly to the unique fixed point of T.

Main results
Now we prove the main results of this note, In the sequel, we always assume that E is a real uniformly smooth Banach space.
Theorem 2.1.Let K be a nonempty closed convex subset of E. Assume that T 1 + T 2 : K → K is a continuous and strongly pseudocontractive mapping, where T 1 : Then the Ishikawa iterative sequence generated from an arbitrary x 1 ∈ K by (IS1), converges strongly to the unique fixed point of Proof.The existence of a fixed point follows from Deimling [4].Let q be a fixed point of T 1 + T 2 .Since T 1 + T 2 is strongly pseudocontractive, thus for all x, y ∈ K, where k ∈ (0,1).Then we may get that q must be unique fixed point of (2.1), we have (2.3) , and {(x n+1 − q)/(1 + x n − q )} are all bounded, using that J is uniformly continuous on bounded subset, hence A n → 0 as n → ∞ and D n → 0 as n → ∞.Applying Lemma 1.1, we obtain (2.4) Again using Lemma 1.1, we obtain 4 Approximation of fixed point mappings in Banach spaces (2.5) Furthermore, we have the following estimates to a part of (2.4): where E n = (3L + M)A n α n (1 + Lβ n ).Substituting (2.5) and (2.6) in (2.4), we have where Hence, we may choose a large positive integer N such that for all n ≥ N, Thus the above inequality (2.7) yields (2.9) By Lemma 1.2 we see that as x n − q → 0 as n → ∞.The proof of theorem is completed.
Remark 2.2.Concrete the following example: let E = (−∞,+∞), K = [0,+∞), where x = |x|, x ∈ E. Let T 1 : K → K be defined by T 1 x = x/3, and let T 2 : K → K be defined Xue Zhiqun 5 by (2.10)  Proof.By virtue of Remark 2.3, the equation Tx = f has unique solution.Set S 1 x = x − T 1 x, S 2 x = −T 2 x, x ∈ E. Then S 1 is Lipschitz, S 2 has the bounded range operator, and Sx = S 1 x + S 2 x + f .Hence S is a continuous and strongly pseudocontractive mapping.We obtain directly the conclusion from Theorem 2.1.
[1] is Lipschitz, T 2 has the bounded range, and T 1 + T 2 is strongly pseudocontractive mapping.But T 1 + T 2 neither is Lipschitzian nor has a bounded range.Remark 2.3.Theorem 2.1 contains a good number of the known results as its special cases.In particular, if the mapping T considered here satisfies one of the following assumptions: (i) T : K → K is a Lipschitzian; (ii) T has the bounded range, then T satisfied the conditions of Theorem 2.1.Remark 2.4.In[1], Bogin proved that T is strongly pseudocontractive if and only if (I − T) is strongly accretive, where I denotes the identity operator.It is well known that if T is continuous and strongly accretive, then T is surjective, so that, for any given f ∈ E, the equation Tx = f has unique solution.Let {α n } ∞ n=1 and {β n } ∞ n=1 be two real sequences [0,1] in satisfying the conditions: (i) α n ,β n → 0 as n → ∞; Theorem 2.5.Assume that T = T 1 + T 2 : E → E is a continuous strongly accretive operator, where T 1 : E → E is Lipschitz, T 2 : E → E has the bounded range operator.For any given f ∈ E, define S : E → E by Sx = f − Tx + x for all x ∈ E.