Abstract

The following main results have been given. (1) Let 𝐸 be a 𝑝-uniformly convex Banach space and let 𝑇𝐸𝐸 be a (𝑝1)-𝐿-Lipschitz mapping with condition 0<(𝑝𝐿/𝑐2)1/(𝑝1)<1. Then 𝑇 has a unique generalized duality fixed point 𝑥𝐸 and (2) let 𝐸 be a 𝑝-uniformly convex Banach space and let 𝑇𝐸𝐸 be a 𝑞-𝛼-inverse strongly monotone mapping with conditions 1/𝑝+1/𝑞=1, 0<(𝑞/(𝑞1)𝑐2)𝑞1<𝛼. Then 𝑇 has a unique generalized duality fixed point 𝑥𝐸. (3) Let 𝐸 be a 2-uniformly smooth and uniformly convex Banach space with uniformly convex constant 𝑐 and uniformly smooth constant 𝑏 and let 𝑇𝐸𝐸 be a 𝐿-lipschitz mapping with condition 0<2𝑏/𝑐2<1. Then 𝑇 has a unique zero point 𝑥. These main results can be used for solving the relative variational inequalities and optimal problems and operator equations.

1. Introduction and Preliminaries

Let 𝐸 be a real Banach space with the dual 𝐸 and let 𝑇 be an operator from 𝐸 into 𝐸. Firstly, for 𝑝2, we consider the variational inequality problem of finding an element 𝑥𝐸 such that 𝑇𝑥,𝑥𝑥0,𝑥𝑥(𝑝1)2.(1.1) Taking 𝑝=2, the problem (1.1) becomes the following variational inequality problem of finding an element 𝑥𝐸 such that 𝑇𝑥,𝑥𝑥0,𝑥𝑥.(1.2) Secondly, for 𝑝2, we consider the optimal problem of finding an element 𝑥𝐸 such that 𝑥(𝑝1)𝑇𝑥2=min𝑥𝐸𝑥(𝑝1)𝑇𝑥2.(1.3) Taking 𝑝=2, the problem (1.3) becomes the following optimal problem of finding an element 𝑥𝐸 such that 𝑥𝑇𝑥2=min𝑥𝐸(𝑥𝑇𝑥)2.(1.4) Thirdly, for 𝑝2, we consider the operator equation problem of finding an element 𝑥𝐸 such that 𝑇𝑥,𝑥=𝑇𝑥𝑝=𝑥𝑝(𝑝1).(1.5) Taking 𝑝=2, the problem (1.5) becomes the following operator equation problem of finding an element 𝑥𝐸 such that 𝑇𝑥,𝑥=𝑇𝑥2=𝑥2.(1.6) Finally, we consider the operator equation problem of finding an element 𝑥𝐸 such that 𝑇𝑥=0.(1.7)

Let 𝐸 be a real Banach space with the dual 𝐸. Let 𝑝 be a given real number with 𝑝>1. The generalized duality mapping 𝐽𝑝 from 𝐸 into 2𝐸 is defined by 𝐽𝑝(𝑥)=𝑓𝐸𝑥,𝑓=𝑓𝑝,𝑓=𝑥𝑝1,𝑥𝐸,(1.8) where , denotes the generalized duality pairing. In particular, 𝐽=𝐽2 is called the normalized duality mapping and 𝐽𝑝(𝑥)=𝑥𝑝2𝐽(𝑥) for all 𝑥0. If 𝐸 is a Hilbert space, then 𝐽=𝐼, where 𝐼 is the identity mapping. The duality mapping 𝐽 has the following properties: (i)if 𝐸 is smooth, then 𝐽 is single-valued;(ii)if 𝐸 is strictly convex, then 𝐽 is one-to-one;(iii)if 𝐸 is reflexive, then 𝐽 is surjective;(iv)if 𝐸 is uniformly smooth, then 𝐽 is uniformly norm-to-norm continuous on each bounded subset of 𝐸;(v)if 𝐸 is uniformly convex, then 𝐽 is uniformly continuous on each bounded subsets of 𝐸 and 𝐽 is single-valued and also one-to-one.For more details, see [1].

In this paper, we firstly present the definition of duality fixed point for a mapping 𝑇 from 𝐸 into its dual 𝐸 as follows.

Let 𝐸 be a Banach space with a single-valued generalized duality mapping 𝐽𝑝𝐸𝐸. Let 𝑇𝐸𝐸. An element 𝑥𝐸 is said to be a generalized duality fixed point of 𝑇 if 𝑇𝑥=𝐽𝑝𝑥. An element 𝑥𝐸 is said to be a duality fixed point of 𝑇 if 𝑇𝑥=𝐽𝑥.

Example 1.1. Let 𝐸 be a smooth Banach space with the dual 𝐸, and let 𝐴𝐸𝐸 be an operator, then an element 𝑥𝐸 is a zero point of 𝐴 if and only if 𝑥 is a duality fixed point of 𝐽+𝜆𝐴 for any 𝜆>0. Namely, the 𝑥 is a duality fixed point of 𝐽+𝜆𝐴 for any 𝜆>0 if and only if 𝑥 is a fixed point of 𝐽𝜆=(𝐽+𝜆𝐴)1𝐽𝐸𝐸 (if 𝐴 is maximal monotone, then 𝐽𝜆 is, namely, the resolvent of 𝐴).

Example 1.2. In Hilbert space, the fixed point of an operator is always duality fixed point.

Example 1.3. Let 𝐸 be a 𝑝-uniformly convex Banach space with the dual 𝐸, then any element of 𝐸 must be the generalized duality fixed point of the generalized normalized duality mapping 𝐽𝑝.

Conclusion 1. If 𝑥 is a generalized duality fixed point of 𝑇, then 𝑥 must be a solution of variational inequality problem (1.1).

Proof. Suppose 𝑥 is a generalized duality fixed point of 𝑇, then 𝑇𝑥,𝑥𝐽=𝑝𝑥,𝑥=𝐽𝑝𝑥𝑝=𝑇𝑥𝑝=𝑥𝑝(𝑝1).(1.9) Obverse that 𝑇𝑥,𝑥𝑥=𝑇𝑥,𝑥𝑇𝑥,𝑥𝑇𝑥𝑝𝑇𝑥𝑥=𝑇𝑥𝑇𝑥𝑝1=𝑥𝑇𝑥𝑥(𝑝1)2𝑥0(1.10) for all 𝑥𝑥(𝑝1)2.

Taking 𝑝=2, we have the following result.

Conclusion 2. If 𝑥 is a duality fixed point of 𝑇, then 𝑥 must be a solution of variational inequality problem (1.2).

Conclusion 3. If 𝑥 is a generalized duality fixed point of 𝑇, then 𝑥 must be a solution of the optimal problem (1.3). Therefore, 𝑥 is also a solution of operator equation problem (1.5).

Proof. If 𝑥 is a generalized duality fixed point of 𝑇, then 𝑇𝑥=𝐽𝑝𝑥, so that 𝑇𝑥,𝑥𝐽=𝑝𝑥,𝑥=𝐽𝑝𝑥𝑝=𝑇𝑥𝑝=𝑥𝑝(𝑝1).(1.11) All conclusions are obvious.

Take 𝑝=2, we have the following result.

Conclusion 4. If 𝑥 is a duality fixed point of 𝑇, then 𝑥 must be a solution of the optimal problem (1.4). Therefore, 𝑥 is also a solution of operator equation problem (1.6).
Let 𝑈={𝑥𝐸𝑥=1}. A Banach space 𝐸 is said to be strictly convex if for any 𝑥,𝑦𝑈, 𝑥𝑦 implies (𝑥+𝑦)/2<1. It is also said to be uniformly convex if for each 𝜀(0,2], there exists 𝛿>0 such that for any 𝑥,𝑦𝑈, 𝑥𝑦𝜀 implies (𝑥+𝑦)/2<1𝛿. It is well known that a uniformly convex Banach space is reflexive and strictly convex. And we define a function 𝛿[0,2][0,1] called the modulus of convexity of 𝐸 as follows: 𝛿(𝜀)=1𝑥+𝑦2.𝑥=𝑦=1,𝑥𝑦𝜀(1.12) It is well known that 𝐸 is uniformly convex if and only if 𝛿(𝜀)>0 for all 𝜀(0,2]. Let 𝑝 be a fixed real number with 𝑝2. Then 𝐸 is said to be 𝑝-uniformly convex if there exists a constant 𝑐>0 such that 𝛿(𝜀)𝑐𝜀𝑝 for all 𝜀[0,2]. For example, see [2, 3] for more details. The constant 1/𝑐 is said to be uniformly convexity constant of 𝐸.
A Banach space 𝐸 is said to be smooth if the limit lim𝑡0𝑥+𝑡𝑦𝑥𝑡(1.13) exists for all 𝑥,𝑦𝑈. It is also said to be uniformly smooth if the above limit is attained uniformly for 𝑥,𝑦𝑈. One should note that no Banach space is 𝑝-uniformly convex for 1<𝑝<2; see [4] for more details. It is well known that the Hilbert and the Lebesgue 𝐿𝑞(1<𝑞2) spaces are 2-uniformly convex and uniformly smooth. Let 𝑋 be a Banach space and let 𝐿𝑞(𝑋)={Ω,Σ,𝜇;𝑋},1<𝑞 be the Lebesgue-Bochner space on an arbitrary measure space (Ω,Σ,𝜇). Let 2𝑝< and let 1<𝑞𝑝. Then 𝐿𝑞(𝑋) is 𝑝-uniformly convex if and only if 𝑋 is 𝑝-uniformly convex; see [3].
Let 𝜌𝐸[0,)[0,) be the modulus of smoothness of 𝐸 defined by 𝜌𝐸1(𝑡)=sup2(.𝑥+𝑦+𝑥𝑦)1𝑥𝑈,𝑦𝑡(1.14) A Banach space 𝐸 is said to be uniformly smooth if 𝜌𝐸(𝑡)/𝑡0 as 𝑡0. Let 𝑞>1. A Banach space 𝐸 is said to be 𝑞-uniformly smooth, if there exists a fixed constant 𝑐>0 such that 𝜌𝐸(𝑡)𝑐𝑡𝑞. It is well known that 𝐸 is uniformly smooth if and only if the norm of 𝐸 is uniformly Fréchet differentiable. If 𝐸 is 𝑞-uniformly smooth, then 𝑞2 and 𝐸 is uniformly smooth, and hence the norm of 𝐸 is uniformly Fréchet differentiable, in particular, the norm of 𝐸 is Fréchet differentiable. Typical examples of both uniformly convex and uniformly smooth Banach spaces are 𝐿𝑝, where 𝑝>1. More precisely, 𝐿𝑝 is min{𝑝,2}-uniformly smooth for every 𝑝>1.

Lemma 1.4 (see [5, 6]). Let 𝐸 be a 𝑝-uniformly convex Banach space with 𝑝2. Then, for all 𝑥,𝑦𝐸, 𝑗(𝑥)𝐽𝑝(𝑥) and 𝑗(𝑦)𝐽𝑝(𝑦), 𝑐𝑥𝑦,𝑗(𝑥)𝑗(𝑦)𝑝𝑐𝑝2𝑝𝑥𝑦𝑝,(1.15) where 𝐽𝑝 is the generalized duality mapping from 𝐸 into 𝐸 and 1/𝑐 is the 𝑝-uniformly convexity constant of 𝐸.

Lemma 1.5. Let 𝐸 be a 𝑝-uniformly convex Banach space with 𝑝2. Then 𝐽𝑝 is one-to-one from 𝐸 onto 𝐽𝑝(𝐸)𝐸 and for all 𝑥,𝑦𝐸, 𝑝𝑥𝑦𝑐21/(𝑝1)𝐽𝑝(𝑥)𝐽𝑝(𝑦)1/(𝑝1),(1.16) where 𝐽𝑝 is the generalized duality mapping from 𝐸 into 𝐸 with range 𝐽𝑝(𝐸), and 1/𝑐 is the 𝑝-uniformly convexity constant of 𝐸.

Proof. Let 𝐸 be a 𝑝-uniformly convex Banach space with 𝑝2, then 𝐽=𝐽2 is one-to-one from 𝐸 onto 𝐸. Since 𝐽𝑝(𝑥)=𝑥𝑝2𝐽(𝑥), then 𝐽𝑝(𝑥) is single valued. From (1.5) we have 𝑥𝑦,𝐽𝑝(𝑥)𝐽𝑝𝑐(𝑦)𝑝𝑐𝑝2𝑝𝑥𝑦𝑝,(1.17) which implies that 𝐽𝑥𝑦𝑝(𝑥)𝐽𝑝𝑐(𝑦)𝑝𝑐𝑝2𝑝𝑥𝑦𝑝.(1.18) That is 𝐽𝑝(𝑥)𝐽𝑝𝑐(𝑦)𝑝𝑐𝑝2𝑝𝑥𝑦𝑝1.(1.19) Hence 𝑝𝑥𝑦𝑐21/(𝑝1)𝐽𝑝(𝑥)𝐽𝑝(𝑦)1/(𝑝1).(1.20) Then (1.6) has been proved. Therefore, from (1.6) we can see, for any 𝑥,𝑦𝐸, that 𝐽𝑝(𝑥)=𝐽𝑝(𝑦) implies that 𝑥=𝑦.

2. Duality Contraction Mapping Principle and Applications

Let 𝐸 be a Banach space with the dual 𝐸. An operator 𝑇𝐸𝐸 is said to be 𝑝-𝐿-Lipschitz, if 𝑇𝑥𝑇𝑦𝐿𝑥𝑦𝑝,𝑥,𝑦𝐸,(2.1) where 𝐿(0,+),𝑝[1,+) are two constants. If 𝑝=1, the operator 𝑇 is said to be 𝐿-Lipschitz.

Theorem 2.1 (generalized duality contraction mapping principle). Let 𝐸 be a 𝑝-uniformly convex Banach space and let 𝑇𝐸𝐸 be a (𝑝1)-𝐿-Lipschitz mapping with condition 0<(𝑝𝐿/𝑐2)1/(𝑝1)<1. Then 𝑇 has a unique generalized duality fixed point 𝑥𝐸 and for any given guess 𝑥0𝐸, the iterative sequence 𝑥𝑛+1=𝐽𝑝1𝑇𝑥𝑛 converges strongly to this generalized duality fixed point 𝑥.

Proof. Let 𝐴=𝐽𝑝1𝑇, then 𝐴 is a mapping from 𝐸 into itself. By using Lemma 1.5, we have 𝐽𝐴𝑥𝐴𝑦=𝑝1𝑇𝑥𝐽𝑝1𝑝𝑇𝑦𝑐21/(𝑝1)𝑇𝑥𝑇𝑦1/(𝑝1)𝑝𝑐21/(𝑝1)𝐿𝑥𝑦𝑝11/(𝑝1)𝑝𝐿𝑐21/(𝑝1)𝑥𝑦(2.2) for all 𝑥,𝑦𝐸, where 0<(𝑝𝐿/𝑐2)1/(𝑝1)<1. By using Banach's contraction mapping principle, there exists a unique element 𝑥𝐸 such that 𝐴𝑥=𝑥. That is, 𝑇𝑥=𝐽𝑝𝑥, so 𝑥 is a generalized unique duality fixed point of 𝑇. Further, the Picard iterative sequence 𝑥𝑛+1=𝐴𝑥𝑛=𝐽𝑝1𝑇𝑥𝑛 (𝑛=0,1,2,) converges strongly to this generalized duality fixed point 𝑥.

Taking 𝑝=2, we have the following result.

Theorem 2.2 (duality contraction mapping principle). Let 𝐸 be a 2-uniformly convex Banach space and let 𝑇𝐸𝐸 be a 𝐿-Lipschitz mapping with condition 0<(2𝐿/𝑐2)<1. Then 𝑇 has a unique duality fixed point 𝑥𝐸 and for any given guess 𝑥0𝐸, the iterative sequence 𝑥𝑛+1=𝐽1𝑇𝑥𝑛 converges strongly to this duality fixed point 𝑥.

From Conclusions 14 and Theorem 2.1, we have the following result for solving the variational inequality problems (1.1) and (1.2), the optimal problems (1.3) and (1.4), and the operator equation problems (1.5) and (1.6).

Theorem 2.3. Let 𝐸 be a 𝑝-uniformly convex Banach space and let 𝑇𝐸𝐸 be a (𝑝1)-𝐿-Lipschitz mapping with condition 0<(𝑝𝐿/𝑐2)1/(𝑝1)<1. Then the variational inequality problem (1.1) (the optimal problem (1.2) and operator equation problem (1.3)) has solutions and for any given guess 𝑥0𝐸, the iterative sequence 𝑥𝑛+1=𝐽𝑝1𝑇𝑥𝑛 converges strongly to a solution of the variational inequality problem (1.1) (the optimal problem (1.3) and the operator equation problem (1.5)).

Taking 𝑝=2, we have the following result.

Theorem 2.4. Let 𝐸 be a 2-uniformly convex Banach space, let 𝑇𝐸𝐸 be a (𝑝1)-𝐿-Lipschitz mapping with condition 0<(2𝐿/𝑐2)<1. Then the variational inequality problem (1.2) (the optimal problem (1.4) and operator equation problem (1.6)) has solutions and for any given guess 𝑥0𝐸, the iterative sequence 𝑥𝑛+1=𝐽1𝑇𝑥𝑛 converges strongly to a solution of the variational inequality problem (1.2) (the optimal problem (1.4) and the operator equation problem (1.6)).

Theorem 2.5 (generalized duality Mann weak convergence theorem). Let 𝐸 be a p-uniformly convex Banach space which satisfying Opial's condition, let 𝑇𝐸𝐸 be a (𝑝1)-𝐿-Lipschitz mapping with nonempty generalized duality fixed point set. Assume 0<(𝑝𝐿/𝑐2)1/(𝑝1)1, and the real sequence {𝛼𝑛}[0,1] satisfies the condition 𝑛=0𝛼𝑛(1𝛼𝑛)=+. Then for any given guess 𝑥0𝐸, the generalized Mann iterative sequence 𝑥𝑛+1=1𝛼𝑛𝑥𝑛+𝛼𝑛𝐽𝑝1𝑇𝑥𝑛(2.3) converges weakly to a generalized duality fixed point of 𝑇.

Proof. Letting 𝐴=𝐽1𝑇, by using Lemma 1.4, we have 𝐽𝐴𝑥𝐴𝑦=1𝑇𝑥𝐽12𝑇𝑦𝑐2𝑇𝑥𝑇𝑦2𝐿𝑐2𝑥𝑦𝑥𝑦,(2.4) for all 𝑥,𝑦𝐸. Hence 𝐴 is a nonexpansive mapping from 𝐸 into itself. In addition, we have 𝑥𝑛+1=1𝛼𝑛𝑥𝑛+𝛼𝑛𝐽1𝑇𝑥𝑛=1𝛼𝑛𝑥𝑛+𝛼𝑛𝐴𝑥𝑛.(2.5) By using the well-known result, we know that the sequence {𝑥𝑛} converges weakly to a fixed point 𝑥 of 𝐴(𝐴𝑥=𝑥). This point 𝑥 is also a duality fixed point of 𝑇(𝑇𝑥=𝐽𝑥).

Take 𝑝=2, we have the following result.

Theorem 2.6 (duality Mann weak convergence theorem). Let 𝐸 be a 2-uniformly convex Banach space which satisfy Opial's condition and let 𝑇𝐸𝐸 be a 𝐿-Lipschitz mapping with nonempty duality fixed point set. Assume 0<2𝐿/𝑐21, and the real sequence {𝛼𝑛}[0,1] satisfies the condition 𝑛=0𝛼𝑛(1𝛼𝑛)=+. Then for any given guess 𝑥0𝐸, the generalized Mann iterative sequence 𝑥𝑛+1=1𝛼𝑛𝑥𝑛+𝛼𝑛𝐽1𝑇𝑥𝑛(2.6) converges weakly to a duality fixed point of 𝑇.

Theorem 2.7 (duality Halpern strong convergence theorem). Let 𝐸 be a p-uniformly convex Banach space which satisfying Opial's condition, let 𝑇𝐸𝐸 be a (𝑝1)-𝐿-Lipschitz mapping with nonempty generalized duality fixed point set. Assume 0<(𝑝𝐿/𝑐2)1/(𝑝1)1, and the real sequence {𝛼𝑛}[0,1] satisfies the condition:(C1): lim𝑛𝛼𝑛=0;(C2): 𝑛=0𝛼𝑛=;(C3): lim𝑛(𝛼𝑛+1𝛼𝑛)/𝛼𝑛+1=0 or lim𝑛(𝛼𝑛/𝛼𝑛+1)=1.Let 𝑢,𝑥0 be given, then iterative sequence 𝑥𝑛+1=𝛼𝑛𝑢+1𝛼𝑛𝐽𝑝1𝑇𝑥𝑛,(2.7) converges strongly to a generalized duality fixed point of 𝑇.

Proof. Let 𝐴=𝐽𝑝1𝑇, then 𝐴 is a mapping from 𝐸 into itself. By using Lemma 1.5, we have 𝐽𝐴𝑥𝐴𝑦=𝑝1𝑇𝑥𝐽𝑝1𝑝𝑇𝑦𝑐21/(𝑝1)𝑇𝑥𝑇𝑦1/(𝑝1)𝑝𝑐21/(𝑝1)𝐿𝑥𝑦𝑝11/(𝑝1)𝑝𝐿𝑐21/(𝑝1)𝑥𝑦(2.8) for all 𝑥,𝑦𝐸, where 0<(𝑝𝐿/𝑐2)1/(𝑝1)1. Hence 𝐴 is a nonexpansive mapping from 𝐸 into itself. In addition, we have 𝑥𝑛+1=𝛼𝑛𝑢+1𝛼𝑛𝐽𝑝1𝑇𝑥𝑛=𝛼𝑛𝑢+1𝛼𝑛𝐴𝑥𝑛.(2.9) By using the well-known result of Xu [7, Theorem  2.3], we know that the iterative sequence {𝑥𝑛} converges strongly to a fixed point of nonexpansive mapping 𝐴. Hence the sequence {𝑥𝑛} converges strongly to a generalized duality fixed point of 𝑇.

Theorem 2.8. Letting 𝐻 be a Hilbert space, then one has its uniformly convexity constant 1/𝑐2/2, that is 𝑐2.

Proof. If 𝑐>2. For any 𝑥𝑦, by using Lemma 1.4, we have 𝐽𝑥𝑦=1𝑥𝐽1𝑦2𝑐2𝑥𝑦<𝑥𝑦.(2.10) This is a contradiction.

3. Fixed Point Theorem of Inverse Strongly Monotone Mappings

Definition 3.1. Letting 𝐸 be a Banach space, the mapping 𝑇𝐸𝐸 is called 𝑞-𝛼-inverse strongly monotone, if 𝑇𝑥𝑇𝑦,𝑥𝑦𝛼𝑇𝑥𝑇𝑦𝑞,𝑥,𝑦𝐸.(3.1)

Lemma 3.2. Let 𝐸 be a Banach space and let 𝑇𝐸𝐸 be a 𝑞-𝛼-inverse strongly monotone mapping. Then 𝑇 is 1/(𝑞1)(1/𝛼)(1/(q1))-Lipschitz.

Proof. Let 𝑇𝐸𝐸 be a 𝑞-𝛼-inverse strongly monotone mapping, that is, 𝑇𝑥𝑇𝑦,𝑥𝑦𝛼𝑇𝑥𝑇𝑦𝑞,𝑥,𝑦𝐸.(3.2) It follows from the above inequality that 𝛼𝑇𝑥𝑇𝑦𝑞𝑇𝑥𝑇𝑦,𝑥𝑦𝑇𝑥𝑇𝑦𝑥𝑦,𝑥,𝑦𝐸,(3.3) which leads to 𝑇𝑥𝑇𝑦𝑞1𝛼𝑇𝑥𝑇𝑦𝑥𝑦,𝑥,𝑦𝐸.(3.4) Further 𝑇𝑥𝑇𝑦𝑞11𝛼𝑥𝑦,𝑥,𝑦𝐸,(3.5) and hence 1𝑇𝑥𝑇𝑦𝛼1/(𝑞1)𝑥𝑦1/(𝑞1),𝑥,𝑦𝐸.(3.6) Then 𝑇 is 1/(𝑞1)(1/𝛼)1/(𝑞1)-Lipschitz.

Theorem 3.3 (fixed point theorem of inverse strongly monotone mappings). Let 𝐸 be a 𝑝-uniformly convex Banach space and let 𝑇𝐸𝐸 be a 𝑞-𝛼-inverse strongly monotone mapping with conditions 1/𝑝+(1/𝑞)=1, 0<(𝑞/(𝑞1)𝑐2)𝑞1<𝛼. Then 𝑇 has a unique generalized duality fixed point 𝑥𝐸 and for any given guess 𝑥0𝐸, the iterative sequence 𝑥𝑛+1=𝐽𝑝1𝑇𝑥𝑛 converges strongly to this generalized duality fixed point 𝑥.

Proof. Letting 𝐴=𝐽𝑝1𝑇, then 𝐴 is a mapping from 𝐸 into itself. By using Lemma 1.5 and Lemma 3.2, we have 𝐽𝐴𝑥𝐴𝑦=𝑝1𝑇𝑥𝐽𝑝1𝑝𝑇𝑦𝑐21/(𝑝1)𝑇𝑥𝑇𝑦1/(𝑝1)𝑝𝑐21/(𝑝1)1𝛼1/(𝑞1)𝑥𝑦1/(𝑞1)1/(𝑝1)=𝑝𝑐21/(𝑝1)1𝛼=𝑞𝑥𝑦(𝑞1)𝑐2𝑞11𝛼𝑥𝑦(3.7) for all 𝑥,𝑦𝐸. It follows from the condition 0<(𝑞/(𝑞1)𝑐2)𝑞1<𝛼 that 0<(𝑞/(𝑞1)𝑐2)𝑞1(1/𝛼)<1. By using Banach's contraction mapping principle, there exists a unique element 𝑥𝐸 such that 𝐴𝑥=𝑥. That is, 𝑇𝑥=𝐽𝑝𝑥, so 𝑥 is a generalized unique duality fixed point of 𝑇. Further, the Picard iterative sequence 𝑥𝑛+1=𝐴𝑥𝑛=𝐽𝑝1𝑇𝑥𝑛 (𝑛=0,1,2,) converges strongly to this generalized duality fixed point 𝑥.

Taking 𝑝=2, we have the following results.

Lemma 3.4. Let 𝐸 be a Banach space and let 𝑇𝐸𝐸 be a 2-𝛼-inverse strongly monotone mapping. Then 𝑇 is (1/𝛼)-Lipschitz.

Theorem 3.5 (fixed point theorem of inverse strongly monotone mappings). Let 𝐸 be a 2-uniformly convex Banach space, let 𝑇𝐸𝐸 be a 2-𝛼-inverse strongly monotone mapping with condition 0<2/𝑐2<𝛼. Then 𝑇 has a unique duality fixed point 𝑥𝐸 and for any given guess 𝑥0𝐸, the iterative sequence 𝑥𝑛+1=𝐽1𝑇𝑥𝑛 converges strongly to this generalized duality fixed point 𝑥.

4. Application for Zero Point of Operators

Lemma 4.1 (see [8]). Let 𝐸 be a 𝑝-uniformly smooth Banach space with uniformly smooth constant 𝑏. Then 𝐽𝑥𝐽𝑦𝑏𝑥𝑦𝑝1,𝑥,𝑦𝐸.(4.1)

Theorem 4.2. Let 𝐸 be a 2-uniformly smooth and uniformly convex Banach space with uniformly convex constant 𝑐 and uniformly smooth constant 𝑏 and let 𝑇𝐸𝐸 be a 𝐿-lipschitz mapping with condition 0<2𝑏/𝑐2<1. Then 𝑇 has a unique zero point 𝑥 and for any given guess 𝑥0𝐸, the iterative sequence 𝑥𝑛+1=𝐽1(𝐽+𝑟𝑇)𝑥𝑛 converges strongly to this zero point 𝑥.

Proof. Let 𝐴=𝐽1(𝐽+𝑟𝑇), then 𝐴 is a mapping from 𝐸 into itself. By using Lemma 1.5 and Lemma 4.1, we have 𝐽𝐴𝑥𝐴𝑦=1(𝐽+𝑟𝑇)𝑥𝐽12(𝐽+𝑟𝑇)𝑦𝑐2=2(𝐽+𝑟𝑇)𝑥(𝐽+𝑟𝑇)𝑦𝑐22(𝐽𝑥𝐽𝑦)𝑟(𝑇𝑥𝑇𝑦)𝑐22(𝐽𝑥𝐽𝑦+𝑟𝑇𝑥𝑇𝑦)𝑐2(𝑏+𝑟𝐿)𝑥𝑦(4.2) for all 𝑥,𝑦𝐸. Observing the condition 0<2𝑏/𝑐2<1, it follows that, there exists a positive number 𝑟>0 such that 0<2/𝑐2(𝑏+𝑟𝐿)<1. By using Banach's contraction mapping principle, there exists a unique element 𝑥𝐸 such that 𝐴𝑥=𝑥. That is, (𝐽+𝑟𝑇)𝑥=𝐽𝑥 which implies 𝑟𝑇𝑥=0, so 𝑥 is a zero point of 𝑇. Further, the Picard iterative sequence 𝑥𝑛+1=𝐴𝑥𝑛=𝐽1(𝐽+𝑟𝑇)𝑥𝑛 (𝑛=0,1,2,) converges strongly to this zero point 𝑥.

Remark 4.3. Under the conditions of Theorem 4.2, we know that the operator equation 𝑇𝑥=0 has a unique solution which can be computed by the iterative scheme 𝑥𝑛+1=𝐴𝑥𝑛=𝐽1(𝐽+𝑟𝑇)𝑥𝑛 (𝑛=0,1,2,) starting any given guess 𝑥0𝐸.

Acknowledgment

This project is supported by the National Natural Science Foundation of China under Grant (11071279).