Strong Convergence Theorems for Nonexpansive Semigroups and Variational Inequalities in Banach Spaces

Question 1. Do Plubtieng and Punpaeng’s results hold for the nonexpansive semigroups in a Banach space?

Lemma 2.1 (Goebel and Reich [10], Proposition 5.3). Let C be a nonempty closed convex subset of a strictly convex Banach space X and T : C → C a nonexpansive mapping with F(T) ≠ ∅. Then F(T) is closed and convex.

Lemma 2.2 (see Xu [11].)In a smooth Banach space X there holds the inequality

Lemma 2.3 (Browder [12]). Let E be a uniformly convex Banach space, K a nonempty closed convex subset of E, and T : K → E a nonexpansive mapping. Then I − T is demi closed at zero.

Lemma 2.4 (see [8], Lemma 2.7.)Let C be a nonempty bounded closed convex subset of a uniformly convex Banach space X, and let 𝒮 = {T(s) : 0 ≤ s < ∞} be a nonexpansive semigroup on C such that F(𝒮) ≠ ∅. For x ∈ C and t > 0. Then, for any 0 ≤ h < ∞,

Lemma 2.5 (Xu [11, Lemma 2.6]). Assume that X has a weakly continuous duality map J_φ with gauge φ, for all x, y ∈ X, there holds the inequality

Lemma 2.6 (Xu [6]). Assume {α_n} is a sequence of nonnegative real numbers such that

where {γ_n} is a sequence in (0,1) and {δ_n} is a sequence in R such that

• (i)

  $$;

• (ii)

  limsup _n→∞δ_n/γ_n ≤ 0 or $$.

Then lim _n→∞α_n = 0.

Lemma 2.7 (see [14], Lemma 1.)Let C be a nonempty closed convex subset of a Banach space X with a uniformly Gâteaux differentiable norm and {x_n} a bounded sequence of E. If z₀ ∈ C, then

if and only if

Lemma 2.8 (Song and Xu [4, Proposition 3.1]). Let X be a reflexive strictly convex Banach space with a uniformly Gâteaux differentiable norm, and C a nonempty closed convex subset of X. Suppose {x_n} is a bounded sequence in C such that lim _n→∞∥x_n − Tx_n∥ = 0, an approximate fixed point of nonexpansive self-mapping T on C. Define the set

If F(T) ≠ ∅, then C^*∩F(T) ≠ ∅.

Theorem 3.1. Let X be a uniformly convex Banach space that has a weakly continuous duality map J_φ with gauge φ, and let C be a nonempty closed convex subset of X. Let 𝒮 = {T(s) : 0 ≤ s < ∞} be a nonexpansive semigroup from C into itself such that F(𝒮) = ⋂_s>0F(T(s)) ≠ ∅ and f : C → C a contraction mapping with the contractive coefficient α ∈ [0,1). Suppose {λ_t} _0<t<1 is a net of positive real numbers such that $$, the sequence {x_t} is given by the following equation:

Then {x_t} converges strongly to $$ as t → 0⁺, where $$ is the unique solution in F(𝒮) of the variational inequality

Proof. Note that F(𝒮) is a nonempty closed convex set by Lemma 2.1. We first show that {x_t} is bounded. Indeed, for any fixed p ∈ F(𝒮), we have

It follows that Thus {x_t} is bounded, so are {f(x_t)} and {T(s)x_t} for every 0 ≤ s < ∞. Furthermore, we note that for every 0 ≤ s < ∞. On the one hand, we observe that for every t > 0. On the other hand, let z₀ ∈ F(S) and D = {z ∈ C : ∥z − z₀∥ ≤ ∥f(z₀) − z₀∥}, then D is a nonempty closed bounded convex subset of C which is T(s)-invariant for each 0 ≤ s < ∞ and contains {x_t}. It follows by Lemma 2.4 that Hence, by (3.5)–(3.7), we obtain for every 0 ≤ s < ∞. Assume $$ is such that t_n → 0 as n → ∞. Put $$, $$, we will show that {x_n} contains s subsequence converging strongly to $$, where $$. Since {x_n} is a bounded sequence, there is a subsequence $$ of {x_n} which converges weakly to $$. By Lemma 2.3, we have $$. For each n ≥ 1, we have Thus, by Lemma 2.5, we obtain This implies that In particular, we have Now observing that $$ implies $$. And since $$ is bounded, it follows from (3.12) that Hence $$.

Next, we show that $$ solves the variational inequality (3.2). Indeed, for q ∈ F(𝒮), it is easy to see that

However, we note that Thus, we get that for t ∈ (0,1) and q ∈ F(𝒮) Taking the limit through $$, we obtain This implies that since J_φ(x) = (φ(∥x∥)/∥x∥)J(x) for x ≠ 0.

Finally, we show that the net {x_t} convergence strong to $$. Assume that there is a sequence {s_n}⊂(0,1) such that $$, where s_n → 0. we note by Lemma 2.3 that $$. It follows from the inequality (3.18) that

Interchange $$ and $$ to obtain Adding (3.19) and (3.20) yields We must have $$ and the uniqueness is proved. In a summary, we have shown that each cluster point of {x_t} as t → 0 equals $$. Therefore $$ as t → 0.

Theorem 3.2. Let X be a uniformly convex Banach space with a uniformly Gâteaux differentiable norm and C be a nonempty closed convex subset of X. Let 𝒮 = {T(s) : 0 ≤ s < ∞} be a nonexpansive semigroup from C into itself such that F(𝒮) = ⋂_s>0F(T(s)) ≠ ∅ and f : C → C a contraction mapping with the contractive coefficient α ∈ [0,1). Suppose $$ is a net of positive real numbers such that $$, the sequence {x_t} is given by the following equation:

Then {x_t} converges strongly to $$ as t → 0⁺, where $$ is the unique solution in F(𝒮) of the variational inequality

Proof. We include only those points in this proof which are different from those already presented in the proof of Theorem 3.1. As in the proof of Theorem 3.1, we obtain that there is a subsequence $$ of {x_n} which converges weakly to $$. For each n ≥ 1, we have

Thus, we have Therefore,

We claim that the set {x_n} is sequentially compact. Indeed, define the set

By Lemma 2.8, we found $$. Using Lemma 2.7 we get that From (3.26), we get that is Hence, there exists a subsequence $$ of {x_n} converges strongly to $$ as k → ∞.

Next we show that $$ is a solution in F(𝒮) to the variational inequality (3.23). In fact, for any fixed x ∈ F(𝒮), there exists a constant M > 0 such that ∥x_n − x∥ ≤ M, then

Therefore, Since the duality mapping J is single valued and norm topology to weak* topology uniformly continuous on any bounded subset of a Banach space X with a uniformly Gâteaux differentiable norm, we have Taking limit as j → ∞ in two sides of (3.32), we get

Finally we will show that the net {x_t} convergence strong to $$. This section is similar to that of Theorem 3.1.

Theorem 4.1. Let X be a uniformly convex Banach space that has a weakly continuous duality map J_φ with gauge φ and C be a nonempty closed convex subset of X. Let 𝒮 = {T(s) : 0 ≤ s < ∞} be a nonexpansive semigroup from C into itself such that F(𝒮) = ⋂_s>0F(T(s)) ≠ ∅ and f : C → C a contraction mapping with the contractive coefficient α ∈ [0,1). Let {α_n} and {β_n} be the sequence in (0,1) which satisfies α_n + β_n < 1, lim _n→∞α_n → 0, lim _n→∞β_n → 0 and $$, and {s_n} is a positive real divergent sequence such that lim _n→∞s_n → ∞. If the sequence {x_n} defined by x₀ ∈ C and

Then {x_n} converges strongly to $$ as n → ∞, where $$ is the unique solution in F(𝒮) of the variational inequality

Proof. Note that F(𝒮) is a nonempty closed convex set. We first show that {x_n} is bounded. Let q ∈ F(𝒮). Thus, we compute that

Therefore, {x_n} is bounded, {f(x_n)} and {T(s)x_n} for every 0 ≤ s < ∞ are also bounded.

Next we show ∥x_n − T(h)x_n∥ → 0 as n → ∞. Notice that

Put z₀ = P_F(𝒮)x₀ and D = {z ∈ C : ∥z − z₀∥ ≤ ∥x₀ − z₀∥ + 1/(1 − α)∥f(z₀) − z₀∥}. Then D is a nonempty closed bounded convex subset of C which is T(s)-invariant for each s ∈ [0, ∞) and contains {x_n}. So without loss of generality, we may assume that 𝒮 = {T(s) : 0 ≤ s < ∞} is a nonexpansive semigroup on D. By Lemma 2.4, we get

for every h ∈ [0, ∞). On the other hand, since {x_n}, {f(x_n)}, and {T(s)x_n} are bounded, using the assumption that lim _n→∞α_n → 0, lim _n→∞β_n → 0, and (4.5) into (4.4), we get that and hence

We now show that

Let $$, where t and λ_t satisfies the condition of Theorem 3.1. Then it follows from Theorem 3.1 that $$ and $$ be the unique solution in F(𝒮) of the variational inequality (3.2). Clearly $$ is a unique solution of (4.2). Take a subsequence $$ of {x_n} such that Since X is uniformly convex and hence it is reflexive, we may further assume that $$. Moreover, we note that p ∈ F(𝒮) by Lemma 2.3 and (4.7). Therefore, from (4.9) and (3.17), we have That is (4.8) holds.

Finally we will show that $$. For each n ≥ 0, we have

An application of Lemma 2.6, we can obtain $$, hence $$. That is, {x_n} converges strongly to a fixed point $$ of 𝒮. This completes the proof.

Theorem 4.2. Let X be a uniformly convex Banach space with a uniformly Gâteaux differentiable norm and C be a nonempty closed convex subset of X. Let 𝒮 = {T(s) : 0 ≤ s < ∞} be a nonexpansive semigroup from C into itself such that F(𝒮) = ⋂_s>0F(T(s)) ≠ ∅ and f : C → C a contraction mapping with the contractive coefficient α ∈ [0,1). Let {α_n} and {β_n} be the sequence in (0,1) which satisfies α_n + β_n < 1, lim _n→∞α_n → 0, lim _n→∞β_n → 0, and $$, and {s_n} is a positive real divergent sequence such that lim _n→∞s_n → ∞. If the sequence {x_n} defined by x₀ ∈ C and

Then {x_n} converges strongly to $$ as n → ∞, where $$ is the unique solution in F(𝒮) of the variational inequality

Proof. We also show only those points in this proof which are different from that already presented in the proof of Theorem 4.1. We now show that

Let $$, where t and λ_t satisfies the condition of Theorem 3.2. Then it follows from Theorem 3.2 that $$ and $$ is the unique solution in F(𝒮) of the variational inequality (3.23). Clearly $$ is a unique solution of (4.13). Take a subsequence $$ of {x_n} such that Since X is uniformly convex and hence it is reflexive, we may further assume that $$. Moreover, we note that p ∈ F(𝒮) by Lemma 2.3 and (4.7). Therefore, from (4.15) and (3.23), we have That is, (4.14) holds.

Finally we will show that $$. For each n ≥ 0, by Lemma 2.2, we have

which implies that where $$, δ_n : = 2(1 − α)α_n/(1 − α_nα), and $$. It is easily to see that δ_n → 0, $$ and limsup _n→∞γ_n ≤ 0 by (4.14). Finally by using Lemma 2.6, we can obtain {x_n} converges strongly to a fixed point $$. This completes the proof.

Theorem 5.1. Let X be a uniformly convex Banach space that has a weakly continuous duality map J_φ with gauge φ and C be a nonempty closed convex subset of X. Let 𝒮 = {T(s) : 0 ≤ s < ∞} be a nonexpansive semigroup from C into itself such that F(𝒮) = ⋂_s>0F(T(s)) ≠ ∅ and f : C → C a contraction mapping with the contractive coefficient α ∈ [0,1). Let {α_n} be the sequence in (0,1) which satisfies lim _n→∞α_n → 0 and $$, and {s_n} is a positive real divergent sequence such that lim _n→∞s_n → ∞. If the sequence {x_n} defined by x₀ ∈ C and

Then {x_n} converges strongly to $$ as n → ∞, where $$ is the unique solution in F(𝒮) of the variational inequality

Proof. Taking β_n = 0 in the in Theorem 4.1, we get the desired conclusion easily.

Theorem 5.2. Let X be a uniformly convex Banach space with a uniformly Gâteaux differentiable norm and C be a nonempty closed convex subset of X. Let 𝒮 = {T(s) : 0 ≤ s < ∞} be a nonexpansive semigroup from C into itself such that F(𝒮) = ⋂_s>0F(T(s)) ≠ ∅ and f : C → C a contraction mapping with the contractive coefficient α ∈ [0,1). Let {α_n} be the sequence in (0,1) which satisfies lim _n→∞α_n → 0 and $$, and {s_n} is a positive real divergent sequence such that lim _n→∞s_n → ∞. If the sequence {x_n} defined by x₀ ∈ C and

Then {x_n} converges strongly to $$ as n → ∞, where $$ is the unique solution in F(𝒮) of the variational inequality

Proof. Taking β_n = 0 in the in Theorem 4.2, we get the desired conclusion easily.

Corollary 5.3 (Reich [2]). Let C be a nonempty closed convex subset of a real Hilbert space H. Let 𝒮 = {T(s) : 0 ≤ s < ∞} be a strongly continuous semigroup of nonexpansive mapping on C such that F(𝒮) is nonempty. Let {α_n} and {β_n} be sequences of real numbers in (0,1) which satisfies α_n + β_n < 1, lim _n→∞α_n → 0, lim _n→∞β_n → 0, and $$. Let f be a contraction of C into itself with a coefficient α ∈ [0,1) and {s_n} be a positive real divergent sequence such that lim _n→∞s_n → ∞. Then the sequence {x_n} defined by x₀ ∈ C and

Then {x_n} converges strongly to $$, where $$ is the unique solution in F(𝒮) of the variational inequality

or equivalent $$, where P is a metric projection mapping from H into F(𝒮).