Definition 1 [6]. Let $$ be a nonempty subset of a Banach space $$ A mapping $$ is said to satisfy condition (E_μ) on $$ if there exists μ ≥ 1 such that for all $$

We say that Ψ satisfies condition (E) on $$ whenever Ψ satisfies (E_μ) for some μ ≥ 1.

Definition 2 [8]. Let $$ be a Banach space. A mapping $$ is said to be b-enriched nonexpansive mapping if ∃b ∈ [0, ∞) such that:

Definition 3. Let $$ be a Banach space and $$ a nonempty subset of $$ A mapping $$ is said to be Suzuki-enriched nonexpansive mapping if there exists b ∈ [0, ∞) such that for all $$:

Example 1 [4]. Consider the mapping $$ defined by $$, where $$. The set of fixed point of Ψ is {1}, and we can say that F(Ψ) = {1}. Moreover, Ψ is a nonexpansive mapping that satisfies the $$-enriched condition. However, at ϑ = 1 and $$, Ψ fails to satisfy condition (E).

Example 2 [5]. Let $$ with the usual norm. Define $$ by the following equation:

Then, Ψ satisfies condition (E). However at ϑ = 2.5 and ϱ = 3, Ψ is not b-enriched nonexpansive mapping for any b ∈ [0, ∞).

Theorem 1 [15]. Let $$ be a uniformly convex Banach space. Then for any d > 0, ε > 0, and $$ with ‖ζ‖ ≤ d, ‖ϱ‖ ≤ d, ‖ζ − ϱ‖ ≥ ε, there exists a δ > 0 such that:

Definition 4 [16]. A Banach space $$ satisfies Opial property if for every weakly convergent sequence {ζ_n} with weak limit $$, it holds:

for all $$ with ζ ≠ ϱ.

Definition 5 [17]. Suppose $$ is a nonempty subset of a Banach space $$. Let ζ be an element in $$ such that there exists a point ϱ in $$ satisfying the following condition: for any $$, ‖ϱ − ζ‖ ≤ ‖z − ζ‖. In this case, we refer to ϱ as a metric projection of ζ onto $$ and denote it by $$. If $$ exists and is uniquely determined for all $$, then we call the mapping $$, the metric projection onto $$.

Definition 6 [17]. A mapping $$ is said to be quasinonexpansive if:

for all $$ and z ∈ F(Ψ).

Proposition 1 [6]. Let $$ be a nonempty subset of a Banach space $$ If $$ is a mapping satisfying condition (E) with F(Ψ) ≠ ∅ then Ψ is quasinonexpansive.

Definition 7 [18]. Let $$ be a nonempty convex subset of a Banach space $$ and $$ be a mapping. A mapping $$ is said to be an α-Krasnosel’skiĭ mapping associated with Ψ if there exists α ∈ (0, 1) such that:

for all $$

Definition 8 [19]. Let $$ be a nonempty subset of a Banach space $$ A mapping $$ is called asymptotically regular if:

Lemma 1 (Browder [20]; demiclosedness principle). Suppose $$ is a nonempty subset of a Banach space $$ that satisfies the Opial property. Let $$ be a mapping that satisfies condition (E). Consider a sequence ζ_n  in  $$ such that ζ_n weakly converges to ζ, and $$ Then, we have Ψ(ζ) = ζ, which implies that I − Ψ is demiclosed at zero.

Proof. The proof directly follows from [6, Theorem 1].

Lemma 2 (Berinde [8]). Let $$ be a nonempty convex subset of a Banach space $$, and $$ be a mapping. Define $$ as follows:

for all  $$  and  λ ∈ (0, 1).  Then, F(S) = F(Ψ).

Definition 9. Let $$ be a Banach space and $$ be a nonempty subset of $$ We define a mapping $$ an (E)-enriched nonexpansive mapping if ∃b ∈ [0, ∞) and M ∈ [1, ∞) such that:

Remark 1. If $$ is a nonempty subset of $$ and $$ is (E)-enriched nonexpansive, and there exists a sequence {ζ_n} in $$ such that ‖ζ_n − Ψ(ζ_n)‖ → 0. Such a sequence is called almost fixed point sequence (a.f.p.s.) for Ψ.

Proposition 2. Let $$ be a Suzuki-enriched nonexpansive mapping with any b ∈ [0, ∞). Then, Ψ is an (E)-enriched nonexpansive mapping for any b ∈ [0, ∞) and M = 2b + 3.

Proof. We assume that Ψ is a Suzuki-enriched nonexpansive mapping. Then, $$ implies the following equation:

Now, we show that either:

Arguing by contradiction, we suppose that:

By the triangle inequality, we get the following equation:

By Equation (13), we get the following equation:

which is a contradiction. Consequently, Equation (14) holds. Therefore, from Equation (14), we have either:

In the first case, we have the following equation:

In the other case, we have the following equation:

Therefore in both the cases, we get the following equation:

Example 3. Let $$ be the set of real numbers equipped with the standard norm and $$ a subset of $$. Let $$ be a mapping defined by the following equation:

First, we show that Ψ is (E)-enriched nonexpansive mapping. For this, we consider the following nontrivial cases:

• Case (1).

  If ζ ≤ 3 and ϱ = 4, then the following equation is obtained:

  $$ ()

• Case (2).

  If ζ > 3 and ϱ = 4, then the following equation is obtained:

  $$ ()

• Case (3).

  If ζ = 4 and ϱ ≠ 4, then the following equation is obtained:

  $$ ()

Moreover, for ζ = 3 and ϱ = 4, we get the following equations:

and

Thus, Ψ is not a Suzuki-enriched nonexpansive mapping with any b ∈ [0, ∞).

Theorem 2. Let $$ be a Banach space and $$ a nonempty subset of $$ Let $$ be a mapping. If:

• (a)

  Ψ is an (E)-enriched nonexpansive mapping on $$

• (b)

  there exists an a.f.p.s. {ζ_n} for Ψ  in  $$ such that {ζ_n} converges weakly to a point z in $$, and

• (c)

  $$ satisfies the Opial property.

Then, Ψ(z) = z.

Proof. By the definition of mapping Ψ, we have the following equation:

for all $$. Take, $$ and put $$ in Equation (28), then the above inequality is equivalent to the following equations: and

Define the mapping S as follows:

Thus, the following equation is obtained:

Then, from Equation (30), we get the following equation:

for all $$ Thus, S is a mapping satisfying condition (E). From Equation (32), it follows that if {ζ_n} is an a.f.p.s. for Ψ then {ζ_n} is an a.f.p.s. for S. Thus, all the assumptions of [6, Theorem 1] are satisfied and S has a fixed point in $$, that is, z ∈ F(S). From Lemma 2, F(S) = F(Ψ) ≠ ∅. This completes the proof.

Corollary 1. Let $$ be a Banach space and $$ a nonempty weakly compact subset of a Banach space $$ Suppose that $$ satisfies the Opial property. Let $$ be an (E)-enriched nonexpansive mapping on $$ Then, Ψ has a fixed point in $$ if and only if Ψ admits an a.f.p.s.

Theorem 3. Let $$ be a Banach space and $$ a nonempty compact subset of a Banach space $$ Let $$ be an (E)-enriched nonexpansive mapping on $$ Then, Ψ has a fixed point in $$ if and only if Ψ admits an a.f.p.s.

Proof. By following the proof of Theorem 1, we can construct a mapping S that satisfies condition (E). Therefore, all the conditions of [6, Theorem 2] are satisfied, and we can guarantee that S has at least one fixed point. Using Lemma 2, we can conclude that F(S) = F(Ψ) ≠ ∅. This completes the proof.

Theorem 4. Let $$ be a uniformly convex in every direction (or UCED) Banach space and $$ a nonempty weakly compact convex subset of $$  Let  $$ be a mapping. If:

• (a)

  Ψ is an (E)-enriched nonexpansive mapping on $$ and

• (b)

  $$,

then Ψ admits a fixed point.

Proof. In view of Theorem 1 and [6, Theorem 3], one can complete the proof.

Theorem 5. Let $$ be a Banach space and $$ be an (E)-enriched nonexpansive mapping. For a given $$ and $$, if the sequence of iterates $$ converges strongly to ζ^†, where Ψ_α is the α-Krasnosel’skiĭ mapping associated with Ψ, then ζ^† ∈ F(Ψ).

Proof. Using the same technique as in Theorem 1, one can define a mapping $$ as follows:

and S is a mapping satisfying condition (E). For a given $$ and γ ∈ (0, 1), one can define a sequence {ζ_n} as follows:

Using the definition of S, we have the following equation

Take $$, then:

Since $$ strongly converges to a point ζ^† in $$ Thus, all assumptions of [21, Theorem 1] are satisfied, and ζ^† ∈ F(S). But F(S) = F(Ψ). This completes the proof.

Theorem 6. Let $$ be a nonempty closed convex subset of a uniformly convex Banach space $$ and $$ an (E)-enriched nonexpansive mapping with F(Ψ) = {ζ^†}. Assume that the mapping I − Ψ is demiclosed at zero. Then for each $$, the sequence of iterates $$ converges weakly to ζ^†, where Ψ_α is the α-Krasnosel’skiĭ mapping associated with Ψ and $$

Proof. Using the same technique as in Theorem 1, one can define a mapping $$ such that S is a mapping satisfying condition (E). In fact, the demiclosedness of I − Ψ and I − Ψ_α at zeros are equivalent. Further, demiclosedness of I − Ψ and I − S at zeros are equivalent. Keeping [21, Theorem 2.2] in mind, one can complete the proof.

Theorem 7. Let $$ be a nonempty closed convex subset of a uniformly convex Banach space $$ which has the Opial property. Let $$ be an (E)-enriched nonexpansive mapping with F(Ψ) ≠ ∅. Then for each $$ and $$, the sequence of iterates $$ converges weakly to a fixed point of Ψ.

Proof. From [21, Theorem 2.2] and Theorem 6, one can complete the proof.

Definition 10. Let $$ be a Banach space and $$ a nonempty subset of $$. A mapping $$ is said to be b-enriched quasinonexpansive mapping if there exists b ∈ [0, ∞) such that for all $$ and ϱ ∈ F(Ψ) ≠ ∅:

Proposition 3. Let $$ be an (E)-enriched nonexpansive mapping with any b ∈ [0, ∞), M ∈ [1, ∞), and F(Ψ) ≠ ∅. Then, Ψ is a b-enriched quasinonexpansive mapping for any b ∈ [0, ∞).

Proof. Let ϱ ∈ F(Ψ) and $$ we have the following equation:

Thus, Ψ is a b-enriched quasinonexpansive mapping.

Example 4 [6]. Let $$ be a subset of $$. Let $$ be a mapping defined by the following equation:

Clearly, F(Ψ) = {0} ≠ ∅. It can be seen that:

for all ζ ∈ [−1, 1] and Ψ is a b-enriched quasinonexpansive mapping with b = 0.

On the other hand, let $$ and $$ for all $$ we get the following equation:

as n → ∞. Hence, Ψ is not an (E)-enriched nonexpansive mapping for any b ∈ [0, ∞).

Theorem 8. Let $$ be a nonempty convex subset of a uniformly convex Banach space $$ If $$ is a b-enriched quasinonexpansive mapping with F(Ψ) ≠ ∅, Then, the α-Krasnosel’skiĭ mapping Ψ_α for $$ is asymptotically regular.

Proof. By the definition of b-enriched quasinonexpansive mapping, we have the following equation:

for all $$ and ϱ ∈ F(Ψ). Take $$ and put $$ in Equation (43), then the above inequality is equivalent to the following equation:

Define the mapping S as follows:

From Lemma 2, F(S) = F(Ψ). Then, from Equation (44), we get the following equation:

for all $$ and ϱ ∈ F(S). Thus, $$ is a quasinonexpansive mapping. And, we obtain the following equation:

Let $$ For each $$ define ϱ_n+1 = Ψ_α(ϱ_n). Thus, the following equations are obtained:

and

In order to prove that Ψ_α is asymptotically regular, it suffices to prove that $$ Since F(Ψ) ≠ ∅, let ζ₀ ∈ F(Ψ). Since S is a quasinonexpansive mapping and F(S) = F(Ψ):

for all $$ From Equation (45), the following equations are obtained: and

Now:

From Equation (52), the following equation is obtained:

Take $$, then β ∈ (0, 1). From Equation (50), the following equation is obtained:

Therefore, the sequence {‖ζ₀ − ϱ_n‖} is bounded by u₀ = ‖ζ₀ − ϱ₀‖. If $$ for any $$ then from Equation (56), ϱ_n → ζ₀ as n → ∞. If ϱ_n ≠ ζ₀ for all $$ take the following conditions:

If $$ and from Equation (55), we have the following equation:

Since $$ is a uniformly convex Banach space, by using the definition of uniformly convex space with ‖z_n‖ ≤ 1, $$, and the following equation:

Noting that modulus of convexity δ(ε) is a nondecreasing function of ε, we obtain the following equation:

From Equations (60) and (62), the following equation is obtained:

Using induction in the above inequality, it follows that:

We shall prove that $$ Arguing by contradiction, consider that ‖S(ϱ_n) − ϱ_n‖ does not converge to zero. Then, there exists a subsequence $$ of {ϱ_n} such that $$ converges to η > 0. Since δ(⋅) ∈ [0, 1] is nondecreasing and $$ we have $$ for all $$ From Equation (63) and for sufficiently large k, we have the following equation:

Making k → ∞, it follows that $$ By Equation (46), we get $$ and $$ as k → ∞, which is a contradiction.

If $$ then $$ because β ∈ (0, 1). Now:

and by the uniform convexity of $$ we get the following equation:

Using induction in the above inequality, we get the following equation:

Using the similar argument as in the previous case, it can be easily shown that ‖S(ϱ_n) − ϱ_n‖ → 0 as n → ∞. Therefore, in both cases, ‖S(ϱ_n) − ϱ_n‖ → 0 as n → ∞. From Equation (47), ‖Ψ(ϱ_n) − ϱ_n‖ → 0 as n → ∞, Ψ_α is asymptotically regular and this completes the proof.

Remark 2. The above theorem is a generalization of [18, Theorem 1] for a more general class of mappings.

Theorem 9. Let $$ be a nonempty closed subset of a Banach space $$ and $$ a quasinonexpansive mapping. Assume that $$ is strictly convex and $$ is a convex compact subset of $$ If S is continuous then for any $$α ∈ (0, 1), the α-Krasnosel’skiĭ process $$ converges to some ζ^∗ ∈ F(S).

Proof. Following the same line of proof of [7, Theorem 5], one can complete the proof.

Theorem 10. Let $$ and $$ be the same as in Theorem 6. Let $$ be a b-enriched quasinonexpansive mapping with F(Ψ) ≠ ∅.  If  Ψ  is continuous then for any  $$$$ the α-Krasnosel’skiĭ process $$ converges to some ζ^∗ ∈ F(Ψ).

Proof. Following the same proof technique as in Theorem 4, we can define a mapping $$ as follows:

and S is a quasinonexpansive mapping. For given $$, β ∈ (0, 1), we can define a sequence {ζ_n} as follows:

From Theorem 6, the sequence {ζ_n} converges to some ζ^∗ ∈ F(S). Using the definition of S, we have the following equation:

Take $$, then, we obtain the following equation:

Hence, $$ strongly converges to a fixed point of S. But F(S) = F(Ψ). This completes the proof.

Theorem 11. Let $$ be a nonempty closed convex subset of a uniformly convex Banach space $$ Let $$ be a quasinonexpansive mapping with F(S) ≠ ∅ and P the metric projection from $$ into F(S). Then for each $$, the sequence {PSⁿ(ζ)} converges to some ϱ ∈ F(S).

Proof. Following the same line of proof of [7, Theorem 6], one can complete the proof.

Theorem 12. Let $$$$, and P be same as in Theorem 8. Let $$ be a b-enriched quasinonexpansive mapping with F(Ψ) ≠ ∅. Then for each $$, $$ the α-Krasnosel’skiĭ process $$ converges to some ϱ ∈ F(Ψ).

Proof. In view of Theorems 4 and 8, one can complete the proof.