Definition 2.1 (see [19].)The linear space $$ is called a special space of formal power series (or in short (ssfps), if it shows the following settings:

• (1)

   e^(m) ∈ ℋ, for all m ∈ ℕ₀, where $$.

• (2)

  If g ∈ ℋ and $$, for every y ∈ ℕ₀, then h ∈ ℋ.

• (3)

  Suppose h ∈ ℋ; then h_[.] ∈ ℋ, with $$ and [b/2] marks the integral part of b/2.

Definition 2.2 (see [19].)A subspace ℋ_ψ of ℋ is said to be a premodular (ssfps), if there is a function ψ : ℋ⟶[0, ∞) that verifies the next conditions:

• (i)

  For h ∈ ℋ, we have ψ(h) ≥ 0 and h = θ⇔ψ(h) = 0, where θ is the zero function of ℋ.

• (ii)

  Suppose h ∈ ℋ and ϖ ∈ ℝ; then there is q ≥ 1 with ψ(ϖh) ≤ |ϖ|qψ(h).

• (iii)

  Let f, g ∈ ℋ; then there is A ≥ 1 such that ψ(f + g) ≤ A(ψ(f) + ψ(g)).

• (iv)

  Suppose $$, for every b ∈ ℕ₀; then ψ(f) ≤ ψ(g).

• (v)

  There is K₀ ≥ 1 so that ψ(f) ≤ ψ(f_[.]) ≤ K₀ψ(f).

• (vi)

  $$, where $$ indicates the space of finite formal power series; that is, for $$, we have l ∈ ℕ₀ with $$.

• (vii)

  One has ξ > 0 with ψ(λe⁽⁰⁾) ≥ ξ|λ|ψ(e⁽⁰⁾), where λ ∈ ℝ.

Definition 2.3 (see [24].)A function $$ is said to be an s-number, if the sequence $$, for all $$, shows the following settings:

• (a)

  If $$, then ‖B‖ = s₀(B) ≥ s₁(B) ≥ s₂(B) ≥ …≥0.

• (b)

  s_b+a−1(B₁ + B₂) ≤ s_b(B₁) + s_a(B₂), for every $$, b, a ∈ ℕ₀.

• (c)

  The inequality s_a(ABD) ≤ ‖A‖s_a(B)‖D‖ holds, if $$, $$, and $$, where $$ and $$ are arbitrary Banach spaces.

• (d)

  Suppose $$ and λ ∈ ℝ; then s_a(λA) = |λ|s_a(A).

• (e)

  Let rank(A) ≤ b; then s_b(A) = 0, whenever $$.

• (f)

  Assume that I_λ denotes the identity mapping on the λ-dimensional Hilbert space $$; then s_r≥λ(I_λ) = 0 or s_r<λ(I_λ) = 1.

Definition 2.4. (see [7]) A class U⊆L is said to be an operator ideal if every vector $$ shows the following settings:

• (i)

  F⊆U, where F is the ideal of all finite rank operators between any arbitrary Banach spaces.

• (ii)

  $$ is linear space on ℝ.

• (iii)

  If $$, $$, and $$, then $$.

Definition 2.5. (see [10]) A function g : U⟶[0, ∞) is called a prequasinorm on the ideal U if it shows the following settings:

• (1)

  For each $$, g(A) ≥ 0 and g(A) = 0⇔A = 0.

• (2)

  One has M ≥ 1 with g(βA) ≤ M|β|g(A), for all β ∈ ℝ and $$.

• (3)

  One has K ≥ 1 with g(A₁ + A₂) ≤ K[g(A₁) + g(A₂)], for every $$.

• (4)

  There is  C ≥ 1 so that if $$, $$, and $$, then g(DBA) ≤ C‖D‖g(B)‖A‖, where $$ and $$ are normed spaces.

Definition 2.6. (see [19])

• (a)

  The prequasinormed (ssfps) η on $$ is said to be η-convex, if

•  

  η(εg + (1 − ε)h) ≤ εη(g) + (1 − ε)η(h), for all ε ∈ [0,1] and $$.

• (b)

  $$ is η-convergent to $$, if and only if lim_y⟶∞η(h^(y) − h) = 0. If the η-limit exists, then it is unique.

• (c)

  $$ is η-Cauchy, if lim_x,y⟶∞η(h^(x) − h^(y)) = 0.

• (d)

  $$ is η-closed, if lim_y⟶∞η(h^(y) − h) = 0, where $$; then h ∈ Y.

• (e)

  $$ is η-bounded, if δ_η(Y) = sup{η(g − h) : g, h ∈ Y} < ∞.

• (f)

  The η-ball of radius l ≥ 0 and center g, for all $$, is defined as

  $$ (3)

• (g)

  A prequasinormed (ssfps) η on ℋ_w((r_y)) verifies the Fatou property, if for every sequence $$ with $$ and every $$,

  $$ (4)

Lemma 2.7. (see [19]) The function $$, for all $$, verifies the Fatou property, when (r_n) ∈ mi_↗∩ℓ_∞.

Lemma 2.8. (see [19]) If (r_n) ∈ mi_↗∩ℓ_∞, then the following settings hold:

• (1)

  The function space $$ is a prequasiclosed and Banach (ssfps), with

  $$ (5)

• (2)

  The class $$ is a prequasi-Banach and closed operator ideal, where $$, where $$, for every z ∈ ℂ.

Lemma 2.9. The following inequalities will be utilized in the continuation:

• (i)

  [25] If b ≥ 2 and for each f, g ∈ ℂ, then

  $$ (6)

• (ii)

  [26] Let 1 < b ≤ 2, and for every f, g ∈ ℂ with |f| + |g| ≠ 0; then

  $$ (7)

• (iii)

  [27] Assume b_y > 0 and f_y, g_y ∈ ℂ, for each y ∈ ℕ₀; then $$ where K = max{1, sup_yb_y}.

Definition 3.1 (see [4], [29].)We define the prequasinorm ψ’s uniform convexity type behavior as follows:

• (1)

  [30] Suppose λ > 0 and β > 0. Let

  $$ (8)

•  

  When $$, we put

  $$ (9)

•  

  When $$, we put V₁(λ, β) = 1. The function ψ holds the uniform convexity (UC) if for each λ > 0 and β > 0, we have V₁(λ, β) > 0. Observe that, for all λ > 0, then $$, for very small β > 0.

• (2)

  [28] The function ψ verifies (UUC) if for every γ ≥ 0 and β > 0, there is ζ₁(γ, β) with

  $$ (10)

• (3)

  [28] Suppose λ > 0 and β > 0. Let

  $$ (11)

•  

  When $$, we put

  $$ (12)

•  

  When $$, we place V₂(λ, β) = 1. The function ψ satisfies (UC 2) if for every λ > 0 and β > 0, one has V₂(λ, β) > 0. Observe that, for each λ > 0, $$, for very small β > 0.

• (4)

  [28] The function ψ verifies (UUC 2) if for all γ ≥ 0 and β > 0, there is ζ₂(γ, β) with

  $$ (13)

• (5)

  [30] The function ψ is strictly convex (SC), if for all $$ so that ψ(f) = ψ(g) and ψ((f + t)/2) = (ψ(f) + ψ(g))/2, we get f = g.

Theorem 3.2. The function $$, for all $$, is (UUC2), if (r_a) ∈ mi_↗∩ℓ_∞ with r₀ > 1.

Proof. Let the condition be satisfied, b > 0, and a > p ≥ 0. Suppose $$ so that

From the definition of ψ, we have

and this implies b ≤ 2. Consequent, let Q = {x ∈ ℕ₀ : 1 < r_x < 2} and P = {x ∈ ℕ₀ : r_x ≥ 2} = ℕ₀∖Q. For every $$, we get $$. From the setup, one has ψ_P(f − g/2) ≥ ab/2 or ψ_Q(f − g/2) ≥ ab/2. Assume first ψ_P(f − g/2) ≥ ab/2. By using Lemma 2.9, condition (i), we obtain

This explains

As

by adding inequalities 2 and 3, and from inequality 1, we have

This gives

Next, suppose ψ_Q((f − g)/2) ≥ ab/2. Set B = (b/4)^K,

As B ≤ 1 and the power function is convex,

Since ψ_Q((f − g)/2) ≥ ab/2, we get

For any d ∈ Q₂, we have

By Lemma 2.9, condition (ii), we have

Hence,

This investigates

Since

by adding inequalities (27) and (28), one has

Since

by adding inequalities (29) and (30) and from inequality 1, we obtain

This implies

It is clear that

By using inequalities 4 and 9 and Definition 3.1, we put

Therefore, we have V₂(a, b) > ζ₂(p, b) > 0, and we conclude that ψ is (UUC2).We will examine the property (R) of the prequasinormed (ssfps) $$ in this second part.

Theorem 3.3. Let (r_a) ∈ mi_↗∩ℓ_∞ with r₀ > 1; then the next setups are satisfied:

• (1)

  Assume that $$, Λ ≠ ∅, ψ-closed and ψ-convex, where $$, for all $$. Suppose $$ so that

  $$ (35)

•  

  Hence, one has a unique λ ∈ Λ with d_ψ(f, Λ) = ψ(f − λ).

• (2)

  $$ satisfies the property (R). This means that, for every decreasing sequence $$ of ψ-closed and ψ-convex nonempty subsets of $$ such that $$, for some $$, then $$.

Proof. Suppose the setups are satisfied. To show (1), let f ∉ Λ as Λ is ψ-closed. Then, one has A : = d_ψ(f, Λ) > 0. So, for every p ∈ ℕ₀, we have g_p ∈ Λ with ψ(f − g_p) < A(1 + 1/p). Assume {g_p/2} is not ψ-Cauchy. Therefore, one obtains a subsequence {g_h(p)/2} and b₀ > 0 so that ψ((g_h(p) − g_h(q))/2) ≥ b₀, for all p > q ≥ 0. Furthermore, one has V₂(A(1 + 1/p), b₀/2A) > ξ: = β₂(A(1 + 1/p), b₀/2A) > 0, for each p ∈ ℕ₀. As

and

Under p > q ≥ 0, we get

Therefore,

with q ∈ ℕ₀. If we let q⟶∞, we get

We have a contradiction. Then {g_p/2} is ψ-Cauchy. As $$ is ψ-complete, {g_p/2}ψ converges to some g. For all q ∈ ℕ₀, we have the sequence {g_p + g_q/2}ψ that converges to g + g_q/2. As Λ is ψ-closed and ψ-convex, one gets g + g_q/2 ∈ Λ. Surely g + g_q/2ψ converges to 2g, so 2g ∈ Λ. For λ = 2g and using Theorem 2.7, since ψ satisfies the Fatou property, we get

Therefore, ψ(f − λ) = d_ψ(f, Λ). As ψ is (UUC2), so it is SC, which implies that λ is unique. To show (2), let $$, for some p₀ ∈ ℕ₀. $$ is increasing. Let $$. Suppose A > 0. Else f ∈ Λ_p, for every p ∈ ℕ₀. By using Part (1), we have one point g_p ∈ Λ_p with d_ψ(f, Λ_p) = ψ(f − g_p), for every p ∈ ℕ₀. A consistent proof will show that {g_p/2}ψ converges to some $$. When {Λ_p} are ψ-convex, decreasing, and ψ-closed, we get $$.

This third part discusses the prequasinormed structure’s ψ-normal structure feature (ssfps) $$.

Definition 3.4. $$ satisfies the ψ-normal structure property if for all nonempty ψ-bounded, ψ-convex, and ψ-closed subset Λ of $$ did not decrease to one point, we have f ∈ Λ with

Theorem 3.5. If (r_a) ∈ mi_↗∩ℓ_∞ with r₀ > 1, then $$ holds the ψ-normal structure property, where $$, for every $$.

Proof. Assume the setups are satisfied. Theorem 3.2 explains that ψ is (UUC2). Let Λ be a ψ-bounded, ψ-convex, and ψ-closed subset of $$ not decreased to unique point. Hence, δ_ψ(Λ) > 0. Let A = δ_ψ(Λ). Suppose f,  g ∈ Λ with f ≠ g. So ψ((f − g)/2) = b > 0. For all λ ∈ Λ, one obtains ψ(f − λ) ≤ A and ψ(g − λ) ≤ A. Since Λ is ψ-convex, one has (f + g)/2 ∈ Λ. Hence,

For all λ ∈ Λ,

Definition 4.1. An operator $$ is called a Kannan ψ-Lipschitzian, if there is κ ≥ 0, so that

For all $$, one has the following:

• (1)

  If κ ∈ [0, 1/2), then the operator $$ is said to be Kannan ψ-contraction.

• (2)

  If κ = 1/2, then the operator $$ is said to be Kannan ψ-nonexpansive.

A vector $$ is called a fixed point of $$, when $$.

Theorem 4.2. If (r_a) ∈ mi_↗∩ℓ_∞ and $$ is Kannan ψ-contraction mapping, where $$, for all $$, then $$ has a unique fixed point.

Proof. Let the setups be satisfied. For every $$, then $$. Since $$ is a Kannan ψ-contraction mapping, we have

Therefore, for every p, q ∈ ℕ₀ with q > p, then we get

So $$ is a Cauchy sequence in $$. As the space $$ is prequasi-Banach (ssfps). Therefore, there is $$ such that $$. To prove that $$, by Theorem 2.7, ψ holds the Fatou property, and we have

Hence, $$. Then g is a fixed point of $$. To show that the fixed point is unique, assume we have two different fixed points $$ of $$. Then, one has

Therefore, f = g.

Corollary 4.3. Let (r_a) ∈ mi_↗∩ℓ_∞ and $$ be Kannan ψ-contraction mapping, where $$, for all $$; then $$ has one and only one fixed point g with $$.

Proof. It is obvious, so it is omitted.

Definition 4.4. Assume $$ is a prequasinormed (ssfps) and $$. The operator $$ is called ψ sequentially continuous at $$, if and only if when $$, $$.

Theorem 4.5. Let (r_a) ∈ mi_↗∩ℓ_∞ with r₀ > 1 and $$, where $$, for all $$. The point $$ is the unique fixed point of $$, if the following conditions are satisfied:

• (a)

  $$ is Kannan ψ-contraction mapping.

• (b)

  $$ is ψ sequentially continuous at $$.

• (c)

  One has $$ with the sequence of iterates $$ having a subsequence $$ converging to g.

Proof. Suppose the settings are verified. If g is not a fixed point of $$, then $$. By conditions (b) and (c), we have

Since the mapping $$ is Kannan ψ-contraction, one can see

Since p_i⟶∞, one has a contradiction. Hence, g is a fixed point of $$. To explain that the fixed point g is unique, suppose we have two different fixed points $$ of $$. Therefore, one gets

So, g = b.

Theorem 4.6. Assume $$ is an increasing, and $$, where $$, for all $$. The point $$ is the only fixed point of $$, if the following conditions are satisfied:

• (a)

  $$ is Kannan ψ-contraction mapping.

• (b)

  $$ is ψ sequentially continuous at $$.

• (c)

  One has $$ so that the sequence of iterates $$ has a subsequence $$ converging to g.

Proof. Let the conditions be verified. If g is not a fixed point of $$, then $$. By conditions (b) and (c), we have

As the operator $$ is Kannan ψ-contraction, one can see

Since p_i⟶∞, one obtains a contradiction. Hence, g is a fixed point of $$. To explain that the fixed point g is unique, assume we have two different fixed points $$ of $$. Then, one gets

So, g = b.

Example 4.7. Pick up $$, where $$, for all $$ and

For all $$ with ψ(f₁), ψ(f₂) ∈ [0,1), we have

For all $$ with ψ(f₁), ψ(f₂) ∈ [1, ∞), we have

For all $$ with ψ(f₁) ∈ [0,1) and ψ(f₂) ∈ [1, ∞), we have

Hence, $$ is Kannan ψ-contraction mapping. By Theorem 2.7, the function ψ satisfies the Fatou property. By Theorem 4.2, the map $$ has a unique fixed point $$.

Let $$ with lim_n⟶∞ ψ(f⁽ⁿ⁾ − f⁽⁰⁾) = 0, where $$ and ψ(f⁽⁰⁾) = 1. Since the prequasinorm ψ is continuous, one gets

Therefore, $$ is not ψ sequentially continuous at f⁽⁰⁾. Then, the map $$ is not continuous at f⁽⁰⁾.

If $$, for all $$. For all $$ with ψ(f₁), ψ(f₂) ∈ [0,1), one obtains

For all $$ with ψ(f₁), ψ(f₂) ∈ [1, ∞), we have

For all $$ with ψ(f₁) ∈ [0,1) and ψ(f₂) ∈ [1, ∞), we have

Therefore, the map $$ is Kannan ψ-contraction mapping and $$

Obviously, $$ is ψ sequentially continuous at $$ and $$ has a subsequence $$ converging to θ. By Theorem 4.5, the point $$ is the unique fixed point of $$.

Example 4.8. Assume $$, where $$, for all $$ and

For all $$ with ψ(f₁), ψ(f₂) ∈ [0,1), we have

For all $$ with ψ(f₁), ψ(f₂) ∈ [1, ∞), we have

For all $$ with ψ(f₁) ∈ [0,1) and ψ(f₂) ∈ [1, ∞), we have

Hence, $$ is Kannan ψ-contraction mapping. From Theorem 2.7, the function ψ satisfies the Fatou property. By Theorem 4.2, the map $$ has a unique fixed point $$.

Let $$ with lim_n⟶∞ψ(f⁽ⁿ⁾ − f⁽⁰⁾) = 0, where $$ and ψ(f⁽⁰⁾) = 1. Since the prequasinorm ψ is continuous, we have

Therefore, $$ is not ψ sequentially continuous at f⁽⁰⁾. So, the map $$ is not continuous at f⁽⁰⁾.

Suppose $$, for all $$. For all $$ with ψ(f₁), ψ(f₂) ∈ [0,1), we have

For all $$ with ψ(f₁), ψ(f₂) ∈ [1, ∞), we have

For all $$ with ψ(f₁) ∈ [0,1) and ψ(f₂) ∈ [1, ∞), we have

So, the map $$ is Kannan ψ-contraction mapping and $$

Obviously, $$ is ψ sequentially continuous at $$ and $$ has a subsequence $$ converging to θ. By Theorem 4.5, the point $$ is the unique fixed point of $$.

Example 4.9. Let $$, where $$, for all $$ and

For all $$ with $$, we have

For all $$ with $$ and then for any ε > 0, we have

For all $$ with $$ and $$, we have

Hence, $$ is Kannan ψ-contraction mapping. Clearly, $$ is ψ sequentially continuous at $$, and there is $$ with $$ such that the sequence of iterates $$ has a subsequence $$ converging to 1/17. By Theorem 4.5, the map $$ has one fixed point $$. Note that $$ is not continuous at $$.

If $$, $$. For all $$ with $$, we have

For all $$ with $$ and then for any ε > 0, we have

For all $$ with $$ and $$, we have

So, $$ is Kannan ψ-contraction mapping. By Theorem 2.7, the function ψ satisfies the Fatou property. By Theorem 4.2, the map $$ has a unique fixed point $$.

Example 4.10. Let $$, where $$, for all $$ and

For all $$ with $$, we have

For all $$ with $$ and then for any ε > 0, we have

For all $$ with $$ and $$, we have

Hence, $$ is Kannan ψ-contraction mapping. Evidently, $$ is ψ sequentially continuous at $$ and there is $$ with $$ such that the sequence of iterates $$ has a subsequence $$ converging to 1/3z. By Theorem 4.5, the map $$ has one fixed point $$. Note that $$ is not continuous at $$.

If $$, $$. For all $$ with $$, we have

For all $$ with $$ and then for any ε > 0, we have

For all $$ with $$ and $$, we have

Hence, $$ is Kannan ψ-contraction mapping. By Theorem 2.7, the function ψ satisfies the Fatou property. By Theorem 4.2, the map $$ has a unique fixed point $$.

Lemma 5.1. Allow the prequasinormed (ssfps) $$ to validate the (R) and ψ-quasi-normal properties. If Ξ is a nonempty ψ-bounded, ψ-convex, and ψ-closed subset of $$, suppose $$ is a Kannan ψ-nonexpansive mapping. For y > 0, let $$. Let

Proof. As $$, this gives Ξ_y ≠ ∅. As the ψ-balls are ψ-convex and ψ-closed, one has Ξ_y is a ψ-closed and ψ-convex subset of Ξ. To prove that Ξ_y ⊂ G_y, suppose f ∈ Ξ_y. If $$, we get f ∈ G_y. Otherwise, suppose $$. Let

From the definition of λ, $$. So, $$, and we have $$. Assume η > 0. Hence, there is w ∈ G_y so that $$. Then,

Since η is randomly positive, we obtain $$, so we have f ∈ G_y. As $$, one has $$; this indicates that Ξ_y is $$-invariant, consequent to prove that δ_ψ(Ξ_y) ≤ y, as

For every f, g ∈ G_y, let f ∈ G_y. Hence, $$. The definition of Ξ_y provides $$. Hence, $$. So, one has ψ(g − w) ≤ y, for every g, w ∈ Ξ_y, this means δ_ψ(Ξ_y) ≤ y. This completes the proof.

Theorem 5.2. Let the prequasinormed (ssfps) $$ verify the ψ-quasi-normal property and the (R) property. Assume Ξ is a nonempty, ψ-convex, ψ-closed, and ψ-bounded subset of $$. Suppose $$ is a Kannan ψ-nonexpansive mapping. Hence, $$ has a fixed point.

Proof. Suppose $$ and d_t = d₀ + 1/t, for every t ≥ 1. By using the definition of d₀, we have $$, with t ≥ 1. Assume $$ is described as in Lemma 5.1. Obviously, $$ is a decreasing sequence of nonempty ψ-bounded, ψ-closed, and ψ-convex subsets of Ξ. The property (R) proves that $$. Assume f ∈ Ξ_∞; one can see $$, for every t ≥ 1. Suppose p⟶∞; we have $$; this gives $$. Therefore, $$. We have d₀ = 0. Else, d₀ > 0; this gives that $$ fails to have a fixed point. Let $$ be defined as in Lemma 5.1. As $$ misses to have a fixed point and $$ is $$-invariant, so $$ has more than one point, which implies, $$. By the ψ-quasinormal property, there is $$ so that

Theorem 5.3. If (r_a) ∈ mi_↗∩ℓ_∞ with r₀ > 1, Ξ is a nonempty, ψ-convex, ψ-closed, and ψ-bounded subset of $$, where $$, for every $$, and $$ is a Kannan ψ-nonexpansive mapping. Then, $$ has a fixed point.

Example 5.4. Let $$ with $$where $$ and $$, for all $$. In view of example 4.8, the map $$ is Kannan ψ-contraction mapping. This implies Kannan ψ-nonexpansive mapping. Evidently, Ξ is a nonempty, ψ-convex, ψ-closed, and ψ-bounded subset of $$. By Theorem 5.3, the map $$ has one fixed point (f = θ) in Ξ.

Notations 6.1. [19]

Definition 6.2. If $$ and $$ are Banach spaces, a prequasinorm Ψ on the ideal $$, where Ψ(W) = ψ(f_s) and $$ converge for any z ∈ ℂ, satisfies the Fatou property if for every sequence $$ with lim_a⟶∞Ψ(W_a − W) = 0 and any $$,

Theorem 6.3. Suppose $$ and $$ are Banach spaces. The prequasinorm $$, for all $$ does not satisfy the Fatou property, if (r_a) ∈ mi_↗∩ℓ_∞.

Proof. Let the condition be satisfied and $$ with lim_p⟶∞Ψ(W_p − W) = 0. By Theorem 2.8, the space $$ is a prequasiclosed ideal, and then $$. Hence, for all $$, we have

Hence, Ψ does not satisfy the Fatou property.

Definition 6.4. Suppose $$ and $$ are Banach spaces. For the prequasinorm Ψ on the ideal $$, where Ψ(W) = ψ(f_s), where $$ converges for any z ∈ ℂ, an operator $$ is called a Kannan Ψ-Lipschitzian, if there is κ ≥ 0, so that

for all $$. An operator G is said to be

• (1)

  Kannan Ψ-contraction, when κ ∈ [0, 1/2).

• (2)

  Kannan Ψ-nonexpansive, when κ = 1/2.

Definition 6.5. Suppose $$ and $$ are Banach spaces. For the prequasinorm Ψ on the ideal $$, where Ψ(W) = ψ(f_s), where $$ converges for any z ∈ ℂ, $$ and $$. The operator G is said to be Ψ sequentially continuous at B, if and only if when $$, $$.

Theorem 6.6. Suppose $$ and $$ are Banach spaces. Let (r_a) ∈ mi_↗∩ℓ_∞ and $$, where $$, for all $$. The point $$ is the unique fixed point of G, if the following conditions are satisfied:

• (a)

  G is Kannan Ψ-contraction mapping.

• (b)

  G is Ψ sequentially continuous at a point $$.

• (c)

  There is $$ so that the sequence of iterates {G^pB} has a subsequence $$ converging to A.

Proof. Let the conditions be verified. If A is not a fixed point of G, then GA ≠ A. From conditions (b) and (c), we have

Since G is Kannan Ψ-contraction mapping, one can see

As p_i⟶∞, we have a contradiction. Therefore, A is a fixed point of G. To show that the fixed point A is unique. Let us have two different fixed points $$ of G. Therefore, one has

So, A = D.

Example 6.7. Suppose $$ and $$ are Banach spaces; $$, where $$, for every $$ and

For all $$ with Ψ(W₁), Ψ(W₂) ∈ [0,1), we have

For all $$ with Ψ(W₁), Ψ(W₂) ∈ [1, ∞), we have

For all $$ with Ψ(W₁) ∈ [0,1) and Ψ(W₂) ∈ [1, ∞), we have

Hence, G is Kannan Ψ-contraction mapping and $$

Evidently, G is Ψ sequentially continuous at the zero operator $$ and {G^pW} has a subsequence $$ converging to Θ. By Theorem 6.6, the zero operator $$ is the only fixed point of G. Assume $$ with $$, where $$ and Ψ(W⁽⁰⁾) = 1. Since the prequasinorm Ψ is continuous, one obtains

Therefore, G is not Ψ sequentially continuous at W⁽⁰⁾. Then, the map G is not continuous at W⁽⁰⁾.

Example 6.8. If $$ and $$ are Banach spaces, $$, where $$, for every $$ and

For all $$ with Ψ(W₁), Ψ(W₂) ∈ [0,1), we have

For all $$ with Ψ(W₁), Ψ(W₂) ∈ [1, ∞), we have

For all $$ with Ψ(W₁) ∈ [0,1) and Ψ(W₁) ∈ [1, ∞), we have

Hence, G is Kannan Ψ-contraction mapping and $$

Obviously, G is Ψ sequentially continuous at the zero operator $$ and {G^pW} has a subsequence $$ converging to Θ. By Theorem 6.6, the zero operator $$ is the only fixed point of G. Suppose $$ with $$, where $$ and Ψ(W⁽⁰⁾) = 1. Since the prequasinorm Ψ is continuous, one gets

Therefore, G is not Ψ sequentially continuous at W⁽⁰⁾. Then, the map G is not continuous at W⁽⁰⁾.

Theorem 7.1. The summable equation (10) has one solution in $$, if $$, f : ℕ₀ × ℂ⟶ℂ, $$, $$, and for every a ∈ ℕ₀, we have κ ∈ [0, 1/2), with

Proof. Let the setups be verified. Consider the mapping $$ defined by (11). We have

According to Theorem 4.2, one obtains a unique solution of equation (10) in $$.

Example 7.2. Assume the function space $$, where $$, for all $$. Consider the summable equation

where q > 0 and i² = −1 and let $$ defined by

It is easy to see that

By Theorem 7.1, the summable equation (114) has one solution in $$.

Example 7.3. Given the function space $$, where $$, for all $$, consider the summable equation (12). It is easy to see that

By Theorem 7.1, the summable equation (114) has one solution in $$.

Example 7.4. Given the function space $$, where $$, for all $$, consider the summable equation (114) with a ≥ 2 and let W : Ξ⟶Ξ, where $$, defined by

Clearly, Ξ is a nonempty, ψ-convex, ψ-closed, and ψ-bounded subset of $$. It is easy to see that

By Theorem 7.1 and Theorem 5.3, the summable equation (114) with a ≥ 2 has a solution in Ξ.

Theorem 7.5. The summable equation (120) has one solution in $$, if the following conditions are satisfied:

• (a)

  $$, f : ℕ₀ × [0, ∞)⟶ℂ, $$, $$, and for every a ∈ ℕ₀, one has κ ∈ [0, 1/2), with

  $$ (122)

• (b)

  W is Ψ sequentially continuous at a point $$.

• (c)

  There is $$ so that the sequence of iterates {W^pB} has a subsequence $$ converging to D.

Proof. Suppose the settings are verified. Consider the mapping $$ defined by (16). We have

In view of Theorem 6.6, one obtains a unique solution of (120) at $$.

Example 7.6. Assume the function space $$, where $$, for all $$.

Consider the nonlinear difference equation

where $$, i² = −1, and let $$ defined by

It is easy to see that

By Theorem 7.1, the nonlinear difference equation (124) has one solution in $$.

Example 7.7. Given the function space $$, where $$, for all $$, consider the nonlinear difference equation (17). It is easy to see that

By Theorem 7.1, the nonlinear difference equation (124) has one solution in $$.

Example 7.8. Given the function space $$, where $$, for all $$, consider the nonlinear difference equation (124) with a ≥ 1 and let W : Ξ⟶Ξ, where $$, defined by

Clearly, Ξ is a nonempty, ψ-convex, ψ-closed, and ψ-bounded subset of $$. It is easy to see that

By Theorem 7.1 and Theorem 5.3, the nonlinear difference equation (124) with a ≥ 1 has a solution in Ξ.