Lemma 1 (see [23].)Suppose $$ for all $$, then

where K = max{1, sup_qτ_q}.

Theorem 2. If (τ_q) ∈ ℓ_∞ and τ_a > 1, for all $$, then

Proof.

Let us indicate ϑ, the zero function of $$ and the space of finite formal power series by $$, i.e, when $$, then there is $$ so that $$. Nakano [28] introduced the concept of modular vector spaces.

Definition 3. Suppose $$ is a vector space. A function $$ is said to be modular, if the next conditions hold

• (a)

  If $$, then h(g) ≥ 0 and g = ϑ⟺h(g) = 0

• (b)

  h(ηg) = h(g) holds, for all $$ and |η| = 1

• (c)

  The inequality h(αg + (1 − α)f) ≤ h(g) + h(f) satisfies, for all $$ and α ∈ [0, 1]

Definition 4 (see [29].)The space $$ is said to be a special space of formal power series (or in short ssfps), if it verifies the following settings:

• (1)

  $$, for every $$, where $$

• (2)

  For all $$ and $$, for every $$, then $$

• (3)

  If $$ then $$, where $$ and [p/2] indicates the integral part of p/2

Definition 5 (see [29].)A subspace $$ of the ssfps is said to be a premodular ssfps, if there is a function $$ verifies the following conditions:

• (i)

  If $$, then h(g) ≥ 0 and g = ϑ⇔h(g) = 0

• (ii)

  When $$ and λ ∈ ℂ, then there are Q ≥ 1 such that h(λf) ≤ |λ|Qh(f)

• (iii)

  Suppose $$, then there are P ≥ 1 such that h(f + g) ≤ P(h(f) + h(g))

• (iv)

  Suppose $$, for all $$, then h(f) + h(g)

• (v)

  There are P₀ ≥ 1 such that h(f) ≤ h(f_[⋅] ≤ P₀h(f))

• (vi)

  The closure of $$

• (vii)

  There are ξ > 0 so that h(λe⁽⁰⁾) ≥ ξ|λ|h(e⁽⁰⁾), where λ ∈ ℂ

Clearly, the concept of premodular vector spaces is more general than modular vector spaces, an example of premodular vector space but not modular vector space.

Example 1. The function $$ is a premodular (not a modular) on the vector space $$. As for every $$, one has

Example 2. The function $$ is a premodular (modular) on the vector space $$.

Definition 6 (see [29].)A subspace $$ of the ssfps is said to be a prequasinormed ssfps, if there is a function $$ verifies the following conditions:

• (i)

  If $$, then the h(g) ≥ 0 and g = ϑ⟺h(g) = 0

• (ii)

  When $$ and λ ∈ ℂ, then there are Q ≥ 1 such that h(λf) ≤ |λ|Qh(f)

• (iii)

  Suppose $$ then there are P ≥ 1 such that h(f + g) ≤ P(h(f) + h(g))

Theorem 7 (see [30].)All premodular ssfps $$ is a prequasinormed ssfps.

Theorem 8 (see [30].)All quasinormed (ssfps) is a prequasinormed (ssfps).

Definition 9.

• (a)

  The function h on $$ is said to be h-convex, if

for every α ∈ [0, 1] and $$

• (b)

  $$ is h-convergent to $$, if and only if, lim_q⟶∞h(g_q − g) = 0. When the h-limit exists, then it is unique

• (c)

  $$ is h-Cauchy, if lim_q,r⟶∞h(g_q − g_r) = 0

• (d)

  $$ is h-closed, when for all h-converges, $$ to g, then g ∈ Γ

• (e)

  $$ is h-bounded, if δ_h(Γ) = sup{h(f − g): f, g ∈ Γ} < ∞

• (f)

  The h-ball of radius ε ≥ 0 and center f, for every $$, is described as

• (g)

  A prequasinorm h on $$ holds the Fatou property, if for every sequence $$ under lim_q⟶∞h(g^q − g) = 0 and all $$, one has h(f − g) ≤ sup_rinf_q≥rh(f − g^q)

Theorem 10. $$, where $$. for all $$, is a premodular (ssfps), when $$ with τ₀ > 1.

Proof. Evidently, h(f) ≥ 0 and h(f) = 0⟺ = ϑ.

Let $$. One has $$ with

As $$, hence from conditions (1-i) and (1-ii), one has $$ is linear. Also $$, for all $$, since

There is $$ with h(αf) ≤ Q|α|h(f) for all $$ and α ∈ ℂ

Assume |f_q| ≤ |g_q|, for all $$ and $$. One finds

then $$.

Obviously, from (58).

Let $$, we get

then $$.

From (59), we obtain $$.

Evidently the closure of $$.

There is $$, for α ≠ 0 or σ > 0, for α = 0 with h(αe⁽⁰⁾) ≥ σ|α|h(e⁽⁰⁾).

Theorem 11. If $$ with τ₀ > 1, then $$ is a prequasi-Banach (ssfps), where

for every $$.

Proof. According to Theorems 10 and 7, the space $$ is a prequasinormed (ssfps). Assume $$ is a Cauchy sequence in $$, hence for every ε ∈ (0, 1), one has $$ such that for all l, m ≥ l₀, one gets

Theorem 12. Suppose $$ with τ₀ > 1, then $$ is a pre-quasi closed (ssfps), where

for every $$.

Proof. According to Theorems 10 and 7, the space $$ is a prequasinormed (ssfps). Assume $$ and lim_l⟶∞h(f^l − f⁰) = 0, then for all, ε ∈ (0, 1), there is $$ such that for all l ≥ l₀, we obtain

which implies $$, as ℂ is a complete space. Therefore, $$ is a convergent sequence in ℂ, for fixed $$. So $$, for fixed $$. Since, h(f⁰) ≤ h(f^l − f⁰) + h(f^l) < ∞. So, $$.

Theorem 13. The function

holds the Fatou property, when $$ with τ₀ > 1, for all $$.

Proof. Let $$ such that $$ Since $$ is a pre-quasi closed space, one has $$. For all $$, one gets

Theorem 14. The function

does not hold the Fatou property, for all $$, when (τ_q) ∈ ℓ_∞ and τ_q > 1, for all $$.

Proof. Let $$ so that lim_r⟶∞h(g^r − g) = 0. Since $$ is a pre-quasi closed space, one gets $$. For every $$, we obtain

Example 3. For $$, the function

is a norm on $$.

Example 4. The function

is a prequasinorm (not a quasinorm) on $$.

Example 5. The function

is a prequasinorm (not a quasinorm) on $$.

Example 6. The function

is a prequasinorm, quasinorm, and not a norm on $$, for 0 < d < 1.

Definition 15. The function $$ is said to be lower semicontinuous at $$ if $$, where $$, where V(G⁽⁰⁾) is a neighborhood system of G⁽⁰⁾.

Definition 16. The function $$ is said to be proper, when

Theorem 17. Suppose $$ and Ξ is a h-closed subset of $$, and Ψ₁ : Ξ⟶(−∞, ∞] is a proper, h-lower semicontinuous function with inf_G∈ΞΨ₁(G) > −∞. If γ > 0, {ϖ_q} ⊂ (0, ∞), and G⁽⁰⁾ ∈ Ξ so that Ψ₁(G⁽⁰⁾) ≤ inf_G∈ΞΨ₁(G) + γ. One gets {G^(q)} ∈ Ξ which h-converges to some G^(γ), and

• (i)

  h(G^(γ) − G^(q)) ≤ γ/2^qϖ₀, with $$

• (ii)

  when G ≠ G^(γ), then

Proof. If S(G⁽⁰⁾) = {G ∈ Ξ : Ψ₁(G) + ϖ₀h(G − G⁽⁰⁾) ≤ Ψ₁(G⁽⁰⁾)}. Since G⁽⁰⁾ ∈ S(G⁽⁰⁾), then $$. As Ψ₁ is h-lower semicontinuous, h holds the Fatou property and Ξ is h-closed, then S(G⁽⁰⁾) is h-closed. Take G⁽¹⁾ ∈ S(G⁽⁰⁾) and

Choose

As S(G⁽⁰⁾), we get $$ and h-closed. Suppose that one has built {G⁽⁰⁾, G⁽¹⁾, G⁽²⁾, ⋯, G^(q)} and {S(G⁽⁰⁾), S(G⁽¹⁾), S(G⁽²⁾), ⋯, S(G^(q))}. Next, take G^{(q + 1)} ∈ S(G^(q)) and

Let

hence we form by induction, the sequences {G^(q)} and {S(G^(q))}. Fix $$. Suppose W ∈ S(G^(q)). One obtains then

As {S(G^(q))} is decreasing with G^(q) ∈ S(G^(q)), for all $$, one gets

with $$. This implies {G^(q)} is h -Cauchy. Since $$ is h -Banach space; hence, {G^(q)} has h–limits G^(γ) and $$. Since G^{(q + 1)} ∈ S(G^(q)), we can see hence, $$ is decreasing. After, let G ≠ G^(γ). One gets $$ with $$, with q ≥ m, i.e.,

Since G^(γ) + ∈S(G^(q)), with q ≥ m, we get

Put q⟶∞ in the previous inequality, then

This gives

Theorem 18. Suppose $$ and Ξ is a h-closed subset of $$. By taking γ > 0 and {ϖ_n} and $$. If H : Ξ⟶Ξ is a mapping and there is a function Ψ₁ : Ξ⟶(−∞, ∞] holds a proper and h-lower semicontinuous with inf_G∈ΞΨ₁(G) > −∞ and

• (1)

  h(H(G) − Y) − h(G − Y) ≤ h(H(G) − Y), for any G, Y ∈ Ξ

• (2)

  h(H(G) − G) ≤ Ψ₁(G) − Ψ₁(H(G)), with G ∈ Ξ

Proof. As $$, one has Ψ₂≔ωΨ₁ is also proper, h-lower semicontinuous and bounded from below. If G ∈ Ξ, one gets

As inf_G∈ΞΨ₂(G) > −∞, one obtains G⁽⁰⁾ ∈ Ξ with Ψ₂(G⁽⁰⁾) < inf_G∈ΞΨ₂(G) + γ. From Theorem 17, there is {G^(q)} which h-converges to some G^(γ) ∈ Ξ, and

for every G ≠ G^(γ). Assume that H(G^(γ)) ≠ G^(γ), we have then

From condition (40), then

The inequality (40) implies that

This is a contradiction, hence H(G^(γ)) = G^(γ).

Definition 19 (see [33].)An s-number function is a mapping $$ that sorts every $$ unique sequence $$ validates the following settings:

• (a)

  ‖V‖ = s₀(V) ≥ s₁(V) ≥ s₂(V) ≥ ⋯≥0, for all $$

• (b)

  s_l+d−1(V₁ + V₂) ≤ s_l(V₁) + s_d(V₂), for all $$ and $$

• (c)

  s_d(VYW) ≤ ‖V‖s_d(Y)‖W‖, for all $$, $$, and $$, where Δ₀ and Λ₀ are arbitrary Banach spaces

• (d)

  when $$ and $$, then s_d(γV) = |γ|s_d(V)

• (e)

  suppose rank (V) ≤ d, then s_d(V) = 0, for each $$

• (f)

  s_l≥q(I_q) = 0 or s_l<q(I_q) = 1, where I_q denotes the unit map on the q-dimensional Hilbert space $$

Some examples of s-numbers are as follows:

• (1)

  The qth Kolmogorov number, described by d_q(X), is marked by

• (2)

  The qth approximation number, described by α_q(X), is marked by

Definition 20 (see [10].)Assume $$ is the class of all bounded linear mappings within any two arbitrary Banach spaces. A subclass $$ of $$ is said to be a mappings’ ideal, when all $$ verifies the following conditions:

• (i)

  $$, where Γ marks Banach space of one dimension

• (ii)

  The space $$ is linear over $$

• (iii)

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

Notations 21 (see [30].)

Theorem 22. (see [29].)Suppose $$ is a (ssfps), then $$ is mappings’ ideal.

According to Theorems 10 and 22, one concludes the next theorem.

Theorem 23. Suppose $$ with τ₀ > 1, then $$ is a mappings’ ideal.

Definition 24 (see [34].)A function $$ is said to be a pre-quasi norm on the ideal $$, if it verifies the following setups:

• (1)

  Let $$, H(V) ≥ 0, and H(V) = 0, if and only if, V = 0

• (2)

  we have Q ≥ 1 so as to H(αV) ≤ D|α|H(V), for every $$ and $$

• (3)

  we have P ≥ 1 so that H(V₁ + V₂) ≤ P[H(V₁) + H(V₂)], for each V₁, $$

• (4)

  we have σ ≥ 1 when $$, $$, and $$ then H(YXV) ≤ σ‖Y‖H(X)‖V‖

Theorem 25 (see [35].)Every quasinorm on the ideal $$ is a prequasinorm on the same ideal.

Theorem 26. If $$ with τ₀ > 1, then H is a prequasinorm on $$, so that H(Z) = h(f_s), where $$ and $$.

Proof.

• (1)

  When $$, H(X) = h(f_s) ≥ 0, and H(X) = h(f_s) = 0, if and only if, s_n(X) = 0, for all $$; if and only if, X = 0

• (2)

  There is Q ≥ 1 with H(εX) ≤ h(εf_s) ≤ Q‖ε‖H(X) for every $$ and ε ∈ ℂ

• (3)

  One has PP₀ ≥ 1 so that for $$; hence, there are $$ with $$ and $$. Therefore, for $$, we have KK₀ ≥ 1 so that

• (4)

  We have ϱ ≥ 1 if $$, $$, and Z ∈ L(Λ, Λ₀); hence, there is $$ with $$. Then, for $$, one has

Theorem 27. Suppose $$ with τ₀ > 1 one has $$ is a prequasi-Banach mappings’ ideal.

Proof. Suppose $$ is a Cauchy sequence in $$. As $$, hence, there is $$ with $$ for very $$, then

hence, $$ is a Cauchy sequence in $$ is a Banach space, so there exists $$ so that $$ and since $$, for all $$ and $$ is a premodular (ssfps), hence, one can see We obtain $$, hence $$.

Theorem 28. If $$ with τ₀ > 1, one has $$ is a prequasiclosed mappings’ ideal.

Proof. Suppose $$, for all $$ and $$, hence, there is $$ with $$, for all $$, there is ς > 0 and as $$, one has

Definition 29. A prequasinorm H on the ideal $$ verifies the Fatou property if for every $$ so that $$ and $$, one gets

Theorem 30. Suppose $$ with τ₀ > 1, then $$ does not verify the Fatou property.

Proof. Assume $$ with $$ Since $$ is a prequasiclosed ideal, then $$. So for every $$, one has

Theorem 31. $$, if $$ with τ₀ > 1. But the converse is not necessarily true.

Proof. As $$ for every $$ and $$ is a linear space. Suppose Z ∈ F(Δ, Λ) with rank (Z) = m, where $$, hence $$ with $$, one has $$. Therefore, $$. Assume $$ we have $$. As h(g_α) < ∞, assume ρ ∈ (0, 1), then there is $$ with $$, for some d ≥ 1, where $$. Since (α_q(Z)) is decreasing, we have

Hence, there is $$ so that rank(Y) ≤ 2q₀ and

since $$ with τ₀ > 1, we have

Therefore, one has

As $$, hence $$, where $$. In view of inequalities (58)–(61), one has

Theorem 32. Suppose $$ with $$, for all $$, hence

Proof. Let $$, hence $$, where $$. One gets

then $$ this implies $$. After, if we choose $$ with $$, we have $$ such that

Theorem 33. Assume $$ with τ₀ > 1, hence $$ is minimum.

Proof. Let $$, one has η > 0 so that H(Z) ≤ η‖Z‖, where

for all $$. According to Dvoretzky’s theorem [36], with $$, we get quotient spaces Δ/Y_r and subspaces M_r of Λ which can be transformed onto $$ by isomorphisms V_r and X_r with $$ and $$. If I_r is the identity map on $$, T_r is the quotient map from Δ onto Δ/Y_r and J_r is the natural embedding map from M_r into Λ.

Theorem 34. If $$ with τ₀ > 1, hence $$ is minimum.

Lemma 35 (see [10].)If $$ and B ∉ ϒ(Δ, Λ), then $$ and $$ with MBDI_b = I_b, with $$.

Theorem 36 (see [10].)In general, we have

Theorem 37. Let $$ with $$, for all $$, hence

Proof. Assume $$ and $$. By using Lemma 35, we have $$ and $$ so that ZXYI_b = I_b, hence with $$, one has

Corollary 38. Assume $$ with $$, for all $$, hence,

Proof. Evidently, as $$.

Definition 39 (see [10].)A Banach space Δ is said to be simple, if there is an unique nontrivial closed ideal in $$.

Theorem 40. Let $$ with τ₀ > 1, hence $$ is simple.

Proof. Let $$ and $$. From Lemma 35, there exist $$ with ZXYI_b = I_b, which gives that $$. Then, $$; hence, $$ is simple Banach space.

Notations 41.

$$

Theorem 42. Assume $$ with τ₀ > 1, hence,

Proof. Let $$, hence $$, where $$ and ‖X − λ_x(X)I‖ = 0, with $$. We have X = λ_x(X)I, for all $$, so

with $$. One gets $$; hence, $$. Next, suppose $$. Hence, $$. One gets

Definition 43 (see [37].)A sequence $$, is said to be ε-separated sequence for some ε > 0, if

Definition 44. [37]. If k ≥ 2 is an integer, a Banach space $$ is said to be k-nearly uniformly convex (k-NUC) when for all ε > 0 one has δ ∈ (0, 1) so that for every sequence $$, with sep(g_p) ≥ ε, we have $$.

Such that

Definition 45 [38]. A function h is said to be hold the δ₂-condition (h ∈ δ₂), if for any ε > 0, there exists a constant k ≥ 2 and a > 0 such that,

Theorem 46 ( (see [38], Lemma 2.1.)Suppose $$, then for any L > 0 and ε > 0 one has δ > 0 with |h(f + g) − h(f)| < ε, $$, with h(f) ≤ L and h(g) ≤ δ.

Theorem 47. Pick an $$ with τ₀ > 1, then for any L > 0 and ε > 0 one has δ > 0 with |h(f + g) − h(f)| < ε, for every $$, so that h(f) ≤ L and h(g) ≤ δ.

Proof. Since $$ with τ₀ > 1, then $$. According to Theorem 46, the proof follows.

We denote $$ and $$ for the unit sphere and the unit ball of $$, respectively.

Theorem 48. Suppose $$ with τ₀ > 1, then is k-NUC, for any integer k ≥ 2.

Proof. Assume ε ∈ (0, 1) and $$, where $$ so that sep(f_n) ≥ ε. For all $$, suppose $$, where $$. As for all $$, $$, from the diagonal method, one has a subsequence $$ of (f_n) with $$ converges for all $$, 0 ≤ i ≤ m. One obtains an increasing sequence of positive integers (t_m) so that $$. Therefore, one has a sequence of positive integers $$ with r₀ < r₁ < r₂ < ⋯, so that

for all $$. For constant integer k ≥ 2, assume $$ from Theorem 47, one gets δ > 0 with

Definition 49 (see [39].)A Banach space $$ holds the uniform Opial property, if for all ε > 0 one has γ > 0 so that for every weakly null sequence $$ and $$ so that h(f) ≥ ε, then

Definition 50 (see [40].)For a bounded subset $$, the set-measure of noncompactness defined by

Definition 51 (see [41], [42].)The ball-measure of noncompactness is defined by

Definition 52 (see [43].)For a subset $$ is said to be α-minimal if α(C) = α(E), for any infinite subset C of E.

Definition 53 (see [43].)The packing rate of a Banach space $$ is denoted by $$, and the formula defines it

where $$ and $$ are defined as the supremum and the infimum, respectively, of the set

Definition 54 (see [41].)The function Δ is said to be the modulus of noncompact convexity, if for every ξ > 0 define

Definition 55 (see [39].)A Banach space $$ is said to be hold property (L), when $$.

Definition 56. An operator $$ is said to be a h-contraction, if one gets α ∈ [0, 1) with h(Vg − Vf) ≤ αh(g − f), for all $$. The operator V is said to be h-non-expansive, when α = 1. An element $$ is said to be a fixed point of V, when V(g) = g..

Theorem 57 (see [39].)

• (1)

  Suppose a Banach space $$ holds property (L), then it has the fixed point property, i.e., for every nonexpansive self-mapping of a nonempty, closed, bounded, convex subset has a fixed point

• (2)

  A Banach space $$ holds property (L), if and only if, it is reflexive and has the uniform Opial property

Theorem 58. Suppose $$ with τ₀ > 1, then $$ has the uniform Opial property.

Proof. Let ε > 0 one finds a positive number ε₀ ∈ (0, ε) with

If $$ and h(f) ≥ ε., one has $$ with

Therefore, one gets

Also, one has

if

For any weakly null sequence $$ in virtue of $$ for i = 0, 1, 2, ⋯, one has $$ with

for m > m₀. One can see if m > m₀. For $$, one obtains

Combining this with the previous inequality, one has

Corollary 59. If $$ with τ₀ > 1, then $$ has the property (L) and the fixed point property.

Definition 60. $$ holds the h-normal structure property, if and only if, for every nonempty h-bounded, h-convex, and h-closed subset Γ of $$ not decreased to one point, one has f ∈ Γ with

Definition 61 (see [44].)The weakly convergent sequence coefficient of a Banach space $$, denoted by $$, is defined as follows:

where

Theorem 62 (see [45].)A reflexive Banach space $$ such that $$ has the normal structure property.

Theorem 63. If $$ with τ₀ > 1, then $$ holds the h-normal structure property.

Proof. Take any ε > 0 and an asymptotic equidistant sequence $$ with $$ and let v₁ = f₁. One has $$ with

As f_n⟶0 coordinate-wise, one gets $$ with

For n ≥ n₂, put $$, one gets i₂ > i₁ with

As f_n(i)⟶0 coordinate-wise, one obtains $$ with

For n ≥ n₃. By induction, one has a subsequence {v_n} of {f_n} with

Take

for n = 2, 3, ⋯. So,

For every $$ so that n ≠ m, one can see

which gives A({f_n}) = A({v_n}) ≥ A({z_n}) − 4ε. Take u_n = z_n/‖z_n‖, for n = 2, 3, ⋯. Then,

On the other hand,

for any $$ with n ≠ m. Therefore,

By the arbitrariness of ε > 0, we have from the relations (112), (113), and (115) that

such that

Take $$ large enough such that

where $$ One gets for that is A_n({u_n}) ≥ (2 − ε)^1/K. Note that

Therefore,

with $$ and n ≠ m. Therefore, $$ and, by the arbitrariness of ε > 0, one has $$. From Theorems 48 and 62, then, the function space $$ has the h-normal structure property.

Theorem 64 (see [46].)If $$ is reflexive Banach space with the uniform Opial property, one has $$.

Theorem 65. If $$ with τ₀ > 1, then $$.

Proof. Since $$ is reflexive Banach space with the uniform Opial property, one obtains

Theorem 66. If $$ with τ₀ > 1 and $$ is h-contraction mapping, where $$, for every $$, then $$ has a unique fixed point.

Proof. Let the setups be satisfied. For every $$, then $$. As $$ is a h-contraction mapping, one gets

So, for all $$ so that q > p, one has

Therefore, $$ is a Cauchy sequence in $$. Since the space $$ is prequasi-Banach (ssfps). One gets $$ with $$, to prove that $$. According to Theorem 13, h verifies the Fatou property; one can see

so $$. Then, g is a fixed point of $$. To prove that the fixed point is unique, let us have two different fixed points $$ of $$. One obtains

So, f = g.

Example 7. Assume

where for every $$ and V(g) = g/4.

Since for all $$, one gets

So V is h-contraction. Assume V : Γ⟶Γ with V(g) = g/4, where

Since V is h-contraction. So, it is h-nonexpansive. By Corollary 59, V holds a fixed point ϑ in Γ.

Theorem 67. The summable equations (132) have only one solution in $$ if $$$$$$$$ assume there is κ ∈ ℂ so that $$ and for every $$, we have

Proof. Let the setups be verified. Consider the mapping $$ defined by (133). We have

Example 1. Assume the function space $$, where

for all $$.

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

Example 2. Given the function space $$, where

for all $$. Consider the summable equations (137) with a ≥ 2 and let W : Ξ⟶Ξ, where $$, defined by

Example 3. Assume the function space $$, where

for all $$.

Consider the non-linear difference equations,

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

It is easy to see that

By Theorem 67, the nonlinear difference equations (144) have one solution in $$.

Example 4. Given the function space $$, where $$, for all $$. Consider the non-linear difference equations (144) with a ≥ 1 and let W : Ξ⟶Ξ, where $$, defined by

Example 5. The summable equations (132) have a solution in $$ if

Evidently, we have

By Theorem 18, one gets a solution of equation (132) in $$.

Example 6. The summable equations (132) have a solution in $$, if

Clearly, we have

By Theorem 18, one gets a solution of equation(132) in $$.

Assume Ω is the set of all closed and bounded intervals on the real line $$. For t = [t₁, t₂] and g = [g₁, g₂] in Ω, suppose

Define a metric ρ on Ω by

Matloka [48] showed that ρ is a metric on Ω, and (Ω, ρ) is a complete metric space.

Definition 68. A fuzzy number g is a fuzzy subset of $$, i.e., a mapping $$ which verifies the following four settings:

• (a)

  g is fuzzy convex, i.e., for $$ and α ∈ [0, 1], g(αx + (1 − α)y) ≥ min{g(x), g(y)}

• (b)

  g is normal, i.e., there is $$ such that g(y₀) = 1

• (c)

  g is an upper semicontinuous, i.e., for all α > 0, g⁻¹([0, x + α)) for all x ∈ [0, 1] is open in the usual topology of $$

• (d)

  The closure of $$ is compact

Recall that the β-level set of a fuzzy real number g, 0 < β < 1 indicated by g^β is defined as

The set of every upper semicontinuous, normal, convex fuzzy number, and is compact and is denoted by $$. The set $$ can be embedded in $$, if we define $$ by

Consider the summable equations of fuzzy reals (132) and assume $$ defined by

where $$ is defined by $$ For more details about the fuzzy numbers and their properties, see Zadeh [49].

Theorem 69. The summable equations (132) have an unique solution in $$ if $$$$$$$$ assume there is κ ∈ ℂ so that $$ and for every $$, we have

Proof. Let the setups be verified. Consider the mapping $$ defined by (157). We have