Fixed Point and Common Fixed Point Theorems for Generalized Weak Contraction Mappings of Integral Type in Modular Spaces

Theorem 1.1. Let (X, d) be a complete metric space, α ∈ [0,1), f : X → X a mapping such that for each x, y ∈ X,

where φ : ℝ⁺ → ℝ⁺ is a Lebesgue integrable which is summable, nonnegative, and for all ε > 0, $$. Then, f has a unique fixed point z ∈ X such that for each x ∈ X, lim_n→∞fⁿx = z.

Definition 1.2. Let X be a vector space over ℝ(or ℂ). A functional ρ : X → [0, ∞] is called a modular if for arbitrary f and g, elements of X satisfy the following conditions:

• (1)

  ρ(f) = 0 if and only if f = 0;

• (2)

  ρ(αf) = ρ(f) for all scalar α with |α | = 1;

• (3)

  ρ(αf + βg) ≤ ρ(f) + ρ(g), whenever α, β ≥ 0 and α + β = 1. If we replace (3) by

• (4)

  ρ(αf + βg) ≤ α^sρ(f) + β^sρ(g), for α, β ≥ 0,  α^s + β^s = 1 with an s ∈ (0,1], then the modular ρ is called s-convex modular, and if s = 1,  ρ is called convex modular.

Definition 1.3. A modular ρ is said to satisfy the Δ₂-condition if ρ(2f_n) → 0 as n → ∞, whenever ρ(f_n) → 0 as n → ∞.

Definition 1.4. Let X_ρ be a modular space. Then,

• (1)

  the sequence (f_n) _n∈ℕ in X_ρ is said to be ρ-convergent to f ∈ X_ρ if ρ(f_n − f) → 0, as n → ∞,

• (2)

  the sequence (f_n) _n∈ℕ in X_ρ is said to be ρ-Cauchy  if ρ(f_n − f_m) → 0, as n, m → ∞,

• (3)

  a subset C of X_ρ is said to be ρ-closed if the ρ-limit of a ρ-convergent sequence of C always belong to C,

• (4)

  a subset C of X_ρ is said to be ρ-complete if any ρ-Cauchy sequence in C is ρ-convergent sequence and its is in C,

• (5)

  a subset C of X_ρ is said to be ρ-bounded if δ_ρ(C) = sup {ρ(f − g); f, g ∈ C} < ∞.

Definition 1.5. Let C be a subset of X_ρ and T : C → C an arbitrary mapping. T is called a ρ-contraction if for each f, g ∈ X_ρ there exists k < 1 such that

Definition 1.6. Let X_ρ be a modular space, where ρ satisfies the Δ₂-condition. Two self-mappings T and f of X_ρ are called ρ-compatible if ρ(Tfx_n − fTx_n) → 0 as n → ∞, whenever {x_n} _n∈ℕ is a sequence in X_ρ such that fx_n → z and Tx_n → z for some point z ∈ X_ρ.

Theorem 2.1. Let X_ρ be a ρ-complete modular space, where ρ satisfies the Δ₂-condition. Let c, l ∈ ℝ⁺, c > l and T, f : X_ρ → X_ρ are two ρ-compatible mappings such that T(X_ρ)⊆f(X_ρ) and

for all x, y ∈ X_ρ, where φ : [0, ∞)→[0, ∞) is a Lebesgue integrable which is summable, nonnegative, and for all ε > 0, $$ and ϕ : [0, ∞)→[0, ∞) is lower semicontinuous function with ϕ(t) > 0 for all t > 0 and ϕ(t) = 0 if and only if t = 0. If one of T or f is continuous, then there exists a unique common fixed point of T and f.

Proof. Let x ∈ X_ρ and generate inductively the sequence {Tx_n} _n∈ℕ as follow: Tx_n = fx_n+1. First, we prove that the sequence {ρ(c(Tx_n − Tx_n−1))} converges to 0. Since,

This means that the sequence $$ is decreasing and bounded below. Hence, there exists r ≥ 0 such that If r > 0, then $$. Taking n → ∞ in the inequality (2.2) which is a contradiction, thus r = 0. This implies that

Next, we prove that the sequence {Tx_n} _n∈ℕ is ρ-Cauchy. Suppose {cTx_n} _n∈ℕ is not ρ-Cauchy, then there exists ε > 0 and sequence of integers {m_k}, {n_k} with m_k > n_k ≥ k such that

We can assume that

Let m_k be the smallest number exceeding n_k for which (2.5) holds, and

Since θ_k ⊂ ℕ and clearly θ_k ≠ ∅, by well ordering principle, the minimum element of θ_k is denoted by m_k and obviously (2.6) holds. Now, let α ∈ ℝ⁺ be such that l/c + 1/α = 1, then we get

Using the Δ₂-condition and (2.4), we obtain

It follows that From (2.8) and (2.11), we also have which is a contradiction. Hence, {cTx_n} _n∈ℕ is ρ-Cauchy and by the Δ₂-condition, {Tx_n} _n∈ℕ is ρ-Cauchy. Since X_ρ is ρ-complete, there exists a point u ∈ X_ρ such that ρ(Tx_n − u) → 0 as n → ∞. If T is continuous, then T²x_n → Tu and Tfx_n → Tu as n → ∞. Since ρ(c(fTx_n − Tfx_n)) → 0 as n → ∞, by ρ-compatible, fTx_n → Tu as n → ∞. Next, we prove that u is a unique fixed point of T. Indeed, Taking n → ∞ in the inequality (2.13), we have which implies that ρ(c(Tu − u)) = 0 and Tu = u. Since T(X_ρ)⊆f(X_ρ), there exists u₁ such that u = Tu = fu₁. The inequality, as n → ∞, yields and, thus, which implies that, u = Tu₁ = fu₁ and also fu = fTu₁ = Tfu₁ = Tu = u (see [25]). Hence, fu = Tu = u. Suppose that there exists w ∈ X_ρ such that w = Tw = fw and w ≠ u, we have $$ and which is a contradiction. Hence, u = w and the proof is complete.

Corollary 2.2 (see [23].)Let X_ρ be a ρ-complete modular space, where ρ satisfies the Δ₂-condition. Suppose c, l ∈ ℝ⁺, c > l and T, h : X_ρ → X_ρ are two ρ-compatible mappings such that T(X_ρ)⊆h(X_ρ) and

for some k ∈ (0,1), where φ : ℝ⁺ → ℝ⁺ is a Lebesgue integrable which is summable, nonnegative, and for all ε > 0, $$. If one of h or T is continuous, then there exists a unique common fixed point of h and T.

Corollary 2.3 (see [23].)Let X_ρ be a ρ-complete modular space, where ρ satisfies the Δ₂-condition. Suppose c, l ∈ ℝ⁺, c > l and T, h : X_ρ → X_ρ are two ρ-compatible mappings such that T(X_ρ)⊆h(X_ρ) and

where φ : ℝ⁺ → ℝ⁺ is a Lebesgue integrable which is summable, nonnegative, and for all ε > 0, $$ and ψ : ℝ⁺ → ℝ⁺ is a nondecreasing and right continuous function with ψ(t) < t for all t > 0. If one of h or T is continuous, then there exists a unique common fixed point of h and T.

Theorem 3.1. Let X_ρ be a ρ-complete modular space, where ρ satisfies the Δ₂-condition. Let c, l ∈ ℝ⁺, c > l and T : X_ρ → X_ρ be a mapping such that for each x, y ∈ X_ρ,

where φ : [0, ∞)→[0, ∞) is a Lebesgue integrable which is summable, nonnegative, and for all ε > 0, $$ and ϕ : [0, ∞)→[0, ∞) is lower semicontinuous function with ϕ(t) > 0 for all t > 0 and ϕ(t) = 0 if and only if t = 0. Then, T has a unique fixed point.

Proof. First, we prove that the sequence {ρ(c(Tⁿx − Tⁿ⁻¹x))} converges to 0. Since,

it follows that the sequence $$ is decreasing and bounded below. Hence, there exists r ≥ 0 such that If r > 0, then $$, taking n → ∞ in the inequality (3.2) which is a contradiction, thus r = 0. So, we have Next, we prove that the sequence {Tⁿ(x)} _n∈ℕ is ρ-Cauchy. Suppose {cTⁿ(x)} _n∈ℕ is not ρ-Cauchy, there exists ε > 0 and sequence of integers {m_k}, {n_k} with m_k > n_k ≥ k such that We can assume that Let m_k be the smallest number exceeding n_k for which (3.5) holds, and Since θ_k ⊂ ℕ and clearly θ_k ≠ ∅, by well ordering principle, the minimum element of θ_k is denoted by m_k and obviously (3.6) holds. Now, let α ∈ ℝ⁺ be such that l/c + 1/α = 1, then we get

Using the Δ₂-condition and (3.4), we obtain

From (3.8) and (3.11), we have which is a contradiction. Hence, {cTⁿ(x)} _n∈ℕ is ρ-Cauchy and again by the Δ₂-condition, {Tⁿ(x)} _n∈ℕ is ρ-Cauchy. Since X_ρ is ρ-complete, there exists a point u ∈ X_ρ such that ρ(Tⁿx − u) → 0 as n → ∞. Next, we prove that u is a unique fixed point of T. Indeed, Since ρ(Tⁿx − u) → 0 as n → ∞, we obtain which implies that So, we have Thus ρ(c/2(u − Tu)) = 0 and Tu = u. Suppose that there exists w ∈ X_ρ such that Tw = w and w ≠ u, we have $$ and which is a contradiction. Hence, u = w and the proof is complete.

Corollary 3.2. Let X_ρ be a ρ-complete modular space, where ρ satisfies the Δ₂-condition. Let f : X_ρ → X_ρ be a mapping such that there exists an λ ∈ (0,1) and c, l ∈ ℝ⁺ where l < c and for each x, y ∈ X_ρ,

where φ : [0, ∞)→[0, ∞) is a Lebesgue integrable which is summable, nonnegative, and for all ε > 0, $$. Then, T has a unique fixed point in X_ρ.

Corollary 3.3 (see [24].)Let X_ρ be a ρ-complete modular space where ρ satisfies the Δ₂-condition. Assume that ψ : ℝ⁺ → [0, ∞) is an increasing and upper semicontinuous function satisfying ψ(t) < t for all t > 0. Let φ : [0, ∞)→[0, ∞) be a Lebesgue integrable which is summable, nonnegative, and for all ε > 0, $$ and let f : X_ρ → X_ρ be a mapping such that there are c, l ∈ ℝ⁺ where l < c,

for each x, y ∈ X_ρ. Then, T has a unique fixed point in X_ρ.