Some New Generalization of Darbo ’ s Fixed Point Theorem and Its Application on Integral Equations

In this article, we propose some new fixed point theorem involving measure of noncompactness and control function. Further, we prove the existence of a solution of functional integral equations in two variables by using this fixed point theorem in Banach Algebra, and also illustrate the results with the help of an example.


Introduction
Integral equations play a significant role in real-world problems.Fixed point theory and measure of noncompactness are useful tools in solving different types of integral equations which we come across in different real life situations.In solving functional integral equations, Schauder and Darbo's fixed point theorems play a significant role.We refer (see [1][2][3][4][5][6][7][8][9][10][11][12][13][14][15]) for application of fixed point theorems and measure of noncompactness for solving differential and integral equations.
In this article using the concept of control function and measure of noncompactness we have proved some new fixed point theorems.Further, we have also applied this theorem to study the existence of solution of functional integral equations in Banach algebra and also with the help of an example we have verified our results.
Let Ē be a real Banach space with the norm . .Let B(a, b) be a closed ball in Ē centered at a and with radius b.If X is a nonempty subset of Ē then by X and Conv X we denote the closure and convex closure of X, respectively.Moreover, let M Ē denote the family of all nonempty and bounded subsets of Ē and N Ē its subfamily consisting of all relatively compact sets.We denote by R the set of real numbers and R + = [0, ∞) .
The following definition of a measure of noncompactness given in [3].
is called a measure of non-compactness in Ē if it satisfies the following conditions: The family ker µ is said to be the kernel of measure µ.Observe that the intersection set Y ∞ from (vii) is a member of the family ker µ.
For a bounded subset S of a metric space X, the Kuratowski measure of noncompactness is defined as [9] where diam (S i ) denotes the diameter of the set S i , that is diam (S i ) = sup {d(x, y) : x, y ∈ S i } .
The Hausdorff measure of noncompactness for a bounded set S is defined as

Definition 2 ([3]
).Let X be a nonempty subset of a Banach space Ē and T : X → Ē is a continuous operator transforming bounded subset of X to bounded ones.We say that T satisfies the Darbo condition with a constant k with respect to measure µ provided µ(TY) ≤ kµ(Y) for each Y ∈ M Ē such that Y ⊂ X.
We recall following important theorems: Theorem 1 (Shauder [16]).Let D be a nonempty, closed and convex subset of a Banach space Ē.Then every compact, continuous map T : D → D has at least one fixed point.
Then S has a fixed point.
In order to establish our fixed point theorem, we need some of the following related concepts.Khan et al. [17] used a control function which they called an altering distance function.
If we take ψ 1 (t) = λt for all t ≥ 0, λ ∈ [0, 1) and ψ 2 (t) = t then we obtain the following function η 2 (t, s) = λs − t for all t, s ∈ R + is in the class of functions Ẑ.If s ≤ t then η 2 (t, s) < 0. Definition 5. Let F be the class of all functions G : R + × R + → R + satisfying the following conditions: G is continuous and nondecreasing.
For example G(a, b) = a + b.

Main Result
Theorem 3. Let C be a nonempty, bounded, closed and convex subset of a Banach space Ē.Also T : C → C is continuous and φ : R + → R + is continuous and nondecreasing functions.Suppose that if for any where µ is an arbitrary measure of noncompactness and η ∈ Ẑ and G ∈ F. Then T has at least one fixed point in C.
If there exists a natural number m such that µ(C m ) = 0 then C m is compact.By Schauder's fixed point theorem we conclude that T has a fixed point.
So we assume that µ(C n ) > 0 for some n ≥ 0 i.e., G(µ Since t n < s n for all n ≥ 0 and lim which is a contradiction.Thus we conclude α = 0 i.e., lim n=1 C n is nonempty, closed and convex subset of C and C ∞ is invariant under T. Thus Schauder's theorem implies that T has a fixed point in C ∞ ⊆ C.This completes the proof.Theorem 4. Let C be a nonempty, bounded, closed and convex subset of a Banach space Ē.Also T : C → C is continuous and φ : R + → R + is continuous and nondecreasing functions.Suppose that if for any where µ is an arbitrary measure of noncompactness and η ∈ Ẑ.Then T has at least one fixed point in C.

Proof
where µ is an arbitrary measure of noncompactness and η ∈ Ẑ.Then T has at least one fixed point in C.
Proof.The result follows by taking G(a, b) = a + b and φ ≡ 0 in Theorem 3.

Theorem 6.
Let C be a nonempty, bounded, closed and convex subset of a Banach space Ē and T : C → C is a continuous function.Suppose ψ 1 and ψ 2 be two altering distance functions such that ψ 1 (t) < t ≤ ψ 2 (t) for all t > 0 and a constant 0 < γ < 1 such that for all X ⊆ C, and a ≤ µ(X) ≤ b we have ψ 2 (µ(T(X))) ≤ ψ 1 (γµ(X)) where µ is an arbitrary measure of noncompactness.Then T has at least one fixed point in C.

Theorem 7.
Let C be a nonempty, bounded, closed and convex subset of a Banach space Ē and T : C → C is a continuous function.Suppose for any 0 < a < b < ∞ there exists 0 < γ(a, b) < 1 such that for all X ⊆ C, and a ≤ µ(X) ≤ b we have µ(T(X)) ≤ γ(a, b)µ(X), where µ is an arbitrary measure of noncompactness.
Then T has at least one fixed point in C.

Application
In this article, we shall work in the space E = C([0, 1] × [0, 1]) which consists of the set of real continuous on [0, 1] × [0, 1].The space E is equipped with the norm The space E has the Banach algebra structure.
Let X be a fixed nonempty and bounded subset of the space E = C([0, 1] × [0, 1]) and for x ∈ X and > 0, denote by ω(x, ) the modulus of the continuity function x i.e., Similar to [5] it can be shown that the function ω 0 is a measure of non-compactness in the space In this part we are going to study the existence of the solution of the following integral equation We consider the following assumptions The function G : Let u : The function F : Proof.Let us consider the operators F and T defined on C(I × I) as follows where t, s ∈ I.
From assumptions (1) to (3) we infer Tx is continuous on I × I for x ∈ C(I × I).Thus T maps C(I × I) into itself.Also for t, s ∈ I we get Then we have Therefore the operator T maps B r into itself.Next we have to prove that T is continuous on B r .Let {x n } be a sequence in B r such that x n → x.For every t, s ∈ I, we have where > 0 and Let us consider an nonempty subset X of B r and x ∈ X then for a fixed > 0 and t 1 , t 2 , s 1 , where Since G, F and u are uniformly continuous on I × I, I × I × [−r, r] × [− Ū, Ū] and I × I × I × [−r, r] respectively therefore, we get, ω(G, ) → 0, ω(F, ) → 0 and ω(u, ) → 0 as → 0. Thus we obtain ω 0 ( TX) ≤ Kω 0 (X).
This implies T is a contraction operator on B r with respect to ω 0 .Thus by Theorem  Here K = 1 4 and M = 0.The inequality in the assumption (4) has the following form 1 + r 2 < r.
For r = 3 we observe that all the assumption from (1)-( 4) of Theorem 8 are satisfied.Thus applying the Theorem 8 we conclude that the Equation ( 2) has at least one solution in C(I × I).

.Theorem 5 .
The result follows by taking G(a, b) = a + b in Theorem 3. Let C be a nonempty, bounded, closed and convex subset of a Banach space Ē and T : C → C is a continuous function.Suppose that if for any 0