1 Introduction

Fixed point theory is one of the most developed and applicable branches of nonlinear mathematical analysis. It based on the well-known principle that every operator equation can be transformed into a fixed point problem, and vice versa.

In essence, the fixed point theory has two branches, namely the metric fixed point theory and topological fixed point theory. Topological fixed point theory is the most important branch of nonlinear analysis. It has a strong and useful history of more than a century. In 1912 Brouwer proved his well-known fixed point theorem. Later in 1922, Banach proved the most versatile result known as Banach contraction principle [38]. The most applicable generalization of Brouwer’s theorem was presented in 1930 by Schauder using compact operators. Banach and Schauder’s results remain most celebrated results in fixed point theory. The following two results of Brouwer and Banach were established in 1912 and 1922, respectively, and are given as follows.

Theorem BR

(Brouwer, [31]) Every compact convex nonempty subset B of \(\mathbb{R} ^{n}\) has a fixed-point property.

Theorem BC

(Banach, [31]) Any contraction mapping of a complete nonempty metric space \(\Omega _{1}\) into \(\Omega _{1}\) has a unique fixed point in \(\Omega _{1}\).

Both theorems are crucial in the existence theory of differential and integral equations. In the literature, there are many generalizations of these results. The most famous generalization of Theorem BR is Schauder’s fixed point theorem.

Theorem SH

(Schauder’s second theorem, [31]) Let \(\Omega _{1}\) be a nonempty convex subset of a normed space L and S be a continuous operator of \(\Omega _{1}\) into a compact set \(B\subseteq \Omega _{1}\). Then S has a fixed point.

A variety of generalizations of these results can be seen in [8, 13, 15, 18, 20, 21, 33, 35], and in the references therein.

Writing the physical problems into mathematical form produces mathematical equations like differential, integral, linear, and nonlinear equations. All these equations can be solved by fixed point techniques. Before solving them, the existence theory for the fixed points of operators plays very a important role. A number of results regarding differential and integral equations in connection with their existence theory can be seen in [1, 2, 6, 9, 11, 17, 23, 24, 28, 32].

In 1958, while studying the existence theory of neutral and delayed differential equations, it was observed that the solution might be expressed as a sum of compact and contractive operators. Working on this idea, Krasnoselskii proved his fixed point results for the sum of compact and contractive operators. The importance and applications of such theorems for the existence of solutions to the equation \(x=Ax+Bx\) can be seen in [10, 12, 14, 25–27, 30, 36, 37].

The Krasnoselskii fixed point theorem can be stated as:

Theorem 1

([31]) Consider a Banach space X and let M be its nonempty convex closed subset. Suppose A and B map M into X such that

(1) \(y,\varkappa \in M\) implies \(Ay +B\varkappa \in M\);

(2) B is a contraction mapping;

(3) A is compact and continuous.

Then there is \(x\in M\) such that \(Ax+Bx=x\).

This Krasnoselskii’s theorem is a generalization of Schauder’s fixed point theorem and Banach contraction principle, as we can see by taking \(B=O\) and \(A=O\), respectively.

To weaken the compactness condition used in Schauder’s fixed point theorem, Darbo and Sadovskii [38, p. 500] generalized Schauder’s fixed point theorem by introducing k-set contractive and condensing operators (noncompact operators) in the following way.

Theorem 2

(Darbo) Assume that

(i) the mapping \(T:M\subseteq X\rightarrow M\) is k-set contractive,

(ii) M is a nonempty, convex, bounded, and closed subset of a Banach space X.

Then there exists \(p\in M\) such that \(p=Tp\).

Remark 3

Since every compact operator is k-set contractive with \(k=0\), Darbo’s theorem is a generalization of Schauder’s fixed point theorem.

Theorem 4

(Sadovskii) Assume that

(i) \(T:M\subseteq X\rightarrow M\) is a condensing operator,

(ii) M is a nonempty, convex, bounded, and closed subset of a Banach space X.

Then T has a fixed point.

Remark 5

Since every k-set contractive operator with \(0\leq k<1\) is condensing, Sadovskii’s theorem is an extension of Darbo’s theorem. For more about the above discussed results, one can see the related monographs and articles [3, 4, 16, 34].

Since Krasnoselskii combined the results of Banach and Schauder, in the above results Darbo and Sadovskii generalized Schauder’s theorem. The next result was established by Reich [29], which is a generalization of Kannan fixed point theorem and Banach contraction principle.

Theorem 6

([29]) Consider a complete metric space X with metric d on it and let \(T:X\rightarrow X\) be a mapping with the following property:

$$ d(Tq,Tp)\leq a_{1}d(q,Tq)+a_{2}d(p,Tp)+a_{3}d(q,p), \quad q,p\in X, $$

where \(a_{1}\), \(a_{2}\), \(a_{3}\) are nonnegative and satisfy \(a_{1}+a_{2}+a_{3}<1\). Then T has a unique fixed point.

Remark 7

Letting \(a_{1}=a_{2}=0\) gives Banach fixed point theorem, and taking \(a_{1}=a_{2}\), \(a_{3}=0\) gives Kannan fixed point theorem [19].

To combine the generalized Reich contraction and k-set contractive mappings in the form of operator equation \(x=Ax+Bx\) and generalize Reich, Kannan, Darbo, Sadovskii, and Krasnoselskii type fixed point results, we need the following definitions.

Definition 8

([5, 7]) Let X be a Banach space and \(\mathit{B}(X)\) be the collection of all bounded subsets of X. A mapping μ of \(\mathit{B}(X)\) into \([0,+\infty )\) is called a measure of noncompactness if the following conditions hold for all \(E,F\in \mathit{B}(X)\):

(1) \(\mu ( F ) =\mu ( \overline{F} ) \);

(2) \(\mu ( F\cup E ) =\max \{ \mu ( F ) ,\mu ( E ) \}\);

(3) \(\mu ( F ) =0 \Leftrightarrow F\) is precompact.

The following conditions can also be deduced:

(4) \(\mu ( E+F ) \leq \mu (E)+\mu ( F ) \);

(5) \(E\subseteq F\) implies \(\mu ( E ) \leq \mu ( F ) \).

Definition 9

([37]) Let X be a Banach space and \(T:M\subseteq X\longrightarrow X\) be a mapping. Then T is called k-set contractive if the following conditions hold for any bounded subset E of M:

(1) T is continuous and bounded;

(2) \(\mu ( T(E) ) \leq k\mu ( E ) \).

Also T is strictly k-set contractive if

(1) T is k-set contractive;

(2) \(\mu ( T(E) ) < k\mu ( E ) \) with \(\mu ( E ) \neq 0\).

Finally, T is condensing if T is strictly 1-set contractive.

Now we define generalized Kannan contractions in Banach spaces.

Definition 10

Consider a Banach space X and let M be its nonempty subset. Suppose A and B map M into X. Then B is called a generalized Kannan contraction if

$$ \bigl\Vert By-By^{\prime } \bigr\Vert \leq \alpha \bigl( \bigl\Vert Ax-(I-B)y \bigr\Vert + \bigl\Vert Ax-(I-B)y^{\prime } \bigr\Vert \bigr), $$

with \(\alpha <\frac{1}{2}\) and all \(x,y,y^{\prime }\in M\).

Remark 11

If \(A=O\) (the zero operator), we obtain a Kannan contraction.

To investigate the novelty of the above definition, the following example is important.

Example 12

Let \(B: \mathbb{R} \rightarrow \mathbb{R} \) be the mapping defined by

$$ Bx= \textstyle\begin{cases} \frac{x}{3}&\text{if }0\leq x< 1, \\ \frac{1}{6}&\text{if }x=1, \\ 0&\text{elsewhere}.\end{cases} $$

Then with usual metric induced from the usual norm \(\vert \cdot \vert \) on \(\mathbb{R} \), we have

$$ \frac{1}{9}=d \biggl( B ( 0 ) ,B \biggl( \frac{1}{3} \biggr) \biggr) =\frac{1}{2} \biggl[ d \bigl( 0,B ( 0 ) \bigr) +d \biggl( \frac{1}{3},B \biggl( \frac{1}{3} \biggr) \biggr) \biggr] =\frac{1}{9} $$

so the condition of a Kannan mapping fails, but if we define \(Ax=\frac{2}{3}\), then the condition of a generalized Kannan contraction holds.

The above example and definition motivated us to define the following.

Definition 13

Consider a Banach space X and let M be its nonempty subset. Suppose A and B map M into X. Then B is called a generalized Reich contraction if

$$ \bigl\Vert By-By^{\prime } \bigr\Vert \leq a_{1} \bigl\Vert Ax-(I-B)y \bigr\Vert +a_{2} \bigl\Vert Ax-(I-B)y^{\prime } \bigr\Vert +a_{3} \bigl\Vert y-y^{\prime } \bigr\Vert , $$

for nonnegative numbers \(a_{1}\), \(a_{2}\), and \(a_{3}\) with \(a_{1}+a_{2}+a_{3}<1\) and all \(y,y^{\prime },x\in M\).

Remark 14

Taking \(A=O\) (the zero operator), we obtain a Reich contraction.

Lemma 15

([37]) Let M be a subset of a Banach space X and A be a Lipschitz mapping of M into X such that

$$ \Vert Ax-Ay \Vert \leq k \Vert x-y \Vert $$

for \(x,y\in M\). Then \(\mu (A(E))\leq k\mu (E)\) holds for every bounded subset E of M.

2 Main results

The following theorem shows that a unique solution exists for the operator equation \(x=Ax+Bx\) if A is a contraction and B is a generalized Reich contraction.

Theorem 16

Let X be a Banach space and M be its nonempty closed subset. Consider the mappings \(A:M\rightarrow X\) and \(B:M\rightarrow X\) such that

(1) \(\Vert By-By^{\prime } \Vert \leq a_{1} \Vert Av-(I-B)y \Vert +a_{2} \Vert Av-(I-B)y^{\prime } \Vert +a_{3} \Vert y-y^{\prime } \Vert \) with \(a_{1}+a_{2}+a_{3}<1\) and all \(y,y^{\prime },v\in M\);

(2) \(\Vert Ax-Ax^{\prime } \Vert \leq k \Vert x-x^{ \prime } \Vert \) with \(k<\frac{1-a_{3}}{1+a_{2}}\);

(3) \(Ap+Bq\in M\) for all \(p,q\in M\).

Then there exists a unique \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Proof

Fix \(x\in M\) and define \(H:M\rightarrow M\) by \(H(y)=Ax+By\). Using (1), and since condition (1) holds for all \(v\in M\), we have

$$ \begin{aligned} \bigl\Vert H(y)-H\bigl(y^{\prime }\bigr) \bigr\Vert &= \bigl\Vert By-By^{\prime } \bigr\Vert \\ &\leq a_{1} \bigl\Vert Ax-(I-B)y \bigr\Vert +a_{2} \bigl\Vert Ax-(I-B)y^{ \prime } \bigr\Vert +a_{3} \bigl\Vert y-y^{\prime } \bigr\Vert \\ &=a_{1} \bigl\Vert y-H(y) \bigr\Vert +a_{2} \bigl\Vert y^{\prime }-H\bigl(y^{ \prime }\bigr) \bigr\Vert +a_{3} \bigl\Vert y-y^{\prime } \bigr\Vert . \end{aligned} $$

This shows that H is a Reich contraction. Hence by [29, Theorem 3], there is a unique \(Gx\in M\) that such that \(Gx=Ax+B(Gx)\). Now

$$ \begin{aligned} \bigl\Vert Gx-Gx^{\prime } \bigr\Vert ={}& \bigl\Vert Ax+B(Gx)-\bigl(Ax^{ \prime }+B\bigl(Gx^{\prime }\bigr) \bigr) \bigr\Vert \\ ={}& \bigl\Vert \bigl(B(Gx)-B\bigl(Gx^{\prime }\bigr)\bigr)- \bigl(Ax^{\prime }-Ax\bigr) \bigr\Vert \\ \leq{}& a_{1} \bigl\Vert Ax-(I-B)Gx \bigr\Vert +a_{2} \bigl\Vert Ax-(I-B)Gx^{ \prime } \bigr\Vert \\ &{} +a_{3} \bigl\Vert Gx-Gx^{\prime } \bigr\Vert + \bigl\Vert Ax^{ \prime }-Ax \bigr\Vert \\ ={}&a_{2} \bigl\Vert Ax-(I-B)Gx^{\prime } \bigr\Vert +a_{3} \bigl\Vert Gx-Gx^{\prime } \bigr\Vert + \bigl\Vert Ax^{\prime }-Ax \bigr\Vert \\ ={}&(a_{2}+1) \bigl\Vert Ax-Ax^{\prime } \bigr\Vert +a_{3} \bigl\Vert Gx-Gx^{ \prime } \bigr\Vert . \end{aligned} $$

Thus

$$ \bigl\Vert Gx-Gx^{\prime } \bigr\Vert \leq \biggl( \frac{a_{2}+1}{1-a_{3}}\biggr) \bigl\Vert Ax-Ax^{\prime } \bigr\Vert \leq k\biggl( \frac{a_{2}+1}{1-a_{3}}\biggr) \bigl\Vert x-x^{\prime } \bigr\Vert . $$

Thus \(G:M\rightarrow M\) is a contraction. Using Banach contraction principle, there is an \(x\in M\) such that \(Gx=x\). Since for this \(x\in M\) there is a \(Gx\in M\) such that \(Gx=Ax+B(Gx)\), this implies that \(x=Gx=Ax+B(Gx)=Ax+Bx \).

For uniqueness, suppose \(s,s^{\prime }\in M\) are such that \(s=As+Bs\) and \(s^{\prime }=As^{\prime }+Bs^{\prime }\). Then

$$\begin{aligned} \bigl\Vert s-s^{\prime } \bigr\Vert &= \bigl\Vert (As+Bs)-\bigl(As^{ \prime }+Bs^{\prime }\bigr) \bigr\Vert \\ &= \bigl\Vert \bigl(Bs-Bs^{\prime }\bigr)-\bigl(As^{\prime }-As \bigr) \bigr\Vert \\ &\leq \bigl\Vert Bs-Bs^{\prime } \bigr\Vert + \bigl\Vert As-As^{ \prime } \bigr\Vert \\ &\leq a_{1} \bigl\Vert As-(I-B)s \bigr\Vert +a_{2} \bigl\Vert As-(I-B)s^{ \prime } \bigr\Vert +a_{3} \bigl\Vert s-s^{\prime } \bigr\Vert +k \bigl\Vert s-s^{\prime } \bigr\Vert \\ &=a_{2} \bigl\Vert As-As^{\prime } \bigr\Vert +a_{3} \bigl\Vert s-s^{ \prime } \bigr\Vert +k \bigl\Vert s-s^{\prime } \bigr\Vert \\ &\leq a_{2}k \bigl\Vert s-s^{\prime } \bigr\Vert +a_{3} \bigl\Vert s-s^{ \prime } \bigr\Vert +k \bigl\Vert s-s^{\prime } \bigr\Vert \\ &=(a_{2}k+a_{3}+k) \bigl\Vert s-s^{\prime } \bigr\Vert . \end{aligned}$$

This means that

$$ \bigl(1-(a_{2}k+a_{3}+k)\bigr) \bigl\Vert s-s^{\prime } \bigr\Vert \leq 0, $$

showing that \(s=s^{\prime }\). □

Remark 17

If we substitute \(A=O\) in conditions (1), (2), and (3) of Theorem 16, we obtain the well-known theorem of Reich (see [29, Theorem 3]).

Corollary 18

Let X be a Banach space and M be a nonempty closed subset of X. Consider mappings \(A:M\rightarrow X\) and \(B:M\rightarrow X\) such that

(1) \(\Vert By-By^{\prime } \Vert \leq \alpha ( \Vert Ax-(I-B)y \Vert + \Vert Ax-(I-B)y^{\prime } \Vert )\) with \(\alpha <\frac{1}{2}\) and for all \(x,y,y^{\prime }\in M\);

(2) \(\Vert Ax-Ax^{\prime } \Vert \leq k \Vert x-x^{ \prime } \Vert \) with \(k<\frac{1}{1+\alpha }\);

(3) \(Ap+Bq\in M\) for all \(p,q\in M\).

Then there exists a unique \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Remark 19

Letting \(A=O\) in conditions (1), (2), and (3), we obtain Kannan fixed point theorem (see [19, p. 406]). The above corollary is a combined form of contraction mapping and generalized Kannan contraction mapping.

Corollary 20

Let X be a Banach space and M be a nonempty closed subset of X. Consider mappings \(A:M\rightarrow X\) and \(B:M\rightarrow X\) such that

(1) \(\Vert By-By^{\prime } \Vert \leq \alpha \Vert Av-(I-B)y \Vert \) with \(\alpha <1\) and for all \(v,y,y^{\prime }\in M\);

(2) \(\Vert Ax-Ax^{\prime } \Vert \leq k \Vert x-x^{ \prime } \Vert \) with \(k<1\);

(3) \(Ap+Bq\in M\) for all \(p,q\in M\).

Then there is a unique \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Corollary 21

Let X be a Banach space and M be a nonempty closed subset of X. Consider mappings \(A:M\rightarrow X\) and \(B:M\rightarrow X\) such that

(1) \(\Vert By-By^{\prime } \Vert \leq c \Vert y-y^{ \prime } \Vert \) with \(c<1\) and for all \(y,y^{\prime }\in M\);

(2) \(\Vert Ax-Ax^{\prime } \Vert \leq k \Vert x-x^{ \prime } \Vert \) with \(k<1-c\);

(3) \(Ap+Bq\in M\) for all \(p,q\in M\).

Then there is a unique \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

In the next result, with the help of Sadovskii fixed point theorem and Reich contraction theorem, we find that a solution of the operator equation \(x=Ax+Bx\) exists, where A is a strictly k-set contractive mapping and B is a generalized Reich contraction.

Theorem 22

Consider a Banach space X and let M be a subset of X such that M is closed, nonempty, convex, and bounded. Consider mappings \(A:M\rightarrow X\) and \(B:M\rightarrow X\) such that

(1) \(\Vert By-By^{\prime } \Vert \leq a_{1} \Vert Ax-(I-B)y \Vert +a_{2} \Vert Ax-(I-B)y^{\prime } \Vert +a_{3} \Vert y-y^{\prime } \Vert \) for \(a_{1}+a_{2}+a_{3}<1\) and for all \(x,y^{\prime },y\in M\);

(2) A is strictly \((\frac{1-a_{3}}{1+a_{2}})\)-set contractive mapping;

(3) \(Ax+By\in M\).

Then there exists \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Proof

Fix \(Ax\in A(M)\) and define \(H:M\rightarrow M\) by \(H(y)=Ax+By\). Using (1),

$$ \begin{aligned} \bigl\Vert H(y)-H\bigl(y^{\prime }\bigr) \bigr\Vert &= \bigl\Vert By-By^{\prime } \bigr\Vert \\ &\leq a_{1} \bigl\Vert Ax-(I-B)y \bigr\Vert +a_{2} \bigl\Vert Ax-(I-B)y^{ \prime } \bigr\Vert +a_{3} \bigl\Vert y-y^{\prime } \bigr\Vert \\ &=a_{1} \bigl\Vert y-H(y) \bigr\Vert +a_{2} \bigl\Vert y^{\prime }-H\bigl(y^{ \prime }\bigr) \bigr\Vert +a_{3} \bigl\Vert y-y^{\prime } \bigr\Vert . \end{aligned} $$

This shows that H is a Reich contraction. Hence by [29, Theorem 3], there is a unique \(G(Ax)\in M\) such that \(G(Ax)=Ax+B(G(Ax))\). Now

$$\begin{aligned} \bigl\Vert G(Ax)-G\bigl(Ax^{\prime }\bigr) \bigr\Vert ={}& \bigl\Vert Ax+B\bigl(G(Ax)\bigr)-\bigl(Ax^{\prime }+B \bigl(G\bigl(Ax^{\prime }\bigr)\bigr)\bigr) \bigr\Vert \\ ={}& \bigl\Vert \bigl(B\bigl(G(Ax)\bigr)-B\bigl(G\bigl(Ax^{\prime }\bigr) \bigr)\bigr)-\bigl(Ax^{\prime }-Ax\bigr) \bigr\Vert \\ \leq{}& \bigl(a_{1} \bigl\Vert Ax-(I-B)G(Ax) \bigr\Vert +a_{2} \bigl\Vert Ax-(I-B)G\bigl(Ax^{ \prime }\bigr) \bigr\Vert \\ &{} +a_{3} \bigl\Vert G(Ax)-G\bigl(Ax^{\prime }\bigr) \bigr\Vert + \bigl\Vert Ax^{\prime }-Ax \bigr\Vert \bigr) \\ ={}&a_{2} \bigl\Vert Ax-(I-B)G\bigl(Ax^{\prime }\bigr) \bigr\Vert \\ &{} +a_{3} \bigl\Vert G(Ax)-G\bigl(Ax^{\prime }\bigr) \bigr\Vert + \bigl\Vert Ax^{\prime }-Ax \bigr\Vert \\ ={}&(a_{2}+1) \bigl\Vert Ax-Ax^{\prime } \bigr\Vert +a_{3} \bigl\Vert G(Ax)-G\bigl(Ax^{ \prime }\bigr) \bigr\Vert . \end{aligned}$$

Thus

$$ \bigl\Vert (G\circ A)x-(G\circ A)x^{\prime } \bigr\Vert \leq \biggl( \frac{a_{2}+1}{1-a_{3}}\biggr) \bigl\Vert Ax-Ax^{\prime } \bigr\Vert . $$

This shows that \(G\circ A\) is a continuous function of M into X.

From (2), the above inequality, and Lemma 15, we deduce

$$ \mu \bigl(G\circ A(N)\bigr)=\mu \bigl(G\bigl(A(N)\bigr)\bigr)\leq \biggl( \frac{a_{2}+1}{1-a_{3}}\biggr)\mu \bigl(A(N)\bigr)< \mu (N)). $$

Using Sadovskii fixed point theorem, there is \(x\in M\) such that \((G\circ A)x=x\). Also for \(Ax\in A(M)\) there is a unique \(G(Ax)\) such that \(G(Ax)=Ax+B(G(Ax))\), therefore \(x=G(Ax)=Ax+B(G(Ax))=Ax+Bx\). □

Remark 23

If we take \(A=O\), we obtain Reich contraction theorem which is a generalization of Banach contraction principle and Kannan contraction theorem (see [22, p. 400]). Also taking \(B=O\) with \(b=c=0\), we get Sadovskii fixed point theorem which is a generalization Schauder’s fixed point theorem.

Corollary 24

Consider a Banach space X and let M be a subset of X such that M is closed, nonempty, convex, and bounded. Consider mappings \(A:M\rightarrow X\) and \(B:M\rightarrow X\) such that

(1) B is a contraction mapping for \(c<1\);

(2) A is a strictly \((1-c)\)-set contractive mapping;

(3) \(Ax+By\in M\).

Then there is \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Proof

Letting \(a_{2}=0=a_{1}\) and \(a_{3}=c\) in Theorem 22, we obtain the above corollary. □

Remark 25

Since every compact operator is a strictly \((1-c)\)-set contractive mapping, Corollary 24 and Theorem 22 are generalizations of Krasnoselskii’s fixed point theorem.

Remark 26

Theorem 22 is a generalization of Theorem 2.11 in [37].

Corollary 27

Consider a Banach space X and let M be a subset of X such that M is closed, nonempty, convex, and bounded. Consider mappings \(A:M\rightarrow X\) and \(B:M\rightarrow X\) such that

(1) \(\Vert By-By^{\prime } \Vert \leq \alpha ( \Vert Ax-(I-B)y \Vert +b \Vert Ax-(I-B)y^{\prime } \Vert )\) for \(\alpha <\frac{1}{2}\) and for all \(x,y,y^{\prime }\in M\);

(2) A is a strictly \((\frac{1}{1+\alpha })\)-set contractive mapping;

(3) \(Ax+By\in M\).

Then there exists \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Corollary 28

Consider a Banach space X and let M be a subset of X such that M is closed non-empty convex and bounded. Consider the mappings \(A:M\longrightarrow X\) and \(B:M\longrightarrow X\) such that

(1) \(\Vert By-By^{\prime } \Vert \leq b \Vert Ax-(I-B)y^{ \prime } \Vert \) for \(b<1\), all x \(y,y^{\prime }\in M\);

(2) A is condensing mapping of M into X;

(3) \(Ax+By\in M\).

Then there exists \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Suppose A and B are two operators of M into X. By using a measure of noncompactness, it makes sense to impose a condition on the operators A and B such that

$$ \mu \bigl(A(E)+B(F)\bigr)< \mu (E)\quad \text{for all }E\subseteq F\text{ with }\mu (E) \neq 0. $$

For example, if A and B are compact operators, then

$$ \mu \bigl(A(E)+B(F)\bigr)\leq \mu (A(E)+\mu \bigl(B(F)\bigr)=0< \mu (E)\quad \text{for all }E \subseteq F\text{ with }\mu (F)\neq 0. $$

Also, if A is a condensing operator and \(B(F)\) lies in a compact subset of X, then

$$ \mu \bigl(A(E)+B(F)\bigr)\leq \mu (A(E)+\mu \bigl(B(F)\bigr)=\mu (A(E)< \mu (E) $$

for all \(E\subseteq F\) with \(\mu (E)\neq 0\).

Theorem 29

Consider a Banach space X and let M be a subset of X such that M is closed, nonempty, convex, and bounded. Consider mappings \(B:M\rightarrow X\) and \(A:M\rightarrow X\) such that

(1) \(\Vert By-By^{\prime } \Vert \leq a_{1} \Vert Ax-(I-B)y \Vert +a_{2} \Vert Ax-(I-B)y^{\prime } \Vert +a_{3} \Vert y-y^{\prime } \Vert \) for \(a_{1}+a_{2}+a_{3}<1\) and for all \(x,y,y^{\prime }\in M\);

(2) \(\mu (A(E)+B(M))<\mu (E)\) for all \(E\subseteq M\) with \(\mu (E)\neq 0 \);

(3) \(Ax+By\in M\) for all \(x,y\in M\), where \(A:M\rightarrow X\) is a continuous mapping.

Then there is \(x\in M\) such that \(x=Ax+Bx\).

Proof

Fix \(Ax\in A(M)\) and define \(H:M\rightarrow M\) by \(H(y)=Ax+By\). Using (1),

$$ \begin{aligned} \bigl\Vert H(y)-H\bigl(y^{\prime }\bigr) \bigr\Vert &= \bigl\Vert By-By^{\prime } \bigr\Vert \\ &\leq a_{1} \bigl\Vert Ax-(I-B)y \bigr\Vert +a_{2} \bigl\Vert Ax-(I-B)y^{ \prime } \bigr\Vert +a_{3} \bigl\Vert y-y^{\prime } \bigr\Vert \\ &=a_{1} \bigl\Vert y-H(y) \bigr\Vert +a_{2} \bigl\Vert y^{\prime }-H\bigl(y^{ \prime }\bigr) \bigr\Vert +a_{3} \bigl\Vert y-y^{\prime } \bigr\Vert . \end{aligned} $$

Thus by [29, Theorem 3], there is a unique \(G(Ax)\in M\) that such that \(G(Ax)=Ax+B(G(Ax))\). Now

$$ \begin{aligned} \bigl\Vert G(Ax)-G\bigl(Ax^{\prime }\bigr) \bigr\Vert ={}& \bigl\Vert Ax+B\bigl(G(Ax)\bigr)-\bigl(Ax^{\prime }+B \bigl(G\bigl(Ax^{\prime }\bigr)\bigr)\bigr) \bigr\Vert \\ ={}& \bigl\Vert \bigl(B\bigl(G(Ax)\bigr)-B\bigl(G\bigl(Ax^{\prime }\bigr) \bigr)\bigr)-\bigl(Ax^{\prime }-Ax\bigr) \bigr\Vert \\ \leq{}& \bigl(a_{1} \bigl\Vert Ax-(I-B)G(Ax) \bigr\Vert +a_{2} \bigl\Vert Ax-(I-B)G\bigl(Ax^{ \prime }\bigr) \bigr\Vert \\ &{}+a_{3} \bigl\Vert G(Ax)-G\bigl(Ax^{\prime }\bigr) \bigr\Vert + \bigl\Vert Ax^{\prime }-Ax \bigr\Vert \bigr) \\ ={}&a_{2} \bigl\Vert Ax-(I-B)G\bigl(Ax^{\prime }\bigr) \bigr\Vert \\ &{}+a_{3} \bigl\Vert G(Ax)-G\bigl(Ax^{\prime }\bigr) \bigr\Vert + \bigl\Vert Ax^{\prime }-Ax \bigr\Vert \\ ={}&(a_{2}+1) \bigl\Vert Ax-Ax^{\prime } \bigr\Vert +a_{3} \bigl\Vert G(Ax)-G\bigl(Ax^{ \prime }\bigr) \bigr\Vert . \end{aligned} $$

Thus

$$ \bigl\Vert (G\circ A)x-(G\circ A)x^{\prime } \bigr\Vert \leq \biggl( \frac{a_{2}+1}{1-a_{3}}\biggr) \bigl\Vert Ax-Ax^{\prime } \bigr\Vert . $$

This shows that \(G\circ A\) is a continuous function of M into M. Using (2),

$$ \mu \bigl((G\circ A)E\bigr))=\mu \bigl(G\bigl(A(E)\bigr)\bigr)=\mu \bigl(A(E)+B \bigl(G\bigl(A(E)\bigr)\bigr)\bigr)\leq \mu \bigl(A(E)+B(M)\bigr)< \mu (E). $$

Using Sadovskii fixed point theorem, there is \(x\in M\) such that \((G\circ A)x=x\). Also for \(Ax\in A(M)\), there is a unique \(G(Ax)\) such that \(G(Ax)=Ax+B(G(Ax))\), therefore \(x=Ax+Bx\). □

Corollary 30

Consider a Banach space X and let M be a subset of X such that M is closed, nonempty, convex, and bounded. Consider mappings \(B:M\rightarrow X\) and \(A:M\rightarrow X\) such that \(B(M)\) lies in a compact subset of X and

(1) \(\Vert By-By^{\prime } \Vert \leq a_{1} \Vert Ax-(I-B)y \Vert +a_{2} \Vert Ax-(I-B)y^{\prime } \Vert +a_{3} \Vert y-y^{\prime } \Vert \) for \(a_{1}+a_{2}+a_{3}<1\) and for all \(y^{\prime },y,x\in M\);

(2) A is a condensing mapping;

(3) \(Aq+Bp\in M\) for all \(q,p\in M\).

Then there exists \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Proof

Since A is a condensing mapping and \(B(M)\) lies in a compact subset of X,

$$ \mu \bigl(A(E)+B(M)\bigr)\leq \mu \bigl(A(E)\bigr)+\mu \bigl(B(M)\bigr)=\mu \bigl(A(E)\bigr)< \mu (E). $$

Hence, by Theorem 29, the proof is complete. □

Corollary 31

Consider a Banach space X and let M be a subset of X such that M is closed, nonempty, convex, and bounded. Consider mappings \(B:M\rightarrow X\) and \(A:M\rightarrow X\) such that

(1) \(Ax+By\in M\);

(2) B is a contraction mapping and \(B(M)\) lies in a compact subset of X;

(3) A is a condensing mapping.

Then there exists \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Proof

Putting \(a_{1}=a_{2}=0\) in Corollary 30, we obtain the required result. □

Remark 32

Corollary 31 is a variant of Krasnoselskii fixed point theorem and Theorem 2.6 in [37].

Corollary 33

Consider a Banach space X and let M be a subset of X such that M is closed, nonempty, convex, and bounded. Consider mappings \(B:M\rightarrow X\) and \(A:M\rightarrow X\) such that

(1) A is a k-set contractive mapping for \(0\leq k<1\);

(2) B is a contraction mapping and \(B(M)\) lies in a compact subset of X;

(3) \(Ax+By\in M\).

Then there exists \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Proof

Since every k-set contractive mapping for \(0\leq k<1\) is condensing, by Corollary 31, we have the required result. □

Remark 34

If \(B=O\) (the zero operator) in Corollary 33, we obtain Darbo fixed point theorem.

Corollary 35

Consider a Banach space X and let M be a subset of X such that M is closed, nonempty, convex, and bounded. Consider mappings \(B:M\rightarrow X\) and \(A:M\rightarrow X\) such that

(1) \(Ax+By\in M\);

(2) B is a contraction mapping and \(B(M)\) lies in a compact subset of X;

(3) A is a compact operator.

Then there exists \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Proof

As every compact operator is k-set contractive for \(0\leq k<1\), by Corollary 33, we have a fixed point for the operator equation \(A\varkappa +B\varkappa =\varkappa \). □

Remark 36

If \(B=O\) (the zero operator), we obtain Schauder’s fixed point theorem (see [38, p. 56]).

Theorem 37

Consider a Banach space X and let M be a subset of X such that M is closed, nonempty, convex, and bounded. Consider mappings \(B:M\rightarrow X\) and \(A:M\rightarrow X\) such that

(1) \(\Vert By-By^{\prime } \Vert \leq q \Vert Ax-(y-By^{\prime }) \Vert \) for all \(x,y,y^{\prime }\in M\) with \(q<\frac{1}{2}\);

(2) \(\mu (A(E)+B(M))<\mu (E)\) for all \(E\subseteq M\) with \(\mu (E)\neq 0\).

Then there is \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

Proof

Fix \(x\in M\) and define a mapping \(H:M\rightarrow M\) by \(H(y)=Ax+By\). Now (1) implies that

$$ \begin{aligned} \bigl\Vert H(y)-H\bigl(y^{\prime }\bigr) \bigr\Vert &= \bigl\Vert By-By^{\prime } \bigr\Vert \\ &\leq q \bigl\Vert y-\bigl(Ax+By^{\prime }\bigr) \bigr\Vert \\ &=q \bigl\Vert y-H\bigl(y^{\prime }\bigr) \bigr\Vert . \end{aligned} $$

Thus by [22, p. 394], there is a unique \(Gx\in M\) that such that \(Gx=Ax+B(Gx)\), where \(G:M\rightarrow M\) is a function. Now

$$\begin{aligned} \bigl\Vert Gx-Gx^{\prime } \bigr\Vert &= \bigl\Vert Ax+B(Gx)-\bigl(Ax^{ \prime }+B\bigl(Gx^{\prime }\bigr) \bigr) \bigr\Vert \\ &= \bigl\Vert \bigl(B(Gx)-B\bigl(Gx^{\prime }\bigr)\bigr)- \bigl(Ax^{\prime }-Ax\bigr) \bigr\Vert \\ &\leq \bigl\Vert B(Gx)-B\bigl(Gx^{\prime }\bigr) \bigr\Vert +\Vert Ax^{ \prime }-Ax)\Vert \\ &\leq q \bigl\Vert Ax^{\prime }-\bigl(Gx-B\bigl(Gx^{\prime }\bigr) \bigr) \bigr\Vert + \bigl\Vert Ax^{\prime }-Ax \bigr\Vert \\ &=q \bigl\Vert Gx-\bigl(Ax^{\prime }+B\bigl(Gx^{\prime }\bigr) \bigr) \bigr\Vert + \bigl\Vert Ax-Ax^{\prime } \bigr\Vert \\ &=q \bigl\Vert Gx-Gx^{\prime } \bigr\Vert + \bigl\Vert Ax-Ax^{ \prime } \bigr\Vert . \end{aligned}$$

Thus

$$ \bigl\Vert Gx-Gx^{\prime } \bigr\Vert \leq \frac{1}{1-q} \bigl\Vert Ax-Ax^{\prime } \bigr\Vert . $$

This shows that G is continuous.

Using (2), we deduce that

$$ \mu \bigl(G(E)\bigr)=\mu \bigl(A(E)+B\bigl(G(E)\bigr)\bigr)\leq \mu \bigl(A(E)+B(M)\bigr)< \mu (E). $$

By Sadovskii fixed point theorem, there is \(x\in M\) such that \(Gx=x\). Also for \(x\in M\), there is a unique Gx such that \(Gx=Ax+B(Gx)\), therefore \(x=Ax+Bx\). □

Corollary 38

Consider a Banach space X and let M be a subset of X such that M is closed, nonempty, convex, and bounded. Consider mappings \(B:M\rightarrow X\) and \(A:M\rightarrow X\) such that \(B(M)\) lies in a compact subset of X and

(1) \(\Vert By-By^{\prime } \Vert \leq q \Vert Ax-(y-By^{\prime }) \Vert \) for all \(x,y,y^{\prime }\in M\) with \(q<\frac{1}{2}\);

(2) A is a condensing mapping.

Then there is \(\varkappa \in M\) such that \(A\varkappa +B\varkappa =\varkappa \).

3 Application

Let \(X=C [ 0,1 ] \), the Banach space of all continuous functions defined on \([ 0,1 ] \) with \(\Vert \cdot \Vert _{\infty }\). Consider the following nonlinear integral equation;

figure a

where \(g: [ 0,1 ] \times \mathbb{R} \rightarrow \mathbb{R} \) and \(c: [ 0,1 ] \rightarrow \mathbb{R} \) are continuous.

As

$$ \frac{1}{ ( 1-a ) }=1+\frac{a}{ ( 1-a ) }, $$

equation (I) can be written as

$$ y ( t ) = \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds+ \frac{a}{ ( 1-a ) } \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds+ \biggl( \frac{c ( y ) -a}{1-a} \biggr) y ( t ). $$

We decompose the above integral equation into a sum of two operators

$$ y ( t ) =Ay ( t ) +By ( t ), $$

where

$$ Ay ( t ) = \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds $$

and

$$ B \bigl( y ( t ) \bigr) =\frac{a}{ ( 1-a ) } \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds+ \biggl( \frac{c ( y ) -a}{1-a} \biggr) y ( t ) . $$

Let us assume the following conditions first:

(C1) \(\vert g ( s,y ( s ) ) -g ( s,y^{ \prime } ( s ) ) \vert \leq \lambda \vert y ( s ) -y^{\prime } ( s ) \vert \),

(C2) \(c+2a<1\) and \(\lambda +c\leq 1\), where \(\max \vert c ( t ) \vert =c\).

(C3) \(\max \vert g ( s,0 ) \vert =\beta \).

First, we prove (3) of Theorem 16 and for this we define

$$ S= \biggl\{ y\in X: \Vert y \Vert \leq r\text{, where }r \geq \frac{2\beta }{a ( 1-c ) } \biggr\} $$

and consider for \(y\in S\),

$$\begin{aligned} \bigl\vert Ay ( t ) +By ( t ) \bigr\vert ={}& \biggl\vert \frac{1}{ ( 1-a ) } \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds+ \biggl( \frac{c ( y ) -a}{1-a} \biggr) y ( t ) \biggr\vert \\ \leq{}& \biggl\vert \frac{1}{ ( 1-a ) } \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds- \frac{1}{ ( 1-a ) } \int _{0}^{t}g ( s,0 ) \,ds+ \frac{1}{ ( 1-a ) }\int _{0}^{t}g ( s,0 ) \,ds \biggr\vert \\ &{}+ \biggl\vert \biggl( \frac{c ( y ) -a}{1-a} \biggr) y ( t ) \biggr\vert \\ \leq{}& \frac{1}{ ( 1-a ) } \biggl\vert \int _{0}^{t} \bigl\{ g \bigl( s,y ( s ) \bigr) -g ( s,0 ) \bigr\} \,ds \biggr\vert \\ &{}+ \biggl\vert \frac{1}{ ( 1-a ) } \int _{0}^{t}g ( s,0 ) \,ds \biggr\vert + \biggl\vert \biggl( \frac{c ( y ) -a}{1-a} \biggr) y ( t ) \biggr\vert \\ \leq{}& \frac{1}{1-a} \int _{0}^{t}\lambda \bigl\vert y ( t ) \bigr\vert \,ds+ \frac{1}{ ( 1-a ) } \int _{0}^{t} \bigl\vert g ( s,0 ) \bigr\vert \,ds+ \biggl( \frac{c-a}{1-a} \biggr) \bigl\vert y ( t ) \bigr\vert . \end{aligned}$$

Now taking the supremum and using (C2) and (C3), we have

$$ \bigl\Vert Ay ( t ) +By ( t ) \bigr\Vert \leq \frac{1}{1-a}\beta + \biggl( \frac{\lambda +c-a}{1-a} \biggr) r\leq r, $$

proving (3).

We will show that the operators A and B satisfy (2) and (1) of our Theorem 16.

Claim: B satisfies (2) of Theorem 16.

For this, consider

$$\begin{aligned} B \bigl( y ( t ) \bigr) -B \bigl( y^{\prime } ( t ) \bigr) =& \biggl\{ \frac{a}{ ( 1-a ) } \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds+ \biggl( \frac{c ( y ) -a}{1-a} \biggr) y ( t ) \biggr\} \\ &{}- \biggl\{ \frac{a}{ ( 1-a ) } \int _{0}^{t}g \bigl( s,y^{\prime } ( s ) \bigr) \,ds+ \biggl( \frac{c ( y^{\prime } ) -a}{1-a} \biggr) y^{\prime } ( t ) \biggr\} , \end{aligned}$$

which gives

$$\begin{aligned} ( 1-a ) \bigl\{ B \bigl( y ( t ) \bigr) -B \bigl( y^{\prime } ( t ) \bigr) \bigr\} =& \biggl\{ a \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds+ \bigl( c ( y ) -a \bigr) y ( t ) \biggr\} \\ &{}- \biggl\{ a \int _{0}^{t}g \bigl( s,y^{\prime } ( s ) \bigr) \,ds+ \bigl( c ( y ) -a \bigr) y^{ \prime } ( t ) \biggr\} , \end{aligned}$$

further implying

$$\begin{aligned} \bigl\{ B \bigl( y ( t ) \bigr) -B \bigl( y^{\prime } ( t ) \bigr) \bigr\} =& \biggl\{ a \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds+ \bigl( c ( y ) -a \bigr) y ( t ) +aBy ( t ) \biggr\} \\ &{}- \biggl\{ a \int _{0}^{t}g \bigl( s,y^{\prime } ( s ) \bigr) \,ds+ \bigl( c ( y ) -a \bigr) y^{ \prime } ( t ) +aBy^{\prime } ( t ) \biggr\} \\ =&a \biggl\{ \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds- ( I-B ) y ( t ) \biggr\} +c ( y ) y ( t ) \\ &{}-a \biggl\{ \int _{0}^{t}g \bigl( s,y^{\prime } ( s ) \bigr) \,ds- ( I-B ) y^{\prime } ( t ) \biggr\} -c \bigl( y^{\prime } \bigr) y^{\prime } ( t ). \end{aligned}$$

Thus we have

$$ \begin{aligned} \bigl\vert B \bigl( y ( t ) \bigr) -B \bigl( y^{\prime } ( t ) \bigr) \bigr\vert ={}& \biggl\vert a \biggl\{ \int _{0}^{t}g \bigl( s,y ( s ) \bigr) \,ds- ( I-B ) y ( t ) \biggr\} +c ( y ) y ( t ) \\ &{}-a \biggl\{ \int _{0}^{t}g \bigl( s,y^{ \prime } ( s ) \bigr) \,ds- ( I-B ) y^{\prime } ( t ) \biggr\} -c \bigl( y^{\prime } \bigr) y^{\prime } ( t ) \biggr\vert \\ \leq{}& a \bigl\vert Ay ( t ) - ( I-B ) y ( t ) \bigr\vert +a \bigl\vert Ay^{\prime } ( t ) - ( I-B ) y^{\prime } ( t ) \bigr\vert \\ &{}+ \bigl\vert c ( y ) \bigr\vert \bigl\vert y ( t ) -y^{\prime } ( t ) \bigr\vert . \end{aligned} $$

Finally,

$$ \bigl\Vert By-By^{\prime } \bigr\Vert \leq a \bigl\Vert Ay- ( I-B ) y \bigr\Vert +a \bigl\Vert Ay^{\prime }- ( I-B ) y^{ \prime } \bigr\Vert +c \bigl\Vert y-y^{\prime } \bigr\Vert , $$

which shows that B satisfies (1).

Claim: A satisfies (2).

For this, we write

$$ \bigl\Vert Ay ( t ) -Ay^{\prime } ( t ) \bigr\Vert \leq \biggl\vert \int _{0}^{t} \lambda \,ds \biggr\vert \bigl\Vert y-y^{\prime } \bigr\Vert \leq \lambda \bigl\Vert y-y^{\prime } \bigr\Vert , $$

where \(\lambda <\frac{1-c}{1+a}\).

Hence all the conditions of Theorem 16 are satisfied, so we obtain \(w\in S\) such that \(w=Aw+Bw\), which is a solution of the integral equation (I).

We summarize all the above in the form of a theorem as follows.

Theorem 39

If g satisfies (C1)–(C3), then a solution to the integral equation (I) exists in S.

In the next example, we apply our main Theorem 16 to obtain an approximate solution of a given nonlinear integral equation.

Example 40

Consider a special case of the integral equation

figure b

namely

figure c

which has an exact solution \(y ( t ) =e^{-t}\), \(t\in [ 0,1 ] \), for \(\lambda =\frac{8}{10c}=4\) and \(c=\frac{2}{10}\). Clearly, \(2a+c<1\) for \(a=0.2\), so using these values, (C2) is satisfied. From (I) and (J), one can see that

$$ \bigl\vert g \bigl( t,y ( t ) \bigr) -g \bigl( t,y^{ \prime } ( t ) \bigr) \bigr\vert \leq \frac{1}{\lambda e^{-1/2}} \bigl\vert y ( t ) -y^{\prime } ( t ) \bigr\vert , $$

which implies that

$$ \bigl\Vert g \bigl( t,y ( t ) \bigr) -g \bigl( t,y^{ \prime } ( t ) \bigr) \bigr\Vert \leq \frac{\sqrt{e}}{4} \bigl\Vert y-y^{\prime } \bigr\Vert $$

holds since \(\frac{\sqrt{e}}{4}<1\). Therefore all the conditions of the above theorem are satisfied. Now we consider the iterative procedure

$$ y_{n+1} ( t ) = \int _{0}^{t} \frac{\sqrt{y_{n} ( s ) }y_{n} ( t ) }{\lambda e^{-s/2}}\,ds+ \biggl( \frac{ \{ 0.2-ct \} -0.2}{0.8} \biggr) y_{n} ( t ), $$

with an initial approximation \(y_{0} ( t ) =1-t\).

The graphs of exact and approximate solutions and absolute error are given below after two iterations.

It can be seen that after two iterations the absolute error is not significant.

Figure 1
figure 1

Show graphs of exact and approximate solutions

Full size image
Figure 2
figure 2

Shows the absolute error obtained from exact and approximate solutions

Full size image

4 Conclusion

In this article, Kannan and Reich contractions are generalized and used to obtain different variants of Krasnoselskii’s fixed point theorem for compact and noncompact operators. Examples and applications in the existence theory of integral equations are given to validate our presented work.