The following theorem was proved by Ben Amar and GarciaFalset [13], and its more general form was presented by Agarwal et al. [14] is a variant of the Sadovskii fixedpoint theorem for the classes of operators which satisfy (\mathcal{A}1).
Theorem 3.1 (see [[13], Theorem 3.1] or [[14], Theorem 2.1 and Corollary 2.2])
Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Assume that T:M\to M is continuous and satisfies (\mathcal{A}1). If T is ωcondensing, then it has a fixed point in M.
Our purpose here is to establish a fixedpoint theorem for the sum of a ωcontractive operator and an φnonlinear contractive operator.
Theorem 3.2 Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Suppose that A:M\to X and B:X\to X are two operators such that

(i)
A is a continuous ωαcontraction with \alpha \in [0,1), and A satisfies (\mathcal{A}1),

(ii)
B is an φnonlinear contraction with \phi (r)<(1\alpha )r for r>0, and B satisfies (\mathcal{A}2),

(iii)
(x=Bx+Ay,y\in M)\Rightarrow x\in M.
Then there is a point x\in M such that Ax+Bx=x.
Proof By Lemma 2.9, IB has a continuous inverse defined on X, and then J:={(IB)}^{1}A is well defined on M. Once we prove that J has a fixed point in M, the proof is achieved.
For any y\in M, according to Lemma 2.9 there exists a unique x\in X such that Ay=xBx. The hypothesis (iii) shows that x\in M, which implies that A(M)\subset (IB)(M) and, therefore, J(M)\subset M.
Obviously, the compound operator J is continuous since A and {(IB)}^{1} is continuous and by Lemma 2.5, J satisfies (\mathcal{A}1). Now by referring to the formula
{(IB)}^{1}A=A+B{(IB)}^{1}A\phantom{\rule{1em}{0ex}}(\mathit{\text{i.e.}},J=A+BJ),
for every subset S of M with \omega \{S\}>0, we have
Since A is ωαcontractive and B satisfies \omega \{BJ(S)\}\le \phi (\omega \{J(S)\}) by Lemma 2.8, we have
\omega \{J(S)\}\le \omega \{A(S)\}+\omega \{BJ(S)\}\le \alpha \omega \{S\}+\phi \left(\omega \{J(S)\}\right).
(3.1)
Now, if \alpha =0, inequality (3.1) becomes \omega \{J(S)\}\le \phi (\omega \{J(S)\}), which implies that \omega \{J(S)\}=0. Otherwise, by recalling the assumption that \phi (r)<(1\alpha )r for r>0, inequality (3.1) becomes
\omega \{J(S)\}<\alpha \omega \{S\}+(1\alpha )\omega \{J(S)\},\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\omega \{J(S)\}<\omega \{S\}.
In both cases, J is shown to be ωcondensing. Now the use of Theorem 3.1 achieves the proof. □
Remark 3.3 It should be noticed to the following particular cases:

(1)
If we take B=0, then we return the above theorem back to [[7], Theorem 2.2], which is an extension of the Darbo fixedpoint theorem for ωcontractive operators.

(2)
If we take \alpha =0 and the function \phi (r)=\beta r (0\le \beta <1) in the above theorem, we obtain a result which was [[7], Theorem 2.3].

(3)
If we only take the function \phi (r)=\beta r (0\le \beta <1\alpha) in the above theorem, we obtain the following Corollary 3.4, which is a new fixedpoint theorem for the sum of two operators.

(4)
If we only take \alpha =0 in the above theorem, we obtain the following Corollary 3.6, which is the new version of Krasnosel’skiitype fixedpoint theorems.
Corollary 3.4 Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Suppose that A:M\to X and B:X\to X are two operators such that

(i)
A is a continuous ωαcontraction with \alpha \in [0,1), and A satisfies (\mathcal{A}1),

(ii)
B is a strict contraction with \beta \in [0,1\alpha ), and B satisfies (\mathcal{A}2),

(iii)
(x=Bx+Ay,y\in M)\Rightarrow x\in M.
Then there is a point x\in M such that Ax+Bx=x.
Remark 3.5 The above corollary is a variant and supplement of [[6], Theorem 3.3], in which the authors demand that the operators A and B are weakly sequentially continuous.
Corollary 3.6 Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Suppose that A:M\to X and B:X\to X are two operators such that

(i)
A is a continuous, A(M) is relatively weakly compact and A satisfies (\mathcal{A}1),

(ii)
B is an φnonlinear contraction, and B satisfies (\mathcal{A}2),

(iii)
(x=Bx+Ay,y\in M)\Rightarrow x\in M.
Then there is a point x\in M such that Ax+Bx=x.
Remark 3.7 The above corollary is a variant and supplement of [[18], Theorem 2.1], in which the author demands that the operators A and B are weakly sequentially continuous, and B is strictly contractive.
Now, on the basis of Corollary 3.6, we prove the following fixedpoint theorem for the sum of a weaklystrongly continuous operator and a nonexpansive operator.
Theorem 3.8 Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Suppose that A:M\to X and B:X\to X are two operators such that

(i)
A is weaklystrongly continuous, and A(M) is relatively weakly compact,

(ii)
B is nonexpansive and ωcondensing,

(iii)
IB is demiclosed,

(iv)
if \lambda \in (0,1) and x=\lambda Bx+Ay for some y\in M, then x\in M.
Then there is a point x\in M such that Ax+Bx=x.
Remark 3.9

(1)
Recall that an operator T:\mathcal{D}(T)\subset X\to X is said to be demiclosed if for any sequence {({x}_{n})}_{n\in \mathbb{N}} in D(T) that {x}_{n}\rightharpoonup x and T{x}_{n}\to y, then x\in \mathcal{D}(T) and Tx=y.

(2)
The assumption (iv) in the above theorem was first introduced in [19], it is slight different with [[10], Corollary 3.1] and [[11], Theorem 2.1].

(3)
In [[11], Theorem 2.1] and [[18], Theorem 2.4], the following condition is required: if {({x}_{n})}_{n\in \mathbb{N}} is a sequence of M such that {({x}_{n}T{x}_{n})}_{n\in \mathbb{N}} is weakly convergent, then {({x}_{n})}_{n\in \mathbb{N}} has a weakly convergent subsequence. In the above theorem, we replaced it with the ωcondensing of B.
Proof of Theorem 3.8 For each \lambda \in (0,1), the operators A and λB fulfill the conditions of Corollary 3.6 and, therefore, there is a point {x}_{\lambda}\in M such that {x}_{\lambda}=\lambda B{x}_{\lambda}+A{x}_{\lambda}. Now choose a sequence {({\lambda}_{n})}_{n\in \mathbb{N}}\subset (0,1) such that {\lambda}_{n}\to 1. Consequently, there exists a sequence {({x}_{n})}_{n\in \mathbb{N}}\subset M such that
{x}_{n}={\lambda}_{n}B{x}_{n}+A{x}_{n}.
Let S=\{{x}_{n}:n\in \mathbb{N}\}. We claim that S is relatively weakly compact. Suppose that it is not the case, by assumption (i) and (ii), we have
\begin{array}{rcl}\omega \{S\}& =& \omega \{{x}_{n}:n\in \mathbb{N}\}=\omega \{{\lambda}_{n}B{x}_{n}+A{x}_{n}:n\in \mathbb{N}\}\\ \le & {\lambda}_{n}\omega \{B(S)\}+\omega \{A(S)\}={\lambda}_{n}\omega \{B(S)\}<\omega \{S\}.\end{array}
This contradiction tells us that the sequence {({x}_{n})}_{n\in \mathbb{N}} has a weakly convergent subsequence, i.e., there exists {({x}_{{n}_{k}})}_{k\in \mathbb{N}} such that {x}_{{n}_{k}}\rightharpoonup x\in M. By assumption (i), we have A{x}_{{n}_{k}}\to Ax, and then (I{\lambda}_{{n}_{k}}B){x}_{{n}_{k}}\to Ax. Since {({x}_{{n}_{k}})}_{k\in \mathbb{N}} is contained in bounded set M, and B maps M into a bounded set (B is nonexpansive), then \parallel B{x}_{{n}_{k}}\parallel is norm bounded. Thus, we have (1{\lambda}_{{n}_{k}})\parallel B{x}_{{n}_{k}}\parallel \to 0. Moreover, we have
\begin{array}{rcl}\parallel (IB){x}_{{n}_{k}}Ax\parallel & \le & \parallel (IB){x}_{{n}_{k}}(I{\lambda}_{{n}_{k}}B){x}_{{n}_{k}}\parallel +\parallel (I{\lambda}_{{n}_{k}}B){x}_{{n}_{k}}Ax\parallel \\ =& (1{\lambda}_{{n}_{k}})\parallel B{x}_{{n}_{k}}\parallel +\parallel (I{\lambda}_{{n}_{k}}B){x}_{{n}_{k}}Ax\parallel \to 0,\end{array}
that is, (IB){x}_{{n}_{k}}\to Ax. By assumption (iii), we have (IB)x=Ax, and then the proof is achieved. □
If the Banach space X is reflexive, then B is always ωcondensing on M (see, e.g., [[15], p.251]). Moreover, if we supposed that X is uniformly convex Banach space, then IB:M\to X is demiclosed (see, e.g., [[20], pp.476478]). Thus, we obtain the following consequence.
Corollary 3.10 Let M be a nonempty, bounded, closed and convex subset of a uniformly convex Banach space X. Suppose that A:M\to X and B:X\to X are two operators such that

(i)
A is weaklystrongly continuous,

(ii)
B is nonexpansive,

(iii)
if \lambda \in (0,1) and x=\lambda Bx+Ay for some y\in M, then x\in M.
Then there is a point x\in M such that Ax+Bx=x.
In order to use the above results on the whole space, we first prove the following result.
Theorem 3.11 Let X be a Banach space X. Assume that the operator T:X\to X be continuous ωcondensing and satisfies (\mathcal{A}1). Then either

(a)
equation x=Tx has a solution, or

(b)
the set \{x\in X:x=\lambda T(x)\} is unbounded for some \lambda \in (0,1).
Proof Choose an arbitrary R>0. Define for each x\in X
\rho (x)=\{\begin{array}{cc}x,\hfill & \parallel x\parallel \le R,\hfill \\ \frac{R}{\parallel x\parallel}x,\hfill & \parallel x\parallel >R.\hfill \end{array}
Clearly, ρ is a continuous retraction of X on {\mathbf{B}}_{R}. Thus, we can define the mapping {T}_{\rho}:{\mathbf{B}}_{R}\to {\mathbf{B}}_{R} by {T}_{\rho}x=\rho (Tx).
Since T and ρ are continuous, obviously {T}_{\rho} is also continuous. Furthermore, since T satisfies (\mathcal{A}1) and ρ is continuous, hence {T}_{\rho} also satisfies (\mathcal{A}1). We next claim that \omega \{{T}_{\rho}(S)\}<\omega \{S\} for any S\subset {\mathbf{B}}_{r} with \omega \{S\}>0.
Indeed, {T}_{\rho}({\mathbf{B}}_{r})=\rho (T({\mathbf{B}}_{r})). For any x\in {\mathbf{B}}_{r}, there are two possibilities:

(1)
\parallel Tx\parallel \le R; in this case, \rho (Tx)=Tx\in T(S)\subset co(T(S)\cup \{\mathbf{0}\}).

(2)
\parallel Tx\parallel >R; in this case, \rho (Tx)=\frac{R}{\parallel Tx\parallel}Tx=\frac{R}{\parallel Tx\parallel}Tx+(1\frac{R}{\parallel Tx\parallel})\cdot \mathbf{0}\in co(T(S)\cup \{\mathbf{0}\}).
The above argument yields {T}_{\rho}(S)\subset co(T(S)\cup \{\mathbf{0}\}).
Now, by using the properties of the measure of weak noncompactness and properties of T, we have that
\omega \{{T}_{\rho}(S)\}\le \omega \{T(S)\}<\omega \{S\},
as claimed, that is, {T}_{\rho} is a ωcondensing.
The above argument shows that {T}_{\rho}:{\mathbf{B}}_{r}\to {\mathbf{B}}_{r} is under the conditions of Theorem 3.1 and thus we have that there exists {x}_{0}\in {\mathbf{B}}_{r} such that {x}_{0}={T}_{\rho}{x}_{0}. Indeed, we obtain the following results:

(a)
if T{x}_{0}\in {\mathbf{B}}_{R}, then {x}_{0}={T}_{\rho}{x}_{0}=\rho (T{x}_{0})=T{x}_{0}, that is, T has a fixed point; otherwise,

(b)
if T{x}_{0}\notin {\mathbf{B}}_{R}, then {x}_{0}={T}_{\rho}{x}_{0}=\rho (T{x}_{0})=\frac{R}{\parallel T{x}_{0}\parallel}T{x}_{0}, that is, {x}_{0} is a solution of the equation x=\lambda Tx for \lambda =\frac{R}{\parallel T{x}_{0}\parallel}\in (0,1) and \parallel {x}_{0}\parallel =R.
Consequently, if there is no x\in {\mathbf{B}}_{R} such that x=Tx for any R>0, then the above arguments show that the set of solutions of equation x=\lambda T(x) is unbounded for some \lambda \in (0,1). □
We are now in a position to prove the main result on the whole space.
Theorem 3.12 Let X be a Banach space. Suppose that A,B:X\to X are two operators such that

(i)
A is a continuous ωαcontraction with \alpha \in [0,1), and A satisfies (\mathcal{A}1),

(ii)
B is an φnonlinear contraction with \phi (r)<(1\alpha )r for r>0, and B satisfies (\mathcal{A}2),

(iii)
function φ satisfies {lim}_{r\to +\mathrm{\infty}}[r\phi (r)]=+\mathrm{\infty}.
Then, either

(a)
the equation x=Ax+Bx has a solution, or

(b)
the set \{x\in X:x=\lambda B(x/\lambda )+\lambda A(x)\} is unbounded for some \lambda \in (0,1).
Remark 3.13 Obviously, assumption (iii) in the above theorem is unnecessary whenever \alpha \ne 0.
Proof of Theorem 3.12 As in the proof of Theorem 3.2, it can be seen that the compound operator J:={(IB)}^{1}A is well defined from X into X. Clearly, J is continuous and J satisfies (\mathcal{A}1) by Lemma 2.5.
Let us prove that J maps bounded set into a bounded set. For any bounded set S such that u,v\in {(IB)}^{1}A(S), there exist x,y\in S such that u={(IB)}^{1}Ax and v={(IB)}^{1}Ay, that is, uBu=Ax and vBv=Ay. Thus, by assumption (ii) and the boundness of A(S), we have
\parallel uv\parallel \phi (\parallel uv\parallel )\le \parallel AxAy\parallel \le diam(A(S))<+\mathrm{\infty}.
(3.2)
Suppose that J(S)={(IB)}^{1}A(S) is unbounded, i.e., there exist sequences {({u}_{n})}_{n\in \mathbb{N}} and {({v}_{n})}_{n\in \mathbb{N}} such that \parallel {u}_{n}{v}_{n}\parallel \to +\mathrm{\infty}, and then \parallel {u}_{n}{v}_{n}\parallel \phi (\parallel {u}_{n}{v}_{n}\parallel )\to +\mathrm{\infty} by assumption (iii). This is a contradiction with diam(A(S))<+\mathrm{\infty} in (3.2) and, therefore, J(S) is bounded.
It is similar to that of Theorem 3.2 to prove that \omega \{J(S)\}<\omega \{S\} for every bounded set S in \mathcal{D}(T) with \omega \{S\}>0. Now, by using Theorem 3.11 for operator J, we obtain that either

(a)
the equation x=Ax+Bx has a solution which is the solution of equation x=Jx, or

(b)
the set \{x\in X:x=\lambda B(x/\lambda )+\lambda A(x)\}=\{x\in X:x=\lambda {(IB)}^{1}Ax\} is unbounded for some \lambda \in (0,1).
□
Corollary 3.14 Let X be a Banach space. Suppose that A,B:X\to X are two operators such that

(i)
A is a continuous ωαcontraction with \alpha \in [0,1), and A satisfies (\mathcal{A}1),

(ii)
B is a strict contraction with \beta \in [0,1\alpha ), and B satisfies (\mathcal{A}2).
Then, either

(a)
the equation x=Ax+Bx has a solution, or

(b)
the set \{x\in X:x=\lambda B(x/\lambda )+\lambda A(x)\} is unbounded for some \lambda \in (0,1).
Corollary 3.15 Let X be a Banach space. Suppose that A,B:X\to X are two operators such that

(i)
A is a continuous, A(M) is relatively weakly compact, and A satisfies (\mathcal{A}1),

(ii)
B is an φnonlinear contraction, and B satisfies (\mathcal{A}2),

(iii)
function φ satisfies {lim}_{r\to +\mathrm{\infty}}[r\phi (r)]=+\mathrm{\infty}.
Then, either

(a)
the equation x=Ax+Bx has a solution, or

(b)
the set \{x\in X:x=\lambda B(x/\lambda )+\lambda A(x)\} is unbounded for some \lambda \in (0,1).