The following theorem was proved by Ben Amar and Garcia-Falset [13], and its more general form was presented by Agarwal et al. [14] is a variant of the Sadovskii fixed-point theorem for the classes of operators which satisfy ().
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 is continuous and satisfies (). If T is ω-condensing, then it has a fixed point in M.
Our purpose here is to establish a fixed-point theorem for the sum of a ω-contractive operator and an φ-nonlinear contractive operator.
Theorem 3.2 Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Suppose that and are two operators such that
-
(i)
A is a continuous ω-α-contraction with , and A satisfies (),
-
(ii)
B is an φ-nonlinear contraction with for , and B satisfies (),
-
(iii)
.
Then there is a point such that .
Proof By Lemma 2.9, has a continuous inverse defined on X, and then is well defined on M. Once we prove that J has a fixed point in M, the proof is achieved.
For any , according to Lemma 2.9 there exists a unique such that . The hypothesis (iii) shows that , which implies that and, therefore, .
Obviously, the compound operator J is continuous since A and is continuous and by Lemma 2.5, J satisfies (). Now by referring to the formula
for every subset S of M with , we have
Since A is ω-α-contractive and B satisfies by Lemma 2.8, we have
(3.1)
Now, if , inequality (3.1) becomes , which implies that . Otherwise, by recalling the assumption that for , inequality (3.1) becomes
In both cases, J is shown to be ω-condensing. 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 , then we return the above theorem back to [[7], Theorem 2.2], which is an extension of the Darbo fixed-point theorem for ω-contractive operators.
-
(2)
If we take and the function () in the above theorem, we obtain a result which was [[7], Theorem 2.3].
-
(3)
If we only take the function () in the above theorem, we obtain the following Corollary 3.4, which is a new fixed-point theorem for the sum of two operators.
-
(4)
If we only take in the above theorem, we obtain the following Corollary 3.6, which is the new version of Krasnosel’skii-type fixed-point theorems.
Corollary 3.4 Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Suppose that and are two operators such that
-
(i)
A is a continuous ω-α-contraction with , and A satisfies (),
-
(ii)
B is a strict contraction with , and B satisfies (),
-
(iii)
.
Then there is a point such that .
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 and are two operators such that
-
(i)
A is a continuous, is relatively weakly compact and A satisfies (),
-
(ii)
B is an φ-nonlinear contraction, and B satisfies (),
-
(iii)
.
Then there is a point such that .
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 fixed-point theorem for the sum of a weakly-strongly 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 and are two operators such that
-
(i)
A is weakly-strongly continuous, and is relatively weakly compact,
-
(ii)
B is nonexpansive and ω-condensing,
-
(iii)
is demiclosed,
-
(iv)
if and for some , then .
Then there is a point such that .
Remark 3.9
-
(1)
Recall that an operator is said to be demiclosed if for any sequence in that and , then and .
-
(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 is a sequence of M such that is weakly convergent, then has a weakly convergent subsequence. In the above theorem, we replaced it with the ω-condensing of B.
Proof of Theorem 3.8 For each , the operators A and λB fulfill the conditions of Corollary 3.6 and, therefore, there is a point such that . Now choose a sequence such that . Consequently, there exists a sequence such that
Let . We claim that S is relatively weakly compact. Suppose that it is not the case, by assumption (i) and (ii), we have
This contradiction tells us that the sequence has a weakly convergent subsequence, i.e., there exists such that . By assumption (i), we have , and then . Since is contained in bounded set M, and B maps M into a bounded set (B is nonexpansive), then is norm bounded. Thus, we have . Moreover, we have
that is, . By assumption (iii), we have , and then the proof is achieved. □
If the Banach space X is reflexive, then B is always ω-condensing on M (see, e.g., [[15], p.251]). Moreover, if we supposed that X is uniformly convex Banach space, then is demiclosed (see, e.g., [[20], pp.476-478]). 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 and are two operators such that
-
(i)
A is weakly-strongly continuous,
-
(ii)
B is nonexpansive,
-
(iii)
if and for some , then .
Then there is a point such that .
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 be continuous ω-condensing and satisfies (). Then either
-
(a)
equation has a solution, or
-
(b)
the set is unbounded for some .
Proof Choose an arbitrary . Define for each
Clearly, ρ is a continuous retraction of X on . Thus, we can define the mapping by .
Since T and ρ are continuous, obviously is also continuous. Furthermore, since T satisfies () and ρ is continuous, hence also satisfies (). We next claim that for any with .
Indeed, . For any , there are two possibilities:
-
(1)
; in this case, .
-
(2)
; in this case, .
The above argument yields .
Now, by using the properties of the measure of weak noncompactness and properties of T, we have that
as claimed, that is, is a ω-condensing.
The above argument shows that is under the conditions of Theorem 3.1 and thus we have that there exists such that . Indeed, we obtain the following results:
-
(a)
if , then , that is, T has a fixed point; otherwise,
-
(b)
if , then , that is, is a solution of the equation for and .
Consequently, if there is no such that for any , then the above arguments show that the set of solutions of equation is unbounded for some . □
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 are two operators such that
-
(i)
A is a continuous ω-α-contraction with , and A satisfies (),
-
(ii)
B is an φ-nonlinear contraction with for , and B satisfies (),
-
(iii)
function φ satisfies .
Then, either
-
(a)
the equation has a solution, or
-
(b)
the set is unbounded for some .
Remark 3.13 Obviously, assumption (iii) in the above theorem is unnecessary whenever .
Proof of Theorem 3.12 As in the proof of Theorem 3.2, it can be seen that the compound operator is well defined from X into X. Clearly, J is continuous and J satisfies () by Lemma 2.5.
Let us prove that J maps bounded set into a bounded set. For any bounded set S such that , there exist such that and , that is, and . Thus, by assumption (ii) and the boundness of , we have
(3.2)
Suppose that is unbounded, i.e., there exist sequences and such that , and then by assumption (iii). This is a contradiction with in (3.2) and, therefore, is bounded.
It is similar to that of Theorem 3.2 to prove that for every bounded set S in with . Now, by using Theorem 3.11 for operator J, we obtain that either
-
(a)
the equation has a solution which is the solution of equation , or
-
(b)
the set is unbounded for some .
□
Corollary 3.14 Let X be a Banach space. Suppose that are two operators such that
-
(i)
A is a continuous ω-α-contraction with , and A satisfies (),
-
(ii)
B is a strict contraction with , and B satisfies ().
Then, either
-
(a)
the equation has a solution, or
-
(b)
the set is unbounded for some .
Corollary 3.15 Let X be a Banach space. Suppose that are two operators such that
-
(i)
A is a continuous, is relatively weakly compact, and A satisfies (),
-
(ii)
B is an φ-nonlinear contraction, and B satisfies (),
-
(iii)
function φ satisfies .
Then, either
-
(a)
the equation has a solution, or
-
(b)
the set is unbounded for some .