We consider a monotone increasing operator in an ordered Banach space having and as a strong super- and subsolution, respectively. In contrast with the well-studied case , we suppose that . Under the assumption that the order cone is normal and minihedral, we prove the existence of a fixed point located in the order interval .
MSC: 47H05, 47H10, 46B40.
Keywords:fixed point theorems in ordered Banach spaces
then f has a fixed point in . It is a natural question whether this result can be extended to the case of ordered Banach spaces. A number of fixed point theorems with assumptions of type (1) are well known; see, e.g., [, Section 2.1]. However, to the best of our knowledge, fixed point theorems with assumptions of type (2) have not been known so far. In the present note, we prove the following fixed point theorem of this type.
Theorem 1LetXbe a real Banach space with an order coneKsatisfying
(a) Khas a nonempty interior,
(b) Kis normal and minihedral.
Theorem 1 generalizes an idea developed by the present authors in , where the existence of solutions to a certain nonlinear integral equation of Hammerstein type has been shown.
Obviously, a cone K is minihedral if and only if for any pair , , bounded below in order there exists the greatest lower bound . If a minihedral cone has a nonempty interior, then any pair is bounded above in order. Hence, and exist for all .
By the Kakutani-Krein brothers theorem [, Theorem 6.6] a real Banach space X with an order cone K satisfying assumptions (a) and (b) of Theorem 1 is isomorphic to the Banach space of continuous functions on a compact Hausdorff space Q. The image of K under this isomorphism is the cone of nonnegative continuous functions on Q.
Proof For any , the maps and are continuous; see, e.g., Corollary 3.1.1 in . Due to the continuity of T, it follows immediately that is continuous as well. The operator is monotone increasing since inf and sup are monotone increasing with respect to each argument. Therefore, for any , we have
Let be an arbitrary sequence in . Since T is compact, has a subsequence converging to some . From the continuity of , it follows that the sequence converges to , thus, proving that the range of is relatively compact. □
Proof Due to , there is a such that . The preimage of under the continuous mapping contains a ball . Hence, holds for all . By the same argument, for all . Choosing sufficiently small, we can achieve that .
From it follows that there is an element such that . Assume that . Then we have . However, in view of the Kakutani-Krein brothers theorem, implies . Thus, it follows that and, therefore, . Similarly one shows that . □
The main tool for the proof of Theorem 1 is Amann’s theorem on three fixed points (see, e.g., [, Theorem 7.F and Corollary 7.40]):
Theorem 4LetXbe a real Banach space with an order cone having a nonempty interior. Assume there are four points inX,
Recall that the operator is called image compact if it is continuous and its image is a relatively compact set.
We choose , , , , where is as in Lemma 3. Since the cone K is normal, by Theorem 1.1.1 in , is norm bounded. Thus, is image compact.
The authors declare that they have no competing interests.
All authors contributed equally. All authors read and approved the final manuscript.
The authors thank H.-P. Heinz for useful comments. This work has been supported in part by the Deutsche Forschungsgemeinschaft, Grant KO 2936/4-1.