In this section, we give the definitions and some properties of the integral operators corresponding to the problem (B), by applying the basic properties of the spaces and X which are given in the previous section.
Throughout this paper, let satisfy the log-Hölder continuity condition (2.2). We define an operator by
(3.1)
for any where denotes the pairing of X and its dual and the Euclidean scalar product on , respectively.
The following estimate, which can be found in [17], plays a key role in obtaining the homeomorphism of the operator J.
Lemma 3.1 For any , the following inequalities hold:
From Lemma 3.1, we can obtain the following topological result, which will be needed in the main result. Compared to the case of being constant (see [11]), the following result is hard to prove because it has complicated nonlinearities.
Theorem 3.2 Let (w1) and (w2) be satisfied. The operator is homeomorphism onto with a bounded inverse.
Proof Let and be operators defined by
Then the operators , are bounded and continuous. In fact, for any , let in X as . Then there exist a subsequence and functions v, in for such that as , and for all and for almost all . Without loss of generality, we assume that for . Then we have
(3.2)
and
(3.3)
and the integrands at the right-hand sides in (3.2) and (3.3) are dominated by some integrable functions. Since in X as , we can deduce that and as for almost all . Therefore, the Lebesgue dominated convergence theorem tells us that in and in as , that is, , are continuous on X. Also it is easy to show that these operators are bounded on X.
Using the continuity for the operators and on X, we finally show that J is continuous on X. From Hölder’s inequality, we have
for all . Hence we get
(3.4)
and the right-hand side in (3.4) converges to zero as . Therefore the operator J is continuous on X.
For any u in X with , it follows that
for some positive constant C. Thus we get
as and therefore the operator J is coercive on X.
Denote
Set
and
(Of course, if both the sets and are nonempty, then by the continuity of .) It is clear that
(3.5)
By using Lemma 3.1 and (3.5), we find that J is strictly monotone on X. The Browder-Minty theorem hence implies that the inverse operator exists and is bounded; see Theorem 26.A in [28].
Next we will show that is continuous on . Assume that u and v are any elements in X with . According to Lemma 3.1, we have
and
for almost all and for some positive constants and . Integrating the above inequalities over Ω and using Lemma 2.2, we assert that
(3.6)
for some positive constants , , and . For almost all , the following inequalities hold:
(3.7)
and
(3.8)
where we put and use the shortcuts
Hence using Lemma 3.1, we assert that
for some positive constant . From Hölder’s and Minkowski’s inequalities, and the inequality
(3.9)
for any positive numbers a, b, r, and s, it follows that
(3.10)
Applying Lemma 2.3 and Minkowski’s inequality,
for any where α is either or . In a similar way,
for any where β is either or . It follows from (3.7)-(3.10) and Lemma 2.2 that
where γ is either or and is positive constant. So
(3.11)
for some positive constant . Consequently, it follows from (3.6) and (3.11) that
(3.12)
for some positive constants and where . For each , let be any sequence in that converges to h in . Set and with . We obtain from (3.12)
Since is bounded in X and in as , it follows that converges to u in X. Thus, is continuous at each . This completes the proof. □
From now on we deal with the properties for the superposition operator induced by the function f in (B). We assume that the variable exponents are subject to the following restrictions:
for almost all . Assume that:
-
(F1)
satisfies the Carathéodory condition in the sense that is measurable for all and is continuous for almost all .
-
(F2)
For each bounded interval , there are a function and a nonnegative constant such that
for almost all and all .
-
(F3)
f satisfies the following inequality:
where and for each .
-
(F4)
There exist a function and a locally bounded function with such that
for almost all and all .
Under assumptions (F1) and (F2), we can define an operator by
(3.13)
and an operator by
(3.14)
for any .
For our aim, we need some properties of the operators F and G. In contrast with [23], we give a direct proofs for the continuity and compactness of F and G without using a continuity result on superposition operators.
Theorem 3.3 If (w1), (w2), and (F1)-(F3) hold, then the operator is continuous and compact. Also the operator is continuous and compact.
Proof Let be an operator defined by
Then for fixed , the operator is bounded and continuous. In fact, for any , let in X as . Then there exist a subsequence and functions v, in for such that and as , and and for all and for almost all . Suppose that we can choose such that implies that . For , we have
and (F2) implies that the integrand at the right-hand side is dominated by an integrable function. Since the function f satisfies a Carathéodory condition, we obtain as for almost all . Therefore, the Lebesgue dominated convergence theorem tells us that in as . We conclude that in as and thus is continuous on X. The boundedness of follows from (F2), Minkowski’s inequality, and the imbedding continuously (see Theorem 2.11 in [23]) as follows:
(3.15)
for all and for some positive constant d.
Minkowski’s inequality and (3.12) imply in view of (F3) that
for all and for all . This shows that for any bounded subset , the family is equicontinuous at each . Hence it follows from the continuity of that Ψ is continuous on , on observing the following relation:
Moreover, Ψ is bounded. Indeed, if and are bounded, we have to verify that is bounded. We may assume that is compact. By the equicontinuity and the compactness of , we can find finitely many numbers such that for every there is an integer with
Since is bounded for each , Minkowski’s inequality hence implies that is bounded.
Recall that the embedding is continuous and compact (see e.g. [8]) and so the adjoint operator given by
is also compact. As F can be written as a composition of with Ψ, we conclude that F is continuous and compact on . The operator G is continuous and compact because G can be regarded as a special case of F. This completes the proof. □
The analog of the following result can be found in [23]. However, our growth condition described in assumption (F4) is slightly different from that of [23].
Lemma 3.4 Let assumptions (w1), (w2), (F1) and (F4) be fulfilled. Then the operator has the following property:
Proof Let . Choose a positive constant R such that for all . Since b is locally bounded, there is a nonnegative constant such that for all . Let with . Set . Without loss of generality, we may suppose that
By assumption (F4), Lemma 2.5 and the continuous imbedding , we obtain that
where and are positive constants. It follows from Hölder’s inequality that
for all with , where is a positive constant. Consequently, we get
□
Recall that a real number μ is called an eigenvalue of (E) if the equation
has a solution in X which is different from the origin.
The following lemma is a consequence about nonlinear spectral theory and its proof can be found in [23]. For the case that is a constant, this assertion has been obtained by using the Furi-Martelli-Vignoli spectrum; see Theorem 4 of [29] or Lemma 27 of [20].
Lemma 3.5 Suppose that assumptions (w1) and (w2) are fulfilled. If μ is not an eigenvalue of (E), then we have
(3.16)