Abstract

Given a bounded domain with a Lipschitz boundary and , we consider the quasilinear elliptic equation in complemented with the generalized Wentzell-Robin type boundary conditions of the form on . In the first part of the article, we give necessary and sufficient conditions in terms of the given functions , and the nonlinearities , , for the solvability of the above nonlinear elliptic boundary value problems with the nonlinear boundary conditions. In other words, we establish a sort of “nonlinear Fredholm alternative” for our problem which extends the corresponding Landesman and Lazer result for elliptic problems with linear homogeneous boundary conditions. In the second part, we give some additional results on existence and uniqueness and we study the regularity of the weak solutions for these classes of nonlinear problems. More precisely, we show some global a priori estimates for these weak solutions in an -setting.

Dedicated to the 70th birthday of Jerome A. Goldstein

1. Introduction

Let , , be a bounded domain with a Lipschitz boundary and consider the following nonlinear boundary value problem with nonlinear second-order boundary conditions:where , , for some constant , is either or , and , are monotone nondecreasing functions such that . Moreover, is the -Laplace operator, , and , are given real-valued functions. Here, denotes the usual -dimensional Lebesgue measure in and denotes the restriction to of the -dimensional Hausdorff measure. Recall that coincides with the usual Lebesgue surface measure since has a Lipschitz boundary, and denotes the normal derivative of in direction of the outer normal vector . Furthermore, is defined as the generalized -Laplace-Beltrami operator on ; that is, , . In particular, and become the well-known Laplace and Laplace-Beltrami operators on and , respectively. Here, for any real-valued function ,where denotes the directional derivative of along the tangential directions at each point on the boundary, whereas denotes the tangential gradient at . It is worth mentioning again that when in (1), the boundary conditions are of lower order than the order of the -Laplace operator, while, for , we deal with boundary conditions which have the same differential order as the operator acting in the domain . Such boundary conditions arise in many applications, such as phase-transition phenomena (see, e.g., [1, 2] and the references therein), and have been studied by several authors (see, e.g., [37]).

In [4], the authors have formulated necessary and sufficient conditions for the solvability of (1) when , by establishing a sort of “nonlinear Fredholm alternative” for such elliptic boundary value problems. We shall now state their main result. Defining two real parameters , bythis result reads that a necessary condition for the existence of a weak solution of (1) is thatwhile a sufficient condition iswhere denotes the range of , , and denotes the interior of the set .

Relation (4) turns out to be both necessary and sufficient if either of the sets or is an open interval. This particular result was established in [4, Theorem  3], by employing methods from convex analysis involving subdifferentials of convex, lower semicontinuous functionals on suitable Hilbert spaces. As an application of our results, we can consider the following boundary value problem:which is only a special case of (1) (i.e., , , and ). According to [4, Theorem  3] (see also (5)), this problem has a weak solution ifwhich yields the result of Landesman and Lazer [8] for . This last condition is both necessary and sufficient when the interval is open. This was put into an abstract context and significantly extended by Brézis and Haraux [9]. Their work was much further extended by Brézis and Nirenberg [10]. The goal of the present article is comparable to that of [4] since we want to establish similar conditions to (5) and (7) for the existence of solutions to (1) when , with main emphasis on the generality of the boundary conditions.

Recall that and are given by (3). Let be the interval . Our first main result is as follows (see Section 4 also).

Theorem 1. Let    be odd, monotone nondecreasing, continuous function such that . Assume that the functions satisfyfor some constants , . If is a weak solution of (1) (in the sense of Definition 30 below), thenConversely, ifthen (1) has a weak solution.

Our second main result of the paper deals with a modified version of (1) which is obtained by replacing the functions and in (1) by and , respectively, and also allowing , to depend on . Under additional assumptions on , and under higher integrability properties for the data , the next theorem provides us with conditions for unique solvability results for solutions to such boundary value problems. Then, we obtain some regularity results for these solutions. In addition to these results, the continuous dependence of the solution to (1) with respect to the data can be also established. In particular, we prove the following.

Theorem 2. Let all the assumptions of Theorem 1 be satisfied for the functions , . Moreover, for each , assume that , as , and , as , respectively.(a)Then, for every with there exists a unique weak solution to problem (1) (in the sense of Definition 40 below) which is bounded.(b)Let , , be such that for some constants . Then, the weak (bounded) solution of problem (1) depends continuously on the data . Precisely, let us indicate by the unique solution corresponding to the data , for each . Then, the following estimate holds: for some nonnegative function , , which can be computed explicitly.

We organize the paper as follows. In Section 2, we introduce some notations and recall some well-known results about Sobolev spaces, maximal monotone operators, and Orlicz type spaces which will be needed throughout the article. In Section 3, we show that the subdifferential of a suitable functional associated with problem (1) satisfies a sort of “quasilinear” version of the Fredholm alternative (cf. Theorem 20), which is needed in order to obtain the result in Theorem 1. Finally, in Sections 4 and 5, we provide detailed proofs of Theorems 1 and 2. We also illustrate the application of these results with some examples.

2. Preliminaries and Notations

In this section we put together some well-known results on nonlinear forms, maximal monotone operators, and Sobolev spaces. For more details on maximal monotone operators, we refer to the monographs [1115]. We will also introduce some notations.

2.1. Maximal Monotone Operators

Let be a real Hilbert space with scalar product .

Definition 3. Let be a closed (nonlinear) operator. The operator is said to be(i)monotone if for all one has (ii)maximal monotone if it is monotone and the operator is invertible.

Next, let be a real reflexive Banach space which is densely and continuously embedded into the real Hilbert space , and let be its dual space such that .

Definition 4. Let be a continuous map.(a)The map is called a nonlinear form on if for all one has , that is, if is linear and bounded in the second variable.(b)The nonlinear form is said to be(i)monotone if ;(ii)hemicontinuous if , ;(iii)coercive, if .

Now, let be a proper, convex, lower semicontinuous functional with effective domain The subdifferential of the functional is defined by By a classical result of Minty [13] (see also [12, 14]), is a maximal monotone operator.

2.2. Functional Setup

Let be a bounded domain with a Lipschitz boundary . For , we let be the first-order Sobolev space; that is,Then , endowed with the norm is a Banach space, where we have set Since has a Lipschitz boundary, it is well-known that there exists a constant such thatwhere if and if . Moreover the trace operator initially defined for has an extension to a bounded linear operator from into where if and if . Hence, there is a constant such thatThroughout the remainder of this article, for , we let

If , one has thatthat is, the space is continuously embedded into . For more details, we refer to [16, Theorem  4.7] (see also [17, Chapter  4]).

For , we define the Sobolev space to be the completion of the space with respect to the norm where we recall that denotes the tangential gradient of the function at the boundary . It is also well-known that is continuously embedded into where if and if . Hence, for , there exists a constant such that

Let denote the -dimensional Lebesgue measure and let the measure on be defined for every measurable set by For , we define the Banach space endowed with the norm if , and If , we will simply denote .

Identifying each function with , we have that is a subspace of .

For , we endow with the norm while is endowed with the norm It follows from (21) and (22) that is continuously embedded into , with and given by (23), for . Moreover, by (21) and (26), is continuously embedded into .

2.3. Musielak-Orlicz Type Spaces

For the convenience of the reader, we introduce the Orlicz and Musielak-Orlicz type spaces and prove some properties of these spaces which will be frequently used in the sequel (see Section 5).

Definition 5. Let be a complete measure space. We call a function a Musielak-Orlicz function on if(a) is nontrivial, even, and convex for , a.e. ;(b) is vanishing and continuous at for , a.e. ;(c) is left continuous on ;(d) is -measurable for all ;(e).

The complementary Musielak-Orlicz function is defined by It follows directly from the definition that for (and hence for all )

Definition 6. We say that a Musielak-Orlicz function satisfies the -condition if there exists a set of -measure zero and a constant such that for all and every .
We say that satisfies the -condition if there is a set of -measure zero and a constant such that for all and all .

Definition 7. A function is called an -function if
(i) is even, strictly increasing, and convex;(ii) if and only if ;(iii) and .We say that an -function satisfies the -condition if there exists a constant such that and it satisfies the -condition if there is a constant such that For more details on -functions, we refer to the monograph of Adams [18, Chapter  VIII] (see also [19, Chapter  I], [20, Chapter  I]).

Remark 8. For an -function , we let be its left-sided derivative. Then is left continuous on and nondecreasing. Let be given by Then As before for all Moreover, if or then we have equality; that is,The function is called the complementary -function of . It is also known that an -function satisfies the -condition if and only iffor some constant and for all , where is the left-sided derivative of .

Lemma 9. Let be an -function which satisfies the -condition with the constant and let be its complementary -function. Then satisfies the -condition with the constant .

Proof. We have Since for all and and are decreasing, we get, for , that Now let . Then for Hence, .

Corollary 10. Let be a Musielak-Orlicz function such that is an -function for , a.e. on . If satisfies the -condition, then satisfies the -condition.

Definition 11. Let be a Musielak-Orlicz function. Then the Musielak-Orlicz space associated with is defined by where On this space we consider the Luxemburg norm defined by

Proposition 12. Let be a Musielak-Orlicz function which satisfies the -condition. Then

Proof. If satisfies the -condition, then there exists a set of measure zero such that for every there exists for all and all . Let be fixed. For there exists satisfying the above inequality. We will show that whenever . Assume that and let be such that . Then for all . If we assume that the last inequality does not hold, then and this clearly contradicts the definition of . Therefore, we must haveFrom (53), (56), we obtain The proof is finished.

Corollary 13. Let be a Musielak-Orlicz function such that is an -function for , a.e. on . If its complementary -function satisfies the -condition, then satisfies the -condition and

2.4. Some Tools

For the reader’s convenience, we report here below some useful inequalities which will be needed in the course of investigation.

Lemma 14. Let and . Then, there exists a constant such thatIf , then there exists a constant such that

Proof. The proof of (60) is included in [21, Lemma  I.4.4]. In order to show (59), one only needs to show that the left hand side is nonnegative, which follows easily.

The following result which is of analytic nature and whose proof can be found in [22, Lemma  3.11] will be useful in deriving some a priori estimates of weak solutions of elliptic equations.

Lemma 15. Let be a nonnegative, nonincreasing function such that there are positive constants and () such that Then with .

3. The Fredholm Alternative

In what follows, we assume that is a bounded domain with Lipschitz boundary . Let satisfy for some constant . Let be the real Hilbert space . Then, it is clear that is isomorphic to with equivalent norms.

Next, let and be fixed. We define the functional by settingwhere the effective domain is given by .

Throughout the remainder of this section, we let . The following result can be obtained easily.

Proposition 16. The functional defined by (62) is proper, convex, and lower semicontinuous on .

The following result contains a computation of the subdifferential for the functional .

Remark 17. Let and let . Then, by definition, and, for all , we have Let , , and set above. Dividing by and taking the limit as , we obtain thatwhere we recall that Choosing with (the space of test functions) and integrating by parts in (64), we obtain Therefore, the single-valued operator is given by

Since the functional is proper, convex, and lower semicontinuous, it follows that its subdifferential is a maximal monotone operator.

In the following two lemmas, we establish a relation between the null space of the operator and its range.

Lemma 18. Let denote the null space of the operator . Then that is, consists of all the real constant functions on .

Proof. We say that if and only if (by definition) is a weak solution ofA function is said to be a weak solution of (70), if, for every , there holdsLet with . Then it is clear that .
Conversely, let . Then, it follows from (71) that Since is bounded and connected, this implies that is equal to a constant. Therefore, and this completes the proof.

Lemma 19. The range of the operator is given by

Proof. Let . Then there exists such that . More precisely, for every , we haveTaking , we obtain that . HenceLet us now prove the converse. To this end, let be such that . We have to show that ; that is, there exists such that (73) holds, for every . To this end, consider It is clear that is a closed linear subspace of and therefore is a reflexive Banach space. Using [23, Section  1.1], we have that the norm defines an equivalent norm on . Hence, there exists a constant such that for every Define the functional by It is easy to see that is convex and lower semicontinuous on (see Proposition 16). We show now that is coercive. By exploiting a classical Hölder inequality and using (78), we haveObviously, this estimate yieldsTherefore, from (81), we immediately get This inequality implies that and this shows that the functional is coercive. Since is also convex and lower semicontinuous, it follows from [24, Theorem  3.3.4] that there exists a function which minimizes . More precisely, for all , ; this implies that for every and every HenceUsing the Lebesgue Dominated Convergence, easy computation shows thatChanging to in (86) gives thatfor every . Now, let . Writing with and using the fact that , we obtain, for every , that Therefore, . Hence, and this completes the proof of the lemma.

The following result is a direct consequence of Lemmas 18 and 19. This is the main result of this section.

Theorem 20. The operator satisfies the following type of “quasilinear” Fredholm alternative:

4. Necessary and Sufficient Conditions for Existence of Solutions

In this section, we prove the first main result (cf. Theorem 1) for problem (1). Before we do so, we will need the following results from maximal monotone operators theory and convex analysis.

Definition 21. Let be a real Hilbert space. Two subsets and of are said to be almost equal, written as , if and have the same closure and the same interior, that is, and

The following abstract result is taken from [9, Theorem  3 and Generalization in p. 173-174].

Theorem 22 (Brézis-Haraux). Let and be subdifferentials of proper convex lower semicontinuous functionals and , respectively, on a real Hilbert space with , and let be the subdifferential of the proper, convex lower semicontinuous functional ; that is, . Then In particular, if the operator is maximal monotone, then and this is the case if .

4.1. Assumptions and Intermediate Results

Let us recall that the aim of this section is to establish some necessary and sufficient conditions for the solvability of the following nonlinear elliptic problem:where are fixed. We also assume that () satisfy the following assumptions.

Assumption 23. The functions are odd, monotone nondecreasing, and continuous and satisfy .

Let be the inverse of . We define the functions () byThen it is clear that , are even, convex, and monotone increasing on , with , for each . Moreover, since are odd, we have , for all and , with a similar relation holding for as well. The following result whose proof is included in [19, Chap.  I, Section  1.3, Theorem  3] holds.

Lemma 24. The functions and satisfy (43) and (44). More precisely, for all If or , then we also have equality; that is,

We note that, in [19], the statement of Lemma 24 assumed that , are -functions in the sense of Definition 7. However, the conclusion of that result holds under the weaker hypotheses of Lemma 24.

Define the functional by with the effective domain

Lemma 25. Let satisfy Assumption 23. Then the functional is proper, convex, and lower semicontinuous on .

Proof. It is routine to check that is convex and proper. This follows easily from the convexity of and the fact that . To show the lower semicontinuity on , let be such that in and for some constant . Since in , then there is a subsequence, which we also denote by , such that a.e. on and , a.e. on . Since are continuous (thus, lower semicontinuous), we have By Fatou’s Lemma, we obtain Hence, is lower semicontinuous on .

We have the following result whose proof is contained in [19, Chap.  III, Section  3.1, Theorem  2].

Lemma 26. Let satisfy Assumption 23 and assume that there exist constants such thatThen is a vector space.

Let the operator be defined by

We have the following result.

Lemma 27. Let the assumptions of Lemma 26 be satisfied. Then the subdifferential of and the operator coincide; that is, for all

Proof. Let and . Then, by definition, and, for every , we get Let , with and . Then by Lemma 26, . Now, dividing by and taking the limit as , we obtainChanging to in (105) gives that In particular, if with , we have and this shows that . Similarly, one obtains that . We have shown that and Conversely, let and set . Since , from (94) and (101), it follows that Hence, . Let . Using Lemma 24, we obtainand similarlyThereforeBy definition, this shows that . We have shown that and . This completes the proof of the lemma.

Next, we define the functional by Note that for while for We have the following result.

Lemma 28. Let the assumptions of Lemma 26 be satisfied. Then the subdifferential of the functional is given by In particular, if, for every , the function , then

Proof. We calculate the subdifferential . Let ; that is, , , and for every , we have Proceeding as in Remark 17 and the proof of Lemma 27, we obtain that Noting that is also a single-valued operator (which follows from the assumptions on and ), we easily obtain (116), and this completes the proof of the first part.
To show the last part, note that it is clear that always holds. To show the converse inclusion, let us assume that, for every , the function . Then it follows from (68), (102) (since ), and (116) that and This completes the proof.

The following lemma is the main ingredient in the proof of Theorem 31 below.

Lemma 29. Let and set . ThenIn particular, if, for every , the function , then

Proof. By Remark 17 and Lemmas 27 and 28, the operators , , and are subdifferentials of proper, convex, and lower semicontinuous functionals , , and , respectively, on . Hence, , , and are maximal monotone operators. In particular, if , for every , then, by Lemma 28, one has . Now, the lemma follows from the celebrated Brézis-Haraux result in Theorem 22.

4.2. Statement and Proof of the Main Result

Next, let be given by (114) if and by (115) if .

Definition 30. Let . A function is said to be a weak solution of (93), if , , , if , andfor every with , if .

Recall that and . We also define the average of with respect to the measure , as follows: where . Now, we are ready to state the main result of this section.

Theorem 31. Let satisfy Assumption 23 and assume that the functions satisfy (101). Let . The following hold:(a)Suppose that the nonlinear elliptic problem (93) possesses a weak solution. Then(b)Assume thatThen the nonlinear elliptic problem (93) has at least one weak solution.

Proof. We show that condition (125) is necessary. Let and let be a weak solution of . Then, by definition, for every , (123) holds. Taking in (123) yields Henceand so (125) holds. This completes the proof of part (a).
We show that condition (126) is sufficient.
(i) First, let , where By definition, one has that since for some constant function on and for some constant function on . Let be such that (126) holds. We must show . By (126), we may choose such that Then, for , we have with First, , since Obviously, . Hence, it is readily seen that (ii) Next, denote by the open ball in of center and radius . Since there exists such that the open ball Since the mapping from into is continuous, then there exists such that for all satisfying . It finally follows from part (i) above that contains an -ball in centered at . ThereforeConsequently, problem (93) is (weakly) solvable for every function , if (126) holds. This completes the proof of the theorem.

Remark 32. It is important to remark that, in order to prove Theorem 31, we do not require that should belong to , for every In particular, only the assumption (121) was needed. However, if this happens, then we get the much stronger result in (122) which would require that the nonlinearities satisfy growth assumptions at infinity.

We conclude this section with the following corollary and some examples.

Corollary 33. Let the assumptions of Theorem 31 be satisfied. Let . Assume that at least one of the sets , is open. Then the nonlinear elliptic problem (93) possesses a weak solution if and only if (126) holds.

Remark 34. Similar results to Theorem 31 and Corollary 33 were also obtained in [4, Theorem  4.4], but only when

4.3. Examples

We will now give some examples as applications of Theorem 31. Let be fixed.

Example 35. Let or be equal to , where , . Note that . It is easy to check that satisfies all the conditions of Assumption 23 and that the function satisfies (101). Then, it follows that problem (93) is solvable for any , .

Example 36. Consider the case when in (93); that is, consider the following boundary value problem: Then, by Theorem 31, this problem has a weak solution if which yields the classical Landesman-Lazer result (see (7)) for and .

Example 37. Let us now consider the case when and , where is a continuous, odd, and nondecreasing function on such that . The problem has a weak solution ifLet us now choose in (140). Then, it is easy to check that is monotone increasing on and that it satisfies , , for some constant . Therefore, (141) becomes the necessary and sufficient condition

5. A Priori Estimates

Let be a bounded Lipschitz domain with boundary . Recall that , , and with , for some constant . We consider the nonlinear elliptic boundary value problem formally given bywhere and for some . If , then the boundary conditions in (144) are of Robin type. Existence and regularity of weak solutions for this case have been obtained in [25] for (see also [26] for the linear case) and for general in [27]. Therefore, we will concentrate our attention to the case only; in this case, the boundary condition in (144) is a generalized Wentzell-Robin boundary condition. For the sake of simplicity, from now on we will also take .

5.1. General Assumptions

Throughout this section, we assume that the functions and satisfy the following conditions.

Assumption 38. for , a.e. if , and , a.e. if .

Since are strictly increasing for , a.e. if , and , a.e. if , then they have inverses which we denote by (cf. also Section 4). We define the functions , , and , by Then, it is clear that, for , a.e. if , and , a.e. if , and are differentiable, monotone, and convex with Furthermore, is an -function and is its complementary -function. The function is then the complementary Musielak-Orlick function of in the sense of Young (see Definition 5).

Assumption 39. We assume, for , a.e. if , and , a.e. if , that and satisfy the ()-condition in the sense of Definition 7.

It follows from Assumption 39 that there exist two constants such that for , a.e. if , and , a.e. if , and for all

Next, let Since and satisfy the -condition, it follows from [18, Theorem  8.19] that and , endowed, respectively, with the norms are reflexive Banach spaces. Moreover, by [18, Section  8.11, p. 234], the following generalized versions of Hölder’s inequality will also become useful in the sequel

5.2. Existence and Uniqueness of Weak Solutions of Perturbed Equations

Let Then, for every , endowed with the norm is a reflexive Banach space. Recall that . Throughout the following, we denote by the dual of .

Definition 40. A function is said to be a weak solution of (144), if for every provided that the integrals on the right-hand side exist. Here

Lemma 41. Assume Assumptions 38 and 39. Let and be fixed. Then the functional belongs to . Moreover, is strictly monotone, hemicontinuous, and coercive.

Proof. Let be fixed. It is clear that is linear. Let . Then, exploiting (150), we obtainwhere This shows , for every .
Next, let . Then, using (59) and the fact that are monotone nondecreasing, that is, , for all , we obtainThis shows that is monotone. Estimate (157) also shows that for all with , that is, or . Thus, is strictly monotone.
The continuity of the norm function and the continuity of , imply that is hemicontinuous.
Finally, since and satisfy the -condition, from Proposition 12 and Corollary 13, it follows that Consequently, we deducewhich shows that is coercive. The proof of the lemma is finished.

The following result is concerned with the existence and uniqueness of weak solutions to problem (144).

Theorem 42. Assume Assumptions 38 and 39. Let , , and , where and . Then, for every , there exists a unique function which is a weak solution to (144).

Proof. Let denote the duality between and . Then, from Lemma 41, it follows that, for each , there exists such that for every . Hence, this relation defines an operator , which is bounded by (155). Exploiting Lemma 41 once again, it is easy to see that is monotone and coercive. It follows from Browder’s theorem (see, e.g., [28, Theorem  5.3.22]) that . Therefore, for every there exists such that ; that is, for every Since and with dense injection, by duality, we have . Since is bounded and , we obtain that This shows the existence of weak solutions. The uniqueness follows from the fact that is strictly monotone (cf. Lemma 41). This completes the proof of the theorem.

Corollary 43. Let the assumptions of Theorem 42 be satisfied. Let(a)Let , , , and . Then, for every , there exists a function which is the unique weak solution to (144).(b)Let , , , and . Then, for every , there exists a function which is the unique weak solution to (144).

Proof. We first prove (a). Let and and let and , where and are given by (164). Let and . Since and with dense injection, then, by duality, , where and . Since , we have that Hence, for every , there exists such that for every The uniqueness of the weak solution follows again from the fact that is strictly monotone.
In order to prove the second part, we use the embeddings , and proceed exactly as above. We omit the details.

5.3. Properties of the Solution Operator of the Perturbed Equation

In the sequel, we establish some interesting properties of the solution operator to problem (144). We begin by assuming the following.

Assumption 44. Suppose that , , satisfy the following conditions:

Theorem 45. Assume Assumptions 38, 39, and 44. Let and let be the continuous and bounded operator constructed in the proof of Theorem 42. Then is injective and, hence, invertible and its inverse is also continuous and bounded.

Proof. First, we remark that since for , a.e. if , and , a.e. if , it follows from (167) that, for all ,Let and . Then, exploiting (60), (169), and the ()-condition, we obtainThis implies that , for all with (that is, , or ). Therefore, the operator is injective and, hence, exists. Since for every from the coercivity of (see (160)), it is not difficult to see thatThus, is bounded.
Next, we show that is continuous. Assume that is not continuous. Then there is a sequence with in and a constant such thatfor all . Let and . Since is a bounded sequence and is bounded, we have that is bounded in . Thus, we can select a subsequence, which we still denote by , which converges weakly to some function . Since strongly in and converges weakly to zero in , we deduceFrom (170) and (174), it follows that while Therefore, strongly in . Since is continuous and it follows from the injectivity of that . This shows that which contradicts (173). Hence, is continuous. The proof is finished.

Corollary 46. Let the assumptions of Theorem 45 be satisfied. Let , and be as in (164) and let be the continuous and bounded operator constructed in the proof of Theorem 42.(a)If , , , and , then is continuous and bounded. Moreover, is compact for every and .(b)If , , , and , then the operator is continuous and bounded. Moreover, is compact for every and .

Proof. We only prove the first part. The second part of the proof follows by analogy and is left to the reader. Let , , , and and let . Proceeding exactly as in the proof of Theorem 45, we obtain Hence, the operator is bounded. Finally, using the facts that , is continuous and , we easily deduce that is continuous.
Now, let and . Since the injection is compact, then, by duality, the injection is compact for every and . This, together with the fact that is continuous and bounded, implies that is compact for every and .
It remains to show that is also compact as a map into for every and . Since is bounded, we have to show that the image of every bounded set is relatively compact in for every and . Let be a sequence in . Let . Since is bounded, then the sequence is bounded. Since is compact as a map into , it follows that there is a subsequence such that . We may assume that in and, hence, in . It remains to show that in . Let and . Since is bounded in , a standard interpolation inequality shows that there exists such that As converges in , it follows from the preceding inequality that is a Cauchy sequence in and therefore converges in . Hence, is compact for every and . The case follows from the fact that and the proof is finished

5.4. Statement and Proof of the Main Result

We will now establish under what conditions the operator maps boundedly and continuously into . The following is the main result of this section.

Theorem 47. Let the assumptions of Theorem 45 be satisfied.(a)Suppose and . Let Let and be such that for every function Then there is a constant such that (b)Suppose . Let Let and satisfy (182). Then there is a constant such that

Proof. Let satisfy (182). Let be a real number and set Let and and . Clearly and . We claim that there exists a constant such thatfor all . Using the definition of the form , we haveSincewe can rewrite (188) as follows:Exploiting inequality (60), from (190) and (169), we deduce thatwhere are the constants from (169). Using (167) and the fact that are strictly increasing, for , we have Multiplying this inequality by , , yieldsSimilarly, for Hence, multiplying this inequality by , we getfor all . Hence, on account of (193) and (195), from (191) we obtain the required estimate of (187).
(a) To prove this part, note that from Definition 40 it is clear thatLet and with and let be any -measurable set. We claim that there exists a constant such that, for every and , we havewhere and are such that and . In fact, note that if and , are such that and if , then by Hölder’s inequalitySince , (198) follows immediately from (200) and claim (198) is proved. Next, it follows from (198) that where we recall that and . Therefore, for every which together with estimate (187) yields the desired inequalityIt follows from (196) and (203) that for every Hence, for every , . Using the fact , we obtain, for every , that Let . Then and on the inequality holds. Thereforewhich shows thatLet and ThenChoosing , from (209) we haveTherefore, combining (207) with (210), we get Setting in Lemma 15, on account of (211), we can find a constant (independent of ) such that This shows that , where . Hence, we have , , a.e. on , so that which completes the proof of part (a).
(b) To prove this part, instead of (196) and (198), one uses and , where and are such that and , and the embedding . The remainder of the proof follows as in the proof of part (a).

We conclude this section with the following example.

Example 48. Let , be a strictly positive and -measurable function and let Then, it is easy to verify that satisfies Assumptions 38, 39, and 44 (see, e.g., [25, Example  4.17]).

Conflicts of Interest

The authors declare that they have no conflicts of interest.