On compact and bounded embedding in variable exponent Sobolev spaces and its applications

For a weighted variable exponent Sobolev space, the compact and bounded embedding results are proved. For that, new boundedness and compact action properties are established for Hardy’s operator and its conjugate in weighted variable exponent Lebesgue spaces. Furthermore, the obtained results are applied to the existence of positive eigenfunctions for a concrete class of nonlinear ode with nonstandard growth condition.

In light of the mentioned results on problem (1), Theorem 3.6 turns out to be actual, since it states λ 1 = 0 for the eigenvalue problem (2) (since for any λ > 0, there exist a solution of the eigenvalue problem). According to [7], if p − > 1, then there are a sequence of discreet eigenvalues λ n with λ ∞ = lim sup λ n = ∞ and λ 1 = lim inf λ n ≥ 0 of the eigenvalue problem: div |∇u| p(x)−2 ∇u + λ|u| p(x)−2 u = 0 in ⊂ R N , which implies that the list eigenvalue may be equal to zero. In [7], it was proved that this problem may has λ 1 = 0 provided that there exists an open set U ⊂ and a point x 0 ∈ U , such that p(x 0 ) < (or >) p(x) for all x ∈ ∂ . Note, the list eigenvalue of the problem (1) is positive in the case of constant exponent (or according to [8], for one-dimensional case with monotony variable exponent p(x)).

Notation, definitions
For 1 < p < ∞, the p denotes conjugate number, 1 p + 1 p = 1; for p = ∞, the p = 1, and for p = 1, the p = ∞. The notations p + = sup t∈(0,l) p(t) and p − = inf t∈(0,l) p(t) are used to denote essential maximum and minimum values of a measurable function p(·). χ E -denotes the characteristic function of set E. C, C 1 , C 2 , . . . denote different constants, the values of which are not essential and may be varied in each appearance.
Denote H f (x) = x 0 f (t)dt-the Hardy operator and H * f (x) = l x f (t) dt-its conjugate. We say that the function g : (0, l) → (0, ∞) is almost increasing (decreasing) if there exists a constant C > 0, such that for any 0 < t 1 Define the following variable exponent spaces that will be used in this paper. For a function f (x) and the exponent p(x), define the modular The variable exponent Lebesgue space L p(·) (0, l) is a space of measurable functions f : (0, l) → R n with finite norm: (0, l) denotes a Sobolev space of absolutely continuous functions f : (0, l) → R, f (0) = 0 endowed with a norm: (0, l) a Sobolev space of absolutely continuous functions on (0, l) with y(0) = y(l) = 0 and having a finite norm: where d(x) = min{x, l − x}. Since xl − x 2 for 0 < x < l is equivalently to ld(x), sometimes, we may use expression lx − x 2 in place of ld(x). Definition Consider the eigenvalue problem: where b(x) is a positive bounded measurable function on (0, l).
We say that the function y = y(x) is a solution of the preceding problem if y ∈W 1, p(·) β (0, l) and for any v ∈W

Main results
Following main results are obtained in this paper. and Assume that p be monotony increasing near origin and there exists ε > 0, such that the function x a.d. on a little δ-neighborhood of origin.
For any absolutely continues function, y : (0, l) → R with y(0) = 0 Theorem 3.1 immediately gives the inequality: i.e., the following assertion takes place. Then, the identity operator maps boundedly space of functions y ∈W ). Moreover, the norm of mapping is estimated by a constant depending on p(·), q(·), ε, δ, β.   (4) and (5). Let p be monotone increasing near origin and decreasing near l. In addition, assume that there exists ε > 0, such that x

near origin and a.i. near l on a little δ-neighborhood.
Then, for all absolutely continuous functions y : where the constant C > 0 depends on p(·), q(·), β, δ, ε. (4) and (5). Let p(·) be a monotone increasing near origin, and decreasing near l. Assume that there exists ε > 0, such that the function +ε be a.d. near origin, and a.i. near l on a little δ-neighborhood. Then, the identity operator maps The proof of Theorem 3.5 is similarly to the proof of Theorem 3.4.
The following assertion takes place for the eigenvalue problem (2).
and the real number β satisfies (5). Assume that p(x) increases near origin and decreases near l. Furthermore,

Proof of the results
To start the proof of Theorem 3.1, we need on the next assertion. origin. There exists ε > 0, such that the function x Then, it holds In addition, using t, If y > t using increasing p, 1 p also will be increasing. Since 1 Assume Derive estimation for every summand in (9). For this purpose, get an estimation for the proper modular: dx.
Using the assumption on β and almost decreasing of x Notice, applying a.d. of x β− 1 p (x) +ε , and Lemma 4.1 it has been used that x p (t) +ε for 2 −n−1 x < t ≤ 2 −n x and 0 < x < l. Therefore, and applying Hölder's inequality from (10), it follows Applying Lemma 4.1 and estimate (11), it follows from (10) that Therefore It has been proved that Inserting (12) in (9), we get Theorem 3.1 has been proved.

Proof of Theorem 3.3
To proof Theorem 3.3, we apply the approaches, e.g., in [5,6]. Insert the operators: As it was stated in [5], P 3 is a limit of finite rank operators, while P 2 is a finite rank operator. From Theorem 3.1, it follows that This completes the proof of Theorem 3.3.

Proof of Theorem 3.4 Notice, the inequality
where C > 0 depends on l, β, p(·), q(·). The boundedness in L q(·) (0, l) for the first summand in the right hand side follows from Theorem 3.2, while the boundedness of the second summand easily can be derived using the assertion of Theorem 3.1, i.e., we need to show the inequality: To prove this inequality is the same to show that for any positive measurable function g : (0, l) → (0, ∞) . Using the definition of variable exponent norm, we have On the other hand changing the variable y = l − t: . Now, since the functionsp,q satisfies all conditions of Theorem 3.1, we get Note, we have used that the condition β < 1 − 1 p − is the same condition β < 1 − 1 p − . This completes the proof of inequality: Proof of Theorem 3.6 To prove this assertion, we shall use the well-known mountain pass theorem approaches. Set E =W 1, p(·) β (0, l). Define the functional dx.
Using the standard argues (see, e.g., [19]), it is not difficult to see that the functional has Gateaux derivative and I λ ∈ C 1 (E, R). It means I λ ∈ E * , and I λ : E → E * continuous. Furthermore, for ∀v ∈ E Palais-Smale condition. Show that Palais-Smale (PS) condition is satisfied for the problem (2). Let {y n } ∈ E be a sequence satisfying the conditions: To show PS condition, we should prove the sequence {y n } ∈ E is compact, i.e., contains a converging in E subsequence y n k → y ∈ E.
To show it, first, establish the boundedness of {y n } in E. Using 1), it follows On the other hand, using condition 2), I λ (y n ) E * = o(1) as n → ∞. It means In particular, inserting v = y n , we get Inserting this, it follows Using Young's inequality and p − > 1 from here, it follows This completes the boundedness of {y n } in E.
Applying well-known fact, there exists a weak convergent subsequence y n k → y in E. Denote it again y n . It follows from the compact embedding Theorem 3.3 that a strong convergence y n → y in L Now, we are ready to show the strong convergence y n → y in E . For this, insert v = y n − yin (14): From this, since y n → y in L q(·),β− 1 p (·) − 1−ε q(·) (0, l), and using Holder's inequality, it follows where also has been used Theorem 3.4 and the estimate (15), to assert the bounded ness {y n } in From this, we infer l 0 (lx − x 2 ) βp(x) y n p(x)−2 y n − y n p(x)−2 y (y n − y )dx Since y n → y weakly in E, it holds This ensures that In the next, we will apply the following two inequalities: for p(x) ≥ 2 and y n p(x)−2 y n − y n p(x)−2 y y − y ≥ γ 2 ( p) y n − y 2 y n 2− p + y n 2− p for 1 < p(x) ≤ 2 . Then, for the case p(x) ≥ 2, we get As to the case 1 < p(x) ≤ 2, we have Using Young's inequality from here, it follows that Therefore where M does not depend on n ∈ N . This and the above estimation together with Young's inequality yield: (1). Therefore, y n → y in E. Now, we are ready to apply the mountain pass theorem. Notice our argues before based on the contrary assumption y n − y E 0. Under it, the estimate was established: (xl − x 2 ) β y n − y p − L p(·) (0,l) = o(1) + o(1) (xl − x 2 ) β (y n − y ) L p(·) (0,l) .
Therefore, using assumption p − > 1 and Young's inequality, we come to the conclusion: (xl − x 2 ) β y n − y L p(·) (0,l) = o(1), i.e., y n → y in E strongly. This completes the proof of PS-property. Mountain pass theorem. Apply the Mountain pass theorem to show the existence of solution for the problem (2).