Embeddings into Orlicz Spaces for Functions from Unbounded Irregular Domains

We study Sobolev functions defined in unbounded irregular domains in the Euclidean n-space. We show that there exist embeddings into suitable Orlicz spaces from the space Lp1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^1_p$$\end{document}, 1≤p<n\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$1\le p <n$$\end{document}. It turns out that the corresponding Orlicz function depends on the geometry of the domain. The results are sharp for L11\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^1_1$$\end{document}-functions.


Introduction
In this paper we study inequalities the constants for the corresponding inequalities so that their constants do not blow up as diam(D i ) → ∞.
Although embeddings for functions defined in bounded irregular domains have been studied systematically, see for example [13,16], unbounded irregular domains seem to have been studied less, we refer to [10,13].
A classical example of an embedding into an Orlicz space for Sobolev functions from the Sobolev space W 1,n is in [18]. But also, if the domain is irregular then an Orlicz space can be a natural target space for functions defined in L 1 p as in [6,8]. For papers where an Orlicz space is a target space when the functions come from another Orlicz space we refer to [3,4].
To be more precise, we assume that bounded domains D i are ϕ-John domains, that is, every point can be connected to a central point of the domain by a flexible cone of the type {(x, x ) ∈ R×R n−1 : |x | < ϕ(x)}. Here the function ϕ satisfies weak Orlicztype conditions, we refer to Sect. 2. We showed in [7,Theorem 4.4,Theorem 3.5] that every u ∈ L 1 p (D i ), can be estimated pointwise almost everywhere by the modified Riesz potential of its gradient |u(x) − u D i | ≤ CˆD i |∇u(y)| ϕ(|x − y|) n−1 dy, (1.2) and the modified Riesz potential can be estimated pointwise by the maximal operator where H is an N -function. This is a generalization of Hedberg's method [9, Lemma, Theorem 1]. In the present paper we modify the definition of ϕ-John domain so that for t ≥ 1 the function ϕ grows linearly, we refer to (1.4). This definition keeps the class of uniformly bounded ϕ-John domains invariant but makes it possible to control the constants in (1.2) and (1.3) when diam(D i ) → ∞. A proper control of the constants is essential, since bounded domains should engulf the given unbounded domain and the required result for the unbounded domain is obtained as a limit of the results to the engulfing bounded domains. Then, we show that N -function H can be calculated from the geometry of the domain. The following theorem tells which kind of N -functions we are interested in. These N -functions can encode and reveal the geometry of the domain. Theorem 1.1 Let 1 ≤ p < n. Let the continuous, strictly increasing function ϕ : [0, ∞) → [0, ∞) be such that ϕ(0) = lim t→0 + ϕ(t) = 0 and suppose that ϕ satisfies the 2 -condition and the inequality ϕ(t 1 ) Assume that there exists α ∈ [1, n/(n − 1)) such that t α /ϕ(t) is increasing for t > 0. If where the implicit constant depends only on n and p.
By Theorem 1.1 we prove as an intermediate step the Sobolev-type inequality (1.1) for functions defined in bounded ϕ-John domains D i , in Theorem 4.1 (1 < p < n) and Theorem 4.2 ( p = 1). These results seem to be new and they recover some known results when p = 1. By using these bounded domains' results we obtain our main result for unbounded domains. Theorem 1.2 Assume that the function ϕ satisfies the conditions (1)- (5), with C ϕ = 1 in the condition (4), from the beginning of Sect. 2. Assume that there exists α ∈ [1, n/(n − 1)) such that t α /ϕ(t) is increasing for t > 0. Let the function ψ be defined as in (1.4). Let D in R n , n ≥ 2, be an unbounded domain that satisfies the following conditions: Let 1 ≤ p < n. Let H be an N -function from Theorem 1.1. Then there exits a constant C such that the inequality holds for every u ∈ L 1 p (D). Here the constant C depends only on n, p, C 2 H , C 2 ϕ , c J , and diam(D 1 ).
We give examples in Example 4.5. Finally in Sect. 5 we show that the target space cannot be a Lebesgue space in general.
We write Now, if ϕ satisfies the conditions (1)-(5), then ψ does, too, and the constant in (4) is the same for the functions ϕ and ψ, that is is an open ball centered at x with a radius r > 0 and the function ψ is defined as in (2.1).
The set Cig E(a, b) is called a cigar with core E joining a and b. We point out that if D is a ϕ-cigar John domain with ϕ(t) = t p , p ≥ 1, then it is a ϕ-cigar John domain with ϕ(t) = t q for every q ≥ p. For the case ψ(t) = ϕ(t) = t for all t ≥ 0, in Definition 2.1, we refer to [17, 2.1] and [15, 2.11 and 2.13]. Note that it is crucial that the length of the curve does not depend on the distance between the end points a and b. In bounded uniform domains the length of the cigar depends on |a − b| but they are much more regular than our cigar John domains, see [15].
If D is a bounded domain then the following definition from [7, Definition 4.1] for a ψ-John domain gives an equivalent definition to a bounded ϕ-cigar John domain.

Definition 2.2
A bounded domain D in R n , n ≥ 2 , is a ψ-John domain if there exist a constants 0 < α ≤ β < ∞ and a point x 0 ∈ D such that each point x ∈ D can be joined to x 0 by a rectifiable curve γ : [0, (γ )] → D, parametrized by its arc length, such that γ (0) = x, γ ( (γ )) = x 0 , (γ ) ≤ β, and The point x 0 is called a John center of D and γ is called a John curve of x.
Note that when diam(D) → ∞, then α → ∞ with the same speed as diam(D).
Proof Assume first that D is a ψ-John domain with a John center x 0 . Let a, b ∈ D and let the John curves γ 1 and γ 2 connect them to x 0 , respectively. We may assume that a, b ∈ D\B(x 0 , dist(x 0 , ∂ D)), since inside the ball the points can be connected by two straight lines going via the center of the ball B(x 0 , dist(x 0 , ∂ D)). Let E be a curve from a to b given by γ 1 and γ 2 . Then, and thus D is a ϕ-cigar John domain. Assume then that D is a ϕ-cigar John domain. Let us carefully choose a suitable John center so that the center is not too close to the boundary of D. Let x, y ∈ D such that |x − y| ≥ 1 2 diam(D). Let E be a core of a John cigar that connects x and y. Then the length of E is at least 1 2 From now on this r and the point x 0 are fixed in this proof. If a ∈ B(x 0 , 2r ), then it can be clearly joint to x 0 by a line segment and the claim is clear.
For every a ∈ D\B(x 0 , 2r ) there exists a curve E such that the cigar Cig E(a, x 0 ) ⊂ D (Fig. 1). Let (E) be the length of E, then (E) ≤ 2 or by Definition 2.1 and (2.1) Note that the total length of E is at least 2r and the length of E inside the ball B(x 0 , r ) is at least r and thus for the points in E ∩ ∂ B(x 0 , r ) the distance to the boundary is at least ψ(r /2)/c J . Let us choose that Since r ≤ (E) ≤ β and ψ is increasing, we have M ≥ 1.
Let z 0 ∈ E be the first point from a that satisfies z 0 ∈ ∂ B(x 0 , r ). We denote by γ the function so that E is parametrized by its arc length such that γ (0) = a, γ (t 0 ) = z 0 and γ ( (E)) = x 0 . We replace E[z 0 , x 0 ] by the radius of the ball B(x 0 , r ), if needed. This new arc is written as E . Note that (E ) ≤ (E).

This inequality and the inclusion
Thus, by (2.4) This means that we may choose α = ψ(r /2) . By using (2.3) we obtain the final α. To be sure that α ≤ β we may choose β to be larger if it is necessary. Thus, D is a ψ-John domain with α and β given in (2.2).

Pointwise Estimates
We proceed to prove pointwise estimates for domains which are not classical John domains.
We note that by the condition (4) of ϕ We recall a covering lemma from [7,Lemma 4.3] which is valid for a bounded ϕ-John domain.
Proof The proof is in [7,Lemma 4.3]. We recall only the proof of the inequality ). Let γ be a John curve joining x to x 0 , its arc length written as l. We write B 0 = B x 0 , 1 4 dist(x 0 , ∂ D) and consider the balls B 0 and By the Besicovitch covering theorem, there is a sequence of closed balls B 1 , B 2 , . . . and B 0 that cover the set {γ (t) : t ∈ [0, l]}\{x} and have a uniformly bounded overlap depending on n only. We write By the fact that ϕ is an increasing function and by the definition of ψ-John domain we obtain .
Let us suppose then that i ≥ 1.
, then the fact that ϕ is increasing and the definition of a ψ-John domain give We recall the following definitions. Let G be an open set of R n . We denote the Lebesgue space by L p (G), 1 ≤ p < ∞. By L 1 p (G), 1 ≤ p < ∞, we denote those locally integrable functions whose first weak distributional derivatives belongs to holds. Here C = c n, c J , C ϕ , C 2 ϕ , ϕ(1), min diam(D), 1 .
We recall the definitions of N -functions and Orlicz spaces. Continuity and lim t→0 + H (t) t = 0 yield that H (0) = 0.

Convexity yields that H (t)
t ≤ H (s) s for 0 < t < s and thus H is a strictly increasing function.
By the notation f g we mean that there exists a constant C > 0 such that f (x) ≤ Cg(x) for all x. The notation f ≈ g means that f g f . Two N -functions H and K are equivalent, which is written as The Orlicz space L H (G) equipped with the Luxemburg norm is a Banach space. Let G in R n be an open set. Assume that f ∈ L 1 (G). The centered Hardy-Littlewood maximal operator is defined as where the function f χ G is understood to be zero in the complement of G. We recall the following theorem from [7, Theorem 3.5] which is applied to the function f χ G .

, ∞) be a continuous function and let H : [0, ∞) → [0, ∞) be an N-function satisfying the 2 -condition. Suppose that there exists a finite constant C H such that the inequality
holds for all t > 0. Let G in R n be an open set. If f L p (G) ≤ 1, then there exists a constant C such that the inequality holds for every x ∈ G. Here the constant C depends on n, p, C ϕ , C H , and the 2 -constants of ϕ and H only.
Our goal is to find a formula which would give all suitable functions H . Examples of some of these functions were given in [7, Section 6].
Here we do the preparations to find H . Assume that there exists α ∈ [1, n/(n − 1)) such that t α /ϕ(t) is increasing for t > 0. This yields that t α /ψ(t) is increasing, too. Under this condition inequality (3.3) holds: Since

Let us define the functions h and δ such that
Then,

Fig. 2 The function F is not necessary convex
If we choose and assume that the inverse function of F −1 exists, that is (F −1 ) −1 =: F exists, then and thus

F h(δ(t))t + ψ(δ(t)) 1−n (δ(t))
Unfortunately, there is a problem with this function F to be a suitable function H ; namely, the function F is not necessary convex. For example, if n = 2, ϕ(t) = t 3 2 , and p = 1.9, then the function F is not convex, see Fig. 2. The angle at the point (1, F −1 (1)) comes from the angle of ψ at the point (1, ψ(1)). Our main theorem, Theorem 1.1 in Introduction, corrects this point: we show that there exists an Nfunction H that is equivalent with F.

Proof of Theorem 1.1 Let us write that
for t > 0 and F −1 (0) = 0. Let us first show that F −1 is strictly increasing. We recall that if ϕ satisfies condition (4), then ψ does too, and the constant is the same for both functions. We have Since p < n the function t → t 1 p − 1 n is strictly increasing. Since the function t → t − 1 n is strictly decreasing, condition (4) with C ϕ = 1 yields that t → (t − 1 n )/ψ(t −1/n ) is strictly increasing. These together yield that F −1 is strictly increasing.
This yields that the function F exists and is strictly increasing. Let us show that lim t→0 + F −1 (t) = 0. Since p < n we obtain Let us show that lim t→∞ F −1 (t) = ∞. Since t/ϕ(t) is decreasing, by the condition (4), and by p < n we obtain We have shown that F −1 : [0, ∞) → [0, ∞) is bijective. Let us then study the condition Since F −1 is a strictly increasing bijection, inequality (3.6) is equivalent to Since t α /ϕ(t) is increasing, then ϕ(t)/t α is decreasing and ψ(t)/t α is decreasing, too. We note that 1 − α(n−1) n > 0, since α < n n−1 . We obtain and thus inequality (3.6) holds.
Hästö has shown in [11, increasing, then f is equivalent to a convex function. We obtain that F is equivalent to a convex function H . Here the implicit constant depends only on the constant in the 2 -condition, that is, it depends only on n and p.
Using lim t→0 + F −1 (t) = 0 and the bijectivity, we obtain and thus also lim t→0 + H (t) t = 0. This gives that H is right continuous at the origin. Since F satisfies 2 -condition so does H and thus it is finite everywhere. Thus by convexity the function H is continuous on [0, ∞).
Since ϕ(t)/t α is decreasing and α < n n−1 , we obtain Since the functions F and H are equivalent, this yields that lim t→∞ Thus we have shown that the function H satisfies the conditions (N1)-(N3).

Remark 3.6 Later it is crucial that
for 0 < t ≤ 1. Namely, for every ϕ the function H satisfies H (t) ≈ t np n− p whenever 0 < t ≤ 1.  As a corollary we obtain from Theorems 3.3 and 3.8:   ∂ D)). Assume that ∇u L p (D) ≤ 1. Corollary 3.9 yields that H u(x) − u B ≤ C(M|∇u|(x)) p , where the constant C depends on n, p, C 2 H , C 2 ϕ , c J , and min{1, diam(D)} only. By integrating over D and using the fact that the maximal operator is bounded whenever 1 < p < n, we obtain that where H * is the conjugate function of H and C is the constant in Hölder's inequality.
By the triangle inequality we have By the (1, 1)-Poincaré inequality in a ball B, [5,Section 7.8], there exists a constant C(n) such that We continue to estimate the right hand side of inequality (4.2) where j 0 = log(C(n)|B| 1 n −1 ) . Assume first that diam(D) is so large that j 0 ≤ −2. When t < 1, then ψ(t −1/n ) = ϕ(1)t −1/n by (2.1) and thus Thus for t < 1 we obtain that H (t) ≈ t n n−1 . This yields that holds for each D i and all u ∈ L 1 p (D). The constant C does not blow up when the diameter of D i tends to infinity. In the case 1 < p < n this is clear. In the case p = 1, we refer to the discussion after (4.4) in the proof of Theorem 4.2. The constant depends on D 1 but this does not cause a problem.
When ∇u L p (D) ≤ 1 inequality (4.7) yields that there exists a constant C < ∞ such that the inequalityˆD holds; here the constant C is independent of i.
Let us write u i = u D i . The triangle inequality yields that Since D i satisfies inequality (4.7), we have u ∈ L H (D 1 ) ⊂ L 1 (D 1 ) and thus the second term is finite. Again, by inequality (4.7) we obtain that Thus the real number sequence (u i ) is bounded and hence there exists a convergent subsequence (u i j ) and b ∈ R such that u i j → b.
Since H is continuous, lim j→∞ χ D i j H (|u(x) − u i j |) = χ D H (|u(x) − b|). Fatou's lemma and the modular form of (4.7) yield that This yields that there exists a constant C such that the inequality u − b L H (D) ≤ C holds for every u ∈ L 1 p (D) with ∇u L p (D) ≤ 1. The claim follows by applying this inequality to the function u/ ∇u L p (D) . (a) R n , n ≥ 2. (b) (x , x n ) ∈ R n : x n ≥ 0 and |x | < ψ(x n ) .
The undounded domain G constructed in Sect. 5, illustrated in Fig. 3.

Lebesgue Space Cannot be a Target Space
In this section we give an example which shows that for certain unbounded ϕ-cigar John domains the target space cannot be a Lebesgue space. The idea is that at near the infinity the target space should be L np/(n− p) but local structure of the domain may not allow so good integrability. We assume a priori that the function ϕ has the properties (1)- (5). Later on we give extra conditions to the function ϕ.
We construct a mushrooms-type domain. Let (r m ) be a decreasing sequence of positive real numbers converging to zero. Let Q m , m = 1, 2, . . . , be a closed cube in R n with side length 2r m . Let P m , m = 1, 2, . . . , be a closed rectangle in R n which has side length r m for one side and 2ϕ(r m ) for the remaining n − 1 sides. Let Q be the first quarter of the space i.e. all coordinates of the points in Q are positive. We attach Q m and P m together creating 'mushrooms' which we then attach, as pairwise disjoint sets, to the side {(0, x 2 , . . . , x n ) : x 2 , . . . , x n > 0} of Q so that the distance from the mushroom to the origin is at least 1 and at most 4, see Fig. 3. We assumed that the function ϕ has the properties (1)-(5), but we have to assume here also that ϕ(r m ) ≤ r m . We need copies of the mushrooms. By an isometric mapping we transform these mushrooms onto the side {(x 1 , 0, . . . , x n ) : x 1 , x 3 , . . . , x n > 0} of Q and denote them by Q * m and P * m . So again the distance from the mushroom to the origin is at least 1 and at most 4. We define (5.1) See Fig. 3. We omit a short calculation which shows that G is a ϕ-cigar John domain.
Thus we have proved the following remarks.

Remark 5.1
Let ϕ satisfy (1)- (5), and assume that lim t→0 + t/ϕ(t) = ∞. Let G be the unbounded ϕ-cigar John domain constructed in (5.1). Let 1 ≤ p < n. Then there do not exist numbers q ∈ R and C ∈ R such that the inequality could hold for all u ∈ L 1 p (G).