Sharp Adams type inequalities in Lorentz-Sobole space

: This article addresses several sharp weighted Adams type inequalities in Lorentz-Sobolev spaces by using symmetry, rearrangement and the Riesz representation formula. In particular, the sharpness of these inequalities were also obtained by constructing a proper test sequence.


Introduction
Sharp Moser-Trudinger inequality and its high-order form (which is called Adams inequality) have received a lot of attention due to their wide applications to problems in geometric analysis, partial differential equations, spectral theory and stability of matter [2,3,5,[8][9][10][11][12][24][25][26][27].This paper is concerned with the problem of finding optimal Adams type inequalities in Lorentz-Sobolev space.
The Trudinger inequality, which can be seen as the critical case of the Sobolev imbedding, was first obtained by Trudinger [30].More precisely, Trudinger employed the power series expansion to prove that there exists β > 0, such that sup ∥∇u∥ n n ≤1,u∈W 1,n 0 (Ω) Ω exp(β|u| where Ω ⊂ R n is a bounded smooth domain and W 1,p 0 (Ω) denotes the usual Sobolev space on Ω, i.e., the completion of C ∞ 0 (Ω)(the space of all functions being infinity-times continuously differential in Ω with compact support) with the norm Let Ω ⊂ R n be an open domain with finite measure.It is well known that for a positive integer k < n and 1 ≤ p < n k , the Sobolev space W k,p 0 (Ω) embeds continuously into L np n−kp (Ω), but in the borderline case p = n k , W k, n k 0 (Ω) ⊊ L ∞ (Ω), unless k = n.For the case k = 1, Yudovich [31] and Trudinger [30] have shown that n−1 of (1.1) by using the technique of the symmetry and rearrangement in [20].Theorem A. [20] Let Ω ⊂ R n be an open domain with finite measure.Then, there exists a sharp constant The constant β n is sharp in the sense that the above inequality can no longer hold with some C 0 independent of f if β > β n .
Theorem A has been extended in many directions, one of which states that sup n−1 , plays an important role in analysis, where ω n−1 is the surface measure of the unit ball in R n .In fact, the constant β n is sharp in the sense that if β > β n , the supremum is infinity.
Since the Polyá-Szegö inequality, on which the technique of the symmetry and rearrangement depends, is not valid on the high-order Sobolev space, many challenges arise in the research of high-order Trudinger-Moser inequalities.In 1988, Adams [1] utilized the method of representative formulas and potential theory to establish the sharp Adams inequalities on bounded domains.Theorem B. [1] Let Ω be an open and bounded set in R n .If m is a positive integer less than n, then there exists a constant where Furthermore, the constant β(n, m) is best possible in the sense that for any β > β(n, m), the integral can be made as large as possible.In the case of Sobolev space with homogeneous Navier boundary conditions W m, n m N (Ω), the Adams inequality was extended by Cassani and Tarsi in [6].It is easy to check that W m, n m N (Ω) contains W m, n m 0 (Ω) as a closed subspace.
Adimurthi and Sandeep proved a singular Moser-Trudinger inequality with the sharp constant in [2].Since then, Moser's results for the first order derivatives and Adams' result for the high order derivatives were extended to the unbounded domain case.Earlier research of the Moser-Trudinger inequalities on the whole space goes back to Cao's work in [7].Later, Li and Ruf [19,23] improved Cao's result and established the following result where proof relies on the rearrangement argument and the Polyá-Szegö inequality.For more on the rearrangement argument, see [21,29].In 2013, Lam and Lu [17] used a symmetrization-free approach to give a simple proof for the sharp Moser-Trudinger inequalities in W  16]).For more applications of the symmetrization-free method, see also [18,32].The Adams type inequality on W m, n m 0 (Ω) when Ω has infinite volume and m is an even integer was studied recently by Ruf and Sani in [22].
In [22], Ruf and Sani used the norm , which is equivalent to the standard Sobolev norm In particular, if u ∈ W m, n m 0 (Ω) or u ∈ W m, n m (R n ), then ∥u∥ W m, n m ≤ ∥u∥ m,n .Since Ruf and Sani only considered the case when m is even, it leaves an open question if Ruf and Sani , s result is still right when m is odd.Recently, the authors of [17] solved the problem and proved the results of Adams type inequalities on unbounded domains when m is odd.
We notice that when Ω has infinite volume, the usual Moser-Truding inequality become meaningless.In the case |Ω| = +∞, a modified Moser-Truding type inequality was established in [13].
Furthermore, for all β ≤ (1 where ϕ(t) = e t − n−2 j=0 t j j! and L n (R n ; |x| −s dx) denotes the weighted Lebesgue space endowed with the norm When α = 0, Ruf in [23] and Li-Ruf in [19] proved the above modified Moser-Truding type inequality in R 2 .Such type of inequality on unbounded domains in the subcritical case (β < β n , α = 0) was first established by Cao in [7] for n = 2 and Adachi Tanaka in [4] for n ≥ 3 in high dimension.
In this paper, we will consider some sharp Adams type inequalities in Lorentz-Sobolev space W α n m ,q (Ω ⊆ R n ) with q n (If q = n, the Lorentz norm becomes the L n (R n ) domain norm).Let 1 < p < +∞ and 1 ≤ q < +∞.Then we recall the Lorentz space L p,q (R n ) as: It is well known that ∥ • ∥ * p,q is not a norm, and is a norm for any p and q.However, they are equivalent in the sense that ∥ψ∥ p,q ≤ ∥ψ∥ * p,q ≤ C(p, q)∥ψ∥ p,q .
Theorem 1.Let m ≤ n be an integer, 0 ≤ α < n, 1 < q < +∞ and A be a positive real number.Then for any bounded domain Additionally, the constant .
(2) sup Additionally, the constant β n,m,q is sharp in the sense that the supremum is infinity if β > β n,m,q .
For the unbounded domain, we take R n for example to have the following inequalities.
Theorem 2. Let m, q, α be the same as in Theorem 1. Then we have where ] and β n,m,q is sharp in the sense that the supremum is infinity if β > β n,m,q .

Proofs of the main results
We begin this section with some preparations which are necessary for the proofs of our main results.
for every t > 0. Its distribution function d f (t) and its decreasing rearrangement f * are defined by respectively.Now, define f ♯ : R n → R by where v n is the volume of the unit ball in R n .Then for every continuous increasing function Ψ : [0, +∞) → [0, +∞), it follows from [14] that Since f * is nonincreasing, the maximal function of f * , which is defined by is also nonincreasing and f * ≤ f * * .For more properties of the rearrangement, we refer the reader to [14,28].
Lemma 2.1.Let 0 < α ≤ 1, 1 < p < ∞ and a(s, t) be a non-negative measurable function on Then there is a constant c Proof.The integral in (2.1) can be written as where F α (t) ≤ λ and E αλ = Ω e αλ|u| n n−1 dx.We first show that there is a constant C = C(p, b, α) > 0 such that F α (t) ≥ −C for all t ≥ 0. To do so, we claim that if E αλ ∅, then λ ≥ −C, and furthermore that if t ∈ E αλ , then there are A 1 > 0 and B 1 > 0 such that In fact, if E αλ ∅, and t ∈ E αλ , then Hence the desired result can be obtained by repeating the argument as in the proof of [1, Lemma 1].
The second is to prove that |E αλ | ≤ A|λ| + B for constants A and B depending only on p, b and α, which is straightforward via modifying the argument of [1, Lemma 1].Thus, we complete the proof of Lemma 2.1.

Lemma 2.2. [15]
There exists a constant K n,m depending only on m and n such that for all u ∈ W α n m ,q (R n ) and 1 < q ≤ +∞.Having disposed of the above lemmas, we can now turn to the proofs of Theorems 1 and 2.

Proof of Theorem 1
It follows from O'Neil's lemma [21] that for all t ≥ 0, Since G m is radial and decreasing, ).Therefore, by taking and using the Hardy-Littlewood inequality, we find and It's easy to check that when 0 < s < t, a(s, t) ≤ 1.This, along with Lemma 2.1 gives +∞ 0 exp[−F(s)]ds ≤ C 0 .Therefore, we have obtained Next, we show the sharpness of β n,m,q according to Adams method in [1].The equivalent form of Theorem 1( 1) is 1 )dx ≤ C m,n,q .
We need to prove that n n is the best one for Ω = B (the unit ball centered at the origin).
Here the infimum is taken over all f > 0 vanishing on the complement of B, and G m * f (x) ≥ 1 on E. It follows from the proof of [1, Theorem 2] that for any ε > 0, one can find 0 < r < 1 small enough such that Then the domain of h * (t) is (r n ω n−1 n , ∞), where Consequently, This gives Finally, a simple computation yields which complete the proof of (1).The statement (2) can be proved similarly as that of (1), we only pay attention to the difference arguments as follows.The Hardy-Littlewood inequality shows that where What is left is to show the sharpness of (1 − α n )β n,m,q , which also inspired by [1].Since the equivalent form of (2) is we only need to prove that n is the best one for Ω = B. Similarly analysis as that of (1), we choose f ≥ 0 such that G m * f ≥ 1 for x ∈ B r with 0 < r < 1, it follows from (1) that , and E is a compact subset of B, where the infimum is taken over all f ≥ 0 vanishing on the complement of B, and G m * f (x) ≥ 1 on E. Analysis similar as that of (1), for any ε > 0, we can choose 0 < r < 1 small enough such that Consequently, we get This shows , which gives as desired.

Proof of Theorem 2
For any u ∈ W m n m ,q (R n ) with ∥(I − ∆) m 2 u∥ n m ,q ≤ 1, set A(u) = ∥u∥ w n m ,q and Ω = {x ∈ R n : |u| > A(u)}.Then it is clear that A(u) ≤ 1.By the property of the rearrangement, we know that for any t ∈ [0, |Ω|), u * (t) > ∥u∥ w n m ,q .
The second inequality of Theorem 2 can be proved similarly via Theorem 1 and the above arguments, we omit its proof here.