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Sharp Singular Trudinger–Moser Inequalities in Lorentz–Sobolev Spaces

  • Guozhen Lu and Hanli Tang EMAIL logo
Published/Copyright: June 18, 2016

Abstract

In this paper, we first establish a singular (0<β<n) Trudinger–Moser inequality on any bounded domain in n with Lorentz–Sobolev norms (Theorem 1.1). Next, we prove the critical singular (0<β<n) Trudinger–Moser inequality on any unbounded domain in n with Lorentz–Sobolev norms (Theorem 1.2). Then, we set up a subcritical singular (0<β<n) Trudinger–Moser inequality on any unbounded domain in n with Lorentz–Sobolev norms (Theorem 1.3). Finally, we establish the subcritical nonsingular (β=0) Trudinger–Moser inequality on any unbounded domain in n with Lorentz–Sobolev norms (Theorem 1.5). The constants in all these inequalities are sharp. In [9], for the proof of Theorem 1.2 in the nonsingular case β=0, the following inequality was used (see [17]):

u ( r ) - u ( r 0 ) 1 n w n 1 / n r r 0 U ( s ) s 1 / n d s s ,

where U(x) is the radial function built from |u| on the level set of u, i.e.,

| u | > t | u | d x = 0 | { | u | > t } | U ( s ) d s .

The construction of such U uses the deep Fleming–Rishel co-area formula and the isoperimetric inequality and is highly nontrivial. Moreover, this argument will not work in the singular case 0<β<n. One of the main novelties of this paper is that we can avoid the use of this deep construction of such a radial function U (see remarks at the end of the introduction). Moreover, our approach adapts the symmetrization-free argument developed in [19, 21, 23], where we derive the global inequalities on unbounded domains from the local inequalities on bounded domains using the level sets of the functions under consideration.

MSC 2010: 42B35; 42B37

1 Introduction

The Trudinger–Moser inequalities can be considered as the limiting case of Sobolev inequalities. They were established by Trudinger [40] (see also [41, 35]). In 1971, Moser [33], sharpening the Trudinger inequality in [35, 40, 41], proved the following theorem.

Theorem A

Let Ω be a domain with finite measure in an n-dimensional Euclidean space n, n2. Then there exists a sharp constant

α n = n ( n π n / 2 Γ ( n 2 + 1 ) ) 1 / ( n - 1 )

such that

(1.1) 1 | Ω | Ω exp ( α | f | n / ( n - 1 ) ) 𝑑 x c 0 <

for any ααn and any fC0(Ω) with Ω|f|ndx1. This constant αn is sharp in the sense that if α>αn, then the above inequality can no longer hold with some c0 independent of f.

Trudinger–Moser’s result has been studied and extended in many directions. For instance, we refer the reader to [10, 32, 25] for the sharp Trudinger–Moser inequality with mean value zero, to [3] for the singular Trudinger–Moser inequality and to [12] for the sharp affine Trudinger–Moser inequalities. Also, for the sharp Trudinger–Moser inequalities on the Heisenberg group and CR spheres, and on compact Riemannian manifolds, etc., see [5, 13, 14, 22] and [16, 26, 27], respectively. We also refer to the articles of Chang and Yang [11] for applications of such inequalities to geometric analysis and of Lam and Lu [20] for descriptions of applications of such inequalities to nonlinear PDEs.

When Ω has infinite volume, the subcritical Trudinger–Moser type inequalities for unbounded domains were proposed by Cao [7] when n=2 and do Ó [6] for the general case n2. These results were sharpened later by Adachi and Tanaka [1] in order to determine the best constant.

Theorem B

Theorem B ([1])

For any α<αn, there exists a positive constant Cn,α such that

n ϕ n ( α | u | n / ( n - 1 ) ) 𝑑 x C n , α u n n

for all uW1,n(n), un1, where

ϕ n ( t ) = e t - j = 0 n - 2 t j j ! .

The constant αn is sharp in the sense that the supremum is infinity when ααn.

We note that in the above theorem, we only impose the restriction uLn(n)1. The method in [1] requires a symmetrization argument which is not available in many other non-Euclidean settings. A symmetrization-free argument was developed to prove the subcritical Trudinger–Moser inequality on the Heisenberg group by Lam, Lu and Tang [23].

Recently, Lam, Lu and Zhang proved in [24] the precise asymptotic estimates for the following supremum.

Theorem C

Let

n 2 , α n = n ( n π n / 2 Γ ( n 2 + 1 ) ) 1 / ( n - 1 ) , 0 β < n , 0 α < α n .

Denote

A T ( α , β ) = sup u n 1 1 u n n - β n ϕ n ( α ( 1 - β n ) | u | n / ( n - 1 ) ) d x | x | β .

Then there exist positive constants c=c(n,β) and C=C(n,β) such that when α is close enough to αn, we have

c ( n , β ) ( 1 - ( α α n ) n - 1 ) ( n - β ) / n A T ( α , β ) C ( n , β ) ( 1 - ( α α n ) n - 1 ) ( n - β ) / n .

Moreover, the constant αn is sharp in the sense that AT(αn,β)=.

The upper bound in the above estimates for the subcritical case was obtained by an argument inspired by the work of Lam and the authors [19, 21, 23], where a local Trudinger–Moser inequality on the level sets of the functions under consideration can lead to a global one on the entire spaces, without a priori knowing the validity of the critical inequality.

The above inequality fails in the critical case α=αn. So it is natural to ask in which case the above inequality is true when α=αn. This is done by Ruf [36], and Li and Ruf [28] by using the restriction of the full norm uLn(n)+uLn(n)1.

Theorem D

Theorem D (see [36, 28])

For all 0ααn, we have

sup u 1 n ϕ n ( α | u | n / ( n - 1 ) ) 𝑑 x < ,

where

u = ( n ( | u | n + | u | n ) 𝑑 x ) 1 / n .

Moreover, this constant αn is sharp in the sense that if α>αn, then the supremum is infinity.

Surprisingly, Lam, Lu and Zhang [24] have shown that the subcritical Moser–Trudinger inequality in [1] and the critical Trudinger–Moser inequality in [36, 28] are actually equivalent. We will provide another proof of the sharp critical Trudinger–Moser inequality using the subcritical one, and vice versa. Furthermore, we have shown the following precise relationship between the suprema in critical and subcritical Trudinger–Moser inequalities.

Theorem E

Let n2, 0β<n, a>0 and b>0. Denote

M T a , b ( β ) = sup u n a + u n b 1 n ϕ n ( α n ( 1 - β n ) | u | n / ( n - 1 ) ) d x | x | β , M T ( β ) = M T n , n ( β ) .

Then MTa,b(β)< if and only if bn. The constant αn is sharp. Moreover, we have the following identity:

M T a , b ( β ) = sup α ( 0 , α n ) ( 1 - ( α α n ) a ( n - 1 ) / n ( α α n ) b ( n - 1 ) / n ) ( n - β ) / b A T ( α , β ) .

In particular, MT(β)< and

M T ( β ) = sup α ( 0 , α n ) ( 1 - ( α α n ) n - 1 ( α α n ) n - 1 ) ( n - β ) / n A T ( α , β ) .

Such critical Trudinger–Moser inequalities on unbounded domains on the Heisenberg group have been established by Lam and Lu [19] where symmetrization is not available.

As far as the existence of extremal functions of Moser’s inequality is concerned, the first result was due to the work of Carleson and Chang [8], in which they proved that the supremum

sup u W 0 1 , n ( Ω ) , Ω | u | n d x 1 1 | Ω | Ω exp ( α n | u | n / ( n - 1 ) ) 𝑑 x

can be achieved when Ωn is an Euclidean ball. This result came as a surprise because it has been known that the Sobolev inequality does not have extremal functions supported on any finite ball. Subsequently, existence of extremal functions has been established on arbitrary domains by Flucher in [15] and Lin in [30], and on Riemannian manifolds by Li in [26, 27].

We now return to the discussions on the sharp Trudinger–Moser inequalities in Lorentz norms.

An improvement of (1.1) by exploiting a Lorentz space has been addressed in [4, 18]. We recall that ψLp,q(n), 1<p<+,1q<+, if the quantity

ψ p , q = ( 0 + [ ψ ( t ) t 1 / p ] q d t t ) 1 / q

is finite, where ψ(t) is the decreasing rearrangement of ψ (see Section 2 for more details). As we know, Lp,p(n)=Lp(n) and if qr, then Lp,r(n)Lp,q(n). Alvino, Ferone and Trombetti [4] established the Trudinger–Moser inequality on bounded domain in Lorentz–Sobolev space (see also [18] for the case 1<qn). They set up the following theorem.

Theorem F

Let Ω be a domain with finite measure in the Euclidean space n, n2, and 1<q. Then there exist a constant c0>0 and a sharp constant αn,q=(nvn1/n)q/(q-1) (where vn is the measure of the unit ball) such that

1 | Ω | Ω exp ( α | f | q / ( q - 1 ) ) 𝑑 x c 0 <

for any ααn,q and any fC0(Ω) with un,q1. This constant αn is sharp in the sense that if α>αn, then the above inequality can no longer hold with some c0 independent of f.

What is the situation when Ω is an unbounded domain? This question has been studied by Cassani and Tarsi in [9]. Indeed the following is shown in [9].

Theorem G

Let 1<q< and n2. Then there exist a constant Cn,q such that for any 0<ααn,q and any uC0(n), the following inequality holds:

sup u n , q q + u n , q q 1 n Φ [ α | u | q / ( q - 1 ) ] 𝑑 x C n , q ,

where

Φ ( t ) = e t - j = 0 k 0 t j j ! , k 0 = [ ( q - 1 ) n q ] .

The constant αn,q is sharp in the sense that if α>αn,q, then the supremum will become infinite.

In this paper, we will consider some sharp singular Trudinger–Moser type inequalities in a Lorentz–Sobolev space with qn. When q=n, the Lorentz norm un,q becomes the Ln norm and in this case the inequalities have been already established [1, 36, 28]. First, we will prove the following singular Trudinger–Moser inequality on a bounded domain in Lorentz–Sobolev spaces.

Theorem 1.1

Let Ω be a bounded domain in n (n2), αn,q=(nvn1/n)q/(q-1) and 0β<n. Then there exists a constant C(n,q,β) such that for any 0ααn,q,

sup u C 0 ( Ω ) , u n , q 1 1 | Ω | 1 - β / n Ω exp [ ( 1 - β n ) α | u | q / ( q - 1 ) ] | x | β 𝑑 x C ( n , q , β ) .

The constant αn,q is sharp in the sense that if α>αn,q, then the supremum will become infinite.

The nonsingular case β=0 was proved in [4, 18].

Next, we will consider the following sharp singular critical Trudinger–Moser inequality on any unbounded domain in n in the spirit of [36, 28].

Theorem 1.2

Let 1<q<+ and 0<β<n. Then there exist a constant C(n,q,β) such that for any 0<ααn,q and any uW01Ln,q(n), the following inequality holds:

sup u n , q q + u n , q q 1 n Φ [ ( 1 - β n ) α | u | q / ( q - 1 ) ] | x | β 𝑑 x C ( n , q , β ) ,

where

Φ ( t ) = e t - j = 0 k 0 t j j ! , k 0 = [ ( q - 1 ) n q ] .

The constant αn,q is sharp in the sense that if α>αn,q, then the supremum will become infinite.

The nonsingular case β=0 was proved by Cassani and Tarsi [9], and our approach is substantially different from theirs. See more remarks at the end of this section.

We note that the inequality in the nonsingular case β=0 in [9] was proved only under the restriction of the full norm un,qq+un,qq1. Then, it is interesting to ask whether either the singular or nonsingular subcritical Trudinger–Moser inequalities on unbounded domains in n holds only under the restriction of Lorentz–Sobolev norms un,q1.

In this paper, we will also address this issue. We will first establish the sharp subcritical singular Trudinger–Moser inequality in the spirit of Adachi and Tanaka [1].

Theorem 1.3

Let 1<q<+ and 0<β<n. Then there exists a constant C(n,q,β) such that for any 0<α<αn,q and any uW01Ln,q(n) with un,q1, the following inequality holds:

n Φ [ ( 1 - β n ) α | u | q / ( q - 1 ) ] | x | β 𝑑 x C ( n , q , β ) u n , q n - β ,

where

Φ ( t ) = e t - j = 0 k 0 t j j ! , k 0 = [ ( q - 1 ) n q ] .

The restriction α<αn,q is sharp in the sense that if ααn,q, then there exists a sequence {uk}1+W01Ln,q(n) with ukn,q=1 such that

1 u k n , q n - β n Φ [ ( 1 - β n ) α | u k | q / ( q - 1 ) ] | x | β 𝑑 x + .

Remark 1.4

The proof of Theorem 1.3 does not give rise to the sharp constant in the case β=0, since in our proof the function 1/|x|β can help us to remedy the integrability issue at infinity for β>0. So we will treat the subcritical nonsingular case when β=0 separately below by using a different argument.

Theorem 1.5

Let 1<q<+. Then there exist a constant C(n,q) such that for any 0<α<αn,q and any uW01Ln,q(n) with un,q1, the following inequality holds:

n Φ ( α | u | q / ( q - 1 ) ) 𝑑 x C ( n , q ) u n , q n ,

where

Φ ( t ) = e t - j = 0 k 0 t j j ! , k 0 = [ ( q - 1 ) n q ] .

The restriction α<αn,q is sharp in the sense that if ααn,q, then the inequality can no longer hold with some C(n,q) independent of u.

The following remarks are in order. First of all, as we have pointed out earlier, nonsingular (i.e., β=0) Trudinger–Moser type inequalities with Lorentz norms have been obtained on bounded domains by Alvino, Ferone and Trombetti [4], and by Hudson and Leckband [18], and on unbounded domains by Cassani and Tarsi [9] in the critical case. Our results extend to the singular case 0<β<n. Our Theorem 1.1 considers the case of bounded domains and Theorem 1.2 takes care of the unbounded domains. Second, our results also address the subcritical case α<αn,q for both the singular (0<β<n) and nonsingular (β=0) case. Moreover, the proofs of these two cases are different. See Theorems 1.3 and 1.5. Third, the proofs of Theorems 1.2, 1.3 and 1.5 are essentially different from that in [9]. Therein, the proof of Theorem 1.2 in the nonsingular case β=0 is based crucially on the inequality

u ( r ) - u ( r 0 ) v ( r ) ,

where v is such that |v| is dominated by |u| in an appropriate sense. In fact, the following result from [17] has been used

u ( r ) - u ( r 0 ) 1 n w n 1 / n r r 0 U ( s ) s 1 / n d s s ,

where U(x) is the radial function built from |u| on the level set of u, i.e.,

| u | > t | u | d x = 0 | { | u | > t } | U ( s ) d s .

The construction of such U uses the deep Fleming–Rishel co-area formula the isoperimetric inequality and is highly nontrivial. In particular, such optimal Fleming–Rishel type co-area formula and the isoperimetric inequality are still not available in other non-Euclidean settings such as the Heisenberg group. An essential difference between our proof of the critical Trudinger–Moser inequality with Lorentz norms and the proof given by Cassani and Tarsi [9] is that we do not use this inequality involving the function U built from |u| on the level set of u at all. Instead, we will consider the integral on n by dividing it into two parts using the level set of u, inspired by an idea of Lam and the first author for Moser–Trudinger inequalities on the Heisenberg group [19] and Adams inequalities on high order Sobolev spaces [21]. The essence of this argument is that we can establish global Trudinger–Moser inequalities on unbounded domains from local Trudinger–Moser inequalities on bounded domains. This argument to derive global inequalities from the local ones using the level sets works in several other settings such as hyperbolic spaces [31], Riemannian manifolds, etc., and can be used for both critical and subcritical Moser–Trudinger inequalities [19, 23].

The organization of the paper is as follows. In Section 2, we introduce some preliminaries about the Lorentz–Sobolev spaces. In Section 3, we will give a singular Trudinger–Moser inequality on bounded domains with the Lorentz–Sobolev space norms (Theorem 1.1). In Section 4, we will prove the singular critical Trudinger–Moser inequality for unbounded domains (Theorem 1.2). In Section 5, we obtain the best constant for the singular subcritical Trudinger–Moser inequality on unbounded domains (Theorem 1.3). Finally, in Section 6, we will give subcritical Trudinger–Moser inequality in the nonsingular case on unbounded domains with the Lorentz–Sobolev space norms (Theorem 1.5)

2 Preliminaries

Let f:n be a function such that

| { x n : | f ( x ) | > t } | = { x n : | f ( x ) | > t } 𝑑 x < +

for every t>0. Its distribution function is defined by

d f ( t ) = | { x : | f ( x ) | > t } | ,

and its decreasing rearrangement f is defined by

f ( s ) = sup { t > 0 : d f ( t ) > s } .

Now, define f:n by

f ( x ) = f ( v n | x | n ) ,

where vn is the volume of the unit ball in n. Then, from [29], for every continuous increasing function Ψ:[0,+)[0,+), we have that

n Ψ ( f ) 𝑑 x = n Ψ ( f ) 𝑑 x .

Since f is nonincreasing, the maximal function f of the rearrangement of f, defined by

f := 1 s 0 s f 𝑑 t for s 0 ,

is also nonincreasing and ff. For more details on the rearrangement we refer the reader to [29, 37, 39].

The Lorentz spaces Lp,q, with 1<p<+ and 1q<+, is defined to consist of all functions ψ such that

ψ p , q = ( 0 + [ ψ ( t ) t 1 / p ] q d t t ) 1 / q < .

It is well known that when q>p the quantity ψp,q is not a norm. But the quantity

ψ p , q = ( 0 + [ ψ ( t ) t 1 / p ] q d t t ) 1 / q

is a norm for any p and q. It is easy to prove, by using Hardy’s inequalities (see [37]), that these two quantities are equivalent in the sense that

ψ p , q ψ p , q C ( p , q ) ψ p , q .

As the classic Sobolev spaces are built up from Lebesgue spaces, similarly one can introduce Lorentz–Sobolev spaces which consist of functions having weak derivatives belonging to the Lorentz space Lp,q(n). For more details on Lorentz spaces, we refer the reader to the book by Stein and Weiss [37, Chapter 5, Section 3].

In particular, let 1<p<+ and let W01Ln,q(n) be the closure of

{ u C 0 ( n ) : u 1 , ( n , q ) < + } ,

under the norm

u 1 , ( n , q ) q = u n , q q + u n , q q .

We recall two important lemmas which will be used in our proof. One is the well-known Strauss radial lemma [38].

Lemma 2.1

If uLp,q(n), 1<p<+, 1q<+, then

u ( t ) ( q p ) 1 / q u p , q t 1 / p .

Lemma 2.2

Let Ω be a bounded domain and uC0(Ω). Then

u ( t ) 1 n v n 1 / n [ t | Ω | | u | ( r ) r 1 / n d r + 1 t 1 - 1 / n 0 t | u | ( r ) d r ] .

The proof of this lemma follows easily by combining the representation formula (see, e.g., [2, Lemma 2]) and O’Neil’s lemma for rearrangement of convolution operators [34] (see the explicit statement listed in the proof of [2, Theorem 2]).

3 Singular Trudinger–Moser Inequality on Bounded Domains with Lorentz–Sobolev Norms

In this section, we will prove Theorem 1.1. To do so, we will apply some ideas used in [4, 2].

First, we state a lemma which was originally proved in [2] and the current version is taken from [21]. This is a key tool for proving the theorem.

Lemma 3.1

Let 0<α1, 1<p<, and a(s,t) be a non-negative measurable function on (-,)×[0,] such that (a.e)

a ( s , t ) 1 when 0 < s < t    𝑎𝑛𝑑    sup t > 0 ( - 0 + t a ( s , t ) p d s ) 1 / p = b < .

If

- ϕ ( s ) p 𝑑 s 1 for ϕ 0 ,

then there exists a constant c0=c0(p,b) such that

0 e - F α ( t ) 𝑑 t c 0 , 𝑤ℎ𝑒𝑟𝑒 F α ( t ) = α t - α ( - a ( s , t ) ϕ ( s ) 𝑑 s ) p .

Now we are ready to prove Theorem 1.1.

Proof.

Using Lemma 2.2, we have

u ( t ) 1 n v n 1 / n [ t | Ω | | u | ( r ) r 1 / n d r + 1 t 1 - 1 / n 0 t | u | ( r ) d r ] .

Then

u ( | Ω | e - s ) | Ω | 1 / n n v n 1 / n [ 0 s | u | ( | Ω | e - r ) e - r / n d r + e s ( 1 - 1 / n ) s + | u | ( | Ω | e - r ) e - r d r ] .

Now set

Φ ( r ) = { | Ω | 1 / n | u | ( | Ω | e - r ) e - r / n if r 0 , 0 if r < 0 ,

and

a ( r , s ) = { 0 if r 0 , 1 if 0 < r s , e ( n - 1 ) ( s - r ) / n if s < r < + .

With this notation, we have

u ( | Ω | e - s ) 1 n v n 1 / n - + Φ ( r ) a ( r , s ) 𝑑 r .

Then we will use the arrangement argument. By simple calculations, we know that

( | x | - β ) ( t ) = ( t v n ) - β / n

and

( exp [ ( 1 - β n ) α | u | q / ( q - 1 ) ] ) ( t ) = exp [ ( 1 - β n ) α ( u ( t ) ) q / ( q - 1 ) ] .

For ααn,q, by the Hardy–Littlewood inequality, we have

1 | Ω | 1 - β / n Ω exp [ ( 1 - β n ) α | u | q / ( q - 1 ) ] | x | β 𝑑 x
1 | Ω | 1 - β / n 0 | Ω | exp [ ( 1 - β n ) α ( u ( t ) q / ( q - 1 ) ) ] ( t v n ) - β / n 𝑑 t
= 1 | Ω | 1 - β n 0 + exp [ ( 1 - β n ) α | u ( e - s | Ω | ) | q / ( q - 1 ) ] ( e - s | Ω | v n ) - β / n e - s | Ω | 𝑑 s
= v n β / n 0 + exp [ ( 1 - β n ) α | u ( e - s | Ω | ) | q / ( q - 1 ) ] e - s ( 1 - β / n ) 𝑑 s
v n β / n 0 + exp [ ( 1 - β n ) ( α ( q - 1 ) / q n v n 1 / n - + Φ ( r ) a ( r , s ) 𝑑 r ) q / ( q - 1 ) - ( 1 - β n ) s ] 𝑑 s
v n β / n 0 + exp [ ( 1 - β n ) ( - + Φ ( r ) a ( r , s ) 𝑑 r ) q / ( q - 1 ) - ( 1 - β n ) s ] 𝑑 s
v n β / n 0 + exp [ - F 1 - β / n ( s ) ] 𝑑 s ,

where

F 1 - β / n ( s ) = ( 1 - β n ) s - ( 1 - β n ) ( - + Φ ( r ) a ( r , s ) 𝑑 r ) q / ( q - 1 ) .

Since a(r,s)1 for a.e 0<r<s and

- 0 a ( r , s ) q / ( q - 1 ) 𝑑 r + s + a ( r , s ) q / ( q - 1 ) 𝑑 r = n ( q - 1 ) ( n - 1 ) q < + ,

we have

- + Φ q ( r ) 𝑑 r = 0 + ( | Ω | 1 / n | u | ( | Ω | e - r ) e - r / n ) q 𝑑 r
= 0 | Ω | [ | u | ( t ) t 1 / n ] q d t t
= u n , q q
1 .

Now using Lemma 3.1, we obtain

0 + exp [ - F 1 - β / n ( s ) ] 𝑑 s c 0 .

Therefore, when ααn,q,

1 | Ω | 1 - β / n Ω exp [ ( 1 - β n ) α | u | q / ( q - 1 ) ] | x | β 𝑑 x C ( n , q , β ) .

Now, let us prove that the constant αn,q is sharp. We can assume that Ω is a ball with |Ω|=1. Set

u k ( x ) = { k ( q - 1 ) / q ( n v n 1 / n ) - 1 if 0 | x | ( 1 v n e - k ) 1 / n , k ( q - 1 ) / q ( n v n 1 / n ) - 1 - ln ( v n | x | n ) k if ( 1 v n e - k ) 1 / n < | x | ( 1 v n ) 1 / n , 0 if | x | > ( 1 v n ) 1 / n .

By calculation, we have

| u k | ( x ) = { 0 if 0 | x | ( 1 v n e - k ) 1 / n , k - 1 / q v n 1 / n | x | if ( 1 v n e - k ) 1 / n < | x | ( 1 v n ) 1 / n , 0 if | x | > ( 1 v n ) 1 / n

and

| u k | ( s ) = { 0 if 1 - e - k s , k - 1 / q ( s + e - k ) 1 / n if 0 s < 1 - e - k .

Then

u k n , q q = 1 k 0 1 - e - k ( s s + e - k ) q / n d s s = 1 k 1 e k ( 1 - 1 y ) q / n d y y - 1 .

If we let

a k = u k n , q = ( 1 k 1 e k ( 1 - 1 y ) q / n d y y - 1 ) 1 / q ,

then limk+ak=1.

Now, choose wk(x)=uk(x)ak. Then wk(x)C0+(Ω) and wkn,q=1. But when α>αn,q,

Ω exp [ α ( 1 - β n ) ] u q / ( q - 1 ) | x | β | x | ( 1 v n e - k ) 1 / n exp [ α ( 1 - β n ) ( a k - 1 k ( q - 1 ) / q ( n v n 1 / n ) - 1 ) q / ( q - 1 ) ] | x | β 𝑑 x
= exp [ α α n , q ( 1 - β n ) a k q / ( 1 - q ) k ] | x | ( 1 v n e - k ) 1 / n 1 | x | β 𝑑 x
exp [ ( 1 - β n ) k ( α α n , q a k q / ( 1 - q ) - 1 ) ]
+ as k + .

Therefore, we have proved that the constant αn,q is sharp. ∎

4 Critical Singular Trudinger–Moser Inequality on Unbounded Domains with Lorentz–Sobolev Norms

In this section, we will prove Theorem 1.2.

Proof of Theorem 1.2.

By the standard density argument, it suffices to prove that for any 0<β<n and uC0+(n) with u0 satisfying un,qq+un,qq1, there exists a constant C(n,q,β) such that

n Φ [ ( 1 - β n ) α n , q | u | q / ( q - 1 ) ] | x | β 𝑑 x C ( n , q , β ) .

Set

A ( u ) = 2 - 1 / [ q ( q - 1 ) ] u n , q and Ω = { x n : u > A ( u ) } .

Then it is clear that A(u)<1. By the property of rearrangement, we know that for any t[0,|Ω|],

(4.1) u ( t ) > 2 - 1 / [ q ( q - 1 ) ] u n , q .

At the same time, using the radial Lemma 2.1, we have

(4.2) u ( t ) ( q n ) 1 / q u n , q t 1 / n ,

Combining inequality (4.1) and (4.2), we have

t 2 n / [ q ( q - n ) ] ( q n ) n / q for any t [ 0 , | Ω | ] .

Therefore,

| Ω | 2 n / [ q ( q - n ) ] ( q n ) n / q .

Now, we write

n Φ [ ( 1 - β n ) α n , q | u | q / ( q - 1 ) ] | x | β 𝑑 x = I 1 + I 2 ,

where

I 1 = Ω Φ [ ( 1 - β n ) α n , q | u | q / ( q - 1 ) ] | x | β 𝑑 x , I 2 = n Ω Φ [ ( 1 - β n ) α n , q | u | q / ( q - 1 ) ] | x | β 𝑑 x .

First, let us estimate I2. Since nΩ{u(x)<1}, we see that

I 2 { u 1 } 1 | x | β j = k 0 + 1 α n , q j ( 1 - β n ) j j ! | u | j q / ( q - 1 ) d x
j = k 0 + 1 α n , q j ( 1 - β n ) j j ! { u 1 } 1 | x | β | u | n 𝑑 x
= C n , q , β ( { u 1 , | x | 1 } 1 | x | β | u | n 𝑑 x + { u 1 , | x | > 1 } 1 | x | β | u | n - β / 2 𝑑 x )
C n , q , β ( { | x | 1 } 1 | x | β 𝑑 x + { | x | > 1 } | u | n - β / 2 | x | β 𝑑 x )
C ( n , q , β ) ( 1 + { | x | > 1 } | u | n - β / 2 | x | β 𝑑 x ) .

Since

( χ { | x | > 1 } | x | β ) ( t ) = ( 1 + t v n ) - β / n ,

by the Hardy–Littlewood inequality and the radial lemma, we have

{ | x | > 1 } | u | n - β / 2 | x | β 𝑑 x 0 + ( u ( t ) ) n - β / 2 ( 1 + t v n ) - β / n 𝑑 t
( q n ) n - β / 2 q 0 + u n , q n - β / 2 t 1 - β / ( 2 n ) ( 1 + t v n ) - β / n 𝑑 t
( q n ) n - β / 2 q u n , q n - β / 2 ( 0 1 1 t 1 - β / ( 2 n ) 𝑑 t + 1 + v n β / n t 1 - β / ( 2 n ) + β / n 𝑑 t )
= C ( n , q , β ) .

Thus, we have proved that I2C(n,q,β).

Next, to estimate I1, we set v(x)=u(x)-A(u) in Ω, thus vC0(Ω). Moreover,

| u | q / ( q - 1 ) ( | v | + A ( u ) ) q / ( q - 1 )
| v | q / ( q - 1 ) + q q - 1 2 q / ( q - 1 ) - 1 ( | v | q / ( q - 1 ) - 1 A ( u ) + | A ( u ) | q / ( q - 1 ) )
| v | q / ( q - 1 ) + q q - 1 2 q / ( q - 1 ) - 1 | v | q / ( q - 1 ) | A ( u ) | q q + q q - 1 2 q / ( q - 1 ) - 1 ( q - 1 q + | A ( u ) | q / ( q - 1 ) )
= | v | q / ( q - 1 ) ( 1 + 2 1 / ( q - 1 ) q - 1 | A ( u ) | q ) + C q ,

where we used Young’s inequality and the following elementary inequality:

( a + b ) p a p + p 2 p - 1 ( a p - 1 b + b p ) for all p 1 and a , b 0 .

Let

w ( x ) = ( 1 + 2 1 / ( q - 1 ) q - 1 | A ( u ) | q ) ( q - 1 ) / q v ( x ) in Ω .

Then wC0(Ω) and |u|q/(q-1)|w|q/(q-1)+Cq. Moreover, we have

w = ( 1 + 2 1 / ( q - 1 ) q - 1 | A ( u ) | q ) ( q - 1 ) / q v .

When q2,

w n , q q / ( q - 1 ) = ( 1 + 2 1 / ( q - 1 ) q - 1 | A ( u ) | q ) u n , q q / ( q - 1 )
( 1 + 2 1 / ( q - 1 ) q - 1 | A ( u ) | q ) ( 1 - u n , q q ) 1 / ( q - 1 )
( 1 + u n , q q q - 1 ) ( 1 - u n , q q q - 1 )
(4.3) 1 .

Here we used the inequality (1-x)p1-px for all 0x1, 0<p1.

On the other hand, when 1<q<2,

w n , q q = ( 1 + 2 1 / ( q - 1 ) q - 1 | A ( u ) | q ) q - 1 u n , q q
( 1 + 2 1 / ( q - 1 ) | A ( u ) | q ) ( 1 - u n , q q )
( 1 + u n , q q ) ( 1 - u n , q q )
(4.4) 1 .

Here we used the inequality (1+x)p1+px for all x0, 0<p1.

By inequality (4.3) and (4.4), we know that un,q1. So we can use Theorem 1.1 to estimate I1 as follows:

I 1 Ω exp ( α n , q ( 1 - β n ) | u | q / ( q - 1 ) ) | x | β 𝑑 x
Ω exp ( α n , q ( 1 - β n ) | w | q / ( q - 1 ) + C q ) | x | β 𝑑 x
C β | Ω | 1 - β / n
C n , β , q ,

since we already know that |Ω|2n/[q(q-n)](qn)n/q.

Now, let us prove that the constant αn,q is sharp. Set

u k ( x ) = { ( 1 - c k ) 1 / q k ( q - 1 ) / q ( n v n 1 / n ) - 1 if 0 | x | ( 1 v n e - k ) 1 / n , ( 1 - c k ) 1 / q k ( q - 1 ) / q ( n v n 1 / n ) - 1 - ln ( v n | x | n ) k if ( 1 v n e - k ) 1 / n < | x | ( 1 v n ) 1 / n , 0 if | x | > ( 1 v n ) 1 / n ,

where

c k = 1 k 1 e k ( ( 1 - 1 y ) q / n 1 y - 1 - 1 y ) 𝑑 y .

As k+, since

1 k 1 e k ( 1 - 1 y ) q / n 1 y - 1 𝑑 y = O ( 1 k ) + 1 k 2 e k ( 1 - 1 y ) q / n 1 y - 1 𝑑 y

and

2 e k c y ( y - 1 ) 𝑑 y 1 k 2 e k [ ( 1 - 1 y ) q / n - 1 ] 1 y - 1 𝑑 y 2 e k C y ( y - 1 ) 𝑑 y ,

we have that

1 k 2 e k ( 1 - 1 y ) q / n 1 y - 1 𝑑 y = 1 k 2 e k 1 y - 1 𝑑 y + O ( 1 k ) = 1 + O ( 1 k ) .

Then

1 k 1 e k ( 1 - 1 y ) q / n 1 y - 1 𝑑 y = 1 + O ( 1 k )

and

c k = O ( 1 k ) .

By calculation, we have

| u k | ( s ) = { ( 1 - c k ) 1 / q k ( q - 1 ) / q ( n v n 1 / n ) - 1 if 0 s e - k , ( 1 - c k ) 1 / q k ( q - 1 ) / q ( n v n 1 / n ) - 1 - ln s k if e - k s < 1 , 0 if s 1 ,
| u k | ( x ) = { 0 if 0 | x | ( 1 v n e - k ) 1 / n , ( 1 - c k ) 1 / q k - 1 / q v n 1 / n | x | if ( 1 v n e - k ) 1 / n < | x | ( 1 v n ) 1 / n , 0 if | x | > ( 1 v n ) 1 / n

and

| u k | ( s ) = { 0 if 1 - e - k s , ( 1 - c k ) 1 / q k - 1 / q ( s + e - k ) 1 / n if 0 s < 1 - e - k .

We also have

u k n , q q = ( 1 - c k ) 1 k 0 1 - e - k ( s s + e - k ) q / n d s s = ( 1 - c k ) k 1 e k ( 1 - 1 y ) q / n d y y - 1

and

u k n , q q = ( 1 - c k ) [ 0 e - k ( k ( q - 1 ) / q ( n v n 1 / n ) - 1 s 1 / n ) q d s s + e - k 1 ( k ( q - 1 ) / q ( n v n 1 / n ) - 1 - ln s k ) q s q / n d s s ]
= ( 1 - c k ) [ n q ( n v n 1 / n ) - q k q - 1 e - k q / n + ( n v n 1 / n ) - q 1 k 0 k s q e - s q / n 𝑑 s ] .

Then

u k n , q q + u k n , q q
= ( 1 - c k ) [ 1 k 1 e k ( 1 - 1 y ) q / n d y y - 1 + n q ( n v n 1 / n ) - q k q - 1 e - k q / n + ( n v n 1 / n ) - q 1 k 0 k s q e - s q / n 𝑑 s ]
= ( 1 - c k ) [ 1 + 1 k 1 e k ( ( 1 - 1 y ) q / n 1 y - 1 - 1 y ) 𝑑 y + n q ( n v n 1 / n ) - q k q - 1 e - k q / n + ( n v n 1 / n ) - q 1 k 0 k s q e - s q / n 𝑑 s ]
= ( 1 - c k ) ( 1 + c k + o ( c k ) )
= 1 + O ( 1 k ) .

Now, choose

w k ( x ) = u k ( x ) ( u k ( x ) n , q q + u k n , q q ) 1 / q .

Then wkW01Ln,q(n) and wk(x)n,qq+wkn,qq=1. But as k+, for α>αn,q,

n Φ [ α ( 1 - β n ) u q / ( q - 1 ) ] | x | β 𝑑 x | x | ( 1 v n e - k ) 1 / n Φ [ α ( 1 - β n ) u q / ( q - 1 ) ] | x | β 𝑑 x
| x | ( 1 v n e - k ) 1 / n Φ [ α ( 1 - β n ) ( 1 - c k 1 + O ( 1 k ) ) 1 / ( q - 1 ) ( k ( q - 1 ) / q ( n v n 1 / n ) - 1 ) q / ( q - 1 ) ] | x | β 𝑑 x
Φ [ α α n , q ( 1 - β n ) ( 1 - c k 1 + O ( 1 k ) ) 1 / ( q - 1 ) k ] exp ( ( 1 - β n ) e - k )
exp [ ( 1 - β n ) k ( α α n , q ( 1 - c k 1 + O ( 1 k ) ) 1 / ( q - 1 ) - 1 ) ]
+ .

Thus, we have proved that the constant αn,q is sharp. ∎

5 Subcritical Singular Trudinger–Moser Inequality on Unbounded Domains with Lorentz–Sobolev Norms

In this section, we will prove Theorem 1.3.

Proof of Theorem 1.3.

By the standard density argument, it suffices to prove that for any 0<β<n, 0<α<αn,q and uC0(n) with u0 satisfying un,q1, there exists a constant C(n,q,β) such that

n Φ [ ( 1 - β n ) α | u | q / ( q - 1 ) ] | x | β 𝑑 x C ( n , q , β ) u n , q n - β .

Set

Ω = { x n : u > 1 } .

Then, by the property of rearrangement, we know that u(t)>1 for any t[0,|Ω|]. At the same time, using the radial lemma, we have

u ( t ) ( q n ) 1 / q u n , q t 1 / n ,

and thus

t ( q n ) n / q u n , q n for any t [ 0 , | Ω | ] .

Therefore,

| Ω | ( q n ) n / q u n , q n .

Now, we write

n Φ [ ( 1 - β n ) α | u | q / ( q - 1 ) ] | x | β 𝑑 x = I 1 + I 2 ,

where

I 1 = Ω Φ [ ( 1 - β n ) α | u | q / ( q - 1 ) ] | x | β 𝑑 x , I 2 = n Ω Φ [ ( 1 - β n ) α | u | q / ( q - 1 ) ] | x | β 𝑑 x .

First, we estimate I2. Since nΩ{u(x)<1}, we see that

I 2 { u 1 } 1 | x | β j = k 0 + 1 α j ( 1 - β n ) j j ! | u | j n / ( n - 1 ) d x
j = k 0 + 1 α n , q j ( 1 - β n ) j j ! { u 1 } 1 | x | β | u | n 𝑑 x
= C n , q , β ( { u 1 , | x | u n , q } 1 | x | β | u | n 𝑑 x + { u 1 , | x | > u n , q } 1 | x | β | u | n - β / 2 𝑑 x )
C n , q , β ( { | x | u n , q } 1 | x | β 𝑑 x + { | x | > u n , q } | u | n - β / 2 | x | β 𝑑 x )
C ( n , q , β ) ( u n , q n - β + { | x | > u n , q } | u | n - β / 2 | x | β 𝑑 x ) .

Since

( χ { | x | > u n , q } | x | β ) ( t ) = ( u n , q n + t v n ) - β / n ,

by the Hardy–Littlewood inequality and the radial lemma, we have

{ | x | > u n , q n } | u | n - β / 2 | x | β 𝑑 x 0 + ( u ( t ) ) n - β / 2 ( u n , q n + t v n ) - β / n 𝑑 t
( q n ) n - β / 2 q 0 + u n , q n - β / 2 t 1 - β / ( 2 n ) ( u n , q n + t v n ) - β / n 𝑑 t
( q n ) n - β / 2 q u n , q n - β / 2 ( 0 u n , q n u n , q - β t 1 - β / ( 2 n ) 𝑑 t + u n , q n + v n β / n t 1 - β / ( 2 n ) + β / n 𝑑 t )
= C ( n , q , β ) u n , q n - β .

So, we have proved that I2C(n,q,β)un,qn-β.

Set v(x)=u(x)-1 in Ω. Then vC0(Ω) and gvn,q1. By Theorem 1.1, we have

1 | Ω | 1 - β / n Ω exp ( α ( 1 - β n ) | v ( x ) | q / ( q - 1 ) ) | x | β 𝑑 x C n , q , β .

Now, put ε=αn,qα(1-βn)-1>0. Using the elementary inequality

( a + b ) p ε b p + ( 1 - ( 1 + ε ) - 1 / ( p - 1 ) ) 1 - p a p for any a , b , ε > 0 and p > 1 ,

we have, in Ω, that

| u ( x ) | q / ( q - 1 ) = ( 1 + v ( x ) ) q / ( q - 1 ) ( 1 + ε ) | v | q / ( q - 1 ) + ( 1 - 1 ( 1 + ε ) q - 1 ) 1 / ( 1 - q ) .

Set

C ε = ( 1 - 1 ( 1 + ε ) n - 1 ) 1 / ( 1 - n ) .

Hence,

I 1 = Ω Φ n ( α ( | u | ) q / ( q - 1 ) ) | x | β 𝑑 x
Ω exp ( α ( | u | ) q / ( q - 1 ) ) | x | β 𝑑 x
= Ω exp ( α ( | v + 1 | ) q / ( q - 1 ) ) | x | β 𝑑 x
Ω exp ( α ( 1 + ε ) | v | q / ( q - 1 ) + α C ε ) | x | β 𝑑 x
e α C ε Ω exp ( α n , q ( 1 - β n ) | v | q / ( q - 1 ) ) | x | β 𝑑 x
C n , q , β | Ω | 1 - β n
C α , β u n , q n ,

since we already know that |Ω|(qn)n/qun,qn.

Now, we prove that the restriction α<αn,q is optimal. Set

u k ( x ) = { C k k ( q - 1 ) / q ( n v n 1 / n ) - 1 if 0 | x | ( 1 v n e - k ) 1 / n , C k k ( q - 1 ) / q ( n v n 1 / n ) - 1 - ln ( v n | x | n ) k if ( 1 v n e - k ) 1 / n < | x | ( 1 v n ) 1 / n , 0 if | x | > ( 1 v n ) 1 / n ,

where

C k = ( 1 k 1 e k ( 1 - 1 y ) q / n d y y - 1 ) - 1 / q .

By calculation, we have Ck1 and Ckq/(q-1)k-k0 as k+. Therefore,

| u k | ( s ) = { C k k ( q - 1 ) / q ( n v n 1 / n ) - 1 if 0 s e - k , C k k ( q - 1 ) / q ( n v n 1 / n ) - 1 - ln s k if e - k s < 1 , 0 if s 1 ,
| u k | ( x ) = { 0 if 0 | x | ( 1 v n e - k ) 1 / n , C k k - 1 / q v n 1 / n | x | if ( 1 v n e - k ) 1 / n < | x | ( 1 v n ) 1 / n , 0 if | x | > ( 1 v n ) 1 / n

and

| u k | ( s ) = { 0 if 1 - e - k s , C k k - 1 / q ( s + e - k ) 1 / n if 0 s < 1 - e - k .

We also have

u k n , q q = C k q k 0 1 - e - k ( s s + e - k ) q / n d s s = C k q k 1 e k ( 1 - 1 y ) q / n d y y - 1 = 1

and

u k n , q q = C k q [ 0 e - k ( k ( q - 1 ) / q ( n v n 1 / n ) - 1 s 1 / n ) q d s s + e - k 1 ( k ( q - 1 ) / q ( n v n 1 / n ) - 1 - ln s k ) q s q / n d s s ]
C k q ( k q - 1 e - k q / n + 1 k 0 k s q e - s q / n 𝑑 s )
1 k as k + .

Thus, we have ukW01Ln,q(n), ukn,q=1 and ukn,q=O(1k1/q). Moreover,

n Φ [ α ( 1 - β n ) u q / ( q - 1 ) ] | x | β 𝑑 x = | x | ( 1 v n e - k ) 1 / n Φ [ α ( 1 - β n ) u q / ( q - 1 ) ] | x | β 𝑑 x + ( 1 v n e - k ) 1 / n < | x | ( 1 v n ) 1 / n Φ [ α ( 1 - β n ) u q / ( q - 1 ) ] | x | β 𝑑 x
| x | ( 1 v n e - k ) 1 / n Φ [ α ( 1 - β n ) ( C k k ( q - 1 ) / q ( n v n 1 / n ) - 1 ) q / ( q - 1 ) ] | x | β 𝑑 x
- j = 1 k 0 ( 1 v n e - k ) 1 / n < | x | ( 1 v n ) 1 / n α j ( 1 - β n ) j | C k k ( q - 1 ) / q ( n v n 1 / n ) - 1 ln ( v n | x | n ) k | j | x | β 𝑑 x
Φ [ α α n , q ( 1 - β n ) C k q / ( q - 1 ) k ] | x | ( 1 v n e - k ) 1 / n 1 | x | β 𝑑 x + j = 1 k 0 C j k j / q
exp [ ( 1 - β n ) k ( α α n , q C k q / ( q - 1 ) - 1 ) ]
1 as k + ,

when α=αn,q. But,

1 u k n , q n - β n Φ [ ( 1 - β n ) α n , q | u k | q / ( q - 1 ) ] | x | β 𝑑 x + .

Therefore, we proved that the restriction α<αn,q is optimal. ∎

6 Subcritical Trudinger–Moser Inequality on Unbounded Domains with Lorentz–Sobolev Norms

In this section, we will prove Theorem 1.5.

Proof of Theorem 1.5.

By the standard density argument, it suffices to prove that for any 0<α<αn,q and uC0(n) with u0 satisfying un,q1, there exists a constant C(n,q) such that

n Φ ( α | u | q / ( q - 1 ) ) 𝑑 x C ( n , q ) u n , q n .

By symmetrization, we have

n Φ ( α | u | q / ( q - 1 ) ) 𝑑 x = 0 + Φ ( α | u | q / ( q - 1 ) ) 𝑑 t .

Now set R=(qn)n/qun,qn. Using the radial lemma, we have

u ( R ) ( q n ) 1 / q u n , q R 1 / n = 1 .

Now, we write

0 + Φ ( α | u | q / ( q - 1 ) ) 𝑑 t = I 1 + I 2 ,

where

I 1 = 0 R Φ ( α | u | q / ( q - 1 ) ) 𝑑 t , I 2 = R + Φ ( α | u | q / ( q - 1 ) ) 𝑑 t .

First, we estimate I2. Since k0=[q-1qn], we have qq-1j>n when jk0+1. Using the radial lemma yields

I 2 = j = k 0 + 1 α j j ! R + | u ( t ) | q / ( q - 1 ) j d t
j = k 0 + 1 α j j ! R + [ ( q n ) 1 / q u n , q t 1 / n ] q / ( q - 1 ) j 𝑑 t
j = k 0 + 1 ( α ( q n ) 1 / ( q - 1 ) ) j j ! u n , q q / ( q - 1 ) j R + [ t - 1 / n ] q / ( q - 1 ) j 𝑑 x
= j = k 0 + 1 ( α ( q n ) 1 / ( q - 1 ) ) j j ! u n , q q / ( q - 1 ) j R 1 - q / [ n ( q - 1 ) ] j q j ( q - 1 ) n - 1
C ( n , q ) u n , q n .

Next, let us estimate I1. Set w(t)=[u(t)-u(R)]χ[0,R](t). Then w(t)0. Now, put ε=αn,qα-1>0. Using the elementary inequality

( a + b ) p ε b p + ( 1 - ( 1 + ε ) - 1 / ( p - 1 ) ) 1 - p a p for any a , b , ε > 0 and p > 1 ,

we have, when 0t<R, that

| u ( t ) | q / ( q - 1 ) = ( w ( t ) + u ( R ) ) q / ( q - 1 ) ( 1 + ε ) | w ( t ) | q / ( q - 1 ) + ( 1 - 1 ( 1 + ε ) q - 1 ) 1 / ( 1 - q ) ( u ( R ) ) q / ( q - 1 ) .

Since u(R)1, we have

I 1 = 0 R Φ ( α | u | q / ( q - 1 ) ) 𝑑 t
0 R exp ( α | u | q / ( q - 1 ) ) 𝑑 t
0 R exp [ α ( 1 + ε ) | w ( t ) | q / ( q - 1 ) + α C ε ( u ( R ) ) q / ( q - 1 ) ] 𝑑 t
(6.1) e α C ε 0 R exp [ α n , q | w ( t ) | q / ( q - 1 ) ] 𝑑 t .

Using Lemma 2.2 yields

w ( t ) = u ( t ) - u ( R ) 1 n v n 1 / n [ t R | u | ( r ) r 1 / n d r + 1 t 1 - 1 / n 0 t | u | ( r ) d r ] .

Then

w ( R e - s ) R 1 / n n v n 1 / n [ 0 s | u | ( R e - r ) e - r / n d r + e s ( 1 - 1 / n ) s + | u | ( R e - r ) e - r d r ] .

Now set

Φ ( r ) = { R 1 / n | u | ( R e - r ) e - r / n if r 0 , 0 if r < 0

and

a ( r , s ) = { 0 if r 0 , 1 if 0 < r s , e ( n - 1 ) ( s - r ) / n if s < r < + .

With this notation we have

w ( R e - s ) 1 n v n 1 / n - + Φ ( r ) a ( r , s ) 𝑑 r .

Then

0 R exp [ α n , q | w ( t ) | q / ( q - 1 ) ) ] d t = 0 + exp [ α n , q | w ( e - s R ) | q / ( q - 1 ) ] e - s R d s
R 0 + exp [ ( α n , q ( q - 1 ) / q n v n 1 / n - + Φ ( r ) a ( r , s ) 𝑑 r ) q / ( q - 1 ) - s ] 𝑑 s
R 0 + exp [ ( - + Φ ( r ) a ( r , s ) 𝑑 r ) q / ( q - 1 ) - s ] 𝑑 s
(6.2) R 0 + exp [ - F 1 ( s ) ] 𝑑 s ,

where

F 1 ( s ) = s - ( - + Φ ( r ) a ( r , s ) 𝑑 r ) q / ( q - 1 ) .

Since a(r,s)1 for a.e 0<r<s, and

- 0 a ( r , s ) q / ( q - 1 ) 𝑑 r + s + a ( r , s ) q / ( q - 1 ) 𝑑 r = n ( q - 1 ) ( n - 1 ) q < + ,

we have

- + Φ q ( r ) 𝑑 r = 0 + ( R 1 / n | u | ( R e - r ) e - r / n ) q 𝑑 r = 0 R [ | u | ( t ) t 1 / n ] q d t t u n , q q 1 .

Now using Lemma 3.1, we obtain

(6.3) 0 + exp [ - F 1 - β / n ( s ) ] 𝑑 s c 0 .

So by inequalities (6.1), (6.2) and (6.3), we have

I 1 C ( n , q ) R = C ( n , q ) u n , q n .

To prove the restriction α<αn,q is optimal, we can take the same uk(x) as in Section 5, i.e.,

u k ( x ) = { C k k ( q - 1 ) / q ( n v n 1 / n ) - 1 if 0 | x | ( 1 v n e - k ) 1 / n , C k k ( q - 1 ) / q ( n v n 1 / n ) - 1 - ln ( v n | x | n ) k if ( 1 v n e - k ) 1 / n < | x | ( 1 v n ) 1 / n , 0 if | x | > ( 1 v n ) 1 / n ,

where

C k = ( 1 k 1 e k ( 1 - 1 y ) q / n d y y - 1 ) - 1 / q .

By almost the same calculation, we can prove that the constant αn,q is sharp, so we omit the details. ∎

Award Identifier / Grant number: DMS-1301595

Award Identifier / Grant number: 11371056

Funding statement: The research of the first author was partly supported by a US NSF grant DMS-1301595 and a Simons Fellowship from the Simons Foundation. The research of the second author was partly supported by a Youth Scholars Program of Beijing Normal University (no. 2014NT30) and a grant of NNSF of China (no.11371056).

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Received: 2015-12-31
Revised: 2016-04-29
Accepted: 2016-05-03
Published Online: 2016-06-18
Published in Print: 2016-08-01

© 2016 by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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