A sharp inequality of Trudinger–Moser type and extremal functions in H1,n(Rn)

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ScienceDirect J. Differential Equations 258 (2015) 4062–4101 www.elsevier.com/locate/jde

A sharp inequality of Trudinger–Moser type and extremal functions in H 1,n(Rn) ✩ João Marcos do Ó ∗ , Manassés de Souza Department of Mathematics, Federal University of Paraíba, 58051-900 João Pessoa, PB, Brazil Received 10 January 2014; revised 21 January 2015 Available online 12 February 2015

Abstract We prove a sharp form of the Trudinger–Moser inequality for the Sobolev space H 1,n (Rn ). The sharpness comes from the introduction of an extra factor unn in the classical Trudinger–Moser inequality. Let  (α) :=

Φ ◦ να (u) dx,

sup {u∈H 1,n (Rn ):u1,n =1}

Rn

 ti n 1/(n−1) |u|n/(n−1) . The main results read: (1) for where Φ(t) := et − n−1 i=0 i! and να (u) := βn (1 + αun ) 0 ≤ α < 1 we have (α) < ∞, (2) for α > 1, (α) = ∞ and (3) moreover, we prove that for 0 ≤ α < 1, an extremal function for (α) exists. © 2015 Elsevier Inc. All rights reserved. MSC: 35J60; 35J92; 46E35; 46E30 Keywords: Limiting Sobolev inequalities; Trudinger–Moser inequality; Sharp constants; Extremal function; Blow-up analysis

✩ Research partially supported by the National Institute of Science and Technology of Mathematics INCT-Mat, CAPES and CNPq. * Corresponding author. Fax: +55 83 3216 7487. E-mail address: [email protected] (J.M. do Ó).

http://dx.doi.org/10.1016/j.jde.2015.01.026 0022-0396/© 2015 Elsevier Inc. All rights reserved.

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1. Introduction This paper is concerned with the problem of finding optimal Trudinger–Moser type inequality for the Sobolev space H 1,n (Rn ) and the existence of extremal functions. Sharp Trudinger–Moser inequality plays an important role in geometric analysis and partial differential equations and continues to be a source of inspiration to many researches in recent years. In order to motivate our work, let us present a brief history of the main results on this class of problems. Let Ω ⊂ Rn (n ≥ 2) be a smooth bounded domain. The classical Sobolev theorem states that the imbedding H01,n (Ω) → Lq (Ω) is continuous for all q : 1 ≤ q < ∞, but H01,n (Ω) → L∞ (Ω) as one can see by taking u(r) = ln(ln(4R/r)) for some R > 0 small enough such that B2R (0) ⊂ Ω, where (without loss of generality) we assume 0 ∈ Ω (cf. [3]). In this limiting case the optimal imbedding is an Orlicz space imbedding, which was proved by V.I. Yudovich [37], S.I. Pohozhaev [28], J. Peetre [27], N.S. Trudinger [35] and J. Moser [25]. More precisely, when Ω is a bounded domain, using the Dirichlet norm ∇un (equivalent to the Sobolev norm in H01,n (Ω)) they proved that 

n/(n−1)

eβ|u|

sup

dx ≤ Cn |Ω|,

(1.1)

{u∈H01,n (Ω):∇un =1} Ω 1/(n−1)

for any β ≤ βn := nωn−1 , where |Ω| denotes the Lebesgue measure of a set Ω in Rn and ωn−1 is the measure of the unit sphere in Rn . Moreover, βn is the best constant in the following sense: the integral on the left actually is finite for any positive β, but if β > βn it can be made arbitrarily large by an appropriate choice of u and the supremum is +∞. However, P.-L. Lions [23] proved that an inequality like (1.1) holds along certain sequences with a constant larger than βn ; more precisely, if (uk ) ⊂ H01,n (Ω), ∇uk n = 1 and uk u ≡ 0 in H01,n (Ω). Then  sup k

ep|uk |

n/(n−1)

dx < ∞

(1.2)

Ω

provided that p<

βn . (1 − ∇unn )1/(n−1)

To complete this analysis, the following results were proposed by Adimurthi and O. Druet [4] for the case n = 2 and by Y. Yang [36] for the case n ≥ 3:  {u∈H01,n (Ω):∇un =1}

n 1/(n−1) |u|n/(n−1)

eβn (1+αun )

sup

 dx

< ∞ if 0 ≤ α < λ1 (Ω) = ∞ if α ≥ λ1 (Ω)

Ω

where λ1 (Ω) = inf{∇unn : u ∈ H01,n (Ω) and un = 1}. The Trudinger–Moser inequalities for unbounded domains were proposed by D.M. Cao [7] for the case n = 2 and J.M. do Ó [14] and R. Panda [26] for general case n ≥ 2. Precisely, if

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u ∈ H 1,n (Rn ), ∇un ≤ 1, and un ≤ M, then for β < βn , there exists a constant C(M, β) > 0 such that 

  Ψ β|u|n/(n−1) dx ≤ C(M, β),

Rn

 ti where Ψ (t) := et − n−2 i=0 i! . See also S. Adachi and K. Tanaka [2] for a related result. All these results treat the subcritical case β < βn , later B. Ruf [29] for the case n = 2, Y. Li and B. Ruf [21] and Adimurthi and Y. Yang [1] for the general case n ≥ 3 studied the critical case β = βn and proved that the result of J. Moser can be fully extended to unbounded domains if the Dirichlet norm ∇un is replaced by the full Sobolev norm u1,n := (∇unn + unn )1/n . More precisely, one has: 



sup

Ψ β|u|

{u∈H 1,n (Rn ):u1,n =1}

Rn

n/(n−1)



 dx

<∞ =∞

if β ≤ βn if β > βn .

In recent work J.M. do Ó et al. [12] proved a result on the weak compactness of the Trudinger– Moser functional defined in H 1,n (Rn ), which is a version of the Concentration–Compactness Principle due to P.-L. Lions (1.2) for unbounded domains. More precisely, if (uk ) is a sequence in H 1,n (Rn ) with uk 1,n = 1 and supposing that uk u ≡ 0 in H 1,n (Rn ), then for all 0 < p < βn (1 − un1,n )−1/(n−1) we have  sup k

  Ψ p|uk |n/(n−1) dx < ∞.

(1.3)

Rn

Another interesting question about Trudinger–Moser inequalities is whether extremal function exists or not. The first result in this direction belongs to Carleson and Chang [8] who proved that if Ω ⊂ Rn is the ball B1 (0), then the supremum in (1.1) is achieved when β ≤ βn . Later, M. Struwe [32] studied the existence of extremal functions for a class of nonsymmetric domains. He obtained a sufficient condition for this class of domains in R2 using blow-up analysis. M. Flucher [17] introduced another method, the conformal rearrangement, and derived an isoperimetric inequality, which implies the existence of extremal functions to any smooth bounded domain in R2 . At last, K. Lin [22] generalized the existence of extremal function to any smooth bounded domain in Rn (n ≥ 2). It should be mentioned that the existence of extremal for (1.1) corresponds to the existence of solutions to an associated Euler–Lagrange equation involving critical growth. Thus, this class of problems is harder or more difficult than subcritical ones and the lack of compactness makes the proofs more involved, as one can see in very intricate analysis given in the papers [8,17,32]. For works related to (1.1) and applications, we refer to [5,9,10,15,16,33] and references therein. 1.1. Statement of main results In the present paper, based on the concentration–compactness (1.3), our first result can be stated as follows.

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Theorem 1.1. For any α ∈ [0, 1), we have  (α) :=

Φ ◦ να (u) dx < ∞,

sup {u∈H 1,n (Rn ):u1,n =1}

(1.4)

Rn

where Φ(t) := et −

n−1 i  t i=0

and

i!

 1/(n−1) n/(n−1) |u| . να (u) := βn 1 + αunn

Moreover, this estimate is sharp: for any α ∈ (1, +∞), the supremum in (1.4) is +∞. As an immediate consequence, we obtain Corollary 1.1. For any α ∈ [0, 1)  Ψ ◦ να (u) dx < ∞.

sup {u∈H 1,n (Rn ):u1,n =1}

Rn

Furthermore, on the attainability of (α) we shall prove. Theorem 1.2. For any α ∈ [0, 1), the supremum (α) is attained, that is, there exists u ∈ H 1,n (Rn ) such that u1,n = 1 and  (α) =

Φ ◦ να (u) dx. Rn

We emphasize that the result stated in Corollary 1.1 fully extends [4, Theorem 1] and [36, Theorem 1.1] for the space H 1,n (Rn ), while Theorem 1.2 partially extends [36, Theorem 1.2] because here we consider in Theorem 1.2 the modified function Φ(t) which is obtained for Ψ (t) by subtraction of the term corresponding to the Ln -norm. This allows us to gain the compactness necessary to prove the attainability of the supremum in (1.4). Moreover, we complement the Trudinger–Moser inequality proved by B. Ruf [29] for the case n = 2, Y. Li and B. Ruf [21] and Adimurthi and Y. Yang [1] for n ≥ 3. It should be mentioned that similar results have been obtained by the authors in [11] on the whole plane for the space  EV := u ∈ H

1,2



R

2





 :

V (x)u dx < ∞ 2

R2

when the potential V (x) is coercive and radially symmetric. It is well known that under these hypotheses the embedding EV → Ls (R2 ) is compact for all 2 ≤ s < ∞. We organize this paper as follows: In Section 2, we prove the existence of maximizers for a sequence of subcritical Trudinger–Moser energy functionals and consequently we obtain a maximizing sequence of radially symmetric and radially decreasing functions for the supremum (α) when α ∈ [0, 1). In Section 3 we prove that the supremum 1.4 is finite and attained for α ∈ [0, 1).

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The proof is established using the method of blow-up analysis combined with a Lions-type concentration compactness lemma and Carleson–Chang procedure. In Section 4, we construct a sequence of functions to show that for any α ∈ (1, +∞), the supremum in (1.4) is +∞, and list some open questions. 2. Maximizers for subcritical Trudinger–Moser energy functionals In this section we prove the existence of maximizers for a sequence of subcritical Trudinger– Moser energy functionals and consequently we obtain a maximizing sequence of radially symmetric and radially decreasing functions for the supremum (α) when α ∈ [0, 1). Proposition 2.1. For any α ∈ [0, 1), there exists a maximizing sequence for (α), that is, (uk ) ⊂ H 1,n (Rn ) with uk 1,n = 1 and  (α) = lim

k→+∞

Φ ◦ να,δk (uk ) dx,

(2.1)

Rn

where

1/(n−1)  |uk |n/(n−1) να,δk (uk ) := (βn − δk ) 1 + αuk nn δk  0 as k → +∞.

Moreover, we can choose uk radially symmetric and radially decreasing. We have divided the proof of Proposition 2.1 into a sequence of lemmata. Let Ω be a smooth bounded domain in Rn and consider the space H01,n (Ω) endowed with the Sobolev norm 1/n  uΩ = ∇unn + unn . Denote by μ1 (Ω) the first eigenvalue of the problem 

 −n u + un−1 , H01,n (Ω)

with μ1 (Ω) =

unΩ , n u∈H 1,n (Ω)\{0} uLn (Ω) inf

(2.2)

0

where n u = div(|∇u|n−2 ∇u) is the n-Laplacian and uLn (Ω) = ( Ω |u|n dx)1/n . Note that μ1 (Ω) ≥ 1. Next we prove that the supremum of the Trudinger–Moser functional in the unit ball of H01,n (Ω) is finite and attained in the subcritical case. Theorem 2.1. For any α ∈ [0, μ1 (Ω)) and δ ∈ (0, βn ), we have  Lα,δ (Ω) =

sup {u∈H01,n (Ω):uΩ =1} Ω

Φ ◦ να,δ (u) dx < ∞,

(2.3)

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where 1/(n−1) n/(n−1)  να,δ (u) := (βn − δ) 1 + αunn |u| . Moreover, (2.3) is attained, that is, there exists a positive function u ∈ C 1,θ (Ω) such that  uΩ = 1

and

Lα,δ (Ω) =

Φ ◦ να,δ (u) dx. Ω

Proof. We first observe that the regularity for any extremal function of (2.3) follows from elliptic regularity theory applied to the associated Euler–Lagrange equation. Next, we prove the attainability of (2.3). Let (uj ) ⊂ H01,n (Ω) be a maximizing sequence for (2.3), that is, uj Ω = 1 and  Φ ◦ να,δ (uj ) dx = Lα,δ (Ω).

lim

j →+∞ Ω

We may assume uj ≥ 0, because Φ ◦ να,δ (u) = Φ ◦ να,δ (|u|) for all u ∈ H01,n (Ω). Since uj Ω = 1, taking subsequence, we have uj u

weakly in H01,n (Ω),

uj → u

almost everywhere in Ω,

uj → u

strongly in Ln (Ω).

Assertion 1. The function u is an extremal for Lα,δ (Ω). It is clear that Assertion 1 follows if we have uΩ = 1. Suppose not, that is, uΩ < 1. Since 0 ≤ α < μ1 (Ω) and 1 + ξ ≤ 1/(1 − ξ ) if 0 ≤ ξ < 1, by using the definition of μ1 (Ω), we obtain 1 + αunLn (Ω) < 1 + μ1 (Ω)unLn (Ω) ≤ 1 + unΩ ≤

1 . 1 − unΩ

Thus,  1/(n−1) lim (βn − δ) 1 + αuj nLn (Ω) <

j →+∞

βn (1 − unΩ )1/(n−1)

and choosing q > 1 sufficiently close to 1 satisfying 1/(n−1)  q (βn − δ) 1 + αunLn (Ω) <

βn , (1 − unΩ )1/(n−1)

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by (1.3), we obtain 

  Φ ◦ να,δ (uj )q dx ≤

Ω



  1/(n−1)  Φ q (βn − δ) 1 + αuj nLn (Ω) |uj |n/(n−1) dx ≤ C(q),

Ω

which together with Hölder’s inequality implies  uj Φ ◦ να,δ (uj ) dx ≤ C1 .

(2.4)

Ω

Since Φ ◦να,δ (u) ∈ L1 (Ω), given  > 0, there exists a δ > 0 such that for any A ⊂ Ω measurable,  Φ ◦ να,δ (u) dx ≤ 

if |A| ≤ δ.

(2.5)

A

Moreover, since u ∈ L1 (Ω), we can find M1 > 0 such that 

  x ∈ Ω : u(x) ≥ M1  ≤ δ.

(2.6)

Taking M = max{M1 , C1 /}, we can write       Φ ◦ να,δ (uj ) dx − Φ ◦ να,δ (u) dx  ≤ I j + I j + I j , 1 2 3   Ω

Ω

where 

j

I1 =

Φ ◦ να,δ (uj ) dx

{x∈Ω:uj (x)≥M}

j I2

  = 



 Φ ◦ να,δ (uj ) dx −

{x∈Ω:uj (x)
{x∈Ω:|u(x)|


I3 =

  Φ ◦ να,δ (u) dx 

Φ ◦ να,δ (u) dx.

{x∈Ω:u(x)≥M}

By (2.4), j I1

 =

Φ ◦ να,δ (uj ) dx

{x∈Ω:uj (x)≥M}



= {x∈Ω:uj (x)≥M}

uj Φ ◦ να,δ (uj ) C1 dx ≤ ≤ . uj M

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Thanks to the dominated convergence theorem, j I2

  = 



  Φ ◦ να,δ (u) dx  → 0.

 Φ ◦ να,δ (uj ) dx −

{x∈Ω:uj (x)
{x∈Ω:u(x)
Using (2.5) and (2.6), j I3

 =

Φ ◦ να,δ (u) dx ≤ .

{x∈Ω:u(x)≥M}

Therefore, from the above estimates we can conclude that  Φ ◦ να,δ (u) dx = Lα,δ (Ω).

(2.7)

Ω

On the other hand, since uΩ < 1 we obtain   unLn (Ω) 1/(n−1) un/(n−1) 1/(n−1) n/(n−1)  (βn − δ) 1 + αunLn (Ω) u < (βn − δ) 1 + α , n/(n−1) unΩ u Ω

which implies 

 Φ ◦ να,δ (u) dx <

Ω

Ω

    unLn (Ω) 1/(n−1) un/(n−1) Φ (βn − δ) 1 + α dx. n/(n−1) unΩ u Ω

This leads to a contradiction with (2.7), which completes the proof. 2 For easy reference we introduce the notation:  S(Rk ) = v

 ∈ H01,n (BRk ) :



|∇v| + |v| n

n



 dx = 1 ,

BRk

where Rk  +∞ as k → +∞ and BRk is the ball of Rn centered at origin with radius Rk . Since any function in S(Rk ) can be extended by zero outside of BRk , we have S(Rk ) ⊂ S(Rk+1 ) ⊂ H 1,n (Rn ). Thus, Lα,δk (BRk ) ≤ Lα,δk+1 (BRk+1 ) where δk  0. Therefore, using that μ1 (BRk ) ≥ 1 and taking Ω = BRk in Theorem 2.1, we can state: Corollary 2.1. For α ∈ [0, 1) and k ∈ N, there exists uk ∈ S(Rk ) ∩ C 1,θ (B Rk ) positive such that  Lα,δk (BRk ) =

 Φ ◦ να,δk (uk ) dx =

BRk

Moreover, (Lα,δk (BRk ))k∈N is an increasing sequence.

Φ ◦ να,δk (uk ) dx. Rn

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2.1. Proof of Proposition 2.1 Let η ∈ C0∞ (Rn , [0, 1]) be a cut-off function such that  η(x) =

if x ∈ B1 , if x ∈ Rn \ B2 .

1, 0,

Then, given any ϕ ∈ H 1,n (Rn ) with ϕ1,n = 1, we have 

      ∇ η(x/L)ϕ n + η(x/L)ϕ n dx → 1,

τ n (L) :=

as L → +∞.

(2.8)

Rn

First note that, using an argument of density, it is sufficient to prove that (2.8) holds for all ϕ ∈ C0∞ (Rn ). Let R > 0 such that supp(ϕ) ⊂ BR , and then, taking L > R, we get τ (L) = ϕ1,n = 1. ϕ Next, for a fixed L and Rk > 2L, using that η( L· ) τ (L) 1,n = 1 and choosing a nonnegative function uk ∈ S(Rk ) ∩ C 1,θ (B Rk ) an extremal function for Lα,δk (BRk ) (cf. Corollary 2.1), we get 

 Φ ◦ να,δk

      ϕ ϕ x Φ ◦ να,δk η dx ≤ dx τ (L) L τ (L)

BL

B2L



≤

Φ ◦ να,δk (uk ) dx,

BRk

which together with the Fatou Lemma implies 

 Φ ◦ να

  ϕ Φ ◦ να,δk (uk ) dx. dx ≤ lim inf k→+∞ τ (L) Rn

BL

Then, letting L → +∞, 

 Φ ◦ να (ϕ) dx ≤ lim inf k→+∞

Rn

Φ ◦ να,δk (uk ) dx. Rn

Hence  (α) ≤ lim inf k→+∞

Φ ◦ να,δk (uk ) dx Rn

which together with  Φ ◦ να,δk (uk ) dx ≤ (α)

lim sup k→+∞

implies that (2.1) holds.

Rn

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If u∗k denotes the Schwarz symmetrization of u, then by a well-known theorem of Polya and Szegö, 



 ∗ n  ∗ n  ∇u  + u  dx ≤

τkn :=

k



 |∇uk |n + |uk |n dx = 1.

k

BRk

(2.9)

BRk

Since 

  Φ ◦ να,δk u∗k dx =

BRk

 Φ ◦ να,δk (uk )dx,

(2.10)

BRk

to complete the proof of Proposition 2.1, it is sufficient to show that τk = 1. Using (2.9) and (2.10), we get 

 Φ ◦ να,δk (uk /τk ) dx ≥

BRk

Φ ◦ να,δk (uk ) dx.

(2.11)

Φ ◦ να,δk (uk ) dx.

(2.12)

BRk

Since u∗k /τk ∈ S(Rk ), we have 

 Φ ◦ να,δk (uk /τk ) dx ≤

BRk

BRk

From (2.10), (2.11) and (2.12), we get 

 Φ ◦ να,δk (uk /τk ) dx =

BRk

  Φ ◦ να,δk u∗k dx,

BRk

which implies that τk = 1 and then, we can assume that uk is radially symmetric and radially decreasing function. 3. Blow-up analysis In this section, by using blow-up analysis, we analyze the behavior of the maximizers uk described in Proposition 2.1 for any α ∈ [0, 1). The proof is inspired by the previous works [4,8, 11,21,36] and our analysis is composed of several lemmas. Let (uk ) ⊂ H 1,n (Rn ) be a maximizing sequence for (α) as in Proposition 2.1; we may assume that uk u in H 1,n (Rn ). Next we state the main result of this section. Proposition 3.1. The function u is an extremal for (α), that is,  (α) = lim

k→+∞

 Φ ◦ να,δk (uk ) dx =

Rn

Φ ◦ να (u) dx. Rn

(3.1)

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It is clear that (3.1) follows if u1,n = 1. Thus, from now on we assume that u1,n < 1. In this case, the proof of (3.1) is very involved and will be done is several steps in the next subsections. For simplicity of notation, we write ck = uk (0)

(maximum point).

(3.2)

We have the following two cases to analyze: (a) (ck ) is a bounded sequence or (b) ck  +∞ (up to a subsequence). One can see that this maximizing sequence (uk ) for (α) satisfies the following Euler–Lagrange equation: ⎧ βk 1/(n−1)   n/(n−1)  ⎪ n−1 ⎪ + γk un−1 Φ αk uk ⎨ −n uk + uk = uk k λk u  = 1, u > 0 ⎪ k 1,n k ⎪ ⎩ uk = 0

in BRk , in BRk , on ∂BRk .

(3.3)

For easy reference let us introduce the notations ⎧   n 1/(n−1) ⎪ , ⎪ ⎪ αk = (βn − δk ) 1 + αuk n ⎪ n ⎪ ⎪  1 + αu k ⎪β = n ⎪ , k ⎪ ⎪ 1 + 2αuk nn ⎪ ⎨ α , γk = ⎪ 1 + 2αuk nn ⎪ ⎪  ⎪ ⎪ ⎪ n/(n−1)   n/(n−1)  ⎪ ⎪ α dx. = u Φ u λ k k ⎪ k k ⎪ ⎪ ⎩ B

(3.4)

Rk

First, we prove that the sequence (λk ) is bounded away from zero, more precisely: Lemma 3.1. infk λk > 0. Proof. Assume by contradiction that λk → 0. Thus, by using the elementary inequality Φ(t) ≤ Φ  (t)t for t ≥ 0, we have  lim

k→+∞

 n/(n−1)  Φ αk uk dx ≤ βn (1 + α)1/(n−1) lim



k→+∞

BRk

which implies that (α) = 0, which is impossible.

n/(n−1)

uk BRk

2

Lemma 3.2. The sequence (uk ) converges strongly to u in Ln (Rn ).

 n/(n−1)  Φ  αk uk dx = 0,

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Proof. It is sufficient to prove that there exists C > 0 such that  |uk |n dx ≤ C/R,

(3.5)

Rn \BR

for R > 0 sufficiently large. Let ψR ∈ C ∞ (Rn , [0, 1]) such that |∇ψR |∞ ≤ C/R and  ψR (x) =

1, if x ∈ Rn \ B2R ; 0, if x ∈ BR .

Using ϕ = uk ψR as a test function in (3.3), we have 

 ψR |∇uk | dx +

Rn

=

βk λk

 uk |∇uk |

n

n−2

∇uk ∇ψR dx +

Rn



n/(n−1)

ψR uk

 n/(n−1)  dx + γk Φ  αk uk

Rn



ψR |uk |n dx R2

ψR |uk |n dx. Rn

Thus,  (1 − γk ) Rn

C βk ψR |uk | dx ≤ uk n ∇uk n−1 + n R λk



n/(n−1)

n

ψR uk

  Φ  αk u2k dx.

(3.6)

Rn

On the other hand, note that 

n/(n−1)

ψR uk

j  ∞   αk n/(n−1)  dx = Φ  αk uk ψR |uk |n(j +1)/(n−1) dx. j! j =n−1

Rn

(3.7)

Rn

By using the Radial Lemma (cf. [6] or [31]), we can write   uk (ρ) ≤



n

1/n

ωn−1

uk n

1 Cn ≤ , ρ ρ

for all ρ > 0.

Hence, 

n(j +1)/(n−1)

|uk |

dx

(j +1)/(n−1) ≤ Cn ωn−1

Rn \BR

+∞ n(j +1) ρ (n−1)− (n−1) dρ.

(3.8)

R

But, for R > 1 we have +∞ n(j +1) ρ (n−1)− (n−1) dρ = R

1 n(j +1) n−1

·

−n R

1 n(j +1) n−1 −n

≤

1 . R

(3.9)

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Consequently, from (3.7), (3.8) and (3.9), we get 

n/(n−1)

ψR uk Rn

 C n/(n−1)  dx ≤ . Φ  αk uk R

Combining (3.4), (3.6) and (3.10), we obtain that (3.5) holds.

(3.10)

2

3.1. Case (a): (ck ) is a bounded sequence The aim of this subsection is to prove the first part of Theorem 1.1 if α ∈ [0, 1) and Theorem 1.2, in the case that (ck ) is bounded, which are consequences of the following lemma. Lemma 3.3. Suppose (ck ) is bounded, then (α) is attained. Proof. Setting      n/(n−1)  − Φ ◦ να (u) dx  Ik :=  Φ αk uk Rn

we obtain the estimate      n/(n−1)  − Φ ◦ να (u) dx  Ik ≤  Φ αk uk BL

   + 

Φ



 

  n/(n−1)  dx  +  αk uk



Rn \BL

  Φ ◦ να (u) dx .

Rn \BL

1,θ Since (ck ) is bounded, standard elliptic estimates yield uk → u in Cloc (Rn ) for some θ ∈ (0, 1). Thus, using that un = limk→∞ uk n and uniform convergence, we obtain

       Φ αk un/(n−1) − Φ ◦ να (u) dx  → 0. k  

(3.11)

BL

Now, we recall that  2  lim Φ αk t n/(n−1) t −n /(n−1) < ∞.

t→0+

Since uk is radially decreasing function, we have  unk (L)|BL | ≤ BL

unk dx ≤ 1.

(3.12)

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

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Thus, we can find L > 0 such that uk (x) ≤  for all x ∈ Rn \ BL , and using (3.12), 



 n/(n−1)  dx ≤ C Φ αk uk

Rn \BL

n2 /(n−1) uk dx

 ≤

Rn \BL

n2

unk dx ≤ C n−1 −n .

(3.13)

Rn

On the other hand, for L > 0 sufficiently large, we have   Φ ◦ να (u) dx  ≤ .

    

(3.14)

Rn \BL

From (3.11), (3.13) and (3.14), and taking  sufficiently small, we obtain        Φ αk un/(n−1) − Φ ◦ να (u) dx  → 0. k  

(3.15)

Rn

Thus, u is an extremal function for (α).

2

3.2. Case (b): ck  +∞ The aim of this section is to use blow-up analysis to study the case (b): ck  +∞ as k → ∞.

(3.16)

In order to perform a blow-up procedure, we define rkn =

λk n/(n−1) αk cn/(n−1) βk ck e k

(3.17)

and σk = βn − δk , where δk  0 (see Proposition 2.1). Lemma 3.4. Assume that (3.16) holds, then rkn e

σk n/(n−1) 2 ck

→ 0, and consequently rk → 0.

Proof. By the definition of rk and λk , we have rkn e

σk n/(n−1) 2 ck

−n/(n−1) −(αk − σ2k )ckn/(n−1) = βk−1 ck e



n/(n−1)

uk

 n/(n−1)  dx. Φ  αk uk

Rn

Given L > 0, we can write rkn e

σk n/(n−1) 2 ck

−n/(n−1) −(αk − σ2k )ckn/(n−1) = βk−1 ck e



n/(n−1)

uk BL

−n/(n−1) −(αk − σ2k )ckn/(n−1) + βk−1 ck e



 n/(n−1)  dx Φ  αk uk

n/(n−1)

uk Rn \BL

 n/(n−1)  dx. (3.18) Φ  αk uk

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J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

Note that 

n/(n−1)

uk



j ∞   αk n/(n−1)  dx = Φ  αk uk j! j =n−1

Rn \BL

|uk |n(j +1)/(n−1) dx.

(3.19)

Rn \BL

Again by the Radial Lemma (cf. [6] and [31]), we have 

  uk (ρ) ≤

1/n

n ωn−1

1 uk n , ρ

for all ρ > 0,

which together with (3.19) implies 

n/(n−1)

uk

 n/(n−1)  dx ≤ C1 (L). Φ  αk uk

(3.20)

Rn \BL

Now, using the fact that     σk n/(n−1) σk n/(n−1) − αk − ≤ − αk − (x) for all x ∈ Rn , ck uk 2 2 we get 

n/(n−1) −(αk − σ2k )unk /(n−1) αk un/(n−1) uk e e k

 dx =

BL

Since

BL

n/(n−1)

uk

e

σk 2

n/(n−1)

uk

dx.

BL

|∇(uk − uk (L))+ | dx ≤ 1, using the classical Trudinger–Moser inequality, we get 

eβn [(uk −uk (L))

+ ]n/(n−1)

dx ≤ C(L).

BL

Note that for any p < βn , there exists C(p) > 0 such that n/(n−1)

puk

 + n/(n−1) ≤ βn uk − uk (L) + C(p),

which together with (3.21) implies that 

n/(n−1)

epuk

dx ≤ C(L, p).

BL

Therefore, there exists C2 = C2 (L) > 0 such that 

n/(n−1)

uk BL

e

σk 2

n/(n−1)

uk

dx ≤ C2 .

(3.21)

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

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Consequently, 

n/(n−1)

uk

 n/(n−1)  dx ≤ C2 , Φ  αk uk

BL

which together with (3.18) and (3.20) implies rkn e

σk n/(n−1) 2 ck

−n/(n−1)

≤ C2 (L)βk−1 ck

−n/(n−1) −(αk − σ2k )ckn/(n−1)

+ C1 (L)βk−1 ck

Since βk is bounded and ck  +∞, the lemma follows.

e

.

2

Lemma 3.5. Suppose (3.16) holds, then uk 0 weakly in H 1,n (Rn ). Moreover, we have αk → βn , βk → 1 and γk → α. Proof. We know that uk u weakly in H 1,n (Rn ) and by Lemma 3.2, uk → u strongly in Ln (Rn ). Suppose u ≡ 0, then using that 0 ≤ α < 1 we have 1 + αuk nn → 1 + αunn ≤ 1 + un1,n <

1 . 1 − un1,n

By (1.3), Φ  (αk uk ) is bounded in Lr (Rn ) for some r > 1 sufficiently close to 1, which together with (3.3) implies that n uk is bounded in Lr0 (BR ) for some r0 > 1 sufficiently close to 1 and R > 0. Employing the elliptic estimates to (3.3), one gets that (uk ) is uniformly bounded, which contradicts (3.16). Therefore u ≡ 0, and consequently by (3.4), αk → βn , βk → 1 and γk → α. 2 n/(n−1)

In order to continue our analysis, we define the blow-up functions

ψk (x) = ck−1 uk (rk x)  1/(n−1)  ϕk (x) = ck uk (rk x) − ck

(3.22)

where ψk and ϕk are defined on Ωk = {x ∈ Rn : rk x ∈ B1 }. By (3.3) and a direct computation we get ⎧ n/(n−1)   (rk x)−ck ) ⎨ − ψ (x) = c−n ψ 1/(n−1) eαk (un/(n−1) k + O rkn ckn in Ωk n k k k n/(n−1)   ⎩ − ϕ (x) = ψ 1/(n−1) eαk (un/(n−1) (rk x)−ck ) k in Ωk . + O rkn ckn n k k

(3.23)

Consequently, by Lemmas 3.4 and 3.5, and using similar argument to due Y. Yang [36, Section 4] and elliptic estimates (see [30,34]), we have

1,θ  n  R , ψk → 1, in Cloc 1,θ  n  ϕk → ϕ, in Cloc R ,

(3.24)

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and ϕ satisfies 

n

−n ϕ = e n−1 βn ϕ in Rn ϕ(0) = 0 = sup ϕ.

(3.25)

Moreover, we have ⎧     ⎪ ωn−1 1/(n−1) n/(n−1) n−1 ⎪ ⎪ |x| ⎪ ⎨ ϕ(x) = − βn log 1 + n  n ⎪ ⎪ e n−1 βn ϕ dx = 1. ⎪ ⎪ ⎩

(3.26)

Rn

In the next lemmas we use the notation uA k = min{uk , ck /A}

for A > 1.

Lemma 3.6. Assume that (3.16) holds, then   A n  A n  ∇u  + u  dx ≤ 1 . lim sup k k A k→∞

(3.27)

Rn

Proof. To obtain (3.27), we write   A n  A n  ∇u  + u  dx = 1 − J1,k + J2,k − J3,k + J4,k k k

(3.28)

Rn

where            ck + n ck + n−1  J1,k := uk dx  + uk − A ∇ uk − A Rn

   ck + n−1 J2,k := uk dx uk − A Rn



J3,k :=

unk dx

[uk >ck /A]



J4,k :=

 A n u  dx. k

[uk >ck /A]

In order to estimate Ji,k for i = 1, . . . , 4, we note that   n    x ∈ Rn : uk (x) ≥ ck  ck =  A  A



[uk ≥ck /A]



ck A

n

 dx ≤ [uk ≥ck /A]

unk dx ≤ 1,

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

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which implies that   n    x ∈ Rn : uk (x) ≥ ck  ≤ A → 0 as k → ∞. n   A ck Thus, we can find a sequence ρk  0 such that 

ck x ∈ R : uk (x) ≥ A n

 ⊂ Bρk . p

Using that uk u weakly in H 1,n (Rn ), then uk converges strongly in Lloc (Rn ) for any p > 1. Therefore, for any R > 0, there exists h ∈ Lp (BR ) such that uk (x) ≤ |h(x)| almost every in BR . Hence, by the Lebesgue dominated convergence theorem, we have  lim

k→∞

 A p u  dx = 0 k

[uk >ck /A]

and  k→∞

p

uk dx = 0.

lim

[uk >ck /A]

Therefore, J2,k → 0, J3,k → 0 and J4,k → 0. Now, testing Eq. (3.3) with (uk − ck /A)+ and setting Lk = rk L, we obtain   J1,k =

uk − Rn

ck A

 

≥

uk −

+ 

ck A

 βk 1/(n−1) αk un/(n−1) uk e k dx + ok (1) λk

+ 

 βk 1/(n−1) αk un/(n−1) uk e k dx + ok (1), λk

BLk

where ok (1) → 0 as k → ∞. Performing the change of variables x = rk y, we get   BLk

ck uk − A

+ 

 βk 1/(n−1) αk un/(n−1) u e k dx λk k

     n/(n−1) ck + βk 1/(n−1) αk uk (rk y) = u (rk y)e uk (rk y) − rkn dy A λk k BL

     n/(n−1) n/(n−1) ck + ck−1 1/(n−1) αk (uk (rk y)−ck ) ψ (y)e ck ψk (y) − rkn dy, = A rkn k BL

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J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

which implies that 

Aψk (x) − 1 1/(n−1) n αk ϕk (1+Ok ((ψk −1)2 )) e n−1 dx + ok (1). ψk A

J1,k ≥ BL

This inequality together with (3.24) gives 

A−1 lim inf J1,k ≥ k→∞ A

n

e n−1 βn ϕ dx. BL

Letting L → +∞, we get lim inf J1,k ≥ k→∞

A−1 . A

(3.29)

Using that J2,k → 0, J3,k → 0, J4,k → 0, (3.28) and (3.29), we get  lim sup k→∞

   A 2  A 2  ∇u  + u  dx ≤ 1 − 1 − 1 + ok (1) = 1 + ok (1). k k A A

R2

2

Hence, the proof is complete.

Corollary 3.7. Assume that (3.16) holds, then for any c > 0, up to a subsequence, 



 |∇uk |n + |uk |n dx = 0 .

lim

k→+∞

[uk ≤c]

Proof. Letting c > 0, using that ck  +∞ and uA k = min{uk , ck /A}, there exists k0 ∈ N such that ck /A > c for all k ≥ k0 , thus uA (x) = u (x) in {x ∈ Rn : uk (x) ≤ c} for all k ≥ k0 , and k k consequently 

|∇uk | + |uk |

lim

k→+∞



n

n



 dx = lim

k→+∞

[uk ≤c]

[uk ≤c]

 A n  A n  ∇u  + u  dx ≤ 1 , k k A

where in the last estimate we have used (3.27). Letting A → +∞, then for any c, we have  lim

k→+∞

and the proof is complete.

2

[uk ≤c]



 |∇uk |n + |uk |n dx = 0,

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

4081

Remark 3.1. Using that uk is radially symmetric and radially decreasing we see that

x ∈ Rn : uk (x) ≤ c = Rn \ Br(c) ,

for some r = r(c) > 0. Thus, assuming (3.16), for any δ > 0 it holds 



 |∇uk |n + |uk |n dx = 0.

lim

k→+∞

Rn \Bδ

Lemma 3.8. Suppose (3.16) holds, then  lim

k→+∞



 n/(n−1)  dx ≤ lim Φ αk uk



L→+∞ k→∞

Rn

n/(n−1)

eαk uk

lim

 − 1 dx

BLk

≤ lim sup k→∞

λk n/(n−1) ck

(3.30)

and consequently λk → +∞ and ck

n/(n−1)

sup

ck

< ∞.

λk

k

(3.31)

Proof. Notice that   n/(n−1)  dx Φ αk uk Rn



 n/(n−1)  dx + Φ αk uk

= [uk ≤ck /A]

 ≤ Rn

Thus, we have 



 n/(n−1)  dx Φ αk uk

[uk >ck /A]

  λk An/(n−1) n/(n−1) dx + n/(n−1) Φ αk |uA k| ck

 n/(n−1)  dx ≤ Φ αk uk

Rn





n/(n−1)

uk

Rn

Φ  (αk uk λk

n/(n−1)

  n/(n−1)   dx + An/(n−1) Φ αk uA k

Rn

)

dx.

λk . n/(n−1) ck

(3.32)

Since uk is radial and decreasing, we can find L such that uk ≤ 1 on Rn \ BL , then uA k = uk on Rn \ BL . Hence by Corollary 3.7 and Radial Lemma (cf. [6] and [31]), given p > 0 there exists C(p) > 0 such that  lim

k→∞

Rn \BL

 n/(n−1)    dx ≤ lim C(p) Φ pαk uA k k→∞



Rn \BL

unk dx = 0.

(3.33)

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By Lemma 3.6, we have  lim sup k→∞

Rn

 A n  A n  ∇u  + u  dx ≤ 1 < 1 k k A

which together with Trudinger–Moser inequality (see (1.1)) implies  sup k

  + n/(n−1)  dx < ∞ for any p  < A1/(n−1) . Φ p  αk uA k − uk (L)

BL

Indeed, for p  < A1/(n−1) we have  + n/(n−1)  lim p  αk  uA < βn . k − uk (L) 1,n

k→∞

Consequently, we get 

 n/(n−1)    dx < ∞ Φ pαk uA k

sup k

for any p < A,

BL

since for any p < p   n/(n−1)   + n/(n−1)    p uA ≤ p   uA + C p, p  . k k − uk (L) n Now, using that uA k → 0 almost everywhere in R , by the same argument as in the proof of [11, Theorem 1], we get

 lim

k→∞

 n/(n−1)    dx = 0. Φ pαk uA k

(3.34)

BL

Hence from (3.32), (3.33) and (3.34), we get  lim sup k→∞

 n/(n−1)  dx ≤ lim sup An/(n−1) Φ αk uk k→∞

Rn

λk . n/(n−1) ck

Letting A → 1, we see that (3.30) holds. Now, if (λk /ck ) is bounded, since ck  +∞, then by (3.30), (α) ≤

λk n/(n/(n−1)) ck

→ 0, n/(n−1)

which is impossible. Using the same argument we can see that supk ck

/λk < ∞.

2

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

Lemma 3.9. Suppose (3.16) holds, then we have  lim

η

k→∞

Rn

βk 1/(n−1)   n/(n−1)  dx = η(0) ck uk Φ αk uk λk

  for any η ∈ C0∞ Rn .

Proof. Suppose supp(η) ⊂ Bρ and fix A > 1. Setting Lk = rk L, we can estimate  η

βk 1/(n−1)   n/(n−1)  dx ≤ I1,k + I2,k + I3,k , ck uk Φ αk uk λk

Bρ

where  I1,k =

η(x)

[uk ≥ck /A]\BLk



I2,k = BLk

η(x)

βk 1/(n−1)   n/(n−1)  dx, ck uk Φ αk uk λk



I3,k =

βk 1/(n−1)   n/(n−1)  dx, ck uk Φ αk uk λk

η(x)

[uk
βk 1/(n−1)   n/(n−1)  dx. ck uk Φ αk uk λk

To complete the proof, it is enough to show that: lim I1,k = lim I3,k = 0

k→∞

k→∞

and

lim I2,k = η(0).

k→∞

First we note that  |I1,k | ≤ η∞ [uk ≥ck /A]\BLk

Using that λk =



n/(n−1)

BRk

uk

Φ  (αk uk

n/(n−1)

  |I1,k | ≤ Aβk η∞ 1 −

βk 1/(n−1)   n/(n−1)  dx. ck uk Φ αk uk λk

) dx, we obtain  1 n/(n−1)   n/(n−1)  dx , uk Φ αk uk λk

BLk

which together with (3.17) and (3.22) implies 

1 n/(n−1)   n/(n−1)  dx uk Φ αk uk λk

BLk

 = BL

  1 n/(n−1) n/(n−1) uk (rk x)Φ  αk uk (rk x) rkn dx λk

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J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

1 = βk

 

uk (rk x) ck

n/(n−1)

n/(n−1)

eαk (uk

n/(n−1)

(rk x)−ck

)

dx + ok (1)

BL

=

1 βk



n/(n−1)

ψk

n

2

(y)e n−1 αk ϕk (y)(1+O(ψk (y)−1) ) dy + ok (1).

BL

Since βk → 1, letting k → ∞ and using (3.24), we get    n |I1,k | ≤ Aη∞ 1 − e n−1 βn ϕ dx , BL

which implies that    n lim |I1,k | ≤ Aη∞ 1 − e n−1 βn ϕ dx .

k→∞

BL

Therefore, making L → ∞ and using (3.26),    n β ϕ n lim |I1,k | ≤ lim Aη∞ 1 − e n−1 dx = 0.

k→+∞

L→+∞

BL

Now, setting y = rk x and using (3.17), (3.24) and (3.26), we have  I2,k =

η(y)

BLk

βk 1/(n−1) αk un/(n−1) ck uk e k dy + ok (1) λk



n

e n−1 βn ϕ dx + ok (1) = η(0) + ok (1),

= η(0) BL

where ok (1) → 0 as k → ∞. Note that  I3,k =

η(x)

[uk
≤ η∞ βk

≤ η∞ βk

ck λk ck λk

βk 1/(n−1)   n/(n−1)  dx ck uk Φ αk uk λk 

1/(n−1)

uk

 n/(n−1)  dx Φ  αk uk

[uk


[uk
 n/(n−1)  1/(n−1)  dx. |uA Φ αk uk k|

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

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By Lemma 3.6, Hölder’s inequality and Radial Lemma,  sup k

 A 1/(n−1)   n/(n−1)  u  dx ≤ C Φ αk u k

k

for any p < A.

(3.35)

[uk
Then we get I3,k ≤ Cη∞ βk

ck . λk

From (3.31) and (3.35), we conclude that I3,k → 0 as k → ∞, which completes the proof.

2

Next, using an argument similar to that due to Struwe in [33], we prove the following result. Lemma 3.10. Suppose (3.16) holds, then for any R > 0 and 1 < p < n, 1/(n−1)

ck

uk G

weakly in H 1,p (BR ),

where G satisfies the following equation in distributional sense, −n G + Gn−1 = δ0 + αGn−1 .

(3.36)

Furthermore, 1/(n−1)

ck 1/(n−1)

Proof. Setting Uk = ck

uk → G

 1,θ  n R \ {0} . in Cloc

(3.37)

uk and using (3.3), we have

−n Uk + Ukn−1 =

βk 1/(n−1)   n/(n−1)  + γk Ukn−1 ck uk Φ αk uk λk

in BRk .

(3.38)

We first prove that for any R > 0 and 1 < p < n, 



 |∇Uk |p + |Uk |p dx ≤ C(p, R).

BR

We split the proof into two cases: α = 0 and 0 < α < 1. Case 1: α = 0. In this case, we have γk = 0 for all k and −n Uk + Ukn−1 =

βk 1/(n−1)   n/(n−1)  in BRk . ck uk Φ αk uk λk

Setting Ωt = {0 ≤ Uk ≤ t} and Ukt = min{Uk , t} ∈ H01,n (BRk ),

(3.39)

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J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

by Lemma 3.9 for k sufficiently large, we obtain 

     ∇U t n + U t n dx ≤ k

 Ukt

k

Ωt

βk 1/(n−1)   n/(n−1)  dx ≤ Ct. ck uk Φ αk uk λk

BRk

Let η ∈ C0∞ (Rn , [0, 1]) be a radially symmetric cut-off function such that  η(x) =

1, 0,

if x ∈ BR ; if x ∈ Rn \ B2R .

Now consider t > 0 sufficiently large such that Rn \ Ωt ⊂ BR . Thus, there exists C1 (R) > 0 such that    t n ∇ ηU  dx ≤ C1 (R)t. k B2R

Let ρk > 0 be such that Uk (ρk ) = t . Then using the estimates above we have   inf |∇v|n dx : v ∈ H01,n (B2R ) and v|Bρk ≡ t ≤ C1 (R)t.

(3.40)

B2R

M. Struwe [33] proved that this infimum is attained by the function

Vk (x) =

⎧ |x| ) ⎨ −t log( 2R ⎩

log( 2R ρ )

in B2R \ Bρk ,

k

in Bρk .

t

Computing ∇Vk nLn (B2R ) and using (3.40), we have ρk ≤ 2Re−C1 t

for some C1 > 0.

Consequently   {x ∈ B2R : Uk ≥ t} = |Bρ | ≤ ωn−1 2n R n e−A(R)t , k n where A(R) is a positive constant depending only on R. Let m1 ∈ N such that t ≤ m1 , then for any δ < A(R),  e BR

δUk

∞    {m ≤ Uk ≤ m + 1}eδ(m+1) dx ≤ m=0

≤ eδm1 |BR | + C1 (R)

∞  m=m1

e−(A(R)−δ)m eδ ≤ C(R).

(3.41)

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

4087

Then, considering the weak version of (3.3) with the test function   1 + 2(Uk (x) − Uk (R))+ ϕ(x) = log , 1 + (Uk (x) − Uk (R))+ we get 

|∇Uk |n dx (1 + Uk − Uk (R))(1 + 2Uk − 2Uk (R))

BR

 ≤ log 2

 βk 1/(n−1)   ck uk Φ αk u2k dx − λk

BR



BR

  1 + 2(Uk − Uk (R))+ log Ukn−1 dx 1 + (Uk − Uk (R))+

≤ C.

(3.42)

Given 1 < p < n, by Young’s inequality we have  |∇Uk |p dx BR

  ≤ BR

 

≤ BR

 p/(n−p)  |∇Uk |n + (1 + Uk )(1 + 2Uk ) dx (1 + Uk − Uk (R))(1 + 2Uk − 2Uk (R))  |∇Uk |n δUk + Ce dx, (1 + Uk − Uk (R))(1 + 2Uk − 2Uk (R))

which together with (3.41) and (3.42) implies  |∇Uk |p dx ≤ C(p, R).

(3.43)

BR

We claim that (Uk ) is bounded in L1 (BR ). Suppose, contrary to our claim that Uk L1 (BR ) → +∞ (up to a subsequence). Thus, setting wk = Uk /Uk L1 (BR ) and using (3.3) we obtain −n wk + wkn−1 =

βk 1 1/(n−1)   n/(n−1)  in BRk , ck uk Φ αk uk Uk L1 (BR ) λk

(3.44)

which together with Lemma 3.9 implies that for k sufficiently large n wk ∈ L1 (BR ). Therefore, using the same argument as above, we can conclude that for any R > 0 and 1 < p < n,  |∇wk |p dx ≤ C(p, R), BR

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J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

from which together with Gagliardo–Nirenberg inequality we can find θ (p) ∈ (0, 1) such that 1−θ(p)

θ(p)

wk Lp (BR ) ≤ C1 (p, R)wk L1 (B ) ∇wk Lp (BR ) ≤ C(p, R). R

Thus (wk ) is bounded in H 1,p (BR ) and consequently, up to a subsequence, weakly in H 1,p (BR )

wk w

and wk → w

strongly in L1 (BR ).

Consequently, from (3.44) and Lemma 3.9 using similar argument as in [13, Lemma 4], up to a subsequence, 

 |∇wk |

n−2

∇wk ∇φ dx →

BR

|∇w|n−2 ∇w∇φ dx,

∀φ ∈ H 1,p (BR ).

(3.45)

BR

Therefore, taking limit in (3.44) and using (3.45) and Lemma 3.9, we get 



 |∇w|n−2 ∇w∇φ + w n−1 φ dx = 0,

∀φ ∈ H 1,p (BR ).

BR

Consequently, we obtain w = 0, which contradicts in fact that wL1 (BR ) = 1. Hence Uk L1 (BR ) ≤ C.

(3.46)

Therefore, by (3.43), (3.46) and Gagliardo–Nirenberg inequality, we have  |Uk |p dx ≤ C(p, R). BR

Thus (Uk ) is bounded in H 1,p (BR ). Case 2: 0 < α < 1. First we claim that (Uk ) is bounded in L1 (BR ). Suppose by contradiction that this is not the case. So, up to a subsequence, Uk L1 (BR ) → +∞. Thus, setting wk = Uk /Uk L1 (BR ) and using (3.3) we obtain −n wk + wkn−1 =

βk 1 1/(n−1)   n/(n−1)  + γk wkn−1 ck uk Φ αk uk Uk L1 (BR ) λk

in BRk , (3.47)

which together with Lemma 3.9 implies that n wk ∈ L1 (BR ). Therefore, using the same argument as in Case 1, we can conclude that for any R > 0 and 1 < p < n,  |∇wk |p dx ≤ C(p, R). BR

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4089

Using the Gagliardo–Nirenberg inequality there exists θ (p) ∈ (0, 1) such that 1−θ(p)

θ(p)

wk Lp (BR ) ≤ C1 (p, R)wk L1 (B ) ∇wk Lp (BR ) ≤ C(p, R). R

Thus (wk ) is bounded in H 1,p (BR ). Hence, up to a subsequence, weakly in H 1,p (BR ) and

wk w

wk → w

strongly in L1 (BR ).

Then, using Lemma 3.9 and taking the limit in (3.47) and using the same argument as in Case 1, we obtain     |∇w|n−2 ∇w∇φ + w n−1 φ dx = α w n−1 φ dx, ∀φ ∈ H 1,p (BR ), BR

BR

which together with the fact that α < 1 and R > 0 is arbitrary implies that w = 0. This contradicts wL1 (BR ) = 1. Hence Uk L1 (BR ) ≤ C.

(3.48)

Therefore, using (3.38) and the same argument as in Case 1, we can conclude that, for any 1 < p < n, there exists C(p, R) such that  |∇Uk |p dx ≤ C(p, R), BR

which together with (3.48) and Gagliardo–Nirenberg inequality implies  |Uk |p dx ≤ C(p, R). BR

Thus (Uk ) is bounded in H 1,p (BR ). Therefore, combining Case 1 and Case 2, we obtain (3.39). Thus, Uk G

weakly in H 1,p (BR ) for any R and 1 < p < n.

Considering the weak version of (3.38) with φ ∈ C0∞ (BR ), we have 



|∇Uk |

n−2

∇Uk ∇φ

 + φUkn−1 dx

BR

 =

βk 1/(n−1)   n/(n−1)  dx + γk φ ck uk Φ αk uk λk

BR

BR

Letting k → ∞, by Lemma 3.9,  BR





 |∇G|n−2 ∇G∇φ + φ Gn−1 dx = φ(0) + α



BR

φ Gn−1 dx.

φUk dx.

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Hence G satisfies the following equation in a distributional sense, −n G + Gn−1 = δ0 + αGn−1 . In order to prove the regularity of Uk , we start fixing r > 0 such that r < 2r < 3r < R and taking a cut-off function η ∈ C0∞ (BR \ Br , [0, 1]) such that η ≡ 1 on B3r \ B2r . Using Remark 3.1 and estimate          ∇(ηuk )n dx = η∇uk + uk ∇ηn dx ≤ C |∇uk |n + |uk |n dx, BR \Br

BR \Br

BR \Br

we obtain   ∇(ηuk ) → 0 as k → ∞, n

(3.49)

n/(n−1)

which implies that eαk (ηuk ) is bounded in Ls (BR \ Br ) for any s > 1. Indeed, using (3.49) n/(n−1) there exists k0 ∈ N such that sαk ∇(ηuk )n < βn for all k ≥ k0 and s > 1. Hence, by the Trudinger–Moser inequality (1.1), we have  esαk (ηuk )

n/(n−1)

 dx =

BR \Br

n/(n−1)

e

sαk ∇(ηuk )n

ηu

( ∇(ηuk )n )n/(n−1) k

dx ≤ C(R, n).

BR \Br

Consequently, n/(n−1)

eαk uk

is bounded in Ls (B3r \ B2r ),

∀s > 1.

(3.50)

Using (3.50), given 1 < p < n, we have  B3r \B2r

     βk  p  ck u1/(n−1) Φ  αk un/(n−1)  ≤ C1 k k λ  k

sp Uk

1/s  

B3r \B2r

e

s  αk uk

n/(n−1)

1/s 

B3r \B2r

≤ C(R, n), where 1/s + 1/s  = 1 with sp < n and s > 1. Thus, since −n Uk =

βk 1/(n−1)   n/(n−1)  − (1 − γk )Ukn−1 ck uk Φ αk uk λk

in BRk ,

we obtain n Uk ∈ Lp (B3r \ B2r ) for any 1 < p < n. Thus, applying the elliptic estimates there exists C(R, n) > 0 such that Uk C 1,θ (B3r \B5r/2 ) ≤ C(R, n). Hence, by the Ascoli–Arzelà theorem, we have Uk → G in C 1,θ (B3r \B5r/2 ). Since R is arbitrary we conclude that (3.37) holds. 2

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

4091

3.2.1. An upper bound of (α) To derive an upper bound to (α), similar to [18–20,24,36], we need the following result due to Carleson and Chang (cf. [8]): Lemma 3.11. Let B be the unit ball in Rn and (vk ) a sequence in H01,n (B) with ∇vk Ln (B) = 1. If vk 0 weakly in H01,n (B), then 



lim sup k→∞

eβn |vk |

n/(n−1)

 − 1 dx ≤ |B|e1+1/2+1/3+···+1/(n−1) .

B

Proposition 3.12. Assume that (3.16) holds, then (α) ≤

ωn−1 βn A0 +1+1/2+1/3+···+1/(n−1) e n

for any α ∈ [0, 1), where A0 is a constant depending only on α. Proof. Slightly modifying the proof in Y. Yang [36, Proposition 5.1], we can prove G(x) = −

n log |x| + A0 + gα (x), βn

where A0 is a constant depending only on α, gα (0) = 0, gα is continuous at 0, and gα ∈ 1 (Rn \ {0}). One can refer to [18] for details. Cloc Now, note that by (3.3), Lemma 3.10 and by an argument similar to [21, Lemma 3.8] 



|∇uk | + |uk | n

n



dx =

Rn \Bδ

=

 

1



|∇G| + |G|

n/(n−1)

ck

n

n



 dx + ok (1)

Rn \Bδ

 

1

n−2 ∂G

G|∇G|

n/(n−1)

ck

∂Bδ

∂ν



 dσ + α

|G| dx + ok (1) . n

Rn \Bδ

Consequently, we obtain 



|∇uk |n + |uk |n

Rn \Bδ

=

1 n/(n−1)

ck





  n  1 −n n log δ + A0 + gα (ξ ) + αGn + O δ log δ + ok (1) , βn

for some ξ ∈ ∂Bδ . By continuity of gα , we have 



|∇uk | + |uk |

Rn \Bδ

n

n



dx =

1 n/(n−1)

ck



 1 −n n log δ + A0 + αGn + oδ (1) + ok (1) . βn

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J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

Set    |∇uk |n + |uk |n dx − unk dx

 

 τk :=



|∇uk |n dx = 1 − Rn \Bδ

Bδ

1

=1−



n/(n−1)

ck

Bδ

 1 log δ −n + A0 + αGnn + oδ (1) + ok (1) + ok (1), βn

which together with Lemma 3.11 implies that  k→∞



1/n n/(n−1) )

eβn (wk /τk

lim sup

 ωn−1 1+1/2+1/3+···+1/(n−1) − 1 dx ≤ δ n , e n

(3.51)

Bδ

where bk = uk (δ) and wk = (uk − bk )+ . Note that wk ∈ H01,n (Bδ ) and wk 0 weakly in W01,n (Bδ ). Note that n/(n−1)

αk uk

 1/n n/(n−1) ≤ βn wk /τk  + log δ −n + βn A0 + oδ (1) + ok (L) 1/(n−1)

Indeed, using that ck get

n/(n−1)

αk uk

on Brk L . (3.52)

1,θ 1,θ uk → G in Cloc (Rn \ {0}), uk n → 0 and ϕk → ϕ in Cloc (Rn ), we

 1/(n−1) ≤ βn 1 + αuk nn (wk + bk )n/(n−1) n/(n−1)

= βn w k

+

βn α n 1/(n−1) + ok (1). Gnn + αk bk wk n−1 n−1

(3.53)

From the estimates above we get  n/(n−1)

βn wk

≤ βn

wk

n/(n−1) −

1/n

τk

  1 1 log δ −n + A0 + oδ (1) + ok (1) . n − 1 βn

(3.54)

Similarly we have 1/(n−1)

bk wk

=

1 log δ −n + A0 + oδ (1) + ok (1). βn

Combining (3.53)–(3.55), we obtain (3.52) on Brk L . Thus, denoting Lk = Lrk we obtain by (3.52) that  e BLrk

n/(n−1)

αk uk

dx ≤ δ

−n βn A0 +oδ (1)+ok (L)



1/n n/(n−1) )

eαk (wk /τk

e

dx

BLk

= δ −n eβn A0 +oδ (1)+ok (L)



BLk



1/n n/(n−1) )

eαk (wk /τk

 − 1 dx + ok (1)

(3.55)

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101



≤ δ −n eβn A0



1/n n/(n−1) )

eαk (wk /τk

4093

 − 1 dx.

BLk

It follows by (3.51) that 

n/(n−1)

k→∞

dx ≤

eαk uk

lim sup

ωn−1 βn A0 +1+1/2+1/3+···+1/(n−1) . e n

(3.56)

BLk

There exists k0 ∈ N such that {x ∈ Rn : uk (x) ≥ ck /A} ⊂ BLrk for k ≥ k0 , thus from (3.34), we have  lim sup k→∞

Φ

n/(n−1)  dx αk uk





  n/(n−1)   dx = 0. Φ αk uA k

= lim sup k→∞

Bδ \BLk

(3.57)

Bδ \BLk

On the other hand, by Radial Lemma (cf. [31]) there exists C = C(δ) > 0 such that 

 n/(n−1)  dx ≤ Cuk nn . Φ αk uk

Rn \Bδ

Since uk nn → 0 as k → ∞, we have 

 n/(n−1)  dx = 0. Φ αk uk

lim sup k→∞

Rn \Bδ

Note that for δ > Lrk , we have 

 n/(n−1)  dx ≤ Φ αk uk

Rn

 e

BLrk

n/(n−1)

αk uk



+

 dx +

 n/(n−1)  dx Φ αk uk

Bδ \BLrk

 n/(n−1)  dx. Φ αk uk

Rn \Bδ

From (3.56), (3.57) and (3.58), we get (α) ≤ which proves the lemma. 2

ωn−1 βn A0 +1+1/2+1/3+···+1/(n−1) , e n

(3.58)

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J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

3.2.2. The supremum (α) is finite and it is attained if α ∈ [0, 1) We proved in Section 3.2.1 (Proposition 3.12) that if ck = sup uk = uk (0)  +∞, Rn

then (α) ≤

ωn−1 βn A0 +1+1/2+···+1/(n−1) . e n

(3.59)

Using an argument similar to [21, Section 5] (see also [20,36]), we construct a family of test functions v ∈ H 1,n (Rn ), with v 1,n = 1 and satisfying  Φ ◦ να (v ) dx > Rn

ωn−1 βn A0 +1+1/2+···+1/(n−1) , e n

for  > 0 sufficiently small. This leads to a contradiction with (3.59) and consequently the blowup does not occur. Let ⎧ ⎨ C − (n−1) log(1+cn |x/|n/(n−1) )+B if |x| ≤ L βn C 1/(n−1) v (x) = ⎩ G(|x|) if |x| > L, C 1/(n−1)

where G is the Green function given in Lemma 3.10, cn = ωn−1 /n and C, B, L are functions of  (which will be defined later by (3.61), (3.62) and (3.64)) satisfying (i) L → +∞, C → +∞ and L → 0 as  → 0; (n−1) log(1+cn Ln/(n−1) )+B βn C 1/(n−1) log L

(ii) C − (iii)

C n/(n−1)

=

G(L) ; C 1/(n−1)

→ 0 as  → 0.

We use the normalization of v to obtain information on B, C and L. First we have 



 |∇v |n + |v |n dx =

Rn \BL

=



1 C n/(n−1)





 |∇G|n + |G|n dx

Rn \BL

 

1



n−2 ∂G

G(L)|∇G|

C n/(n−1) ∂BL

∂ν

 dσ + α

 |G| dx . n

Rn \BL

Thus 



|∇v | + |v |

Rn \BL

n

n



dx =

1 C n/(n−1)



 n n log(L) + A0 + αGn + o (1) . βn

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

4095

By a standard calculation using an argument similar to [20,21,36]  |∇v |n dx = BL

   n−1   log 1 + cn Ln/(n−1) − 1 + 1/2 + 1/3 + · · · + 1/(n − 1) n/(n−1) βn C +

1 C n/(n−1)

  O L−n/(n−1) +

1 C n/(n−1)

  O (L)n logn (1/L) log L .

It is easy to check that 

  |v |n dx = O (L)n C n log L .

BL

Consequently, 



 |∇v |n + |v |n dx

Rn

=

    1 −(n − 1) 1 + 1/2 + · · · + 1/(n − 1) + βn A0 + βn αGnn βn C n/(n−1)     1 + (n − 1) log 1 + cn Ln/(n−1) − log(L)n + φ, n/(n−1) βn C

where   φ = O (L)n C n log L + (L)n logn (L) + L−n/(n−1) + o (1) . Setting v 1,n = 1, we obtain   βn C n/(n−1) = −(n − 1) 1 + 1/2 + · · · + 1/(n − 1) + βn A0 + βn αGnn   + (n − 1) log 1 + cn Ln/(n−1) − log(L)n + φ.

(3.60)

Hence   βn C n/(n−1) = −(n − 1) 1 + 1/2 + · · · + 1/(n − 1) + βn A0 + βn αGnn + log cn − log  n + φ.

(3.61)

By (ii) we have   βn C n/(n−1) − (n − 1) log 1 + cn Ln/(n−1) + B = βn G(L), then from (3.60), we get   B = −(n − 1) 1 + 1/2 + · · · + 1/(n − 1) + φ.

(3.62)

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J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

Now, we estimate the integral 

eβn (1+αv n )

n 1/(n−1) |v |n/(n−1) 

dx.

BL

On BL similar to [21, Section 5] we can check that 

eβn (1+αv n )

n 1/(n−1) |v |n/(n−1) 

 dx ≥

BL

eβn C

n/(n−1) −n log(1+c

n |x/|

n/(n−1) )− n B n−1

dx.

BL

Then using (3.61), we get 

eβn (1+αv n )

n 1/(n−1) |v |n/(n−1) 

dx ≥

BL

ωn−1 βn A0 +1+1/2+···+1/(n−1) e n   + O (L)n C n log L + L−n/(n−1) + (L)n logn (L) .

Thus  Φ ◦ να (v ) dx ≥ BL

ωn−1 βn A0 +1+1/2+···+1/(n−1) e n   + O (L)n C n log L + L−n/(n−1) + (L)n logn (L) + o (1).

Moreover, on Rn \ BL we have the estimate  Rn \BL

βn Φ ◦ να (v ) dx ≥ n n!

 Rn \BL

   G(x)n/(n−1) n    C 1/(n−1)  dx

and thus we get  Φ ◦ να (v ) dx ≥ Rn

ωn−1 βn A0 +1+1/2+···+1/(n−1) e n +

βnn n!C n/(n−1) 



  G(x)n/(n−1) n dx

Rn \BL

 + O (L)n C n log L + L−n/(n−1) + (L)n logn (L) + o (1). (3.63) Now taking L() = − log , we have L() → 0 as  → 0. We then need to prove that there exists a C = C() which solves Eq. (3.61). Setting   z(t) = −βn t n/(n−1) − (n − 1) 1 + 1/2 + · · · + 1/(n − 1) + βn αGnn + βn A0 + log cn − log  n + φ

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

4097

since, for  small,  −

z

2 log  n βn

n/(n−1)  <0

and  −

z

1 log  n 2βn

n/(n−1)  >0

then there exists t ∈ ((− 2β1n log  n )n/(n−1) , (− β2n log  n )n/(n−1) ) such that z(t ) = 0. Thus, we defined C, and it satisfies βn C n/(n−1) = − log  n + O (1).

(3.64)

Therefore, as  → 0, we have log L C n/(n−1)

→0

and then (L)n C n+n/(n−1) log L + C n/(n−1) L−n/(n−1) + C n/(n−1) (L)n logn (L) → 0. Therefore (i), (ii) and (iii) hold and we can conclude from (3.63) that for  > 0 sufficiently small  Φ ◦ να (v ) dx > Rn

ωn−1 βn A0 +1+1/2+···+1/(n−1) . e n

4. The supremum (α) is infinity if α > 1 Consider φ1,R a positive eigenfunction associated with μ1 (BR ) the first eigenvalue of (−n u + un−1 , H01,n (BR )) with φ1,R Ln (BR ) = 1 and μ1 (BR ) ≥ α, for a suitable R > 0. Next, we consider the Moser sequence

wk (x) =

1

⎧ (n−1)/n ⎨ (log k)

1/n ωn−1 ⎩

log (1/|x|) (log k)1/n

0

if |x| < 1/k if 1/k ≤ |x| ≤ 1 if |x| > 1.

Notice that wk ∈ H01,n (B1 ), supp(wk ) = B 1 , ∇wk Ln (B1 ) = 1 and wk 0 in H01,n (B1 ). Consider vk = wk + tk φ1,R such that (tk ) is a sequence satisfying (T1 ) (T2 ) (T3 ) (T4 )

tk > 0 for all k ∈ N tk → 0 as k → ∞ ξk = tk (log k)1/n → +∞ as k → ∞ tkn (log k)1/n → 0 as k → ∞

4098

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

(take for example tk = (log k)−1/(n+1) ). The aim is to prove that  Φ(ηk ) dx → +∞ as k → ∞,

(4.1)

Rn

where    n/(n−1) vk nn 1/(n−1) vk η k = βn 1 + α . vk n1,n vk 1,n For that purpose, we estimate vk n1,n and vk nn . First, we have n/2  |∇wk + tk ∇φ1,R |n = |∇wk |2 + 2tk ∇wk · ∇φ1,R + tk2 |∇φ1,R |2 n/2  = |∇wk |2 + 2tk |∇wk ||∇φ1,R | cos θ + tk2 |∇φ1,R |2  n/2 |∇wk |2 |∇wk | = tkn |∇φ1,R |n 1 + 2 . cos θ + 2 tk |∇φ1,R | tk |∇φ1,R |2 Thus, using the inequality 

1 + 2t cos θ + t 2

n/2

  ≤ 1 + t n + nt cos θ + C t 2 + t n−1

for t ≥ 0, uniformly in θ,

we obtain  ∇wk + tk ∇φ1,R nn

≤ 1 + tkn ∇φ1,R nn 

n−2 

+ Ok tk

+ ntkn−1

|∇φ1,R |n−2 ∇φ1,R ∇wk dx

BR

+ Ok (tk ).

(4.2)

On the other hand, vk nn =

n    n j =0

j

j



n−j j φ1,R dx.

wk

tk

BR

Consequently,  vk nn = tkn φ1,R nn + ntkn−1

n−1 φ1,R wk dx + ok (1).

BR

From (4.2) and (4.3), we obtain  vk n1,n

≤ 1 + μ1 (BR )tkn 

n−2 

+ Ok tk

+ ntkn−1



 n−1 |∇φ1,R |n−2 ∇φ1,R ∇wk + φ1,R wk dx

BR

+ Ok (tk ) + ok (1).

(4.3)

J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

Let  −1/n Z(x) = ωn−1 log(1/|x|) 0

if |x| ≤ 1 if |x| > 1.

Then by the Lebesgue dominated convergence theorem, we obtain vk n1,n

≤ 1 + μ1 (BR )tkn



+ nμ1 (BR )tkn ξk−1

φ1,R Z(x) dx B1

    + ok tkn ξk−1 + Ok tkn−2 + Ok (tk ) + ok (1), where ok (tkn ξk−1 ) → 0 as k → ∞. Moreover, vk nn = tkn + ntkn ξk−1



  n−1 φ1,R Z(x) dx + ok tkn ξk−1 + ok (1).

B1

Thus 1+α

vk nn ≥ 1 + αtkn + nαtkn ξk−1 vk n1,n



    n−1 φ1,R Z(x) dx + o tkn ξk−1 + Ok tk2n + ok (1).

B1

Hence        n vk nn 1 n−1 n −1 1+α ≥ 1 + α − μ (B ) t + nt ξ φ Z(x) dx 1 R k k k 1,R vk n1,n vk n1,n B1

    + ok tkn ξk−1 + Ok tk2n + ok (1). Since α ≥ μ1 (BR ), we get  1+α

     1 vk nn ≥ 1 + ok tkn ξk−1 + Ok tk2n . n n vk 1,n vk 1,n

Thus, using that n/(n−1)

βn vk

 n/(n−1) = βn wk +

 n 1/(n−1) wk tk φ1,R + ok (1), n−1

we obtain, on B1/k  βn

vk nn 1+α vk n1,n

1/(n−1) 

vk vk 1,n

n/(n−1) ≥ n log k +

n2 1/n ω ξk φ1,R + ok (ξk ). n − 1 n−1

4099

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J.M. do Ó, M. de Souza / J. Differential Equations 258 (2015) 4062–4101

Hence by (T3 ) and using that φ1,R (0) > 0, we obtain 



 eηk − 1 dx → +∞ as k → ∞,

B1/k

and (4.1) follows, which completes the proof. Open questions. Here we have collected some interesting open questions for this class of Trudinger–Moser inequality: 1. 2. 3. 4.

Does a similar statement of Theorem 1.1 hold for the case α = 1? Does a similar statement of Theorem 1.2 hold for the case α = 1? Does a similar statement of Theorem 1.2 hold for the nonlinear term Ψ (t)? Similar question to (1)–(2) can be formulated when we replace Φ(t) by Ψ (t) in the supremum given in (1.4).

Remark 4.1. In our argument to prove Theorem 1.1 and Theorem 1.2 it was crucial Lemma 3.2, where we proved the compactness of the maximizing sequence. We emphasize that it was important the fact that 0 ≤ α < 1 and we do not know how to prove similar result in the case α = 1. Acknowledgment The authors would like to thank Dr. Yunyan Yang for valuable suggestions and comments on this work. References [1] Adimurthi, Y. Yang, An interpolation of Hardy inequality and Trudinger–Moser inequality in RN and its applications, Int. Math. Res. Not. IMRN 13 (2010) 2394–2426. [2] S. Adachi, K. Tanaka, Trudinger type inequalities in RN and their best exponents, Proc. Amer. Math. Soc. 128 (2000) 2051–2057. [3] D.R. Adams, J.F. Fournier, Sobolev Spaces, second ed., Pure Appl. Math., vol. 140, Elsevier/Academic Press, Amsterdam, 2003. [4] Adimurthi, O. Druet, Blow-up analysis in dimension 2 and a sharp form of Trudinger–Moser inequality, Comm. Partial Differential Equations 29 (2004) 295–322. [5] Adimurthi, M. Struwe, Global compactness properties of semilinear elliptic equations with critical exponential growth, J. Funct. Anal. 175 (2000) 125–167. [6] H. Berestycki, P.-L. Lions, Nonlinear scalar field equations. I. Existence of ground state, Arch. Ration. Mech. Anal. 82 (1983) 313–345. [7] D.M. Cao, Nontrivial solution of semilinear elliptic equation with critical exponent in R2 , Comm. Partial Differential Equations 17 (1992) 407–435. [8] L. Carleson, A. Chang, On the existence of an extremal function for an inequality of J. Moser, Bull. Sci. Math. 110 (1986) 113–127. [9] D.G. de Figueiredo, J.M. do Ó, B. Ruf, On an inequality by N. Trudinger and J. Moser and related elliptic equations, Comm. Pure Appl. Math. 55 (2002) 135–152. [10] D.G. de Figueiredo, J.M. do Ó, B. Ruf, Elliptic equations and systems with critical Trudinger–Moser nonlinearities, Discrete Contin. Dyn. Syst. 30 (2011) 455–476. [11] M. de Souza, J.M. do Ó, A sharp Trudinger–Moser type inequality in R2 , Trans. Amer. Math. Soc. 366 (2014) 4513–4549. [12] J.M. do Ó, M. de Souza, E. de Medeiros, U.B. Severo, An improvement for the Trudinger–Moser inequality and applications, J. Differential Equations 256 (2014) 1317–1349.

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