# A modified singular Trudinger–Moser inequality

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(1)

## inequality

### Yamin Wang

1*

*Correspondence: 18811219726@163.com 1School of Mathematics, Renmin University of China, Beijing, P.R. China

Abstract

Let

### Ω

⊂R2be a smooth bounded domain,W1,2

0 (

### Ω

) be the standard Sobolev space. Assuming certain conditions on a functiong:R→R, we derive a modiﬁed singular Trudinger–Moser inequality, which was originally established by Adimurthi and Sandeep (Nonlinear Diﬀer. Equ. Appl. 13:585–603,2007), namely,

sup uW01,2(Ω),∇u2≤1

Ω

1 +g(u)

e

4π(1–γ)u2

|x|2γ dx, (1)

where 0 <

### γ

< 1. Following Yang and Zhu (J. Funct. Anal. 272:3347–3374,2017), we prove that the extremal functions for the supremum in (1) exist. The proof is based on a blow-up analysis.

MSC: 46E35

Keywords: Singular Trudinger–Moser inequality; Blow-up analysis; Extremal function

1 Introduction

LetΩ be a smooth bounded domain inR2, andW1,2

0 (Ω) be the completion ofC0∞(Ω)

under the normuW1,2 0 (Ω)= (

Ω|∇u|

2dx)1/2. For 1p< 2, the standard Sobolev

embed-ding theorem states thatW01,p(Ω)Lq(Ω) for all 1 <q2p/(2 –p); while ifp> 2, we have

W01,p(Ω)C0(Ω). As a borderline of the Sobolev embeddings, the classical Trudinger– Moser inequality [21–23,26,33] says

sup

u∈W1,2

0 (Ω),∇u2≤1

Ω

eαu2

dx< +∞, ∀α≤4π. (2)

Moreover, these integrals are still ﬁnite for anyα> 4π, but the supremum is inﬁnity. Here and in the sequel, for any real numberq≥1, · qdenotes theLq(Ω)-norm with respect to the Lebesgue measure.

A functionu0is called an extremal function for the Trudinger–Moser inequality (2) if

u0belongs toW01,2(Ω),∇u02≤1 and

Ω

eαu20dx= sup

u∈W1,2

0 (Ω),∇u2≤1

Ω

eαu2dx.

(2)

An interesting question on Trudinger–Moser inequalities is whether or not extremal func-tions exist. The existence of extremal funcfunc-tions for (2) was obtained by Carleson–Chang [5] whenΩis a unit ball, and by Struwe [24] whenΩis close to the ball in the sense of mea-sure. Then Flucher [12] extended this result whenΩis a general bounded smooth domain inR2. Later, Lin [16] generalized the existence result whenΩis an arbitrary dimensional

domain. For recent developments, we refer the reader to Yang [28].

Using a rearrangement argument and a change of variables, Adimurthi–Sandeep [2] generalized the Trudinger–Moser inequality (1) to a singular version as follows:

sup

u∈W1,2

0 (Ω),∇u2≤1

Ω

e4π(1–γ)u2

|x|2γ dx<∞. (3)

This inequality is also sharp in the sense that all integrals are still ﬁnite whenα> 1 –γ, but the supremum is inﬁnity. Clearly, ifγ= 0, (3) reduces to (1). Following the lines of Flucher [12], in Csato and Roy [9], they adopt the concentration–compactness alternative by Li-ons [17] and deduced that the existence of extremals for such singular functionals. Later, (3) was extend to the entireRN by Adimurthi and Yang [4]. Meanwhile, Souza and do Ó modiﬁed the singular to another version inRN in [10]. WhenΩis the unit ballB, (3) was improved by Yuan and Zhu [32]. Similarly, an analog is also be proved by Yuan and Huang by using the method of symmetrization in [31]. Such singular Trudinger–Moser inequal-ities play an important role in the study of partial diﬀerential equations and conformal geometry; see [2,4,10,14,27] and [6] for details.

Recently, using a method of energy estimates in [19], Mancini–Martinazzi [20] reproved Carleson–Chang’s result. For applications of this method, we refer the reader to Yang [29]. Using the same idea, they proved that the supremum

sup

u∈W1,2

0 (B),∇u2≤1

B

1 +g(u)e4πu2dx (4)

can be achieved for certain smooth functiong:R→R, whereBis a unit ball. On the other hand, in Yang and Zhu [30], one studied the following singular form:

sup

u∈W1,2

0 (Ω),∇u1,α≤1

Ω

eβu2

|x|2γ dx, (5)

and they veriﬁed there exists some function u0 to achieve this supremum for any β <

4π(1 –γ), where

u1,α=

Ω

|∇u|2dxα

Ω

u2dx

1/2

,

andαsatisﬁes

α< inf

u∈W1,2 0 (Ω),u≡0

u2 2 u2

2

.

(3)

We are aim to prove two main results: One is to explain the new supremum is ﬁnite; the other is to discuss the existence of extremals for such functionals. In our proof, unlike the previous energy estimate procedure in [19,20,29], we mainly employ the method of blow-up analysis as in [11,14,15,18] to prove the supremum in the following (9) can be achieved. Based on Mancini–Martinazzi [20] (see pages 3 and 4), we assume the function

gin (9) satisﬁes

gC1(R), inf

R g> –1, g(–t) =g(t),

lim

|t|→∞t

2g(t) = 0, g(t) > 0 (t> 0). (6)

In the proof, we derive

=

1

λε

1 +g() +

g()

8π(1 –γε)

uεe4π(1–γε)u 2

ε= 1

λε

1 +h()

uεe4π(1–γε)u 2

ε

for someλε∈R, where we set

h(t) :=g(t) + g (t)

8π(1 –γε)t, t∈R\ {0}. (7)

We further assume

inf

(0,+∞)h(t) > –1, (0,+sup)h(t) < +∞, and t→∞limt

2h(t) = 0. (8)

Comparing the conditions on the function g in Mancini–Martinazzi [20], one can see some diﬀerences. In this note, we assumeg(t) > 0 (∀t> 0), which is used in the Lemma4. Moreover, the assumptions in (6) and (8) implies thatlim|t|→∞g(t) = 0 in [20]. Our main conclusion can be stated as the following two theorems, respectively.

Theorem 1 LetΩbe a smooth bounded domain inR2and W1,2

0 (Ω)be the usual Sobolev

space.Let0 <γ < 1be ﬁxed.Suppose gC1(R)satisﬁes the hypotheses in(6)and(8).Then

the supremum

Λ4π(1–γ):= sup

u∈W1,2

0 (Ω),∇u2≤1

Ω

1 +g(u)e

4π(1–γ)u2

|x|2γ dx<∞. (9)

Theorem 2 LetΩbe a smooth bounded domain inR2and W1,2

0 (Ω)be the usual Sobolev

space.Let0 <γ < 1be ﬁxed.Suppose gC1(R)satisﬁes the hypotheses in(6)and(8).Then,

for anyβ≤4π(1 –γ),the supremum

sup

u∈W1,2

0 (Ω),∇u2≤1

Ω

1 +g(u)e

βu2

|x|2γ dx

can be attained by some function u0∈W01,2(Ω)∩Cloc1 (Ω\ {0})∩C0(Ω).

(4)

variation. We derive a diﬀerent Euler–Lagrange equation on which the analysis is per-formed. The essential problem is the presence of the functiong. To meet the necessary of our proof, we assumegsatisﬁes certain conditions. Then following Yang and Zhu [30], we deﬁne maximizing sequences of functions by using a more delicate scaling. The exis-tence of singular term|x|–2γ with 0 <γ < 1 causes exact asymptotic behavior of certain

maximizing sequence near the blow-up point. Unlike in [28], we employ two diﬀerent classiﬁcation theorems of Chen and Li [7,8] to get the desired bubble. And our method in dealing with the bubble is also diﬀerent from Yang–Zhu [30] because of the functiong. We refer to Adimurthi and Druet [1], Carleson–Chang [5], Li [15], Struwe [24], Adimurthi and Struwe [3], Iula and Mancini [13], Yang [28], Lu and Yang [18], respectively.

2 Proof of Theorem1

We divide the proof into several steps as follows.

2.1 Existence of maximizers for

### Λ

4π(1–γε)and the Euler–Lagrange equation

In this subsection, we shall prove that maximizers for the subcritical singular Trudinger– Moser functionals exist.

Proposition 3 For any0 <ε< 1–β,there exists some uεW01,2(Ω)∩C1loc(Ω\{0})∩C0(Ω)

satisfyingu2= 1and

Ω

1+g()

e4π(1–γε)u2

ε

|x|2γ dx=Λ4π(1–γε):= sup

u∈W1,2 0 (Ω),

∇u2≤1

Ω

1+g(u)e

4π(1–γε)u2

|x|2γ dx. (10)

Proof This is based on a direct method of variation. For any 0 <β< 1, let 0 <ε< 1 –γ be ﬁxed. We take a sequence of functionsujW01,2(Ω) satisfying∇uj2≤1 and, asj→ ∞,

lim

j→∞

Ω

1 +g(uj)e

4π(1–γε)u2j

|x|2γ dx=Λ4π(1–γε). (11)

Sinceujis bounded inW01,2(Ω), there exists someW01,2(Ω) such that up to a

subse-quence, assuming

ujuε weakly inW01,2(Ω),

uj strongly inLp(Ω),∀p≥1,

uj a.e. inΩ.

Since

0≤

Ω

|∇uε|2dx≤lim sup

j→∞

Ω

|∇uε|2dx

1

2

Ω

|∇uj|2dx

1 2

,

we have 0≤ ∇2≤1. Note that

Ω

(uj) 2

dx=

Ω

|∇uε|2dx

Ω

|∇uj|2+oj(1)

≤1 –

Ω

(5)

Following Hölder’s inequality, for any 1 <p≤1γ,δ> 0,w> 1 andw=w/(w– 1), we have

Ω

1 +g(uj)p 1

|x|2γpe

4π(1–γε)pu2j

dxC

Ω

1

|x|2γpe

4π(1–γε)p(1+δ)w(uj)2dx 1

w

×

Ω

1

|x|2γpe

4π(1–γε)p(1+41δ)wu2εdx

1 w

. (13)

Whenp, 1 +δandsare suﬃciently close to 1, we have

(1 –γε)p(1 +δ)w+γwp< 1. (14)

Combining (12), (13) and (14), we have by the singular Trudinger–Moser inequality (3)

1 +g()

|x|–2γe4π(1–γε)u2ε is bounded inLp(Ω),

for somep> 1. Note that

1 +g(uj)e

4π(1–γε)u2j

|x|–2γ

1 +g()

e4π(1–γε)u2ε

|x|–2γC|x|–2γe4π(1–γε)u2j +e4π(1–γε)u2εu2

ju2ε,

+|x|–2γmax g(uj),g()

|ujuε|e4

π(1–γε)u2j

. (15)

Sinceujstrongly inLp(Ω) for anyp> 1, in view of (6) and (8), we can conclude

from (15) that

Ω

1 +g(uj)

|x|–2γe4π(1–γε)u2j

dx

Ω

1 +g()

|x|–2γe4π(1–γε)u2

εdx,

asj→ ∞. This together with (11) immediately leads to (10). Obviously≡0. If∇2<

1, set=∇u

ε2, then we obtain∇2= 1. Since 0≤<and≡0, it follows from

(6) that

Ω

1 +g()

e4π(1–γε)u2ε

|x|2γ dx<

Ω

1 +g()

e4π(1–γε)u2ε

|x|2γ dxΛ4π(1–γε),

which contradicts (10). Consequently,∇2= 1 holds. Furthermore, one can also check

that|uε|attains the supremumΛ4π(1–γε). Thus,can be chosen so that≥0. It is not

diﬃcult to see thatsatisﬁes the following Euler–Lagrange equation:

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

=λ–1ε |x|–2γ(1 +h())uεe4π(1–γε)u 2

ε inΩ⊂R2,

≥0, ∇2= 1 inΩ⊂R2,

λε=

Ω|x|

–2γ(1 +h(u

ε))u2εe4π(1–γε)u 2

εdx,

(16)

(6)

2.1.1 The case when uεis uniformly bounded inΩ

The proof of Theorem2will be ended if we can ﬁnd someu0∈W01,2(Ω)∩C1loc(Ω\ {0})∩

C0(Ω) satisfying∇u02= 1 and

Ω

1 +g(u0)

e4π(1–γ)u2 0

|x|2γ dx= sup

u∈W1,2

0 (Ω),∇u2≤1

Ω

1 +g(u)e

4π(1–γ)u2

|x|2γ dx. (17)

Sinceis bounded inW01,2(Ω), we assume without loss of generality

uεu0 weakly inW01,2(Ω),

u0 strongly inLp(Ω),∀p≥1,

u0 a.e. inΩ.

(18)

Let=() =maxΩuε. Ifis bounded, for anyuW01,2(Ω) withu≥0,∇u02= 1,

together with Lebesgue dominated convergence theorem gives

Ω

1 +g(u)e

4π(1–γ)u2

|x|2γ dx=εlim0

Ω

1 +g()

e4π(1–γε)u2

|x|2γ dx

≤lim ε→0

Ω

1 +g()

e4π(1–γε)u2ε

|x|2γ dx

=

Ω

1 +g(u0)

e4π(1–γ)u20

|x|2γ dx. (19)

By the arbitrariness ofuW01,2(Ω), we conclude thatu0is the desired maximizer when

is uniformly bounded in Ω. Applying elliptic estimates to its Euler–Lagrange

equa-tion, one can deduce thatu0∈W01,2(Ω)∩Cloc1 (Ω\ {0})∩C0(Ω). And then (17) follows

immediately.

2.2 Blowing up analysis

In this subsection, as in [1,17], we will use the blow-up analysis to understand the asymp-totic behavior of the maximizers. Assume =()→ ∞and we distinguish two

cases to proceed.

Case 1.Ifu0≡0, the supremum in (9) can be attained byu0without diﬃculty. And the

proof will just be divided into several simple steps.

Step 1. A similar estimate as in (13), one can easily check that (1+g())

|x|2γ e4

π(1–γε)u2ε is

bounded inLp(Ω) (p> 1).

Step 2.By the mean value theorem and the Hölder inequality, we have

lim ε→0

Ω

|x|–2γe4π(1–γε)u2εdx=

Ω

(7)

Step 3.Based on the above steps, one can easily check that

Ω

1 +g()

|x|–2γe4π(1–γε)u2ε1 +g(u

0)

|x|–2γe4π(1–γ)u20dx

g(u0) + 1

Ω

|x|–2γe4π(1–γε)u2ε|x|–2γe4π(1–γ)u20dx

+

Ω

|x|–2γe4π(1–γε)u2εg(u

ε) –g(u0)dx

=(1).

Thus, we arrive at the conclusion that

lim ε→0

Ω

1 +g()

|x|–2γe4π(1–γε)u2

εdx=

Ω

1 +g(u0)

|x|–2γe4π(1–γ)u2 0dx.

This together with (17) gives the desired result.

Case 2.Ifu0≡0, in view of Eq. (16), it is important to ﬁgure out whetherλεhas a positive

lower bound or not. For this purpose, we have the following.

Lemma 4 Letλεbe as in(16).Then we havelim infε→0λε> 0.

Proof By an inequalityet21 +t2et2 fort0, it follows from (6) and (7) that

λε

1 4π(1 –γε)

Ω

1 +h()

(e4π(1–γε)u2ε– 1)

|x|2γ dx

≥ 1

4π(1 –γε)

Ω

1 +g()

e4π(1–γε)u2ε

|x|2γ dx

Ω

(1 +g()) |x|2γ dx

+

Ω

g()

8π(1 –γε)|x|2γu ε

e4π(1–γε)u2ε– 1dx

≥ 1

4π(1 –γε)

Ω

1 +g()

e4π(1–γε)u2

ε

|x|2γ dx

Ω

(1 +g()) |x|2γ dx

.

This together with (10) leads to

lim inf ε→0 λε

1 4π(1 –γ)

Λ4π(1–γ)–

Ω

(1 +g(0))

|x|2γ dx

> 0.

Or equivalently, we have

1

λε

C. (20)

Therefore, λ1

ε is uniformly bounded inΩ. This ends the proof of the lemma.

2.2.1 Energy concentration phenomenon

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Proposition 5 For the function sequence{uε},we have uε0weakly in W01,2(Ω)and

→0strongly in Lq(Ω)for any q> 1.Moreover,|∇uε|2dx δ0weakly in a sense of

mea-sure,whereδ0is the usual Dirac measure centered at the point0.

Proof Since∇2= 1, we have the same assumptions as in (18). Observe that

Ω

(u0)2dx= 1 –

Ω

|∇u0|2dx+o(1). (21)

Supposeu0≡0. In view of (21) and an obvious analog of (13), it follows that

1 +g()

|x|–2γe4π(1–γε)u2

ε is bounded inLq(Ω),

for someq> 1. Then applying elliptic estimates to (18), one can deduce thatis bounded

in W02,q(Ω). Together with Sobolev embedding results, we conclude is bounded in

ε→ ∞. Thereforeu0≡0 and (21) becomes

Ω

|∇uε|2dx= 1 +(1). (22)

We next prove|∇uε|2dx δ

x0. If the statements were false, suppose|∇uε|

2dx ηin a

sense of measure. In view ofη=δx0, there existsr0> 0 such that

lim ε→0

Br0(x0)

|∇uε|2dxη+ 1

2 < 1.

In view of (22) andu0≡0, we can choose a cut-oﬀ functionφC01(Br0(x0)), which is equal

to 1 onBr0/2(x0), then it follows that

lim sup ε→0

Br0(x0)

(φuε) 2

dx< 1.

By the singular Trudinger–Moser inequality (3), one sees that (1 +g(φuε))e

4π(1–γε)(φuε)2

|x|2γ is

bounded inLr(B

r0(x0)) for somer> 1. Applying elliptic estimates to (16), one getsis

uniformly bounded inΩ, which contradicts→ ∞again. Therefore|∇uε|2dx δx0.

Moreover, we get→0 inC1loc(Ω\ {0,x0})∩C0loc(Ω\ {x0}).

In fact, we have x0 = 0. Set r0 =|x0|/2. Note that λ–1ε |x|–2

γ(1 +h(u

ε))uεe4π(1–γε)u 2

ε

is bounded in Lq1(B

r0(0)) for some q1 > 1. When |x|>r0, by the classical Trudinger–

Moser inequality (2), we recognizeλ–1

ε |x|–2γ(1 +h())uεe4π(1–γε)u 2

εis bounded inLq2(Ω\

Br0(0)) for some q2 > 1. Choose q =min{q1,q2} > 1, and we conclude λ–1ε |x|–2γ(1 +

h())uεe4π(1–γε)u 2

εis bounded inLq(Ω). Then the elliptic estimate on the Euler–Lagrange

equation (16) implies thatis bounded, which also makes a contradiction. Thus, we

com-plete the proof of the proposition.

2.2.2 Asymptotic behavior of uεnear the concentration point

Let

=

λεc–1ε e–2

(9)

For any 0 <δ< 1 –γ, in view of (8), we have by using the Hölder inequality and the singular Trudinger–Moser inequality (3),

λε=

Ω

|x|–2γ1 +h()

u2εe4

π(1–γε)u2εdx

e4π δc2ε

Ω

|x|–2γ1 +h()

u2εe4

π(1–γεδ)u2εdx

Ce4π δc2ε

for some constantCdepending only onδ. This leads to

r2εe4π μc2

εCc–2

ε e

4π(δ+μ)e–4π(1–γε)c2

ε0, for0 <μ< 1 –γ, (24)

asε→0. To characterize the blow-up behavior more exactly, we need to divide the process into two cases as in [30].

Case 1.–1/(1–γ)C.

LetΩε={x∈R2:+r1/(1– γ)

ε xΩ}. Deﬁne two blow-up sequences of function onΩε

as

ζε(x) =c–1ε

+1/(1–γ)x

, ϑε(x) =

+r1/(1–ε γ)x

.

A direct computation shows

ζε(x) =c–2ε x+r–1/(1– γ)

ε

–2γ

1 +h()

ζεe4π(1–γε)(u 2

εc2ε) inΩε, (25)

ϑε(x) =x+–1/(1–γ) –2γ

1 +h()

ζεe4π(1–γε)(1+ζε)ϑε inΩε. (26)

We now investigate the convergence behavior ofζε(x) andϑε(x). Assumelimε→0r–1/(1– γ)

ε ×

= –x¯. From (24), we have→0 obviously. ThusΩε→R2asε→0. In view of|ζε(x)| ≤

1 andζε(x)→0 inxΩε\ {¯x}asε→0, we have by elliptic estimates thatζε(x)→ζ(x)

inCloc1 (R2\ {¯x})∩Cloc0 (R2), whereζ is a bounded harmonic function inR2. Observe that

ζ(x)≤lim supε0ζε(x)≤1 andζ(0) = 1. It follows from the Liouville theorem thatζ ≡1

onR2. Thus, we have

ζε→1 inC1loc

R2\ {¯x}C0 loc

R2 (27)

asε→0. Note also that

ϑε(x)≤ϑε(0) = 0 for allxΩε(x).

In view of (27), we conclude by applying elliptic estimates to (26) that

ϑεϑ inC1loc

R2\ {¯x}C0 loc

R2, (28)

(10)

Observe that

ζε(x) =

(+r1/(1–ε γ)x)

→1 inC1loc(BR\B1/R), (29)

asε→0. Sety=+r1/(1– γ)

ε xwith|xx¯| ≤R, and then we have

|y| ≤r1/(1–ε γ)|xx¯|++r1/(1–ε γ)x¯≤2Rr1/(1– γ)

ε .

Since–1/(1–γ)C, chooseRbig enough such that

xr–1/(1–γ)

ε R.

This together with (29) leads to

lim ε→0

(r1/(1–ε γ)x)

L∞(BR\B1/R(x¯))

=lim ε→0

(+r1/(1– γ)

ε (xr–1/(1–ε γ)))

L∞(BR\B1/R(x¯))

= 1.

Combining with Fatou’s lemma, we obtain

BR\B1/R(x¯)

|xx¯|–2γe8π(1–γ)ϑdx

≤lim sup ε→0

BR\B1/R(x¯)

x+r–1/(1–γ)

ε

–2γ

e4π(1–γε)(1+ζε)ϑεdx

≤lim sup ε→0

1

λε

B

2Rr1/(1–ε γ)\B1 2Rr

–1/(1–γ)

ε

(0)

1 +h() u2

ε(y) |y|2γ e

4π(1–γε)u2ε(y)dy

≤1. (30)

Passing to the limitR→ ∞, we have

R2|xx¯| –2γ

e8π(1–γ)ϑdx≤1.

The uniqueness theorem obtained in [3] implies that

ϑ(x) = – 1 4π(1 –γ)log

1 + 1

1 –γ|xx¯|

2(1–γ)

. (31)

Letx= 0, and then

ϑ(0) =lim

(11)

Thus, it follows from (31) thatx¯= 0. Namely,

ϑ(x) = – 1 4π(1 –γ)log

1 + 1 1 –γ|x|

2(1–γ)

. (32)

Furthermore, we can get

R2|

x|–2γe8π(1–γ)ϑdx= 1. (33)

Case 2. r–1/(1–ε γ)→+∞. Set

Ωε= x∈R2:+rε|xε|γxΩ

.

Denote the blowing up functions onΩε

αε(x) =c–1ε

+rε|xε|γx

, βε(x) =

+rε|xε|γx

.

In view of (16),αε(x) is a distributional solution to the equation

αε(x) =(u) inΩε, (34)

where

=c–2ε |xε|2 γ

+rε|xε|γx –2γ

1 +h()

αεe4π(1–γε)c 2

ε(α2ε–1).

Since–1/(1–γ)→+∞, we have|xε|2γ|+rε|xε|γx|–2γ = 1 +(1) clearly. Since|αε(x)| ≤

1, we obtainis bounded inLp(p> 1) according to (8). Elliptic estimates and embedding

α(x) = 0 inR2.

Note thatα≤1 andα(0) = 1. Thus, together with the Liouville theorem, we obtainα≡1. Also we have

βε=|xε|2γxε+rε|xε|γx –2γ

1 +h()

αεe4π(1–γε)βε(αε+1) inΩε. (35)

Applying elliptic estimates to (35), we conclude that βεβ inCloc1 (R2), whereβ is a

distributional solution to

⎧ ⎨ ⎩

β(0) = 0 =supβ,

β= –e8π(1–γ)β inR2. (36)

For 0 <β< 1, (36) follows from Chen and Li [6] thatβsatisﬁes

R2e 8π(1–γ)β

dx≥ 1

(12)

Using a suitable change of variabley=+rε|xε|γx, for anyR> 0, we have

BR(x¯)

e8π(1–γ)βdx=lim ε→0

BR(0)

1 +h()

e4π(1–γε)(u2

ε(+|xε|γx)–c2ε)dx

≤lim ε→0

1

λε

BRrε||γ()

1 +h() u2

ε(y) |y|2γ e

4π(1–γε)u2

ε(y)dy

≤1, (37)

which leads to a contradiction. Thus, it is impossible forCase 2to happen.

2.2.3 Convergence away from the concentration point

To understand the convergence behavior away from the blow-up pointx0= 0, we need to

investigate howcεuεconverges. Similar to [1,15], deﬁne,τ=min{τcε,uε}, then we have

the following.

Lemma 6 For any0 <τ< 1,we have

lim ε→0

Ω

|∇,τ|2dx=τ.

Proof Observe that/= 1 +(1) inBRr1/(1–γ)

ε (). For any 0 <τ < 1, it follows from

Eq. (16) and the divergence theorem that

Ω

|∇,τ|2dx=

1

λε

Ω

,τuε |x|2γ

1 +h()

e4π(1–γε)u2

εdx

≥ 1

λε

B

Rr1/(1–ε γ)

()

,τuε |x|2γ

1 +h()

e4π(1–γε)u2εdx+oε(1)

=τ

BR(0)

(1 +h())e4π(1–γε)(u 2

ε(+r1/(1–ε γ)y)–c2ε)

|y+–1/(1–γ)xε|2γ

dy+(1).

Hence

lim inf ε→0

Ω

|∇,τ|2dxτ

BR(0)

e8π(1–γ)ϑdy, R> 0.

In view of (33), passing to the limitR→+∞, we obtain

lim inf ε→0

Ω

|∇,τ|2dxτ. (38)

Note that

(τcε)+ 2

=∇(τcε)+· ∇ onΩ

and

(τcε)+=

1 +(1)

(1 –τ) inBRr1/(1–γ)

(13)

Testing Eq. (16) by (τcε)+, for any ﬁxedR> 0, simple computation shows that

Ω

(τcε)+ 2

dx=

Ω

(τcε)+

λε|x|2γ

1 +h()

e4π(1–γε)u2εdx

B

Rrε1/(1–γ)()

(τcε)+

(1 +h())

λε|x|2γ e

4π(1–γε)u2

εdx

=1 +(1)

(1 –τ)

BR(0)

ζε

1 +h()

e4π(1–γε)ϑε2dx.

By passing to the limitε→0, we get

lim inf ε→0

Ω

(τcε)+ 2

dx≥(1 –τ)

BR(0)

e8π(1–γ)ϑdx= 1 –τ. (39)

Since|∇,τ|2+|∇(τcε)+|2=|∇uε|2almost everywhere, it follows that

Ω

(τcε)+ 2

dx+

Ω

|∇,τ|2dx=

Ω

|∇uε|2dx= 1 +(1). (40)

Therefore, we end the proof of this lemma together with (38), (39) and (40).

The following estimate is a byproduct of Lemma 6and will be employed in the next section.

Lemma 7 We have

lim ε→0

Ω

|x|–2γ1 +g(u ε)

e4π(1–γε)u2

εdx=1 +g(0)

Ω

|x|–2γdx+lim ε→0

λε

c2 ε

. (41)

Proof Let 0 <τ< 1 be ﬁxed. By the deﬁnition of,τ, we can get

τcε

1 +g()

e4π(1–γε)u2ε

|x|2γ dx

1 +g(0)

Ω

1

|x|2γ dx

Ω

1 +g(,τ)

e4π(1–γε)u2ε,τ

|x|2γ dx

1 +g(0)

Ω

1

|x|2γ dx

Ω

g(,τ) –g(0)

e4π(1–γε)u2

ε,τ

|x|2γ dx+1 +g(0)

Ω

(e4π(1–γε)u2

ε,τ – 1)

|x|2γ dx. (42)

Combining Lemma6and Proposition5, we see that,σ converges to 0 inC1loc(Ω\ {0}) as

ε→0. Then from (3), one can deduce that

Ω

e4π(1–γε)u2ε,τ

|x|2γ g(,τ) –g(0)dx=(1). (43)

According to the Hölder inequality and the Lagrange theorem, we have

Ω

1

|x|2γ

e4π(1–γε)u2ε,τ– 1dx=o

(14)

Inserting (43) and (44) into (42), one has

lim ε→0

τcε

1 +g()

e4π(1–γε)u2ε

|x|2γ dx=

1 +g(0)

Ω

1

|x|2γ dx. (45)

Moreover, we calculate

>τcε

1 +g()

e4π(1–γε)u2

ε

|x|2γ dx

≤ 1

τ2

>τcε

u2 ε

c2 ε

1 +g()

e4π(1–γε)u2ε

|x|2γ dx

≤ 1

τ2

λ2ε

c2 ε

. (46)

Combining (45) and (46), we obtain

lim ε→0

Ω

(1 +g())e4π(1–γε)u 2

ε

|x|2γ dx

1 +g(0)

Ω

1

|x|2γ dx+

1

τ2lim infε→0

λ2ε

c2 ε

.

It follows by lettingτ→1 that

lim ε→0

Ω

(1 +g())e4π(1–γε)u 2

ε

|x|2γ dx

1 +g(0)

Ω

1

|x|2γ dx≤lim infε0

λ2ε

c2 ε

. (47)

On the other hand, in view of (16), we estimate

Ω

1 +g()

e4π(1–γε)u2ε

|x|2γ dx

1 +g(0)

Ω

1

|x|2γ dx

Ω u2 ε c2 ε

1 +g()

e4π(1–γε)u2ε

|x|2γ

1 +g(0) 1

|x|2γ

dx

=λε

c2 ε – 1 c2 ε Ω

(1 +g(0))u2ε |x|2γ dx

1

c2 ε

Ω

uεg()

8π(1 –γε)|x|2γe

4π(1–γε)u2εdx.

Thus, by Proposition5and (6), (8), one can check that

lim sup ε→0

λ2 ε

c2 ε

≤lim ε→0

Ω

|x|–2γ1 +g()

e4π(1–γε)u2εdx1 +g(0)

Ω

|x|–2γdx. (48)

In view of (47) and (48), we complete the proof of Lemma7.

Corollary 8 Ifθ< 2,thenλε

cθε → ∞asε→0.

Proof In contrast, we haveλε/c2ε →0 asε→0. For anyνW 1,2

0 (Ω) with∇ν2≤1,

clearly, it is impossible for (41) to hold sinceν≡0.

Lemma 9 For any functionφC01(Ω),we have

lim ε→0

Ω

1 +h()

λ–1ε cεuε|x|–2γe4π(1–γε)u 2

(15)

Proof To see this, letφC1

0(Ω) be ﬁxed. Write for simplicity

ωε=

1 +h()

λ–1ε cεuε|x|–2γe4π(1–γε)u 2

ε.

Clearly

Ω

ωεφdx=

{uε<τcε}

ωεφdx+

{uετcε}\B

R1/(1–rε γ)()

ωεφdx

+

{uετcε}∩B

R1/(1– γ)

()

ωεφdx. (50)

Given 0 <τ < 1, we estimate the three integrals on the right hand of (50), respectively. Note that→0 inLq(∀q> 1). This together with Lemma6and Corollary8gives

{uε<τcε}

ωεφdxλ–1ε

sup Ω

φ1 +h()

{uε<τcε}

uε|x|–2γe4π(1–γε)u2ε,τdx

–1ε

{uε<τcε}

uε|x|–2γe4π(1–γε)u2ε,τdx

=(1). (51)

Now we consider inB

R1/(1– γ)()⊂ {x

Ω|τcε}for suﬃciently smallε> 0. One can

deduce from (33) that

{uετcε}∩B

R1/(1–rε γ)()

ωεφdx=φ(0)

1 +(1)

BR\1/R(0)

|x|–2γe8π ϑdx

=φ(0)1 +(1) +oR(1)

. (52)

On the other hand, we calculate

{uετcε}\B

R1/(1–rε γ)()

ωεφdx

C

τ

1 –

B

R1/(1–rε γ)()

u2 ε

λε

e4π(1–γε)u2

ε

|x|2γ dx

=C

τ

1 –

BR(0)

e8π(1–γ)ϑ |x|2γ dx

.

Hence, we derive by (33) that

lim

R→∞εlim→0

{uετcε}\B

R1/(1–r γ)

ε

()

ωεφdx= 0. (53)

Inserting (51)–(53) to (50), we conclude (49) ﬁnally.

In particular, we propose, by lettingφ= 1,

ωε(x) =

1 +h()

λ–1ε cεuε|x|–2γe4π(1–γε)u 2

ε is bounded inL1(Ω), (54)

(16)

We now prove thatcεuε converges to a Green function in distributional sense when

ε→0, whereδ0stands for the Dirac measure centered at 0. More precisely, we have

Lemma 10 cεuεG in C1loc(Ω\ {0})and weakly in W01,q(Ω)for all1 <q< 2,where

GC1(Ω\ {0})is a distributional solution satisfying the following:

⎨ ⎩

G=δ0 inΩ,

G= 0 on∂Ω. (55)

Moreover,G takes the form

G(x) = – 1

2πlog|x|+A0+ξ(x), (56)

whereξ(x)∈C1(Ω)and A0is a constant depending on0.

Proof By Eq. (16),cεuεis a distributional solution to

(cεuε) =ωε inΩ. (57)

It follows from (54) thatωε is bounded inL1(Ω). Using the argument in Struwe ([25],

Theorem 2.2), one concludes thatcεuεis bounded inW01,q(Ω) for all 1 <q< 2. Hence, we

can assume, for any 1 <q< 2,r> 1, that

cεuεG weakly inW01,q(Ω),

cεuεG strongly inLr(Ω).

Testing (57) byφC01(Ω), we deduce

Ω

∇(cεuε)∇φdx=

Ω

φλ–1ε cεuε

1 +h()

|x|–2γe4π(1–γε)u2

ε.

Letε→0 and it yields by (55)

Ω

Gφdx=φ(0),

which implies that –G=δ0in a distributional sense. Since(G+21πlog|x|)∈Lp(Ω) for

anyp> 2, (56) follows from the elliptic solution immediately. Applying elliptic estimates to Eq. (57), we arrive at the conclusion

cεuεG inC1loc

Ω\ {0}. (58)

(17)

2.3 Upper bound calculates by means of capacity estimate

In this subsection, we aim to derive an upper bound of the integralsΩ(1 +g())|x|–2γ×

e4π(1–γε)u2εdx. Analogous to the one obtained in [15], we mainly use the capacity estimate.

Now choose a properδto ensure thatB2δΩ, and then construct a new function space

Mε(ρε,σε) = u|uW1,2

Bδ()\BRr1/(1–γ)

ε ()

:u|∂Bδ()=ρε,u|∂B

Rr1/(1–ε γ)

()=σε

where

ρε= sup Bδ()

, σε= inf B

Rrε1/(1–γ)

()

.

Deﬁne

Λε= inf

u∈Mε(ρε,σε)

Bδ()\B

Rrε1/(1–γ)

()

|∇u|2dx.

Clearly, the inﬁmumΛεcan be attained by the sequenceuk∈Mask→ ∞. By the proof of the Poincaré inequality, we infer thatukis bounded inW01,2(Ω). Without loss of generality,

there exists some functiontW1,2(Ω) such that up to a subsequence. Ask→ ∞, we have

uktweakly inW1,2(Ω),uktinLploc(Ω) for anyp> 0 andukta.e. inΩ. Besides, fort∈Mε(ρε,σε), we have

Bδ()\B

Rr1/(1–ε γ)()

|∇t|2dx≤ lim

k→∞

Bδ()\B

Rr1/(1–ε γ)()

|∇uk|2dx=Λε

and

Λε

Bδ()\B

Rr1/(1–ε γ)

()

|∇t|2dx.

Through the method of variation, we see that there exists some harmonic functiont(x) to reach theΛεwhich satisﬁes the following:

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

t= 0 inBδ()\BRr1/(1–γ)

ε (),

t|∂Bδ()=ρε,

t|∂B

Rrε1/(1–γ)

()=σε.

(59)

Obviously, the solution of (59) can be expressed as

t(x) =alog|xx0|+b.

One can check that

⎧ ⎪ ⎨ ⎪ ⎩

a= σερε

logδ–logRrε1/(1–γ)

,

b=σεlogRrε1/(1–γ)–ρεlogδ

logRrε1/(1–γ)–logδ

.

(18)

Thus,t(x) can be expressed as

t(x) = σε(logδ–log|xxε|) –ρε(logRr

1/(1–γ)

ε –log|xxε|) logδ–logRrε1/(1–γ)

.

With a direct computation, it is easy to check that

Bδ()\B

Rr1/(1–ε γ)()

|∇t|2dx= 2π(σερε)

2

logδ–logRr1/(1–ε γ)

. (61)

According to (23), we have

logδ–logRrε1/(1–γ)=logδ–logR+

2π(1 –γε)c2 ε

1 –γ

1 2(1 –γ)log

λε

c2 ε

. (62)

Furthermore, Lemma10and (31) show that

σε=+

1

– 1 4π(1 –γ)log

1 + π 1 –γR

2(1–γ)

+o(1)

(63)

and

ρε=

1

– 1

2πlogδ+A0+o(1)

, (64)

whereo(1)→0 by lettingε→0 andδ→0 in succession. Setuε=max{ρε,min{,σε}}.

Fromuε∈Mε(ρε,σε), one can easily check that

Bδ()\B

Rr1/(1–ε γ)

()

|∇t|2dx=Λε

Bδ()\B

Rr1/(1–ε γ)

()

uε 2

dx. (65)

Observe that|∇uε| ≤ |∇uε|a.e. inBδ()\BRr1/(1–γ)

ε () ifεis suﬃciently small. Thus, it

follows

Bδ()\B

Rr1/(1–ε γ)

()

uε 2

dx

Bδ()\B

Rr1/(1–ε γ)

()

|∇uε|2dx. (66)

In view of (61), (65) and (66), it can be inferred that

2π(σερε)2≤

1 –

Ω\Bδ()

|∇uε|2dx

B

Rr1/(1–ε γ)()

|∇uε|2dx

×logδ–logRr1/(1–ε γ)

(19)

SincecεuεGinCloc1 (Ω\{0}), we obtain the conclusion through integrating by parts:

Ω\Bδ()

|∇uε|2dx= 1

c2 ε

Ω\Bδ()

|∇Gε|2dx

= –1

c2 ε

Ω\Bδ()

GG dx+

Bδ()

G∂G

∂ν ds

= –1

c2 ε

1

2πlogδA0+(1) +(1)

. (68)

Observe thatϑεϑinCloc1 (R2\ {0}), and

=

ϑε(x)

+ inBRr1/(1–γ)

ε (). (69)

A direct computation shows that

BR(0)

|∇ϑ|2dx=

R

0

2π

4(1 –γ)2(1 + π

1–γ|r|2(1–γ))2

r–4γdr

= 1 4π(1 –γ)log

π

1 –γ +

1

2πlogR

1

4π(1 –γ)+O

1

R2(1–γ)

. (70)

Then it follows from (69) and (70) that

B

Rrε1/(1–γ)()

|∇uε|2dx= 1

c2 ε

B

Rr1/(1–ε γ)()

ϑε(x) 2

dx

= 1

c2 ε

BR(0)

ϑ(y)2dy+(1)

= 1 4πc2

ε(1 –γ)

log π

1 –γ +

1 2πc2

ε

logR– 1

4πc2 ε(1 –γ)

+o(1)

c2 ε

.

This together with (62)–(64) and (68), we obtain

–2πA0–

log(1 +1–πγR2(1–γ))

1 –γ ≤–2logR+

(1 –logλε

c2ε –log

π 1–γ)

2(1 –γ) +o(1). Hence,

lim sup ε→0

λε

c2 ε

π

1 –γe

4π(1–γ)A0+1.

In view of Lemma7, we arrive at the conclusion

Λ4π(1–γ)≤

1 +g(0)

Ω

|x|–2γdx+ π

1 –γe

(20)

2.4 Completion of the proof of Theorem1

As a consequence, if→ ∞, it follows from (71) thatΛ4π(1–γ)is bounded. Otherwise, we

can ﬁnd the extremal functionu0which satisﬁes (17). Therefore, necessarily

sup

u∈W1,2

0 (Ω),∇u2≤1

Ω

1 +g(u)e

4π(1–γ)u2

|x|2γ dx<∞.

3 Proof of Theorem2

3.1 Test function computation

Similar to [30], we construct a blow-up sequenceφεW01,2(Ω) with∇φε2= 1. For

suf-ﬁciently smallε> 0, there exists

Ω

|x|–2γ1 +g(φε)

e4π(1–γ)φ2εdx>1 +g(0)

Ω

|x|–2γdx+ π 1 –γe

4π(1–γ)A0+1. (72)

Then we will ﬁnd (72) is a contradiction to (71), so thathas to be bounded, which means

the blow-up cannot take place. Furthermore, Theorem2follows immediately from what we have proved according to the elliptic estimates. For this purpose we set

φε(x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

c+1

c(–

1

4π(1–γ)log(1 + π 1–γ

|x|2(1–γ)

ε2(1–γ) ) +b), forx∈BRε,

Gξ η

c , forx∈B2\BRε,

G

c, forxΩ\B2,

(73)

whereηC01(B2) is a cut-oﬀ function satisfyingη= 1 onBRε, and|∇η| ≤R2ε. AndGis

given as in (56).bandcare constants which depend only onε, to be determined later. To ensureφεW01,2(Ω), we let

c+1

c

– 1 4π(1 –γ)log

1 + π 1 –γ

|x|2(1–γ)

ε2(1–γ)

+b

=1

c

– 1

2πlog+A0

,

2πc2= –logε– 2πb+ 2πA0+

1 2(1 –γ)log

π

1 –γ +O

1

R2(1–γ)

. (74)

Now we calculate

B

|∇φε|2dx=

BR

|x|2–4γ

4c2(1 –γ)2(1 + π 1–γ|x|2–4

γ)2dx

=

π

1–γR2–2γ

0

t dt

4πc2(1 –γ)(1 +t)2dt

= 1 4πc2(1 –γ)

log π

1 –γ – 1 +logR

2–2γ +O

1

R2–2γ

(21)

On the other hand

Ω\B

|∇φε|2dx= 1

c2

Ω\B

|∇G|2dx+

B2\B

(ξ η)2dx

– 2

B2\B

G∇(ξ η)dx

= 1

c2

Ω\B

GG dx

B

G∂G

∂ν ds

+

B2\B

(ξ η)2dx– 2

B2\B

G∇(ξ η)dx

.

Observe thatξ(x) =O(|x|) asx→0. Sinceηis a cut-oﬀ function, it yields|∇(ξ η)|=O(1) asε→0. Then we have

B2\B

(ξ η)2dx=OR2ε2,

B2\B

G∇(ξ η)dx=O(),

which together with (56) leads to

Ω\B

|∇φε|2dx= 1

c2

– 1

2πlog() +A0+O()

. (76)

Combining (75) and (76), a delicate but straightforward calculation shows

Ω

|∇φε|2dx= 1

c2

–logε 2π

1 4π(1 –γ)+

1 4π(1 –γ)log

π

1 –γ +A0+O

1

R2–2γ

.

Put∇φε2= 1. It yields

c2=A0–

1 2πlogε+

1 4π(1 –γ)log

π

1 –γ

1

4π(1 –γ)+O

1

R2–2γ

. (77)

Together with (74) and (77), we are led to

b= 1

4π(1 –γ)+O

1

R2–2γ

. (78)

For allx∈BRε, it follows from (77) and (78) that

4π(1 –γ)φ2ε≥4π(1 –γ)c2+ 8π(1 –γ)b– 2log

1 + π|x|

2(1–γ)

(1 –γ)ε2(1–γ)

= 1 + 4π(1 –γ)A0+log

π

1 –γ – 2(1 –γ)logε

– 2log

1 + π|x|

2(1–γ)

(1 –γ)ε2(1–γ)

+O

1

R2–2γ

. (79)

Note thatφε(x)

c L∞(BRε)→1 by passing to the limitε→0. Whenr, there exists

φε(x)

c

=1 +–log(1 +π r2 ε2) +b

c2

(22)

asε→0. Sinceφε(x)∼cinBRεandg(c) =o(c12), we concludeg(φε(ξε)) =o(c12) asε→0,

whereξε∈BRε. Combining with the mean value theorem, it follows from (79) that

B

1 +g(φε)

e4π(1–γ)φε2 |x|2γ dx=

1 +gφε(ξε)

B

e4π(1–γ)φε2 |x|2γ dx

1 +o

1

c2

π

1 –γe

1+4π(1–γ)A0+O( 1 R2–2γ)

× R

0

2πr1–2γ

(1 +1–πγr2–2γ)2dr

= π (1 –γ)e

1+4π(1–γ)A0+O

1

R2–2γ

+o

1

c2

. (80)

Furthermore,cG

ε≥0 a.e. inΩ\BRε, by using the inequalitye

tt+ 1,t0, we estimate

Ω\B

1 +g(φε)

e4π(1–γ)φε2 |x|2γ dx

Ω\B2

1 +g(φε)

1 + 4π(1 –γ)φ2 ε |x|2γ dx

Ω

1 +g(0)|x|–2γdx+O()2–2γlog2()

+4π(1 –γ)

c2

Ω

1 +g(0)|x|–2γG2dx+O(Rε)2–2γ. (81)

Observe that

O()2–2γ=O()2–2γlog2()=O

1

R2–2γ

.

This together with (80) and (81) yields

Ω

1 +g(φε)

e4π(1–γ)φε2 |x|2γ dx

≥1 +g(0)

Ω

|x|–2γdx+ π (1 –γ)e

4π(1–γ)A0+1

+4π(1 –γ)

c2

Ω

(1 +g(0))G2 |x|2γ dx+O

1

R2–2γ

+o

1

c2

. (82)

Recalling (77) and the choiceR= –logε1/(1–γ), one can deduce that 1

R2–2γ =o(c12).

There-fore, we conclude from (82) that

Ω

1 +g(φε)

e4π(1–γ)φε2 |x|2γ dx>

1 +g(0)

Ω

|x|–2γdx+ π

1 –γe

4π(1–γ)A0+1.

References

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