introduction - new york university

A REMARK ON THE CONCENTRATION COMPACTNESSPRINCIPLE IN CRITICAL DIMENSION

FENGBO HANG

Abstract. We prove some re�nements of concentration compactness princi-ple for Sobolev space W 1;n on a smooth compact Riemannian manifold ofdimension n. As an application, we extend Aubin�s theorem for functions onSn with zero �rst order moments of the area element to higher order momentscase. Our arguments are very �exible and can be easily modi�ed for functionssatisfying various boundary conditions or belonging to higher order Sobolevspaces.

1. Introduction

Let n � 2, � Rn be a bounded open subset with smooth boundary. In [M], itis showed that for any u 2W 1;n

0 () n f0g,Z

exp

0@an jujn

n�1

krukn

n�1Ln()

1A dx � c (n) jj : (1.1)

Here jj is the volume of and

an = n��Sn�1�� 1

n�1 : (1.2)��Sn�1�� is the volume of Sn�1 under the standard metric. For convenience, we willuse this notation an throughout the paper. (1.1) can be viewed as a limit case ofthe Sobolev embedding theorem.To study extremal problems related to (1.1), a concentration compactness theo-

rem [Ln, Theorem 1.6 on p196, Remark 1.18 on p199] was proved. For n = 2, theargument is elegant. The approach is recently applied in [CH] on smooth Riemannsurfaces to deduce re�nements of the concentration compactness principle (see [CH,Section 2]). These re�nements are crucial in extending Aubin�s classical theorem onS2 for functions with zero �rst order moments of the area element (see [A, Corollary2 on p159]) to higher order moments cases, motivated from similar inequalities onS1 (see [CH, GrS, OPS, W]).For n � 3, due to the subtle analytical di¤erence between weak convergence in

L2 and Lp, p 6= 2, [Ln, p197] has to use special symmetrization process to gain thepointwise convergence of the gradient of functions considered. Unfortunately, aspointed out in [CCH], this argument is not su¢ cient to derive [Ln, Remark 1.18].More accurate argument is presented in [CCH]. More recently, [LLZ] deduce similarresults on domains in Heisenberg group without using symmetrization process.The main aim of this note is to extend the analysis in [CH] from dimension 2

to dimensions at least 3. In particular we will prove re�nements of concentrationcompactness principle (Proposition 2.1 and Theorem 2.1). Our approach also doesnot use symmetrization process and is more close to [CH, Ln]. It can be easily

1

2 FENGBO HANG

modi�ed for functions satisfying various boundary conditions or belonging to higherorder Sobolev spaces.Throughout the paper, we will assume (Mn; g) is a smooth compact Riemannian

manifold with dimension n � 2. For an integrable function u on M , we denote

u =1

� (M)

ZM

ud�: (1.3)

Here � is the measure associated with the Riemannian metric g. The Moser-Trudinger inequality (see [F]) tells us that for every u 2W 1;n (M) n f0g with u = 0,we have Z

M

exp

0@an jujn

n�1

krukn

n�1Ln(M)

1A d� � c (M; g) : (1.4)

Here an is given in (1.2).It follows from (1.4) and Young�s inequality that for any u 2 W 1;n (M) with

u = 0, we have the Moser-Trudinger-Onofri inequality

log

ZM

enud� � �n kruknLn + c1 (M; g) : (1.5)

Here

�n =

�n� 1n

�n�11

jSn�1j : (1.6)

We will use this notation �n throughout the paper.In Section 2 we will derive some re�nements of concentration compactness prin-

ciple in dimension n. These re�nements will be used in Section 3 to extend Aubin�stheorem on Sn to vanishing higher order moments case. In Section 4, we discussmodi�cations of our approach applied to higher order Sobolev spaces and Sobolevspaces on surfaces with nonempty boundary.Last but not least, we would like to thank the referee for careful reading of the

manuscript and many valuable suggestions.

2. Concentration compactness principle in critical dimension

As usual, for u 2W 1;n (M), we denote

kukW 1;n(M) =�kuknLn(M) + kruk

nLn(M)

� 1n

:

We start with a basic consequence of Moser-Trudinger inequality (1.4). It shouldbe compared with [CH, Lemma 2.1].

Lemma 2.1. For any u 2W 1;n (M) and a > 0, we haveZM

eajujn

n�1d� <1: (2.1)

Proof. We could use the same argument as in the proof of [CH, Lemma 2.1]. Insteadwe modify the approach a little bit so that it also works for higher order Sobolevspaces. Without losing of generality, we can assume u is unbounded. Let " > 0 bea tiny number to be determined, we can �nd v 2 C1 (M) such that

ku� vkW 1;n(M) < ":

CONCENTRATION COMPACTNESS PRINCIPLE IN CRITICAL DIMENSION 3

We denote w = u� v, thenu = v + w = v + w + w � w:

It follows thatjuj � kvkL1 + jwj+ jw � wj :

Hencejuj

nn�1 � 2 1

n�1 (kvkL1 + jwj)n

n�1 + 21

n�1 jw � wjn

n�1 :

We get

eajujn

n�1 � e21

n�1 a(kvkL1+jwj)n

n�1e2

1n�1 ajw�wj

nn�1

� e21

n�1 a(kvkL1+jwj)n

n�1ean

jw�wjn

n�1

krwkn

n�1Ln

when " is small enough. It follows from (1.4) thatZM

eajujn

n�1d� � c (M; g) e2

1n�1 a(kvkL1+jwj)

nn�1

<1:

�

Now we are ready to prove a localized version of [Ln, Theorem 1.6]. The readershould compare it with [CH, Lemma 2.2].

Proposition 2.1. Assume ui 2 W 1;n (Mn) such that ui = 0, ui * u weakly inW 1;n (M) and

jruijn d�! jrujn d�+ � (2.2)

as measure. If x 2 M and p 2 R such that 0 0,

supi

ZBr(x)

eanpjuijn

n�1d� <1: (2.3)

Herean = n

��Sn�1�� 1n�1 : (2.4)

Proof. Fix p1 2�p; � (fxg)�

1n�1�, then

� (fxg) < 1

pn�11

: (2.5)

We can �nd a " > 0 such that

(1 + ")� (fxg) < 1

pn�11

(2.6)

and(1 + ") p < p1: (2.7)

Let vi = ui � u, then vi * 0 weakly in W 1;n (M), vi ! 0 in Ln (M). For any' 2 C1 (M), we have

kr ('vi)knLn� (k'rvikLn + kvir'kLn)

n

� (k'ruikLn + k'rukLn + kvir'kLn)n

� (1 + ") k'ruiknLn + c (n; ") k'ruknLn + c (n; ") kvir'k

nLn :

4 FENGBO HANG

It follows that

lim supi!1

kr ('vi)knLn

� (1 + ")

�ZM

j'jn d� +ZM

j'jn jrujn d��+ c (n; ") k'ruknLn

= (1 + ")

ZM

j'jn d� + c1 (n; ")ZM

j'jn jrujn d�:

Note that we can �nd a ' 2 C1 (M) such that 'jBr(x)= 1 for some r > 0 and

(1 + ")

ZM

j'jn d� + c1 (n; ")ZM

j'jn jrujn d� < 1

pn�11

:

Hence for i large enough, we have

kr ('vi)knLn <1

pn�11

:

In other words,

kr ('vi)kn

n�1Ln <

1

p1:

We have ZBr(x)

eanp1jvi�'vijn

n�1d� �

ZM

eanp1j'vi�'vijn

n�1d�

�ZM

eanj'vi�'vij

nn�1

kr('vi)kn

n�1Ln d�

� c (M; g) :

Next we observe that

juijn

n�1

= j(vi � 'vi) + u+ 'vijn

n�1

� (1 + ") jvi � 'vijn

n�1 + c (n; ") jujn

n�1 + c (n; ") j'vijn

n�1 ;

henceeanjuij

nn�1 � e(1+")anjvi�'vij

nn�1

ec(n;")jujn

n�1ec(n;")j'vij

nn�1

:

Since e(1+")anjvi�'vijn

n�1 is bounded in Lp11+" (Br (x)), ec(n;")juj

nn�1 2 Lq (Br (x)) for

any 0 < q < 1 (by Lemma 2.1), ec(n;")j'vijn

n�1 ! 1 as i ! 1 and p11+" > p, it

follows from Holder inequality that eanjuijn

n�1 is bounded in Lp (Br (x)). �Corollary 2.1. Assume ui 2 W 1;n (M) such that ui = 0 and kruikLn � 1. Wealso assume ui * u weakly in W 1;n (M) and

jruijn d�! jrujn d�+ � (2.8)

as measure. Let K be a compact subset of M and

� = maxx2K

� (fxg) : (2.9)

(1) If � < 1, then for any 1 � p < ��1

n�1 ,

supi

ZK

eanpjuijn

n�1d� <1: (2.10)

(2) If � = 1, then � = �x0 for some x0 2 K, u = 0 and after passing to asubsequence,

eanjuijn

n�1 ! 1 + c0�x0 (2.11)

as measure for some c0 � 0.

Proof. First assume � < 1. For any x 2 K, we have

1 � p < ��1

n�1 � � (fxg)�1

n�1 : (2.12)

By the Proposition 2.1 we can �nd rx > 0 such that

supi

ZBrx (x)

eanpjuijn

n�1d� <1: (2.13)

We have an open coveringK �

[x2K

Brx (x) ;

hence there exists x1; � � � ; xN 2 K such that

K �N[k=1

Brk (xk) :

Here rk = rxk . For any i,ZK

eanpjuijn

n�1d� �

NXk=1

ZBrk(xk)

eanpjuijn

n�1d�

�NXk=1

supj

ZBrk(xk)

eanpjuj jn

n�1d�

< 1:Next assume � = 1, then for some x0 2 K, � (fx0g) = 1. SinceZ

M

jrujn d�+ � (M) � 1;

we see u must be a constant function and � = �x0 . Using u = 0, we see u = 0.After passing to a subsequence, we can assume ui ! 0 a.e. For any r > 0, it follows

from the �rst case that eanjuijn

n�1 is bounded in Lq (MnBr (x0)) for any q < 1,hence eanjuij

nn�1 ! 1 in L1 (MnBr (x0)). This together with the factZ

M

eanjuijn

n�1d� � c (M; g)

implies that after passing to a subsequence, eanjuijn

n�1 ! 1 + c0�x0 as measure forsome c0 � 0. �Theorem 2.1. Assume � > 0, mi > 0, mi !1, ui 2W 1;n (Mn) such that ui = 0and

log

ZM

enmiuid� � �mni : (2.14)

We also assume ui * u weakly in W 1;n (M), jruijn d�! jrujn d�+� as measureand

enmiuiRMenmiuid�

! � (2.15)

6 FENGBO HANG

as measure. Let �x 2M : � (fxg) � ��1n �

= fx1; � � � ; xNg ; (2.16)

here

�n =

�n� 1n

�n�11

jSn�1j ; (2.17)

then

� =NXi=1

�i�xi ; (2.18)

here �i � 0 andPN

i=1 �i = 1.

Proof. Assume x 2 M such that � (fxg) < ��1n �, then we claim that for somer > 0, � (Br (x)) = 0. Indeed we �x p such that

�1

n�1n ��

1n�1 0,ZBr(x)

eanpjuijn

n�1d� � c; (2.20)

here c is a positive constant independent of i. By Young�s inequality we have

nmiui � anp juijn

n�1 +�nm

ni

pn�1; (2.21)

hence ZBr(x)

enmiuid� � ce�nm

ni

pn�1 :

It follows that RBr(x)

enmiuid�RMenmiuid�

� ce

��n

pn�1��mni :

In particular,

� (Br (x)) � lim infi!1

RBr(x)


= 0:

We get � (Br (x)) = 0. The claim is proved.Clearly the claim implies

� (Mn fx1; � � � ; xNg) = 0: (2.22)

Hence � =PN

i=1 �i�xi ; with �i � 0 andPN

i=1 �i = 1. �

It is worth pointing out that the arguments for Proposition 2.1 and Theorem 2.1can be easily modi�ed to work for functions satisfying various boundary conditionsor belonging to higher order Sobolev spaces. We will discuss these examples inSection 4.


3. A generalization of Aubin inequality to higher order momentscase on Sn

Here we will extend Aubin�s inequality for functions on Sn with zero �rst ordermoments of the area element (see [A, Corollary 2, p159]) to higher order momentscases. For n = 2, this is done in [CH].First we introduce some notations. For a nonnegative integer k, we denote

Pk = Pk�Rn+1

�=�all polynomials on Rn+1 with degree at most k

;(3.1)

�Pk =

�Pk�Rn+1

�=

�p 2 Pk :

ZSnpd� = 0

�: (3.2)

Here � is the standard measure on Sn.

De�nition 3.1. For m 2 N, let

Nm (Sn)= fN 2 N : 9x1; � � � ; xN 2 Sn and �1; � � � ; �N 2 [0;1) s.t. �1 + � � �+ �N = 1

and for any p 2�Pm, �1p (x1) + � � �+ �Np (xN ) = 0.

�= fN 2 N : 9x1; � � � ; xN 2 Sn and �1; � � � ; �N 2 [0;1) s.t. for any p 2 Pm,

�1p (x1) + � � �+ �Np (xN ) =1

jSnj

ZSnpd�:

�,

and

Nm (Sn) = minNm (Sn) : (3.3)

As in [CH], every choice of �1; � � � ; �N and x1; � � � ; xN in De�nition 3.1 corre-sponds to an algorithm for numerical integration of functions on Sn (see [Co, HSW]for further discussion).

Theorem 3.1. Assume u 2W 1;n (Sn) such thatRSn ud� = 0 (here � is the standard

measure on Sn) and for every p 2�Pm,

RSn pe

nud� = 0, then for any " > 0, we have

log

ZSnenud� �

��n

Nm (Sn)+ "

�kruknLn + c (m;n; ") : (3.4)

Here Nm (Sn) is de�ned in De�nition 3.1 and

�n =

�n� 1n

�n�11

jSn�1j :

Proof. Let � = �nNm(Sn) + ". If the inequality is not true, then there exists vi 2

W 1;n (Sn) such that vi = 0,RSn pe

nvid� = 0 for all p 2�Pm and

log

ZSnenvid�� krviknLn !1

as i!1. In particularRSn e

nvid�!1. Since

log

ZSnenvid� � �n krviknLn + c (n) ;

8 FENGBO HANG

we see krvikLn !1. Let mi = krvikLn and ui =vimi, then mi !1, kruikLn =

1 and ui = 0. After passing to a subsequence, we have

ui * u weakly in W 1;n (Sn) ;

log

ZSnenmiuid�� mn

i ! 1;

jruijn d� ! jrujn d�+ � as measure;enmiuiR

Sn enmiuid�

! � as measure.

Let �x 2 Sn : � (fxg) � ��1n �

= fx1; � � � ; xNg ; (3.5)

then it follows from Theorem 2.1 that

� =NXi=1

�i�xi : (3.6)

Moreover �i � 0,PN

i=1 �i = 1 and

NXi=1

�ip (xi) = 0

for any p 2�Pm. Hence N 2 Nm (Sn). It follows from � (Sn) � 1 and (3.5) that

��1n �N � 1;We get

� � �nN� �nNm (Sn)

:

This contradicts with the choice of �. �

Indeed, what we get from the above argument is the following

Theorem 3.2. If u 2 W 1;n (Sn) such thatRSn ud� = 0 (here � is the standard

measure on Sn) and for every p 2�Pm,��Z

Snpenud�

�� b (p) ; (3.7)

here b (p) is a nonnegative number depending only on p, then for any " > 0,

log

ZSnenud� �

��n

Nm (Sn)+ "

�kruknLn + c (m;n; b; ") : (3.8)


�n =

�n� 1n

�n�11

jSn�1j :

We remark that the constant �nNm(Sn) +" is almost optimal in the following sense:

If a � 0 and c 2 R such that for any u 2W 1;n (Sn) with u = 0 andRSn pe

nud� = 0

for every p 2�Pm, we have

log

ZSnenud� � a kruknLn + c; (3.9)

then a � �nNm(Sn) . This claim can be proved almost the same way as the argument

in [CH, Lemma 3.1]. For reader�s convenience we sketch the proof here.First we note that we can rewrite the assumption as for any u 2W 1;n (Sn) withR

Sn penud� = 0 for every p 2

�Pm, we have

log

ZSnenud� � a kruknLn + nu+ c: (3.10)

Assume N 2 N, x1; � � � ; xN 2 Sn and �1; � � � ; �N 2 [0;1) s.t. �1 + � � �+ �N = 1and for any p 2

�Pm, �1p (x1) + � � � + �Np (xN ) = 0. We will prove a � �n

N .The above remark follows. Without losing of generality we can assume �i > 0 for1 � i � N and xi 6= xj for 1 � i < j � N .For x; y 2 Sn, we denote xy as the geodesic distance between x and y on Sn. For

r > 0 and x 2 Sn, we denote Br (x) as the geodesic ball with radius r and center xi.e. Br (x) = fy 2 Sn : xy < rg.Let � > 0 be small enough such that for 1 � i < j � N , B2� (xi) \B2� (xj) = ;.

For 0 < " < �, we let

�" (t) =

8<:nn�1 log

�" ; 0 < t < ";

nn�1 log

�t ; " < t < �;

0; t > �:

(3.11)

If b 2 R, then we write

�";b (t) =

8<: �" (t) + b; 0 < t < �;b�2� t

�

�; � < t < 2�;

0; t > 2�:(3.12)

Let

v (x) =NXi=1

�"; 1n log �i (xxi) ; (3.13)

then calculation showsZSnenvd� =

��Sn�1�� n2

n�1 "�n

n�1 +O

�log

1

"

�(3.14)

as "! 0+.For p 2

�Pm, using

NXi=1

�ip (xi) = 0; (3.15)

we can show ZSnenvpd� = O

�log

1

"

�(3.16)

as "! 0+.To get a test function satisfying orthogornality condition, we need to do some

corrections. Let us �x a base of�Pm��Sn, namely p1jSn ; � � � ; pljSn , here p1; � � � ; pl 2

�Pm. We �rst claim that there exists 1; � � � ; l 2 C1c

Snn

N[i=1

B2� (xi)

!such that

10 FENGBO HANG

the determinant

det

�ZSn jpkd�

�1�j;k�l

6= 0: (3.17)

Indeed, �x a nonzero smooth function � 2 C1c

Snn

N[i=1

B2� (xi)

!, then �p1; � � � ; �pl

are linearly independent. It follows that the matrix�ZSn�2pjpkd�

�1�j;k�l

is positive de�nite. Then j = �2pj satis�es the claim.It follows from (3.17) that we can �nd �1; � � � ; �l 2 R such thatZ

Sn

0@env + lXj=1

�j j

1A pkd� = 0 (3.18)

for k = 1; � � � ; l. Moreover

�j = O

�log

1

"

�(3.19)

as "! 0+. As a consequence we can �nd a constant c1 > 0 such thatlX

j=1

�j j + c1 log1

"� log 1

": (3.20)

We de�ne u as

enu = env +

lXj=1

�j j + c1 log1

": (3.21)

Note this u will be the test function we use to prove our remark.

It follows from (3.18) thatRSn e

nupd� = 0 for all p 2�Pm. Moreover using (3.14)

and (3.19) we seeZSnenud� =

��Sn�1�� n2

n�1 "�n

n�1 +O

�log

1

"

�(3.22)

=��Sn�1�� n2

n�1 "�n

n�1 (1 + o (1)) ;

hence

log

ZSnenud� =

n

n� 1 log1

"+O (1) (3.23)

as "! 0+. Calculation shows

u = o

�log

1

"

�(3.24)

and ZSnjrujn d� =

�n

n� 1

�n ��Sn�1��N log 1"+ o

�log

1

"

�: (3.25)

We plug u into (3.10) and get

n

n� 1 log1

"��

n

n� 1

�n ��Sn�1��Na log 1"+ o

�log

1

"

�:

Divide log 1" on both sides and let "! 0+, we see a � �nN .


It is clear that N1 (Sn) = 2. Hence Aubin�s theorem [A, Corollary 2 on p159]follows from Theorem 3.1.

Lemma 3.1. N2 (Sn) = n+ 2.

Corollary 3.1. Assume u 2 W 1;n (Sn) such thatRSn ud� = 0 (here � is the stan-

dard measure on Sn) and for every p 2�P2,��Z

Snpenud�

�� b (p) ;

here b (p) is a nonnegative number depending only on p, then for any " > 0, wehave

log

ZSnenud� �

��nn+ 2

+ "

�kruknLn + c (n; b; ") : (3.26)

We will prove Lemma 3.1 after some preparations. In RN , we have the hyper-plane

H =�x 2 RN : x1 + � � �+ xN = 1

: (3.27)

Here x1; � � � ; xN are the coordinates of x. Let e1; � � � ; eN be the standard base ofRN and

y =1

N(e1 + � � �+ eN ) : (3.28)

We denote

� =

(x 2 H : jx� yj =

rN � 1N

): (3.29)

Note that � is N � 2 dimensional sphere with radiusq

N�1N .

Lemma 3.2. For any p 2 P2�RN�, we have

1

j�j

Z�

pdS =1

N

NXi=1

p (ei) : (3.30)

Here dS is the standard measure on �, and j�j is the measure of � under dS.

Proof. For 1 � i � N , we have

1

j�j

Z�

xidS =1

j�j

Z�

x1dS:

Moreover1

j�j

Z�

(x1 + � � �+ xN ) dS = 1;

hence 1j�jR�x1dS =

1N . It follows that for 1 � i � N ,

1

j�j

Z�

xidS =1

N: (3.31)

To continue we observe that for 1 � i � N ,

1

j�j

Z�

x2i dS =1

j�j

Z�

x21dS:

12 FENGBO HANG

On the other hand,

1

j�j

Z�

NXi=1

�xi �

1

N

�2dS =

N � 1N

;

developing the identity out we get for 1 � i � N ,

1

j�j

Z�

x2i dS =1

N: (3.32)

At last we claim for 1 � i < j � N ,

1

j�j

Z�

xixjdS = 0: (3.33)

Indeed this follows from

1

j�j

Z�

xixjdS =1

j�j

Z�

x1x2dS;

1

j�j

Z�

(x1 + � � �+ xN )2 dS = 1;

and (3.32). Lemma 3.2 follows from (3.31), (3.32) and (3.33). �

Now we are ready to prove Lemma 3.1.

Proof of Lemma 3.1. Assume N 2 N2 (Sn), we claim N � n+ 2. If this is not thecase, then N < n+ 2. We can �nd x1; � � � ; xN 2 Sn, �1; � � � ; �N � 0 such that

�1p (x1) + � � �+ �Np (xN ) =1

jSnj

ZSnpd�

for every p 2 P2�Rn+1

�. In particular

�1 + � � �+ �N = 1;

�1x1 + � � �+ �NxN = 0:

Let

V = span fx1; � � � ; xNg ;then dimV � N � 1 � n. Hence we can �nd a nonzero vector � 2 V ?. Letp (x) = (� � x)2, then

0 = �1p (x1) + � � �+ �Np (xN ) =1

jSnj

ZSnpd� > 0:

A contradiction. Hence N2 (Sn) � n+ 2.On the other hand, it follows from Lemma 3.2 and dilation, translation and

orthogornal transformation that the vertex of a regular (n+ 1)-simplex embeddedin the unit ball, namely x1; � � � ; xn+2, satis�es

1

n+ 2p (x1) + � � �+

1

n+ 2p (xn+2) =

1

jSnj

ZSnpd�

for every p 2 P2�Rn+1

�. Hence N2 (Sn) � n+ 2. �

4. Further discussions

In this section we show that our approach above can be modi�ed without muche¤ort for higher order Sobolev spaces and Sobolev spaces on surfaces with nonemptyboundary. For reader�s convenience we carefully write down theorems in everycase considered, although they seem a little repetitive. Another reason is thesestatements are of interest themselves.

4.1. W s;ns (Mn) for even s. Recall (Mn; g) is a smooth compact Riemannianmanifold with dimension n. Let s 2 N be an even number strictly less than n, wehave the usual Sobolev space W s;ns (M) with norm

kukW s; n

s (M)=

sX

k=0

ZM

��Dku��ns d�! s

n

(4.1)

for u 2 W s;ns (M). Here Dk is the di¤erentiation associated with the naturalconnection of g. Standard elliptic theory tells us

kukW s; n

s� c (M; g)

� � s2u Lns+ kuk

Lns

�: (4.2)

We also have the Poincare inequality

ku� ukLns� c (M; g)

� s2u Lns

(4.3)

for u 2W s;ns (M).Let

as;n =n

jSn�1j

�n2 2s�

�s2

��n�s2

� ! nn�s

: (4.4)

Here � is the usual Gamma function i.e.

� (�) =

Z 1

0

t��1e�tdt (4.5)

for � > 0. The Moser-Trudinger inequality (see [BCY, F]) tells us that for everyu 2W s;ns (M) n f0g with u = 0, we haveZ

M

exp

0@as;n jujn

n�s � s2u nn�s

Lns

1A d� � c (M; g) : (4.6)

It follows from (4.6) and Young�s inequality that for any u 2W s;ns (M) with u = 0,we have the Moser-Trudinger-Onofri inequality

log

ZM

enud� � �s;n � s

2u ns

Lns+ c1 (M; g) : (4.7)

Here

�s;n = s

�n� sn

��Sn�1��n�ss

��n�s2

��n2 2s�

�s2

�!ns

: (4.8)

Again we start from a basic qualitative property of functions in W s;ns (M).

Lemma 4.1. For any u 2W s;ns (M) and a > 0, we haveZM

eajujn

n�sd� <1: (4.9)

14 FENGBO HANG

This follows from (4.6) exactly in the same way as the proof of Lemma 2.1.Indeed the argument there is written in a way working for higher order Sobolevspaces as well.Next we will derive an analogy of Proposition 2.1 for W s;ns (M).

Proposition 4.1. Assume s 2 N is an even number strictly less than n, ui 2W s;ns (Mn) such that ui = 0, ui * u weakly in W s;ns (M) and�� s

2ui��ns d�! �� s

2u��ns d�+ � (4.10)

as measure. If x 2 M and p 2 R such that 0 0,

supi

ZBr(x)

eas;npjuijn

n�sd� <1: (4.11)

Here

as;n =n

jSn�1j

�n2 2s�

�s2

��n�s2

� ! nn�s

: (4.12)

Proof. Fix p1 2�p; � (fxg)�

sn�s�, then

� (fxg) < 1

pn�ss

1

: (4.13)


(1 + ")� (fxg) < 1

pn�ss

1

(4.14)

and

(1 + ") p < p1: (4.15)

Let vi = ui � u, then vi * 0 weakly in W s;ns (M), vi ! 0 in W s�1;ns (M). For any' 2 C1 (M), we have

�� s2 ('vi)

�� '� s

2 vi��+ c s�1X

k=0

��Ds�k'�� Dkvi

��

��'� s2ui��+ ��'� s

2u��+ c s�1X

k=0


�� :Hence � s

2 ('vi) ns

Lns

� '� s

2ui Lns+ '� s

2u Lns+ c

s�1Xk=0


�� Lns

!ns

� (1 + ") '� s

2ui ns

Lns+ c (")

'� s2u ns

Lns+ c (")

s�1Xk=0


�� ns

Lns:

Letting i!1,

lim supi!1

� s2 ('vi)

ns

Lns

� (1 + ")

�ZM

j'jns d� +

ZM

j'jns�� s

2u��ns d��+ c (") '� s

2u ns

Lns

= (1 + ")

ZM

j'jns d� + c1 (")

ZM

j'jns�� s

2u��ns d�:

We can �nd a ' 2 C1 (M) such that 'jBr(x)= 1 for some r > 0 and

(1 + ")

ZM

j'jns d� + c1 (")

ZM

j'jns�� s

2u��ns d� < 1

pn�ss

1

:

Hence for i large enough, we have � s2 ('vi)

ns

Lns<

1

pn�ss

1

:

In other words, � s2 ('vi)

nn�s

Lns<1

p1:

We have ZBr(x)

eas;np1jvi�'vijn

n�sd� �

ZM

eas;np1j'vi�'vijn

n�sd�

�ZM

e

as;nj'vi�'vij

nn�s � s

2 ('vi) nn�s

Lns d�

� c (M; g) :


juijn

n�s

= j(vi � 'vi) + u+ 'vijn

n�s

� (1 + ") jvi � 'vijn

n�s + c (") jujn

n�s + c (") j'vijn

n�s ;

henceeas;njuij

nn�s � e(1+")as;njvi�'vij

nn�s

ec(")jujn

n�sec(")j'vij

nn�s

:

Since e(1+")as;njvi�'vijn

n�s is bounded in Lp11+" (Br (x)), ec(")juj

nn�s 2 Lq (Br (x)) for

any 0 < q <1 (by Lemma 4.1), ec(")j'vijn

n�s ! 1 as i!1 and p11+" > p, it follows

from Holder inequality that eas;njuijn

n�s is bounded in Lp (Br (x)). �Proposition 4.1 together with a covering argument implies the following corollary,

just like the proof of Corollary 2.1.

Corollary 4.1. Assume ui 2W s;ns (M) such that ui = 0 and � s

2ui Lns� 1. We

also assume ui * u weakly in W s;ns (M) and�� s2ui��ns d�! �� s

2u��ns d�+ � (4.16)


� = maxx2K

� (fxg) : (4.17)

16 FENGBO HANG

(1) If � < 1, then for any 1 � p < ��s

n�s ,

supi

ZK

eas;npjuijn

n�sd� <1: (4.18)


eas;njuijn

n�s ! 1 + c0�x0 (4.19)


We are ready to derive an analogy of Theorem 2.1 for W s;ns (M).

Theorem 4.1. Let s 2 N be an even number strictly less than n. Assume � > 0,mi > 0, mi !1, ui 2W s;ns (Mn) such that ui = 0 and

log

ZM

enmiuid� � �mnsi : (4.20)

We also assume ui * u weakly in W s;ns (M),�� s

2ui��ns d� ! �� s

2u��ns d� + � as

measure andenmiuiR

Menmiuid�

! � (4.21)

as measure. Let �x 2M : � (fxg) � ��1s;n�

= fx1; � � � ; xNg ; (4.22)

here

�s;n = s

�n� sn


��n�s2

��n2 2s�

�s2

�!ns

; (4.23)

then

� =NXi=1

�i�xi ; (4.24)


i=1 �i = 1.

Proof. Assume x 2 M such that � (fxg) < ��1s;n�, then we claim for some r > 0,� (Br (x)) = 0. Indeed we �x p such that

�s

n�ss;n ��

sn�s < p < � (fxg)�

sn�s ; (4.25)


eas;npjuijn

n�sd� � c; (4.26)

here c is a positive constant independent of i. By Young�s inequality we have

nmiui � as;np juijn

n�s +�s;nm

nsi

pn�ss

; (4.27)

hence ZBr(x)

enmiuid� � ce

�s;nm

nsi

pn�ss :




� ce

�s;n

pn�ss

��!m

nsi

:

In particular,


RBr(x)


= 0:

We get � (Br (x)) = 0. The claim is proved.Clearly the claim implies

� (Mn fx1; � � � ; xNg) = 0: (4.28)

Hence � =PN


i=1 �i = 1. �

The analogy of Theorem 3.2 is the following

Theorem 4.2. Let s 2 N be an even number strictly less than n. If u 2W s;ns (Sn)such that

RSn ud� = 0 (here � is the standard measure on S

n) and for every p 2�Pm,��Z

Snpenud�

�� b (p) ; (4.29)


log

ZSnenud� �

��s;n

Nm (Sn)+ "

� � s2u ns

Lns+ c (m;n; b; ") : (4.30)


�s;n = s

�n� sn


��n�s2

��n2 2s�

�s2

�!ns

: (4.31)

Proof. Let � = �s;nNm(Sn) + ". If the inequality (4.30) is not true, then there exists

vi 2W s;ns (Sn) such that vi = 0,��ZSnpenvid�

�� b (p) ;

for all p 2�Pm and

log

ZSnenvid��

� s2 vi ns

Lns!1

as i!1. In particularRSn e

nvid�!1. By (4.7),

log

ZSnenvid� � �s;n

� s2 vi ns

Lns+ c (n) ;

18 FENGBO HANG

hence � s

2 vi Lns! 1. Let mi =

� s2 vi Lnsand ui =

vimi, then mi ! 1, � s

2ui Lns= 1 and ui = 0. After passing to a subsequence, we have

ui * u weakly in W s;ns (Sn) ;

log

ZSnenmiuid�� m

nsi ! 1;�� s

2ui��ns d� !

�� s2u��ns d�+ � as measure;

enmiuiRSn e

nmiuid�! � as measure.

Let �x 2 Sn : � (fxg) � ��1s;n�

= fx1; � � � ; xNg ; (4.32)

then it follows from Theorem 4.1 that

� =

NXi=1

�i�xi : (4.33)

Moreover �i � 0,PN

i=1 �i = 1 and

NXi=1

�ip (xi) = 0

for any p 2�Pm. It follows that

��1s;n�N � 1;and N 2 Nm (Sn). We get

� � �s;nN

� �s;nNm (Sn)

:

This contradicts with the choice of �. �

Similar as before, the constant �s;nNm(Sn)+" is almost optimal. It is also interesting

to compare Theorem 4.2 with [BCY, lemma 4.3] and [CY2, lemma 4.6].

4.2. Paneitz operator in dimension 4. Let�M4; g

�be a smooth compact Rie-

mannian manifold with dimension 4. The Paneitz operator is given by (see [CY2,HY])

Pu = �2u+ 2div (Rc (ru; ei) ei)�2

3div (Rru) : (4.34)

Here e1; e2; e3; e4 is a local orthonormal frame with respect to g. The associated Qcurvature is

Q = �16�R� 1

2jRcj2 + 1

6R2: (4.35)

In 4-dimensional conformal geometry, P and Q play the same roles as �� andGauss curvature in 2-dimensional conformal geometry.For u 2 C1 (M), let

E (u) =

ZM

Pu � ud� (4.36)

=

ZM

�(�u)

2 � 2Rc (ru;ru) + 23R jruj2

�d�:

By this formula, we know E (u) still makes sense for u 2 H2 (M) =W 2;2 (M).In [CY2], it is shown that if P � 0 and kerP = fconstant functionsg, then for

any u 2 H2 (M) n f0g with u = 0,ZM

e32�2 u2

E(u) d� � c (M; g) : (4.37)

In particular

log

ZM

e4ud� � 1

8�2E (u) + c1 (M; g) : (4.38)

Note that 32�2 = a2;4 and 18�2 = �2;4.

We want to remark that if P � 0 and kerP = fconstant functionsg, then forany u 2 H2 (M) with u = 0,

kuk2L2 � c (M; g)E (u) : (4.39)

On the other hand, it follows from standard elliptic theory (or the Bochner�s iden-tity) that

kuk2H2 � c (M; g)�k�uk2L2 + kuk

2L2

�� c (M; g)

�E (u) + kuk2H1

�� c (M; g)E (u) +

1

2kuk2H2 + c (M; g) kuk2L2

� 1

2kuk2H2 + c (M; g)E (u) :

We have used the interpolation inequality in between. It follows that

kuk2H2 � c (M; g)E (u) : (4.40)

It is also worth pointing out that on the standard S4, P � 0 and kerP =fconstant functionsg. Moreover in [Gur, GurV], some general criterion for suchpositivity condition to be valid were derived.

Proposition 4.2. Let�M4; g

�be a smooth compact Riemannian manifold with

P � 0 and kerP = fconstant functionsg. Assume ui 2 H2 (M) such that ui = 0and E (ui) � 1. We also assume ui * u weakly in H2 (M) and

(�ui)2d�! (�u)

2d�+ � (4.41)


� = maxx2K

� (fxg) : (4.42)

(1) If � < 1, then for any 1 � p < 1� ,

supi

ZK

e32�2pu2i d� <1: (4.43)


e32�2u2i ! 1 + c0�x0 (4.44)

20 FENGBO HANG

Proof. Since ui * u weakly in H2 (M), we see ui ! u in H1 (M). It follows thatu = 0 and

E (ui)! E (u) + � (M) :

Since E (ui) � 1, we getE (u) + � (M) � 1:

By the assumption on P we know E (u) � 0 and E (u) = 0 if and only if u = 0, hence� (M) � 1. With these facts at hand, Proposition 4.2 follows from Proposition 4.1by the same argument as in the proof of Corollary 2.1. �

On the standard S4, we have P = �2 � 2� and

E (u) =

ZS4

�(�u)

2+ 2 jruj2

�d� � k�uk2L2

for u 2 H2�S4�. It follows from this inequality and Theorem 4.2 that

Proposition 4.3. If u 2 H2�S4�such that

RS4 ud� = 0 (here � is the standard

measure on S4) and for every p 2�Pm,��Z

S4penud�

�� b (p) ; (4.45)


log

ZS4e4ud� �

�1

8�2Nm (S4)+ "

�ZS4

�(�u)

2+ 2 jruj2

�d�+ c (m; b; ") : (4.46)

Here Nm�S4�is de�ned in De�nition 3.1.

In view of Proposition 4.3 and the recent proof of sharp version of Aubin�sMoser-Trudinger inequality on S2 in [GuM], it is tempting to conjecture that foru 2 H2

�S4�with

RS4 ud� = 0 andZ

S4xie

4ud� (x) = 0 for 1 � i � 5, (4.47)

we have

log

�1

jS4j

ZS4e4ud�

�� 1

16�2

ZS4

�(�u)

2+ 2 jruj2

�d�: (4.48)

In a recent work [Gu], some progress has been made toward proving (4.48) foraxially symmetric functions.

4.3. W s;ns (Mn) for odd s. Let s 2 N be an odd number strictly less than n.Denote

as;n =n

jSn�1j

�n2 2s�

�s+12

��n�s+1

2

� ! nn�s

: (4.49)

The Moser-Trudinger inequality (see [F]) tells us that for every u 2W s;ns (M) n f0gwith u = 0, Z

M

exp

0B@as;n jujn

n�s r� s�12 u nn�s

Lns

1CA d� � c (M; g) : (4.50)

This implies the Moser-Trudinger-Onofri inequality

log

ZM

enud� � �s;n

r� s�12 u ns

Lns+ c1 (M; g) : (4.51)

Here

�s;n = s

�n� sn


��n�s+1

2

��n2 2s�

�s+12

�!ns

: (4.52)

Proposition 4.4. Assume s 2 N is an odd number strictly less than n, ui 2W s;ns (Mn) such that ui = 0, ui * u weakly in W s;ns (M) and��r� s�1

2 ui

��ns d�! ��r� s�12 u��ns d�+ � (4.53)

as measure. If x 2 M and p 2 R such that 0 0,

supi

ZBr(x)

eas;npjuijn

n�sd� <1: (4.54)

Here

as;n =n

jSn�1j

�n2 2s�

�s+12

��n�s+1

2

� ! nn�s

: (4.55)

Since the proof of Proposition 4.4 is almost identical to the proof of Proposition4.1, we omit it here. The same will happen to Corollary 4.2 and Theorems 4.3, 4.4below.

Corollary 4.2. Assume ui 2W s;ns (M) such that ui = 0 and r� s�1

2 ui

Lns� 1.

We also assume ui * u weakly in W s;ns (M) and��r� s�12 ui

��ns d�! ��r� s�12 u��ns d�+ � (4.56)


� = maxx2K

� (fxg) : (4.57)

(1) If � < 1, then for any 1 � p < ��s

n�s ,

supi

ZK

eas;npjuijn

n�sd� <1: (4.58)


eas;njuijn

n�s ! 1 + c0�x0 (4.59)as measure for some c0 � 0.

Theorem 4.3. Let s 2 N be an odd number strictly less than n. Assume � > 0,mi > 0, mi !1, ui 2W s;ns (Mn) such that ui = 0 and

log

ZM

enmiuid� � �mnsi : (4.60)

We also assume ui * u weakly in W s;ns (M),��r� s�1

2 ui

��ns d�! ��r� s�12 u��ns d�+�

as measure andenmiuiR

Menmiuid�

! � (4.61)

22 FENGBO HANG

as measure. Let �x 2M : � (fxg) � ��1s;n�

= fx1; � � � ; xNg ; (4.62)

here

�s;n = s

�n� sn


��n�s+1

2

��n2 2s�

�s+12

�!ns

; (4.63)

then

� =NXi=1

�i�xi ; (4.64)


i=1 �i = 1.

Theorem 4.4. Let s 2 N be an odd number strictly less than n. If u 2W s;ns (Sn)such that

RSn ud� = 0 (here � is the standard measure on S

n) and for every p 2�Pm,��Z

Snpenud�

�� b (p) ; (4.65)


log

ZSnenud� �

��s;n

Nm (Sn)+ "

� r� s�12 u ns

Lns+ c (m;n; b; ") : (4.66)


�s;n = s

�n� sn


��n�s+1

2

��n2 2s�

�s+12

�!ns

: (4.67)

4.4. Functions on compact surfaces with boundary. In this subsection, weassume

�M2; g

�is a smooth compact surface with nonempty boundary and a

smooth Riemannian metric g.

4.4.1. Functions with zero boundary value. We denote H10 (M) = W 1;2

0 (M). Itfollows from [CY1] that for every u 2 H1

0 (M) n f0g,ZM

e4� u2

kruk2L2 d� � c (M; g) : (4.68)

As a consequence,

log

ZM

e2ud� � 1

4�kruk2L2 + c1 (M; g) : (4.69)

Lemma 4.2. For u 2 H10 (M) and a > 0, we haveZ

M

eau2

d� <1: (4.70)

Proof. Indeed, �x " > 0 tiny, we can �nd v 2 C1c (M) such that kru�rvkL2 < ".With this v and (4.68) at hands, we can proceed exactly the same way as in theproof of Lemma 2.1. �

Proposition 4.5. Assume ui 2 H10 (M) such that ui * u weakly in H1

0 (M), and

jruij2 d�! jruj2 d�+ � (4.71)

as measure. If x 2M , p 2 R satis�es 0 0,

supi

ZBr(x)

e4�pu2i d� <1: (4.72)

Proof. Fix p1 2�p; 1

�(fxg)

�, then

� (fxg) < 1

p1: (4.73)


(1 + ")� (fxg) < 1

p1(4.74)

and(1 + ") p < p1: (4.75)

Let vi = ui � u, then vi * 0 weakly in H10 (M), vi ! 0 in L2 (M). For any

' 2 C1 (M), we have

kr ('vi)k2L2� (k'rvikL2 + kvir'kL2)

2

� (k'ruikL2 + k'rukL2 + kvir'kL2)2

� (1 + ") k'ruik2L2 + c (") k'ruk2L2 + c (") kvir'k

2L2 :

It follows that

lim supi!1

kr ('vi)k2L2

� (1 + ")

�ZM

'2d� +

ZM

'2 jruj2 d��+ c (") k'ruk2L2

= (1 + ")

ZM

'2d� + c1 (")

ZM

'2 jruj2 d�:

We can �nd a ' 2 C1 (M) such that 'jBr(x)= 1 for some r > 0 and

(1 + ")

ZM

'2d� + c1 (")

ZM

'2 jruj2 d� < 1

p1:

Hence for i large enough,

kr ('vi)k2L2 <1

p1:

Note that 'vi 2 H10 (M). We haveZ

Br(x)

e4�p1v2i d� �

ZM

e4�p1('vi)2

d�

�ZM

e4�

('vi)2

kr('vi)k2L2 d�

� c (M; g) :

24 FENGBO HANG


u2i = (vi + u)2 � (1 + ") v2i + c (")u2;

hencee4�u

2i � e4�(1+")v

2i ec(")u

2

:

Since e4�(1+")v2i is bounded in L

p11+" (Br (x)), ec(")u

2 2 Lq (Br (x)) for any 0 < q <

1 (by Lemma 4.2), and p11+" > p, it follows from Holder inequality that e4�u

2i is

bounded in Lp (Br (x)). �

Corollary 4.3. Assume ui 2 H10 (M) such that kruikL2 � 1. We also assume

ui * u weakly in H10 (M) and

jruij2 d�! jruj2 d�+ � (4.76)


� = maxx2K

� (fxg) : (4.77)

(1) If � < 1, then for any 1 � p < 1� ,

supi

ZK

e4�pu2i d� <1: (4.78)


e4�u2i ! 1 + c0�x0 (4.79)


With Proposition 4.5, we can derive Corollary 4.3 exactly in the same way asthe proof of Corollary 2.1.

Theorem 4.5. Let�M2; g

�be a smooth compact Riemann surface with nonempty

boundary. Assume � > 0, mi > 0, mi !1, ui 2 H10 (M) and

log

ZM

e2miuid� � �m2i : (4.80)

We also assume ui * u weakly in H10 (M), jruij

2d� ! jruj2 d� + � as measure

ande2miuiR

Me2miuid�

! � (4.81)

as measure. Let

fx 2M : � (fxg) � 4��g = fx1; � � � ; xNg ; (4.82)

then

� =NXi=1

�i�xi ; (4.83)


i=1 �i = 1.

Theorem 4.5 follows from Proposition 4.5 by the same arguments as in the proofof Theorem 2.1.

4.4.2. Functions with no boundary conditions. Let H1 (M) =W 1;2 (M). It followsfrom [CY1] that for every u 2 H1 (M) n f0g with u = 0, we haveZ

M

e2� u2

kruk2L2 d� � c (M; g) : (4.84)

As a consequence,

log

ZM

e2ud� � 1

2�kruk2L2 + c1 (M; g) : (4.85)

Lemma 4.3. For any u 2 H1 (M) and a > 0, we haveZM

eau2

d� <1: (4.86)

Using (4.84), the same argument as the proof of Lemma 2.1 implies Lemma 4.3.

Proposition 4.6. Let�M2; g

�be a smooth compact Riemann surface with non-

empty boundary. Assume ui 2 H1 (M) such that ui = 0, ui * u weakly in H1 (M)and

jruij2 d�! jruj2 d�+ � (4.87)

as measure. Given x 2M and p 2 R such that 0 0,

supi

ZBr(x)

e4�pu2i d� <1: (4.88)

(2) If x 2 @M , then for some r > 0,

supi

ZBr(x)

e2�pu2i d� <1: (4.89)

Proof. Fix p1 2�p; 1

�(fxg)

�, then

� (fxg) < 1

p1: (4.90)


(1 + ")� (fxg) < 1

p1(4.91)

and

(1 + ") p < p1: (4.92)

Let vi = ui � u, then vi * 0 weakly in H1 (M), vi ! 0 in L2 (M). For any' 2 C1 (M), we have

kr ('vi)k2L2� (k'rvikL2 + kvir'kL2)

2

� (k'ruikL2 + k'rukL2 + kvir'kL2)2

� (1 + ") k'ruik2L2 + c (") k'ruk2L2 + c (") kvir'k

2L2 :

26 FENGBO HANG

It follows that

lim supi!1

kr ('vi)k2L2

� (1 + ")

�ZM

'2d� +

ZM

'2 jruj2 d��+ c (") k'ruk2L2

= (1 + ")

ZM

'2d� + c1 (")

ZM

'2 jruj2 d�:

If x 2 Mn@M , then we can �nd a ' 2 C1c (M) such that 'jBr(x)= 1 for some

r > 0 and

(1 + ")

ZM

'2d� + c1 (")

ZM

'2 jruj2 d� < 1

p1:

Hence for i large enough, we have

kr ('vi)k2L2 <1

p1:

Note that 'vi 2 H10 (M). We haveZ

Br(x)

e4�p1v2i d� �

ZM

e4�p1('vi)2

d�

�ZM

e4�

('vi)2

kr('vi)k2L2 d�

� c (M; g) :


u2i = (vi + u)2 � (1 + ") v2i + c (")u2;

hence

e4�u2i � e4�(1+")v

2i ec(")u

2

:

Since e4�(1+")v2i is bounded in L

p11+" (Br (x)), ec(")u

2 2 Lq (Br (x)) for any 0 < q <

1 (by Lemma 4.2), and p11+" > p, it follows from Holder inequality that e4�u

2i is

bounded in Lp (Br (x)).If x 2 @M , then (4.84) and similar arguments as in the proof of Proposition 2.1

tells us

supi

ZBr(x)

e2�pu2i d� <1:

�

Corollary 4.4. Assume ui 2 H1 (M) such that ui = 0 and kruikL2 � 1. We alsoassume ui * u weakly in H1 (M) and

jruij2 d�! jruj2 d�+ � (4.93)


�0 = maxx2Kn@M

� (fxg) ; (4.94)

�1 = maxx2K\@M

� (fxg) ; (4.95)

(1) If �1 < 1, then for any 1 � p < minn2�0; 1�1

o,

supi

ZK

e2�pu2i d� <1: (4.96)

(2) If �1 = 1, then � = �x0 for some x0 2 @M , u = 0 and after passing to asubsequence,

e2�u2i ! 1 + c0�x0 (4.97)


Proof. First assume �1 < 1. We claim for any x 2 K, there exists a rx > 0 suchthat

supi

ZBrx (x)

e2�pu2i d� <1:

Once this claim is proved, we can deduce

supi

ZK

e2�pu2i d� <1

by the covering argument in the proof of Corollary 2.1.If x 2 Kn@M , then

p <2

�0� 2

� (fxg) :

Hence p2 <

1�(fxg) . It follows from Proposition 4.6 that for some r > 0,

supi

ZBr(x)

e2�pu2i d� = sup

i

ZBr(x)

e4��p2u

2i d� <1:

If x 2 K \ @M , thenp <

1

�1� 1

� (fxg) :

It follows from Proposition 4.6 again that for some r > 0,

supi

ZBr(x)

e2�pu2i d� <1:

This proves the claim.The case when �1 = 1 can be handled the same way as in the proof of Corollary

2.1. �

Theorem 4.6. Let�M2; g

�be a smooth compact Riemann surface with nonempty

boundary. Assume � > 0, mi > 0, mi !1, ui 2 H1 (M) such that ui = 0 and

log

ZM

e2miuid� � �m2i : (4.98)

We also assume ui * u weakly in H1 (M), jruij2 d� ! jruj2 d� + � as measureand

e2miuiRMe2miuid�

! � (4.99)

as measure. Let

fx 2Mn@M : � (fxg) � 4��g [ fx 2 @M : � (fxg) � 2��g (4.100)

= fx1; � � � ; xNg ;

28 FENGBO HANG

then

� =NXi=1

�i�xi ; (4.101)


i=1 �i = 1.

Proof. Assume x 2Mn@M with � (fxg) < 4��, then we claim that for some r > 0,� (Br (x)) = 0. Indeed we �x p such that

1

4��< p <

1

� (fxg) ; (4.102)


e4�pu2i d� � c; (4.103)

here c is a positive constant independent of i. We have

2miui � 4�pu2i +m2i

4�p; (4.104)

hence ZBr(x)

e2miuid� � cem2i

4�p :


e2miuid�RMe2miuid�

� ce(1

4�p��)m2i :

In particular,


RBr(x)

e2miuid�RMe2miuid�

= 0:

We get � (Br (x)) = 0.If x 2 @M with � (fxg) < 2��, then similar argument shows for some r > 0,

� (Br (x)) = 0.Clearly these imply

� (Mn fx1; � � � ; xNg) = 0: (4.105)

Hence � =PN


i=1 �i = 1. �

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Courant Institute, New York University, 251 Mercer Street, New York NY 10012E-mail address : [email protected]

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