thesis

PRESCRIBING SCALAR CURVATURE
LIU HONG
(B.Sc., Peking University)
A thesis submitted for the
Degree of Doctor of Philosophy
Supervisor
Professor Xingwang Xu
DEPARTMENT OF MATHEMATICS
NATIONAL UNIVERSITY OF SINGAPORE
2015

Acknowledgements
I would like to thank all people who have helped and inspired me through my thesis.
First and foremost I would like to express the deepest appreciation and sincerest gratitude
to my supervisor, Professor Xingwang Xu, who has supported me during my doctoral study
with his patience and knowledge. Without his guidance, encouragement and persistent help,
this dissertation would not have been possible. I am grateful to be his student, and his
perpetual energy and enthusiasm in research enabled me to develop a good understanding of
research.
I would like to thank Professor G. Tian, Professor P. Yang, Professor A. Chang, Professor
S. T. Yau, Professor R. Schoen and Professor J. Case for their kind help and encouragement.
I thank Professor Fei Han, Professor Lei Zhang, Dr. Hengfei Lu, Dr. Feng Zhou, Dr. Hong
Zhang, Dr. Ruilun Cai, Dr. Jize Yu, Dr. Caihua Luo, Dr. Yuke Li, Dr. Ran Wei, Dr. Liuqin
Yang, Dr. Ying Cui, Dr. Yufei Zhao, Dr. Ruixiang Zhang, Dr. Xiaowei Jia and his girlfriend
Master Ya Sun, Dr. Chen Yang and his wife Master Dan Zheng, Dr. Yan Wang and his
boyfriend Dr. Xiaofei Zhao, Dr. Jinjiong Yu and his girlfriend Master Yile Li, Dr. Hawk,
Master Yuchen Wang, Miss Andrea, Dr. Yuke Li, Master Qiong Liu, Miss Jiayu Zhu and Miss
Xueliang Hu for their assistance, stories and jokes. Thanks, Dr. Han.
I would like to thank my grandmother for her spiritual support. I would like to thank
Mr. Bo Yang, Ms. Zhao Han, Miss Mengqi Hong and their families for their aid in my most
diﬃcult time.
Finally, I would like to thank my father Heng Long Hong and mother Liu Ying Xie for
giving my life in the ﬁrst place for unconditional support and love.
i

Contents
Acknowledgements i
Summary iv
List of Symbols v
1 Introduction 1
2 Elementary estimates and long time existence 3
3 Reduction 18
4 The case u∞ = 0 20
5 Convergence 26
iii

Summary
The prescribing scalar curvature problem originated from the classical Yamabe problem. Since
Yamabe constant is a conformal invariant, we can divide conformal closed manifolds into
positive, negative and zero Yamabe cases. The negative Yamabe case is well understood. So
we focus on and solve the positive case here.
iv

Symbol
Symbol Deﬁnition
N a conformal closed manifold
n dimension of N
χ(N) Euler characteristic of N
Sg, S scalar curvature
[g0] conformal class of g0
Y (N, [g0]) Yamabe constant
Eg[u] Yamabe energy
Sn
standard sphere
expp exponential map from p
Tp(N) tangent space at p
∆ Laplacian operator
gradient operator
2∗ 2n
n−2
V ol volume
∂ partial diﬀerential operator
o(1) tending to zero
v

Chapter 1
Introduction
Prescribing scalar curvature problem is a fundamental problem in modern geometry. This
problem in diffeomorphic class is well understood from the work of J. L. Kazdan and F. W.
Warner [15],[16]. For closed 2-manifolds, Kazdan and Warner proved in 1975 [15] that the
obvious sign condition demanded by the Gauss-Bonnet Theorem:
(a) K positive somewhere if χ(N) > 0,
(b) K changes sign unless K ≡ 0 if χ(N) = 0,
(c) K negative somewhere if χ(N) < 0,
is sufficient for a smooth function K on a given closed 2-manifold N to be the Gaussian
curvature of some metric on N. In case of dimension n ≥ 3, they proved if f is negative
somewhere, then f could be the scalar curvature of some smooth metric in [16]. Furthermore,
if N admits a metric with positive scalar curvature, any smooth function could be a scalar
curvature of some Riemannian metric (see [16]). They also proved if N is non-compact of
dimension n ≥ 3 diffeomorphic to an open submanifold of some closed manifold, then every
smooth function could be a scalar curvature (see [16]).
So we consider the prescribing scalar curvature problem in conformal class which is more
subtle and plays a crucial role in modern conformal geometry. The Yamabe constant of a
conformal manifold (N, [g0]) is defined as
Y (N, [g0]) = inf
g∈[g0]
N
Sgdµg
( N
dµg)
n−2
n
,
where Sg is the scalar curvature of g and µg is the volume density of the metric g. The Yamabe
1

CHAPTER 1. INTRODUCTION 2
energy of a function u on (N, g) of class H1
is defined as
Eg[u] = N
(cn| gu|2
+ Sgu2
)dµg
V ol(N, g)
, cn =
4(n − 1)
n − 2
.
Since Yamabe constant is a conformal invariant, we can divide conformal manifolds into
positive, negative and zero Yamabe cases. The negative Yamabe case is easier and well
understood. So we focus on positive Yamabe case in this paper. In 1976, T. Aubin proved
a perturbation theorem with non-sphere condition. Namely, if (N, g) with positive Yamabe
constant is not conformal to standard sphere, then there is κ > 1 depending on (N, g) such that
any smooth positive function f satisfying sup f ≤ κ inf f is the scalar curvature of a conformal
metric(see [1]). In 1986, J. F. Escobar and R. M. Schoen proved in [12] that for given closed
3-manifold of positive scalar curvature which is not conformally equivalent to standard sphere,
any smooth function which is positive somewhere could be the scalar curvature of a conformal
metric. They treated the case of dimension n ≥ 4 with locally conformally flat restriction.
The most subtle case is standard sphere. In 1991, A. Bahri and J. M. Coron proved in [6]
a result on S3
with an index condition. In the same year, A. Chang and P. C. Yang proved
the perturbation theorem for general Sn
in [10] with an analogous index condition. In 2010,
X. Chen and X. Xu gave a concrete bound for the perturbation theorem using the scalar
curvature flow in [11].
During 2005-2007, S. Brendle provided a new proof of Yamabe problem by constructing
a family of test functions in [3] and [4] which is reminiscent of Schoen’s proof of Yamabe
problem [19]. Both proofs study the role of Green’s functions on manifolds and rely on Yau’s
positive mass theorem(see [22]).
Since Q. A. Ngo and X. Xu [17] have solved the zero Yamabe case, we treat the non-sphere
positive Yamabe case here as the last piece of the jigsaw puzzle. Our main result is:
Theorem 1. Given a closed conformal manifold (N, [g0]) which is not conformally equivalent
to the round sphere and satisfies Y (N, [g0]) > 0. If n ≥ 3, then any positive smooth function
on N could be the scalar curvature of some g ∈ [g0].
It seems that G. Bianchi and E. Egnel [7] have constructed a counter example to Theorem
1. But we have clarified their case in [14].

Chapter 2
Elementary estimates and long time
existence
From now on, we always assume (N, [g0]) is a manifold satisfying the conditions in Theorem
1. Let f be a positive smooth function on N. We can choose g0 in the conformal class such
that the scalar curvature S0 of g0 is a positive constant. Since N is compact, 0 < m ≤ f ≤ M.
We write one conformal metric g as g = u
4
n−2 g0 . To find the desired metric we need to find a
positive solution of the following equation
−cn∆u + S0u = fu
n+2
n−2 .
Consider the scalar curvature flow
ut =
n − 2
4
[α(t)f − S(t)]u
where S(t) is the scalar curvature of g(t) which has standard formula
S = u− n+2
n−2 (−cn∆u + S0u)
where cn = 4(n−1)
n−2
, ∆ = ∆g0 and S0 is the scalar curvature of g0. By multiplying a constant,
we can assume
N
fdµg0 = 1 .
3

CHAPTER 2. ELEMENTARY ESTIMATES AND LONG TIME EXISTENCE 4
From now on, the integration is always on the whole manifold N. First let us define
E[u] := (cn| u|2
+ S0u2
)dµ = Sdµg
Ef [u] :=
E[u]
( fdµg)
n−2
n
where dµg is the volume density of g and dµ = dµg0 , | · | = | · |g0 .
The factor α(t) is chosen such that
0 =
d
dt
fdµg =
d
dt
fu2∗
dµ0 = 2∗
fu
n+2
n−2 utdµ
=
n
2
α f2
dµg −
n
2
fSdµg ,
where 2∗
= 2n
n−2
is the critical exponent and = g0 . Hence the natural choice for this factor
is
α(t) =
fSdµg
f2dµg
.
For p ≥ 1, we define
Fp(g(t)) = |αf − S|p
dµg .
Lemma 1. If a positive smooth function u satisfies
ut =
n − 2
4
(αf − S)u ,
S = u− n+2
n−2 (−cn∆u + S0u) ,
then
d
dt
E[u] =
d
dt
Ef [u] = −
n − 2
2
F2 .
Proof.
d
dt
Ef [u] =
dE
dt
( fdµg)
n−2
n
−
2E[u] fu
n+2
n−2 utdµ
( fdµg)
n−2
n
+1
.
Since
d
dt
(| u|2
) = 2 u, ut
= 2div(ut u) − 2ut∆u ,

Lemma 3.
(αf − S)t = (n − 1)∆g(αf − S) + S(αf − S) + α f .
Proof. By the formula
S = u− n+2
n−2 (−cn∆u + S0u) ,
we have
St = u− n+2
n−2 (−cn∆ut + S0ut) −
n + 2
n − 2
u− 2n
n−2 ut(−cn∆u + S0u)
=
n − 2
4
u− n+2
n−2 {−cn[u∆(αf−S)+2 (αf−S), u +(αf−S)∆u]+S0(αf−S)u}−
n + 2
4
(αf−S)S
= −
n + 2
4
(αf−S)S+
n − 2
4
u− n+2
n−2 (αf−S)(−cn∆u+S0u)−(n−1)u− n+2
n−2 [u∆(αf−S)+2 (αf−S), u ] .
Since
∆g = u− 2n
n−2 det(g0)−1
2 ∂j(u2
det(g0)
1
2 gij
0 ∂i)
= u− 4
n−2 ∆ + 2u− n+2
n−2 u, · ,
we have
St = −(αf − S)S − (n − 1)∆g(αf − S) ,
(αf − S)t = (n − 1)∆g(αf − S) + (αf − S)S + α f .
Since the case of dimension three has been proved by Escobar and Schoen, we always
assume n ≥ 4 from now on.
Lemma 4. F2 is bounded.
Proof.
d
dt
F2 = −
n − 4
2
[cn | (αf−S)|2
gdµg+ |(αf−S)|2
gdµg]−cn | (αf−S)|2
gdµg+
n
2
αf(αf−S)2
dµg
≤
n
2
αf(αf − S)2
dµg since Y (N) > 0
≤ CF2
for some constant C > 0. Thus,
F2(g(t)) ≤ F2(g(0)) + C
t
0
F2(g(s))dµg(s)

≤ F2(g(0)) +
2C
n − 2
E[u0] .
In later estimates, we will constantly utilize positive Yamabe condition.
Lemma 5. There exists a constant C such that α ≤ CF
1
2
2 .
Proof.
|α | = ( f2
dµg)−1
|
d
dt
( fSdµg) − α
d
dt
( f2
dµg)|
≤ C|
d
dt
( fSdµg) − α
d
dt
( f2
dµg)|
≤ C| − (n − 1) f∆g(αf − S)dµg +
n − 2
2
fS(αf − S)dµg −
nα
2
f2
(αf − S)dµg|
≤ C| − (n − 1) (∆gf)(αf − S)dµg| + C1F1 + C2F2 + C3F
1
2
2 .
By Cauchy inequality and boundness of F2, we have
|α | ≤ C| (∆gf)(αf − S)dµg| + CF
1
2
2 .
Moreover,
(∆gf)(αf − S)dµg = (αf − S)(∆f)u− 4
n−2 dµg + 2 (αf − S) u, f u− n+2
n−2 dµg.
By Cauchy inequality,
| (αf − S)(∆f)u− 4
n−2 dµg| ≤ (sup |∆f|)( u
2(n−4)
n−2 dµ)
1
2 F
1
2
2 .
Since 2(n−4)
n−2
< 2n
n−2
, u
2(n−4)
n−2 dµ is bounded.
| (αf − S) u, f u− n+2
n−2 dµg| ≤ sup(| f|)( | u|2
u− 4
n−2 dµ)
1
2 F
1
2
2 .
Hence we only need to bound | u|2
u− 4
n−2 dµ.
If n = 6,
| u|2
u− 4
n−2 dµ = | u|2
u−1
dµ

= −
∆u
√
u
(
√
u ln u)dµ
≤ ( (∆u)2
u−1
dµ)
1
2 ( u(ln u)2
dµ)
1
2 .
Since 2×6
6−2
= 3, we have
u(ln u)2
dµ =
u≥1
u(ln u)2
dµ +
u<1
u(ln u)2
dµ
≤ u3
dµ +
4
e2
dµ
≤ C.
Since S = u−2
(S0u − 5∆u), we have
(∆u)2
u−1
dµ =
1
25
(S2
0u − 2S0Su2
+ S2
u3
)dµ.
Since fu3
dµ = 1, (αf − S)2
u3
dµ and α are bounded, we know the above term is bounded
too and the desired estimate follows.
If n = 6,
(S − S0u− 4
n−2 )u2
dµ =
4(n − 1)(n − 6)
(n − 2)2
| u|2
u− 4
n−2 dµ .
So the result follows from
S0u
2n−8
n−2 dµ ≤ C ,
Su2
dµ ≤ ( S2
dµg)
1
2 ( u
2(n−4)
n−2 dµ)
1
2 ≤ C .
Now we are able to give a lower bound of the scalar curvature S.
Lemma 6.
S − αf ≥ γ, t ≥ 0 ,
where γ := min{S0 − α(0)f, − 2√
3
max{0, α f + α2f2}, −αf}.
Proof. Deﬁne
L = ∂t − (n − 1)∆g + αf − γ ,

then
LS = St − (n − 1)∆gS + (αf − γ)S
= S(S − γ) − (n − 1)∆g(αf) .
By using comparison function w(t) = αf + γ, we have
Lw = αtf − (n − 1)∆g(αf) + (αf − γ)(αf + γ)
≤ −
1
4
γ2
− (n − 1)∆g(αf)
≤ S(S − γ) − (n − 1)∆g(αf)
≤ LS .
But
w(0) ≤ α(0)f + γ ≤ S0 .
Thus we get a lower bound of S(t), t ≥ 0, by the maximum principle for a linear parabolic
differential equation.
The following estimates will help us to achieve the long time existence of the flow.
Lemma 7. For any T > 0, there exists C = C(T) such that
C−1
≤ u(x, t) ≤ C, (x, t) ∈ N × [0, T) .
Proof. From the equation of the flow and lemma 6, we know
u(t) ≤ e−n−2
4
γt
u(0), t ∈ [0, T] .
Define a constant
P = S0 + sup
t∈[0,T]
sup
N
[−(αf + γ)u
4
n−2 ] .
From Lemma 6 again, we have
0 ≤ (S − αf − γ)u
n+2
n−2
= −cn∆u + S0u − (αf + γ)u
n+2
n−2
≤ −cn∆u + Pu .

Hence by the Corollary A.3 in [3], we have
1 = V ol(N, g0) ≤ C inf
N
u(sup
N
u)
n+2
n−2 .
So the assertion follows.
Deﬁne
δ = sup
t∈[0,T]
αf + γ ∞ + 1 .
From Lemma 6, we have
S + δ ≥ αf + γ + |αf + γ| + 1 ≥ 1 .
Lemma 8. For any p > 2, we have
d
dt
(S + δ)p−1
dµg = −
4(p − 2)(n − 1)
p − 1
| g(S + δ)
p−1
2 |2
gdµg
+(p − 1)(p − 2)(n − 1) (S + δ)p−3
g(S + δ), g(αf + δ) gdµg
−(
n
2
+ 1 − p) (S + δ)p−1
(S − αf)dµg − (p − 1)δ [(S + δ)p−2
− (αf + δ)p−2
](S − αf)dµg
−(p − 1)δ (αf + δ)p−2
(S − αf)dµg .
Proof.
d
dt
(S + δ)p−1
dµg =
d
dt
(S + δ)p−1
u
2n
n−2 dµ
=
2n
n − 2
(S + δ)p−1
u
n+2
n−2 utdµ + (p − 1) (S + δ)p−2
Stdµg
=
n
2
(S+δ)p−1
(αf−S)dµg−(1−p) (S+δ)p−2
(S−αf)Sdµg−(p−1)(n−1) (S+δ)p−2
∆g(αf−S)dµg
= −
n
2
(S + δ)p−1
(S − αf)dµg − (1 − p) (S + δ)p−2
(S − αf)(S + δ − δ)dµg
−(p − 1)(n − 1) (S + δ)p−2
divg( g(αf − S))dµg
= −(
n
2
+ 1 − p) (S + δ)p−1
(S − αf)dµg − (p − 1)δ (S + δ)p−2
(S − αf)dµg

−(p−1)(n−1) {divg[(S +δ)p−2
g(αf −S)]−(p−2)(S +δ)p−3
g(S +δ), g(αf −S) g}dµg
= −(
n
2
+ 1 − p) (S + δ)p−1
(S − αf)dµg − (p − 1)δ [(S + δ)p−2
− (αf + δ)p−2
](S − αf)dµg
−(p−1)δ (αf+δ)p−2
(S−αf)dµg+(p−1)(p−2)(n−1) (S+δ)p−3
g(S+δ), g(αf+δ−S−δ) gdµg .
Since
(S + δ)p−3
g(S + δ), g(αf + δ) g − (S + δ)p−3
| g(S + δ)|2
g
= (S + δ)p−3
g(S + δ), g(αf + δ) g −
4
(p − 1)2
| g(S + δ)
p−1
2 |2
g ,
the identity follows.
For any p ≥ 2, we have
d
dt
Fp =
d
dt
|αf −S|p
dµg = p |αf −S|p−2
(αf −S)(αf −S)tdµg +
n
2
|αf −S|p
(αf −S)dµg
= p |αf−S|p−2
(αf−S)[(n−1)∆g(αf−S)+(αf−S)S+α f]dµg+
n
2
|αf−S|p
(αf−S)dµg .
Since
|αf − S|p−2
(αf − S)∆g(αf − S) = divg(|αf − S|p−2
(αf − S) g(αf − S))
− g(|αf − S|p−2
(αf − S)), g(αf − S) g ,
g(|αf − S|p−2
(αf − S)) = (p − 1)|αf − S|p−2
g(αf − S) ,
hence
d
dt
|αf − S|p
dµg = −p(p − 1)(n − 1) |αf − S|p−2
| g(αf − S)|2
gdµg + p |αf − S|p
Sdµg
+pα f|αf − S|p−2
(αf − S)dµg +
n
2
|αf − S|p
(αf − S)dµg
= −p(p − 1)(n − 1) |αf − S|p−2
| g(αf − S)|2
gdµg + p αf|αf − S|p
dµg
pαt f|αf − S|p−2
(αf − S)dµg + (
n
2
− p) |αf − S|p
(αf − S)dµg .

Lemma 9. For any p > max{n
2
, 2}, we have
d
dt
Fp + C1F
2
2∗
pn
n−2
≤ C2(Fp + F
2p−n+2
2p−n
p ) ,
where the positive constants C1, C2 are independent of t.
Proof. Since
cn| g|αf − S|
p
2 |2
g =
p2
(n − 1)
n − 2
|αf − S|p−2
| g(αf − S)|2
g ,
so from the derivative formula we have
d
dt
Fp = −
(p − 1)(n − 2)
p
(cn| g|αf − S|
p
2 |2
g + S|αf − S|p
)dµg
+[p+
(p − 1)(n − 2)
p
] S|αf−S|p
dµg+
n
2
|αf−S|p
(αf−S)dµg+pα f|αf−S|p−2
(αf−S)dµg .
By lemma 5 and H¨older inequality, we have |α | ≤ CF
1
2
2 ≤ CF
1
p
p , and thus
|pα f|αf − S|p−2
(αf − S)dµg| ≤ C|α |Fp−1
≤ CFp .
With the help of positive Yamabe condition, we have
d
dt
Fp ≤ −C( |αf − S|
np
n−2 dµg)
n−2
n + CFp + CFp+1 .
By H¨older inequality and Young inequality we deduce that
Fp+1 = |αf − S|p+1
dµg
≤ ( |αf − S|
np
n−2 dµg)
n−2
2p ( |αf − S|p
dµg)
2p−n+2
2p
≤ ε( |αf − S|
np
n−2 dµg)
n−2
n + C(ε)( |αf − S|p
dµg)
2p−n+2
2p−n .
By choosing suitable ε, the assertion follows.
Lemma 10. For any T > 0, there exists C = C(T) such that
F n2
2(n−2)
≤ C(T), t ∈ [0, T) .

Proof. Since S + δ, αf + δ > 0 and p > 2, we deduce
[(S + δ)p−2
− (αf + δ)p−2
](S − αf) = [(S + δ)p−2
− (αf + δ)p−2
][(S + δ) − (αf + δ)] ≥ 0 .
So we can choose p = n+2
2
> 2 in Lemma 8 to achieve
d
dt
(S + δ)
n
2 dµg ≤ −
4(n − 1)(n − 2)
n
| g(S + δ)
n
4 |2
gdµg
+
n(n − 1)(n − 2)
4
(S + δ)
n−4
2
g(S + δ), g(αf + δ) gdµg −
n
2
δ (αf + δ)
n−2
2 (S − αf)dµg .
From Lemma 6 we have
−
n
2
δ (αf + δ)
n−2
2 (S − αf)dµg ≤ −
n
2
δγ (αf + δ)
n−2
2 dµg
≤ C(T) dµg
≤ C(T) .
By Cauchy inequality and Young inequality we have
| (S + δ)
n−4
2
g(S + δ), g(αf + δ) gdµg| =
4
n
| (S + δ)
n−4
4
g(S + δ)
n
4 , g(αf + δ) gdµg|
≤
4
n
(S + δ)
n−4
4 | g(S + δ)
n
4 |g| g(αf + δ)|gdµg
≤ ε | g(S + δ)
n
4 |2
gdµg + C(ε) | g(αf + δ)|2
g(S + δ)
n−4
2 dµg .
We know from Lemma 7 that u is bounded on [0, T). Hence by using H¨older inequality we
can achieve
| (S+δ)
n−4
2
g(S+δ), g(αf+δ) gdµg| ≤ ε | g(S+δ)
n
4 |2
gdµg+C(T, ε)[ (S+δ)
n
2 dµg]
n−4
n .
Set y(t) = (S + δ)
n
2 dµg. By choosing small ε we can get
dy
dt
+ | g(S + δ)
n
4 |2
gdµg ≤ C2(T)y
n−4
n + C2(T) .
We claim that y is bounded by a constant depending on T.
Since n ≥ 4, (y
4
n ) ≤ C2 + C2y
4−n
n . So y
4
n ≥ 1 ⇒ (y
4
n ) ≤ 2C2 and the claim follows.

From the deﬁnition of y we know that y has the volume of N as a lower bound. Integrating
the inequality
dy
dt
+ | g(S + δ)
n
4 |2
gdµg ≤ C2(T)y
n−4
n + C2(T)
we get
T
0
| g(S + δ)
n
4 |2
gdµgdt ≤ C3(T) .
Due to our positive Yamabe condition, we have
( |S + δ|
n2
2(n−2) dµ)
n−2
n ≤ C (cn| |S + δ|
n
4 |2
+ S0|S + δ|
n
2 )dµ ≤ C4(T) .
So
T
0
( |S + δ|
n2
2(n−2) dµg)
n−2
n dt ≤ C5(T)
T
0
( |S + δ|
n2
2(n−2) dµ)
n−2
n dt ≤ C6(T) .
Since S + δ, αf + δ > 0, we have
|S − αf| = |(S + δ) − (αf + δ)| ≤ max{|S + δ|, C7(T)}
and
|S − αf|p
≤ |S + δ|p
+ |αf + δ|p
.
Since n−2
n
< 1, we have
( |S − αf|
n2
2(n−2) dµg)
n−2
n ≤ ( |S + δ|
n2
2(n−2) dµg)
n−2
n + ( |αf + δ|
n2
2(n−2) dµg)
n−2
n .
Setting p = n2
2(n−2)
in Lemma 9 we get
(log F n2
2(n−2)
) ≤ C + CF
n−2
n
n2
2(n−2)
.
So the assertion follows by integrating above inequality.
By the argument of Proposition 2.6 in [3] or Lemma 2.11 in [11], we achieve the following
inequality:
Lemma 11. For any λ ∈ (0, min{4
n
, 1}), T > 0, there exists C = C(λ, T) > 0 such that
|u(x, t) − u(y, s)| ≤ C[dN (x, y)λ
+ |t − s|
λ
2 ] ,
for any x, y ∈ N, t, s ∈ [0, T), 0 < |t − s| < 1 .

So the long time existence of the scalar curvature ﬂow follows from the standard result of
parabolic equations. For example, one can read Theorem 8.3 and Theorem 8.4 in [24].
Lemma 12.
∞
0
F2dt < ∞.
Proof. Since the ﬂow has long time existence, from lemma 1 we know
∞
0
F2dt =
2
n − 2
(E[u0] − lim
t→∞
E[u(t)]) < ∞ .
Thus the assertion follows.
Lemma 13.
F2 → 0 .
Proof. In the proof of lemma 4, we know d
dt
F2 ≤ CF2. Thus, for any t > tν > 0, we have
F2(t) ≤ F2(tν) +
∞
tν
F2 .
By lemma 12, we can pick a time sequence tν → ∞ such that F2(tν) → 0 and the result
follows.
Now we are going to achieve the convergence of α(t).
Lemma 14.
α = E[u] + o(1), t → ∞.
Proof. Since
d
dt
V ol(N, g(t)) =
n
2
(αf − S)dµg ,
by Lemma 13 and Cauchy inequality, we have d
dt
V ol(N, g(t)) → 0. We also have the identity
d
dt
V ol(N, g(t)) =
n
2
α fdµg −
n
2
E[u] =
n
2
(α − E[u]) .
So the result follows.
Lemma 15. For any p ∈ [2, n
2
],
d
dt
Fp ≤ CFp .

Proof. By the derivative formula in the proof of lemma 9 and the inequalities
|pα f|αf − S|p−2
(αf − S)dµg| ≤ CF
1
p
p Fp−1 ≤ CFp
αf − S ≤ −γ ,
we deduce the desired inequality.
Lemma 16. If 2 ≤ p < n
2
and Fp is integrable over (0, ∞), then Fp+1 is also integrable over
(0, ∞).
Proof. By the formula
(
n
2
− p) |αf − S|p
(S − αf)dµg = p αf|αf − S|p
dµg
+pα f|αf − S|p−2
(αf − S)dµg −
d
dt
Fp −
4(p − 1)(n − 1)
p
| |αf − S|
p
2 |2
gdµg ,
we have
T
0
|αf − S|p
(S − αf)dµgdt ≤ C
T
0
[Fp +
2
n − 2p
Fp(0)]dt ≤ C .
Since αf − S ≤ −γ, we have
T
0
Fp+1dt ≤
T
0
|αf − S|p
(S − αf + 2|γ|)dµgdt ≤ C
for any T > 0. As C is independent of T, the result follows.
Now we have the following fundamental result.
Lemma 17. For any p ∈ [1, n
2
], Fp → 0.
Proof. By lemma 12, lemma 16 and H¨older inequality, Fp is integrable for any p ∈ [2, n
2
], and
the result follows from the same argument in the proof of lemma 13.
Deﬁne
α∞ := lim
t→∞
α(t) ,
E∞ := lim
t→∞
α(t) ,
Ef := inf{Ef [u] : u ∈ H1, u is not constantly 0} .

Since E[u] is monotone, E∞ clearly exists and α∞ = E∞ by lemma 14.
Corollary 1. For any p ∈ [1, n
2
], |S − α∞f|p
dµg → 0.
Proof. It is direct from lemma 14, lemma 17 and the inequality
|S − α∞f|p
≤ 2p−1
(|S − αf|p
+ |α − α∞|p
|f|p
) .

Chapter 3
Reduction
The classical result of T. Aubin in Nonlinear Analysis on Manifolds. Monge-Ampere equations
131, shows that:
Lemma 18. If N is not a standard sphere, then Ef ≤ Y (sup f)−n−2
n := n(n−1)ω
2
n
n (sup f)−n−2
n .
Moreover, if Ef < n(n − 1)ω
2
n
n (sup f)−n−2
n , the prescribing scalar curvature problem of f has
a smooth positive solution.
Therefore, from now on, we only need to consider the case where Ef = n(n−1)ω
2
n
n (sup f)−n−2
n .
For abbreviation, let uν = u(tν), λν = λu(tν ), α(gν) = αν and gν = g(tν). We have the following
compactness result from Brendle [3]:
Lemma 19 (Bubble Decomposition).
After passing to a subsequence if necessary, we can ﬁnd an l ∈ N, a smooth function u∞ and
a sequence of m-tuples (x∗
k,ν, ε∗
ν)1≤k≤l such that:
(1)
4(n − 1)
n − 2
∆u∞ − S0u∞ + α∞fu
n+2
n−2
∞ = 0 ,
where α∞ is the limit of the subsequence.
(2) For any i = j, we have
d(x∗
i,ν, x∗
j,ν)
ε∗
ν
ν→∞
−−−→ ∞ .
(3)
uν − u∞ −
l
k=1
¯u(x∗
k,ν ,ε∗
ν ) H1(N) −→ 0 ,
18

CHAPTER 3. REDUCTION 19
where ¯u(p,ε) are a family of test functions satisfying
lim
ε→0
ε
n−2
2 ¯u(p,ε)(expp(εξ)) = (
1
1 + |ξ|2
)
n−2
2
for all p ∈ N and ξ ∈ TpN.
Proof. Since Ef [u(t)] is decreasing, {uν} is bounded in H1
(N). By passing to a subsequence,
αν tends to a limit α∞ and u∞ is defined as the weak limit of {uν}. Then the argument from
Struwe’s proof of Proposition 2.1 in [20] works. The first and third results are from Struwe
[20], and the second statement is due to Bahri and Coron [5].
The recent work of Q. A. Ngo and X. Xu, [17] also gives a proof of the above lemma in the
special case of scalar curvature flow. A useful corollary is that, when E[u0] < 2
2
n Y (sup f)−n−2
n ,
the bubble number l ≤ 1. The reason is that, if l ≥ 2, then for sufficiently small ε > 0,
we have contradictory inequalities l
2
(sup f)
n−2
2 < l( Y
α∞
)
n
2 ≤ lim supt→∞ l B(x∗
j ,ε)
f
n
2 dµg ≤
(sup f)
n
2
−1
lim supt→∞ fdµg ≤ (sup f)
n−2
n . We also refer the reader to [17] for the detailed
computation.
By the classical strong maximum principle (see Theorem 8.19 in [13]) we can deduce the
following proposition from
−
4(n − 1)
n − 2
∆u∞ + S0u∞ = α∞fu
n+2
n−2
∞ ≥ 0 .
Proposition 1. If u∞ vanishes at one point, then it vanishes everywhere.
From now on,as u∞ itself is a solution in case u∞ > 0, we only consider the case u∞ = 0.

Chapter 4
The case u∞ = 0
From now on, we restrict the initial data u(0) to satisfy Ef [u(0)] < 2
2
n Ef . Thus, the bubble
number l in the bubble decomposition could only be 1 noticing that Ef [u(t)] is decreasing
w.r.t. t, u∞ = 0 and the bubble decomposition is H1 convergence. For every time sequence
tν, let xν
be one maximum point of u(tν). Since N is compact, by passing to a subsequence,
we can assume that xν
→ x∞ ∈ N. Set
kν = uν(xν
)
2
n−2
εν
=
1
kν
where uν = u(tν). We choose the normal coordinate system {ξ} w.r.t. g0 which means
(g0)ij = δij + O(|ξ|2
) ,
dVg0 = (1 + O(|ξ|2
))dξ .
Thus,
∆g0 = ∆ + bi∂i + dij∂2
ij
where
bi = O(|ξ|2
) ,
dij = O(|ξ|2
) .
Now we follow the method from Q. A. Ngo and X. Xu [17] to construct the bubble decom-
position explicitly. First of all, we pick a smooth cut-oﬀ function on Rn
for δ > 0 such
20

CHAPTER 4. THE CASE U∞ = 0 21
that
0 ≤ ηδ ≤ 1 ,
χB(0,δ) ≤ ηδ ≤ χB(0,2δ),
| ηδ| ≤
C
δ
.
ην(ξ) := ηδ(
ξ
kν
) .
We could write down the standard bubble as
Vν(ξ) = ην(ξ)(
n(n − 1)
α∞f(x∞)
)
n−2
4 (
2εν
(εν)2 + |ξ − ξν|2
)
n−2
2 .
Set τn = n(n−2)
n(n−2)+2
and δν = (εν
)τn
. Given a pair (x, ε) ∈ M × [0, ∞), we deﬁne the test
functions as the approximation of the bubble by
V(x,ε)(ξ) = χ(x∞,δν )(
n(n − 1)
α∞f(x∞)
)
n−2
4 (
2ε
ε2 + |ξ − ξx|2
)
n−2
2
:= χ(x∞,δν )(
n(n − 1)
α∞f(x∞)
)
n−2
4 v(x,ε)(ξ)
where ξx is the coordinate of x. Lemma 5.8, 5.9, 5.10 and 5.11 from [22] imply that uν −vν → 0
in H1
sense. By the Sobolev imbedding H1
(N) → L
2n
n−2 (N), we have
1 = lim
ν
fu
2n
n−2
ν dµ
= lim
ν
fV
2n
n−2
ν dµ
= lim
ν
fV
2n
n−2
ν dµ
= lim
ν B(O,2δν )
f(
n(n − 1)
α∞f(x∞)
)
n
2 (
2εν
(εν)2 + |ξ − ξν|2
)n
dξ
= lim
ν B(O,2δν )
f(
n(n − 1)
α∞f(x∞)
)
n
2 (
2εν
(εν)2 + |ξ|2
)n
dξ
= f(x∞)(
n(n − 1)
α∞f(x∞)
)
n
2 ωn .
Thus,
α∞ = f(x∞)
2
n
−1
n(n − 1)ω
2
n
n = f(x∞)
2
n
−1
Y .

For every ν ∈ N, we define Aν = {(x, ε, γ) ∈ (M×R+
×R+
) : d(x, xν
) ≤ εν
, 1
2
≤ ε
εν ≤ 2, 1
2
≤
γ ≤ 2, }. Letting the coordinate of xν be ξν, as in [3], we can find a 3-tuple (xν, εν, γν) ∈ Aν
such that
E[uν − γνV(xν ,εν )] ≤ E[uν − γV(x,ε)]
for all (x, ε, γ) ∈ Aν. Since α is bounded, by passing to a subsequence we assume that αν → α∞
,and thus we have Eν → E∞.
Lemma 20.
uν − γνV(xν ,εν ) H1
ν→∞
−−−→ 0 .
Proof. The assertion follows from the definition of (xν, εν, γν).
Lemma 21. We have
d(xν, xν
) ≤ o(1)εν
,
εν
εν
= 1 + o(1) ,
γν = 1 + o(1) .
In particular, (xν, εν, γν) is an interior point of Aν for sufficiently large ν.
Proof. We have
γνV(xν ,εν ) − V(xν ,εν ) H1 ≤ uν − γνV(xν ,εν ) H1 + uν − V(xν ,εν ) H1 = o(1) .
Thus the assertion follows from lemma 20.
Let us decompose uν into
uν = vν + wν
where
vν = γνV(xν ,εν ) .
From Lemma 20, we know E[wν] = o(1) .
Lemma 22. (1) We have
| V
n+2
n−2
(xν ,εν )wνdµ| ≤ o(1)( |wν|
2n
n−2 dµ)
n−2
2n .

(2) We have
| V
n+2
n−2
(xν ,εν )
ε2
ν − d(xν, x)2
ε2
ν + d(xν, x)2
wνdµ| ≤ o(1)( |wν|
2n
n−2 dµ)
n−2
2n .
(3) We have
| V
n+2
n−2
(xν ,εν )
εν exp−1
xν
(x)
ε2
ν + d(xν, x)2
wνdµ| ≤ o(1)( |wν|
2n
n−2 dµ)
n−2
2n .
Proof. By the definition of (xν, εν, γν), we have
(cn V(xν ,εν ), wν + S0V(xν ,εν )wν)dµ = 0 ,
hence
[cn∆V(xν ,εν ) − S0V(xν ,εν )]wνdµ = 0 .
From the estimate
cn∆V(xν ,εν ) − S0V(xν ,εν ) + α∞fV
n+2
n−2
(xν ,εν ) L
2n
n+2
= o(1) ,
we can conclude that
| V
n+2
n−2
(xν ,εν )wνdµ| ≤ o(1) wν
L
2n
n−2
.
This proves (1). The results in (2) and (3) follow from (1) and Lemma 18.
Lemma 23. If ν is sufficiently large, then we have
n + 2
n − 2
α∞ v
4
n−2
ν w2
νfdµ ≤ (1 − c) (cn| wν|2
+ S0w2
ν)dµ
for some positive constant c independent of ν.
Proof. By the definition of vν and lemma 21, we have
|v
4
n−2
ν − V
4
n−2
(xν ,εν )|
n
2 dµ = o(1) .
Therefore, we only need to prove that
n + 2
n − 2
α∞ V
4
n−2
(xν ,εν )w2
νfdµ ≤ (1 − c) (cn| wν|2
+ S0w2
ν)dµ
for some positive constant c.

Suppose this is not true. Upon rescaling, we obtain a sequence of functions { ˜wν : ν ∈ N}
such that
(cn| ˜wν|2
+ S0 ˜w2
ν)dµ = 1
and
lim
ν→∞
n + 2
n − 2
α∞ V
4
n−2
(xν ,εν ) ˜w2
νfdµ ≥ 1 .
Note that | ˜wν|
2n
n−2 dµ ≤ Y (N)− n
n−2 . Take a sequence {Nν} such that Nν → ∞, Nνεν → 0.
Let Ων = BNν εν (xν) BNν εν (xν). We have
lim
ν→∞
V
4
n−2
(xν ,εν ) ˜w2
νdµ > 0 ,
and
lim
ν→∞ Ων
(cn| ˜wν|2
+ S0 ˜w2
ν)dµ ≤ lim
ν→∞
n + 2
n − 2
α∞ V
4
n−2
(xν ,εν ) ˜w2
νfdµ
We now deﬁne a sequence of functions ˆwν : TMxν → R by
ˆwν(ξ) = ε
n−2
2
ν ˜wν(expxν
(ενξ)), ξ ∈ TMxν .
This sequence satisﬁes
lim
ν→∞ {ξ∈TMxν :|ξ|≤Nν }
cn| ˆwν(ξ)|2
dξ ≤ 1
and
lim
ν→∞ {ξ∈TMxν :|ξ|≤Nν }
| ˆwν(ξ)|
2n
n−2 dξ ≤ Y (N)− n
n−2 .
Hence, if we take the weak limit as ν → ∞, then we obtain a function ˆw : Rn
→ R such that
(
1
1 + |ξ|2
)2
ˆw(ξ)2
dξ > 0 ,
and
| ˆw(ξ)|2
dξ ≤ n(n + 2) (
1
1 + |ξ|2
)2
ˆw(ξ)2
dξ .
Moreover, it follows from Lemma 18 that
(
1
1 + |ξ|2
)
n+2
2 ˆw(ξ)dξ = 0 ,
(
1
1 + |ξ|2
)
n+2
2
1 − |ξ|2
1 + |ξ|2
ˆw(ξ)dξ = 0 ,

(
1
1 + |ξ|2
)
n+2
2
ξ
1 + |ξ|2
ˆw(ξ)dξ = 0 .
Using a result of O. Rey, we conclude that ˆw = 0 (see [18], Appendix D, pp. 49-51). This is
a contradiction.

Chapter 5
Convergence
Lemma 24. Let {tν} be a time sequence tending to infinity. For sufficiently large ν, there
exists a constant C such that
E[uν] − α∞ ≤ C( |S(gν) − α∞f|
2n
n+2 dµgν )
n+2
n .
Proof. Using the identity
S(gν) = −u
− n+2
n−2
ν (cn∆uν − S0uν)
we obtain
E[uν] = (cn| uν|2
+ S0u2
ν)dµ
= (cn| vν|2
+ S0v2
ν)dµ + 2 u
n+2
n−2
ν S(gν)wvdµ − (cn| wν|2
+ S0w2
ν)dµ
= E[vν] + 2 u
n+2
n−2
ν [S(gν) − α∞f]wvdµ
− (cn| wv|2
+ S0w2
v −
n + 2
n − 2
α∞fv
4
n−2
ν w2
v)dµ
+α∞ [−
n + 2
n − 2
v
4
n−2
ν w2
ν + 2(vν + wν)
n+2
n−2 wν]fdµ .
Using Hölder inequality, we obtain
| u
n+2
n−2
ν [S(gν) − α∞f]wνdµ| ≤ ( u
2n
n−2
ν |S(gν) − α∞f|
2n
n+2 dµ)
n+2
2n ( |wν|
2n
n−2 dµ)
n−2
2n
≤ ( |S(gν) − α∞f|
2n
n+2 dµgν )
n+2
2n ( |wν|
2n
n−2 dµ)
n−2
2n
26

CHAPTER 5. CONVERGENCE 27
It follows from Lemma 23 that
(cn| wv|2
+ S0w2
v −
n + 2
n − 2
α∞fv
4
n−2
ν w2
v)dµ ≥ c (cn| wv|2
+ S0w2
v)dµ ,
hence
(cn| wv|2
+ S0w2
v −
n + 2
n − 2
α∞fv
4
n−2
ν w2
v)dµ ≥ c( |wν|
2n
n−2 dµ)
n−2
n .
By the Cauchy inequality,
( u
2n
n−2
2n
n+2 dµ)
n+2
2n ( |wν|
2n
n−2 dµ)
n−2
2n
≤ C( u
2n
n−2
2n
n+2 dµ)
n+2
n +
c
2
( |wν|
2n
n−2 dµ)
n−2
n .
Moreover,
E[vν] − α∞ = Ef [vν]( fv
2n
n−2
ν dµ)
n−2
n − α∞
= (Ef [vν] − α∞)( fv
2n
n−2
ν dµ)
n−2
n + α∞[( fv
2n
n−2
ν dµ)
n−2
n − 1]
≤ (Ef [vν] − α∞)( fv
2n
n−2
ν dµ)
n−2
n +
n − 2
n
α∞[ fv
2n
n−2
ν dµ − 1] .
As calculated in [17] 7.3, we have the estimate
Ef [vν] = Ef [V(xν ,εν )]
≤ α∞ + (ενδ−1
ν )C[−1 + O((ενδ−1
ν )2
)] + O(δ2
ν) .
By lemma 21, we have
Ef [vν] ≤ α∞ + Cε
2(n−2)
n(n−2)+2
ν [−1 + O(ε
4
n(n−2)+2
ν )] + O(ε
2n(n−2)
n(n−2)
ν ) .
Thus, for suﬃciently large ν,
Ef [vν] ≤ α∞ − Cε
2(n−2)
n(n−2)+2
ν + O(ε
2n(n−2)
n(n−2)
ν ) ≤ 0 .
Now we only need to estimate the term
α∞ [−
n + 2
n − 2
v
4
n−2
ν w2
ν+2(vν+wν)
n+2
n−2 wν+
n − 2
n
v
2n
n−2
ν −
n − 2
n
(vν+wν)
2n
n−2 ]fdµ−
c
2
( |wν|
2n
n−2 dµ)
n−2
n .

We have a pointwise estimate due to Brendle [3] page 180,
| −
n + 2
n − 2
v
4
n−2
ν w2
ν + 2(vν + wν)
n+2
n−2 wν +
n − 2
n
v
2n
n−2
ν −
n − 2
n
(vν + wν)
2n
n−2 −
n − 2
n
w
2n
n−2
ν |
≤ C|vν|max(0, 6−n
n−2
)
|wν|min( 2n
n−2
,3)
.
Therefore,
| −
n + 2
n − 2
v
4
n−2
ν w2
ν + 2(vν + wν)
n+2
n−2 wν +
n − 2
n
v
2n
n−2
ν −
n − 2
n
(vν + wν)
2n
n−2 |
≤ C|vν|max(0, 6−n
n−2
)
|wν|min( 2n
n−2
,3)
+
n − 2
n
|w
2n
n−2
ν | ,
α∞ [−
n + 2
n − 2
v
4
n−2
ν w2
ν + 2(vν + wν)
n+2
n−2 wν +
n − 2
n
v
2n
n−2
ν −
n − 2
n
(vν + wν)
2n
n−2 ]fdµ
≤ C |vν|max(0,6−n
n−2
)
|wν|min( 2n
n−2
,3)
dµ +
n − 2
n
|w
2n
n−2
ν |dµ .
If n ≥ 6,
C |vν|max(0, 6−n
n−2
)
|wν|min( 2n
n−2
,3)
dµ +
n − 2
n
|w
2n
n−2
ν |dµ
= C |wν|
2n
n−2 dµ +
n − 2
n
|w
2n
n−2
ν |dµ
= o( |wν|
2n
n−2 dµ)
n−2
n ) .
If n < 6,
C |vν|max(0, 6−n
n−2
)
|wν|min( 2n
n−2
,3)
dµ +
n − 2
n
|w
2n
n−2
ν |dµ
= C |vν|
6−n
n−2 |wν|3
dµ +
n − 2
n
|w
2n
n−2
ν |dµ
≤ C( |vν|
2n
n−2 dµ)
6−n
2n ( |wν|
2n
n−2 dµ)
3(n−2)
2n +
n − 2
n
|w
2n
n−2
ν |dµ
≤ C( |wν|
2n
n−2 dµ)
3(n−2)
2n +
n − 2
n
|w
2n
n−2
ν |dµ .
Since 3n−6
2n
> n−2
n
, the desired inequality holds for suﬃciently large ν.
Corollary 2. For any c > 0 and 0 < γ < 1 there exists t0 > 0 such that, for any t > t0,
E[u] − α∞ ≤ c( |S(g) − α∞f|
2n
n+2 dµg)
n+2
2n
(1+γ)
.

Proof. If this is not true, there exists a γ and a time sequence {tν} such that
E[uν] − α∞ > c( |S(gν) − α∞f|
2n
n+2 dµgν )
n+2
2n
(1+γ)
.
Meanwhile, for suﬃciently large ν, we have
E[uν] − α∞ ≤ C( |S(gν) − α∞f|
2n
n+2 dµgν )
n+2
n .
So we deduce
( |S(gν) − α∞f|
2n
n+2 dµgν )
n+2
2n
(1−γ)
>
c
C
which is contradictory to corollary 1.
Lemma 25. For any 0 < γ < 1, there exists t0 > 0 and C > 0 such that, for any t > t0,
E[u] − α∞ ≤ C( |S − αf|
2n
n+2 dµg)
n+2
2n
(1+γ)
.
Proof. We have
( |S(gν) − α∞f|
2n
n+2 dµgν )
n+2
2n
(1+γ)
= S − α∞f 1+γ
L
2n
n+2
≤ C S − αf 1+γ
L
2n
n+2
+ C α − α∞
1+γ
L
2n
n+2
.
Since
α − α∞ = α − E[u] + E[u] − α∞
= (αf − S)dµg + E[u] − α∞ ,
we have
α − α∞
1+γ
L
2n
n+2
≤ CF1+γ
1 + C(E[u] − α∞)1+γ
,
(E[u] − α∞)1+γ
≤ c1+γ
( |S(gν) − α∞f|
2n
n+2 dµgν )
n+2
2n
(1+δ)(1+γ)
,
(E[u] − α∞)1+γ
= o(( |S(gν) − α∞f|
2n
n+2 dµgν )
n+2
2n
(1+γ)
) .
Thus the desired result follows.
By a similar argument in corollary 2, we can deduce

Corollary 3. For any c > 0 and 0 < γ < 1 there exists t0 > 0 such that, for any t > t0,
E[u] − α∞ ≤ c( |S − αf|
2n
n+2 dµg)
n+2
2n
(1+γ)
.
Proposition 2.
∞
0
( u(t)
2n
n−2 (S − αf)2
dµ)
1
2 dt ≤ C .
Proof. It follows from corollary 3 that there exists 0 < γ < 1, t0 > 0 and C > 0 such that, for
any t > t0,
E[u(t)] − α∞ ≤ C( u(t)
2n
n−2 |S − αf|
2n
n+2 dµ)
n+2
2n
(1+γ)
.
It follows from Lemma 1 that
d
dt
(E[u(t)] − α∞) ≤ −cF2 ≤ −cF
n+2
n
2n
n+2
≤ −c(E[u(t)] − α∞)
2
1+γ , t ≥ t0.
Hence, for any t > t0,
d
dt
[(E[u(t)] − α∞)−1−γ
1+γ ] ≥ c ,
and
(E[u(t)] − α∞)−1−γ
1+γ ≥ ct ,
E[u(t)] − α∞ ≤ Ct− 1+γ
1−γ .
It follows from H¨older inequality that, for any T > t0,
2T
T
( u(t)
2n
n−2 (S − αf)2
dµ)
1
2 dt ≤ [T
2T
T
u(t)
2n
n−2 (S − αf)2
dµdt]
1
2
≤ C[T(E[u(2T)] − E[u(T)])]
1
2
≤ C[T(E[u(2T)] − α∞)]
1
2
≤ CT− γ
1−γ .
Finally, by picking L ∈ N such that 2L
> t0, we have
∞
0
( u(t)
2n
n−2 (S − αf)2
dµ)
1
2 dt =
2L
0
( u(t)
2n
n−2 (S − αf)2
dµ)
1
2 dt
+
∞
k=L
2k+1
2k
( u(t)
2n
n−2 (S − αf)2
dµ)
1
2 dt

≤ C + C
∞
k=L
2− γ
1−γ
k
≤ C .
From Proposition 2, we know that the volume does not concentrate:
Proposition 3. Given any η0 > 0, we can find some r > 0 such that
Bx(r)
u(t)
2n
n−2 dµ ≤ η0 , for all x ∈ N, t ≥ 0 .
Proof. Choose T > 0 such that
∞
T
( u(t)
2n
n−2 (S − αf)2
dµ)
1
2 dt ≤
η0
n
.
From the long time existence of the flow, we can choose r > 0 such that
Bx(r)
u(t)
2n
n−2 dµ ≤ η0 , for all x ∈ N, t ∈ [0, T] .
Then, for t ≥ T we have
Bx(r)
u(t)
2n
n−2 dµ ≤
Bx(r)
u(T)
2n
n−2 dµ +
n
2
∞
T
( u(t)
2n
n−2 (S − αf)2
dµ)
1
2 dt
≤ η0 .
Proposition 4. There are positive constants c, C independent of t such that
c ≤ u(t) ≤ C, for all t ≥ 0 .
Proof. Fix n
2
< q < p < n+2
2
. It follows from Lemma 2 and Lemma 14 that
|S|p
dµg ≤ C
where C is a positive constant independent of t. By Proposition 3, we can find a constant

r > 0 independent of t such that
Bx(r)
dµg ≤ η0 , for all x ∈ N, t ≥ 0 .
It follows from H¨older inequality that
|S|q
dµg ≤ (
Bx(r)
dµg)
p−q
p ( |S|p
dµg)
q
p .
Hence, if we choose η0 small enough, then we have
Bx(r)
u(t)q
dµ ≤ η1 , for all x ∈ N, t ≥ 0 ,
where η1 is the constant in Proposition A.1 [3] and we can conclude that u(t) is uniformly
bounded from above. From Corollary A.3 [3] again, we know that u(t) is uniformly bounded
w.r.t. (x, t) ∞.
Thus, under the assumption that the weak limit u∞ = 0, we have proved u converges
strongly to a nonzero limit which is a contradiction. Thus the weakly limit is never zero
under small initial energy condition which proves Theorem 1.

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