Revista Matematica Iberoamericana
Vol. 15, N.o 1, 1999
Harnack inequalities on a
manifold with positive or
negative Ricci curvature
Dominique Bakry and Zhongmin M. Qian
Summary. Several new Harnack estimates for positive solutions of the
heat equation on a complete Riemannian manifold with Ricci curvature
bounded below by a positive (or a negative) constant are established.
These estimates are sharp both for small time, for large time and for
large distance, and lead to new estimates for the heat kernel of a manifold with Ricci curvature bounded below.
1. Introduction and main results.
The main purpose of this paper is to present several new Harnack
estimates for non-negative solutions of the heat equation on a complete
manifold with Ricci curvature bounded below by a constant which may
be positive or negative. To obtain Harnack inequalities, we rst deduce
gradient estimates, that is upper bounds of the gradient of the logarithm
of a solution of the heat equation by a concave function of the time and
the time derivative of the same quantity. Then, by standard methods,
these bounds lead to Harnack inequalities and then to bounds on the
heat kernel.
In this context, we obtain quite strong Harnack inequalities, which
are improvements of the famous Li-Yau's estimate in 6], 12]. Although
143
144
D. Bakry and Z. M. Qian
our methods are similar in both cases of positive and negative lower
bound on the Ricci curvature, our results are completely new in the
positive case, and are improvements of previous results in the negative
one.
In order to state our results, we rst introduce some basic notations: let M be a complete Riemannian manifold of dimension n, and
let be the Laplace-Beltrami operator. Let u be a positive solution of
the heat equation
( ; @t ) u = 0 (1)
on 0 1) M
and let f = log u. Denote by rf the gradient of the function f and by
ft the time derivative of f .
In 1975, Yau 10] proved a Harnack inequality via Ricci curvature bounds for harmonic functions on a complete manifold. In their
paper 6], Li and Yau have established a sharp Harnack inequality for
parabolic harmonic functions on a complete manifold with non-negative
Ricci curvature. Namely,
(2)
jrf j2 ; ft 2nt for all t > 0 :
They also proved the following gradient estimate for a manifold with
Ricci curvature bounded below by ;K K 0
2
2
(3) jrf j2 ; ft n2t + 2n(;K1) for all t > 0 > 1 :
In his book 6], Davies improved the previous inequality under the same
assumption to the following one
2
2
(4) jrf j2 ; ft n2t + 4n(;K1) for all t > 0 > 1 :
Recently Yau 12] (also see Yau 11]) further established, among other
things, the following gradient estimate: if Ric ;K K 0, then
p
r
(5) jrf j ; ft 2 n K jrf j2 + 2nt + 2 n K + 2nt for all t > 0 :
2
With the method described in Section 4, it is standard to deduce from
this a Harnack inequality close to (12) (see below), but with dierent
Harnack inequalities on a manifold
145
constants. In Appendix A, with the method described in this paper,
and under the same assumption, we will improve this inequality to
p r 2 n nK n
2
for all t > 0 (6) jrf j ; ft n K jrf j + 2 t + 4 + 2 t which yields a Harnack inequality essentially similar to (12) for small
time and large distance.
Let us also mention that Hamilton 8] has obtained a Harnack
inequality for negative curvature manifolds.
The path to obtain Harnack inequalities is to rst establish gradient estimates as (4) or (5). To begin with, let us state the main results
of this paper. To this end, we rst introduce two functions X and Xe
as follows: let K 0, n > 0 be two constants. Then the functions X
and Xe are dened on (0 1) R by
8 nK
>
; 2 +Y
>
>
>
p rn K
>
>
+ nK 4 ; Y
>
>
r
>
p
2
t
>
< cotanh n n K n4K ; Y Y n4K (7) X (t Y ) = > n K
>
; 2 +Y
>
>
r nK
>
p
>
> + nK Y ; 4
(8)
>
r
>
p
>
2
t
: cotan n K Y ; n K Y > n K n
4
4
8 nK
>
>
2 + Yr
>
>
p
nK
>
>
+
n
K
Y
+
>
4 r
>
>
p
>
< cotanh 2nt n K Y + n4K Y ; n4K Xe (t Y ) = > n K
>
>
2 + Yr
>
>
p
>
+ n K ;Y ; n4K
>
>
r
>
p
>
2
t
: cotan n K ;Y ; n K Y < ; n K n
4
4
146
D. Bakry and Z. M. Qian
respectively. Indeed, as we will see later, X and Xe are solutions of
some simple dierential equations. The key inequalities obtained in
this paper are the following gradient estimates
Theorem 1. Suppose the Ricci curvature is bounded below by a constant ;K K 0, then we have :
1) jrf j2 Xe (t ft), on ft ;n K=4.
2) For any Y0 ;n K=4, we have
jrf j2 @ Y Xe (t Y0) (ft ; Y0) + Xe (t Y0) for all t > 0 :
3) There is a universal constant c > 0, such that
jrf j2 Xe (t ft) for all 0 < t Kc :
See Theorem 4 below for the precise value of the constant c.
As a consequence, if Ric ;K K 0, then
;ft 2nt + n4K for all t > 0 which is very close to the best possible one could expect, since (n ;
1) K=4 is the spectral gap of the space form with Ricci curvature ;K .
Indeed we will prove a better but slightly more complicated estimate
than (9).
Then, by standard methods, we deduce from Theorem 1 the following Harnack inequality
2 Z t+s 2 2 u
(
t
x
)
(10) u(t + s y) exp 4 s +
C 4 s2 ; A 4 s2 d t
for all t 0 s > 0, x y 2 M , where = d(x y) is the geodesic distance
from x to y, and
C (t Y ) = Xe (t Y ) ; Y
p r nK
p r nK
n
K
2
t
= 2 + n K Y + 4 cotanh n n K Y + 4 A(t Y ) = @ Y C (t Y )1(C+(@t YC) (;t YY @) Y C (t Y )) :
Y
Harnack inequalities on a manifold
147
In particular, since A(t Y ) > 0 when Y 0, we have
(11)
r
sinh 2 (t + s)pn K 2 + n K !n=2
u(t x) n
4 s2 4
r
u(t + s y)
2 p
2 n K sinh n t n K 4 s2 + 4
2 n K exp 4 s + 2 s :
By an elementary computation, inequality (10) yields the following
u(t x) t + s n=2
t
(12) u(t + s y)
( + pK n s)2 pn K
p
exp
+ 4 min f n K sg 4s
for all s > 0, x y 2 M . We will give an independent proof of this
in Section 4. We have been informed by Professor S. T. Yau that he
already obtained a Harnack inequality in this context, see 15], 14].
As usual, Harnack inequalities lead to lower bounds of the heat
kernel. Let H (t x y) be the heat kernel: the fundamental solution of
the heat equation (1). Then (12) implies that
H (t x y) (4 1t)n=2
( + pK n t)2 pn K
p
exp
;
;
min
f
(13)
n K tg 4t
4
for all (t x y) 2 (0 1) M M , = d(x y). See 3], 8] for a comparison theorem for heat kernels.
Notice that the leading term in (13) for small time is
1 exp ; 2 4t
(4 t)n=2
for large time is
exp ; n4K t for large distance is
2 pn K exp ; 4 t ; 2 :
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D. Bakry and Z. M. Qian
They are all very close to those best possibles we could expect.
For positive Ricci curvature manifolds, we will prove the following
gradient estimate
Theorem 2. Let Ric K K 0. Then
1) jrf j2 X (t ft), on ft n K=4.
2) If t 2=K , then jrf j2 X (t ft) and ;n=(2 t) ft n K=4.
3) If t 2=K , then
jrf j2 X (t ft) on ft Y0 and
jrf j2 @ Y X (t Y0) (ft ; Y0) + X (t Y0) where
on ft > Y0 2
Y0 = 1 + 64 n4K :
It turns out that both jrf j2 and ft are uniformally bounded for
each t 2=K if Ric K > 0.
We could also deduce from this a Harnack inequality, but it takes
a more complicated form than in the negative curvature case, and we
will therefore omit it in this paper.
The main tool used in this paper is the maximum principle, which
plays a fundamental r^ole in Partial Dierential Equations theory, see
for example 5]. Although the basic idea adopted in this paper is to
apply the maximum principle and Bochner identity to some nice test
functions this has been developed in a series of papers by Yau 9], 10],
12], Cheng and Yau 4], Li and Yau 6] etc. (see 7] for more references)
the main diculty with this method relies on the fact that, for any
family of test functions, one gets dierent kind of results, and therefore
the test functions in use are related to the results one is looking for. But
it is not always easy (and indeed quite hard in general) to device what is
the best estimate one could expect from a given dierential inequality.
Our main contribution in this context is to develop a method which
produces the best possible estimates and to show how to construct
good test functions in order to prove the expected estimates via the
maximum principle. This method applies to a more general setting
Harnack inequalities on a manifold
149
than the one described here, and it could be used in dierent contexts
for more general equations.
Let us explain our main idea as following. Everything relies on the
three following equations satised by jrf j2 and ft
(14)
(15)
f = ft ; jrf j2 (e ; @t )ft = 0 and
(e ; @t ) jrf j2 = 2 jHessf j2 + 2 Ric (rf rf ) where e = + 2 rf which is an elliptic operator. See Section 2 for
detail.
(16) comes from the Bochner identity. Therefore, if K is a lower
bound of the Ricci curvature, then we have the following inequality
(17)
(e ; @t ) jrf j2 n2 (f )2 + 2 K jrf j2 :
Inserting (14) into (17), we end up two dierential inequalities
(18)
(e ; @t ) jrf j2 n2 (jrf j2 ; ft )2 + 2 K jrf j2 (e ; @t )ft = 0 :
The main point of this paper is to compare (jrf j2 ft) with the solution
(X Y ) of the following system of dierential equations
; @t X = n2 (X ; Y )2 + 2 KX (19)
; @t Y = 0 with the condition that X (0) = 1. Since Y = constant, we regard it
as a parameter, and write the solution as X = X (t Y ). It is easy to
see that if K 0, then X (t Y ) is the function dened by (7), and if
K 0, then X (t Y ) = Xe (t Y ) with ;K in (8).
In fact, we were not able to prove (and we do not think it is true)
that jrf j2 X (t ft) everywhere, and this comes from the lack of
concavity of the curve Y ;! X (t Y ). What we show is that this
inequality holds on the most part of the curve. More precisely, our
150
D. Bakry and Z. M. Qian
main result asserts that if Ric K , then the most part of the curve
(X (t ft) ft) is above the curve (jrf j2 ft). In other words, jrf j2 X (t ft) for most of the values of ft , and we have a linear upper bound
on the remaining part.
The paper is organised as follows. In Section 2 we establish gradient estimates and some consequences for manifolds with Ricci curvature
bounded below. Section 3 deals with the case of positive Ricci curvature
manifolds. We deduce Harnack inequalities in Section 4. In Section 5,
we describe several extensions to other diusion operators, and, in the
end, we give an improved form of Yau's gradient estimate.
The results obtained in this paper have been announced in 2].
2. Gradient estimates for complete manifolds.
The main purpose of this section is to prove Theorem 1. Thus
throughout this section it will be assumed that Ric ;K , where K 0
is a constant.
Let u be a positive solution of the heat equation
(20)
( ; @t ) u = 0 on 0 1) M
and let f = log u. One can easily see that
(21)
( ; @t ) f = ;;(f f ) where ;(f f ) = jrf j2. In general, if is replaced by any sub-elliptic
dierential operator, we may dene
;(g h) = 21 ((hg) ; hg ; gh) for all g h 2 C 1 (M ) and therefore ;(g h) will stand for hrg rhi.
Dierentiating (21) in t, we obtain the rst fundamental equation
(22)
ft + 2 hrf rfti ; @t ft = 0 :
Then, dene the bilinear operator ;2 by iterating the previous denition
of ;
(23)
;2 (g h) = 21 (;(g h) ; ;(g h) ; ;(g h)) Harnack inequalities on a manifold
151
for all g h 2 C 1 (M ). Using ;2 , we may rewrite the classical Bochner
identity as
(24)
;2(g h) = hHess g Hess hi + Ric (rg rh) (26)
;(f f ) + 2 hrf r;(f f )i ; @t ;(f f ) = 2 ;2(f f ) :
for all g h 2 C 1 (M ). Since Ric ;K and jHess gj (g)2=n,
Bochner identity yields the following curvature-dimension inequality
for all g 2 C 1 (M ) :
(25)
;2 (g g) n1 (g)2 ; K ;(g g) This is the only form in which the Ricci curvature will appear in what
follows. Then, the fundamental remark is that, using (21) and (23) and
the previous denition of ;2 , we get another fundamental equation
For simplicity, introduce a dierential operator: L = + 2 rf ; @t .
Then the basic equations (22) and (26) can be rewritten
Ljrf j2 = 2 ;2(f f ) Lft = 0 :
If we notice that (21) can be rewritten as
(28)
;f = jrf j2 ; ft then the curvature-dimension inequality, implies that
(29)
L jrf j2 n2 (jrf j2 ; ft )2 ; 2 K jrf j2 :
We next look for a smooth function B on (0 1) R such that
jrf j2 ; ft B (t ft) for all t > 0 :
To this end, we set F = jrf j2 ; ft ; B (t ft), and G = t F . By the fact
that Lft = 0, we have
LB (t ft) = ;@t B (t ft) + @ Y2 B (t ft) jrftj2 :
Therefore
LF = L jrf j2 ; LB (t ft)
(30)
= L jrf j2 ; @ Y2 B (t ft) jrftj2 + @t B (t ft)
= 2 ;2(f f ) ; @ Y2 B (t ft) jrftj2 + @t B (t ft) :
(27)
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D. Bakry and Z. M. Qian
Thus if @ Y2 B 0 (which means B is concave in Y ), then
LF n2 (jrf j2 ; ft )2 ; 2 K jrf j2 + @t B (t ft)
= n2 (F + B )2 ; 2 K jrf j2 + @t B (t ft)
(31)
= n2 F 2 + n4 BF + n2 B 2 ; 2 K jrf j2 + @t B (t ft)
4
2
2
= n F + n B ; 2 K F + @t B + n2 B 2 ; 2 K (ft + B ) :
(32)
Hence
LG = ;F + t LF
2 t F 2 + 4 t B ; 2 K t ; 1F
(33)
n
n
; 2 K t (ft + B ) + 2nt B 2 + t @t B (t ft) :
Next we specify the function B , so that
jrf j2 ; ft B (t ft) for all t > 0 :
More precisely, for any Y0 > ;n K=4, we shall produce a function B
depending on the parameter Y0 for which we shall prove the above
upper bound.
To this end, consider the solution C of the dierential equation on
the half line (0 1) with a parameter Y 2 R
(34)
@t C + n2 C 2 ; 2 K (Y + C ) = 0 C (0) = 1 :
Then if Y > ;n K=4, we nd that
(35)
where
(36)
C (t Y ) = Xe (t Y ) ; Y = n2K + 2nt b(t2Y ) cotanh b(t2Y ) r
p
b(t Y ) = 4nt n K Y + n4K :
Harnack inequalities on a manifold
153
It is easily seen that
(37)
so that
2
@ Y b(t Y ) = 4nt n2K b(t1Y ) b eb @ Y C = 4nt @ Y b + e2b@;Y 1b ; 2(ebb@;Y 1)
2
b
2
e
= 2 K t b + b (eb ; 1) ; (eb ; 1)2 :
1
Therefore
(38)
and
(39)
2
n + nK
lim
C
(
t
Y
)
=
2t 2
Y !;nK=4
2K t lim
@
Y C (t Y ) =
3
Y !;nK=4
lim
!1 @ Y C (t Y ) = 0 :
b
Moreover, for each t > 0, the function Y ;! C (t Y ) is concave on the
interval (;n K=4 1).
Taking derivative with respect to Y in (34) we get that
(40)
@t @ Y C + n4 C@ Y C ; 2 K (1 + @ Y C ) = 0 :
Let Y0 > ;n K=4, and take B to be the linearization of C at Y0, i.e.
(41)
B (t Y ) = @ Y C (t Y0) (Y ; Y0 ) + C (t Y0) :
Then
@t B = @t @ Y C (t Y0) (Y ; Y0) + @t C (t Y0) and
2 B 2 = 2 (@ C (t Y ) (Y ; Y ) + C (t Y ))2
0
0
0
n
n Y
= n2 (@ Y C (t Y0) (Y ; Y0 ))2 + n4 C (t Y0) @ Y C (t Y0) (Y ; Y0 )
+ n2 C (t Y0)2 2 K (Y + B ) = 2 K (Y0 + C (t Y0)) + 2 K (Y ; Y0 ) (1 + @ Y C (t Y0)) :
154
D. Bakry and Z. M. Qian
Therefore, by (34) and (40), we get that
(42)
@t B + n2 B 2 ; 2 K (Y + B ) = n2 (@ Y C (t Y0) (Y ; Y0))2 :
Now dene F = jrf j2 ; ft ; B (t ft), where the constant Y0 > ;n K=4
in the denition of the function B . Let G = t F . Then by (33), we have
4B
1
2
2
2
2
LG n t G + n ; 2 K ; t G + t @t B + n B ; 2 K (ft + B )
4
4
2
1
2
= n t G + n @ Y C (t Y0) (ft ; Y0 ) G + n C (t Y0) ; 2 K ; t G
(43)
+ 2nt (@ Y C (t Y0) (ft ; Y0 ))2
4
2
1
2
= n t (G + t @ Y C (t Y0) (ft ; Y0)) + n C (t Y0) ; 2 K ; t G :
However,
4 t C (t Y ) ; 2 K t = b(t Y ) + 2 b(t Y0) 0
0
n
eb(tY0 ) ; 1
and therefore by the elementary inequality
b + eb2;b 1 2 for all b > 0 we have
(44) n4 C (t Y0) ; 2 K ; 1t 1t for all t > 0 Y0 ; n4K :
Hence by (43)
(45)
LG n2t (G + t @ Y C (t Y0) (ft ; Y0 ))2 + 1t G :
If the manifold is compact, consider a point (t0 x) at which the maximum of G on 0 t] M is attained: then, at this point, by the maximum
principle, e (G) 0. Moreover, @G=@t 0 and rG = 0. From these
we conclude that G 0. In this case we have
jrf j2 ; ft @ Y C (t Y0) (ft ; Y0 ) + C (t Y0) :
Harnack inequalities on a manifold
155
In the case where the manifold M is non-compact, we will get the same
conclusion with a slight modication in the arguments. Since Ricci curvature is bounded below, we may use a generalised maximum principle
(see 12], 5]) as following: replace K by any Ke > K in the denition of
the function C . Then the same argument yields the following inequality
LG n2t (G + t @ Y C (t Y0))2 + 1t G + 2 t (Ke ; K ) jrf j2
(46)
1t G + 2 t (Ke ; K ) jrf j2 :
Using then the Li-Yau's estimate (3), we may check that
G(t ) n2 (1 + @ Y C (t Y0))2 + n4K (1 + @ Y C (t Y0))2 @ C (tt Y ) :
Y
0
However,
t
3 :
lim
=
t!0 @ Y C (t Y0 )
2 Ke
Therefore for any t > 0,
sup G < 1 :
0t] M
Thus we can use the generalised maximum principle to the function G
on 0 t] M for any xed t > 0: if sup0t]M G > 0, then we may nd
a point t0 2 0 t] and a sequence of points fxk g 2 M , such that
G(t0 xk ) k1 jrGj(t0 xk ) k1 :
G(0 ) 0, and therefore t0 > 0. Also,
@t G(t0 xk ) 0 klim
!1 G(t0 xk ) = sup G :
0t] M
Hence we have
LG(t0 xk ) k1 + k2 jrf j which together with (46) implies that
1 LG(t x ) ; 2 jrf j
0 k
k
k
t1 G(xk ) + 2 t0 (Ke ; K ) jrf j2 ; k2 jrf j
0
1 :
t1 G(xk ) ; e 1
0
2 (K ; K ) t0 k2
156
D. Bakry and Z. M. Qian
Letting k ;! 1 we get that
0 t1 sup G 0 0t] M
which is a contradiction to the assumption that sup0t]M G > 0.
Therefore G 0. Since Ke > K is arbitrary, we have proved the main
result of this section:
Theorem 3. Let Ric ;K K 0, and let f = log u, where u is a
positive solution of the heat equation. Then
(47)
(48)
nK 4
jrf j2 ; ft 2 3K t ft + n4K + 2nt + n2K
2nt + n2K on ft ; n4K
jrf j2 Xe (t ft) on ft > ;
and
(49)
jrf j2 ; ft Y >inf
;nK=4(@ Y C (t Y0) (ft ; Y0 ) + C (t Y0)) :
0
Proof. We have proved (49). By taking Y0 = ft and noticing that
Xe (t Y ) = Y + C (t Y ) we get (47). Letting Y0 ;! ;n K=4 we get (48).
So we completed the proof.
Remark. In the above proof, we in fact used a \parabolic version" of
Yau's generalised maximum principle. Indeed, if we apply Yau's argument in 12] to the product manifold 0 t] M (with boundary), and use
the Hopf's maximum principle by Hopf (i.e. maximum principle with
boundaries), since 0 t] is compact, we can easily obtain the parabolic
version of the generalised maximum principle.
Corollary 1. Let Ric ;K , K 0, and let f = log u, where u is a
positive solution of the heat equation. Then we have
n nK nK n nK
(50)
;ft 12
+ 4 + 4 2t + 4 :
2
t
1+ 3Kt
Harnack inequalities on a manifold
157
Proof. For any Y0 > ;n K=4, we have
;ft C (t Y10+) ;@ YC0 @(tY YC ()t Y0) :
(51)
Y
0
By letting Y0 ;! ;n K=4, we get the conclusion.
We note that
n 1 n + nK + nK 2t 1+ 2 K t 2t 4
4
3
that is, the upper bound of ;ft would not be better than n=(2 t) for
negative curvature manifolds. However one would expect that the best
upper bound of ;ft should be n=(2 t) + (n ; 1) K=4, as (n ; 1) K=4 is
the spectral gap of the heat semi-group of the constant curvature space
form with Ricci curvature ;K . But we can see that
1 n + nK+ nK
n + (n ; 1) K 2t
4
4
1 + 23 K t 2 t 4
if and only if t (n ; 3)=(2 K ). Therefore, if the dimension of M is
bigger than 3, then our upper bound is even better than the expected
one: n=(2 t) + (n ; 1) K=4, within the time range (0 (n ; 3)=(2 K )).
Corollary 2. Let Ric ;K K 0, and let H (t x y) be the heat
kernel. Then
(52)
n=8
H (t x x) (4 1t)n=2 1 + 23 K t e;nKt=4 for all t > 0, x 2 M .
Proof. By Corollary 1, we have
;@t log (4 t)n=2H = ;@t log H ; 2nt
1 n + nK; n + nK
2t 4
1 + 23 K t 2 t 4
= ; n4K (3 + 12 K t) + n4K :
158
D. Bakry and Z. M. Qian
Using the fact that
n=2
lim
! (4 t) H (t x x) = 1 t 0
and integrating both sides over 0 t], we get the conclusion.
Corollary 3. Let Ric ;K K 0. Then
r
p
(53) jrf j2 ; ft n K ft + n4K + 2nt + n2K and
(54)
on ft ; n K 4
jrf j2 ; ft 23K t ft + n4K + 2nt + n2K :
Proof. We only need to prove the rst inequality. Since
eX (t Y ) Y + n K + n 1 + b 2 2t
2
p r nK
n
K
n
= Y + 2t + 2 + nK Y + 4 (54) follows immediately from Theorem 3.
By estimate (51), we have
;ft 2nt + n4K :
With this estimate, we can prove a better gradient estimate.
Indeed, let C be the solution of the dierential equation (34) on
the half line (0 1) with a parameter Y < ;n K=4 and C (0) = 1.
Then
(55)
C (t Y ) = n2K + 2nt b(t2Y ) cotan b(t2Y )
with
p r
4
t
b(t Y ) = n n K ;Y ; n4K :
Harnack inequalities on a manifold
159
Note that the function C , dened by (35) for Y > ;n K=4 and by (55)
for Y < ;n K=4 and
nK nK n
C t ; 4 = 2 + 2 t is a smooth function on (0 1) R . However, Y ;! C (t Y ) is not
concave on (;1 ;n K=4).
It is easy to see that Xe (t Y ) = Y + C (t Y ) for all Y 2 R .
Up to now, we restricted our attention to the part Y ;n K=4 of
the curve Y ;! C (t Y ). In what follows, we are going to improve the
previous estimate for any value of Y provided that the time t is not too
big.
Let cK be the positive constant
2 1 :
32 K
Then, we have
Theorem 4. Let Ric ;K K 0, and let f = log u, where u is a
positive solution of the heat equation. Then for any 0 < t cK ,
(56)
jrf j2 Xe (t ft) :
In fact, x any 0 < t cK , and let s 2 (0 t]. Let
h n nK n K Y0 2 ; 2 t ; 4 ; 4 and let
B (s Y ) = @ Y C (s Y0) (Y ; Y0) + C (s Y0) :
Dene a test function as usual: F = jrf j2 ; fs ; B (s fs) and G = sF .
Then the same argument as above yields that
(57) LG n2s (G+s @ Y C (s Y0) (fs;Y0))2 + n4 C (s Y0);2 K ; 1s G :
Notice that
4 s C (s Y ) ; 2 K s = b(s Y ) cotan b(s Y0) 0
0
n
2
160
D. Bakry and Z. M. Qian
as Y0 < ;n K=4, where
r
b(s Y0) = 2 s pn K ;Y ; n K :
0
2
n
4
Since s t cK , we have
r
r
b(s Y0) 2 s pn K n 2 t pn K n :
2
n
2t n
2t 4
Therefore, for any s t cK , and
h n nK nK
Y0 2 ; 2 t ; 4 ; 4 we have
p
b(s Y0) cotan b(s2Y0) 2 cos b(s2Y0) 2 :
Hence, for those s and Y0, we have
p
LG n2s (G + s @ Y C (s Y0) (fs ; Y0 ))2 + 2s; 1 G and by applying the maximum principle to G on 0 t] M , we conclude
that
(58)
jrf j2 ; fs @ Y C (s Y0) (fs ; Y0) + C (s Y0) for any 0 < s t cK and
h
Y0 2 ; 2nt ; n4K ; n4K :
In particular if ft ;n K=4, t cK , since ft > ;n=(2 t) ; n K=4, we
can take Y0 = ft in (58) to get that
jrf j2 ; ft C (t ft ) :
Thus we completed the proof.
By the above proof, we also proved in fact the following
Harnack inequalities on a manifold
161
Theorem 5. Let Ric ;K K 0, and let f = log u, where u is a
positive solution of the heat equation. Then for any Y0 ;n K=4,
(59)
jrf j2 ; ft @ Y C (t Y0) (ft ; Y0 ) + C (t Y0) for any
t p r n
:
n
K
8 n K ;Y0 ; 4
Corollary 4. Let Ric ;K K 0, and let f = log u, where u is a
positive solution of the heat equation. Then
1) If 0 < t cK , then ft Y1(t), where Y1(t) is the unique solution
of the equation
Y + C (t Y ) = 0 Y < 0:
2) For any t > 0,
;ft C (t Z1 t+) ;@ ZCt @(tY ZC ()t Zt) Y
where
t
;Zt = n4K + 64 Knt2 :
2
3. Positive curvature manifold.
The goal of this section is to prove Theorem 2. The method follows exactly the same lines as in the previous section, although the
conclusions are quite dierent.
Let M be a Riemannian manifold with dimension n, such that
Ric K , where K is a positive constant.
Let f = log u, and u be a positive solution of the heat equation.
In this case we have Li-Yau's estimate
(60)
jrf j2 ; ft 2nt :
Let U (t Y ) be the solution of the dierential equation on the half line
(0 1) with a parameter Y
(61)
@t U + n2 U 2 + 2 K (Y + U ) = 0 U (0) = 1 :
162
D. Bakry and Z. M. Qian
If Y < n K=4, then
(62) U (t Y ) = X (t Y ) ; Y = ; n2K + 2nt h(t2 Y ) cotanh h(t2Y ) r
with
If Y > n K=4, then
p
h(t Y ) = 4nt n K n4K ; Y :
U (t Y ) = X (t Y ) ; Y = ; n2K + 2nt h(t2Y ) cotan h(t2 Y )
with
Therefore
and
r
p
h(t Y ) = 4nt n K Y ; n4K :
n ; nK lim
U
(
t
Y
)
=
2t 2
Y !nK=4
2K t :
lim
@
Y U (t Y ) = ;
Y !nK=4
3
Moreover, U is a smooth function on (0 1) R , and for any t > 0,
the function Y ;! U (t Y ) is concave on (;1 n K=4). But it is not
concave on (n K=4 1).
For any Y0 < n K=4, we dene a test function G = t F , F =
jrf j2 ; ft ; B (t ft), where
B (t Y ) = @ Y U (t Y0) (Y ; Y0 ) + U (t Y0) :
Then by Bochner inequality, we get that
LG n2t G2 + 4nB + 2 K ; 1t G + t @t B + n2 B 2 + 2 K (ft + B )
(63)
= n2t (G + t @ Y U (t Y0) (ft ; Y0 ))2 + n4 U (t Y0) + 2 K ; 1t G :
By the fact that
4 U (t Y ) + 2 K ; 1 1 0
n
t t
for all Y0 < n4K t > 0 Harnack inequalities on a manifold
163
we conclude by the maximum principle that G 0. Therefore we have
Theorem 6. Let Ric K 0, and let f = log u, where u is a positive
solution of the heat equation. Then
(64)
jrf j2 ; ft Y <nK=
inf 4(@ Y U (t Y0) (ft ; Y0 ) + U (t Y0)) 0
for any t > 0. In particular, we have
(65)
and
(66)
jrf j2 X (t ft) on ft n K 4
jrf j2 ; ft 23 K t n4K ; ft + 2nt ; n2K :
Dene a function V (t Y ) by
V (t Y ) = X (t Y )
= ; n2K + Y
p rn K
p rn K
2
t
+ n K 4 ; Y cotanh n n K 4 ; Y when Y < n K=4, and
V (t Y ) = ; n2K + Y ; 2 K3 t Y ; n4K when Y n K=4. Then we can rewrite the estimates in Theorem 6 to
be
(67)
jrf j2 V (t ft) for all t > 0 :
It is easily seen that there is a unique zero point of V (t Y ) in (;1 0)
for each t > 0, denoted it by Y1 (t). Then by the fact that jrf j2 0,
we have ft Y1 (t).
If n=(2 t) ; n K=4 < 0, that is, if t > 2=K , then there is a unique
zero point of V (t Y ) in (0 n K=4], denoted by Y2(t), and again by the
fact that jrf j2 0, the estimate (67) yields that ft Y2 (t).
164
D. Bakry and Z. M. Qian
If n=(2 t) ; n K=4 > 0 and ;2 K t=3 + 1 < 0, that is, if 3=(2 K ) <
t < 2=K , then we can see that the unique zero point of V (t Y ) in (0 1)
is
n K + 2 K t ; 1;1 n ; n K 4
3
2t 4
and by the same reasoning as the above, we have
n nK
ft n4K + 2 K 1t
; 4 :
2
t
3 ;1
In these two cases, that is, if t > 3=(2 K ), there is a unique maximum
value of V (t Y ), attending at some point in (Y1(t) n K=4], which is
denoted by V0 (t). Then by (67), we have jrf j2 V0 (t).
Thus we have proved the following
Theorem 7. Let Ric K 0, and let f = log u, where u is a positive
solution of the heat equation. Then
(68)
Y1 (t) ft Y2(t) n4K for all t >
2 K
n nK
(69) Y1 (t) ft n4K + 2 K 1t
;
for all t > 3 2t 4
2K
;
1
3
and
(70)
jrf j2 V0 (t) for all t > 23K :
Now let us estimate Y1 (t). To this end, let Ue be the solution of the
dierential equation
@t Ue + n2 Ue 2 + 2 Ke (Y + Ue ) = 0 Ue (0) = 1 :
Then Ue is given by the formula as for U instead of K by Ke .
Let W = t (U ; Ue ). Then for any Y < n K=4, we have
4
1
2
2
;@t W = n t W + n U + 2 K ; t W + 2 (K ; Ke ) (Y + Ue ) t
n2t W 2 + 1t W + 2 (K ; Ke ) (Y + Ue ) t :
Harnack inequalities on a manifold
165
In particular, if K Ke = 0, then Ue = n=(2 t), and
;@t W n2t W 2 + 1t W + 2 K Y + 2nt t :
Applying the maximum principle to W on 0 t] R for any 0 < t (;n=(2 Y )) _ 0, we conclude that W 0 for any t (;n=(2 Y ) _ 0.
Therefore we have proved the following
Proposition 1. If K 0, then for any Y t > 0 such that Y < n K=4,
Y + n=(2 t) 0, we have
U (t Y ) 2nt :
As a consequence, we have
Y1 (t) ; 2nt for all t > 0 :
Our next goal is to bound jrf j2 ; ft for small time t.
Theorem 8. Let Ric K > 0, and let f = log u, where u is a positive
solution of the heat equation. Then for any Y0 > n K=4, we have
(71)
for
jrf j2 ; ft @ Y U (t Y0) (ft ; Y0 ) + U (t Y0) n
0<t p r
:
n
K
8 n K Y0 ; 4
Proof. Let G = t F , F = jrf j2 ; ft ; B (t ft ), where
B (t Y ) = @ Y U (t Y0) (Y ; Y0 ) + U (t Y0) :
Then by (63), we have
4
2
1
2
LG n t (G + t @ Y U (t Y0) (ft ; Y0)) + n U (t Y0) + 2 K ; t G :
However, when Y0 > n K=4 and
n
t p r
n
K
8 n K Y0 ; 4
166
D. Bakry and Z. M. Qian
we have
4 t U (t Y ) + 2 K t
0
n
= h cotan h2
r
r
p
p
= 4nt n K Y0 ; n4K cotanh 2nt n K Y0 ; n4K
p r
2
t
2 cos n n K Y0 ; n4K
2 cos 4
p
= 2:
Therefore
p
LG 2t; 1 G and by the maximum principle, we get the conclusion.
Corollary 5. Let
2
Y0 = 1 + 64 n4K :
Then for any 0 < t 2=K , we have
(72)
jrf j2 ; ft U (t ft) on ft Y0 and
(73)
jrf j2 ; ft @ Y U (t Y0) (ft ; Y0 ) + U (t Y0) :
Therefore, if
E (t Y ) = X (t Y ) = Y + U (t Y ) if Y Y0 and
E (t Y ) = @ Y U (t Y0) (Y ; Y0) + U (t Y0) + Y if Y > Y0 Harnack inequalities on a manifold
then
jrf j2 E (t ft) (74)
167
for all t K2 where Y0 is dened in Corollary 5.
Putting all discussions together, we have the following
Theorem 9. Let Ric K > 0, and let f = log u, where u is a positive
solution of the heat equation. Then we have the following conclusions.
1) If t 2=K , then jrf j2 X (t ft) and ;n=(2 t) ft n K=4.
2) If t 2=K , then we have
jrf j2 X (t ft) on ft Y0 and
jrf j2 ; ft @ Y U (t Y0) (ft ; Y0 ) + U (t Y0) where
on ft > Y0 2n K
Y0 = 1 + 64 4 :
4. Harnack inequalities.
In this section we rst show how to deduce a Harnack inequality
from a gradient estimate, although it is very standard, see 9]. Then
we prove the main Harnack estimates.
The link between Harnack inequalities and gradient estimates is
given in the following
Proposition 2. Let M be a complete Riemannian manifold, and let
f = log u where u is a positive solution of the heat equation. Suppose
that
jrf j2 (t ft) for all t > 0 where : (0 1) R ;! R is a continuous function, then
(75)
(76)
u(t x) exp Z t+s K d u(t + s y)
s
t
168
D. Bakry and Z. M. Qian
where is the geodesic distance between x and y, and
K (t ) =
sup
fY : (tY )0g
; p
(t Y ) ; Y :
Proof. The proof of this proposition is straightforward. Let be a
minimal geodesic joining x and y, so that (0) = y and (1) = x. If
= d(x y), then j_ j = . Dene
p() = ( () (1 ; ) t2 + t1 ) t2 = t + s t1 = t :
Then p(0) = (y t2) and p(1) = (x t1). Set () = f (p()). It is clear
that
f (t1 x) ; f (t2 y) = (1) ; (0) =
Z1
0
_ () d =
Z1
0
(hrf _ i ; s ft) d with t = (1 ; ) t2 + t1 . We end up with
f (t1 x) ; f (t2 y) Z1
( jrf j ; s ft) d
Z s 0
s jrf j ; ft d
Z s K s d :
0
0
From this result and the previous gradient estimates, we may now
prove Harnack inequalities: we shall rst establish the simplest one, for
which the computations are easy: it follows from the gradient estimate
(53).
Theorem 10. Let Ric ;K , K 0, and let u be a positive solution
of the heat equation. Then
u(t x) t + s n=2
u(t + s y)
t
p
( + pn K s)2 pn K n p
o
n
K
(77)
exp
+ 2 min ( 2 ; 1) 2 s 4s
for all s > 0, t 0, x y 2 M , where = d(x y).
Harnack inequalities on a manifold
169
Proof. Let f = log u, and > 0 be xed.
If ft > ;n K=4, then the gradient estimate (53) yields that
r
2
jrf j2 + n4K + 2nt (78)
p
where = ft + n K=4, so that
;ft = n4K ; 2 :
Denote by
u=
s
r
2
+ n4K + 2nt 0 :
Then jrf j u and
r
r
2
;ft = n4K ; u2 ; 2nt ; n4K
r n
p
n
= ;u2 + 2 t + n K u2 ; 2 t
p
;u2 + 2nt + n K u :
Hence in this case we have
p
p
2
(79) jrf j ; ft ( + n K ) u ; u2 + 2nt ( + 4n K ) + 2nt :
If ft ;n K=4, then on one hand the estimate (54) implies that
;ft ;jrf j2 + 2nt + n2K so that
(80)
2
jrf j ; ft 4 + 2nt + n2K
p 2
p
(
+
n
K
)
n
n
K
nK :
=
+
+
;
4
2t 4
2
170
D. Bakry and Z. M. Qian
On other hand, by the estimate (54), we have
jrf j2 ; ft 2 K3 t ft + n4K + 2nt + n2K :
Therefore
jrf j ; ft j rf j ; 12 K jrf j2
1+ 3 t
1 n + nK+ nK
+
4
1 + 23K t 2 t 4
!
2
4 + 1 ; r 1
jrf j
2
K
1+ 3 t
n nK nK
1
+
+
+ 4
1 + 23K t 2 t 4
!r
2
n + nK 4 + 1 ; r 1
2t 2
1 + 23K t
1
+ 2nt + n4K ; n12K
1 + 23K t
r
2K t n + nK
2
3
2t r 2
+ 2nt + n4K
4 +r
1 + 23K t 1 + 1 + 23K t
r
2
4 + n2K + 2nt + n4K
p 2
p
n
K
)
2 ; 1 pn K :
(
+
n
=
+
+
4
2t
2
Hence, if ft + n K=4 < 0, then we have
p 2
(
+
nK ) + n
jrf j ; ft 4
2t
(81)
p
p o
n
p
n
K
+ 2 min ( 2 ; 1) n2 K :
Harnack inequalities on a manifold
171
Therefore, by (79) and (81), we always have the following estimate
p
2
jrf j ; ft ( + 4n K ) + 2nt
(82)
p
p o
np
n
K
+ 2 min ( 2 ; 1) n2 K for any 0.
Using now the previous proposition, we get
f (t1 x) ; f (t2 y)
p
Z 1 ( + pn K s)2 n pn K n p
n K so d 2
;
1)
+
s
+
min
(
4s
2t
2
2
0
which yields the Harnack inequality.
Remark. As we pointed out in the introduction, S. T. Yau mentioned
to us that he obtained a similar Harnack inequality.
Now we turn to prove the main Harnack estimate. Let Ric ;K
for some constant K 0. We have seen that the main point is to
estimate jrf j ; ft for > 0.
For any Y0 > ;n K=4, we have
jrf j2 ; ft C (t Y0) + @ Y C (t Y0) (ft ; Y0 ) and therefore
2
;ft ; 1 + @jrCf j(t Y ) + C (t Y10+) ;@ YC0 @(tY YC ()t Y0) :
Y
0
Y
0
Hence for any > 0, we have
jrf j ; ft ; 1 + @ 1C (t Y ) jrf j2 + jrf j + C (t1 Y+0)@; CY0(tCY(t)Y0)
Y
0
Y
0
2
4 (1 + @ Y C (t Y0)) + C (t Y10+) ;@ YC0 @(tY YC ()t Y0)
Y
0
2
2
(83)
= 4 + @ Y C (t Y0) 4 ; Y0 + C (t Y0)
; 1 +@ Y@C (Ct(Yt0Y) ) (C (t Y0) + @ Y C (t Y0) (0 ; Y0 )) :
Y
0
172
D. Bakry and Z. M. Qian
Letting Y0 = =4 in (83), we get that
(84)
where
2 2
2
jrf j ; ft 4 + C t 4 ; A t 4 A(t Y ) = @ Y C (t Y )1(C+(@t YC) (;t YY @) Y C (t Y )) :
Y
Notice that A(t Y ) 0 when Y 0. Therefore we have proved the
following
Theorem 11. Let Ric ;K K 0, and let u be a positive solution
of the heat equation. Then
2 Z t+s 2 2 u
(
t
x
)
C 4 s2 ; A 4 s2 d (85) u(t + s y) exp 4 s +
t
for any t 0 s > 0 and x y 2 M , where is the geodesic distance
between x and y.
Remark. Although we have the simple fact that
p
r
C (t Y ) n K Y + n4K + 2nt + n2K for any Y > ;n K=4, however, unlike C (t Y ) whose linearization at any
point Y ;n K=4 is an upper bound of jrf j2 ; ft , the linearization of
p
r
n K Y + n4K + n2K + 2nt
at some points may not be an upper bound of jrf j2 ; ft. In this
sense, therefore, the analysis via C (t Y ) is even simpler and yields
much stronger conclusions. This is also the reason why we give an
independent proof of Theorem 10.
Harnack inequalities on a manifold
173
Since
C (t Y ) = n2K + 2nt b(t2Y ) cotanh b(t2Y )
r
Z t+s
t
r
2 t p
p
n
K
n
K
n
K
= 2 + n K Y + 4 cotanh n n K Y + 4 p p
Z (2=n)(t+s) nK Y +nK=4
n
K
n
C ( Y ) d = 2 s + 2
cotanh d
p p
(2=n)t nK Y +nK=4
r
sinh 2 (t + s) pn K Y + n K !
n
4 r
= n2K s + n2 log
p
so that we have the following
sinh n2 t n K Y + n4K
Corollary 6. Let Ric ;K K 0, and let u be a positive solution
of the heat equation. Then
u(t x)
(86) u(t + s y)
r
sinh 2 (t + s) pn K 2 + n K !n=2
n
4 s2 4
E ( s t) 2 p r 2 n K sinh n t n K 4 s2 + 4
where
2 n K Z t+s 2 E ( s t) = exp 4 s + 2 s ;
A 4 s2 d :
t
Applying Corollary 6 to the heat kernel we have
r
2 pn K 2 + n K
!n=2
2
n
4t
4
;1
H (t x y) 2 t p r 2 n K E ( t 0)
4 sinh n n K 4 t2 + 4
(87)
= (4 1t)n=2
rK p
2 + n K t2 !n=2
n
E ( t 0);1 r K p 2
sinh n + n K t2
174
D. Bakry and Z. M. Qian
where
2
E ( t 0);1 = exp ; ; n K t +
4t
2
Z t 2 A 4 t2 d :
0
Remark. Proposition 2 together with the gradient estimates for the
positive Ricci curvature manifolds yields a Harnack inequality. However, its form is quite complicated. Since the upper bound function
(t Y ) is in general nonlinear , we can improve the Harnack inequality in Proposition 2 by varying the time speed, that is, replacing the
straight line joining t and t + s by a curve. Therefore we decided to
write down the explicit Harnack inequalities for positive Ricci curvature
manifolds together with the compact manifold case in a separate paper.
5. Extensions.
The same arguments in previous sections can be applied to the
case when the manifold M with convex boundary @M the second fundamental form of the boundary @M is nonnegative. This is because
of the fact that if @u=@ = 0 on the boundary where denotes the
pointed out normal vector eld then
@ jruj2 = ;2 (ru ru) @
so that we can use the Hopf maximum principle when 0. We only
write down a theorem in this case.
Theorem 12. Let M be a complete Riemannian manifold with a convex
boundary @M , and let u be a positive solution of the heat equation
( ; @t ) u = 0 on 0 1) M @ u = 0 on (0 1) @M :
Let f = log u. Then
(88)
and
jrf j2 ; ft C (t ft) on ft ;
nK 4
jrf j2 ; ft @ Y C (t Y0) (ft ; Y0 ) + C (t Y0) Harnack inequalities on a manifold
175
for all t > 0, Y0 ;n K=4.
There is a further generalisation of our gradient estimates and Harnack inequalities to a general elliptic operator. We only state a result.
Let M be a complete manifold (without or with a convex boundary), and let B = +B be an elliptic operator where B is a C 2 -vector
eld. Assume that B satises a curvature-dimension inequality (see
1])
;2 (g g) m1 (B g)2 ; K ;(g g) for all g 2 C 1 (M ) for some constants m > 0 and K 0, where by denition ;(f g) =
hrf rgi, and
;2 (f g) = 21 (B (fg) ; ;(B f g) ; ;(f B g)) :
This condition is satised if and only if
1 B B ;K Ric ; rsB ; m ;
n
where m n, n = dim M , Ric denotes the Ricci curvature and
rsB ( ) = 21 (hr B i + hr B i) for all 2 TM :
If f = log u, u is a positive solution of the heat equation
(B ; @t ) u = 0 on 0 1) M (in the case that the boundary @M 6= 0, we further assume that u
satises the Neumann boundary condition), then
jrf j2 ; ft C (t ft) on ft ; m4K and
jrf j2 ; ft @ Y C (t Y0) (ft ; Y0) + C (t Y0) for all Y0 ; m4K where C is the solution of the dierential equation
@t C + m2 C 2 ; 2 K (Y + C ) = 0 C (0) = 1 :
176
D. Bakry and Z. M. Qian
6. Appendix.
The goal of this appendix is to give a proof of Yau's estimate (6).
We will use the same notations as in Section 2.
Let F = jrf j2 ; ft ; Q(t jrf j2) for some positive function Q which
will be given later, and G = t F .
It is easily seen that
LF = (1 ; @X Q) L jrf j2 + @t Q ; @X2 Q jrjrf j2j2
= 2 (1 ; @X Q) ;2(f f ) ; @X2 Q jrjrf j2j2 + @t Q and therefore if @X2 Q 0, 1 ; @X Q 0, then we have
LG = ;F + 2 t (1 ; @X Q) ;2 (f f ) ; t @X2 Q jrjrf j2j2 + t @t Q
;F + 2 t (1 ; @X Q) n1 (jrf j2 ; ft )2 ; K jrf j2 + t @t Q :
However jrf j2 ; ft = F + Q, so that
1
2
2
LG ;F + 2 t (1 ; @X Q) n (F + Q) ; K jrf j + t @t Q
2
= 2 t (1 ; @X Q) Fn + ; 1 + 2 t (1 ; @X Q) 2nQ F
Q2
2
+ 2 t (1 ; @X Q) n ; K jrf j + t @t Q :
Thus, if
(89)
(90)
@X2 Q 0 lim
! t Q(t X ) 0 t 0
4 t Q (1 ; @ Q) 1 X
n
1 ; @X Q > 0 and
2
@t Q + 2 (1 ; @X Q) Qn ; KX 0 for all t > 0, X 0, then, if the manifold is compact,
(91)
(92)
jrf j2 ; ft Q(t jrf j2) for all t > 0 :
Harnack inequalities on a manifold
177
Theorem 13. Let Ric ;K for some non-negative constant K , and
let u be a positive solution of the heat equation and f = log u. Then
(93)
p
r
jrf j ; ft n K jrf j2 + 2nt + n4K + 2nt :
2
Proof. For simplicity, let a = n=(2 t) + n K=4. Then
p
Q(t X ) = m X + a + 2nt Therefore
and
@X2 Q = ;
Q2
p
m = nK :
mp
0
4 (X + a) X + a
@t Q + 2 (1 ; @X Q) n ; KX
p
a
2
m
n
m
2
m
X
+
a
m
p
p
; t ; 2
+2 1;
=
t
4t X + a
2 X+a n
mt (2 pa ; m) ; 4 tnpa
2
= mt 24pa a;+mm ; 4 tnpa
= mt t (2 p2an+ m) ; 4 pna t
mt t (42 pn a) ; 4 pna t
> 0:
Thus condition (91) is satised.
Let us now check the condition (90). It is easily seen that
4 Q (1 ; @ Q) = 4 m pX + a ; m + 1 2 ; p m X
n
n
2
t
X +a
!
r
p
n nK nK 1
4
m
m
n 2t+ 4 ; 2 + t 2; r
:
n + nK
2t 4
178
D. Bakry and Z. M. Qian
Therefore, when n=(2 t) 3 n K=4,
1
2; r m
n + nK
2t 4
so that
4 Q (1 ; @ Q) 1 X
n
t
and when n=(2 t) 3 n K=4,
4 Q (1 ; @ Q) 2 m r
1 p
4 >1:
X
n
t
n + nK + nK 3t t
2t 4
2
Hence condition (90) is satised. Therefore we proved Theorem 13 for
compact manifolds. If the manifold is non-compact, we may use the
generalised maximum principle to go through the proof.
Acknowledgements. The authors thank Professor S. T. Yau for his
comments on their paper 2]. The second author was supported by a
grant from the Ministry of Foreign Aairs, France.
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Recibido: 26 de noviembre de 1.997
Dominique Bakry
Laboratoire de Statistique et Probabilites
CNRS
Universite Paul Sabatier
118, route de Narbonne
31062 Toulouse Cedex, FRANCE
[email protected]
and
Zhongmin M. Qian
Department of Mathematics
Imperial College of Science, Technology and Medicine
180 Queen's Gate
London SW7 2BZ, UNITED KINGDOM
and
Shanghai TieDao University, CHINA
[email protected]