SLIDE 1 YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM FOR POSITIVE PSEUDOHARMONIC FUNCTIONS IN A COMPLETE PSEUDOHERMITIAN MANIFOLD
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
- Abstract. In this paper, we …rst derive the sub-gradient estimate for positive pseudohar-
monic functions in a complete pseudohermitian (2n+1)-manifold (M; J; ) which satis…es the CR sub-Laplacian comparison property. It is served as the CR analogue of Yau’s gradient
- estimate. Secondly, we obtain the CR sub-Laplacian comparison theorem in a class of com-
plete noncompact pseudohermitian manifolds. Finally we have shown the natural analogue
- f Liouville-type theorems for the sub-Laplacian in a standard Heisenberg (2n+1)-manifold
(Hn; J; ) :
In [Y1] and [CY], S.-Y. Cheng and S.-T. Yau derived a well known gradient estimate for positive harmonic functions in a complete noncompact Riemannian manifold. Proposition 1.1. ([Y1], [CY]) Let M be a complete noncompact Riemannian m-manifold with Ricci curvature bounded from below by K (K 0): If u (x) is a positive harmonic function on M; then there exists a positive constant C = C(m) such that (1.1) jrf(x)j2 C( p K + 1 R)
- n the ball B (R) with f(x) = ln u(x): As a consequence, the Liouville theorem holds for
complete noncompact Riemannian m-manifolds of nonnegative Ricci curvature.
1991 Mathematics Subject Classi…cation. Primary 32V05, 32V20; Secondary 53C56. Key words and phrases. CR Bochner formula, Subgradient estimate, Liouvile theorem, Pseudohermtian Ricci, Pseudohermitian torsion, Heisenberg group, Pseudohermitian manifold.
Research supported in part by the NSC of Taiwan.
1
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SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
In this paper, by modifying the arguments of [Y1], [CY] and [CKL], we derive a sub- gradient estimate for positive pseudoharmonic functions in a complete noncompact pseudo- hermitian (2n + 1)-manifold (M; J; ) which is served as the CR version of Yau’s gradient
- estimate. Then we prove that the CR analogue of Liouville-type theorem holds for positive
pseudoharmonic functions as well. We …rst recall some notions as in section 2: Let (M; ) be a (2n+1)-dimensional, orientable, contact manifold with contact structure ; dimR = 2n. A CR structure J compatible with is an endomorphism J : ! such that J2 = 1. We also assume that J satis…es the integrability condition ( see next section). A CR structure J can extend to C and decomposes C into the direct sum of T1;0 and T0;1 which are eigenspaces of J with respect to eigenvalues i and i, respectively. A pseudohermitian structure compatible with is a CR structure J compatible with together with a choice of contact form and = ker . Such a choice determines a unique real vector …eld T transverse to which is called the characteristic vector …eld of , such that (T) = 1 and LT = 0 or d(T; ) = 0. Let fT; Z; Z
g be a frame of TM C, where Z is any local frame of T1;0; Z = Z 2 T0;1.
We de…ne Ric and Tor on T1;0 by (1.2) Ric(X; Y ) = R
XY
(1.3) Tor(X; Y ) = i P
;(A
Y
Here X = XZ , Y = Y Z; R
- is the pseudohermitian curvature tensor, R
= R
- is the pseudohermitian Ricci curvature tensor and A is the torsion tensor.
In Yau’s method for the proof of gradient estimates, one can estimate ( (x) jrf(x)j2) for a nonegative cut-o¤ function (x) on B (2R) via Bochner formula and Laplacian com-
- parison. At the end, one has gradient estimate (1.1) by applying the maximum principle
SLIDE 3 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 3
to (x) jrf(x)j2: However in order to derive the CR subgradient estimate, one of di¢cul- ties is to deal with the following CR Bochner formula (Lemma 2.1) which involving a term hJrb'; rb'0i that has no analogue in the Riemannian case. b jrb'j2 = 2
+ 2 hrb'; rbb'i + (4Ric 2 (n 2) Tor) ((rb')C ; (rb')C) + 4 hJrb'; rb'0i : Here
- rH2 ; b, rb are the subhessian, sub-Laplacian and sub-gradient respectively. We
also denote '0 = T'. In order to overcome this di¢culty, we introduce a real-valued function F (x; t; R; b) : M [0; 1] (0; 1) (0; 1) ! R by adding an extra term t (x) f 2
0 (x) to jrbf(x)j2 as
following F (x; t; R; b) = t
0 (x)
- n the Carnot-Carathéodory ball B (2R) with a constant b to be determined. In section 3,
we derive the CR subgradient estimate (1.11) and (1.8) by applying the maximum principle to (x) F(x; t) for each …xed t 2 (0; 1] if the CR sub-Laplacian comparison property (see De…nition 1.2) holds on (M; J; ). De…nition 1.1. Let (M; J; ) be a pseudohermitian (2n + 1)-manifold. A piecewise smooth curve : [0; 1] ! M is said to be horizontal if 0(t) 2 whenever 0(t) exists. The length of is then de…ned by l() = Z 1 h0(t); 0(t)i
1 2
Ldt:
Here h ; iL is the Levi form as in (2.2). The Carnot-Carathéodory distance between two points p, q 2 M is dc(p; q) = inffl() j 2 Cp;qg where Cp;q is the set of all horizontal curves joining p and q. We say M is complete if it is complete as a metric space. We refer to [S] for some details. By Chow connectivity
SLIDE 4
4
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
theorem [Cho], there always exists a horizontal curve joining p and q, so the distance is …nite. Furthermore, there is a minimizing geodesic joining p and q so that its length is equal to the distance dc(p; q). De…nition 1.2. Let (M; J; ) be a complete noncompact pseudohermitian (2n+1)-manifold with (1.4) (2Ric (n 2)Tor) (Z; Z) 2k jZj2 for all Z 2 T1;0, and k is an nonnegative constant. We say that (M; J; ) satis…es the CR sub-Laplacian comparison property if there exists a positive constant C0 = C0(k; n) such that (1.5) br C0(1 r + p k) in the sense of distributions. Here b denote sub-Laplacian and r (x) is the Carnot-Carathéodory distance from a …xed point x0 2 M: Let (M; J; ) be the standard Heisenberg (2n + 1)-manifold (Hn; J; ) : We have R
=
0 and A = 0: Then the following CR sub-Laplacian comparison property holds on (Hn; J; ). Proposition 1.2. ([CTW]) Let (Hn; J; ) be a standard Heisenberg (2n+1)-manifold. Then there exists a constant CHn
1
> 0 (1.6) brHn CHn
1
rHn : Remark 1.1. 1. In the paper [CTW], it was shown tha the CR sub-Laplacian comparison property (1.5) holds in a complete Heisenberg (2n + 1)-manifold (Hn; J; ) : Here we will give an another proof by applying the di¤erential inequality for sub-Laplacian of Carnot- Caratheodory distance as in Lemma 4.1.
SLIDE 5 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 5
- 2. We expect that the method will be able to adapt to derive the CR sub-Laplacian compar-
ison property in a complete noncompact pseudohermitian manifold other than the standard Heisenberg manifold. In particular, (1.5) holds in an asymptotic Heisenberg manifold as well, we will discuss it elsewhere. We refer to Theorem 4.3 as in section 4 for some details. In order to have an analogue of Liouville-type theorem ( see Corollary 1.4 ) for positive pseudoharmonic functions (i.e. bu = 0) in a complete noncompact pseudohermitian (2n + 1)-manifold, we need to show the following sub-gradient estimate for positive pseudoharmonic functions u: Theorem 1.3. Let (M; J; ) be a complete noncompact pseudohermitian (2n+1)-manifold with (2Ric (n 2)Tor) (Z; Z) 2k jZj2 for all Z 2 T1;0, and k 0: Furthermore, we assume that (M; J; ) satis…es the CR sub- Laplacian comparison property (1.5). If u (x) is a positive pseudoharmonic function with (1.7) [b; T]u = 0
- n M. Then for each constant b > 0, there exists a positive constant C2 = C2(k) such that
(1.8) jrbuj2 u2 + bu2 u2 < (n + 5 + 2bk)2 (5 + 2bk)
b + C2 R
- n the ball B (R) of a large enough radius R which depends only on b, k.
As a consequence, let R ! 1 and then b ! 1 with k = 0 in (1.8);we have the following CR Liouville-type theorem. Corollary 1.4. Let (M; J; ) be a complete noncompact pseudohermitian (2n+1)-manifold with (2Ric (n 2)Tor) (Z; Z) 0
SLIDE 6 6
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
for all Z 2 T1;0. Furthermore, we assume that (M; J; ) satis…es the CR sub-Laplacian comparison property. If u (x) is a positive pseudoharmonic function with [b; T]u = 0: Then u (x) is constant. It is shown that (Lemma 2.3) (1.9) [b; T] u = 4 Im[i
n
X
;=1
(A
If (M; J; ) is a complete noncompact pseudohermitian (2n + 1)-manifold with vanishing
[b; T] u = 0: Corollary 1.5. Let (M; J; ) be a complete noncompact pseudohermitian (2n+1)-manifold
- f nonnegative pseudohermitian Ricci curvature tensors and vanishing torsion. Furthermore,
we assume that (M; J; ) satis…es the CR sub-Laplacian comparison property. Then any positive pseudoharmonic function is constant. By applying Corollary 1.5 and Proposition 1.2, we have the following CR Liouville-type theorem for a positive pseudoharmonic function u on (Hn; J; ) with R
= 0 and A = 0:
Corollary 1.6. There does not exist any positive nonconstant pseudoharmonic function in a standard Heisenberg (2n + 1)-manifold (Hn; J; ). Remark 1.2. Koranyi and Stanton ([KS]) proved the Liouville theorem in (Hn; J; ) by a di¤erent method. In general, we have the following weak sub-gradient estimate for positive pseudoharmonic functions in a complete noncompact pseudohermitian (2n + 1)-manifold.
SLIDE 7 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 7
Theorem 1.7. Let (M; J; ) be a complete noncompact pseudohermitian (2n+1)-manifold with (2Ric (n 2)Tor) (Z; Z) 2k jZj2 and (1.10) max fjAj ; jA;
jg k1
for all Z 2 T1;0 and k 0; k1 > 0: Furthermore, we assume that (M; J; ) satis…es the CR sub-Laplacian comparison property. If u (x) is a positive pseudoharmonic function on M. Then there exists a small constant b0 = b0(n; k; k1) > 0 and C3 = C4(k; k1; k2) such that for any 0 < b b0, (1.11) jrbuj2 u2 + bu2 u2 < (n + 5)2 5
b + C3 R
- n the ball B (R) of a large enough radius R which depends only on b.
Remark 1.3. By comparing the Yau’s gradient estimate (1.1), we need an extra assumption (1.10) to obtain the CR subgradient estimate (1.11) due to the natural of sub-Laplacian in pseudohermitian geometry. However, we do obtain an extra estimate on the derivative of pseudoharmonic functions u(x) along the missing direction of characteristic vector …eld T. We brie‡y describe the methods used in our proofs. In section 2, we …rst introduce some basic materials in a pseudohermitian (2n + 1)-manifold. Then we are able to get the CR Bochner-type estimate and derive some key Lemmas. In section 3, let (M; J; ) be a com- plete noncompact pseudohermitian (2n+1)-manifold with the CR sub-Laplacian comparison property, we obtain subgradient estimates for positive pseudoharmonic functions. As a con- sequence, the natural analogue of Liouville-type theorem for the sub-Laplacian holds in a complete noncompact pseudohermitian (2n + 1)-manifold of nonnegative pseudohermitian Ricci curvature tensor and vanishing torsion. In section 4, we give a proof of sub-Laplacian comparison theorem in a standard Heisenberg (2n+1)-manifold (Hn; J; ) ; which is based
SLIDE 8 8
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
- n the work of [W] and [CTW]. In appendix A, we get the CR sub-Laplacian comparison
property (1.6) in (Hn; J; ) by a straightforward computation. Acknowledgments. The …rst author would like to express his thanks to Prof. S.-
- T. Yau for the inspiration, Prof. C.-S. Lin, director of Taida Institute for Mathematical
Sciences, NTU, for constant encouragement and supports, and Prof. J.-P. Wang for his inspiration of sublaplacian comparison geometry. The work would be not possible without their inspirations and supports. Part of the project was done during J. Tie’s visits to Taida Institute for Mathematical Sciences.
- 2. The CR Bochner-Type Estimate
In this section, we derive some key lemmas. In particular, we obtain the CR Bochner-type estimate as in Lemma 2.2. We …rst introduce some basic materials in a pseudohermitian (2n + 1)-manifold (see [L1], [L2] for more details). Let (M; ) be a (2n+1)-dimensional, orientable, contact manifold with contact structure . A CR structure compatible with is an endomorphism J : ! such that J2 = 1. We also assume that J satis…es the following integrability condition: If X and Y are in , then so are [JX; Y ]+[X; JY ] and J([JX; Y ]+[X; JY ]) = [JX; JY ][X; Y ]. Let fT; Z; Z
g be a frame of TM C, where Z is any local frame of T1;0; Z = Z 2 T0;1
and T is the characteristic vector …eld. Then
- ; ;
- , which is the coframe dual to
fT; Z; Z
g, satis…es
(2.1) d = ih ^ for some positive de…nite hermitian matrix of functions (h
). Actually we can always choose
Z such that h
= ; hence, throughout this note, we assume h = .
The Levi form h ; iL is the Hermitian form on T1;0 de…ned by (2.2) hZ; WiL = i
SLIDE 9 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 9
We can extend h ; iL to T0;1 by de…ning
- Z; W
- L = hZ; WiL for all Z; W 2 T1;0. The Levi
form induces naturally a Hermitian form on the dual bundle of T1;0, denoted by h ; iL
, and
hence on all the induced tensor bundles. Integrating the Hermitian form (when acting on sections) over M with respect to the volume form d = ^ (d)n, we get an inner product
- n the space of sections of each tensor bundle. We denote the inner product by the notation
h ; i. For example hu; vi = Z
M
uv d; for functions u and v. The pseudohermitian connection of (J; ) is the connection r on TM C (and extended to tensors) given in terms of a local frame Z 2 T1;0 by rZ =
Z;
rZ
=
;
rT = 0; where are the 1-forms uniquely determined by the following equations: d = ^
+ ^ ;
0 = ^ ; 0 =
+
(2.3) We can write (by Cartan lemma) = A with A = A. The curvature of Tanaka- Webster connection, expressed in terms of the coframe f = 0; ;
g, is
= d! ! ^ ! ; = 0 = 0
0 = 0:
Webster showed that can be written (2.4)
+ W
SLIDE 10 10
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
where the coe¢cients satisfy R
W
= W :
We will denote components of covariant derivatives with indices preceded by a comma; thus write A;. The indices f0; ; g indicate derivatives with respect to fT; Z; Z
derivatives of a scalar function, we will often omit the comma, for instance, u = Zu; u
=
Z
Zu !(Z )Zu:
For a real function u, the subgradient rb is de…ned by rbu 2 and hZ; rbuiL = du(Z) for all vector …elds Z tangent to contact plane. Locally rbu = P
u Z + uZ . We can
use the connection to de…ne the subhessian as the complex linear map (rH)2u : T1;0 T0;1 ! T1;0 T0;1 by (rH)2u(Z) = rZrbu: In particular, jrbuj2 = 2uu; jr2
buj2 = 2(uu + uu):
Also bu = Tr
(u + u ):
Next we recall the following commutation relations ([L1]). Let ' be a scalar function and = be a (1; 0) form, then we have (2.5) ' = '; '
'
ih'0; '0 '0 = A'
;
;0 ;0 = ;
A A; ;
;0
;
= ;A
SLIDE 11 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 11
and (2.6) ; ; = iA iA; ;
ihA
;
;
ih;0 + R
Now we recall a lemma from A. Greenleaf ([Gr]) and also ([CC2]). Lemma 2.1. For a real function ', (2.7) b jrb'j2 = 2
+ 2 hrb'; rbb'i + (4Ric 2 (n 2) Tor) ((rb')C ; (rb')C) + 4 hJrb'; rb'0i ; where (rb')C = '
Z is the corresponding complex (1; 0)-vector of rb'.
Lemma 2.2. For a smooth real-valued function ' and any > 0, we have b jrb'j2
n
P
;=1
n
P
;=1;6=
! + 1
n (b')2 + n'2 0 + 2 hrb'; rbb'i
+
- 4Ric 2 (n 2) Tor 4
- ((rb')C ; (rb')C) 2 jrb'0j2 :
- Proof. Since
j(rH)2'j2 = 2 Pn
;=1('' + '')
= 2 Pn
;=1(j'j2 + j'j2)
= 2(Pn
;=1 j'j2 + Pn ;=1 6=
j'j2 + Pn
=1 j'j2)
and from the commutation relation (2.5) Pn
=1 j'j2
=
1 4
Pn
=1 (j' + 'j2 + '2 0)
=
1 4
Pn
=1 j' + 'j2 + n 4'2 0:
It follows that j(rH)2'j2 = 2(Pn
;=1 j'j2 + Pn ;=1 6=
j'j2) + 1
2
Pn
=1 j' + 'j2 + n 2'2
;=1 j'j2 + Pn ;=1 6=
j'j2) +
1 2n (b')2 + n 2'2 0:
SLIDE 12 12
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
On the other hand, for all > 0 4 hJrb'; rb'0i
jrb'j2 2 jrb'0j2 :
Then the result follows easily from Lemma 2.1.
- De…nition 2.1. ([GL]) Let (M; J; ) be a pseudohermitian (2n + 1)-manifold. We de…ne
the purely holomorphic second-order operator Q by Q' = 2i
n
X
;=1
(A
By apply the commutation relations (2.5), one obtains Lemma 2.3. ([GL], [CKL]) Let ' (x) be a smooth function de…ned on M. Then b'0 = (b')0 + 2
n
X
;=1
That is 2 Im Q' = [b; T] ':
- Proof. By direct computation and the commutation relation (2.5), we have
b'0 = '0 + '0 =
= '0 +
= '0 + '0 + 2 h A'
= (b')0 + 2 h A'
: This completes the proof.
- Let u be a positive pseudoharmonic function and f (x) = ln u (x) : Then
bf = jrbfj2 :
SLIDE 13 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 13
We …rst de…ne V (') =
n
X
;=1
' + A
Lemma 2.4. Let u be a positive pseudoharmonic function with f = ln u. Then bf0 = 2 hrbf; rbf0i + 2V (f) :
bf0 = (bf)0 + 2
n
X
;=1
Since bf = jrbfj2 ; it follows from the commutation relation (2.5) that bf0 = (bf)0 + 2
n
P
;=1
0 + 2 n
P
;=1
2 hrbf0; rbfi + 2
n
P
;=1
f + A
- ff
- :
- Lemma 2.5. Let (M; J; ) be a pseudohermitian (2n + 1)-manifold and u be a positive func-
tion with f = ln u: Suppose that 2 Im Qu = [b; T] u = 0: Then (2.8) V (f) = 0:
SLIDE 14 14
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
(2.9) V (f) =
n
P
;=1
f + A
n
P
;=1
f + A
f + A
n
P
;=1
u uu u2
u
u
u
u u + A;
u2
+ A
u
u
n
P
;=1 1 u
1 2u [b; T] u:
This completes the proof.
- 3. CR Analogue of Yau’s Gradient Estimate
In this section, we will prove main Theorem 1.7 and Theorem 1.3. We …rst recall a real-valued function F (x; t; R; b) : M [0; 1] (0; 1) (0; 1) ! R de…ned by (3.1) F (x; t; R; b) = t
0 (x)
where (x) : M ! [0; 1] is a smooth cut-o¤ function de…ned by (x) = (r (x)) = 8 < : 1; x 2 B (R) 0; x 2 MnB (2R) such that (3.2) C R
1 2 0 0
and (3.3)
R2;
SLIDE 15 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 15
where we denote
@ @r by
0 and r(x) is the Carnot-Carathéodory distance to a …xed point x0:
In the following calculation, the universal constant C might be changed from lines to lines. Proposition 3.1. Let (M; J; ) be a complete noncompact pseudohermitian (2n + 1)-manifold with (3.4) (2Ric (n 2) Tor) (Z; Z) 2k jZj2 for all Z 2 T1;0, where k is an nonnegative constant. Suppose that (M; J; ) satis…es the CR sub-Laplacian comparison property. Then bF
" 4
n
P
;=1
jfaj2 + 4
n
P
;=1;6=
n (bf)2
+
R
4 bt
R jrbfj2 f 2 0 + 4btf0V (f)
i :
- Proof. By CR sub-Laplacian comparison property,
b =
0br
R2 C R
C1
R + C2
R:
First we compute (3.5) b (btf 2
0)
= bt (f 2
0b + bf 2 0 + 2 hrb; rbf 2 0i)
Rf 2 0 + 2f0bf0 + 2 jrbf0j2 + 4f0 hrb; rbf0i
Rf 2 0 + 2f0bf0 + 2 jrbf0j2 4 jf0j jrbj jrbf0j
Rf 2 0 + 2f0bf0 +
2
0]
Rf 2 0 + 2f0bf0 +
2
where we use the Young’s inequality and the inequality (3.2) which implies that 1 jrbj2 C R2:
SLIDE 16 16
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
Second, it follows from assumption (3.4), Lemma 2.2 and (3.5) that bF = t
0)
4
n
P
;=1
jfaj2 + 4
n
P
;=1;6=
n (bf)2 + nf 2 0 + 2 hrbf; rbbfi
2
- k + 1
- jrbfj2 2 jrbf0j2 bCt
R f 2 0 + 2btf0bf0 + 2 bt 2 jrbf0j2
" 4
n
P
;=1
jfaj2 + 4
n
P
;=1;6=
n (bf)2 +
R
0 + 2 hrbf; rbbfi
2
bt
2
Then taking = bt
2 ,
(3.6) bF
" 4
n
P
;=1
jfaj2 + 4
n
P
;=1;6=
n (bf)2 +
R
2
2 bt
- jrbfj2 + 2 hrbf; rbbfi + 2btf0bf0
i : Finally, by Lemma 2.4 2 hrbf; rbbfi + 2btf0bf0 = 2
+ 2btf0 [2 hrbf; rbf0i + 2V (f)] = 2
F
t btf 2
- 4btf0 hrbf; rbf0i + 4btf0V (f)
= 2
t hrbf; rbFi + 2bt hrbf; rb (f 2 0)i 4btf0 hrbf; rbf0i + 4btf0V (f)
= 2
t hrbf; rbFi + 2btf 2 0 hrbf; rbi + 4btf0V (f)
Now by Young’s inequality, we have (3.7) 2 hrbf; rbbfi + 2btf0bf0 = 2
t hrbf; rbFi + 2btf 2 0 hrbf; rbi + 4btf0V (f)
t hrbf; rFi 2btf 2 0 jrbfj jrbj + 4btf0V (f)
t hrbf; rFi 2Cbt R f 2 0 jrbfj
1 2 + 4btf0V (f)
t hrbf; rFi Cbt R f 2 0 Cbt R f 2 0 jrbfj2 + 4btf0V (f) :
SLIDE 17 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 17
Substituting (3.7) into (3.6), bF
" 4
n
P
;=1
jfaj2 + 4
n
P
;=1;6=
n (bf)2
+
R
0 2
2 bt
R f 2 0 jrbfj2 + 4btf0V (f)
i :
Let (M; J; ) be a complete noncompact pseudohermitian (2n + 1)- manifold with (2Ric (n 2) Tor) (Z; Z) 2k jZj2 for all Z 2 T1;0, where k is an nonnegative constant. Suppose that (M; J; ) satis…es the CR sub-Laplacian comparison property. Then for all a 6= 0 (3.8) tb (F)
na2 (F)2 C R (F) + 2t hrb; rbFi 2t2 hrbf; rbFi
+4t22
n
P
;=1
jfaj2 +
n
P
;=1;6=
! +
R
2b
na2 + bC R
+ h 2(1+a)
na2
(F) 2k 4
b
i t jrbfj2 + 4bt33f0V (f) : Proof. By using Proposition 3.1, we …rst compute b (F) = (b) F + 2 hrb; rbFi + bF
RF + 2 hrb; rbFi 2 hrbf; rFi
+t " 4
n
P
;=1
jfaj2 +
n
P
;=1;6=
! + 1
n (bf)2 +
R
2
2 bt
R f 2 0 jrbfj2 + 4btf0V (f)
i
SLIDE 18 18
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
and for each a 6= 0 (bf)2 =
= 1
atF 1 a jrbfj2 1 abtf 2 0 jrbfj22
= 1
atF a+1 a jrbfj2 1 abtf 2
2 =
1 a2t2F 2 +
a+1
a
2 jrbfj4 + 1
a2b2t22f 4
2(a+1)
a2t F jrbfj2 2b a2Ff 2 0 + 2(a+1)bt a2
jrbfj2 f 2
a2t2F 2 2(a+1) a2t F jrbfj2 2b a2Ff 2 0:
Then b (F)
na2tF 2 C RF + 2 hrb; rbFi 2 hrbf; rFi
+4t
n
P
;=1
jfaj2 +
n
P
;=1;6=
! +
R 2b na2F
0 +
na2 F 2kt 4 b
Cb
R
(tf 2
0) + 4bt22f0V (f) :
Hence (3.9) b (F)
na2tF 2 C RF + 2 hrb; rbFi 2 hrbf; rFi
+4t
n
P
;=1
jfaj2 +
n
P
;=1;6=
! +
R
2b
na2 + bC R
+
na2 F 2kt 4 b
Finally, multiply t on the both sides of (3.9) and note that t 1; 1 tb (F)
na2 (F)2 C RF + 2t hrb; rbFi 2t2 hrbf; rbFi
+4t22
n
P
;=1
jfaj2 +
n
P
;=1;6=
! +
R
2b
na2 + bC R
+ h 2(1+a)
na2
(F) 2k 4
b
i t jrbfj2 + 4bt33f0V (f) :
SLIDE 19 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 19
Proposition 3.3. Let (M; J; ) be a complete noncompact pseudohermitian (2n + 1)-manifold with (2Ric (n 2) Tor) (Z; Z) 2k jZj2 for all Z 2 T1;0, where k is an nonnegative constant. Suppose that (M; J; ) satis…es the CR sub-Laplacian comparison property. Let b, R be …xed, and p (t) 2 B (2R) be the maximal point of F for each t 2 (0; 1]. Then at ( p (t) ; t) we have (3.10)
na2 C R
R (F) + 4t22 n
P
;=1
jfaj2 +
n
P
;=1;6=
! +
R
2b
na2 + bC R
+ h 2(1+a)
na2
(F) 2k 4
b C R
i t jrbfj2 + 4bt33f0V (f) :
- Proof. Since (F) (p (t) ; t; R; b) =
max
x2B(2R) (F) (x; t; R; b), at a critical point (p (t) ; t)
- f (F) (x; t; R; b), we have
rb (F) (p (t) ; t; R; b) = 0: This implies that (3.11) Frb + rbF = 0 at (p (t) ; t) : On the other hand, (3.12) b (F) (p (t) ; t; R; b) 0 at (p (t) ; t) : Now we apply (3.11) to 2t hrb; rbFi and 2t2 hrbf; rbFi in (3.8), we can derive the following estimates. (3.13) 2t hrb; rbFi = 2tF jrbj2
R2 F
R F
SLIDE 20 20
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
and (3.14) 2t2 hrbf; rbFi = 2tF hrbf; rbi
R (F)
1 2 jrbfj
R (F)2 C Rt jrbfj2 :
Here we have applied the Young’s inequality for (3.14). Finally, substituting (3.12), (3.13) and (3.14) into (3.8) in Proposition 3.2, and noting that t 1,
na2 C R
R (F) + 4t22 n
P
;=1
jfaj2 +
n
P
;=1;6=
! +
R
2b
na2 + bC R
+ h 2(1+a)
na2
(F) 2k 4
b C R
i t jrbfj2 + 4bt33f0V (f) : This completes the proof.
- Now, we are ready to prove our main theorems.
Proof of Theorem 1.3 :
(3.15) V (f) = 0 by assumption (1.7) and Lemma 2.5. We begin by substituting (3.15) into (3.10) in Proposition 3.3 at the maximum point (p(t); t). Hence (3.16)
na2 C R
R [(F)]
+
R
2b
na2 + bC R
+ h 2(1+a)
na2
(F) 2k 4
b C R
i t jrbfj2 +4t2
02 n
P
;=1
jfaj2 +
n
P
;=1;6=
! :
SLIDE 21 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 21
We claim at t = 1 (3.17) (F) (p (1) ; 1; R; b) < na2 2 (1 + a)
b + C R
- for a large enough R which to be determined later. Here (1 + a) < 0 for some a to be chosen
later (say 1 + a = 5+2bk
n
). We prove it by contradiction. Suppose not, that is (F) (p (1) ; 1; R; b) na2 2 (1 + a)
b + C R
Since (F) (p (t) ; t; R; b) is continuous in the variable t and (F) (p (0) ; 0; R; b) = 0, by Intermediate-value theorem there exists a t0 2 (0; 1] such that (3.18) (F) (p (t0) ; t0; R; b) = na2 2 (1 + a)
b + C R
Now we apply (3.16) at the point (p (t0) ; t0), denoted by (p0; t0). We have by using (3.18) (3.19)
na2 C R
R [(F) (p0; t0)]
+
R
2b
na2 + bC R
+4t2
02 n
P
;=1
jfaj2 +
n
P
;=1;6=
! : Moreover, we compute (3.20) 1
na2 C R
R
h 1
na2 C R na2 2(1+a)
2k + 4
b + C R
R
i = n
1 2(1+a)
b
R
h
na2 2(1+a)
b + C R
1 2(1+a) + 3
io and (3.21)
R
2b
na2 + bC R
R
2b
na2 + bC R na2 2(1+a)
2k + 4
b + C R
R + b (1+a)
b + C R
R
2(1+a)
2k + 4
b + C R
4 1+a + 2bk 1+a
R
h ab
1+a + na2b 2(1+a)
b + C R
i :
SLIDE 22 22
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
Now we choose a such that (1 + a) < 4 + 2bk n and then
4 1 + a + 2bk 1 + a
In particular, we let (3.22) 1 + a = 5 + 2bk n : Then for R = R(b; k) large enough, one obtains 1 na2 C R
R
and
R 2b na2 + bC R
This leads to a contradiction with (3.19). Hence from (3.17) and (3.22) (F) (1; p (1) ; R; b) < (n + 5 + 2bk)2 2 (5 + 2bk)
b + C R
This implies max
x2B(2R)
- jrbfj2 + bf 2
- (x) < (n + 5 + 2bk)2
2 (5 + 2bk)
b + C R
When we …x on the set x 2 B (R), we obtain jrbfj2 + bf 2
0 < (n + 5 + 2bk)2
2 (5 + 2bk)
b + C R
This completes the proof. Next we prove Theorem 1.7. The proof is similar to Theorem 1.3. Proof of Theorem 1.7 :
SLIDE 23 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 23
- Proof. Firstly, we recall (Proposition 3.2) that
(3.23) tb (F)
na2 (F)2 C R (F) + 2t hrb; rbFi 2t2 hrbf; rbFi
+4t22
n
P
;=1
jfaj2 +
n
P
;=1;6=
! +
R
2b
na2 + bC R
+ h 2(1+a)
na2
(F) 2k 4
b
i t jrbfj2 + 4bt33f0V (f) : Now we need to deal with the term 4bt33f0V (f) in (3.23). (3.24) 4bt33f0V (f) = 4bt33f0
n
P
;=1
f + A
n
P
;=1
f + A
f + A
n
P
;=1
j
- f
- +
- A
- jfj jfj
- In (3.24), by Young’s inequality and noting that t 1, 1, we have following estimates:
(3.25) 8bt33 jf0j
n
P
;=1
P
;=1
8k1bt33 jf0j jfj
P
;=1
- 4k1bt33 jfj2 4k1bt33f 2
- 4k1bn2 (t22f 2
0) 4k1bt22 n
P
;=1
jfj2 and (3.26) 8bt33 jf0j
n
P
;=1
jA;
j
n
P
;=1
jf0j
n
P
;=1
2f 2 0 + 1 2
0 4k1bnt33 n
P
=1
0) 2k1bn
SLIDE 24 24
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
and (3.27) 8bt33 jf0j
n
P
;=1
n
P
;=1
1
2 jfj2 + 1 2 jfj2
n
n
P
=1
jfj2 + n
n
P
=1
jfj2 !
- 4k1bnt33 jf0j jrbfj2
- 2k1b2nt33f 2
0 jrbfj2 2k1nt33 jrbfj2
= 2k1b2n
(t22f 2
0) 2k1nt33 jrbfj2 :
Substitute estimates (3.25), (3.26), and (3.27) into (3.23), one obtains tb (F)
na2 (F)2 C R (F) + 2t hrb; rbFi 2t2 hrbf; rbFi
+4t22 " (1 bk1n)
n
P
;=1
jfaj2 +
n
P
;=1;6=
# +
R
2b
na2 + 2b2k1n + bC R
+ h 2(1+a)
na2
(F) 2k 2n (1 + b) k1 4
b
i t jrbfj2 : Next as shown in the same computation as in Proposition 3.3, at the maximal point (p(t); t) (3.28)
na2 C R
R (F)
+4t22 " (1 bk1n)
n
P
;=1
jfaj2 +
n
P
;=1;6=
# +
R
2b
na2 + 2b2k1n + bC R
+ h 2(1+a)
na2
(F) 2k 2n (1 + b) k1 4
b C R
i t jrbfj2 : We claim at t = 1; there exists a small constant b0 = b0(n; k; k1) > 0 such that for any 0 < b b0 (F) (p (1) ; 1; R; b) < na2 2 (1 + a)
b + C R
- if R is large enough which to be determined later. Here (1 + a) < 0 for some a to be chosen
later (say 1 + a = 5
n):
SLIDE 25 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 25
We prove it by contradiction. Suppose not, that is (F) (p (1) ; 1; R; b) na2 2 (1 + a)
b + C R
Since (F) (p (t) ; t; R; b) is continuous in the variable t and (F) (p (0) ; 0; R; b) = 0, by Intermediate-value theorem there exists a t0 2 (0; 1] such that (F) (p (t0) ; t0; R; b) = na2 2 (1 + a)
b + C R
Now we apply (3.28) at the point (p (t0) ; t0), denoted by (p0; t0). We have (3.29) 1
na2 C R
R
h 1
na2 C R na2 2(1+a)
2k + 2n (1 + b) k1 + 4
b + C R
R
i = n
1 2(1+a)
b
R
h
na2 2(1+a)
b + C R
1 2(1+a) + 3
io and (3.30)
R
2b
na2 + 2b2k1n + bC R
- (F) (p0; t0)
- = n 8bk1n2 bC
R
2b
na2 + 2b2k1n + bC R na2 2(1+a)
2k + 2n (1 + b) k1 + 4
b + C R
R + ( na2 2(1+a))( 2b na2 + 2b2k1n)
b + C R
R
2(1+a)
2k + 2n (1 + b) k1 + 4
b + C R
- = fn 8bk1n2 + ( b+a2b2n2k1
(1+a)
)[2k + 2n (1 + b) k1 + 4
b]g
+ C
Rfb + ( b+a2b2n2k1 (1+a)
) +
na2b 2(1+a)[2k + 2n (1 + b) k1 + 4 b + C R]g:
Now we choose a and b such that (3.31) n 8bk1n2 + ( b+a2b2n2k1
(1+a)
)[2k + 2n (1 + b) k1 + 4
b]
= n bf8k1n2 ( 1+a2bn2k1
(1+a)
)[2k + 2n (1 + b) k1] ( 4a2n2k1
1+a )g + 4 1+a
> 0: This can be done by choosing (1 + a) < 4 n
SLIDE 26 26
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
and then choose a small b0 = b0(n; k; k1) > 0 such that for any b b0 n bf8k1n2 (1 + a2bn2k1 (1 + a) )[2k + 2n (1 + b) k1] (4a2n2k1 1 + a )g + 4 1 + a > 0 and (1 bk1n) > 0: In particular, we let 1 + a = 5 n: Then for any 0 < b b0, one obtains 1 na2 C R
R
and
R 2b na2 + 2b2k1n + bC R
for R = R(b; k; k1) large enough. This leads to a contradiction with (3.28). Hence (F) (1; p (1) ; R; b) < na2 (1 + a)
b + C R
This implies for 1 + a = 5
n
max
x2B(2R)
- jrbfj2 + bf 2
- (x) < (n + 5)2
5
b + C R
When we …x on the set x 2 B (R), we obtain jrbfj2 + bf 2
0 < (n + 5)2
5
b + C R
- n B (R). Note that the preceding computation is not valid if F is not smooth at x0. In
this case, we may use a trick due to E. Calabi ( see [W] for details). This completes the proof of Theorem 1.7.
SLIDE 27 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 27
4. The Sub-Laplacian Comparison Theorem In this section, we give the proof of sub-Laplacian comparison theorems on (Hn; J; ). In
- rder to prove Proposition 1.2, we …rst derive the di¤erential inequality for sub-Laplacian of
Carnot-Caratheodory distance. For simplicity, we prove the propositions for n = 1: In the setting, we write Z1 = 1
2(e1ie2)
for real vectors e1; e2: It follows e2 = Je1: Let e1 = Re(1); e2 = Im(1): Then f; e1; e2g is dual to fT; e1; e2g: Now in view of (2.1) and (2.3), we have the following real version of structure equations: d = 2e1 ^ e2; re1 = ! e2; re2 = ! e1; de1 = e2 ^ ! mod ; de2 = e1 ^ ! mod : We also write 'ei = ei' and rb' = 1
2('e1e1 + 'e2e2): Moreover, 'eiej = ejei' rejei'
and b' = 1
2('e1e1 + 'e2e2): Now we can write down the real version of the commutation
relations as in 2.5 and 2.6. (4.1) 'e1e2 'e2e1 = 2'0 '0e1 'e10 = 'e1 Re A11 'e2 Im A11 '0e2 'e20 = 'e1 Im A11 + 'e2 Re A11 'e1e1e2 'e1e2e1 = 2'e10 2'e2W 'e2e1e2 'e2e2e1 = 2'e20 + 2'e1W: For a …xed point p 2 M, we consider the Carnot-Carathéodory distance function rp(x) = r(p; x) from x to p, and we will simply write r(x): The distance function in general is not smooth due to the presence of cut-points. However, it can be seen that it is a Lipschitz function with Lipschitz constant 1. In particular, we have jrrj2 = 1
SLIDE 28 28
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
almost everywhere on M: Though r might not be a C2-function, one can still estimate its sub-Laplacian in the sense of distribution ([BGG], [S]). De…nition 4.1. Let (M; J; ) be a pseudohermitian 3-manifold. We de…ne ([L1]) P' = ('
111 + iA11'1)1 = P' = (P1')1;
which is an operator that characterizes CR-pluriharmonic functions. Here P1' = '
111 +
iA11'1 and P' = (P 1)
Lemma 4.1. Let (M; J; ) be a complete pseudohermitian 3-manifold. Then (4.2) @r(br) + 2(br)2 2r0e1 + 2r2
0 + (W Im A11) = 0
and (4.3) r00e2 + r2
0e1 (Re A11)2 (Re A11)0 = 0:
As a consequence, we have (4.4) 3@r(br) + 2(br)2 + 2r2
0 8 < Pr +Pr; dbr > +(W + 3 Im A11) = 0:
Here dbr = r11 + r
1
Proof. We will follow the method as in ([W]). Let x 2 expp Upnfpg: Here expp is the exponential map due to R. Strichartz ([S]). Let be the minimal geodesic joining p to
- x. As in [S], the CR Gauss lemma implies that one can choose a CR orthonormal frame
fT; e1; e2g along such that rr = e2: Then (4.5) jrrj = re2 = @rr = 1 and (4.6) re1 = 0 and re2e1 = 0 = re2e2:
SLIDE 29
CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 29
Hence (4.7) = (r2
e1 + r2 e2)e1e1 + (r2 e1 + r2 e2)e2e2
= 2(re1re1e1 + re2re2e1)e1 + 2(re1re1e2 + re2re2e2)e2 = 2(r2
e1e1 + re1re1e1e1 + r2 e2e1 + re2re2e1e1)
+2(r2
e1e2 + re1re1e2e2 + r2 e2e2 + re2re2e2e2)
= 2(r2
e1e1 + r2 e2e2 + r2 e2e1 + r2 e1e2 + re2e1e1 + re2e2e2):
On the other hand from (4.1) r2
e1e2 = 4r2 0 ,
r2
e1e1 = 4(br)2:
Moreover re2e1e1 = re1e2e1 2r0e1 = re1e1e2 2re10 + 2re2W 2r0e1 = re1e1e2 4re10 + 2re2W 2re1 Re A11 + 2re2 Im A11 = re1e1e2 4re10 + 2W + 2 Im A11: All these and (4.7) imply = r2
e1e1 + r2 e1e2 + re2e1e1 + re2e2e2
= 4(br)2 + 4r2
0 + (re1e1 + re2e2)e2 4re10 + 2W + 2 Im A11
= 4(br)2 + 4r2
0 + 2@r(br) 4re10 + 2W + 2 Im A11:
That is (4.8) 0 = @r(br) + 2(br)2 2re10 + 2r2
0 + (W + Im A11):
But (4.9) r0e1 re10 = Im A11: Hence 0 = @r(br) + 2(br)2 2r0e1 + 2r2
0 + (W Im A11):
SLIDE 30
30
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
On the other hand, since = (r2
e1 + r2 e2)0
= 2(re1re10 + re2re20); it follows from (4.5) and (4.6) re20 = 0: Now as in (4.7), we have = (r2
e1 + r2 e2)00
= 2(re1re10 + re2re20)0 = 2(r2
e10 + re1re100 + r2 e20 + re2re200)
= 2(r2
e10 + re200):
Compute r2
e10 = (r0e1 + Im A11)2
and re200 = (r0e2 re1 Im A11 re2 Re A11)0 = r0e20 re10 Im A11 re1(Im A11)0 re20 Re A11 re2(Re A11)0 = r0e20 re10 Im A11 (Re A11)0 = (r00e2 r0e1 Im A11 r0e2 Re A11) (r0e1 + Im A11) Im A11 (Re A11)0 = r00e2 2r0e1 Im A11 r0e2 Re A11 (Im A11)2 (Re A11)0: All these imply 0 = r00e2 + r2
0e1 r0e2 Re A11 (Re A11)0 = r00e2 + r2 0e1 (Re A11)2 (Re A11)0:
Finally, by de…nition we compute < Pr +Pr; dbr >= (r111r1 + iA11r1r1) + conjugate: First we note that for Z1 = 1
2(e1 ie2)
r1 = 1 2(re1 ire2) = 1 2i
SLIDE 31 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 31
and then (4.10) iA11r1r1 + conjugate = 1 4iA11 + conjugate = 1 2 Im A11: Secondly, one can derive r111 = 1
8[re1e1e1 + re2e2e1 + re2e1e2 re1e2e2]
+ 1
8i[(re2e1e1 re1e2e1) (re1e1e2 + re2e2e2)]
and r111r1 =
1 16i[re1e1e1 + re2e2e1 + re2e1e2 re1e2e2]
1
16[(re2e1e1 re1e2e1) (re1e1e2 + re2e2e2)]:
Hence (4.11) r111r1 + conjugate = 1
8[(re2e1e1 re1e2e1) (re1e1e2 + re2e2e2)]
= 1
4[@r(br) + r0e1]:
It follow from (4.10) and (4.11) that 4 < Pr +Pr; dbr >= @r(br) + r0e1 + 2 Im A11 and then 3@r(br) + 2(br)2 + 2r2
0 8 < Pr +Pr; dbr > +(W + 3 Im A11) = 0:
- Lemma 4.2. Let (Hn; J; ) be a standard Heisenberg (2n + 1)-manifold. Then there exists
a constant CHn
1
> 0 (4.12) jrHn
00 j CHn 1
(rHn)3
SLIDE 32 32
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
and (4.13) jrHn
0 j CHn 1
(rHn)1: We will give the proof of Lemma 4.2 at the end of this section. Proof of Propositon 1.2 :
- Proof. Note that it is a straightforward from (4.20) to get the CR sub-Laplacian comparison
property (1.6) as in ( [CTW]). For completeness, we sketch the proof in appendix A. Here we give another proof by applying the di¤erential inequality for sub-Laplacian of Carnot-Caratheodory distance as in Lemma 4.1. From A11 = 0 and (4.3), we have (4.14) 0 = r00e2 + r2
0e1:
Note that rr = e2 and re2 = @rr: By integrating both sides of (4.14) with respect to r; it follows from (4.12) that jr0e1j C r2: Now by applying (2.5) and W = 0, @r(br) + 2(br)2 C r2 0: Then by ODE inequality, there exists a constant CH1
2
> 0 br CH1
2
r : Here CH1
2
= 1+p8C+1
4
which solves the equation ([W]) 2m2 m C = 0:
- In view of (4.4), we have the following CR sub-Laplacian comparison theorems under the
assumption of < Pr +Pr; dbr >= O(r2):
SLIDE 33 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 33
Theorem 4.3. Let (M; J; ) be a complete pseudohermitian 3-manifold with W + 3 Im A11 k2 for some constant k2. Suppose that, for some constant l 0 8 < Pr +Pr; dbr > l r2: Then, for any x 2 M where r(x) is smooth, we have br 8 > > > < > > > : m2 pK2 cot(pK2r); k2 > 0
m1 r ;
k2 = 0 m3 pK3 coth(pK3r); k2 < 0 for some positive constants m1; m2; m3 and K2; K3: Moreover it holds on the whole manifold in the sense of distribution. Remark 4.1. In fact, the same method as in the proof of Lemaa 4.2, we can estimate < PrHn + PrHn; dbrHn > 3 8 1 (rHn)2 in a standard Heisenberg (2n + 1)-manifold (Hn; J; ).
- Proof. (i) For k2 = 0, we have
3 @r(br) + 2(br)2 l r2 0: Then br m1 r : Here m1 = 3+
p 9+8l 4
3
2 solves the equation
2m2 3m l = 0: (ii) For k2 > 0, then from (4.4) 3@r(br) + 2(br)2 l r2 + k2 0:
SLIDE 34 34
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
Hence (4.15) 3@r(br) + 2(br)2 + (1 1)k2 0 if r q
l 1k2 with any positive constant 1 < 1: Then by applying Wang’s method again
(4.16) br m2 p K2 cot( p K2r) for some constant m2 and K2 = (1 1)k2: (iii) Similarly for k2 < 0, we have 3@r(br) + 2(br)2 (1 + 2)jk2j 0 and if r q
l 2jk2j with any positive constant 1 < 1
br m3 p K3 coth( p K3r) for some positive constant m3 and K3 = (1 + 2)jk2j:
- Proof of Lemma 4.2 :
- Proof. For simplicity, we prove (4.12) and (4.13) for n = 1: We consider the following two
vector …elds de…ned on R3 with coordinates (x; t) = (x1; x2; t): (4.17) X1 = @ @x1 + 2x2 @ @t and X2 = @ @x2 2x1 @ @t: It is easy to check that (4.18) [X1; X2] = 4 @ @t: The vector …elds X1, X2 and T = 2 @
@t are left-invariant with respect to the “Heisenberg
translation": for (x; t) = (x1; x2; t) and (y; s) = (y1; y2; s) 2 R3, (4.19) (x; t) (y; s) = (x1 + y1; x2 + y2; t + s + 2[x2y1 x1y2]):
SLIDE 35 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 35
Let z = x1 + ix2. Then we denote the vector …elds in terms of complex coordinates as following : Z = 1 2(X1 iX2) = @ @z + i z @ @t and
2(X1 + iX2) = @ @ z iz @ @t: Actually, the above multiplicative law de…nes a group structure on R3 which we call the 1-dimensional Heisenberg group with (x; t)1 = (x; t). From the paper of [CTW] ( also [BGG]), the square of the Carnot-Caratheodory distance [rH1(x; t)]2 from the origin is (4.20) [rH1(x; t)]2 =
sin(2c) 2 kxk2 = (2c)
; where c is the unique solution of (2)kxk2 = jtj in the interval [0; =2) and (z) =
z sin2 z
cot z: Moreover (z) = z2 sin2 z 1 1 + (z) = z2 z + sin2 z sin z cos z; (0) = 2: . Introduce a new parameter = 2c. Then the Carnot-Cartheodory distance between the
- rigin and point (x1; x2; t) can be expressed as
rH1(x; t) =
with ()kxk2 = jtj and 2 [0; ): Denote g() =
and R = x2
1 + x2 2:
We have @rH1 @t = R1=2g0()@ @t and @2rH1 @t2 = R1=2g00() @ @t 2 + R1=2g0()@2 @t2 : Next from t = R(), we compute the derivatives: @ @t = 1 R0() and @2 @t2 = 00() R2(0)3:
SLIDE 36 36
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
This implies @rH1 @t = R1=2 g0() 0() and @2rH1 @t2 = R3=2 1 0() g0() 0() : From the straightforward calculation, we have 1 g0 = f 2 with f = sin2 cos 2(1 cot ): This implies @2rH1 @t2 = f 2 R3=2: But rH1(R; t) =
@2rH1 @t2 = 1 4R3=2 sin2 cos (1 cot ) = 1 4(rH1)3 3 cos sin (1 cot ): Hence j@2rH1 @t2 j C(rH1)3: Similarly, we have j@rH1 @t j C(rH1)1:
Here we derive the CR sub-Laplacian comparison property in a standard Heisenberg (2n+ 1)-manifold with n = 1. We refer to [CTW] for details with n 1. Denote b = 1
2(X2 1 + X2 2): We will compute brH1(x; t) in polar coordinates as following.
b = 1 2( @2 @s2 + 1 s @2 @s@ + 1 s2 @2 @2) + 2 @2 @t@ + 2s2 @2 @t2: Since rH1(x; t) depends only on s = kxk = p x2
1 + x2 2, we have
brH1(x; t) = (1 2 @2 @s2 + 2s2 @2 @t2)rH1(s; t):
SLIDE 37 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 37
From (4.20), we introduce a new variable u = jtj=s2. Then rH1(s; t) := fc(s; u) =
where u satis…es u = () = sin cos sin2 : Hence brH1(s; t) = (1 2 @2 @s2 + 2 s2 @2 @u2)fc(s; u) = 2 s @2 @u2( sin ); where u is given by u = (). Let g() =
dg du = dg d d du and d2g du2 = d2g d2 (d du)2 + dg d d2 du2 : We next compute dg
d, d2g d2, d du and d2 du2.
dg d = sin cos sin2 and d2g d2 = (1 + cos2 ) 2 sin cos sin3 : Next u = () implies 1 = 0()d du; d du = 1 0() and d2 du2 = 00() (0)3 : We now compute 0() and 00() from () = csc2 cot . 02 2 csc2 cot + csc2 = 2 csc2 (1 cot ) and 002 cot (1 cot ) + 2 csc2 ( csc2 cot ) = 2 csc2 [(3 cot2 + 1) 3 cot ]: We …nally compute bfc(s; u) = 2
s d2 du2g().
bfc(s; u) = 2 s d2g d2 (d du)2 + dg d d2 du2
s d2g d2 1 (0)2 dg d 00() (0)3
2 s(0)2 d2g d2 dg d 00() 0()
SLIDE 38 38
SHU-CHENG CHANG1, TING-JUNG KUO2, AND JINGZHI TIE3
We shall compute the term in [: : : ] in term of …rst. d2g d2 dg d 00() 0() =(1 + cos2 ) 2 sin cos sin3 sin cos sin2 2 csc2 [(3 cot2 + 1) 3 cot ] 2 csc2 (1 cot ) =(1 + cos2 ) 2 sin cos sin3 (3 cot2 + 1) 3 cot sin =(1 + cos2 ) 2 sin cos (3 cos2 + sin2 ) + 3 cos sin sin3 =sin cos cos2 sin3 : Hence we have bfc(s; u) = 2 s(0)2 d2g d2 dg d 00() 0()
sin3
2s csc4 (1 cot )2 = (1 cot ) sin cos 2s csc (1 cot )2 = sin2 cos 2s(1 cot ): Since rH1 =
(A.1) brH1 = 1 2rH1 sin2 cos sin cos : We next study the function F() =
sin2 cos 2(sin cos ); where is given by
s2() = t with () = sin cos sin2 : The function F() is smooth on the interval [0; ], decreasing from [0; m] and increasing from [m; ]. m is the unique critical point of F() inside the interval (0; ). F(0) = 3, F(=2) = F() = 0. As s ! 0 with t > 0 …xed, ! and the equation s2() = t implies s2 t sin cos sin2 = 1:
SLIDE 39 CR YAU’S GRADIENT ESTIMATE AND LIOUVILLE THEOREM 39
This shows that ! and sin ( t )1=2s as s ! 0: This implies (A.1) makes sense when s = 0. This corresponds to = . All these imply brH1 3 rH1 in a standard Heisenberg 3-manifold. References
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1Department of Mathematics, National Taiwan University, Taipei 10617, Taiwan, 1Taida
Institute for Mathematical Sciences (TIMS), National Taiwan University, Taipei 10617, Taiwan E-mail address: scchang@math.ntu.edu.tw
2Taida Institute for Mathematical Sciences (TIMS), National Taiwan University, Taipei
10617, Taiwan E-mail address: tjkuo@ntu.edu.tw
3Department of Mathematics, University of Georgia, Athens, GA 30602-7403, U.S.A.
E-mail address: jtie@math.uga.edu
