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0, then u(X)
has the limit f(Q) for almost every Q. The theorem is most easily proved by writing down a formula for the solution, u(X) =
fa
B
Mlx'-y'! and DnB ~ !X=(x',xn):xn>¢(x')lnB.
is called a C 1 domain. If in addition, 'V¢ satisfies a Holder condition
In both the case of the Dirichlet problem and the Neumann problem,
Q with IX-QI < (l+a) dist(X,
0:::
1 l-IXj2 P(X,Q) f(Q) da(Q), where P(X,Q) = - --I n w n IX-Q
The estimate now follows as an easy consequence of the Hardy-Littlewood maximal theorem. An analogous formula holds for the Neumann problem.
we call D a C I,a domain. Notice that the cone r Similarly, r
= l(x',x n ): xn M\x'll satisfies rinBCD. Thus, Lipschitz e
i domains satisfy the interior and exterior cone condition.
The function ¢ satisfying the Lipschitz condition l¢(x') --¢(y')1 ::; n I M!x'-y'l is differentiable almost everywhere and 'V¢ (L OO(R - ), I!'V¢11 00
::;
M.
-
~~_.......
"_E
136
_ ..... -
- - , . . - - - -....
Surface measure
0
G(X,Y) as a function of Y is the potential induced by a unit charge at
is defined for each Borel subset E CaD n B by
o(E)
=
(1+!'V¢(x,)!2/
12
ux(Y) be the harmonic function with boundary values uX(Y)!aD
E*
III
I ,I~I
ao
('V¢(x'), -1)/(1 + !'V¢(x')1 2 )1/2 exists for almost every x'. The unit
On the other hand, if we know G(X,Y), we can formally write down the
normal at Q will be denoted by N . It exists for almost every Q ( Q with respect to do.
ao ,
solution to the Dirichlet problem. In fact, u(X) =
111,1
to recall some formulas from advanced calculus ,and some definitions. We
III
will start with the derivation corresponding to the Dirichlet problem.
III
We first recall the fundamental solution F(X) to Laplace's equation
in Rn :~F = 8, where 1
(n-2) wn IX In F(X)
2
n
=
>2
IXI
'1
1
II
= F*tf!(X) =
f
J[u(Y)~yG(X,Y) :0
=
=
t/J, with
II
I
II
I
'I
(X,Q)-
;;;0 (Q)G(X,Q~
da(Q) ,
f
u(Q)
a~Q G(X,Q)do(Q) ,
derived the formula
Rn It will be convenient to put F(X,Y) ~ F(X-Y). Notice that ~yF(X,Y) =
Green function G(X,Y). It is the function on
=
where the fourth equality follows from Green's formula. Thus, we have
i5 xD
and satisfying ~yG(X,Y) = 8(X-Y), X (D; G(X,Y)
continuous for X =
0, X (D, Y
i
(2)
u(X) =
J :t f(Q)
aD
8(X-Y). The fundamental solution in a bounded domain is known as the I
-i\u(Y)' G(X,Y)]dY
aD
F(X-Y)t/J(Y)dY.
II
u(y)AyG(X,Y)dY =
aD
1
III1II
f D
fIU(Q)
n=2
(C O'(Rn ) ,
W(X)
=
D
It provides a formula for a solution W to the equation ~w
'l
u(Y)o(X,Y)dY
D
where wn is the surface area of the unit sphere in R n . F(X) is the
electrical potential in free space induced by a unit charge at the origin.
tf!
f
= 1 2TT log
1111i
I,
F(X,Y)l
G(X,Y) = F(X,Y)-ux(Y) .
(1)
given in the coordinate system by
In order to motivate the use of the method of layer potentials, we need
IJ
=
Then,
Ix': (x', ¢(x')) (EI.
The unit outer normal to
lilT
Illi
=
The Green function can be
obtained if one knows how to solve the Dirichlet problem. In fact, let dx "
I where E*
ao.
X that is grounded to zero potential on
f
137
ELLIPTIC BOUNDARY VALUE PROBLEMS
CARLOS E. KENIG
(X,Q)do(Q)
Q
Y
(ao.
for the harmonic function u with boundary values f. The problem with
ao ·
138
l i
CARLOS E. KENIG
formula (2) is of course that we don't know G(X,Q). Because of formulas
1 u(X) = (Un
(1) and (2), C. Neumann proposed the formula
w(X) =
f :Q J
f(Q)
-
n
~
1(; I,ll
II
C - IP_Ql n
l
~
+a
and so this
is a COO domain, K is compact from Ck,a(aD) to Ck,a(aD), k=1,2,"',O,,'(t) = (1 +t 2)-n/2, and where the equality holds at
Iz-xl>e
every z at which ¢ is differentiable, i.e. for a.e.z. COROLLARY 2.1.4. (Nz I]Sg) ±(z) = =+:
}
g(z) - K*g(z), where K* is the LEMMA. If a" R n- 1 , a.j ¢(a) , f" C';(R n - 1) and >.. is as in the
L 2(aD, da) adjoint of K.
previous lemma, then The proof of Theorem 2.1.3 a) follows by well-known techniques from the deep theorem of Coifman, McIntosh and Meyer ([2]). THEOREM ([2]). Let
e: R -.. R
be even, and C DO . Let A,B: R n- 1 -.. R
r
be Lipschitz. Let K(z,x) = A(z)-A(x) e B(\z)-i(x)l. Then, the maximal Iz-xl n z-x operator
L
M*g(z) = sup e-..O
I
f
J
K(z,x)g(x)dxl
It-xl>e
is bounded on LP(R n- 1 ), 1 < P < DO, with
l J a-¢(a) - (a-x)· I]¢(x) f(x)dx = wn lla-xl 2 + [¢(x) _a]2]n/2
1 . 1 = 2" slgn(a-¢(a))f(a) - w n
f~ ~
xk-ak >.. (¢(x).a) A.. (x)dx . !x_aln-1 ~ dx k
Moreover, the integral on the right-hand side of the eqwlity is a continu ous function of (a,a) " Rn .
It is easy to see that (at leas t the existence part) of Theorems 2.1.1 and 2.1.2 will follow immediately if we can show that (} I + K) and
IITtl
150
CARLOS E. KENIG
I.~l
-t
1+ K*
ELLIPTIC BOUNDARY VALUE PROBLEMS
are in vertible on L 2(aD, do). This is the result of
derivatives are suitably small at
G. Verchota ([33)). THEOREM 2.1.5.
(±~I+K), (±~
I+K*) are invertible on L 2(aD,do).
~
J\V'u\2 dO =2 1+ K*)
aD
are invertible. III order to do so, we show that if f f L 2(aD, do),
Proof. Observe that div (e n lV'u I2) =
1+ K*\ fll 2 J L (aD,do) ;:::; II (~ 1- K*) fll L 2 (aD,do) ' where the constants of equivalence depend only on the Lipschitz constant M. Let us take
div
this for granted, and show, for example, that} I+K* is invertible. To
gives the "lemma.
II (}
do this, note first that if T=}I+K*, IITfIIL22:CllfIlL2' where C depends only on the Lipschitz constant M. For 0 S t
:s 1,
consider the
a/:
-J;- V'u· V'u + ~.
div V'u
=
2
t
~ V'u· V'u.
V'u' V'u, while Stokes' theorem now
1 / ::;::;l. (l+M 2)12
1
J(:f
LEMMA 2.1.6. Suppose that T t : L 2(R n - 1) .... L 2(R n - 1) satisfy (a) liTtfll 2 2: C Illfll 2
aD
do::; C
J aD
lV'tuI2do.
L
Proof. Let a = en - < Nx,e n > Nx ' so that a is a linear combination of
II
:il
=
lV'ul 2 =
COROLLARY 2.1.9. Let u beasin2.1.8,andlet T 1(x) , T/x) , Tn _ 1 (x) be an orthogonal basis for the tangent plane to aD at (X, ¢(X)). n-l Let lV'tu(x)12 = I \. Also, T 1 (x), T2 (x), "', Tn - 1(x). Then, ()y n 2 lV'ul = (~y + lV'tuI2, and so faD (~y do + faD 1V'tuI2do =
Then, T 1 is invertible. The proof of 2.1.6 is very simple. We are 1 1 thus reduced to proving (2.1.7) 11(-2 I+K*)fll 2 ;:::; 11(-2 I-K*)fll 2 L
(aD,do)
L
In order to prove (2.1.7), we will use the following formula, which goes back to Rellich [30] (also see [28], [29], [27)). LEMMA 2.1.8. Assume that u fLip (0), L\u = 0 in D, and u and its
['
(aD,d)
2 faD (~~)2 do + 2 faD 2, we can find a Lipschitz domain so that
~l~ ~~~~~;~',
· ,· ,~li, ', .," '.
I, c.
.'
Ii
N(vw)(s) :::::
LP(ds) if and only if p
Proof. The boundedness follows from 2.1.3a). Because of the L 2-Neumann
I!
=
.~
t'
(~ 1-· K*)
is not invertible in LP. The example can also be used to
II
1S4
CARLOS E. KENIG
show that
~
1+ K is not always invertible in Lq, when q < 2. In
fact, fix q < 2, and let p satisfy
P~-PO E
=
C
lim G(s+eN)-G(s) E
=
=
~. We
THEOREM 2.2.1. There exists E = E(M) > 0 such that, given f f LP(iD,da), 2-E'S p < "", there exists a unique u harmonic in D, with N(u) f LP(aD, da) such that u converges non-tangentially almost everywhere to f. Moreover, the solution u has the form
In fact, let G(X) be the Green's function of
11(3 with pole at ~*' Then, for s near 0, k(s) E~O
ISS
ELLIPTIC BOUNDARY VALUE PROBLEMS
av
aN
=
u(X)
=
(s) ~ s-I+TT/(3,
where the first inequality follows from the fact that both G and v are
In
<X-Q,N >
J aD
Q
[X-Q'"
g(Q)da(Q),
for some g f LP(ao,da).
positive, and harmonic on B, and 0 on an(3 n B (this is Lemma 5.10
~ 1+ K were invertible on L q(ds). Let
in [19]). Assume now that g
2' 0 f C(a11(3) ' and heX) be the solution of the Dirichlet problem with
data g. Then, h(X*)
gdw
a11 =
13
,
h(X,J
s I~I
£
K[(~ I+~-I(g~ (X*) , where Kis
4
(
h 0 such that given f ( Li(A) , . 1 < P < 2 +E, there exists a harmonic function u, with IIN(\7u)11 p L ~) CII\7tfl/
,
:s
, and such that \7tU = \7tf (a.e.) non-tangentially on A· u
is unique (modulo constants). Moreover, u has the form
u(X)
= UJ n
But this implies that k fLP(ds), a contradiction.
We now turn to the positive results. They are:
J
LPCA)
because of the second formula for h(X*) , and the assumed Lq bounded ness of
tending to 0 at "", with N(\7u) f LP(ao, da), such that NQ \7u(X) con
u(X)
By the mean value property of harmonic functions and Harnack's
principle, we ha ve
LP(ao, da), 2-E 'S P < "", there exists a unique u harmonic in D,
gkds ,
=
the double layer potential. Let U be a ball centered at X*, contained in 11
l
a11(3
13
2 also, by the L -theory, h(X*)
f
verges non-tangentiallya.e. to f(Q). Moreover, u has the form
J J
=
THEOREM 2.2.2. There exists E = E(M) > 0, such that, given
for some g f LP(aD,da).
(2) 1 n-
J aD
g(Q)da(Q) 1 \X_Qlnn
I
156
CARLOS E. KENIG
ELLIPTIC BOUNDARY VALUE PROBLEMS
The case p ~ 2 of the above theorems was discussed in part (a). The
157
THEOREM 2.2.8. Let T be a linear operator such that lITfl!
first part of 2.2.1 (i.e. without the representation formula), is due to B. Dahlberg (1977) ([5]). Theorem 2.2.3 was first proved by G. Verchota (1982) ([33]). The representation formula in 2.2.1, Theorem 2.2.2, and
,and such that for all atoms a,
C Ilf\\ 2
L
2
(A)
O. For R ~ R O =
and Iv(X)I:; C Ilull \ ' IXI 2- n -- v , where v> 0, C > 0 depend L ""(R n B*) only on the ellipticity constants of L. Moreover, a = C f8(X) 'lu(X)· 'It/J(X), where t/J f C ""(R n) , . t/J= 0 for IX I in 28 *, and t/J == 1 for large X. Let us assume for the time being that
u
D;
J,
D
8'lu'lt/J =
= I(x,y): !(x,y)\ < p,
side equals lim E .... O
harmonicity of u,
f
J,
D
Y> ¢(x) +E I,
aD p faDE :
=0
D
E"'O
f
au = lim aN E .... O
E
f
'lu· 'It/J = lim
aD;
E
P
~,
AR
N('lu)2, where, A R =I(x,¢(x)):R ¢(x) + d,
f(~,H
from 1/4 to 1/2 gives
p E
E
E
f e [t/J-l] ~T = lim f [t/J-Il ~N aD p Ul~ E .... 0 aD~,l Ol'l = fa t/Ja - fa a = f =w-. t/Ja = 0, since D D uu
and aDp 2 = aop \ aDp l' Then, lim , , E .... 0 lim
E "'0
faD E p,2
[t/J-l]
~, = fa D [t/J -1] a
UI'I
t/J = 0 on supp a. Moreover,
fD -
8'lu'Vt/J =
fD
u) is
and so a = O. We now show that u (and hence
bounded. We will
assume that n 2: 4 for simplicity. Since Iia II 2 A < C, we know that u(X)
=
(
)
f( Q) . C n fa da(Q), WIth Ilfll 2 < C. Now, for X f D 1 = D IX-Q In - 2 L (A)
l(x,y):y>¢(x)+ll,
I
Nl('Vu)2da:S
A R
~
I
2 l'lul dX :s RC3
u1 / 4 \n l12
J
"-'2 u ,
C 1RO, Jm 2 +E=EJ."".\E-l E.\ h2:.\ I 0
N(Vu) t L l(aD) , and such that vtulao = Vtf a.e. Moreover, u = S(g) ,
H~t,
f
f( Q) non-tangentiallya.e. Moreover, u(X) =
S(g)(X) , g t H~t. Also, ula o t Hi , at(aD). b) Given f t Hi ,at' there
g t
m2 +
~ t H~t(aD).
E J"" .\E-I c) If u is harmonic, and N(Vu) t Ll(aD),
o
J.E.\ m2 d.\ .\lld.\+CE 0
J.
I m>.\ I
2 m d.\
/\
•.-
' . - - - .--:-::::=..;''':'::::''"-:-''':-..,:::::"-------,-=:~.--:-;,-.::::.----
162
CARLOS E. KENIG
DO
lEAl:; Cal!m >AII. Ae-1Cfh>A m 2)dA:; Ce fm 2 +e+
Thus, fm +E:; Ce f O Al+e Ilm>Al\dA+Ce f O 2 e Cfm h . If we now choose eo so that Ce o AI, such that 3QkCEA' and 13Q
f
that h(x):; A, and hence,
2Qk
f2 2: CA21Qk I. For 1:; r 2: 2, let Qk,r =
f.Q
m2 :; Ca k
12Q
f.
< CA21EAI + Ca Im*>A,h01 -
f.
1m*>Al
m2 , which is the claim. Note also
that the same argument gives the estimate IIN(vu)ll p 2: CIIVtullp' 2 < p < 2 + e, and the LP theory is thus completed.
(a) The systems of elastostatics.
In this part we will s ketch the extension of the L 2 results for the Laplace equation to the systems of linear elastostatics on Lipschitz
'V
domains. These results are joint work of B. Dahlberg, C. Kenig and
n xk + Y*
Ak,r
aQk ,r
Bk,r
n\ aQk,r n R+ Ak,r 'V
G. Verchota, and will be discussed in detail in a forthcoming paper ([8]). Here we will describe some of the main ideas in that work. For simplicity
'
= Qk r U A k r U B k r' Note that the height of B r is k dominated by Ca length (Qk)' and that IVu I :; A on A k r' Let m , l be the maximal function of Vu, correspond ing to the domain Qk r (i.e. r ,
,
,
,
J
where the cones are truncated at height ~ f(Qk))' Then, for x
,
f
Qk '
here we restrict our attention to domains D above the graph of a Lipschitz function ¢: R 2 ... R. Let A> 0, /l2': 0 be constants (Lame moduli). We will seek to
Jm~:; f C
f Bk,r
l
~u
(3.1.1)
-->
+ (A + /l) V div u = 0 in D
...
mi:; (using the L 2-theory on Qk,r):;
ul
aQ'k,r
2 IVul da+ c
.
solve the following boundary value problems, where ~ = (ul'u 2,u 3 )
m(x) :; m l (x) + A. Also,
Qk
k
k
(Qk,r)) is a Lipschitz domain, uniformly in k,r. Also, by construction m*(xk):; A. Let
2 m + CA2 IQ k l .
m2 + CA2IQkl. Adding in k, we see that
§3. Systems of equations on Lipschitz domains
of Qk' there exists xk with dist(xk,Qk) ~ length (Qk) , and such that
J 2Q
k
rQk' and Qk,r = l(x,y): x oQk' 0 < Y < r length (Qk)1. Qk,r (and
so that aQk
163
ELLIPTIC BOUNDARY VALUE PROBLEMS
Bya well-known inequality (see [14] for example), 2
~...:
2
-->
aD
=
f
f
L (aD,da)
~ ~~ + (A+/l) V div ~ = 0
J
2 IVul da+c
Ak,r
J J f2:; C
2Q k
IVuI2da+cA2IQkl.
Sk,r
(3.1.2)
l
A(div
in D
~)N + /lIV~+(v~)tlNlaD = f
f
L 2(aD,da).
'ii'-'~';'~~~~~~~,~7j~-----
---'-----::.'~ =~---.-=.-:;;;-"--=-=-"-~
",,~';ii-~,...,
164
...;:.:...
~~~~~_;;:.~·c:":~=:-:=~'~~~f'i£.~"~'~.,,;;;::"~.::c..a C ! 1~,}l2, we would have, if ao = lex, ¢(x)).: ¢: R n- -, R, 1J 1 J t L
166
a a rs· ax-:a us=O ' ox; 1
iJ
In
D • F rom varIa . t'lona 1 conSl'd erahons, . t h e most
J
e,
..
:S Ml, that II'Vtull 2
II'V¢II <Xl
natural boundary conditions are to Dirichlet condition Neumann type condjtions,
f
or the
->h = en'
a~j ~~
dition, and such that ~Il + (A+Il) 'V div Il
~ a~~ ~s i
1J
j
= 0 in D, and with Til =
=
2 =2C f h i
In order to obtain the
(an,da)
f
r l'Vu I2da :S C
an
0 in D if and only if
ill.
L
I~->I a I 2(an,da) .
::;;
II
f
hen e a:j Xi
~' near; ~~ da:S2C ( ~
equivalence between the tangential derivatives and the stress operator we need an identity of the Rellich type. Such identities are available for
In fact, if we take
Thus, !
faD
~ dO' =
an
an
II
L
then we would have
= f r · The interpretation of J problem (2) in this context is that we can find constants a~~, 1 < i, j < 3 , 1J 1 :S r, s :S 3, which satisfy the ellipticity condition and the symmetry con =
ni
(Jl an = f)
~~_.~~~,""--._~---
r 2 !'Vu \ da :S C
Ifl
fan
1/2( fl'Vurl2da, an
?1/2 Jlfl2da an
2 dO'.
general constant coefficient systems (see [29], [27]). For the opposite inequality, observe that, for each r,s,j fixed, the LEMMA 3.1.5 (The Rellich, Payne-Weinberger, Neeas identities). Suppose . a constant vector In . Rn , .. ax us=O 'In D , a rs .. = a s..r , h.. . IS t ha t ax a rs i 1J j 1J J1
a
a
and Il and its derivatives are suitably small at
Jhte
rs a iJ·
an
r
s
1
J
au ax. au dO' = 2 ax.
J au hi
r
ax. 1
an
<Xl.
rs nea ej
Then,
Jh
J
Hence,
au
....
=
Laplace's equation.
2
en' we recover the identity we used before for
fan
I,
e au au S d a. ax.' ax.
rs h rs) eneaij - i nea ej
1
J
.... 2 I . . 2 1/2 .... 2 1/2 l'Vu\ da:S C(fan 'Vtul dO') (fan l'Vul dO') ,and so
an
=
f(h an
c
au
REMARK 1. Note that if we are dealing with the case m = 1, aij and we choose h
J
0'=
I \~~\2da:s f
r s r(h rs h rs h rs) ] 0 eaij - iaej - ja ie ax ' aX = . j i
L
S r au ax . . au ax. d 1
dO'.
Proof. Apply the divergence theorem to the formula
a aXe
rs eneaij
an
S
au ax.
vector hineaej - henea~j is perpendicular to N. Because of Lemma 3.1.5,
an
l'VIl\
2da
:S C
(
J
\'Vt
/2
2
irI
)
an
REMARK 3. In the case in which we are interested, i.e. the case of the systems of elastostatics,
iii
,--'_.
,.,.""
"';.'ii7"'~-==
C"
=-
.......-=... -,~~,~
168
.~
"",=_.
CARLOS E.
" ...~_.
_:"~-':~~~~"~'"'-'--"--'-";;,,;,.;;;=;.;'--"'c;.'"
i
which clearly does not satisfy a~~ t{(S :;;. C ~ IJ J e, t
II
C(fan
1
1,)
It;
1
i \s
(t{).
a~~nknj
av
II
a ie ninr
i
rs rs aue 2 fan (hen e a ij - hi ne aej ) ax,
s
au . ax.
!
do =
J
1
(h
. . do, and for fIxed r, s, J ,
dXj
.~ 2 1 /2 -> 2 1/2 Ivtul do) (fan Ivul do) . Consider now the matrix drs =
/:1 2 1 12 .,.,
M oreover, drs
.
(~\ avlr (au, a~ -
s rs aur au d rt aut st a .. ax. ax. = rsnkakeax . nma mv IJ 1 J e
ST auT
nma me aXe
-
that for t, T,
rs aur au S d rt aiJ' ax. ax. = rsni a ij 1 J
auT tT aut auT ax v - a vrax v ax r
aut . nea sm ax. ek J
au m aX k
rt aut = rsnkakvax v
aUT! rt st tT I aut auT aXe = drsnkakvnma me - ave ax aXe' v v
N
ow, note
ldrsnkak~ma~te - a;e l is perpendicular to N, by
our definition of drs' and the symmetry of a~j : drsnka~ma~~nv - a;rnv = rt d st tT tr d sT tr 8 st tT akvnknv rsamenm - amtm = avJ!lvnk rsamenm -- amtm = tsamrnm-amrnm =
a~enm-a~tm
= O. Therefore,
fan henedrs(fl(~~)"
an
n
do)'
/ 2+
fan
.• 2 IVtul dol.
'
fan
Jl'Vt~12 dJ.j
an
the remark follows.
fan
-> 2 Ivul do'S C
IVtu 12 do'S C
fan
fan
ITu->\2. do, It
-> -> ->t 2 IA(div u)I + If.!vu+Vu II do.
In fact, if this inequality holds, we would clearly have that C
fan Iv~ + v~tl2 do
fan
Iv ii'j2 d
as
(Korn type inequality at the boundary). The Rellich
Payne-Weinberger-Necas identity is, in this case (with h = en ),
fan
j If. -> ->t 2 ,., 2t nn1:2Ivu+vu \ +A(dlV u) \da=2
But then,
fan
-> 2 Ivul do
S C(fan
fan
-> au -> I -> -> t I I dy' IA(div u)N+If.IVu+Vu N do.
-> 2 1/2 -> -> -> t 2 1 /2 Ivul do) (fan IA(div u)N + If.lVu+ VU IN] do) .
The rest of part (a) is devoted to sketching the proof of the above inequality. THEOREM 3.1.6. Let 17 solve ~17+(A+If.)vdivii'=O in D, ii'=S(g), where
do'S
-> 2 = \Vt U "'12 + ,aN 'I~ 12 Ivul I
suffices to show that
IvUI
2
an
REMARK 5. In order to show that
d
tT aut a vr ax
e fixed,
0
,j
Now, as
(a~~ n1,nJ,)-I. This is a strictly positive matrix, since a~~ t tJ.r,r.,.,s :;;. IJ IJ i C 1s
I I -> 2 1/2 -> 2 1/2 do'S C (fan Vtul do) (fan Ivul do) +
f I~l' da J(f l\lt~I' dr'( f ~
~'.
, I vector. Th us, fan heneaij rs dXj aur au s d 0_ < rs - h i nra rs),IS a tangenha eneaij ej
C(fan
auT
hrnedrs{a~~nknj aa~}
2' C, and recall that (drs) and e (a~~nknj) are strictly positive definite matrices. We then see that
J
1
fan
-> We now choose h = en' so that hrn
in the general case, directly from Lemma 3,1.5, by a more complicated
fan henea~j ~~. ~~
rs au s rs au S rs au s I rs - ak·nknJ,n .. a-x - ak·nknJ.ni ax = nia .. i ax i = ni a 1J J j 1 j IJ
is perpendicular to N, and so
ST
REMARK 4. The inequality Ilv~ II 2 'S Cllvt~11 2 holds L can,da) L can ,dO)
algebraic argwnent. In fact, as in Remark 2,
=
S s
rs I ax au = I ni a rs rs I ax' au B ut, f d a,, rs ak,nlfinj .. -a·knkn.nj or 'l,r,S f'lxe, 1 j IJ 1 1 j IJ
!
In
169
s a~ r - akjnknj rs au dV aN
->12d )1/2(f I ->1 2 d a)1/2 . Now, IVt U a an vu
n a rs au S i .. y IJ OA j
11
2 , since the
(
quadratic form involves only the symmetric part of the matrix -> this case, of cour~e au = T u-> = A(div ->. uN + If.IVU-> + VU~I N.
I
.--'--
ELLIPTIC BOUNnARY VALUE PROBLEMS
S 1
--_•. ~
~---
KENIG
aur . 2 If. ~(auj au )2 ax:-' ax. = A(dlV u) + 2" ~ ax. + ax. ' J " J
rs au a ij
'"'··,.C-
:;;:.
g is nice,
Then, there exists a constant C, which depends only
on the Lipschitz constant of ¢ so that
170
CARLOS E. KENIG
ELLIPTIC BOUNDARY VALUE PROBLEMS
J ..
J
. . 2 daSe I'Vul
We now turn to a sketch of the proof of Lemma 3.1.7. ,We will need
I.... ....t II 2 da.
IA(divu)I+p.'Vu+'Vu
an
the following unpublished real variable lemma of G. David ([10]).
an
The proof of the above theorem proce eds in two steps. They are: LEMMA 3.1.7. Let II beas'inTheorem3.1.6. Then,
LEMMA 3.1.10. Let F: R x R n .... R be a function of two variables t ( R, x = (Xl'''' ,x n) t R n . Assume that for each x, the function t f--> F(t,x) is Lipschitz, with Lipschitz constant less than or equal to M, and for
each i, 1
f
N('Vu) 2 da '''''''\~''''''''''''.~-
ELLIPTIC BOUNDARY VALUE PROBLEMS
->
f
.......,"""
~,~..F.""""'~""'''".'~~='1W~'1."""";,.,,,,,,,,",,,,,''f
->
and N(u) f L (aD,da). Moreover, u(X)
Now, assume that Uj is of type II. Note that in this case there exists Q j f aD¢, with dist(Qj'U j ) ~ diam Uj , and such that Ivtr(x)!
LEMMA 3.2.5. Let l(X-Q)g(Q)da(Q).
,.~.,.
179
This is shown by using the following two integral identities.
J
faD
- .'
~ (I(t I+K*) gt2(aD,da)'
J
aD
aD
I
vii' 12 da,
where C depends only on M. A consequence of Corollary 3.2.7 and Lemma 3.2.5, is that, if -cfu = -cfu - p.N
L2 (an,da)
J
2 p da
f aD
aD faD a~.
3.2.5. Then,
In
aD
+2
+ p.v.
au S
aD
LEMMA 3.2.6. Let h,p and u be as
'
1H'(Q) l(p-Q)l g(Q)da(Q)
(Sg)j ±(P) = ±ri(p:gj(P)
S
pnsheaxEda.
an
IIN(VSg)11 L 2 (aD,da) «() = 1>z«() = 1 +2 ( ,
f«() d,Y
(-z
':0
which is the Cauchy Integral Formula for the disc.
L\ is biholomorphic (i.e. holomorphic, one-to-one, and
onto, with a holomorphic inverse). =
Jt y
called a Mobius transformation, has the following properties:
(b) 1>(0)
(1.6)
function of ( which vanishes at O. By (1.4), the second integral
(+Z
-->
ez(f«()/(l-z(» d( (-0
y
symmetry. Recall that if z (L\ is fixed then the function
(a) 1>: L\
t
J
Now the numerator of the second integrand in (1.6) is a holomorphic
e iO , 0 S 0 < 277.
=
~ 2m
REMARK 1. If f is only assumed to be harmonic, then we cannot argue that the second integral in (1.6) vanishes. Instead, a little algebra
z
applied to (1.5) gives
(c) 1>, ¢-1 are smooth on a neighborhood of 11. If f is holomorphic on a neighborhood of ~, define g(t")
=
f o¢(t"). f(z)
Then (1.4) applied to g yields
=1277
f
277
f(ei~ 1~lz12
dO.
le10_z/2
o f(z)
= g(O) =---.!,.
!
2m:r
g(f) dt" t"
=1277i
jf(¢(t"» d C
y Change variables by t"
=
H z)
1>-1«()
=~ ! 2m
Jy
t"
y =
«(-z)/(1-z(). Then
n ()] d(
f«() [(¢-I 1>-1«()
S •
This is the Poisson Integral Formula. REMARK 2. Among bounded domains in e, only the disc (and domains biholomorphic to it) has a transitive group of biholomorphic self maps. (This follows from the Uniformization Theorem; see also [32].) Thus approach I has serious limitations in e 1 . In en the limitations are even more severe. Indeed in Section 7, after we develop a lot of machinery, we shall return to the concept of symmetry in en and gain some new insights.
188
STEVEN G. KRANTZ
Discussion of ll. We need some notation. Recall that in real differential 2
analysis on R
t
we use the basis Xx'
189
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
An arbitrary I-form is written
for the tangent space (i.e. all u(z)
=
(1.8)
a(z)dz + b(z)dz
linear first order differential operators are linear combinations of these) and dx, dy for the cotangent space. We have the pairings
a
< ax' dx >
=
a
< ay' dy >
a
=
and we define the exterior differentials
au aabz dz
1
=
a
< ax ' dy > = < ay' dx > = 0 .
Clearly du
=
au + au.
STOKES'THEOREM.
In complex analysis, it is convenient to define differential operators
A
au = aa
dz,
dZ
dz
A
(1.9)
dz .
Recall
nee Rn
If
is a bounded domain with smooth
boundary and u is a smooth form on
a 1 (aax-lay' . a) (Tz=2 a 1 (aax+lay' . a) az=2
n
then
Ju fdU. =
The motivation for this notation is twofold. First,
-tz=~z=l, az (Tz
an
aaz=l...z=o. z az
In our new notation, if
n
n ~ c1 ~ R 2
and u is a I-form as in (1.8), (1.9),
then Stokes' Theorem becomes Secondly, if f( z)
=
u(z) + iv(z) is a C 1 function, with u and v real
valued, then
~ = 0 ~ (~ =~
and
~ =- ~) ,
Ju= fau+au
(1. 7)
an
n (1.10)
which is the Cauchy-Riemann equations. Thus
.ili. = dZ
holomorphic.
0 means th8t f is
f (~ -;)
We also define
dz
A
dz .
n dz
=
dx + idy, dz
=
dx - idy . Now we can prove
It is immediate that
f in L 2 (an).
Thus we may make H 2 into an inner product space by defining
=~ 2m
ff(')
, -z
an
d'
-~ J«(Jf/JO d( 2m '-z
Ad' .
11
This formula, valid for all f (C l(n), will be valuable later on.
(1.12)
< f,g>
=
ff~da an
for f,g (H 2 (n) .
(1.13)
192
STEVEN G. KRANTZ
BASIC LEMMA. If K C 0 is compact then there is a constant C
193
INTEGRAL FORMULAS IN COMPLEX ANALYSIS =
C(K)
Then the lemma, with K
=
lz I,
rP z
shows that
is continuous. The 2
such that
Riesz Representation Theorem yields a unique kz (H (O) such that sup If(z)1 z(K
s Cllfll
Proof. Fix z ( K. Let El
->
f(z) .
e
aaz j z·J = Lazj z·J = 1, < aaz j , dz·J > and all other pairings are O.
'=
= 1, J
j
=
1,'" ,n ,
= w-' ie titl ZX::W::
=g....._l'~
z=:
z= ....:;:;m:;:w
194
=
~........~
EMMJ:~
STEVEN G. KRANTZ
1'22
,",~~
If a = (al,···,ak)' {3 = ({31,···,{3e) are tuples of non-negative
We define a constant
integers (multi-indices) then we write
a dz = dz
W(n) " ... "dz
al
ak'
cIZ{3 = cIZ
{3I
" '" "dZ
=
(3e'
f
w([) " w«() .
B(O, I)
Here B(z,r) = t( (C n : I( -z 1< rl-
A differential form is written u
195
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
Now we may formulate a generalization of approach II in Section 1.
~~ a ~ a{3
dz a " rrz{3
(2.1) THEOREM (The Cauchy-Fantappie formula).
a,{3
~
bounded domain. Assume that w with smooth coefficients aa{3' (If 0
'S p,q (Z and the sum in (2.1)
Let
(wl,""w n) (C
n ~ Cn oo
be a smoothly
(il x n\~),
Wj =wj(z,(), and
ranges over lal = p, 1{31 ~ q only, then u is called a form of type
n
(p,q).) We then define
~ Wj(z,()
au = ~ aa:.~ dZ j "dz a,{3,j
a
" rrz{3,
'ilI =
U[;J
a,{3,j
By a calculation (or functoriali ty), du = called hoIomorphic if
du
=0
~
. «(j-Zj) = 1 on
nx n\ ~.
j=1
aaa{3 cIZ j " dz a " dzf3 . OZ.
au + au.
J
A C I function f is
O. (Note: this means that
di
=0
If f ( C I (n) is hoI om orphic on
n,
nW~n)
I
f(z) =
0,
j ~ 1,'" ,n, so f is holomorphic in the one variable senSe in each
variable separately.) Finally, we introduce two special forms: if w ~ (wI""'w ) is an n n-tuple of smooth functions then we define the Leray form to be
then for any z
(n
f«() Tf(W) " w«() .
Before proving this result, we make some detailed remarks. REMARK 1. In case n
~ 1,
then w = wI = _1_ (of necessity). So
(-z 1
1
Tf(w) = _1_, w«() = d(, nW(n) = 217i . . I
Tf(w) ~ ~ (-I)J+ w j " dW I " ... "dw j _ 1 "dw j +1 " ." "dw n . j=1
(-z
The Cauchy-Fantappie formula becomes
Likewise
w(w) ~ dW I " ... " dw n .
we have
an
(lZ.
J
~
(2.2)
1
f(z) =1;i
ff«() d( ,
(-z an
which is just Cauchy's formula.
(2.3)
-
'~~~
196
~~-
-""~
.."~'
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
STEVEN G. KRANTZ
REMARK
2. As soon as n::': 2, the condition (2.2) no longer uniquely
If a 1 ,a 2
determines w. However an interesting example is given by
w(z,()
Z
= (w 1(z,0, ... ,w n(z,(» =(ZC
1 ,"',
\1(-zI 2
=
2. Now
a(l
-
1
2
(2) a(B(a ,a ) == 0 ;
(S:
then B(a 1,a 2 ) - B({3I,{32) is a( exact.
Assuming (1), (2) and (3) for the moment, let us complete the proof (note
dw
w2
We claim that B has three key properties: 1 (1) B(a 1,a 2) does not depend on a ;
(3) If al,a 2 ,{3I,{32
= wI _ d(1 + wI _ 2 d(2 ( a(l a(2
-
that (1) is used only to prove (3». Letting
a
awl -) --=-d(2 "d(l" d(2
1= a 2= (WI) (~ 2
{31 = {32 =
Z2-Z 24 dZI + ZI-zl d(2)
1t:-zI
w
a(2
which by direct calculation
= (-
~ €(o)a~(I)"a(a~(2)' ocs 2
Zn-zn)
l'-zl 2
"Tj(w) " 1
J
J
the system is then "over-determined" and a compa tibili ty condition is Now an easy calculation, as in (1), shows that this last equals aA where a1 A
=
det
ai-b i )
: ( a 2
2 2 a 2-b 2
This completes our discussion of the Cauchy-Fantappie' formula.
necessary. For n
=
1 the system is not over-determined.
The three basic considerations about a POE are existence, uniqueness and regularity. It is easy to check that
a is elliptic on functions in the
interior of a given domain; hence, if u exists, it will be smooth whenever f is (we will see this in a more elementary fashion later). So interior regularity is not a problem. Also, since the kernel of (i consists of all holomorphic functions, uniqueness is out of the question. So, for us,
§3. Introduction to the
a problem
One of the principa 1 problems in complex function theory is the con struction of holomorphic functions with specified properties. In one dimension, there are a number of highly developed techniques: Runge and
existence is the main issue. The following example shows that the compatibility condition does not by itself guarantee existence of u. EXAMPLE.
Let
n c; C 2
be given by
Mergelyan theorems, power series, infinite products, integral formulas, and so on. In several variables, these techniques are either unavailable, much less useful, or much less accessible.
n=
(B(0,4)\B(O,2)) U B (2,0),
~)
at = 0
----------"-_.._.,-_ ...
202
-------------~-
.__._------_._-",-----------_.
__
--_._----'::::
._-----.._--_._-------,"---
STEVEN G. KRANTZ INTEGRAL FORMULAS IN COMPLEX ANALYSIS
203
EXAMPLE. Consider the following question for an open domain nee en: /1-//
U
If w == n (3.1 )
v
n IZ n = O! i ~ and if g is holomorphic
on w (in an obvious sense), can we find G
{
holomorphic on n such that Gl w
g ?
=
w
I Let U
=
8
(1,0),~)
and
V 8 ((1,O),~) =
as shown. Let TJ
f
C;(U)
satisfy TJ == 1 on V. Finally, let
If n is the unit ball, then the trivial extension G(zl,z2,",zn) == g(zl,,,,,zn_l'O) will suffice. However if n = 8(0,2)\8(0,1) s:; e 2 then g(z 1,0)
f
Then f is smooth and
Jry
=
holomorphic on s upp (Jry)
n
(l,O),~) n IZ 1=11).
--L
zl1
is we ll-defined and
n. If there existed a u satisfying
(dIt = 0)
au
=
f
on
But u would necessarily be smooth near
(1,0) (since f is) hence h has a singularity at, for instance, (1,0). Thus we have created a function holomorphic on 8(0,4)\8(0,2) which does not continue analytically to (1,0). This contradicts the Haitogs
au
=
Proof. Let 17: en 8 = lz
f
->
n: 17Z I wi.
is not always solvable, let us turn to
an example where it is useful to be able to solve the
a equation.
en be given by 17(zl,···,zn)
n
then the equation
=
(zl, .. ·,zn_l'O). Let
Then 8, ware disjoint relatively closed subsets
of n, so there is a COO function ¢
on n such that ¢ == 1 on a rela
tive neighborhood of wand ¢ == 0 on a relative neighborhood of 8. 'V
Define F on n by 'V
F(z)
au = f
0
f has a smooth solution. Then the answer to (3.1) is "yes."
0
Now that we know that
but could not have an extension G
whenever f is a smooth a-closed (0,1) form on
extension phenomenon (an independent proof of this phenomenon will be given momentarily).
(j)
THEOREM. Suppose that ween is a connected open set such that n since
then the function h '" u -~1 would be holomorphic zl n\(8
1/z 1 is holomorphic on
(else the Hartogs extension phenomenon would be contradicted).
zl-1
aclosed on
=
{¢(Z) . f(17(z))
if
o
else.
Z
f
supp ¢
=
'V
Then F gives a Coo (but certainly notholomorphic) extension of f to n.
....,....,====--;0;;;-.....-....,""",.""....."""'" 204
STEVEN G. KRANTZ
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
We now seek a v such that F + v is holomorphic and F + vl w = f. With this in mind, we take v of the form zn . u and we want
au az = ~ Jz:
t rf ~ J,
2m
or
C
17 is holomorphic on supp ¢ and Zn is holomorphic so all that
-~ JfJ (aeP/~()(I;) dZ
remains is
d¢ . (f o 17)
A
=-2~i Jf(aeP/az~«(+Z)dZAd(
a¢ . (f 17) + ¢ . a(f 017) + (dZ n ) . u + zn . ~ = 0 . 0
( dZ d~
¢«( +z)
C
a(F+v) =0
Now f
205
+ zn . ~
=
2m
0
A
d(
L
D(O,R) or
(- d¢) . (f
~
where D(O,R) is a disc which contains supp ¢. We apply Remark 1 of
17)
(3.2)
zn
The critical fact is that, by construction,
n n lZn =O!
0
ap = 0
a closed.
on D(O,R) to obtain that the last line
in a neighborhood of
so the right-hand side of (3.2) is smooth on
easily checked to be
II in Section 1 to ¢
n.
=
Also it is
¢(z) -
f
~
2m
¢«() J' '0
dJ'
'0 •
-z
JD(O,R)
Thus our hypothesis is satisfied and a 'V
u satisfying (3.2) exists. Therefore F == F +v has all the desired properties.
0
Our two examples show that solving the
a equation is (i) subtle and
(ii) useful. Thus we have ample motivation to prove our next result. LEMMA. Let ¢ (C~(C), k::: 1, and define f = ¢(z)dz. Then
1
u(z) == - 217i
IJ¢(I;) dZ (-Z
C satisfies u ( C k(C) and
~
=
f.
A
d(
The integral vanishes since ¢ = 0 on JD(O,R), hence ~ = f. Observe finally that u (C k by differentiation under the integral sign. 0 REMARK. In general, the u given by the lemma will not be compactly supported. Indeed if
JJ¢
f. 0 and if u were compactly supported (say
u ~ D(O,R) ) then a contradiction arises as follows:
o=
f
ud( =
JD(O,R) The supports of solutions to the
f
¢
dZ
A
d(
f. 0 .
D(O,R)
a problem explain many phenomena in
one and several complex variables. We explore this theme later. Mean Proof. We have
while, contrast the Lemma and Remark with the follOWing result.
---"-_._.
2()j
--- ._-,
-,,---_ ..
STEVEN G. KRANTZ
INTEGRAL FORMULAS
> 1 and let (z) '" ¢ Idz 1 + ... + ¢ndzl be a-closed on en and suppose each ¢j (C~(Cn). Then for any 1 :s i :s n the
THEOREM. Let n
IN
COMPLEX ANALYSIS
au = f.
compactly supported solution to
207
In the present case, one exists
and is given by an integral formula.
function uj(Z) '" __1_
21Ti
satisfies u j
f
Proof. Fix 1 m
=
'-z.
§4. The Hartogs Extension phenomenon and more on the
fy¢j(Zl'""Zj_l",Zj+l,""Zn)
J.. e
We have cited the Hartogs extension phenomenon in the examples of
d' Ad'
J
au
C~(en) and
j
.
=
Section 3. The reader will want to check that the proof of it that we now
Moreover u j
=
u e for all j, e.
:s m :s n.
:s
We need to check that ; j '" ¢j' 1:s m n. If zm j then the result follows from the lemma. If m -/ j then use the
a¢.
compatibility condition ~
aim
-
azm
1 = 21Ti
THEOREM (The Hartogs Extension Phenomenon). Let n c;; en, n
aZ j
n\K is connected. If f is holomorphic on n\K then there is a holo such that F In\K
=
f.
an
and ¢ == 0 in a neighborhood of K. Define
-aZm (z l' ... 'j_l,S,Zj+l'''''Zn) Z
)'
d' d' J '-z, II aif' (ZI""'Zi;~~e'Zi"""'Zn)
J..
rv
de Ade.
J
e
(n\K
if
Z
if
Z (
F(z) =
A
J
l¢(z) . f(z)
o
a¢
1 21Ti
be compact. Assume that
Proof. Choose ¢ (C""(n) such that ¢ == 1 in a neighborhood of
C
= -
n
be a bounded, connected open set. Let K C;;
>1 ,
_m to write
a¢.J
a' uJ(z)
give is independent of those examples.
morphic F on n
a¢ =
a problem
K .
Then F is a C"" extension of f to n, but it is not in general holomor rv
phic. We now seek v such that F + v is holomorphic on rv
F +v!n\K
=
n
(and
f). Thus we seek v satisfying rv
By the lemma, this last equals ¢m' Thus a-uj = . Notice that, if f j, then uj = 0 for large (since then ¢j = 0). Also u i is
e
holomorphic for
ze
ze
or
au =
large (since
is then 0). So, by analytic continuation, u == 0 off a compact set. Next, u j - } == 0 since it is compactly supported and holomorphic. Finally, ui (C~(en) by differen tiation under the integral sign.
a(F+v)=o
0
REMARK. The proof actually shows that u is zero on the unbounded n
component of c( j~1 supp ¢j)' It also shows that there is at most one
a(¢f) +
av = 0
or
"Jv
=
(-a¢»' f
(4.1 )
since f is holomorphic on supp ¢. Now ¢ == 1 near is smooth and compactly supported in
n.
an
so (-
ap) . f
The theorem of Section 3 now
guarantees that there exists a v satisfying (4.1). Moreover, the remark following the theorem guarantees that v == 0 near
an.
Thus F + v is
.
~-~==~--=--==-=.-=-.=--=;;;;;.-..,.~,-:;:~=-
208
tion,
F +v
=
ao, F + v = F =
¢ . f
f. By analytic continua
=
f on U and the result follows.
supp v. The Hartogs extension phenomenon has several interesting consequences: A holomorphie function f in en, n ~ 2, cannot have an isolated singularity. If it did, say at P, then f would be holomorphic on B(P,2E)\B(P,E) for E small hence, by the Hartogs phenomenon, on B(P,2E) , and hence at P. That is a contradiction (ii) A holomorphic function f in en, n ~ 2, cannot have an isolated zero. If it did, say at P, then apply (i) to l/f to obtain a
(iii) If U S; en is open, E S; U, f is holomorphic on U\E, and E is a complex manifold of complex codimension at least 2, then f continues analytically to all of U. To see this, notice that for =
2 the set E is discrete and the result follows from 0). For
n > 2, the result follows from the case n
=
2 by considering
fl (O\E)nE ranging over all two dimensional complex affine spaces
Ec en.
Topic (2) has been discussed vis
a vis the Hartogs phenomenon.
Topics (1) and (3) are more subtle. First note that if
a(u+h)
ai.t
= f then also
f for any holomorphic h. Given f, one cannot expect all u + h to be nice (i.e. bounde d, or L 2 , or C<Xl up to the boundary). How =
does one find a nice solution? An idea from Hodge theory is to study the solution u to ~ = f which
a,
is orthogonal to the kernel of
i.e. whic h is orthogonal to holomorphic
functions. This solution has been studied by Kohn [25], [27], [28], Catlin [3], [4], Greiner-Stein [16], and others. It is often called the Kohn
a equation.
We now briefly review some of what is known about the Kohn solution, and other solutions, to the
a problems on domains in
en. (In this
section we shall take "pseudoconvex" and "strongly pseudoconvex" as undefined terms. These .terms will be discussed in detail in Section 5; for now, a pseudoconvex domain is a domain of existence for the
a operator.
There are essentially
four aspects to this matter:
=
~fjdzj is a
a-closed form with all f j ( L2(0), then there is a u ( L 2(0) with Jud. Also IluIIL2SC(0):flIfjIIL2' See [20], (Exercise: the
(1) Existence of solutions; (2) Support of
a
operator .) (a) If 0 is a bounded pseudoconvex domain in en and f
Now we return to discussion of the
a data and a solutions;
(3) Choosing a good solution, where "good" means smooth or
canonical solution also satisfies this estimate.) (b) If 0 S; en is smoothly bounded and pse~doconvex and f = ~ fjdzj ~s
(4) Estimates and regularity.
a a-closed (0,1) form with all f j (C<Xl(O) then there is a u (C<Xl(n) satisfying = f. See [26]. It is not known whether the canonical
Regarding (4), we have noted that ellipticity considerations imply
solution has this property.
bounded;
that when
av
=
au
g then v is smooth wherever g is. As an exercise, use
(c) If 0 S; en is strongly pseudoconvex with C 2 boundary and if
the theorem of Section 3 to give another proof of this assertion. (Hint: if
f
g is smooth on B(P,r), then let ¢ (C;(B,(P,r)) satisfy ¢ == 1 on
there is a u satisfying
B
(ifJ . J¢. v) + (l-ifJ)(a¢· v)+ ¢. g.)
=
solution or canonical solution to the
contradiction.
n
of Section 3, decomposing f as f
This is all that we shall say about interior regularity.
0
Notice how the hypothesis n > 1 was used in the proof to control
(i)
209
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
STEVEN G. KRANTZ
holomorphic and, near
---=----=-:;-i~~:;:::::.~~~=_____:_=-=----==----==~~~.==___::;c.===-=-~
(p,~).
Let ifJ
u = ¢ . v and f
=
(C;(B~, ~)) a(¢ . v)
=
satisfy ifJ ==1 on B (p,~). Define
(j¢ . v + ¢ . g. ~hen au = f. Apply the theorem
=
~fjdZj is a a-closed (0,1) form with bounded coefficients, then
au =f
and Ilull <Xl 0::: C~llfJ·lI L
<Xl' The L
210
Outline of proof. Let g:
Sibony [39] has shown that there are smooth pseudoconvex domains
Let V be an open neighborhood of P such that
V C
canonical solution has this property. See [11], [17], [22], [16], [351
on which uniform estimates for
211
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
STEVEN G. KRANTZ
f. We conclude this section with an example of how estimates can be
useful. DEFINITION. Let 11 function f: 11
-->
0, fl on H
B(P,e)
is unbounded.
It is useful to be able to construct singular functions. Often we 'can nearly do this in the sense that we can find a neighborhood U of P and a holomorphic function on
un
11 which is singular at P (this is 'called
a local singular function). Then the problem reduces to extending local
0 is sufficiently large and p(z) (e Ap (z)_l)/A then
=
I
n
j,k=l
az.;
aZ'V J
az.az J k
~
j, k=l
(ao
EXERCISE. If P
Z
n
+1 ~ ~ (P)ajak~O But a similar inequality also obtains for ia (j"p(an). Adding the two inequalities yields the result.
(P)WjWk~ClwIZ, VP (ao,VW (j"p(an).
(5.3)
k
§6. Solutions for the
0
a problem
We briefly describe the Hilbert space setup for Hormander's L Z theory of the
See [31] for details.
a problem.
We fix a smoothly bounded 0 C en and introd uce the
notation The rather technical notion of pseudoconvexity is vindicated by the LZ(O)
following deep theorem (see [31]): THEOREM. If 0 C en is smoothly bounded then the following are equivalent: (i)
0 is pseudoconvex
(ii) 0 is a domain of holomorphy
(iii) the equation solvable.
au
=
f, f a
J
a (0,0)-=T
•
LZ
(0)
(0,1)
J(Ol)=S
,
•
LZ
(0)
(0,1)
III
III
III
HI
HZ
H3
The operators T,S are of course unbounded, but they are densely defined closed (p,q) form, is always
:'" """,,;~
""'1'.
':=
..
.,~:::' J~j!""!',
....
216
.;~~~~~~~~··t-;t':7£"L:;;c.~-::~~~~Z-::'''::~'·=''-:::~';;::i:~~~l';~ 0 so small
is
bounded) and one obtains an existence theorem in L 2(p, dx). A leisurely
that n == Iz (en:dist(z, 11) < E! is convex (hence pseudoconvex). Let f
exposition of all these ideas can be found in [31]. We now formulate a
be a smooth, a-closed (0,1) form on n. By Hormander's theorem, there
version of Hormander's result, which we will use freely in what follow:
is a smooth u on n such that
au
=
f. We apply the Bochner-Martinelli
formula to u (which is certainly in C l(n». Thus THEOREM (Hormander). If n t; en is smoothly bounded and pseudo convex, and if f is a
a closed
(0,1) from on
cients, then there exists u (L 2 (Q,dx) such that
n
with L2(n,dx) coeffi
au
=
f.
I I
218
STEVEN G. KRANTZ
I !
u(z)
1 W( ) !U(t';,)1] ( n n
=
an
I
1 - nW(n)
I
J~(t';,)
~z.\
I =
1 nW(n)
J
an
l-z J
u(01] (
10/
I!;,-z
A
n
A
w(t';,)
n
f
- _1_ nW(n)
(6.3)
\t';,-z\ij
1 W()
w(O -
v(z) , n.i(n) {
Jute)" (le~1
A
Ju«()q(w)
w(()
an
Z-z \
1] (
n
I
W(t';,)
A
\t';,-z\2j
A
219
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
A
W«()}
an
J
f(t';,)
1] (
A
Z-z \
w(t';,)
A
I!;,-z Ii}
0'
I - II .
n Let G
f(t';,)
A
1] (
n
Z-Z:.\
!t';,-z\2j
A
w(O .
=
n x [0,1]
and define g(z,!;" A)
form on G. Then I
~/
=
..
,
f
=
(1---A)
u(!;, )1](g)
A
t';,-z + Aw(z,l;,) to be a Jt';,-zI2
w(!;,) .
aG
The first term on the right is not useful, since it involves u, so we will remedy matters by subtracting an appropriate holomorphic function from the right side of (6.3) (see the discussion in Section 4 on choosing a good
By Stokes' theorem, applied on G, this
solution). The Cauchy-Fantappie formalism now comes into play: If
n = Ip < O!,
1
let
=
w(z,t';,)
n
with ¢(z, 0
=
~
j=l
a ::.J'
~
(z, t';,)
d(U(!;,)1](b)
-~ (t';,)) at';,n
=
nW~n)
But d(u(l;,)1](g)
A
we!;,)) = au
J
u(t';,)TJ(w(z, t';,))
= A
h(z). Then certainly
A
1](g)
A
+ u(!;,)d(1](g))
w(t';,)
f
A
1](g)
A
w(l;,) A
w(l;,)
w(O
(this last equality takes advantage of the special algebraic properties of Cauchy-Fantappie forms). So we finally obtain
is well defined and holomorphic in z (because w is).
=u(z) -
w(t';,)) .
(z, t';,)
J
J
A
G
an Now let v(z)
J
(t';,)(z, -t';,J') and observe that
UI,,'
H(z)
=
-~ (t';,) 0(,1
nW(n)
av
=
~
HENKIN'S INTEGRAL FORMULA.
=f
and, by (6.3),
If
n = Ip < 01
is C 2 and convex and
f is a smooth, a-closed (0,1) form on a neighborhood of
function
n
then the
220
STEVEN G. KRANTZ
v(z),
f
!" , S
~ anX[O,t]
fCO
A
"(g)
A
221
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
v(z)
f a~t
-~ «(
WcO = __ 1
2W(2)
an
-ff(O "." (!p_zIZ) ,"-Z ~ "w(~)l
)«(z--zz) +
~ a;z (0 (7
'ot-Zl)
ll>(z, 0 \(-z \ z
"(fl(~)d[l +fz(Od[z) "d~l
"d(2
n
satisfies
av
=
f on
n.
1
Here
I f 1(O('1-Zt)+f Z(O«(Z-Z2) d'l "d'z "d(l "d(z
I(-z
- 2W(2)
g(z,~, A)
w(Z,
0
(I-A) (
=
'-z\
Ip-zl1
n + AW(z,
0, 1 =2W(2)
_± (0 _±a(n (0) a~t ( ll>(Z, 0 , ''', t1>(z, 0
=
1 + 2W(Z)
ll>(z,
0
=
f
fl(OAl(z,Od~l "d(l "d(z
an
and n
14
f
flOAz(z, ()d~z "d~l "d~z
an
a
I aI. (O(Zj -~j) . j=t
J
+zJ(Z) !f1(OB1(z,Od[t "d[z "d(l "d(z
n
The standard reference for Henkin's work is [18]; see also [31]. Similar formulas were derived by Grauert-Lieb [11], Kerzrnan [22], and 0vrelid [35]. Now we assume that
n
is strongly convex, and show how to use
Henkin's formula to obtain uniform estimates for solutions to the problem. For simplicity we work in C 2 only. So f " .,,(g) " w(O
=
r
ag z
£ " Lgt CA dA - g2
ag t
ax.
.l
d'j "d ~t "d
~2
a
+
2W~2)
J
f 2(0 Bz(z,Od[l "d'2 "d(l "d(z .
n
.
In order to prove an inequality of the form
llvll Loa S C\lfl\ L oo ,
After some algebra, and integrating out A, the Henkin formula is it suffices to check that
(6.4)
p;
r"ll~?iiiEiii7
~~'l
222
*iGIi
1lIIl~lIiIn
&i25SGi'i~.".-
is
iImi~:-:"'~"~-'-
~I=~""·""""".~"""--""'~~>,,... ~,
STEVEN G. KRANTZ
!IA/z,')!da(')'Sc,
,.,,.,"
223
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
j=1,2
IRe (z, 01 ~ Cllz-'1
(6.5)
an
Let
2
+ Ip(z)ll.
an.
Let t l be a coordi nate in the complex normal direction at 7TZ (that is, i times the real
and
TTZ
be the normal projection of z to
an.
norma I direction) in . JIBj(Z,')/dV(,) 'SC,
j =1,2
2
I
(6.6)
n
ap, (S)Uj = 0 ~ U
j=l
with the estimates uniform over z
n.
f
(Here da is area measure on
an.)
Recall that
a:J
J
IBI(z,OldV(') 'S J
n
n
for all z
f
n.
2
Then
Re
a
~ ;.
(')Uj
=
0
~
U fTp(n).
J
j=I
_C_ dye,) Iz_~13
:fp(n)
and
By symmetry we check only j = 1. For B 1 , choose R > 0 such that B(z, R)2
f
It follows that
B(z,R)
1m
R
N(z,O n(z,O
fS{Z,t.:)f(t.:)dO(t.:)
_(N(t.:,Z) \ n(t.:,z)}
is a kernel which is less singular than the original Henkin kernel. Thus has the follow ing three properties: (a) S: L 2(an) ... H 2(n)
H _ H* , rather than being a non-isotropic singular integral operator (as is H ), is a smoothing operat or. This observation of Kerzman and Stein is
(b) S is self-adjoint
now exploited as follows. Denote H* -H
(c) S is idempotent.
=
A.
The reproducing properties of Sand H guarantee that
Therefore S is the Hilbert space projection of L 2(an) onto H 2(n).
(1) S
It turns out that the Henkin operator on a strongly pseudoconvex
domain very nearly has properties (a) - (c). First, by a theory of non
=
HS
and
isotropic singular integrals developed especially for boundaries of
(2) H
=
SH .
strongly pseudoconvex domains (see [8], [36]): the Henkin operator Thus H: f
I->
nW~n)
f
f(t.:) 7J(w) "
w(t.:)
(7.1)
an
(3) S
=
S*
=
(HS)*
=
S(H* - H) = SA .
=
S*H*
SH* .
=
Subtracting (2) from (3) gives
(with the w produced from the Fornaess theorem as in Section 6) maps L 2(a11) onto H 2 (n). Also H is idempotent. Now H is not quite self adjoint, but it is nearly so. To see this, one needs to write (7.1) in the
S- H
This is an operator equation on L 2 • We may resubstitute the equation
into itself as follows:
form
1
nW(n)
Jf(}') N(z,t.:) '.:>.nn}' '¥ (z, '.:»
S = H + SA
do(t.:)
an where do is area measure on
an.
this when
{t.:,z)
n
H + (H+SA)A
=
2 H + HA + SA
(7.2)
H + HA + (H+SA)A2 _ H + HA + HA2 + SA3
=
This is a straightforward but tedious
calculation (see [23]). It turns out that N is real. It also turns out that (z,O -
=
vanishes to higher order at z
=
t.:
= ... =
1 H + HA + ... HAk + SAk + .
than does . (Try
is the ball to see how this works - details are in [23].)
2
Now we know that each of the operators HA, HA , ... are smoothing. If we apply both sides of (7.2) to a sequence cPj
(c~(an)
such that cPj ....
0t.:
230
I I
STEVEN G. KRANTZ
in the weak-* topology on
an,
231
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
I
we obtain an equation relating S(z, () B :f
and H(z,(). In particular, Hand S are equal modulo terms which are
I->
less singular. From this fundamental result, many basic mapping proper
fee) K(z, () dV«()
11
ties of S can be determined (see [36]). The basic construction of Kerzman and Stein can be used in other contexts. Let us turn now to one of these: the Bergman kernel. Fix a domain 11 cc en and define
A'(O)
~r
holom",phic on !L
[If I' (z)
dV 01 (ll)
0 .
= !fgdV
Therefore we may set
11
Ilfll
=
f
g.. (z) IJ
a2
= --
az.i]i. 1
If1 2 dV I /2
11
log K(z,z) .
J
Bya calculation (see [31]), the matrix (gi/z)) gives a non-degenerate Kahler metric on 11 (called the Bergman metric) which is invariant under
2 for f,g! A (11). The basic lemma in this context, sup If(z)1 :S CKllfl1 K
for K CC 11, is easily derived from the mean value property for holomor phic functions. As in Section 1, the abstract Hilbert space theory yields a reproducing kernel for A2 which we call the Bergman kernel. Just like the Szego kernel, the Berman kernel (denoted by the letter K) satisfies K(z,() = K«(,z). Thus the associated operator
biholomorphic mappings. In particular it holds that if 11>: 11 1
....
11 2 is
biholomorphic then 11 1 11 2 dist s erg (z,w)=dist s erg (lI>(z),II>(w)). As a result, metric geodesics and curvature are preserved. The Bergman metric and kernel are potentia lly powerful tools in func tion theory, provided we can calculate them. To do so, we exploit the idea of Kerzman and Stein [231 to compare K with the Henkin kernel. However a complication arises: the Henkin integral (7.1) is a boundary
---------,,-.-
232
••- - ; - ' - -... --:.,-'---::-=,...-•.:::-:.---"'.,.--,.,---
..
'-'::'"":::-=.~_::--.--_:--::---:--V:."'''::.~~~._~'''·~~",:,,--:,".-"_~~~--:-::-:'-C:-= ·-c-:·:.::::"",;;;:,~,,,,;::~,,,:,,,,,,~_~'-,",,":~-"";;;;:_--:;:~;';;'-:~:-.:-:T: ::---_::.~_-.,.~_~.'~~--".
STEVEN G. KRANTZ
integral while the Bergman integral is a solid integral. How can we Com pare functions with different domains? What we would like to do is apply Stokes' theorem to the Henkin integral and turn it into an integral over
III
n.
However, for z (n fixed, Henkin's kernel has a singularity at (= z. So Stokes' theorem does not apply.
I
The remedy to this situation is to use an idea developed in [19], [30], [33]: for each fixed z (n, let
I
i I
lfz(()
=
N(z,O cf>n(z, ()
!(an
Now we conclude this paper by coming full eire Ie and discussing once again the topic of symmetry of domains. The reader should consider that, up to now, all of our effort has been directed at obtaining (a), ({3), (y). Now we use those to derive concrete information about symmetries. If
n~
en is a domain, let Aut
n
n1
mappings. If two domains
denote the group of biholomorphic seIf and
n2
are biholomorphic we will write
n1 ~ n2 · THEOREM (Bun Wong [41 D. If
nee
strongly pseudoconvex and if Aut
n
en is smoothly bounded and
acts transitively on
Now construct a smooth extension IJIz of lf to n. The Cauchyz Fantappie formula is still valid with IJIz replacing lf (since the integral z ta kes place on the boundary where IJIz = Ifz). Thus Stokes' theorem can
n ~ ball.
be applied to the new Henkin formula containing IJIz. The resulting solid
the holomorphic sectional curvature tensor
integra I opera tor on L 2(n) can be compared with the Bergman integra I
satisfies
via the program of Kerzman and Stein (details are in [33]). The result is that K(z, ()
=
233
INTEGRAL FORMULAS IN COMPLEX ANALYSIS
IJIz (() + (terms which are less singular) .
->
an.
K(P ) O
=
then
be any fixed point. Let IP j !
K
for the Bergman metric
(constant curvature tensor of the ball).
Thus the Bergman metric curvature tensor is constant on
n.
We now use
As a result, curvature, geodesics, etc. of the Bergman metric may be
THEOREM (Lu Qi-Keng [34]). If M is a complete connected Kahler
calcula ted. Also the dependence of these invariants on deformations of
manifold with the constant holomorphic sectional curvature of the balI
an
then M ~ ball.
can be determined (see [12], [13]; it should be noted that the methods
of [1] or [6] may be used for the deformation study instead of the Kerzman
This theorem, together with (*), completes the proof.
(*)
0
Stein technique). The following are the three principal consequences of these calculations for a smoothly bounded strongly pseudoconvex (a)
As Q) z ->
an,
n:
THEOREM (Greene-Krantz [13 D. If
an
an.
(i)
n~ B
or
(f3) The kernel and the curvature vary smoothly with smooth perturbations of
an.
(y) n, equipped with the Bergman metric, is a complete Riemannian manifold.
is smoothly bounded and if
is COO sufficiently close to the unit baIl B then either
the Bergman metric curvature tensor at z converges
to the constant Bergman metric curvature tensor of the unit ball. The con vergence is uniform over
n
¢oao¢-l
i
Aut no
is a univalent group homomorphism.
the kernel for n is the product of those for the disc and annulus. Now un j where OJ OI. We identify the n n boundary of R~+l with R = R n x lOI, and, given a function f on R ,
we want a function u(x,y) harmonic in R~+l such that u(x,y) .... f(x o) as (x,y) .... (x o' 0). For continuous boundary data, the problem is completely solved by n the Poisson integral formula. Thus suppose f is continuous on R , and f ( L l(R n) + L ""(R n). Set
group are involved in Koranyi's extension of Fatou's theorem to existence Pf(x,y)
of admissible limits of holomorphic functions.
=
Py
* f(x)
=
In each of these settings, there is a naturally given family of first
cny
f
dt
f(t)
R n [lx-tI2 +y2]
2
order linear homogeneous differentia 1 operators, or vector fields. In part II of this paper, we shall see that the general construction of metrics applied
where
to these families of vector fields gives back the natural metric in these n+1
classical settings.
cn r (n~l) /IT 2 =
The discussion of results in these examples will be very brief, but references are given for the complete proofs of all the results. Then:
§l. The isotropic Euclidean metric and the Laplace operator The standard metric on Rn is defined by n
Ix-yl
= [
l l x j - y j\2
(a) Pf is harmonic on R~+l. (b) PE extends continuous ly to the boundary and takes on the boundary values f.
]1/2
(c)
f
Rn
IPf(x,y)\Pdx:S IIfil p for 1 ~ p ~ "".
(d) Suppose u is harmonic on R~+l and
J=l
In this example, the important first order operators are just the partial derivatives with respect to the n variables
r:. ,''', 1x 1
n
. The Laplace
s~p f1u(X,y)!PdX < "" .
244
ALEXANDER NAGEL Then if s >
° and
VECTOR FIELDS AND NONISOTROPIC METRICS
f s (x) = u(x ,s) , P(fs)(x,y)
=
245
We say that a function u(x,y) has a nontangential limit at Xo (R n if and only if for all a > 0, lim u(x,y) exists as (x,y) approaches
u(x,y+s) .
(xo,O) and (x,y) (ra(xO>. In 19CX> Fatou [4] proved: Proofs of these assertions, a long with many other of the results dis
cussed here, can be found in Stein [16], Chapter III.
Assertions (c) and (d) above suggest a generalization of the 'Dirichlet problem to certain classes of discontinuous boundary functions. For
1 :::: p :::: 00, let h p denote the space of functions u(x ,y) harmonic on
R~+l which satisfy
J
sup y>o
THEOREM 1. For 1 :::: p ::::
if u ( h p ' then u has a nontangential
00,
Rn .
limit at almost every point of
A standard modern approach to this theorem involves two rna in esti mates. The first involves the Hardy-Littlewood maximal operator. Let f (Lloc(R n) and set Mf(x o) = sup \B[-l
lu(x,y)IPdx
=
Iluli P
hP
define:
00,
00
(Rn
< ay} .
The very def inition of the maximal operator involves the family of Euclidean balls, and the proof of the crucial estimate (ii) depends on a covering lemma for these balls.
Note that if B(x O' 0) = Ix (Rnllx-xol standard Euclidean metric, then ra(x 0)
=
I (x,y)
< o! are the balls defined by the
( R~+llx ( B(x 0' ay)1 .
Thus the nontangential approach regions in R~+l are really defined in terms of the projection 17(X,y) "height function" h(x,y)
= y,
=
x of R~+l onto the boundary, the
and the family of Euclidean balls on the
boundary. We shall later see that in other examples, natural approach regions can be defined in essentially the same way.
The second basic estimate needed to prove Fatou's theorem involves the nontangentia I supremum of a function defined on R~+l. Thus for any
a > 0 and any v(x,y) defined on R~+l set Nav(xo> =
sup Iv(x,y)1 . (x,y)(ra(x O)
For Poisson integrals, this non-tangential supremum is point-wise domi nated by the Hardy-Littlewood maximal function of the boundary data:
......-..- -
r----.--------~------------·
246
~~~-~"~=~~~.;.~~F':;.~~""'-"-':.~~~~~,-.i.~i~~~ff~~:u; ....~:;.~·.;,,~' ,;.-l~~~,~t.,;:,,;~_;;;,V;~~ifv
---'n:;,'"
ALEXANDER NAGEL
..... .;;:vd:~j;f,;;;~~~~,;~-;;~~,;r££.;;;...;,~k~:~~;;i.;~~';~"";;.-;.~~~~~k';~~,jf,£_i.5i~~~~~~
247
VECTOR FIELDS AND NONISOTROPIC METRICS
THEOREM 3 (Hardy and Littlewood). For a > 0 there exists a constant C a < 00 so that if f f L l(R n) + L OO(Rn) and if u(x,y) = P * f(x), then
y for all x f Rn
Now let u f hp be real valued, and set 0au(XO)
=
lim sup u(x,y) -lim inf u(x,y)
Nau(x) ~ CaW(x) . where the limits are taken as (x,y) approaches (xo,O) and (x,y) These are the two quantitative estimates which underlay the qualita
la(x O)'
Then the following facts are easy to verify:
tive statement of Fatou's theorem. Complete proofs of these results can
(a) 0au(x) ~ 2N a u(x)
be found for example in Stein [16], Chapters I and III. However, since we
(b) 0a(uw)(x)::: 0au(x) + 0av(x)
shall appeal to this kind of argument again, we now recall how Fatou's
f
(c) 0a u(x 0) = 0 if and only if u has a limit within l ix 0)
theorem follows from these two theorems. (d) 0au(x) '" 0 if u = Pf and f is continuous.
Let p,\ .
B
.
B(xi ,K(2k+1)oi ). Then B). C B~ , and so k k 1k
IMf >,\!, and let ICE be any compact sub
!Bxl- l
k
Hence /l(I):s I k=l
type
/l(Bt). But by property (2) of spaces of homogeneous k
x
/l(B* ) < A1+1og 2K(2K+1)
1h _
The balls IB x IX(I cover I, and since I is compact we can find a subcover, which we call B 1 , "', B N . Suppose B j = B(xj,Oj) so B j has "radius" oJ" Choose B i so that 0i ::: o. for all j. We can then 1
1
/l(B.lk) .
= A l/l(B i ) . k
J
inductively choose B i ,"', B i so that 1 k (1) Bi is disjoint from Bi ,"', B i k 1 k-1 B· (2) 0i ~ OJ for all j such that B j is disjoint from B 1.... , 1k_1 1' k In this way we get a finite subsequence B i ' "', B i which are disjoint. 1 m
Thus
/l(~):SA1 ~ Il(B ik ) 0 define
n
2
(4771:) -2"e-1x 1 /4t
if
t> 0
the Newtonian potential N, the fundamental solution E possesses a
0A(X,t) = (Ax,A2t) .
It is easy to check that E(OA(X,t)) = E(Ax,A2t) = ,\-nE(x,t) .
E(x,t)
o
if
t
on Rn+1\{(O,O)!, and LE = 0 in the sense of distribu
tions. (See Folland [5], Chapter 4.) Thus if ¢ (C~(R~+1)
¢(x,t) =
Jf
so that
n
OX>
p((x,t), (y,s)) = (lx-yI4 + (t_s)2)1/4
(41TS)-2"e- 1y12 /4s L¢(x-y ,t-s) dy ds
p(O,\(X,t),OA(y,S)) = Ap((x,t), (y,s)).
o Rn The corresponding family of balls
ff t
-OX>
where L = III I
-~
n
(41T(t-S)-2"e- CX - y )2 !4Ct-s)L¢(y,s)dyds
Rn
- t1 x is the formal adjoint of L.
In order to study the operator f .... E
* f,
we would like to obtain size
estimates on the kernel E and its derivatives analogous to those for the II
Bp((x, t), 0) = I(y,s) ( Rn+1\p((x ,t), (y,s))
< 01
are now ellipsoids of size 0 in the directions of xl'''',x n , and of size 0 2 in the direction of t. Thus !Bp((x,t), 0)1 ~ on+2 .
-
~lr ~~256
- --
••
""'".......
1
III
I' I
domains. This is discussed for example in the monograph by Stein [17].
(7)
(c)
I~ (x,t)1
(d)
/a:: 1j (x,t)/ S CIBp «x,t),0)1 1
:: C !Bp«x,t), 0)1- 1
Here we recall what happens in the case of a model strictly pseudoconvex domain, the generalized upper half space, and its boundary, the Heisenberg
2
group. We let
=
p«O,O), (x, t)).
n=
Thus we obtain estimates for the fundamental solution E(x,t) which
l(Z1'''''Zn'Zn+1)
are exactly analogous to those we have for N(x) , provided we view the operator
1
257
boundary behavior of holomorphic functions in strongly pseudoconvex
III
III
~.,=:;;;:;::;;:-...::
Nonisotropic balls and metrics play an important role in the theory of
(b) IVxE(x,t)1 :S C8IB «x,t),0)/-1 p
where 0
III I
....
,,""'=:;c""'.~~~=="",.""_=,,;.="""""_,~====__~~~~-=_,.__':_.=.;~
§4. The Siegel upper half space and the Heisenberg group
(a) jE(x,t)! :: C 02IBp«x,t), 0)1- 1
1II1
!II
....,=...
VECTOR FIELDS AND NONISOTROPIC METRICS
Illi II
-"~""'''''''-==-------::;;=-~
ALEXANDER NAGEL
Moreover, it follows from the homogeneity of E that we now have: 1: I,
....
-""'"~--==---=-._-~_~-='~-y-=~=."",,--""-"=-"';-'=;"'=~-"'--~..,....:.:~".,.,.,.
-_.-.~",.,..-,,~,=,-..,~~
i
(z,zn+1)
n
en+1IImzn+1 >
I::
~
Izjl2
=
as acting like a second order operator, so in equation (7c)
we loose two powers of 0 rather than one. Recall that We can now use the general theory of spaces of homogeneous type to
n
is the image of the unit ball B
=
under the biholomorphic map
I
z 12
!. n+1!
(W1,···,wn+1)1.~ lwl \2]1 /4
Th Let p(z,zn+l)
'=
2
au,
The function d is invariant under the action of Hn · Thus if
From this it follows that Hn is a group, and
h = (u,r) f H , easy algebraic manipulations show that n
. T h = T h ·h 1 2 1
d(Th(z,t), Th(w ,s»
Imz n + l - Izl 2 be the "height function" for n so
that (z,zn+l)fn if and only if p(z,zn+l) > 0, and (z,zn+l)fan if
n
= p(z,zn+l) .
B«w,s),o) = !(z,t)\d«z,t), (w,s» < 0\
j!
then this invariance property means tha t
1
III
II
'I
II
T(w,s)(B«O,O),
0» . =
(z,t), W= (w,s) and
(u,r) are in H n with
an
d(i',w) < o,d(w;U)< 0
same as Th(O,O). Thus under the identification, H acts on an as n follows: if (z,t) f an and if h = (w,s) f H , then n
then I"
i~
.~.
., J :~ ;
I
• i,
I
=
to itself. If h = (w,s) f H , we n have identified (w,s) with the point (w,s+ijwI 2) f an, and this is the
Thus T h carries n to itself and
li\
=
We claim that d is a pseudometric. For if i'
III i
1.1
d«z,t), (w,s» .
we compute
B«w,s),O)
lill
=
If we define the balls
! < 0
2
2
.
260
ALEXANDER NAGEL
VECTOR FIELDS AND NONISOTROPIC METRICS
Hence
261
Thus the balls B(z,8) are essentially "ellipsoids" of size 8 in the Iz-u I 'S Iz-w I + /w-u I < 28 ,
directions of the complex part of the tangent space to an at z, and of size 0 2 in the orthogonal real direction, and hence in particular
and
IB(Z,o)1 ::(: 02n+2 .
/t-r+2Im(z,u>/ < It-s +2Im1 + Is-r+2Im<w,u>1
Thus the doubling property of the balls is verified, and
'S
an
(or H n ) equipped with the pseudometric d is indeed a space of homogeneous type.
+ 121m «z,u >--<w,u »1 2 28 + 21Im1
We now want to discuss the analogue of Fatoti's theorem for boundary
< 48 2 = (28)2 .
behavior of holomorphic functions in n. This problem was first studied by Koranyi [9] for domains like n, and was later generalized by Stein [17]
Therefore
to general smoothly bounded domains in en. Here we want to emphasize
II
:1 11
IIII!
the role of the nonisotropic balls on the boundary, in analogy with the role of Euclidean balls on Rn in Fatou's theorem.
d(Z,u) ~ 2max (d(z,w), d( w,li»
1
'S 2[d(z,w)+d(Vi',u)].
We begin by defining appropriate nonisotropic approach regions in n. Let 17: n --. an be the projection
What do the corresponding balls B(Z,o) look like? We see that w ( B(z, 0) is essentially equivalent to the pair of inequalities:
' IIII
II
17(z,zn+l) = (z,zn+Cip(z,zn+l» .
Iz-wl < 8
For a> 0 and Vi' = (w,s+ilwI 2 ) (an let
IRe (zn+l-wn+I-2i]-(n+1)
n 1 n 1 with c n ~ 2 - n!/17 + (see Nagel and Stein [11], page 23). In particular, for z = (Z,t) and VI = (w,s) on we have,
an
= C
2 n (i)-(n+1) [(s-t-2Im=~~+~Jb. On q forms, ~q) acts diagonally, and is given by the operator fa where
S(z,w)
a =
n - 2q. The operator C1J is not elliptic but Kohn's fundamental
work [Sa] showed that one can obtain subelliptic estimates for fa' Folland and Stein [6] discovered a fundamental solution for fa' Define:
where 0
=
d(z,w). It is also easy to check
that derivatives of S yield estimates with corresponding negative powers
¢a(z,t)
=
n+a) (n-a) (lzI 2-it)- ( -2- (lzI 2 +it)- -2
Thus S behaves like a singular integral kernel relative to this
family of nonisotropic balls, and from this follows LP and Lipschitz estimates for the operator f
->
Sf (see KoninYi and Vagi [lO]).
Finally, we consider fundamental solutions. On H
x.J
II
on functions,
ZjdZ j ,
and we extend this in the usual way to (O,q) forms on
where
o.
I
j=1
an
of
=
=
o + 2y. dta dX: j
J
Yj =
a
ay. -
a
2x j at' T
n
=
let
a
di
J
THEOREM 10 (Folland and Stein). fa¢a tions, where c a is a constant, and c a
Ie
=
°
caD in the sense of distribu if a
/c ±n, ±(n+2), ±(n-t4) ,
etc.
Thus except for the exceptional values of a, J-¢a is a fundamental a solution for fa' and it is easy to verify that
li.11111
1II1
where we write z j = x j + iy j' These vector fields form a basis for the
I
left invariant vector fields on H . Put n
I
1I
II
1
~I
z·J =!. 2
1 . 1
1
III
(X·-iY.) J
J'
Z.
J
=!. 2
l¢a(z,t)1 :S. Co 2 IB«0,0),o)!-1 where 0
(X.+iY.) J
J
=
d«O,O),(z,t)). One also obtains corresponding estimates for
derivatives of ¢a' so that again there is a complete analogy with the
1
1
II
I
II'
I I
!~
and consider for
a (
C .
estimates (4) for the Newtonian potential.
268
ALEXANDER NAGEL
269
VECTOR FIELDS AND NONISOTROPIC METRICS
In the case of the Heisenberg group, the basic vector fields are
_a__ ~L ax n + 1 ~ ax? . J=1 J
X 1""'X n , Y1""'Yn' and the vector field T which is given "weight" two. We shall later see how the general construction applied to these vector fields gives the nonisotropic pseudometric d.
(C) Let N
Part II. Metrics defined by vector fields Our object in this part of the paper is to outline the construction of metrics from certain families of vector fields. Many details of the argu ments will be omitted, and complete proofs can be found in [13]. In [7J, Hormander studied differentiability along noncommuting vector fields, and used the techniques of exponential mappings and the Campbell-Hausdorff formula. The case of vector fields of type 2 was studied in [111 Balls
y. J
=
q ~ 2n+l and let
a + 2y.--.;;a =..., ox, J
Y n +j =
l::;j::;n
01
J
a
with d. = 1
J
a
ay. - 2xj at
l::;j::;n
J
Y 2n+l = [Yj,Yn + j ] = -4
tt
with d 2n+ 1 = 2 .
Inthis case of course we are dealing with the Heisenberg group, and
reflecting commutation properties of vector fields have also been studied
we shall recover the invariant metric on Hn defined earlier.
by Fefferman and Phong [4a], by Folland and Hung [Sa], and by Sanchez
(D) (An example of Grushin type). Let N = 2, q = 3, and let
Calle [15a]. Let
n c RN
be a connected open set, and let Y1,· .. ,Y q be COO
real vector fields defined on a neighborhood of
n.
of vector fields. (1) For each Xfa, the vectors {Y1(x), ... ,Yq(x)! span RN .
cjk (Coo(Q). Here
~ c~k(x)Y~ de::;d /d k J
where
[X,y].~ XY - YX is the commutator of the two
vector fields.
-i:.
and let d j = 1 for 1:S j < n. In th is J case we shall recover the standard Euclidean metric on R n .
(B) Le t N
with d 1 = d
2
2
2
= 1 and d 3 = 2. This example leads to a metric
appropriate for studying the hypoelliptic operator See Grushin [6a1
= q = n + 1 , let Y j = ~., and Ie t d j = 1 for 1:S j with ¢(j) = Xj'
O'
