A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
E RIC B EDFORD
Levi flat hypersurfaces in C
2with prescribed boundary : stability
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 9, n
o4 (1982), p. 529-570
<http://www.numdam.org/item?id=ASNSP_1982_4_9_4_529_0>
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with Prescribed Boundary: Stability (*).
ERIC BEDFORD
1. - Introduction.
Let
(z, w) == (x + iy, u
+iv)
be coordinates onC2,
and let Dee C X R =~v
=0}
be a domain such that D X iR isstrongly pseudoconvex.
Given afunction
onaD,
we setand
given 0
onD,
we setWe are interested in the
following problem:
(3)
Given acompact
2-manifoldTc C2,
7 find a Levi-flathypersurface
such that
81~
= T.For technical reasons we consider the
following nonparametric
form of(3):
(4)
Given q eC2(8D),
find a such = rp and is Levi flat.A surface of class C2 is said to be Levi
flat
if its Levi form vanishesidentically.
As is wellknown,
a Levi flat surface in C2 of class C2 may be foliatedby complex
manifolds. We will use this characterization of Levi-(*) Research
supported
in partby
the National Science Foundation and a SloanFellowship.
Pervenuto alla Redazione 1’8 Settembre 1981.
flatness, and
we will construct the surfaceF(Ø)
in certain casesby finding complex analytic
disks with boundaries inr(gg).
Because ofglobal
difficultiesinvolving
the diskconstruction,
we cannot show that the surface constructed iseverywhere
of class C2. Thusby
« Levi flat » in(4)
we will mean thatF(Ø)
is
pseudoconvex
from both sides.This
problem
may also be considered as acomplex
Plateauproblem.
This is because the Levi form of a
hypersurf ace ~S
inC2, suitably
normalized(cf. [~]),
isgiven
at pby
the curvature form ofS, averaged
on theunique complex
line inTS~ .
Since this does not involve the full meancurvature,
the
resulting expression, equation (16),
y isdegenerate elliptic
and does notseem to fall within the scope of standard P.D.E.
techniques. (See
the discus-sion of this
problem by
Debiard and Gaveau[8].)
Let L1 =
(Q e 1}
denote the unit disk in C. Our main result is thefollowing
THEOREM 1.
Suppose
that solves(4),
and let usIf F(99,,)
contains noparabolic points
andif
all thecomplex manifolds
inare disks and
closed,
then(4)
is solvablefor all 99
and aDsufficiently
closein C2 to qo and The solution 0 has the
following properties:
(i)
ø ELip (D) ;
J(ii) for
each there exists aunique holomorphic mapping
F =
(z, w) :
which is continuouson d
and such thatand
(iii) if
V is an open subsetof
acomplex variety lying
in then Vmust lie in one
of
the disksof ( ii) ;
(iv) i f
q; EC°° ( aD) ,
then 0 is Coo in aneighborhood of
eachpoint (z,, u1)
ED
such that:(Zl, u1)
is not anelliptic point of aD,
andif (Zl’ u1)
Etor
oneof
the in( ii ) ,
thenI’’( ad )
does not contain ahyperbolic point.
’
We remark that if aD is a
2-sphere,
thenby
Theorem 3.1 of[4],
thecomplex
manifolds contained in are disks and areclosed,
and so thishypothesis
of Theorem 1 isautomatically
satisfied.Of course, we would ¡like to solve
(4)
for all 99 EC2(aD),
in which caseit would suffice to show that the set of cp for which
(4)
can be solved is both open and closed.By
Section4,
this set is closed. Thedifficulty
for us,then,
is to show that this set is also open, and Theorem 1 may be seen asa
step
in that direction.One motivation for
solving (4)
is that(4)
is related to the construction of certainenvelopes
ofholomorphy, i.e.,
the «tin cans >> withtops cut, O(l
in
(4)
in[4].
The surfaceP(O)
sometimes alsogives
apolynomial
hull.THEOREM 2. Let D cc C X R be
smoothly bounded,
let D X iR bestrongly pseudoconvex,
and let aD behomeomorphic
to S2.If Po E Lip (Do), EC2(oDo), is a
stable solutionof (4)
withproperty (ii), (i.e., if (4)
is solvable
for
smallC2-perturbations of
andaDo,
and the solution (P satis-fies ( ii ) ),
then thepolynomial
hultof
Since the solution
given
in Theorem 1 is stable in this sense,it
satisfiesthe
hypotheses
of Theorem 2. Thus Theorems 1 and 2give
afamily
ofexamples
ofpolynomial hulls,
obtainedby solving
ageneralized
Dirichletproblem. They
alsogive
the hullregularity
andcomplex
structure.The condition that a 2-manifold be of the form
(1)
is ratherrestrictive,
but if this condition is
dropped,
severalthings
can go wrong(cf.
Section 2of
[4]).
Theimbedding
of the torus T2 = 3JX 84
c C2 is in some sense«
opposite »
to the 2-manifoldssatisfying ( 1 ) .
Yet if r is close to thisstandard )>
T2,
y aglobal
disk construction is stillpossible (see [1]
and[2]).
The
points
where a 2-manifold is nottotally
real aregenerically
« elliptic )}, « parabolic »,
or« hyperbolic» (see
Section3). Bishop [7]
gavea method of
constructing
disks near anelliptic point
such that the boundaries of the disks lie in .h. In[4],
thehypothesis
that1’(g?)
have nohyperbolic points
wasadopted.
In this case isnecessarily
a2-sphere
with twoelliptic points, and
it was shown in[4]
that a disk construction can bepushed
all the way from one
elliptic point
to the other.For more
general T(p),
the method of[4] permits
us to continue the construction of a1-parameter family
ofcomplex
disks until theboundary
of one of the disks arrives at a
complex tangency. Figure
1 shows the kindof
degeneracy
that wehope
to have in the moregeneral
case: Anelliptic point E corresponds
to a «vanishing » disk,
and ahyperbolic point H
cor-responds
to a «bifurcation». Thepresent
paper,therefore,
studies thebifurcation at a
hyperbolic point
in some detail.In order to illustrate the technical
point
that we will beworking with,
let us consider the case where
F(99)
is a2-sphere
with threeelliptic points
and one
hyperbolic point.
Let 0 be a solution of(4)
with boundaries of disksfilling
as inFigure
1. To discuss thestability
of theproblem (4),
we consider
perturbations F(99’)
of Forinstance,
suppose thaty’
isany smooth function which coincides
with T
in aneighborhood of y
and suchthat the
only singularities
of away from H areelliptic.
Thenfollowing
Fig, I
the
recipe
of[4]
we may start with the curves in aneighborhood
of y and construct disks to fill1~(q~’),
thusending
up at theperturbed elliptic points
..~r’1 ~ .~’2 ,
andE’ 3-
The
possibility
which is not covered in[4]
is that we could have a per- turbationy’ of q
which differsfrom
in aneighborhood
of apoint
A E y.In this case, we may construct disks
starting
at theelliptic points.
Thedisks will be the same as in
Figure
1 untilthey
touch theregion
of the per- turbation(the
shadedregion containing
A inFigure 2).
Oncethey
touchthis
region,
it is acpriori possible
thatthey degenerate
in the mannerdepicted
in
Figure
2. In this case we would not be able to construct disks in theregion
between the curves
yi, y2,
andFig.
2In
Figure
2 the disksbelow y
= y,, areunchanged.
This illustrates the« shadow» effect and
gives
apossible explanation why
the solution is ingeneral
not smooth at ahyperbolic point:
The disks inFigure
1 that arebelow the curve y, do not « see » the
perturbation
atA,
whereas the disks above Y3 are affectedby
theperturbation.
We will show that under the
hypotheses
of Theorem 1 thecurves Y2 and y’ 3
fittogether nicely,
and as aconsequence y’ 1
coincides withy’ 2
Uy’ 37
so we are
essentially
back toFigure
1. The crux of the matter is tostudy
a free
boundary problem
which isequivalent
toshowing
that the curve y inFigure
1 ispiecewise
smooth. Thisproblem
is different from those con-sidered,
forinstance,
in[13]
and[15]
because thenondegeneracy
conditionfails at the
hyperbolic point.
This paper is
organized
as follows. Section 2 is devoted to theproblem
of the
stability
of asingle
disk withboundary
in atotally
real 2-manifoldin C2.
Stability
is determinedby
a« topological »
index. In Section3,
theelementary properties
ofhyperbolic points
arepresented.
In Section4,
somebasic results from the use of barrier functions are assembled. These are
used with more
precision
in Section 5 to control how a disk mayasympto- tically approach
ahyperbolic point.
With thisasymptotic control,
we areable in Section 6 to
adapt
the reflectionargument
ofLewy [15]
forplane
free
boundary problems
to ourdegenerate
case. Thisgives
the needed regu-larity
of y, and y3. Section 7 discusses two forms of « almost »holomorphic flattening T(cp) globally, along
the curve y, V y3. Section 8 ismainly
technicaland combines the results of the
previous
sections to prove Theorem 1.Theorem 2 is
proved
in Section 9.An
appendix
discusses the relation betweenproblem (3)
and aquestion
of
holomorphic flattening.
Anexample
is included to show that the « true »regularity
of 0 lies somewhere betweenLip (D)
andC3(D) .
In Sections 3and 6 it was shown that the
angle
0 == an of theopening of y
at H is deter-mined
by
the second derivatives ofF(99)
at H. In theappendix
it is shown that if a is irrational then y is neverpiecewise
realanalytic
atH, except
in the trivial case.2. -
Stability
of disks.Given a
complex
disk withboundary
in atotally real,
orientable mani- fold .I’ inC2,
thequestion
ariseswhether,
for any smallperturbation
r,of
7~,
there isagain
acomplex
disk near theoriginal
one with itsboundary
in ~’’ . We show in this section that
stability
in this sense is determinedby
a
geometric
index of ,h about the disk.Let Tc C2 be a
smooth, totally
real2-manifold,
and letf : C2, f E Cl ( d ) , f ’
=7~ 0 onj,
be aholomorphic mapping
withf(84 )
We note(cf. [12]
or Theorem 4.5 of[4])
that it followsautomatically
thatf
is as smoothas 7B Let us choose a smooth
holomorphic mapping
g =( g1, g2 ) : d --~
C2such that
Then -P = f -f-
wggives
animbedding
of3
X(2013 ~ e)
into C2. For an arbi-trary
smooth extensionpi
of F to aneighborhood of !
X(2013 ~ ê),
we considerf =
We will
identify
thecomplex
vector(ai
+ibi,
a2+ ib2) E
C2 with thereal vector
(al , b1, a2 , b~ ) E R4.
be thetangent
to 84 X Then there is a vector field Y =
(a, #)
such that .X and Yspan the
tangent
space tof along
84 Xf 01.
Sincef’
istotally 0,
so we find a real function 2 > 0 and an invertible
holomorphic
functionh(z)
such that =
A(z)zPh(z).
As in[4],
we will say that p is the indexof
rabout the disk
f (d ) .
If we make thebiholomorphic change
of coordinates z* = z, w* =h(z)w,
then we may assume that#(z)
=A(z)zP.
Thus X =a/a~
and Y = Re
(a( a/az)
+ span thetangent
space to at adIt follows that we may find a real
p #
0 such thatare
orthogonal
tof along
ad X{01.
We will consider
perturbations
off
of the formwhere and T: - denotes the harmonic
conjugate operator
on 3z).(That is,
if a E then there is a holo-morphic
function A + iA * on LJ withboundary
values a+ iTa,
andA *(0)
=0.)
We note thatalong
M= I(f’
istangen-
tial to
7~
and ==CI’.
Thus agives
the deformation normal to .I~and in the direction of the disk. The term
involving
g will account for the other directionorthogonal
to T.Using equation (5),
we mayidentify
ourspace of
perturbations
off
with the spaceWe will write our
perturbations
of 1-’similarly.
We may choosedefining
functions Q2 for
/’
in aneighborhood
of aJ xfOl
such thatholds on
Mapping
forward viaF
wTe havedefining
functionsr; = @(/fi-I ) , j = 1 , 2 ,
for T.The space of deformations of 1~’ may now be written as
ivhere 6 > 0 will be chosen to be
small,
and U is a smallneighborhood
of tl.A function or E ~ is identified with the surface
For 6
sufficiently small,
is asmall,
smoothperturbation
of .T =T(011
in a
neighborhood
off(84).
Now we consider the
mapping
given by
This
mapping
is of class C’(cf.
Lemma 5.1 of[11]).
We want to find a con-tinuous
mapping
a --+(a(a), b(a)) taking
A intoÐ,
with theproperty
thatFor this we consider the differential dS of S :
~a~ X ~ -~ C’~~x( ad )~, i. e. ,
the differential of S in terms of ac
and b,
with a fixed.By (5),
we see thatthe differential at
(ac, b) = 0, ~ = 0
iswhere
, ~
indicates thepairing
between 1-forms and vectors. Since-> = i~ f ’,
wehave,
forinstance,
yThus, by (6).
Similarly,
we haveThus
Repeating
thisargument,
we haveBy
thesecomputations, (8)
may be written in matrix form asis smooth of class it follows that dS: - is invertible.
Thus,
we mayapply
the usualImplicit
Function Theorem to the map-ping
~S to deduce that if 6 > 0 is chosensufficiently small,
y then for each there exists apair solving (7).
We now summarize ourdiscussion
by stating
a theorem.THEOREM 2.1. Let .fc C2 be a
totally real, orientable,
2-dimensional mani-fold
which is smoothof
classCm+2..If f :
j ---->. C2 is aholomorphic mapping
with
f
EC"’~~‘(d ),
0 «1, f ’
0 orc3, f ( ad )
c rand hasnonnegative
index about
f (d ),
thenf
isstable ; i.e., i f l~’
is a small Cm+2perturbation of .1~,
then there exists
f
closeto f
in nc~ ( d )
withf ( ad )
cP.
EXAMPLE. Let us show that if the index is
negative, then
disks are not.stable. We let
r(- p)
begiven
as theimage
of 84 X(- e, E)
under the map-ping (z, w)
-~(z,
With thisrepresentation
it is clear that the index of1~’(- p)
about the disk d is - p. We may also write.h(- p) _
with
Now let us
replace
Tby
T -E--3,
and let us suppose that there exists a diskFa(~) == (~, 0)
-~-g,,(~),
whereg,,(~)
==g2(~) )
isuniformly small,
andc
I~’a(- p) - (1
-f-- ð = 0 = If this is the case, thenholds
for 1’1 =1,
whichimplies
thatfor J. In
particular, ~ +
~ 0 on J. But ifC 1,
then thiscontradicts Rouché’s
Theorem,
so we conclude that the disk 4 X{01
is notstable in
T(- p).
.3. -
Hyperbolic points.
If M c C2 is a smooth real submanifold of dimension
2,
then .M has acomplex tangent
at p E M if andonly
if there is a linearholomorphic change
of coordinates in a
neighborhood
of p such that p =0,
and M isgiven by
in a
neighborhood
of p, whereNow we consider a
complex tangency
of a 2-manifoldF(qJ) given by (1).
We take a
point
p E and we set p = 0.Further,
we assume that wemay write 3D near 0 in the form
i.e.,
D =fr(z, u) 01,
whereFurther,
y at 0 we may writerp E 02(oD)
aswith It is
easily
seen that the2-plane
is
totally real, i.e.,
not acomplex line,
if andonly
ifThus
lix,
+=
-if311 I
measures the « distance » of thetangent
spaceof
r(cp) at p
= 0 to CP’ insideGr(2, 4).
Now we assume that
b1
= -icx1, i.e., r(cp)
has acomplex tangent
at(0, 0).
We note that if aD cannot be written in the form
(9), Le., if
thetangent
to aD at
(0, 0)
is{(x,
y,u)
E C XR: Re az =01,
then JT isnecessarily totally
real at
(0, 0).
Thus acomplex tangency
ofF(99) always
has the form(9)
and
(11).
For the rest of this section it will be convenient todrop
theterms. Now
T(q?)
has the formIf we make the
change
of coordinatesthen has the form
Now we choose z real
with
such thatWith coordinates z*
== e-iT/2 z,
we havewhere
It is sometimes convenient to
replace
oby
w** = w*+ -1 (Z*)2
so thathas the form
We use the
terminology
ofBishop [7]
and saythat p
ishyperbolic
if  >1, p
iselliptic
if }~1,
and p isparabolic
if 1 = 1. now suppose that Z > 1 and consider families of disks in the(z*, w**)-coordinates,
whoseboundaries lie in an obvious
family
of these isgiven
in terms of a realparameter
Q astw**
=or}.
The disk =01,
which passesthrough
thehyperbolic point,
intersectsF(T)
in the two lineswhere
93 and 0,
are the two solutions ofwith 0
3n/4.
If A =1,
these two solutionscoincide,
but we notethat for A > 1 but A very close to
1,
the solutions aregiven by
Now we consider smooth solutions of
(4).
If is Leviflat,
then the-Levi determinant vanishes
(see equation (10)
of[8]), i. e. ,
The condition of
being
Levi flat isholomorphically invariant,
so we canapply
it to a solution in the coordinates(15).
If we suppress the ** from thecoordinates,
y we see that D isgiven by
the surface is
given by
99 = 0. The solution 0 is notunique
near(z, u)
=(0, 0)
ifA > 1,
so we cannot conclude that 0 =: 0.If we write
0,
then we may evaluate condition(16).
It is easy to check that(16) yields øaex
= 0 at(z, u )
=(0, 0),
and thusyi(0)
= 0. Weconclude, then,
that thegradient
of a solution to(4)
is determined at ahyperbolic point
ofReturning
to theoriginal
coordinates(15),
y wesee that the condition
dv**fdu**
= 0yields
4. -
Regularity
of the solution.In this section we assume that the solution 0 of
(4)
iscontinuous,
andwe derive an a
priori
estimate on the modulus ofcontinuity
of 0.With this we may see that the set of cp E
C(8D)
for which(4)
is solvable is closed.Given a function
f
on a setK,
we define the modulus ofcontinuity
as.where h^ means the smallest concave
function > h; i.e.,
we assume ourmodulus of
continuity
is concave. Given functionsIf2 E C(D),
we w-rite,Now we recall an observation from
[4].
We supposethat q
Ehas barriers
W+, T-, i.e., Vl+ = ~- - g~
onaD,
andT+)
ispseudo-
convex. Now let M c
P+)
be acomplex
manifold with smoothboundary
3M c If at eachpoint (zo, wo)
EM,
we may write .Mlocally
as a
graph w = w ( z) ,
then it follows that(This
is seen because theanalytic
functiondw/dz
takes its maximum modulus at all. There it is boundedby
because MeIf~). )
We will consider solutions 0 of
(4)
such that(19)
for each(zo,
E there is a nonconstantholomorphic
map.f : 4 - C2
such thatIt follows that if (o c is any open subset and if F is
holomorphic
ina
neighborhood
ofCo,
then(Otherwise,
for some 6 >0, -- z2
takes an interior maximum at somepoint (zo, wo)
E w, which contradicts the maximumprinciple
onf(4 ). )
The
assumption (19)
alsogives
thefollowing
maximumprinciple:
(20)
If solve(4)
and(19)
onD,
and if onaD,
thenon D.
(Otherwise,
for some 6 >0, 02 + 3 ]z]2 - Oi
has an interior maximum at some(zo, uo).
But this contradicts the maximumprinciple
on a diskwhere
Ø1
isharmonic. )
LEMMA 4.1. Let
Pl, P2
EC(D)
begiven
withPlieD
=P21eD
= 9 and"assume that
T2)
ispseudoconvex. If
ø EC(Z?)
is a solutionof (4)
satis--fying (19),
thenPROOF. First we note that
by (19)
we haveThis is because each
point
of lies in avariety
with8 V c
T(gJ).
Since{(z, w)
E Dx iR :
v ispseudoconvex,
V must lie,in this
set,
andso ~ Similarly,
0.Let us fix
a E C X R.
It follows thatfor all p E aD such that p
+
a E D. Thus for p ED
is continuous on
D,
and the set v > ispseudoconvex..
It follows as above that 0. Thus for p, p
+
a e DReplacing
ocby - a,
we have(21).
LEMMA 4.2. Let 99 E
C2(aD)
begiven,
and let ø EC(D)
be a solutionof (4) satisfying (19).
Then there is a constant C(depending only
onD)
so thatPROOF. Since D x iR is
strongly pseudoconvex,
there exists a functionr =
r(z, u)
EC2(D)
which isstrongly plurisubharmonic.
Since 99 EC2(oD),.
there
exists ép
EC2(D)
with = cpo We may choose C > 0 so that isplurisubharmonic
on D. Thus 0 E +g~,
- Cr + and so(23)
follows from
(21).
LEMMA 4.3. Let 0 E
C(D)
be a solutionof (4) satisfying (19),
and letq?
If
D is convex withnonvanishing principal curvatures,
then there, is a constantC, depending only
onD,
such thatPROOF.
By
Lemma4.1,
it suffices to show that we can find barriers-PI, P2
above and below such thatIt is
easily
seen that it will suffice to show that there is a lower barrierTll
for the function on
aD,
for any fixedp E aD.
Forgeneral
q, we take a supremum of functionsBy
is convex, so we will take to be theconvex minorant of
- c.~( ~z - p ~ ) .
Nowby
thespecial convexity
assump- tion onaD,
there exists a linearsupporting
functionL(z, u)
for aD at p ,such thatholds for
(z, u )
EaD, and /VLI> k
> 0 where kdepends only
on aD. It follows that,since the
right-hand
side is a convex function.REMARIK. An estimate of this sort was
given
for solutions of thecomplex Monge-Ampère equation by
Gaveau[10]. Using
theargument above,
one~can
modify
theproof
of Theorem 6.2 of[6]
toyield
anonprobabilistic proof
~of Gaveau’s result.
THEOREM 4.4. Let D be convex with
strictly positive
normal curvature.that
~~
EC(D)
solve(4)
andsatisfy (19). Suppose further
that(piled
= CPi convergesuniformly
to 99 EC(D),
and there is a common moduluso f continuity a ( ) for
thefamily
CP2’... .
Then the limit 0 _ =lim
-is taken
uniformly, 0
has modulusof continuity Co)(CA/6),
and 0 solves(4).
PROOF.
By
Lemma4.3,
the functionsøj
all have a common modulus>of
continuity Further,
since convergesuniformly,
we mayassume that it is monotone
increasing. By (20),
theøj
are also monotoneincreasing.
Since there is a common modulus ofcontinuity,
the conver-gence is uniform. It is now clear that the uniform limit 0 solves
(4).
THEOREDZ 4.5.
Suppose
thatoj E C2(D)
solves(4),
and let us setthen there is a
subsequence f 0,,l
c such that 0 =lim
I exists andbelongs
toLip’ (D). Further, 0
solves(4)
andsatis f ies (19)
and has pro-perty (iii) of
Theorem 1.PROOF.
By
Lemma 4.2 and theproof
of Theorem4.4, P
ELip’ (D)
exists and solves
(4).
Now let us consider acomplex
manifoldlying
in thegraph ~(~;).
If we write itlocally
as w =w(z) ,
thenby (18)
we have theestimate
where k
depends only
on D. Let us fix apoint (zo,
Then thereis a =
Ae
suchthat
the manifold~; passing through (zo,
may be written as z.a =f;(z),
wheref;(z)
issingle-valued
andanalytic
on
Further,
E > 0 isindependent of j. Since
isuniformly bounded,
and it follows converges
uniformly
toThus
~~z, f(z)):
is acomplex
disk inI~(~), containing (zo, zvo),
andso
F(tP)
satisfies(19).
The
property (iii)
of Theorem 1 is a consequence of thefollowing lemma,
since we may move to a
regular point
of V.LEMMA 4.6. Let 0 be an
arbitrary Viy V2
are varie-ties in C2 at
(zo, VI nonsingular
andV~1
uV2
c.I’(~),
thenVI
=Vi.
PROOF. We let a be the intersection of the
(z, u)-projections
ofT~1
and
V2.
Let us write theprojection
of as{u
=~(’s)}.
isa harmonic function
vanishing
at(zo,
Thus a ==ul
is a 1-dimen-sional real
analytic
curve. SinceVI U V2
c1’((/»,
it follows thatlies in both
V1 and V 2.
SinceVI
UV2
contains a 1-dimensionalset, Yl
andY2
must have a
component
in common. ThusYl = Y~ .
5. - Barrier estimate.
Now we look at the estimate
(22)
in aneighborhood
of ahyperbolic point.
Let us assume that for a fixed solution0,,
we have upper and lower barriers P- EC1(D) .
Thatis, P+laD
= = cpo, and the domainA(W-, P+)
ispseudoconvex.
We note that thequantity
will be taken to be small.
Let U =
f~
E C:1m ,> 01,
and let us consider aholomorphic mapping f :
U - C2 such thatLet us write
f (R)
= y andy±
=f (R±).
We write z = s + it and set=
(z(C), w(~) ).
If r has the form in(12),
then for zsufficiently small,
It follows
that, taking
thepartial
derivative withrespect to s,
we haveNear
aD, (17) gives
usThus the barrier estimate
(22)
isequivalent
tofor
(z, u) sufficiently
close to 0. If we differentiate(27)
withrespect
to twe obtain
By
theCauchy-Riemann equations, izs
= z, andiws
= wi. Thuswhere the last line is obtained
by substituting vs
from(25). Similarly,
wesee from
(25)
thatLet us consider the vector field
which is defined for Z E C and has a
singular point
at z = 0. Sincei.e.,
0 is ahyperbolic point, Z
has index1 at z = 0. Thus we havewhere (., .)
denotes the usual Euclidean innerproduct
in theplane Similarly,
Thus the estimate
(28)
becomesGeometrically,
y this means thatNow let us such that
tan k = - b,
and let us setThen
(30)
isequivalent
towhere E =
E(x)
is a small numberdepending
on x. Wedepict (31)
geo-metrically
inFigure
3. The vector field .X isdepicted
on the unit circleIzl - 1,
and theangle
about Xcorresponding
to(31)
is indicated.Fig.
3Let us note that we may write
since
tanu - b.
Now thepoint z;
witchargument 8~
satisfiesThus
which
yields
Thus,
asexpected,
the solutions obtained here agree with the solutions ob- tained in Section 3(for
the*-coordinates).
Inparticular,
we conclude thatLet us use the notation
Let
~i, 52>
0 be theangles
indicated inFigure
3. We may choose x > 0(and sufficiently
small thatNow we consider the
possibility
of asimple
curveapproaching z
= 0 whileunder the constraint of
Figure
3. It is clear that asimple
curve y =f (R)
can reach the
origin only
if it remains within-A (622013~Ss-j-~g)
or+ ~ -
~Z , 82
+ x+ 52) when If I
is very small.Similarly,
it canonly
exit
through A(01- ~1, 61-E- ~1)
orA(01 +
n-~51, 01 + n
+61) -
In the
proof
of Theorem1,
the disksatisfying (24)
will arise as a limit of a1-parameter family
ofdisks,
all of which aresubject
to(31).
0let us write
By
theLipschitz
estimate(18),
the disk isgiven
as agraph
overLEMMA 5.1..F’or q > 0
sufficiently
the diskf U11)
isgiven,
nearthe
hyperbolic point (0, 0),
as agraph I(z, w(z)):
z E where thef unction
wis
holomorphic
and w’ is continuous onlJ11.
PROOF. The fact that
f ( Un)
is agraph
over follows from the discus- sion above.Further,
w’ Eby (18). By
thearguments
of[4],
w issmooth at all
points
of For z near0,
we use(25)
to see that forz E
Thus it follows that
and thus
Now we make our essential observation
concerning Figure
3.LEMMA 5.2.
If
y-approaches
0 inside theangle A(82 -- ~2, O’~’
-~-c~2),
then
y+
exitsthrough
theangle A (0, - ~1, 81
-I- and theregion S~~
lieson the side
of
ycontaining
theangle A(01 + O2- ~~ ~ .
PROOF.
By
thepreceding discussion,
it is sufficient to show that there is a contradiction if we assume y exitsthrough
theangle
about81
+ ~.Since
f
preservesorientation,
y theregion
contains theangle running
from
()2
-6,
to81
--~- ~ +6,
andcontaining
theorigin.
Since we havepotted
that 0
62
-81
it follows that thisangle
isgreater
than ~c, and thuswe can find
points Lx cQn
such thatSince
= u(C) ED
for it follows thatThus it follows that
for z in the circle
(32),
and whereh(z)
is the harmonic function withboundary
values zz on the
boundary
of the circle(32).
Now h reaches its minimum atz = 0,
soh x ~ 0
there. Thusby
Lemma5.1,
we maycompute
On the other
hand,
we havesince u
+0( IZ12 )
holds on y.This proves the first
assertion;
the second statement follows becausef
preserves orientation.
We conclude that the
region
is either located as inFigure
4 or isthe mirror
image
of it below the x-agis.Fig.
46. - Reflection
principle.
As in the
previous Section,
we consider acomplex
diskf :
U - C2satisfying (24).
Let us assume thatF(92)
hasexactly
the form(12) (without
any
0(lzI2) terms).
Let be theregion
as in Lemma5.1,
and let ussimply
write S~ =
S2,n.
Let us also writey±
=y~
and U = Thus we mayassume that y
approaches
z = 0through
theangle A(02 - ~2, 82
-f-~2)
and
y+
leaves z = 0through
theangle A ( 81- ~ 1, 6~ -E- ~ 1 ) . Finally,
let uschoose x > 0
sufficiently
small thatfor
some 3 > 0.
This ispossible
since00220130iyr/2.
Now let us use the coordinates of
(15). Suppressing
the*’s,
we havethe surface
given by
By
Lemma5.1,
isgiven
as agraph
of a function g : Q -C,
andThus
is
analytic,
andg(z(I) )
is realfor ~
E R.By
the Schwarz reflectionprinciple,
g(z(C))
isanalytic
in aneighborhood of ~ = 0,
andsince
g(O)
= 0.We will use the
following
result onLipschitz regularity
for conformalmappings.
LEMMA 6.1. Let
star-shaped regions of
theform
where
kj
issmooth,
andIf
F: (VI - W2 is aconformal equivalence,
then _F’ ELip’~ (~.~1),
wherePROOF. At each
point
zo E8mj
we may constructcomparison
domains.After
translating
androtating coordinates,
y we may assume that zo = 0 and thereexists ij>
0(independent
ofzo)
such thatIf
H~(z)
is a function on a), withH; C 0, Hj == 0
on and0,
L1HjE
thenusing
thecomparison
domainsabove,
we havefor
0 y ~. By
a similarcomparison,
we can bound thegrowth
of thenCarath6odory
metric:In this
estimate,
y isapproximately proportional
todist (y,
Since
C1(z; ~)
=02(f(z); f’($) ) ,
it follows thatThus we have
If
Hj
is a function asabove,
thenby (36), H,(z)
is estimableby
a certainpower of the distance to the
boundary.
Since~s(/)
isagain
a function of the form we have"There r
=(1-
2arctane)/(1 +
2 arctane).
It follows now thatf
ELip" (&1)
which is the desired result.
LEMMA 6.2. Let be the
conf ormat rnapping satisfying (24).
Thenz f
x ~ 0 is smallenoitgh,
2ve havef or
somePROOF. Let us consider the conformal
mapping
given by H(z)
---- It follows thatH(y+)
lies in the sectorflarg zl 26,,n/(O,
--0,)I. Furthermore,
since .H preservesangles, Figure 3.
gives
for 0
~
~. We have a similar estimate for - q~
0. Sincez(~)
is.smooth
for ’* 0,
we may shrink U so that the domain m =satisfies the
hypotheses
of Lemma 6.1. Thus we haveH(z(~) )
Ewhere we may take ë’
arbitrarily
smallby choosing x
> 0 small. Itfollows then,
thatThus we have
where v2
=( 1 /c ) ( 6 - 81 ) ( 1- ’ ) 2 by (33).
Now we have the other ine-quality by applying
Lemma 6.1 to the inversemapping f(z)
=H -J(z(C)).
This
gives
the reverseinequality
with~i==(l/~)(0s20130~)(12013~)"B
whichcompletes
theproof.
Let =
X(O)
be the functionhomogeneous
ofdegree
zero such thatIn the *-coordinates of
(15),
which we areusing
thisSection,
the lineslarg z
=6J, j
=1, 2,
inFigure
3correspond
to the lines{arg z
=nj2
+as in
Figure
4. These lines are also theset where X
= 0.It follows from
(35)
and Lemma 6.2 that for- r~ ~: ~ c r~
Thus for some 6 > 0 we have
and thus
: Since
~’(6)
~= 0 for 0 =nj2 +
a, it follows that the curve y- isasymptotic
to
(argz
=()2} (in
the oldcoordinates)
andfor
- ~ ~ s C 0.
Thecorresponding
statement also holds fory+.
LEMMA 6.3. We have the estimate
)g(z) ) C f or
all z E Q and some~~>0.
PROOF. Since
g(z)
is bounded on,~,
it isgiven by
the Poissonintegral
formula of its
boundary
values vanish at z = 0 as in(37).
By using comparison domains,
as in theproof
of Lemma6.1,
we may show that the harmonic functionh(z)
on S~ withboundary
values~z~2+a
is estimatedby 0 C h(z) C c ~z ~ 2 + a’
for some~’ >
0. Thisgives
the desired estimate.Now we use a reflection
argument
motivatedby arguments given
in[13]
and
[15].
Sinceholds for we consider the
complexified equation:
,Solving (40),
we have two solutionsand we define
Clearly .R~
is an anti-conformalmapping. Geometrically,
yBi
reflects Qabout
y±
as indicated inFigure
4. If thenRi(z)
= z, and if z E y,1~2(z)
= z. In factand thus
Ri(y-)
asasymptotic
to the rayfarg
z =~c/2 - Similarly
is