A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
E. L. S TOUT
On the multiplicative Cousin problem with bounded data
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 3
esérie, tome 27, n
o1 (1973), p. 1-17
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WITH BOUNDED DATA
E. L. STOUT
Introduction..
Classically,
themultiplicative
Cousinproblem gives
as data an opencovering
of acomplex
manifold .1~ and afamily
offunctions, fa holomorphic
onV.
such that isholomorphic
and zero-free on
Ya
n and it asks whether there is a function Fholomorphic
onjtf such that for every a, is
holomorphic
and zero-free onFa .
In anearlier paper
[9],
we showed that it ispossible
to solve themultiplicative
Cousin
problem,
on the unitpolydisc
in under thesupplementary
restrictions that the functions and F be
bounded, provided
thecovering V
isrequired
tosatisfy
an additionalgeometric
condition.Once we know the result of
[9],
we are lednaturally
toinquire
intothe
solvability
ofmultiplicative
Cousinproblems,
with boundeddata,
onother domains. The case of the unit ball comes to mind
immediately.
Themethods of
[9],
the one dimensionalCauchy integral
and apinch
of harmo-nic
analysis,
shed nolight
whatsoever on the ball case. In this note weshall show that the
multiplicative
Cousinproblem
is solvable on the ballin
CN and, indeed,
on a somewhat moregeneral
class of domains. We denoteby
the class of bounded domains X inCN
with theproperty
that thereexists a real valued function e of class
e2
defined on aneighborhood
Wof 8X which has the
properties
thatde #
0 fl W =~~ E W : LO (3)
and
H",
the real Hessianof o,
ispositive
definite on W. If we take coor-With this notation in
mind,
we may state our resnlt :Pervenuto alla Redazione il 17 Agosto 1971.
1. Annali della Scuola Norm. Sup di Pzaa,
2
THEOREM. The
multiplicatiroe
Cousinproblem
with bounded data is solvableon every doinain
of
class!BN - Explicitly, if X E ~N , if C19
is anopen cover
,for X, if for
each a,fa
is afunction holomorphic
and bounded onVa if for
all a,~B
EI,
is bounded onVa n V,8
flX,
then there is a boundedholomorphic function
F on X such thatfor
all a, is bounded and bounded awayfrom
zero onYa
f1 X.The theorem admits a
cohomological
formulation. Define a sheafQt
on Xby requiring
that if V is an open subset of~,
thenif is
compact,
thenf
isbounded on g f1
x) (1).
Thus, 9(
is the sheaf of germs oflocally
boundedholomorphic
functionsIts
global
sections arethe bounded holomorphic
functionson X,
but foran
arbitrary
open set Vc X. the elements of need not be boundedon V n X. Note that the stalks
g{;3,;3
EaX,
are neither localrings
norNoetherian.
Also, ck
is not a subsheaf of the sheaf of germs of continuouscomplex
valued functions on X(or X).
Denoteby E
the sheaf ofmultipli-
cative groups of invertible elements of
9~
If V is open inX,
The theorem
simply
asserts thetriviality
of thecohomology
groupH i
for every open cover of
~.
In this
context,
habitsuggests
that we consider the sequence of sheaveswhere Z/
is the constant sheaf ofintegers, j
is theinclusion,
and exp isthe map
f
I--~ e2ni!. This kind of sequence is usedtraditionally
to reducethe
multiplicative
Cousinproblem
to the union of an additiveproblem
anda
topological problem.
Thisprocedure
is notapplicable
in thepresent
situa-tion,
for the sequence1)
is not exact. Of course it is exact over everypoint
ofX,
but it is not exact overboundary points. If i
E not everyelement of
ê3
is of the forme,2ni!,
as is seenalready
in one dimension.If u is a bounded harmonic function in the open unit disc with unbounded harmonic
conjugate v,
then is bounded and so is itsreciprocal,
but it is not of the form eh for any bounded h.
(i) As usual, Õ (8) denotes the space of holomorphic functions on the compleg space ~’,
There is an alternative
cohomological
formulation of the result. Define thesheaf, 1Jb,
oflocally bounded, pluriharmonic
functions on Xby requi- ring
that if is open, then~b (Y ~ =
is real andpluriharmonic
on V n X andu is bounded for every
compact
set RcJ1}
Our theorem is
equivalent
to thevanishing
of ~1(X, C)J, ?b)
for every open coverct9.
The
principal ingredient
of ourproof
is theOauchy-Fantappi£ integral
as
developed by Leray [7] together
with someHP.space theory.
At severalpoints
our discussion isstrongly
influencedby
the work of Henkin[6].
I am indebted to my
colleague
Gunter Lumer for some veryhelpful
discussions of and for the
opportunity
ofseeing
hisComnptes
Rendus note
[8]
inpreprint.
I. THE GEOMETRY OF
5N
DOMAINS. We shall need certainelementary
facts about the
geometry
of domainsbelonging
to the class!BN.
The firstof these observations is the fact that if Q E
5N,
then S~ is convex.Suppose,
toward this
end,
that 0 is definedby
thefunction p
as in the definitionof the class
f5N -
Ifail,
and if B is a ballaround 3o , then e
is aconvex function on B
provided
B is smallenough,
for the Hessianof e
is
supposed
to bepositive
definite near ail.(For
all the information werequire
from thetheory
of convexfunctions,
see[5].)
It follows that B n Q === B : e
(3) 0)
is a convex set. A set which is convex is aneighbor-
hood of every
boundary point
isnecessarily
convex, so Q is a convex setas we asserted. In
fact,
we see that 0 isstrictly
convex.Secondly,
we shall need to know that an S~ E5N
can be definedby
aglobally
defined function. Consider therefore an D E5N
which is definedby
the
e2 function Lo
with domainW,
aneighborhood
of aQ. We havejust
seen that D is convex. Let us suppose, in order to
simplify notation,
that0 E S~. Let
(~
be the Minkowskifunctional
associated with Q so that~?
isthe
nonnegative
functional onC-"
definedby
The
boundary
of R is ae2-manifold,
so it follows thatQ
is ae2
functionon
(0).
That this is so is an exercise in calculus. We will take up the details in a briefappendix
at the end of this paper. The functionalQ
is
positive homogeneous, i.e., Q (t 3)
=tQ (3)
if t :~~!0,
so sinceQ (3)
= 0only if 3
=0,
we find thatdQ
is zero at nopoint
of10). Consequently,
there is a
positive
function h such thaton a
neighborhood
of(That h
ispositive
follows from the fact thatS~ =
13 : Q (~) I]-) Consequently, dQ (3)
= h(3) d~o (3) if 3
E and fromthis we find that for
every 3 E aD,
the linear transformations anddo
have the same null spaces. It follows from2)
that if t =(t1 ~ ... , t2N)E 1R2N,
then
The fact
that e
vanishes on theequality
of the null spaces ofdQ (3)
and
for 3 E aD,
and the fact that the Hessianof e
ispositive
definiteall taken
together
show that ifprovided
not all thetm’s
are zero. It followsthat HQ (3),
the real Hessianof
Q at 3,
satisfiesH Q ( j) (t, t) >
0 for all t E(0)
such that~(3)~)4~
no matter
what 3
E(0)
weconsider,
forQ
ispositive homogeneous.
.,... J . - .
° ’
so the Hessian is
positive
definite on Thefunction
defines Q in the sense of the definition of the class%N
and idof class
e2
onCN B {0).
II. THE
CAUCHY-FANTAPPIÈ
INTEGRAL. The basic tool for what we do is anintegral
formula valid in convex domains. We recall in thepresent
section the form of this
integral ;
its derivation may be found inLeray’s
paper
[7] (2).
The space ~’~ of all
complex hyperplanes (of
codimensionone)
in(:N
may be identified in a natural way with the N dimensional
projective
space~N (C).
Thiscorrespondence
is establishedby associating
to apoint
with
projective
coordinates(~o : ~1:
... :~N)
thenullspace
of the affine form$0 + ~,
Zl+ ... +
Denoteby
· thepairing
of andCN given by
functional on in
particular,
it ishomogeneous.
If~*
E ~ ~ and ro E the condition~*. ro
= 0(or # 0)
admits a well definedmeaning : If ~*
has
projective
coordinates(~o : 8~ :
,.. :~N),
then thevanishing
of the number(~o ~ ~i , ··· ~ ~N) ~ ~
isindependent
of theparticular
choice ofprojective
Icoordinates.
If we define forms
i , 1 ..
and
then
is a well defined
holomorphic
denotes the
variety
I. -
Assume
given
arelatively compact
open convex set Y in(tN. Let
be a
point Y,
and let;*: ax -+ E
be a continuous map which satisfies.. ’" - ¿:;.
If f is
a functionholomorphic
on aneighborhood
ofY,
thenf (3),Q
is(2) Added in proof. Another derivation of the integral formnla for convex domains has been given by L. A.
Aizenberg,
Integral representations of functions which areholomorphic in convex regions of
C"
space, Soviet Math. Dokl. 4 (1963), 1145.a well defined
(2N -1)-form,
and theCauchy-Fantappi6
formula assertsWe are interested in a somewhat more
special
situation. Weimpose
on Y the
hypothesis
that itsboundary
be manifold of classe2
so that forsome real valued
function o
defined and twicecontinuously
differentiableWe define a map
;0’ For this purpose,
let , >
denote thecustomary
inner
product
onand let V Q
(3)
be the vectorWe define
)
dimensionalplane
withequation
A direct calculation shows
that $* (3)
is contained in the realtangent plane T3;
it is the maximalcomplex hyperplane
contained inT3.
The definitionof ~*
shows that theprojective
coordinates(3)
are(Note
that - . , ,Also,
we havethat 8* (3)’ 3
= 0 forall 3
E aY,
and()-==0
for no ro E Y. The second of these isclear,
forI* (3)
lies inand the
convexity
of Y shows that Y andT3
aredisjoint.
It follows that in our case, the
Cauchy-Fantappib integral
assumesthe form
Let us denote the numerator of
(5) by
Yf.Recalling
that d = a-f-
a and"1-
It follows that
where the coefficients combinations of the first and seccnd order derivatives
of e. They
are,therefore,
continuous. If we write Zk =then
is
except
for a constantfactor,
If we
impose
the additionalhypothesis
that the set Y bestrictly
convex, then each of the formsand
is of the form do the element of surface area on
a Y, x
a continuousfunction.
Combining
this observation with thepreceding paragraph
weobtain an
integral representation
d X
k a continuous function.
The formula
(6)
isvalid,
sofar,
under thehypothesis
thatf
be bolo-morphic
on aneighborhood
of Y. We need it for functions whichsatisfy
less
stringent boundary
conditions. Because ourgeometric configuration
isso
simple,
it extendsimmediately
to functions isbolomorphic
inY ),
for every isuniformly approximable
on Yby
functions
holomorphic
on aneighborhood
ofdefine,
for r E(0, 1), fr (3) by
For each
i.e., , f
isholomorphic
on aneighborhood
ofY,
andfr - f uniformly
as r -+ 1. It follows that(6)
holds for functions in s4(Y).
To extend further the class of functions for which the formula
(6) holds,
we need a result of the Fatoutype,
a resultaffirming
the existence ofnontangential
limits for certain classes of functions. The smoothness condition we haveimposed
suffices for us toconclude, by
way of results ofAronszjan
and Smith[1,
Section10]
that if and admits aharmonic
majorant, p b 1,
thenf
hasnontangential
limits at almost everypoint
ofa Y,
almost every understood in the sense of the measure a,and,
moreover, these limits
belong
to LP(do).
This holds inparticular,
for allfunctions
f
forwhich
admits apiurihacrrnonic majorant (3).
Next we
establish,
with anargument
shown usby Lumer,
that certain functionsnecessarily
admitpluriharmonic majorants.
For this purpose, we shall say that a domain in~N
is starshaped
with star center 3o if formeets
ao
in onepoint.
LEMMA 1.
If
bounded starshaped
domain in(ZN,
andif f
E0 (Y )
has bounded real
part, then f Ip
has apluriharmonic majorant for every p >
0.PROOF. We suppose without loss of
generality,
that0 E S~,
andindeed,
that 0 is star center for 0. Let
9X
denote thefamily
of allprobability
measures on the Shilov
boundary
h of 0(with respect
to stl(0))
whichrepresent 0 : p
E~
iff (o)
for(~3). According
to Bochner’sr
general
version of the Rieszconjugation
theorem[3],
there are constantsC,, depending only
on n=1, 2,...
such that foreach ,u E 9N
and each- .. - , , I -"""’rr. - . - . - . -
(3) Added in proof. Since this paper was written, Stein’s book Boundary Behavior of Holontorphic Funotions of Several Complex Variables, Princeton, 1972, has appared. It contains all the results we need on the boundary properties of functions in the Hardy classes.
Thus,
iff =
u+
iv with ubounded,
and iffr (3) -- f (r~)~
0r 1,
thenfor
given n = 1, 2, ... ,
yIt follows now from Theorem 2 of
[8]
that forevery n,
has apluri-
harmonic
majorant,
and as this holds for all n, the lemma isproved.
It follows from this
lemma,
from thequoted
result on the existenceof
nontangential boundary
values and from Theorem 10 of[8]
that theintegral
formula(fi)
retains itsvalidity
for all functions whichhave bounded real
parts provided
we understandby
thef (3)
which appearon the
right
theboundary
function off.
It is in thisgenerality
that wewill use the formula.
III. PROOF 0~’ THE 1’HEOREM. As the first
step
in ourproof
of thetheorem,
we need a lemma onintegrals
ofCauchy type.
We fix attentionon
a Y E ~N
which is definedby
a function ~O as in the definition of the class5N
so thatde
is not zero at anypoint
of ~ Y and the HessianJ?Jis
positive
definite on aneighborhood
of aYe As we have notedabove,
Y isstrictly
convex, so theCauchy- Fantappi6 integral
formula is valid on it.LEMMA 2. With Y
and C)
asjust given,
letf
E(j (Y)
have bouuded realpart,
g~ be a realfunction
on a Y whichsatisfies a Lipschitz
condition.If
then
f rp
ha,s bounded realpart.
PROOF. We assume, as we may without loss of
generality,
that 0 E Y.Under this
hypothesis,
everypoint w +
0 of Y is of the form r roi for someunique r
E[0,1 )
and someunique point Wi E a Y.
If ~! Ea Y,
r E[o,1 ),
we haveAs cp
is bounded andrea~l?
the boundedness of Rej,
can beproved by proving
that theintegral
on theright
in(7),
call itI (r,
is boundeduniformly
in r and w, -According
to Lemma1,
has apluriharmonic
majorant
for everyp > 0,
so thenontangential
limitsof f belong
to LP(dg)
(at
leastConsequently, if p and q
areconjugate indices ~ I
Holders’
inequality
and the boundedness of lcimply
C some constant.
Recalling
that c~ satisfies aLipschitz condition, iv-e
seethat to prove I
bounded,
it suffices to show that ifthen J is bounded
uniformly
in r and w, . This we cando,
at least forcertain values
of p,
values whichdepend
on N.In
proving
Jbounded,
we find itnecessarily
to discuss in some detail the function( m - 3, FLO (3) >.
This function differs from the functionF(z, C)
considered
by
Henkin[6,
pp.603-4] only by
a certain second orderterm,
so much of Henkin’s
analysis applies,
mutatismutandis,
to our situation.If w =
(wl, ... ~ wN), 3
= ... ,z~), 3
near a Y so that g(3)
isdefined,
wehave
We
apply Taylor’s
theorem to the function o,expanding
about thepoint
3,to find .
If we write 1 7 then
We have assumed that the Hessian
.g~
ispositive definite,
so there existsan absolute constant V,
>
0 suchthat I
Thus,
for some absoluteconstant y
>0,
we haveFor
3 E aY
and w EY, m
in the domainof e,
theright
side of(8)
isstrictly positive,
7for g (3)
= 0and g (w)
0. ,At this
point
in ourargument,
we can invoke theanalysis given by
Henkin
[6,
p.608]
to find that there is61
>0,
such thatif 3 E a Y,
thenit is
possible
to introduce real coordinates on the ball ~S(3’ ðt)
centered at the
point 3
such that~(~)===0y ~===ly...~2~ ti
= eand t2
= 3,Ve (3) ~’
The numberb1,
if chosen smallenough,
willwork for every choice
of 3
Fix attention on a and let
J~==(~6~y:~2013~~~),
~’_ = a Y B ~+ .
We write theintegral
as the sum of two inte-grals, J+ (r, wl)
and J-(r, wl)
whichcorrespond, respectively,
tointegration
over
1+
and ~_ . The first of theseintegrals
isbounded, independently
of the choice of ~1 because its
integra~nd
is. To estimate theintegral
J-
(r,
notice that on2:-,
dg c Constt2
...dt2N.
This leads to the esti-mate,
valid for some absolute constant C anduniformly
in ~1,and then use the estimate
(8)
and the choiceof t2
in the t-coordinatesystem
In
(9)
wereplaces I e I by
zero and we pass tospherical
coordinates inthe t2 t3 ,
... ,t2n-space.
These coordinates aregiven by
Thus, (9) implies
where
C (p)
is a constant whichdepends
on po
This is finite for 1 because
N > 2.
The lastintegral
in thestring
ofinequalities (10)
is finiteif p +
2N -2N~ -1 ] -1,
and forfixed N >
2~
this isequivalent to
,,Our
argument
shows that if we choose p at the outset tosatisfy
-,.
then the
integral
J(r, no)
is boundeduniformly
in r andThus,
Ref,
isbounded,
as we were to prove.In the
proof
of ourtheorem,
we shall need adecomposition
lemma.Let us suppose
given
a domain Y EZN,
and let L be a realhyperplane
which intersects
Y,
say thehyperplane x2N == A
where we takecomplex coordinates Xi + iXN+I
... , xN+ iX2N
onCy.
Denoteby YL;6
the s-slabof
Y about
L, i.e.,
the setY : I x2N - A 8).
We fix an E>
0 so smallthat
Y B Y L; 6
consists of twocomponents-
Assume
Y+,
y Y_ to be two members ofSN
whichsatisfy
thefollowing
conditions :
ii) Y+ n
Y_ contains U n Y where U is someneighborhood
inE$N
ofthe
compact
setLEMMA 3. In the
geometric configuration just described, if f-u+iv
is aholomorphic function
onY4 -
with ubounded,
thenf = f+ -~- f
wheref+
and
f-
areholomorphic
onX+
and X-respectively
and both have bounded realpart.
PROOF. Let "P be a function which satisfies a
Lipschitz
con-dition and which has the
properties
thatLet e
be ae2
function on aneighborhood
ofY+ _
which defines thisset as in the first section of the paper. Write
f
as aCaiiehy-Fa-ntappi6
integral:
where
and
The functions
f+
andf-
areholomorphic
and have bounded realparts
onY+ _ , Moreover, f+
andf-
areholomorphic
onsubstantially larger
sets.The function
f+
isholomorphic
on aneighborhood
of the set E- _This is
immediate,
for sincewe have
function of 11) vanishes
only
on thecomplex tangent
toY+- at 3,
and thestrict
convexity
ofY+ _ implies
that thesetangents
are at astrictly positive
distance from the set
if 3
is constrained to lie in 1. It follows thatf +
isholomorphic
on aneighborhood
of J~2013.Similarly,
y if we define~+
tobe
13
E Y:X2N ¿ A -
thenf_
isholomorphic
on aneighborhood
ofJ57~..
The lemma isproved.
The
following
lemma records an obviousgeometric fact,
a fact which will enable us toapply
the lemmajust proved.
LEMMA 4. Let and let II and II’ be real
hyperplanes
in~N
which areparaLlel
and which rneet Y. LetY B
II =Y1
UY2
andY B
Let H be the vpace
of GN
deter-mined
by
IT which containsY1’
There isX,
an elementof r*N
such thatRoughly speaking,
we can flatten Y toward II withoutaltering
thepart
of Y in .g and stillstay
within the class14
PROOF. We can choose real
coordinates xi x2 , .,. , x2N
onGN
in sucha way that n and II’ are the
hyperplanes X2N == 0
and xzN = 1respectively,
and so
that,
moreover, theorigin
lies in Y.Let
Q
be the Minkowski functional associated with Y. As we saw in SectionI,
the function y =QeQ - 1
is ae 2
function on(0)
whichdefines Y as an element
of %
and which haspositive
definite real Hessianat every
point
of(0). Moreover, dy
never vanishes on(0).
Let q
be a real valued function of a real variable such that(1) q;
is. The function P defined
by
is of class
C~2
on and its Hessian ispositive
definite. Since dPnever vanishes on the set
is an element of
SN
with theproperties
we seek.Finally,
y we turn to theproof
of the theorem.The
proof
is achievedby
apatching argument
familiar in this sort of context. Assumegiven X,
and as in the statement ofthe theorem,
but suppose, in order to derive acontradiction,
that no F withthe stated
properties
exists. -Let M = max
ix 2N :
forsome 3
and let ~n be the
corresponding
minimum. Let 8satisfy
0t ’
6
Apply
Lemma 4with X
inplace
of Y and with thehyperplanes
X2N ---- and thehyperplanes
IxN --- 1 ( m - E
inplace
of 1I and II’ to obtain sets X+ and X-2
pas in the conclusion of that lemma.
By
construction the sets X+ and X-satisfy
thegeometric hypotheses
of Lemma 3.We have assumed that the theorem in
false,
and inparticular,
that thefunction F does not exist. This entails the conclusion that our
problem
is not solvable on both X+ and X- : There cannot exist
F’+
and F-in
respectively
such that for all a,F + fa 1
andF - f a -1
are bounded and bounded away from zero onVa
n X+ andYa fl
X rrespectively. Suppose,
to thecontrary,
that such functionsF+
and F-exist. The function
10
= is bounded and bounded away from zerof with Ref bounded.
Decompose f according
to Lemmale ~ - " f-!- " f- n" ,t,." +- h... -4-... v - - f-
It follows that we can define G
by
G .:F+
onX+, G
= F-c f
on X-. The fnnction G is a well defined bounded
holomorphic
function onX,
and its definition shows that onV~a
nX, Of;;1 is
bounded and boundedaway from zero. We have
supposed
that no such function Gexists,
soeither
F+
or F does not exist: Theinsolvability
of ourproblem
on Xforces its
insolvability
on one ofX+, X-,
y say onX+.
The
xzN-width
of~+,
I, e., the numbermax x2N - X’2’ 1,
the maximum taken over allpairs
ofpoints 3" 3"
in~+,
is not more than two thirdsof the
x2N-width
of X. We now treat ~+ as we treatedX, using
the coor-dinate x2N-1 rather than and we find a smaller set X+ on
which the
problem
is not solvable and which has theproperty
that theX2N-l-width
of g++ is not more than two thirds that ofX+.
We iterate this process,
running cyclica,lly through
the real coordinateof and we obtain a
shrinking
sequence of sets on which ourproblem
is not solvable. Since
c19
is an open cover, the sets we obtaineventually
lie in some
element,
sayVa ,
ofCJY,
and onV~a ,
the functionfa
is a solo-tion to the induced
problem. Thus,
we have a contradiction and the theorem isproved.
APPENDIX. ON THE SMOOTHNESS OF MINKOWSKI FUNCTIONALS In this brief
appendix
we take up ageometric
matter which should be well known but for which we are unable to cite a reference.PROPOSITION. Let Dc open set with
aS~
ainanifold of
class
l,
so thatfor
someneighborhood U of aD
and 80nte real valueddefined
andpossessing
continuous derivativesof’
order p inU,
8Q =i x
EU : o (x) = 0 ~ and,
moreover,de
does not vanish onail.
Assu1neThen Minkowski
junctional, Q,
associated with D isof
classep
onPROOF. The
function Q
is continuous.Without loss of
generality,
we can supposethat e
is defined and of classep
on the whole of1Rm
and that it vanishesonly
onaS2.
Define a function 4Y :
1R
X(1R B }0j) by
16
We see,
therefore,
that if is near is not zero :t7~ -
-
x~)
E then thetangent plane,
toBQ
at Xo has theequation
We have assumed that 0 E
S,
so since lies on one side of0 Txo .
- n -
This means that for no
xo in
and it follows thatJ2013-*- ’J
is close
enough
to~~3.
v
We now
apply
theimplicit
function theorem[2].
IfXo E 1Rm,
thenø
(xo , Q (xo))
=0,
andaccording
to theimplicit
functiontheorem,
there isa
neighborhood V of xo
in which there exists aunique
continuous map 1p: V -1R with 1p (xo)
=Q (xo)
and 0(x, y (x))
= 0 if x is in V.The p
foldcontinuous
differentiability
of 0implies
that of y. SinceQ
is continuousand -0
(x, Q (x))
=0,
we find thatQ
has p continuous derivatives near xo .As xo
is anarbitrary
chosenpoint
of1Rm B (0),
we have the stated result.Departernent of Mathematio8 The University of Washington Seattle, Washington
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