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. on p. J
each connected component V of fl = f +c. c. 0 at every point
> 0, so that
sup u( t;.). t;.. au
=
Suppose that ~u>o everywhere and that u(a) = sup u(t;.), a.U, t;.. au for U
CC n ;
let :~~ u(z) = u(zo)'
g(t)=u(xo,t)
Then zo' U and the function
has a maximum at t=yo
(zo=xo+iyo)'
Hence
+{g(y +h) + g(y -h)- 2g(y))
lim h-O
h
0
a
0
~O.
a2u In the "arne way - 2 - (z ) ~ D, so that ~u(zo) ~ 0, contradiction. ax
Proposition "l.
...!...R 2"
(pe e
pe
ie
a Let a. 0 and p(z, S) = P
+(z-a))
i8
for
< p, 0:5. e ~ 2".
1 z-al
a,p
(z,8)
Then P(z,8)? 0
- (z-a)
and if h(e) is a continuous function with h(O) = h(2,,), 2"
lim z-z
[
P(z,9)h(e)d8 = h(S ) 0
0
a
the limit is uniform with re spect to 9 0 Proof.
We may suppose that a = 0, p = I, and that
h{e) =
-00.
Let·f! > 0 and 0' = {z'
E
00,
suppose first that
01 u(z'} < u(z}+' }; 0' is an open
neighborhood of z, and contains a disc
Iz'- zl < Ii.
Let ko be such
Then u(z'} - klz'-zl ~u(z'} 0, then
= {Zl E 01
1z, - z 1 < Ii,
90
u(z} < -c} that
'\.{z} ~ max{-c, M-kli} as before, and utJz} -
-00
as k -
00.
0',
00.
34
Definition 4. function on n.
Let fl be an opcn set in 0::
Suppose that u is ;/.
and u an u. s. c.
on any connected component
-00
We say that u is subharmonic if the following condition is
of fl.
satisfied. For any open U
CC n and any continuous real valued function h
on U which is harmonic in U, if u(z):>, h(z) for z. BU, then u(z) :>. h(z) for z. U.
1.
Remarks.
1= (a, b) C IR, the solutions of Ah =
a where
d2
A = - - are the linear functions h(t) = at + 13. Functions u on I dt 2 such that u(t o ) ~ h(to )' u(t l ) ~ h(t l ) implies u(t):>. h(t) for to:>' t $. tl (where to < tl belong to I and h is a linear function) are precisely the
~
functions.
Subharmonic functions may thus be looked upon
as complex analogues of convex functions. 2.
Definition 4 may be reformulated as follows.
For any open U
C efland any h harmonic and real valued on
U,
the maximum principle holds for u - h. Lemma 2.
Let u be subharmonic on the open set fl Co::.
We
have (a)
The set {z< nl u(z) = -oo}
(b)
If a< fl and p>O is such that {ZE o::/Iz-al
contains no non-empty open set.
j"
!u(a+ pe i9 )ld9
. p}, we have Ken, u(Zo) >
-00
for some zo' K, u(z)
subset of BK.
Let {uk}
= -00
for all z in a nonempty open
be a sequence of continuous functions decreas-
35
decreasing to u on (a neighborhood of) K (Lemma I).
j zrr Pa,
h (z) = k 0
p
'9 {z, 9)ll. {a + pel )d9. I
-00.
As in (a), we have
$. {
rr
o
'9
P
$. C.M
a,p
{z, S)ll. (a + pel ide I
0 such that
ural $
Proof. (i)
i9 2TrI [2Tr 0 u(a+ pe )d9
n
We may suppose that
E
«:
n.
Ilz-al $ R} C
Let {'\.}
o K, 0 < P < R.
~(z)
Then
= '\.(z) ~ u(z)
for
~(z) = in particular, since
(ii)
o
~ Z E
be a sequence of continuous
~(z) =/Tr P (z,8)'\.(a+pe iB )d8, o a,p
functions decreasing to u on K, and E
is connected.
Suppose that u is subharmonic, and suppose that
K = {z
z
for 0 < P < R(a).
is continuous on K, harmonic on K and BK.
!ff o
P (a,B) '" a,p
Hence u(z) ~ ~(z), P
a,p
I
Z E
K.
•
z
(z, 8)u(a + pe i8)d8
'2" '
ural < -
Hence
~
E
I
I [21f i8 u(a+pe )d8. 'IT 0
'2
For the converse, because of Remark 2 after Definition 4 and
Corollary I to Proposition 3, it suffices to prove that if u satisfies condition ( >1
z. U u(a)
=
n,
and suppose that u(z ) > sup u(z), o ZE au
sup u(I;,). 1;,. au
sUE.. u(z).
Z
U.
E
Since u is u. s.c., there exists a.
Clearly, a
z. U
lau,
so that a. U.
Then
0
U
so that
We shall prove that,
if (*) holds, u is constant on the connected component V of U con-
taining al this is impossible since u(a) > sup u( 1;,) 1;,. au Let E E is closed.
= {z
E
V I u(z)
= u(an.
~
sup
u( 1;,).
I;,.av
Then E i~, and since u is u. s. c.
It suffices to prove that
E is open.
Clearly u(a) >
-00.
37
Let b. E and R=R{b»O be so that {z.ctIJz-bJ~R}C V aDd
1
u(a) = u(b) ~ Z'I\'
u{b+pe
i9 0
)~u(a)-f
tIT
u(b + pe
o
i9
)d9 for 0 < p ~ R.
,E>O, then u(b+pe
nonemptyopen subset I of [0, Z'I\'].
i9
If
) 0, u{z) = i f{z) I a
Proof.
is subharmonic on
If f{a) ~ 0, condition (,:,) at a
zeros in a disc
p
is trivial.
about a, then u(z) = I g(z) I where
is holomorphic in D.
n.
If f
has no
g(z) = e a log f(,,)
Hence
g(a) =
21 "
j'211 g(a + pe is )d6, 0
hence uta) = Ig{a)1
(iii)
If u
:s.
j2.".u{a+pe '6 )d6.
1 12'!l' '9 1 2'!l' 0 [g{a tpe 1 )ld9 = 2.". 0
is continuous on
n
and, for some
harmonic on 11 - {a}, then u is subharmonic on Proof. For any E> 0, ue{z) on
n:
= u(z)+e
uf
n,
u
is sub-
n.
loglz-al
condition (",) is obviously verified for
a.
1
at
is subharmonic a, and also at any
40 point
f
a since, in a neighborhood, u and
Hence, for z
f
a,
1 z-a 1
< p,
p small,
u(z) + tlog Iz-al $. Since u and u
Jof" Pa,p (z,e)ue (a+pe i9 )de.
are continuous for z
1o
u(z) ~
log 1 z-a 1 are subharmonic.
f
a, we get, letting
"e
2l!
P (z, e)u(a+ pel )de a,p
for
E-
0,
Iz-al < p , z fa.
By continuity, this holds also for z = a, and the result follows.
Remark.
The same result, with a similar proof, applies when the
single point a is replaced by any countable subset of
n.
Proposition 6 (Hadamand1s three circles theorem). and
n = {z E «: 1 0
0 and, for
Then log M(r) is a convex
10 g M( e t) is a convex function of t.
Let u(z) = sup
log
1f(ze ia)
I.
Then, u is continuous
aElR
on n, and u(z) = log M( 1z I).
By Example (i) above and Corollary 3 to
Proposition 4, u is 9ubharmonic in
n.
Suppose now that log M(r) lO.l (log r) where 1
for
is a linear function,
r = ro ' r l ,
1 (t) = at + f3.
log M(r)~l(log r) for ro < r< r l • u(z) $
h(z)
for
ZE
au, u
0 < ro < r l < R, We have to prove that
Now
= {z
E
nl ro < Izl < r l }.
where h(z) = a log 1z/ + f3 is harmonic in n.
Since u is sub-
harmonic, this inequality holds in U, and our result follows.
41
This result can be written
Remark.
where log r l - log r log r l - log ro The following proposition is one of the main steps in a fundamental theorem of Hartogs.
Proposition 7 (Hartogs). let {~}
= {z
Let R> 0, D
E I[;
J JzJ< R},
be a sequence of subharmonic functions in D.
and
Suppose, in
addition, that the following two conditions are satisfied: (i)
There exists M> 0 such that ~(z) ~ M for all zED and all k.
(ii)
lim u. (z) ~ m k-oo 1
In particu1ar, if (ii) holds for all zED, the conclusion holds for all
r< R. Proof. .. (E)
O.
(a) Let
Then there exists E C [O,21fJ with
denotes Lebesgue measure) and ko such that
< m +e
if 9
I
E,
k ~ ko •
In fact, let
00
Since lim ~(z) ~ rn
for
JzJ
= p, we have
n
k= 1
Ek
=~.
Now
42 Ek C E k +1 • E = Ek . o (b)
Hence. there exists ko such that ",(E k ) < f
Clearly. (a) is satisfied with E
This is the proposition.
If we set C = sup Po have
~(z):s.
.P
Lr~(pel'a )de
:s.
+
MC",(E)
f
~
f
Po .p
+ (m H)
< P.
EI=[O,21T]-E we
(:t.. a)~(pel'a )da
Po
E'
E'
:::. MCe
1z1:s. r
Izl:s. r.and
(z.a). e [ [0. 21T].
C
and ko as obtained here.
We have. for
E
.p
(z. e)(m +e )de
J
since
E'
•
Po ,p
k.2:. ko •
(z. 9)d9 :5.{1TPo (ZI e)de 0 ,p
= Thus m k (r):5. MC l' + m +fRemark. IZ I
let
I
o
• k.2:. k o ' and the result follows.
If h is a function continuous on
< p, and if
1.
1 Z 1 :::. p and harmonic in
lim u. (z) :::. h(z) for all z. 1z 1 = P. we have. for r < P. k-oo I
ct subsets of D' X On' and the a O! are holomorphic on D n' By Cauchy's inequalites (Chapter 1, Proposition 3),
M =
On the other hand, for holomorphic on 0' pact 8ubbets of
I{lnl
sup zeD'XD
n
n
{Zl < C!;nllz.1 < R-o).
is
Hence, in particular, for each
A(zn):> 0 such that
n
V it (
IN
n-l
F rom these two relations, we conclude that
(b)
.... it,,)
so that the series (S) converge", uniformly on cOIn-
la(z)I5.. A (z)p- l a l • O!
V ( j . INn - l •
< R. the function (zl ••• · ' zn_l)
J
Zn' there exists
If(z)l,
Yz o t on .
since
p < R-li .
47
e>
Hence, by Proposition 7, if
0, and r < p there exists ko > 0
80
that we have
laQ'(znll
1
(p - H)
$.
1".1
for
IZnl s, rand lal
Hence, the series {S) converges uniformly for
~ ko •
Izn I $. r,
Since (; > 0 is arbitra:ry, and r < p is arbitl'ary, it follows from Weierstrass' theorelll (chapter I, Proposition 5) tha.t £ is holomorphic on {"E
0
and
rp> 0 be so that
Irp(z)1 < '1 for
'1< IWol < R-'1.
Then f is holomorphic in the neighborhood of (z,w o )
for
Izl
p for all z near zo' where
Irp{zo} - wol < p < R(zo).
vergence of the series for
p is a number
This would imply the uniform con-
z near zo' I w I < p; in particular, f would
have an analytic extension to a neighborhood of the point (zo' rp{zo» since
Irp(zo} -wol < p, which is impossible by assumption.
Thus we
have Lemma 2.
The Hartogs radius
R of the series (Ho) is given by for
Proposition 4.
is harmonic in
I z I < Ii
•
If Ii, '1 and rp are as above, the function
I z I < Ii for any w
o
59 Let u(z. w 0) = - log I . - ",(a + >,(zo - a» is holomorphic in the
= {>. ( Cia + >,(zo - a)
• p}.
Since the limit of holomorphic
6i functions is again holomorphic, it Buffices to prove this for Zo in a dense subset S C p.
We may suppose that
p={lzl 0 being such that
< p}C D.
Consider now / . the k-th iterate of f defined by f1 = f, fk = f . / - 1 •
Suppose that f{z)
i
z.
Let :-< be the smallest integer
such that f(z) = z
+
PN{z)
+ ••.•
One sees easily. by induction on k.
that
67 Since
«
is a map of D into itself, our remark above shows that the
coefficients of kPN(z) are ::;. Rp -N in absolute value.
Since k is
arbitrary, this implies PN ;: 0, a contradiction. Definition Z.
A bounded domain DC en is called a circular
domain if z. D and 9
E
m.
implies that e
Proposition Z (H. Cartan). and f = (f i , ... , in) • Aut D.
i9
z ( D.
Let D be a circular domain in en
Suppose that 0
E
D and that f(O) = O.
Then each f. is linear, i. e. , J
a ..• C. lJ Proof.
For any holomorphic map g:D'" D, we denote by
dg = (dg)O the linear map of en into itself given by
Let 9 ( IR, and k9 k9 is k -9.
Let
tp
E
Aut(D) the map z - e
i9
z.
Then the inverse of
be the inverse of f, and let g
=
k _ 90 tp 0 k9 0 f .
Since k9 and f, hence also k -9 and rp leave the origin invariant, we find that
Now,
dk9 = e i9 id. is a linear map of en into itself which commutes
with any linear map of (;n. Since clearly dk _9 0 dk9
Hence dg
= id. = dtp. df.
= (dk -9 odk9 ) 0 (dtp odf) = Hence g(O)
=0
identity.
and dg
= id.
It follows that the expansion of g in a series of homogeneous polynomials about 0 has the form
68
g(z}
= z + P 2 (z} + •••
so that, by Proposition 1, g(z} '" z. k9°
{=
This can be written {oke
Hence, if £ = (f1 ,···, £n)' we have f.(e J f.(z} = J
i9
for all 9. IR •
z}
= e i9 f.(z}. J
If
a za , we deduce that a
a. INn e
This implies that aa Definition 3.
i9
=0
a
for all 9. IR •
a
if
lal
Ii,
and the result follows.
A holomorphic map f: D - D', DC 4;n, D' C (;m
is called proper if, for any compact K'CD', the set f- 1 (K') is compact in O. Remark. Moreover,
{z)
C
Trivially, an automorphism of a domain 0
is proper.
f:D - 0' is proper if and only if for any sequence
D which has no limit point in D, the sequence {i(z)}
has no
limit point in 0'. We shall now see how these results can be applied to determine all automorphisrns of a polydisc. Proposition 3.
Let 0 = {z. (;nllz.1 J
< 1}. Then, for any
{. Aut{O}, there exists a permutation p: (1, •.• ,n) - (1, .•• ,n) of the integers from 1 to n, real numbers 9 1 , ••• , 9 n < IR , and complex numbers ai, ... ,an ,
Ia. I < J
i, so that
69 Z - a n n) la.l< i. 1-az • J nn
Proof. rJ'
a
Aut(D) for
E
tng f by
0"a,"
that if f(O)
la.1 < 1 and. for Zo [ D. O"z (zo) = O. J
Hence replac-
0
f. a = £(0). we may suppose that f(O)
= O.
Then
= O.
We shall show
then
By Proposition Z. if f
= (£1 •.••• f n ).
we have
n
\(z) =
L
j=1
Moreover. if I z.1 < 1. then J
ak·z. • a k ·• Q; J J J
I\(z)i
.~
r < 1. if we choose Zj
= reI
< 1 (since feD) C D). •
j where akj
= I~j Ie
Hence. for
-l~·
J. we get
n
~lak.l..) = pea + >..(j3-a)) V is connected.
• >..
E
V •
We set • >... V' • >.. ( V-V'
We shall prove below that u is subharmonic in V. follows from chapter 3. Lemma Z that u p(a+ >..(j3-a»= 0 for a. Pf"'I O. fl. U -
~
-00
It then follows
on V. i. e ••
D and>... V. This clearly
implies that p= 0 on 0 {"\ p. hence on O. To prove that u is subharmonic on V. it is sufficient. because of chapter 3. Proposition 4. that u is subharmonic on the open set V· - {>..
E
v'l
u(>..) = -oo}. = V" say.
1
aZu a>..arNow. if f1 •••••
~
aZ
L --
V" we have
On
k
log
,,=1 a>..ar-
2: IJ. =1
1'1'
(aH(j3-a)) I
Z
f1 v
are holomorphic on V" and not simultaneously O.
we have
aZ a>.. aX"
which is
~
0 by Schwarz's inequality.
This proves our assertion. and
with it. (*). Proof of Theorem Z. converging to a point
wO (
Let {w"
J
be a sequence of points of Oz
(aDZ)" U z• Then. there is a subsequence
{v k } so that if '1'. (z) = f.(z.w ). then {
have a distance 2!. I zn I = 1
+" > 1
0
IZII :::'6+£, IZjl:::.e,
sufficiently small (since these points from
0 and B has radius I).
Hence g can be continued holomorphically to a neighborhood of Zo (by chapter 4, Theorem Z) which is absurd since g -
as
CD
z -
z
o
Theorem 3. W = WI
x
W Z' Wj
= I, Z
A is dense and
.
Let D = Dl
X
2 D Z C (au)
p .... V =
Now, if Vo is large enough,
o",v(U)
y
Ii
'if
y
"y(au)
c "v(au).
1£
'I'y(a) , p.
{o"y(Un ('\ P
C V for
v
is
"y
We claim now that
Since q>(a) , P ,,(a)
i
1"\
pJ
and ",)a) - ,,(a), if
y
is
"y(U) byassuITlption.
V {(C n - " y (U)) "
and each of the open sets above is nonempty
pJ
["y(a) belongs to the first,
q>(a) to the second] contradicting the fact that P
Ii
Since
we would have
P = {q> y (U)
{O'l')U)} .... P ~
2:. Yo'
Hence q>v(U) is a relatively compact
On the other hand,
= Ii,
about ",(a) so that
is large enough (which would contradict (~,)
above and so end the proof). large,
Hence there is a neighbor-
Ii.
open set in C n with 8tp (U) C V. v
{8tp (U)} '" P ~
",(I~U).
and a polydisc P
(*)
open, we have
i
and the proposition is proved.
is connected.
Thus
80 Corollary (Hurwitz's theorem). in
.:n
and {f} v
Let rl be an open connected set
a sequence of holomorphic functions on rl, converging
uniformly on compact sets to a holomorphic function f. f)z)
f
Then if
f
0 for all v , and all z, and f is nonconstant, we have fez)
for all
rl.
Z E
Suppose that f(a) = 0, a
Proof. about a.
o
on
D
= {},.
0
Then f
rl E (;
$
since rl
Ia
Let "'v(},.)
f
",(0) = 0, tp(l) = feb)
tp v
p}.
E
= f)a+
be a small polydisc
Let b
E
P,
fib)
f
Let
O.
},.(b-a)), ",(},.)
=
f(a +l..(b-a)).
0, so that tp is nonconstant on D.
Then
Hence for
is also nonconstant, hence an open map of D into
::>
{O}
if
V
I(;
By Proposition 5 above,
is large, a contradiction. (i) => (ii).
Proof of Theorem 4. (i) => (iii).
Let P
Then D is a convex, hence connected
(by chapter I, Proposition 4). f)rl) ~ tp)D)
rl.
(since if it were, f would be
is connected).
+ }"(b-a)
open Bet in (;.
large v ,
0 on P
E
Obvious.
If f e Aut (D) and a e D, and
g = £-1
E
Aut (D), we have
go f = identity, hence (dg)f(a) o(df)a (iii) => (ii).
If (df)
a
= identity,
so that (df)a is invertible.
has a nonzero determinant, then, by Lemma 3,
feD) contains a (nonempty) neighborhood of f(a). hence feD)
(ii) => (iii).
Clearly fin)
CD.
Hence, if (ii) holds, feD)
Let a ( D be so that f(a) = b • D. a subsequence so that {g v}
Let gv =
(1 ,
¢.
aD.
n D I £I.
and let {v k } be
converges uniformly on compact subsets
k
of D to g: D - en (MonteI's theorem, chapter I, Proposition 6).
81
We have g(b) = lim £-1 (£(a» • k-oo "k Moreover, if k is large, £ (a) is close to f(a), hence in a compact -1 f" k
Since
subset of D.
"k converges uniformly on compact subsets
of D, we deduce that
-I
g(b) = lim £"k (f"k(a» k-oo Thus g(b) = a
E
D.
=
lim a = a. k-oo
Let V be a small neighborhood of b.
Then
g(V) lies in a compact subset of D, hence, there is K compact in D 80
(V) C K (k large). Then, for x E V, we have "k (since g (V) C K f(g(z)) = lim f(g (x» = lim f (g (x» k-oo "k k-oo "k "k "k and f -+ f uniformly on K)
that g
"k
= x. Hence (df) ( )0 (dg) g x
x
= identity for
x. V; in particular, det«df) )/0 y
for y. g(V), which proves (iii). It remains to prove that
(iii) ==> (i).
The function j)x) = det(df)x is holomorphic on D and
converges to j(x) = det(df)x' uniformly on compact subsets of D. Moreover, if (iii) holds, j(x) since
£,,'
¢ O. Also j,,(x) f. 0 for all", and all x
Aut (D) (see proof that (i) ==> (iii».
1£ j(x) is constant,
j(x) is obviously never 0, and if j{x) is nonconstant, it is again never
o
by the corollary to Proposition 5 above.
j(x) ,; 0 for all xED.
By Lemma 3, f: D - (;n is an open map and
any x. D is isolated in f- I rex). feD) C U fJD) = D.
Hence, in either case,
It follows, from Proposition 5, that
82 Let {v k } be a subsequence of {v} uniformly on compact subsets of D.
so that g
Then, for x
0
vk
D,
converges {f
(x)} con-
vk
verges to f(x). D, hence lies in a compact subset of D. g(f(x)) = lim g (f (x)) = x, k-co vk vk In particular, det(dg)
y
"I
0 for y
0
feD).
Hence
for all x. D. Hence, repeating our lies in
argument above, we conclude that g(D) CD, so that a compact subset of D for any x • D. f(g(x)) =
Hence
lim fv (Sv (x)) = x • k-co k k
Thus fog = identity, gof = identity, and we conclude that f c Aut(D). We shall now give some applications of this theorem. Proposition 6. compact sets in D.
Let D be a bounded domain in en and K, L Then the set
G(K,L) = {fo Aut (D)I f(K)n L i~} is compact. Proof.
Let {fv}
be a sequence of elements of G(K, L).
passing to a subsequence, we may suppose that f v of D into en (Montel's theorem). a v • K so that f(a v ) = b v ' L. av -
k
a. K, b
vk
... b. L
converges to a map
Since f)K) n L
If {v k }
By
i ;,
there is
is a subsequence so that
(K, L are compact), then f{a) = b, so that
f. Aut (D) by Theorem 4 and since f(a) = b, f. G(K, L).
Since any
sequence of elements in GCK, L) contains a subsequence which converges in G(K, L), this set is compact.
83
If D is a bounded domain, Aut (D) is a locally
Proposition 7. compact group. If K, L
Proof.
o are compact sets in D with K C L , then
a(K, L) is a neighborhood of the identity (by definition of the topology on Aut (D»
which is compact by Proposition 6.
Definition 4. topological space.
Let a
be a topological group and X a (Hausdorff)
We say that a
operates on X if we are given a
continuous map a x X - X, (g, x) -> g. x such that ex = x for all X
E
X and (gg')x If a
= g(g'x)
for all g, g'
£
a, x
E
X.
and X are locally compact, we say that a
acts properly
on X if the map a x X - X x X defined by (g, x) ~ (gx, x) is proper. If a
is discrete and X is locally compact, we say that a
properly discontinuously on X if, for any a hood U of a so that {g Remark.
If. a, X
al g(U) " U
E
I
are locally compact, a
~
E
acts
X, there is a neighbor-
is finite.
acts properly on X if and
only if, for any compact sets K, LeX, the set a(K, L) = {g
E
a 1 g(K) " L
I
~
is compact.
In fact, suppose this condition is satisfied.
Any compact set
in X X X is contained in a set of the form K X K, K ( X compact. The inve rse image of K X K by the map (g, x) ~ (gx, x) is just a(K, K) and so is compact. Conversely, if the map (g, x)........" (gx, x) is proper, then, as above, a(K, K) is compact and G(K, L)
C G(A, A), A = K v L.
84 A discrete group G acts properly discontinuously if and only if
G(K, L) is finite for any compact K, LeX. If D is a bounded domain in C n , Aut (D) acts
Proposition 8. properly on D.
This follows at once by the remark above and Proposition 6. Proposition 9.
A subgroup r ( Aut (D), provided with the dis-
crete topology, acts properly discontinuously on D if and only r discrete subgroup of Aut(D) Proof.
If r
is a
[i. e., a discrete subset].
is a discrete subgroup of Aut (D) and K, L com-
pact in D, then r(K, L)
= {'I
E
r
I
y(K) " L i~}
in Aut (D) by Proposition 6; since r
is relatively compact
is discrete, hence closed, it is
a compact, hence finite, subset of r. Conversely, if r compact in D, element of
acts properly discontinuously and K, L are
o
K C L, then r(K, L) is a neighborhood of the unit
r (since it contains the projection on r of the inverse
o image by the map ('I, x) t---+ (yx, x) of the open set LX U, where U o is open in D and K CUe U C L). Moreover, r(K, L) is finite (since r
acts properly discontinuously).
Hence r
is a discrete subgroup of
Aut (D). Proposition 10.
r C Aut (D)
Let D be a bounded domain in C n and
a discrete subgroup.
the equivalence relation:
Let Djr be the quotient of D by
x - y if there exists 'I
E
yx= y. If
D/r
is compact, r
is finitely generated.
r
such that
85 Let {U v }
Proof.
Uv C
in D such that
natural projection. D/r and
U V"
be a sequence of relatively compact open sets
UIIH •
U UII
= D. Let ,,: D - D/r denote the
Then " is open. so that VII = ,,(U II) is open in
= D/r.
Since V"
is p so that V = D/r.
C
V"H • and D/r
U '1(K).
This implies that D =
U
where
'I' r
p
K=
is compact. there
is compact in D.
P
f'l i •.••• '1 N}
Let Clearly.
-i
'Ii
for which '1(K)"KjI'~.
be the elements of r
is a '1 j • i = 1 ••••• N.
We claim that any 'I' r
written in the form '1= 'Ii •.• '1 . • i!::.i. i 'p l
:n - X is an
S-extension of Po:O - en if, to every f
E
5, there is F f
E
K(X) such
Note that F f is uniquely determined (first on 1"(0) since hence on X by analytic continuation). continuation) of f to X.
If 01" = f,
It is called the extension, (or
91 Let p :rI - C{;n be a connected domain over C{;n o
Definition 2..
C J(.(X).
and S
An S-envelope of holomorphy is an S-extension
p: X - C{;n, '1': rI - X such that the following holds: For any S-extension p':X' - C{;n, 'I":rI - X' of p :rI - C{;n, there o is a holomorphic map u, X' - X such that p' = p oU, 'I' = u 0'1" Ff' = Ffou, for all f < S, where Ff'Ff' X, X'
and
are the extensions of f < S to
respectively. Note that u in (*) is unique (since it is determined on '1"(\"2)
by the equation
uotp'
= (n) which is open in X. chapter 2., Proposition 5, uov on X'.
= identity
on X.
such
= UoiCpl = q>,
Hence, by
Similarly, vou
Thus, u is an isomorphism of X' onto X with p'
= identity
= po u,
'" = uoq>', which is the uniqueness. Theorem I S
C J{(n)
exists.
(Thullen).
The S-envelope of holomorphy of any
92 Proof. 'P = 'P(po' S)
For any Po:rl - C[;n and S C I{(rl), we define a map
Let U be an open neighborhood of a ism onto an open set
Let a • rl and a
(9 (S) as follows.
of 11 into
U
o
C C[;n.
o
= p (a) • C[;n. 0
such that Po I U is an isomorph-
Let.&,.
be the S-germ at a o
5·(PO I U)
-1
,5 < S.
defined
We set
'P(a) = .&,. o One verifies at once that
x' commutes.
Here cp:n - X, cp':rl' - X' are the mappings of
Definition I. Proof.
Let v = cpl. u:n
local isomorphism. u
0
tp
= v.
-+
X'.
Then v is holomorphic and a
We have to show that there is
Consider the map
'" = pl. v: n - en.
u: X
-+
X' so that
Then .:, is again a local
94 isomorphism. If '" = ("'I' ••• ,'" ), the "'J are holomorphic. Let 'I be nilljl the jacobiaA determinant 'II = det{ h~ ), when : . ' for a bolomoprhic J J f on 0 is as in chapter Z, Definition 5. Then, since oj. is a local isomorphism, 'l{x) " 0 for all x (0.
Let 9. be the extension of oj.. J J
= (9 1, ••• , "i'n)' Let H be the extension of '1 to X. &9i H = det{-&--). Moreover, by PropOSition Z, H{x) -I 0 x.
to X, and let 9 Then, clearly, for all x (X. Lemma 3).
J
Hence 9:X- C
n
is a local isomorphism (chapter 5,
Moreover, 90'1' = ",.
Consider now the domains "':0 .... o;n, p':X''''' o;n and v:O"" X'. Let S = {foul ff 3t(O')}
= {Fovl
FE X{X')}.
p': X' .... o;D is the S-envelope of holomorphy of following lemma.) tended to X,
80
that
We claim that
' y < I},
Consider the
j = 0,1.
disjoint union X = Do V Dl and introduce the following equivalence relation in X: 1 < IZll < 2,
(zo,wo )-(zl'w l ) if and only if zo=zl' wo=wl and
a~
IWll < 1;
here (zo,wo ). Do' (zl'w 1 )
be the quotient of X by this relation and p : n o
.. (;2
2
by the inclusion of Do,Dl into (; • Then po:n .. (;
E
D 1•
Let
the map induced
2
is clearly a
Reinhardt domain for which Po is not injective.
For references related to the results of this chapter, see [1],
[10], [19] ,[29].
n
7
DOMAINS OF HOLOMORPHY: CONVEXITY THEORY
Throughout this chapter, Po:o - ct n will be a connected domain over
en. Definition 1.
(a) If f. ~(o) and A is a subset of 0, we
write
II filA
= sup
xEA
If{x) I·
~ 00)
(b) If A is a subset of 0, and S C ){(o) , we set AS = {x
E
01 If{x)1 !S.lIfllA for all f ~
AI
E
S}.
..
If S = ,..{O), we write A = AS' If S is closed under multiplication, then
Remark.
AS = {x
E
01 there exists Mx> 0 such that for all f
E
If{x)l!S. MxllfliA
S}. i\
Proof.
If we denote by
Now, if x. B, and f
E
B the set on the right, we have AS C B.
S, then fP. S, P = 1,2, ••• , so that
I f{x) I P :>. M
x
II fliPA
.
Since M l!p - 1 as p - co, it follows that x
Lemma 1. a
I K,
Let K be compact and M> 0,
there exists f
E
J{{n) such that f{a)=M,lIfII K O.
Then, for
104
Let g. J{ (11) be such that
Proof.
I g(a) I > II gil K'
Then, we
may take f = M(g/g(a))p with a large enough integer p.
f
E
}{(I1).
II f ~ A < en for any
Let A C 11 be a subset such that
Lemma 2.
C 11
Then, there is a compact set K
such that A C
K.
Proof. Suppose the result false.
Let {K } _ 0 1 be a p p- , , •.• o sequence of compact sets in 11 with K C K I' UK = 11. Then, p p+ P
A¢. Kp ,
since
there is x
•
p
A - R. P
By replacing {K } by a subP
sequence, if necessary, we may suppose that x fOE
p
,
Let
~ (11) be such that Ifo(xo )I > I, IIfoliK
and, by induction, let f
p
• J{(I1) be such that
I f (x ) I > p+l +
(1)
P
< I, o
P
~ q=O
I f (x ) I , II f II K < 2 - P . q P P P
00
2: f
Then the series subset of 11,
80
= f converges uniformly on every compact
p=O p that £.
JL (11).
Moreover p-1
CD
If(x ) I ~ p
2:
2:
I f (x ) I I f (x ) I p p q =p+l q P
q =0
If
q
(x ) I p
00
L p+l -
~ q=p+1
by (I).
Now, for 00
Hence
~
q=p+1
I f (x ) I q P
q>p, xp' K. q
I f (x ) I < 1, so that q p
p = 0, I, 2, • •• our hypothesis.
Hence
If q (x p ) I -
f is not bounded on A, hence
p.
II f
q
II K q
(b). in O.
t
Then {x)
Let {x)
(b) = O.
po(y) = po(y').
Then po(Y) = po(y')
E
Let Q{x)
c;(x) n c;(x').
are mapped isomorphically onto
Let y
E
p(x), y' ( P(x')
= po(P(x)),
Q(x')
=
po(P(x')).
Moreover, p(x) n P, P(x') " P
c;(x) ("\ Q, Q(x') n Q respectively,
108
Clearly, if C(x) f'I Q{x') P{x) n P{x')f'I P
f ¢.
-I ¢,
E
p{x)n P{x').
Q{x) f'I C(x') f'I
Hence, by our remark above,
P{x) "p{x') isomorphicallyonto y,y'
then
Q{x)" C(x').
Q-I ¢.
Po
Hence
maps
This implies that
Since po{y) = po{y'), and poIP{x)" P{x') is in-
jective, this implies that y = y', So that polu is injective. Now, P (U)= o
U
Q, where
ZEQ
= {we a;nj Iw-zl < p}, so that
Q
Z
Z
PotU) contains a polydisc about po(a) of radius
R> roo
This implies
that d{a) ~ R > r , a contradiction. o d{P) = O.
Thus
" Let K be a compact subset of nand xo' K.
PropositlOn 1.
Let a = po{x o )' and let V
be a polydi5C about xo' and P = po(V).
g~io{p 0
Then, for any f, J{. (n), if
Ivf l ,
J{{P),thescrics
(z-a)~
L aE I'\n
converges in the polydisc {z Proof. disc of radius
E
([n I I z-a I < d(K)}.
Let 0 < r < d{K). r
about
x
For an,' x ( K, let
I--..J
,md let K' =
C(x) .
Qxi be the polyTh,m K'
is COITl-
x,K pact.
Let M = IIfIlK,.
have
I D"f{x) I
TIy Cauc!">y's inequality applied to
~:-"1.0'!r-I"'I,
x, K,
So
that
il
Q(x), we
1IJ)Q'fIlK~\!·O'!r-jO'I.
"
IIenee, by definition of K, we have
IIDO'fll,,::; M.0'!r- 1aj K
Since x
o
E
the series
"
K, this iITlplics that
L
a
D g{a)· (O'!)
-1
jOO'g{a)1 ::CM.O'!r-lal.
(z-a)
a
converges for
r < d{K) is arbitrary, the result follows.
It follows that
jz-al < r.
Since
10')
Theorem 1 (H. Cartan - P. Thullen). a domain of holomorphy.
and let
Po
= d(f r 0 = d(x 0)'
r
r
In particular,
about ",(xo ). n contains a
about xo' contradicting our assumption that
This proves the theorem.
The same reasoning can be used to prove the following: Let S C Jl(p.) be a subalgebra
Theorem I' (Cartan-Thullen). of
J{(n)
containing the functions
Pi'···' Pn'
closed under differentiation (i. e., f. S
(po = (Pi'···' Pn)) and
=- Daf.
S for all a • INn).
Then, if the natural map of P. into its S-envelope of holomorphy is an d(K) = d(KS) for any compact Ken.
isomorphism, we have Corollary.
If
n
is an open set in C n which is a domain of
holomorphyand Po is the inclusion of n set K
C
P.,
K
in C n , then for any compact
is also compact. A
Proof. it follows that
K
A
is clearly closed in P..
K is closed in
cannot meet ro).
a:;n
'" Moreover, K
"
(since the closure of
K in
Cn
is contained in the polydisc
where p = max K
Moreover, since d(K) = d(K),
II z.11 J
K
,and so is bounded.
Hence
is compact. Theorem 2 (Cartan-Thullen).
Let
p : n - Cn o
that for any compact set Ken, we have d(K) that
J(
(n) separates points of n.
if we denote by p: X p : P. - C n o then
where
C n , ",:n - X
5 = {g}
have the property
> O. Suppose further
Then, there is
g.
J{ (n) such that
the S-envelope of holomorphy of
is the set consisting of the single element g,
'" is an isomorphism.
111 In other words.
n
is the domain of existence of g.
Before starting on the proof. we give a definition. Definition 5.
Let f ( Jl(n). £ 'I O.
Then. if a.
n.
the zero of f at a is defined to be the largest integer k Daf(a) = 0 for all cr. INn with
/0'/ < k.
the order of ~
0 such that
We denote this by w{f.a).
Note that the function a ..... w(f. a) is upper semi-continuous on O.
In particular. lt is bounded above on any compact subset of Proof of Theorem Z. Let {x)
Part I:
We shall prove the following.
be a dense sequence in 0
disc about x
y'
Then. there is g.
n.
and let Py be the maximal poly-
J{ (0)
such that
(a) g has zeros of arbitrarily large order in each P v such that p -1p (E) = E and such
(b) There is a dense set E C 0
o
0
that g separates the points of E. To prove this. we proceed as follows.
We consider the sequence
Pl' PI·P Z• p l ,P Z·P 3 • P l .P Z.P 3 .P 4 •••• (the essential.property being that each P k occurs in this sequence infinitely often). Let
We denote by Q
p
the p-th po1ydisc of this sequence.
{K} be a sequence of compact subsets of 0 p
such that
Then d(K ) > 0 by hypothesis. p
Lemma 7. f
p
•
Hence. by
By Lemma 1. there is
J{ (0) such that F(y)=l. p
p
/IF
p
11K
p
P. and
112
X E
K ,
P
o
Since UK p
= n,
any compact subset of
n
is contained in a Kp' so
IFp(X) I
J{(n). Furthermore, h¢ 0 (since e.g.,
that fE
0, set
U
P(x, r), 0 < r < d(A); here P(x, r) is the closure in 11 of x,A the polydisc of radius r about x. Then, for 0 < r < d(K), L = K(r) A(r)
is compact and we have
"
,..
K(r) C L. Proof.
Note that K(r) is defined since d(K) = ddh.
the inclusion false, and let x o ' f
E
,. K(r),
xo'
"L.
Suppose
Then, there is
J{ (0) so that f(x ) = 1 , o
The function g = 1
~f
I f II
K(r)
< 1 .
is holomorphic in a neighborhood U of
L.
Let p > 0, r < p < d(K) be such that the closure of the polydisc P(a,p) of radius
p about a' K is contained in U.
inequalities that
80
that
It follows from Cauchy's
116
N
gN' gN =
Now, g = lim N-oo
,.
a neighborhood of L
L
p=o
(since
fP
E
~ (0), and the limit is uniform on
/I fIlK(r)
O
E r, 11m Z I !!:. M}, then
Z
suchthat
,.. KIt(O) contains A)...
here r = { tao I O!!:. t S- I} v {tal los- t S- I} • Proof.
Let
S~
E> 0 be sufficiently small, and let
= {z
c:
E
and Sa = Se' n T A.
n
l
Z Z zi +zZ - e(zl + zZ) = 1-8 , z3 = •.• = zn = O}.
We claim that for any f
~ (0), we have
E
If we write z. = X. + iy., we have, on S. , J
J
xI+x Z - E(X IZ +
J
co
X~)
+ f:(YIZ +y;) = 1-£
In particular, ylZ + y; S- 1
~f.
o lO: Xz S- I, it follows that Se on S£
,
XI?' 0, xz?' O. Xl +xZS-l.
Since. moreover 0 S- XI !!:. I.
•
is compact.
Furthermore. Xl +x Z < I
except at the points a o and a l • Suppose now that
I
rI St I
attains a maximum at a point I;, £ Se - r'.
If I;, = (1;,0'1;,1.0' •••• 0), there is a holomorphic function .,,(u) in a con-
nected neighborhood U of u = 1;,0 in «; such that .,,(1;,0) = 1;,1 and the set {(u • .,,(u). 0, •••• On is a neighborhood of I;, on If(u,.,,(u).O ••••• 0)1
Se.
Hence
has a maximum at u = 1;,0 and so is constant on U.
It follows that the set of
Z E
Sf. -
r'
at which I
is open and it is obviously also closed.
Hence
dSe I
has a maximum
Hflls = IIflis n t:
It is easily verified that if ).. XE
r'.
I:
< 1. and E is small enough. for any
A)... there is y= (yl'yZ.O ••••• O) such that X£ SE+iy. Iyl S-
Hence. If(x) I S-,fll(S,; Hy)" M=
zA,
r' .
If we set
K= {ZEc:nl RezE
r.
Ilmzl ~M},
~
•
121 this implies that
..
so that A~ C K. Lemma 10.
Let
ao.al.r.A~.
A be as in Lemma 9.
Let B be
the Wlion of two open convex sets in IRn containing. respectively. the sets {tao I Os.t:;>.l}
and {tall OS.t:;>.l}.
Let n=T B .
Thenany
f. ;K{n) can be extended holomorphically to a neighborhood of T A' Let E be the set of
Proof.
~.
0:;>' ~ ::;. 1. such that there is a
convex neighborhood U of A~ in IRn with the property that any f.
,e (o)
can be extended to T U.
E is clearly open in [0.1] and
(Note that T U" T B is connected.)
o.
E.
Let ~.
Let P = O.
E .
and let U be a convex
neighborhood of A .. such that any f E J{{O) can be extended to TU' Then B' = U U B is connected. and so is U" B. therefore be extended to T B' = 0'. I~ -
"0 1 < r < p.
Also doter) 2: p.
"0
•
K (relative to
By Proposition i. the Taylor series of any f
point a. A
"0
E
0') contains
:Jt{O') about any
converges in the polydisc of radius p about a.
conSidering the iWlctions f : z ... fez y
+ iy),
same is therefore true for any a ETA
It follows that there is a
"0
Hence ~. E. so that E is closed.
lemma is proved.
By con-
y. IRn , one sees that the
convex neighborhood V of A~ such that any f E to TV'
Let .. o < ...
By Lemma 9, there is M> 0 such that if
K = {ZE U;nl Rez=r. Ihnzi :;>.M}, then A
Any f. 1({ 0) can
J.4o')
can be extended
Hence E = [0. i) and the
IZ5 Proof of Proposition 6.
As remarked earlier, it suffices to show
that any f. 3 0 and f. I( (0), there exist linear functions n
~
j=l
and constants a v • ¢ , v ~
k
= 1, ••. , p,
/f(z)- 2..., a e v =I v
k
.z v,J j
.• JR , v,l
such that
Iv(z) / 0 such that the following holds. Let P be a normalized polynomial which is of degree S. d with respect to each variable zl' ••• ' zn'
Let 0 < t < 1 and let
S= S(t,P,K) = {ZE KI Ip(z)1 Then
mrS) where m
~
Ct Z/n •
denotes Lebesgue measure in en.
We shall need the following lemma.
~td}.
134 Lemma 1.
Let
q(t) = t P
+
f
k =1
a t p- k k
be a monic polynomial of degree
p with real coefficients.
Let g(e) = q(cos e), -". S. 9 S. "..
g (9)=
Then
I q(t) I 2:.Z- p
max -1S.tS.+l Proof.
a. , 1R J
Now
(ei8+e-i9) _....e.... ik9 c =c =2- P q Z - 2.; cke 'p -p k=-p
Hence z -p = Izl".
j'"_".
Proof of Proposition 3. that K
C {z
I>-jl < R+l,
, (; I I z I
s.
g(9)e -ipe de I
s. R}.
sup
I g(e) I
-".s.es.".
Part 1.
Case n
sup
= 1.
Let R> 0 be so
Let >-1' ••• ' >- p be the ze ro s of P
and "1' ••• '''q the zeroS with
we count zeros with multiplicity).
I q(t) I.
-1S.t~1
11.1)
with
2:. RH (p+q = d, and
Then
P(z) = a(z - >-1)'" (z - >- )(1 __z_) •.• (1 __z_) , p J.l 1 "q
a ( ([;
d
2:
=
cyZ
Y
,
max Ic) = 1.
v =0 Let k be such that
I c k I = 1.
Clearly, if Ak denotes the
coeffic ient of z k in
we have (since =
(~)(RH)klal.
I>-jl < RH, Ifljl? R+1 > i), 1 = Ickl S.IAkl Since
(~)
lal ?c- d , Since, for I z
I s. R,
we have
S. 2 d , this gives c = 2{R+1). /1 - _z_ I > R:1 ' this implies that J.l 1 -
i35
where T= (R+i).c·t= cit, say.
Now, if cit':::'i, we have
(1) Suppose that T < 1.
Then Td S TP, so that
Then A contains the projection of So onto the real axis JR. motA) be Lebesgue measure of A
relative ~o JR.
a nonempty open set, so that mo (A) > O.
Let
Clearly, A contains
Consider the map '1': JR - JR
defined by
Clearly rp{x).:::. -1
and rp(-R)
= -1,
rp(+R)
= 1.
Moreover, if x < x'
( *) so that, in particular, 'I' is continuous.
Moreover, if I is any open
interval contained in JR - A, 'I' is clearly constant on 1.
Hence
'P(A) = rp(JR) = [-1,+11. In vie\v of (*), we have, for x ( A,
Since rp(A)
= [-1,+11,
Lemma 1 shows that
I
Z-Ps. sup 1'P(x)-'P(a 1 )I ••• I'P(x) - 'P(a p ) s.{ZT/mo{A))P, x(A which gives
motA) S 4T.
In the same way, one shows that the projection A I of So on the imaginary axis has measure
136 Hence
This, and (i), imply that m{5{t, P, K)) Part Z.
~ c 3t Z
,
The general case.
a:
the result proved in Let p{z) =
We proceed by induction; suppose
n-f
2: caz
Q
=
2:
z,flPfl{zn)' z, = (zi'···' zn_i)'
flElN n - i There is then flo such that Pflo is a normalized polynomial. Let
5i={ZE5{t,P,K)IIPI!{zn)l~td/n}, o 5 Z = 5{t,P,K) - 51 • It follows at once from Part i above that
m{5 ) < M· c t Z/ n = c t Z/ n i 3 4' where M is the Lebesgue measure in CCn - i of the projection of K onto
Cn - f • Let Ko be the projection of K onto the zn -axis, K' that onto (;n-f and let
~E
Ko'
Let
(z',~)
O{z')
= P{z', ~)/Pfl
E
5 Z ' and
(~).
o I Pfl (~) I '2 t d / n and the maximum of the absolute values of the o coefficients of 0 is ~ i. Hence Then
By induction hypothesis, there is c S > 0 so that the Lebesgue measure in (;
n-i
of 5 Z, ~
Hence, by Fubini's theorem
137
Hence
Remarks.
(i).
This inequality is essentially the best possible.
(2) For the actual application in view. a much weaker inequality. that can be proved using Jensen's inequality, is sufficient.
We only
need to know that the measure m{S(t,P,K)) - 0 as t - 0 uniformly in P
(in particular, uniformly with respect to deg Pl.
We
have given the best possible inequality above in view of the interest of the methods used (which are due to Bishop). Proposition 4.
Let p : n o
and K a compact set in n.
-
<en be a cOIUlected domain over a;n
Let f . J{{n).
Then, there is 8. 0 0 such that po(x) = a for all v. d({x)) ~ p
and the polydisc P (xv' p) of radius p about x
v
is contained in
K.
A
Proof.
Let L be a compact set in 0
.. L
such that L is not com-
pact.
Let {y) be a sequence of points in
without any limit point
inO.
Since Po= (Pl •.••• Pn) where the PjE ~(O).wehave
Hence. by passing to a subsequence. if necessary. we may suppose that p (y ) .. z o
v
• CC n • Now. by chapter 7. Theorem 1. we have
0
1\
deL) = deL) = 2., > O. Let
e>
0 be small enough. and a. C n • la-zol 0
K
is independent of d,O, if d
~
We now keep D fixed, and set A
P
=
U d>
-P
D and D is large enough,
n co
A
d,O
A =
,
A
p=D
,
P
Then m(A) ~ Further, if z z
I
S d,O'
E
K
,
A, then, there are infinitely many d
Hence, for
Z E
such that
A, there are infinitely many d
.!. d5 max I Cd}( 1 > 9 2 k = 0 " " , D Pk z}
such that
141
we have the following:
(* )
For (z, w) E X, z. A, we have D
(d)
12: c k
k (z)w
.!.d6
1< el
for infinitely many d.
k=O
Moreover,
m:x 1ck{d){zl!
= 1.
Hence, choosing a subsequence of the
d for which (*) is true, we may suppose that m~ Ickl = 1.
=ck{z),
..... c k '
Hence, we obtain from (*) the following:
For z. A, (z, w) ck
c~d){z)
E
X, there exist c k ' C, k = 0, ••• , D, not all 0,
such that
From the definition of X we obtain the following result: Lemma 3. such that if z P -1 (z) "
o
E
There exists a subset A C Q of positive measure A, then f takes at most D values on the set
K.
K
(For, by our remark above, any value of f on p -1 (z) n o D k ~ ckw = 0 , where not all the c k are 0.)
. . . sattsf1es an equatton
k= 0
Let {x} v
be an infinite sequence of points in
K.
Then p( x • p) C v
As in Part 1 of the proof of Theorem Z in
chapter 7, we can find f Let now, for z
E
Kas in Lemma Z.
E
~ (O) such that f( x ) v
+f(x f.L )
if v
+f.L
Q, Yv(z) be the point of P(xv ' p) with po{Y)x» = z.
We assert that the set of z. Q such that f does not separate the points Yv( z) is of measure O. This would contradict Lemma 3 above, and therefore would complete the proof of Oka's theorem.
142 To prove the above assertion. we set g = fo(p Ip(x .p)fl. v
Then gv E J{( Q).
= U
A
I-' ~ v
f
A
1-'. v
\I
0
We set A
{Zf QI g (z)= g (z)}. v f!
1-'. v
Moreover, since g (a) v
A
1-'. v
is an analytic set in
Q. ~ Q.
O. it suffices to show that each A
1-'. v
= f(x v ) ~
Clearly.
= gf! (a).
f(x ) I-'
(I-'
f
v).
To prove that Af is of measure is of measure O.
This is an
immediate consequence of the following lemma. Lemma 4. h
*
O.
Let 0 be an open connected set in Q;n and h E ).((0).
Then. the set
has measure O. Proof of Lemma 4.
It is sufficient to prove that any a. Z has
a neighborhood U such that U" Z is of measure O.
By a linear
change of coordinates, we may suppose that a = 0 and that. for a small enough r > O.
Then. for t:
>
0
small enough. we have h(zl ••.•• Z n-l' zn)
+0
for
Iz.I~&.j=l ••••• n-l.lz I=r. Let U={zEQ;nllz.l 0 and a unique real analytic map g = gX:flo X I - fl,
I = {t
E
IR I
It I < p}
such that
ag~:, t} = X(g(x, t}},
g(x,O}
= x,
X E
flo' t
E
I.
II U is open in IRq and X:fl XU - IR P is real analytic, then for
U0 C C U, there is p > 0 such that the following holds: For". U, set X,,(x} = X(x, ,,}.
map g:fl
o
X I XU
0
- I"l
X E
I"l.
Then there is an analytic
(I = {t. IR I It I < p}}
such that the map g",:l"loX I - fl defined by g,,(x, t} = g(x, t, ,,} Moreover, if t, s, t+s
satisfies
We call g
= gx
E
I, we have
the local one-parameter group associated to the
vector field X. Note that for any f. CCD(fl} , we have
Definition 2.
Let V be a finite dimensional vector space of
vector fields on an open set
n C IR P . We say that V is a Lie algebra
of vector fields if whenever X, Y • V, we have [X, Y]
E
V.
In all that follows, we shall assume that the vector fields belonging to V are real analytic. We shall use the follOwing result from classical Lie theory. Since it appears somewhat difficult to give a reference to the theorem in the form we need, we shall give a proof.
147 Theorem I
(Lie's theorem).
Let V be a finite-dimensional
Lie algebra of real analytic vector fields on an open connected set
n c: lR P •
Let
nee n. o
Then there exists a neighborhood U of 0 in
V and a real analytic map g:Oo X U -
°
with the following properties. Let gul 0 0 (i)
-
n
be the map x
t+
g(x, u).
For u and v sufficiently near 0, there is a unique w
= w(u, v) E
U
such that
(ii)
For u
E
U, the map t ... gtu (t
E
lR, near 0) is the one-parameter
group associated to the vector field u. (In particular, go (iii)
sufficiently near 0, the maps
For u o ' v 0 u
t+
w(u, v 0)
and
v .... w(u o ' v)
are analytic isomorphisms of a neighborhood of 0 of v 0' U o
= identity.)
E
V onto neighborhoods
respectively.
The map g is called the local Lie group of transformations associated to V. Proof.
If a
E
Let 00 C. C 01
C
e0
lm . and let X , ••• , X be a baslS of V.
lRm , we shall denote by X(a) the vector field
X(a) = j
Let p> 0 and 1= {It I
f-= I
a.X j • l
< p} be such that there is a map
.,: 0IXIXU I - 0 ,
U1
= {aE
lRmllajl 0 be an integer such that So = to/p g(x, to)
= fs
• ••
E
I.
Let
We set
fs (x).
~o
p times That g is independent of the choice of p follows from the properties of the local one-parameter group stated in Proposition I.
That g satis-
159 fies the required differential equation follows from the principle of analytic continuation. This map g is called the one-parameter group associated to X. Theorem 2.
Let
be a sequence of elements of G = Aut(D),
{IT)
D being a bounded domain in a;n.
Suppose that {IT)
converges to the
identity element in G and that there exists a sequence {m v} integers, m v -
as
CD
v-
of
with the following property:
00
{X), X)x) = mv(IT)x) - x), of maps of D into a;n
The sequence
converges, uniformly on compact subsets of D, to a map X: D _ a;n. Then X Proof.
is a vector field associated to G. Let no
C ( D and let p>
0 be so small that the local
one-parameter group g:n X I - D o corresponding to X
is defined on no X 1.
be the largest integer!>. m}o.
Then, q
v
-
Let 0 < to < p, and let qv 00
and 0 S. m t - q < 1. v 0 v
We shall prove that there exists a nonempty open set B
c: no
such that
q
if to is small enough, then
lTv
v
converges uniformly on B to the
map gt :x .... g(x,t o )· It follows from Vitali's theorem (chapter I, o qv Proposition 7) and chapter 5, Theorem 4 that IT converges to an v element IT
E
G which, by the principle of analytic continuation satisfies
We have
ag~:, t)
It =0 = X(g(x, 0» = X(x)
,
X En
o
If K is a compact subset of no' with nonempty interior, this implies that (l)
g(x, t) - x
= g(x, t)
- g{x, 0) = t {X(x)
+
c 1 (x, t)}
160 where
El (x, t) .. 0 as t - 0, uniformly for x. K. IT v
(x) -
Now Xly - X as
X
=-
1
my
y -+ 00
to X (x) = X' (x), y qy y
(since 0 ~
my to
Moreover,
X'
q
y
- qy < 1).
=--y- X tomy
y
Hence
t
(1 I)
IT
y
e vex)
where
- 0 as
(x) y -
X
=~ {X(x) qy
00,
+ E y (x)}
uniformly for x. K.
It follows at once
that
(z) where 6 y
-
0 as
y -+ 00.
about xo' K, B C. K.
radius r
16
< M and let B be a ball of o We shall suppose that the ball of
Let M > 0 be so chosen that
1/ X
radius r about any point of B is contained in K. to
0 such that I g(x) - g(y) I ~ C E Ix-y I , x, Y E D~ • Clearly Ix-yl - Ig(x) - g(y) I ~ If(x) - f(y) I ~ Ix-yl + Ig(x) - g(y)l. The assertion follows immediately from this. If ",(el) < p, p = 1, ... , q-1, and if f(x) = 1. {x + IT(x) + ... +.,. q-1(x)} , q
then ",(f) is continuous. Proposition 7. Let D
Proof.
q>{U) is a neighborhood of e
o
C.C DI C nand .,.{,,) = sup
xED f
0
(= identity) in G.
I,,{x}
- x
Let K be a compact symmetric neighborhood of 0, K C U. the result false. "y -
e
as y -
Then, there exists a sequence
<X> ,
CTy
I
I
as above.
Suppose
{CT) ( G - {e},
0 as v'"
Let U0
be a neighborhood of 0 in V, 6
=
U0 C
o K.
00 •
Then
f1(q>(a}} > 0
inf aEK-U o
and
a) = 6
lim inf f1(q>(a} V"'oo aEK-U o since a v - e. a
V
E
Since C v '" 0, it follows that for sufficiently large v,
U • 0
Let t ... h t be the one-parameter group of X (corollary to Proposition 4).
Then, for small t, w
v=-h '1/ \,
(in the notation of Theorem 1). 00,
bv
E
K if
0
T
q>(K}.
E
v
Consider
•
= q>(b).a v '
Then wv = h_i/'\,oq>(a)o O. XE01
Since
This proves
Proposition 7. Theorem 4
(H. Cartan).
The group G = Aut(O) carries the
structure of a Lie group such that the map G X 0 - 0, (CT, x) ....