Several Complex Variables (Chicago Lectures in Mathematics)

defined by (f)

C Xo' a

E

a

.... (aaf zl

en. and P

a polydisc pea. p) and '1'1.·.· .q>n holomorphic

functions in P

defining

morphic on P

and

~ aZj

q>1.a' ••.•

= q>. on p. J

each connected component V of fl = f +c. c.
Then if U

'1- I (U)

Let f be holo-

= {{'I'I.b ••••• 'I'n.b)lbEP}.

is of the form N{P.fl). where

Clearly '1IV is a homeomorphism onto U.

Let X' be a connected component of '1- 1 (X). a covering and X'. X are connected. '1 is a homeomorphism.

Then '1: X' - X is

By the corollary to Proposition I.

25

We now define a holomorphic function { on 0 by f(z) = the value at z of the germ 'l-l.l\I(z),

f~.

X', such that

p(f~)=

z.

Clearly,

a;;-

Sf = gj' J'=l , ••• , n. Theorem 4 (Poincar~-Volterra theorem).

Let X be a topo-

logical space (Hausdorff) with a countable base for its open sets, and Y a co_ected manifold.

Suppose that there is a continuous map f: Y ... X

8uch that f-l(x) is discrete for any x. X, i. e., for any a. f-l(x), . 'there is a neighborhood U of a in Y such that U"

C l (x)

= {a}.

Then Y has a countable base {or its open sets. Proof. in X.

Let {X }

v ,v=l.Z ••.•

Consider the family

follows:

'lit =

{U}

be a countable base of open sets of open sets in Y defined as

U is relatively compact in Y and is a connected component

off-leX ) for some".

"

We assert that

(i)

'Ul

is a base of open sets in Y,

(ii)

1JL

is countable.

Proof of (i).

Let a. Y.

V an open set of Y,

a. V.

Let W be

be a relatively compact open set in Y such that a. W C Vi C V and

Wn f-If(a) I(a) ( f(aW).

= {a}.

Then aw is compact, hence so is f(aW).

Let" be so that f(a). X"

C X - f{aW), and U be the

eounected component of f-I(X ) containing a.

"

Further,

Then U C W (since

otherwise un aw .;. j4 and X,," f{aW):> feU) n f{BW) .;. ;).

In par-

Ikular, U is relatively compact, which proves (i). Proof of (ii). :-

If U.1/t, U is a finite union of open sets homeon

iiftIorphic to open sets in IR. ~perty:

It follows that U has the following

26

(*)

If {V,,) (nA

V f"'I V = ~ if a 13

a'; 13,

then A

(1JI..

there exists

n' =

n,n'

U V,

(a)

UUtk =lit,

(b)

1.ttk

'\,1k

= {V
is countable.

U't -uUtk •

the union being over those V.

nected, 12' = ~, so that

U ~ = 1J(.

(b). 1J(0 is countable.

Suppose 12=

Y,

= {V),

the union being over those V. U~,

are open, and, by definition of the

For each

has the property (*).

V. '\,\-1 with V n V ,; ~.

Let 12 =

U V,

1}t

Define, inductively, 1110

We assert that (a).

with

is countable.

Note that any countable union of sets in Let Vo

V

is a family of nonempty open subsets of

III k' n" n'

'VL k _ l

U

countable.

=~.

Y being con-

n

Then let

V.

it follows from (*) that the family Fy v

Then 12 v 12' = Y,

•

V·1Jt k _1

ponents of f- l (X ) which meet

and

is countable.

of connected com-

Since Vtkc.

Uv

F

v

,

Uk is countable. This proves the theorem.

Remark.

The theorem is true under weaker hypotheses on Y, e. g. ,

that it is connected, locally compact, and locally connected.

One

proves, using the fact that f has disc rete fibers and that X has a countable base, that each point of Y has a countable fundamental system of neighborhoods. above proof applies.

Property (*) is then easily established, and the

27

Corollary.

Any connected component of

Definition 5. OUB

function f: X - e

has a countable base.

Let p: X - en be a domain over en.

A continu-

is called holomorphic (relative to p) if for every

a < X, there is a neighborhood U (i)

CO

such that

pi U is a homeomorphism onto the open set V C en,

(ii) the function f. (p I U) -1 is ho10morphic on V. We define also the derivatives Daf of f by the requirement

D a fo{p IU) -1 = D a (fc>(p 1-1 U) ). Here U is as in (i), and the derivatives on the right are those on V.

ern

Definition 6.

Let p:X _ en, p':X' _ ern be domains over en,

respectively.

A continuous map u,X - X' is called ho10morphic

if, for any open set V' C X' and f'

holomorphic on V', the function

flou isholomorphicon u-I(V'). Most of the the theorems of chapter 1 extend obviously to domains over

Let now fl be a connected open set in C n on

n, and let p: X -

Cn

and f holomorphic

be a connected domain ove r a:;n.

We say that

f can be continued analytically to X if there is a holomorphic map u:fl - X and a holomorphic function g on X such that (i) p. u is the inclusion of

n in a:;n;

(ii) go u = f. Note that g, if it exists, is unique (since g is determined by f on u(n) which is open in X by (i), and we can apply the principle of analytic continuation; see below).

28 Proposition 5 (principle of analytic continuation). p':X' _
be domains over q;n,C m

respectively.

nected, and f, g:X - X' be two holomorphic maps. empty open subset of X, then f =. g.

Let p: X -
Let X be conIf f = g on a non-

[In particular, two holomorphic

functions on X which coincide on a nonempty open subset of X are identical.

1

Proof. borhood of a. f

=g

Let E be the set of a E X such that f = g in a neighE is clearly open.

on E, we have f = g on E.

We show that it is closed. Let bEE and y

o

= f(b)

Since

= g(b).

Let V be an open neighborhood of Yo such that p' IV is a homeomorphism onto an open subset W of q;m.

Let U be an open neigh-

borhood of b such that pi U is a homeomorphism onto a polydisc P in C n , UC f-I(V)

("I

morphic maps of P

g-I(V).

Then plofop-Iand p l ogop-l areholo-

into W which coincide on the open set ptE n U)

which is nonempty since bEE.

Hence (by chapter I, Proposition 1

applied to the components of these maps into Itm) p l ofop-l = p l ogop-l on U.

Since pllv is a homeomorphism, it follows that f = g on U.

Hence U C E, hence bEE and E is closed. Theorem 5.

For any holomorphic function f on a connected

open set fl in q;n, there is a connected domain p: X _It n and a continuation g of f to X with the following property. For any connected domain p': X' - It n and a continuation g' of f to X', there is a holomorphic map

Proof.

9':X' - X such that poll' = p' and

Let X be the connected component of (9 containing fa

where a • fl and fa is the germ defined by f at a.

Let p: X _ It n

29 be the restriction to X of the projection of (fi into C[:n. g:X ... C be defined as follows:

g(

equals the value at z of

c X,

Let

- (9

be the map u(z) =

Clearly u is continuous: since n

and we obtain a map u: n ... X.

is connected,

Clearly p.u = inclusion of

If p':X' - C[:n is another connected domain over C[:n, u':n'" X'

and g':X''''

(!;

define the continuation of f to X', we define a map

.,:X' .... X as follows. Let x'

E

X, a'

= p'(x').

Let U' be a neighborhood of x'

mapped homeomorphically by p' onto an open set U in C[:n, and let h = g'. p'

-1

on U.

Let
a continuous map
C9.

Clearly
connected, this gives uS a map
This gives us

Since X' is

One verifies the properties

etated in the theorem very easily. We shall see later (in chapter 6) how this construction can be generalized to families of holomorphic functions and leads to the theory of domains of holornorphy.

3

SUBHARMONIC FUNCTIONS AND HARTOGS' THEOREM

Let

n

be an open set in «:.

Wewrite z=x+iy, z< «:,x,y
If u i$ a (real or complex valued) function twice continuously

diffe rentiable in n, we write

Definition 1. Remark. harmonic.

u is called harmonic in

n

if Au"

o.

The real and imaginary part of a harmonic function are A holomorphic function f is harmonic (since

Definition 2.

Let s be a real valued function on

- 00 is admitted, not the value + 00). ciple holds for s if, for any U Cc

s(z) Proposition 1. harmonic function on

~

sup

r;,. au

n

(the value

We say that the maximum prin-

n,

we have

s(l;.) ,

z. U.

The maximum principle holds for any real valued

n.

This is a consequence of the following Proposition 2.

The maximum principle holds for any twice

continuously differentiable real valued function u with Au

30

~

o.

31

Proof. of

n.

~u

It suffices to prove the result when

In fact, if u (z) €

~u = ~u+4e

+ elzl 2 ,

= u(z)

u{z) = lim u E (z) ~ lim {sup u (t;.)} f - 0 r - 0 t;.. au e

> 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. <1:, p > 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) =
( PO,l z, S) -

f

o

1f

2 1- z1 is 2

I

tp

is continuous on the circle so that P(z, S)

1

f

/z/

O.

2

e

1f

is

+ z d9

-e-i-e--z

Moreover, for

'S

Let e> O.

Now

Izi

< I,

~J 2m / 12:! t;.-z t;. / t:.1= 1 ...!...

= Re {

J21f P(z,9){tp(e l

P(z, S)h(9)d9 - h(9) =

0

= 1.

1.

< I,

1f

T =[

~

Iz 1

Ie - zI

p(z,e)de = Rc 21f 0

Hence, for

,

0

is

) -
0

Then there exists 5 > 0

90

that /
i9

) -
ie o

)1 < C

for

32

/ e i8 _ e i8 0

< 5.

/

Hence, since P? 0 ,

J2" o

/T / <

P(z, 8)d9 + 2M

where M=sup/q>(e '·8 )/.

it follows that if for

/e

i8

- e

i9 0

/z - e

/ ~ 6.

I

P(z,8)d8 ,

/ e i9 -e180 / ?6 Since

i8 0

1

2 0 " P(z,8)d8=1

/ tends to 0,

and P(z,8) =

P(z,9) tends to 0 uniformly

r-

Im i9 / T / 5. £ , and £ > 0 being arbiz -e 0

Hence

trary, the result follows. Corollary 1. {z <

a: / / z-a I ~

p}

If u

c

is harmonic in n, a

E

n

and

n, then

u(z) =/"P(z,8)u(a+ pe i8 )d9

Iz-a/ < p.

for

o Proof. We may suppose that u is real-valued.

The function

v(z) = { " P(z, 9) uta + p e i9 )d9 is the real part of the holoITlorphic

o

· f unction

harmonic.

1 2" /

"

o

() pe i9+ z - a u (a+Pe I·9 )d8·In i9 pe

Iz-a I <

P, an d so·IS

- (z-a)

The result follows from Propositions 1 and 3.

Corollary 2.

A harmonic function is a real analytic function of

(x,y), z=x+iy. Definition 3. u(n)

f

{-w}.

Let u be a map of n e e into lR OJ {-w}

We say that u

is upper semi-continuous (u. s. c.) if

lim u(z) 5. uta) ; z - a, z f a equivalently, if the set {z

E

so that

nl

u(z) < "'}

V

a.n,

is open in

n

for any a

E

lR.

33 Note that if u is u. s. c., it is bounded above on every compact

K CO, in fact K is contained in the union of the increasing sequence of open sets {z E 01 u(z} < v}, so in one of them. Lemma 1.

Let u be u. s. c. in 0 and bounded above.

there is a sequence {'\.} of continuous functions on 0 decreases to u(z} for all z ~. -00

< '\.(z} <

E

Then

such that '\.(z}

O.

We set '\.(z} = sup {u(J;} - kl r,-zl}, M= sup u(r,). Clearly r,EO r,EO for all z EO.

+00

We have '\.(z} ~ u(z} - klz-zl = u(z}.

Further, since the sequence {u(r,} - kl r,-z[}

decreases as k in-

creases (for fixed t,z), '\.(z} decreases for every z.

Further, if

OJ

Z,Z· E

that Uk is continuous on O. To prove that '\.(z} - u(z} as k u(z} >

-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'}
that M-koli
while u(z,} -klz'-zl < M-koli< u(z} for

Zl

to',

k~ko'

Z'E

Hence, for

k~ko' u(z} ~ '\.(z} < u(z} +E If u(z} =

-00,

and

0' contains a disc

, so that '\.(z} - u(z} as k -

c> 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}

n, then

00 •

o

Proof.

(a)

Suppose the assertion false.

Then, there is a <

(2

and p > 0 such that if K = {z • ell z-a I :>. 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<

Then (Proposition 3), h

~l<

o tinuous on K and is clearly harmonic on K. ~

u{z} if z

is con-

Further, ~{z} = "k{z)

Hence, by definition of subharmonic functions,

3K.

E

Let

Since ll.

decreases to u, and P

k.

(z, 9) is positive,

a,p

0

we conclude that -00

h (z ) = [

< u{z ) $. lim

ok_a> K

0

zrr P

0

a,p

'9 (z, 6)u{a+pe l )dS , 0

contradicting our assumption that u{a + pe i9 ) =

for 9 in a nonempty

-00

open subset of [O,ZrrJ (since u is bounded above on K). Let K

(b).

=

{z

E


I! z-a I :::. p} c n,

and let {"k}

be a sequence

Let M = sup u l (z). z< K o a • K be Let uk = min{"k'O). Let o

of continuous functions decreasing to u on K. Then "k{z):::, M

for z. K.

such that u{zo) >

-00.

As in (a), we have

$. {

rr

o

'9

P

$. C.M

a,p

{z, S)ll. (a + pel ide I<

0

+t"P o

a,p

{z ,S)ll.-{a+pe i9 )d9 0

I<

c so that if u

a,p

e

a,p

(z, 9) 0

min{u,O), we have lim - [ k-oo 0

Since P

= sup P

'Q

Z"I1"

P

a,p

{z, 9)::: 5> 0 for all e, we deduce that 0

Since u is bounded above on K. the result follows.

{z, 9),\{a + pe' }d9 0

36 Proposition 4.

n

function on

n be an open set in

Let

such that u"

-00

I{;

and u an u. s. c.

n.

on any connected component of

Then u is subharmonic if and only if the following condition

is satis-

Cied:

(*)

For any a

n,

E

there exists R = R(a) > 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<), then the maximum principle holds for u. Let U C C sup u(z) >

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

)
Hence, if .. denotes the Lebesgue

measure on the real line f

f

u(a) ~ ZIT .. ([0, Z'I\'] - I)u(a) + ZIT .. (I)[u(a) -!o] < u(a) , Hence u(b + pe i9 ) = u(a) for all 9. [0, Z'I\'].

a contradiction.

Since this

holds for all p, 0 < P < R, it follows that E is open, q. e.d. Remark. for

If u is subharmonic in

(I

and {z.

(;1 Jz-aJ ~ p} C

(I,

then

Jz-aJ < p, we have

~ ('I\' P (z, 9)u(a + pe i9 )d9 • o a,p

u(z)

This is part (i) of the proof of Proposition (5). Corollary 1.. If u l ' subharmonic for

~l~

Corollary Z. Proof. If a ~ ----u(a)

0,

U

z

~~

If u l ' (I

=u,(a) ~ J

are 9ubharmonic,

~l u l

+

~ZuZ

is also

O.

U

z

are subharmonic, so is u = max(u l , u Z).

and u(a) = u,(a), j = I, Z, J

{'II'u.(a + pel'8 )d8

1 'I\' 0

-Z

J

1 {'I\' '8 u(a + pel )d9 0

~,.

'I '

..

for small enough p. Corollary 3.

If {ua } is a family of subharmonic functions in 0,

and if the function

z ........ u(z) is u. s. c. in

(I,

=

then u is subharmonic.

Corollary Z, using the above Remark.)

supu (z) a a (Same reasoning as in

38

Corollary 4.

If u is subharmonic on u{a) =

n

and a (

n,

then

lim u{z). z-a, z'; a

lim u{z) . Since u is u. s. c. , I ~ u{a). z-a, z'; a If 1 < u(a), we have uta + pe i9 ) < I' (where I < I ' < uta»~ for small Proof.

enough p

I =

Let

I 0, contradicting ( *). If u is a continuous function such that u and -u

Corollary 5.

are subharmonic, then u is harmonic Proof.

We see at once that u(z)

I z-a I < p~

p small enough and

here a

=[21T P (z, 9)u(a + pe i9 )d9 o a,p 11 is arbitrary.

E

Corollary 6 (converse of Proposition 3). function such that, for all

d

1

u(a) = 2" then u

If u

is a continuous

P., there exists

R(a) > 0 such that

12TTuta + pe i9 )d9

for 0 < P < R(a),

E

for

0

is harmonic.

Proof.

For then u and -u are subharmonic.

Corollary 7.

If every point of 11 has a neighborhood in which

u is subharmonic, then u is subharmonic on 11. Proposition 5.

Let u be a twice continuously differentiable

real valued function on the open set P. C (;. and only if

~u ~

Proof.

u CC n,

(i)

0

Then u is subharmonic if

everywhere in 11. Suppose that

~u ~

0

on P..

If h is harmonic on

then ~(u

- h) = ~u ~ 0

on U.

Hence, by Proposition 2, the maximum principle holds {or u-h, so that u is subharmonic by Remark 2 after Definition 4.

39

(ii)

Conversely, suppose that u Then ~u{z) < 0 for all

a ' fl. hood of

is subharmonic iLnd that ~u(a) < 0,

z ( U, where

is an open neighbor-

By Part (i), -u is subharmonic on U.

il.

above, u is harmonic on U, so that ~u(a) Examples. (i)

t;

= 0,

By Corollary 5

a contradiction.

nC

Let f be a function holomorphic on the open set

q;.

If f does not vanish identically on any connected component

of fl, then u{z)

= log If{z) I

is subharmonic. Proof. We have to verify condition (',) of Proposition 'i. a < fl.

If ita) = 0, uta) =

in a disc about a

and (':') at a

-00,

in which

is trivial.

f has no zeros, u

holomorphic function (any g with e g

= f),

Let

01

If ita)

0, then

is the real part of a

so is harmonic, in particular

subharmonic. (ii)

For any '" > 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 <

1z 1 < R}.

0< r < R, let M(r) =

sup Izl= r

function of log r , i. e., Proof.

Let f be holomorphic on If(z) I.

n,

Let R> 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<

z

for fixed

Then, if r < p and m k = ~(r) =

with sup

Jz J =

p,

p< R •

~(z), we have

JzJ ~ r

lim ~ ~ rn. k-+a>

In particu1ar, if (ii) holds for all zED, the conclusion holds for all

r< R. Proof. .. (E) <

e (..

·9

uk{pe 1

)

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<

lim Q'k(r):::' O. where Q'k(r) = k-oo

sup

Izl:::.r

{~(z) - h(z)}.

The proof is

identical; we are of course assuming (i). W", have seen in chapter 1. that if f is a holomorphic function on n

an open set 11 C (; • then

(Ie (lz.

.

= 0 for J = 1 ••••• n.

J theorem of Hartogs asserts the converse.

A remarkable

This can be formulated as

follows. Let 11 be an open set

11j •a the open set in ct any function f on

n.

10

ct

n

and a l •.••• an < ct.

We denote by

{z. C! (a l ••••• an_I' z. aj+l' •••• an)

we denote by f. the function on J.a

f. (z) = e(al •...• a. 1.t:.a.+ 1 ••..• a ). J.a J-] n

n.

J.a

E

nl.

For

defined by

43 Theorem of Hartogs.

Let f be a function defined on

n

such that.

for any a I , .••• an_I'
n. .

Then { is holomorphic on

J.a

We may clearly suppose that

n

is holo-

n.

= {z , O.

The

J

proof requires a few prelimina ry propositions, Lemma 3.

n is the polydisc defined above and that

Suppose that

f satisfies the conditions of Hartogs' theorem.

fis bounded on Proof.

I!:'j I

= p.

1£ n

n.

Then { is holOlnorphic on

Suppose in addition that

n.

We remark that { is measurable on the product of circles

= I.

f is continuous. hence measurable.

Assuming the

measurability proved in n-I dimensions, let 9 1 •••• , 9 p be points on the circle with 9 p

9 1 and the length of the arc 9 j 9 j +l < e,

Then, the

function

(open on the right and closed on the left) converges at each point to f(!:.) since f is continuous in sl ~

for fixed

sZ •.•. , Sn'

By induction,

is measurable, hence so is f. Suppose that

If(z) I :s. M.

Let 0 < p < R.

By chapter I,

Proposition Z applied to the case n = I, we have, if

the disc, we may repeat this j = 1t

••• ,

n ,

I z.1 J

< R,

procedure, and we obtain, for

Iz.1 < p, J

44

where the integral is an iterated integral.

I Zj I ~

Now, for

r < p and

I~jl = p, we have ~ 11 (~.- z.)-1 = L....,

J

Z

a

_..n

J

aE IN

and the series converges uniformly for

I~jl = p,

IZjl

~ r

< p. Since

is bounded and measurable, we may multiply this series by integrate term by term.

Ical ~ Mp-Ia l .

moreover, to f for

We obtain, for

I z.1 J

J

!>. r < p,

Hence the above series converges uniformly

!>. r, and the lemma follows.

Lemma 4 (Baire's theorem). and {W } -1 l p p= , ....

n

Iz.1

and

Let W be an open set in !Rn ,

be a sequence of open dense subsets of W.

Then

00

A=

p=l

W

P

Proof.

is dense in W.

Let U 0 C W be a nonempty open subset, and let

VoCCU o ' where Vo is a nonemptyopen set. a nonempty open set since WI empty open set, VIC CUI' and V hence

Then U l =V o "" WI

is open and dense.

Let

By induction on P. let U

p

be a nonempty open subset of U , V ( ( U. P P p P

n V p = n V.p

Since each

Vp

VI = V

p-l

Then

is

be a non-

n

W

V C P

p' VI'

is a nonempty compact set and

p-

45

vp c Vp-I '

nV p

we have

I~.

Clearly, nV

P

c

(nW )n U • P 0

Since U o is arbitrary, the result follows. Remark.

The theorem is true when W is any complete metric space.

We have only to replace the condition V e e U

p

V C p

p

by the conditions

U , diameter (V ) - 0 in the above proof. p p

Lemma 5.

Let 0 = {ZE c:;nll z.1 < R}, let p < R, and let f be a J

function defined on 0

satisfying the conditions of Hartogs' theorem.

Suppose that. for fixed a , n

Ia n I < R,

f(zl' .... zn_l' an) is holomorphic for M(z'}=

sup If(zl ..... z l.a lanls.p nn

iB an open dense set U' C

Q'.

)1.

the function (zl'" •• z

IZj I <

n-

1) .....

R. j = 1 ..... n-l. and let

I)'

z, = (zl ..... z

Then. there

n-

0 ' = {Z'E c:;n-lllz!1 < R, j J

= 1 ••••• n-l}.

such that M(ZI} is bounded on any compact subset of U'. Proof.

Let V' be any nonempty open subset of 0 ' , and let

Vk = {z' E V'I M(ZI} s. k}, k =

1,....

Then

closed in

Ian I ~ p,

the set

V':

for each an'

UVk

V' and Vk is

{z1.vlllf(z"a}l~k} n is closed in V' since Zl ..... f(zl,a n ) is holomorphic by assumption; Vk is the intersection of these sets when an runs over the disc lanl s. p.

k

V' - V

Hence

k ) = ~.

(V' - V

By Lemma 4, at least one contains a nonempty open set W'

is not dense, so that Vi.: o

(for some k o )' when V'

n

Clearly M(z'}

~ ko' z' E W'. Consider

runs over all nonempty open subsets of 0'.

each V', so is dense in 0'.

E

K'.

UW'

U' intersects

Moreover, any compact subset K' of U'

is contained in the union of finitely many W', for Zl

U' =

80

that M( z') is bounded

46 We can now prove Hartogs' theorem. The theorem is trivial if n = 1.

Proof of Hartog'" theorem.

We assume. by induction, that the theorem has been proved for open sets in tt

c n - l • Let n ={z.

r = {z' E cu-lll z.1J

cnll z.1 J

< R} = n' X 0,

< R, j = 1, ••• , n-1} , D =

{zn (c II zn I < R)

Let p < R, 5 < R - p a.nd U' be an open dense subset of £ is bounded on K' X Kp' Kp = {ZE

(Lemma 5).

Let

a'.

bounded on the set 0'

on

= {z • CI:

II" I <: p)

>(

U',

la'i <

e/I zl

n'

such that

$ p} for any compact K' (" {;"'

5. Then i is holorn.orphic

and

C!;n-Illz'_a'i <,,) ,

0 , 0'" (z" n

if " is small enough (Lemma 3).

Hence

the series converges uniformly un comp.>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


The theorem is proved. We proye next a proposition concerning subha.rmonic functions which we shall use in chapter 4. Let Q be a connected open set in cr; and s

PropositionS.

a subharrnomc

function

t

-00

on

n.

Let A = {" • flls(x) "

1£ u is a continuous function on

suppose that A is closed.

Then u

+ Es

n,

Since s is bounded above on compact subsets of

may suppose, replacing s

by s - c, that s S. 0 on n.

15 subharmonic on

n.

and

n.

subharmonic on fI - A, then u is subharmonic all. ~.

-col

n,

we

Let 2> O.

In fact, condition (*) of

Proposition 4 is trivially satisfied at points of A, (since u +c s =

-00

at these pointe) and outside, (*) is satisfied since u + e s is sub harmonic on ,

n - A.

If

aE n

and {z

II z-a I ~ R} C

Q

and

'9

"IT

h(z)

=o

for

Iz-al=R

P R(z,8)u(a+Re ' )de, thenbyProposition.3, a,

(Since 9S,0).

Hence u+csS.h on

u+EsS.h

Iz-als,R.

Letting

£ .... 0 we find that u(z) 5.. h(z;) if But by Lemma Z, (a),

n-A

I z-al

is dense in

$. R,

n.

z

I

A.

Since u and hare con-

48

tinuous, it follows that u{z) $ h{z) for

Iz-a I $

R.

It follows that u is subharmonic.

1£ s, A are as above, and u is continuous on n and

Corollary.

harmonic on n - A, then u is harmonic on n. We may suppose that u is real valued.

We have then only to

apply Proposition 8 to u and -u. Proposition 9.

Let n, s, and A be as in Proposition 8.

U

then u is a bounded, continuous subharmonic function on 0 - A, there is a subharmonic function v on n with v In-A = u. Proof.

We define v on A by

v{a) Let R> 0 be

=

lim

z-a,z; A

80

that

{z

u{z) ,

II z-a I $

{S.[o,h)1 s{a+Re iS )= -co}

[ 21T P

h{z) =

o

a. A, v(z) R} C n.

121T

R(z, S)u{a+Re' )d& =

if z ~ A.

By Lemma 2, (b), the set

is of measure O. °9

a,

= u{z)

0

P

a,

Define h by 0& R{z, &)v{a+Re' )dS •

Then, as in the proof of Proposition 8, h is harmonic in and v-h+~s is subharmonic in Iz-al < R. if

I ~- z I

= R,

Since lim.. (v-h+Es)(Q$O z-~

it follows that v-h +e s $ 0 on

u{z) !!. h{z) for

I z-a I <

R,

I z -a I < R,

z

I

I z-a I < R.

Hence letting

A •

It follows from the definition of v that v{z)!!. h{z),

I z-a I < R,

and v

is thus subharmonic. Corollary.

1£ s, A are as above, any bounded harmonic function

on 0 - A has a unique harmonic extension to O.

49 These results are merely special cases of much more precise results known in potential theory. functions, see [Z5].

For the theory of subharmonic

Hartogs' theorm is proved in [13].

See also [17]

for a presentation of this proof. We have developed the theory of subharmonic functions to a point at which RadO's theorem (chapter 4, Theorem 1) and a generalization needed in chapter 5 (proof of Theorem Z in chapter 5) become rather obvious.

4

HARTOGS' THEOREM ON THE SINGULARITIES OF HOLOMORPHIC FUNC TIONS

The object of this chapter is to prove a theorem due to Hartogs to the effect that the set of singularities of a holomorphic function tends to be an analytic set. Definition I.

Let fl be an open set in «;n and A C fl.

We say

that A is an analytic set if, for all a < fl, there exists a neighborhood U of a

and finitely many holomorphic functions f l , ••• , fp in U such

that A ('\ U = {f l (z) = ••• = Vz) = o}. Proposition I.

Let fl be connected, A an analytic subset of fl.

Then we have the following. (i)

A is closed in fl.

(ii)

If A

(iii)

0 - A is connected.

-I fl,

Proof. such that

fl - A is dense in fl.

(i) By definition, any a < fl has a neighborhood U

A"" U is closed in U.

(ii) Suppose (ii) is false.

This implies that A is closed.

o

Then B = A -I~.

We shall prove that

B is closed in 0, sinct' B is open and 0 connected, this will prove that A = fl, a contradiction. Let a < B, U be an open connected neighborhood of a in 0, and let f l , •.• , fp be holomorphic in 1'2 and such that

50

An

u

= {x ( U I fl (x) = ••• = fp(x) =

B n U.

o}.

Then each fj vanishes on

Hence by the principle of analytic continuation, each f j ", 0

o on U, so that U C A; hence U C A = B; in particular, a ( B, so that B =

B. (iii) It is enough to prove the following:

(*)

Any a ,

n has a connected neighborhood C such that U - A is

connected. In fact, if (*) is proved, and Po - A = U\ U U z where the Uj are open, nonempty, and diSjoint, we have, by (ii), 0 = 0 - A = Since 0

is connected,

1\ u u z .

U1" U z I- 0. Let U be a neighborhood of a

such that U - A is connected.

Then U - A

= (U l - A)

V (U z - A) cannot

be connected unless one of U l ' U z is contained in A which is impossible by (ii). To prove (*), let U be a convex neighborhood of a in which there exist holomorphic functions £1' .•• ' fp Wlth unA= {x. ul fl(x) = ... = fp(x) = o}. V

= b .•

C IlI.x0 + (I -~)'fl • u}.

one g.(lI.) = f.(lI.'X J

J

A'

= {lI..

C IlI.xo

0

+ (l-lI.)x l ) of

+ {1-lI.)x I

o to

I, t -

'Y{t)x o

is a convex set in C(; and at least

V

0 on V.

• unA}

connected and 0,1. V - A',

Let xo,x l < U-A, and

Hence

is discrete

In

V.

Hence V -A' is

If t .... 'Y(t) is an arc in V - A' connecting

+ (l-'Y(t»x l is an arc in U - A connecting Xl to xo'

PrOposition Z (Riemann's continuation theorem). open set in C(;n and A an analytic set such that f be holomorphic on 0 - A and

suppos~

n-A

Let

be an

is dense.

Let

that for any a. A, there exists

a neighborhood U of a in 0 such that fl U-A is bounded. a unique holomorphic function F

n

on P. with FlO - A = f.

Then there is a

52 Proof.

The uniqueness is obvious.

We first prove the theorem when n = 1. We have only to prove that if f is bounded and holomorphic in {O < I Z I < p}. then f can be continued to the disc {I z I < p}.

Now

J

CD

L

f(z) =

-CD

f(z)z

-,,-1 dz

for any 0 < r < p.

Izl=r

For ,,< O. this integral" 0 as r " 0 since f is bounded.

Since a

a" = 0 for" < 0 and the result follows.

is independent of r.

In the general case. let a

E

A. and let V be a connected neigh-

borhood of a so that f IV -A and in which there are holomorphic functions fl ••••• f p with A" V = {x may suppose that h. fl

+O.

Vlfl(x) = •••

E

= fp(x)

= O}.

We

By making a linear change of coordinates

in C n • we may suppose that a = 0 and that h(O ••••• O. zn); 0 in a neighborhood of zn

= O.

There then exists 6> 0 so that

h(O ••••• O.zn);I 0 for 0 < IZnl ~ 6. Let E> 0 be such that h(z'. zn)

f.

We denote (zl ••••• zn_l) by z'.

0 for

sider the function g(z)

= 21Ti

f

Iz'l ~e .Itl

holomorphic for I z'l < e

•

Proposition 2 in chapter 1.)

I zn l = 6.

n E

V-A and so g is

(Note after Corollary 2 to

Moreover. for fixed z'. Iz'l < E. the

function t .... f(z'.t) admits a holomorphic extension to the disc by the case n = 1 of the theorem (since h(z'. zn) that

f.

0 for

zn'" h(z'. zn) has only finitely many zeros in

by Cauchy's formula.

Con-

t -z

6. the point (z'.t) I zn I < Ii.

and

f(z'.t)dt:

Itl=6 note that for

I z'l ~ £

It I < 6

I zn I = 6. so

I zn I < 6).

Hence.

53

g(z) = fez) if z

E

Iz'l
V-A.

Hence f can be continued to a neighborhood of any point of A. The result now follows since the extension is unique. Theorem 1 (Rado's theorem).

Let U.O be two open sets in en.

U CO. and let f be holomorphic in U.

a

E

(au)"

Suppose that for every point

0 we have

z1i!l'a.

U fez) = o.

ZE

Then the function defined by fez) F(z} =

{

if z

E

U

0 onO-U

is holomorphic in O.

This is equivalent to the following. Theorem I'. oce n and let U= on U.

Let f be a continuous function on the open set {ZE

01

f{z)io}.

Suppose that f isholomorphic

Then f is holomorphic on O. Theorem 1

function F

=- Theorem I'.

Given f as in Theorem I'. the

defined in Theorem 1 is

Theorem I' =:> Theorem 1.

= f.

It is enough to check that if f is

given on U and tends to 0 at (aU)" o. then F

is continuous.

This

is obvious. Proof of Theorem I'. n

= 1.

It is enough to prove the theorem when

For. assuming the theorem in the case n

= 1.

if we consider

a 1 ••••• a n _ 1 • e. then the function z ..... f(a 1 ••••• a. 1.z .• a .••••• a} J JJ J n satisfies the conditions of Theorem I' in an open set in C and so is holomorphic.

Hence. by chapter 3. Lemma 3. { is ho10morphic on O.

54 The theorem being local, we may suppose that n

to.

Moreover, we may suppose that f s

is connected.

Let s{zl = loglf(zll.

Then

is subharrnonic in n : Condition (,~) of chapter 3, Proposition 4 is

trivially verified at points of A = {z ( nl F{z) = O} = {z , nl s{z) = -co}; on n-A,

fin-A

s

is harmonic, and condition (~c) is again verified.

is holomorphic, hence h"rrnonic.

Proposition 8, f is harmonic on

differentiable.

Since

~ =0

n,

on

ilZ-

Df

3, Lemma 2, (a)), we have

By chapter 3, Corollary to

in particular continuously

n-

o

a-z

~1oreovcr,

A and

n-

A is dense (chapter

on n, So that f is holomorphic.

The idea of the above proof is due to II. Cartan [9].

It is a simple

matter to elimmate all reference to subharrnonic functions and to prove directly, using chapter 3, Proposition 3, the special case of chapter 3, Proposition 8 needed. Definition 2.

This has been pointed out by E. Heinz [15]. (al A function defined on a subset

seen

is

called holomorphic on S if it is the restriction to S of a function holomorphic on an open set t: (b) Let

n

~

be open,

s"y that f is singular at a

S.

A C E

n.

If f is holomorphic on n-A, we

A if there is no holomorphic function in

a neighborhood U of a whose restriction to U-A is flU-A. These definitions arc not the best ones available, but we give them simply for convenience in stating the main theorem of this chapter.

55 Theorem 2

(Hartogs' continuity theorem).

nected open set in en and let

n

Let f be holomorphic On

Q

X Q.

= {w

•
I

r <

Let

Iwi

n be a con-

< R}, 0 ~ r < R.

Suppose that there is a point a <

n

such that f can be continued holomorphically to a neighborhood of {a} X D, D = {w.

cJlwl

< R}.

Then f can be continued holomorphic-

ally to rl X D. co

Proof.

2:

Let f(z, w) = v

a v (z)w"

be the Laurent expansion

=-00

of f which converges uniformally on compact subsets of rl X Q. the a v are holomorphic on

n.

Now, if r < p < R, there is

Then

£ > 0

so

that f(z, w) can be extended to a holomorphic function F{z, w) on

{ Iz-a I < £ , Iwi {a} X 0).

< p}

(since f can be extended to a neighborhood of

Hence, for

Iz-a I < £

, the a v

the principle of analytic continuation,

a

v

with

=

0 on

< 0, are

v

n

= O.

if v < 0,

90

By tha t

co

f{z, w)

=

L

v =0

a v (z)w V

,(z, w) <

n

X Q.

This latter series then converges uniformly on compact subsets of

n X 0, and the result is proved. Corol~ary.

o < Iw I < Let

Let

R in a:;n x C.

be a function holomo rphic on the set z = 0, Suppose that (0,0) is a singularity of f.

5 > 0 be sufficiently small.

any zoo

I Zo I < E , Let

t> 0 so that for

f cannot be extended holomorphically to any neigh-

borhood of {zo} X 0, 0 = Proof.

Then, there ey_ists

{Iwl

< o}.

e> 0 be so chosen that f is holomorphic on the The corollary follows from

Theorem 2. Our next main theorem is again due to Hartogs.

56 Theorem 3.

cp: n - (; a map.

Let

n

be a connected open set in {;n and

Suppose that

Icp(z) I <

U=IlX D, D= {WE cllwl
R

for all ZEn and let

u)f cp(z)=w}.

Let 1= {{z,w).

Suppose that f is a function holomorphic on U - 1

r.

at every point of

which is singular

Then cp is a holomo rphic function of z.

The proof requires several preliminaries. Let n, cp, f, R, D be as in Theorem 3.

Lemma 1.

Then cp is

continuous. Proof.

en, cp(zo) = woo

Let Zo

Theorem 2, there is to the set {z}

I)

e>

O.

By the corollary to

> 0 so that f cannot be extended holomorphically

{Iw-wol <E}

X

Let

for

Iz-"'ol < Ii.

Since, for

f is holomorphic on {z} X (o - {cp(z)}), it follows that

Iz-zol < Ii

Icp{z) -woi
so that 'i' is continuous. Let

n

be an open set in C and

holornorphic functions on Il.

{f),

v = 0,1, . ••

a sequence of

We consider the series

co

L

(H)

f

v =0

v

(z)w v

,

and suppose that it converges uniformly on compact subsets of a neighborhood of

n

X

{O}.

Definition. function R:1l - lR

The Hartogs radius of the above series is the

+

defined by

co v converges uniformly on a neighborR(z o ) = sup{ Iwll ""' ~ f v (z)w

o

hood of (zo' w) } •

Since the series converges uniformly on a neighborhood of is clear that

R{z ) > 0 for all z o 0

{Q.

n

X {OJ, it

57

If R(z)

Proposition 3.

t-

+ ro, the function

-log R(z) is sub-

harmonic in fl. Proof. there is

a>

To see that -logR(z) is u. s.c., let

r < R(zo).

Then

such that the series converges uniformly in a neighbor-

0

hood of the set

Iz-zol ~o, Iwl :::.r.

Iz-zol !>.o, w= r, hence on

In particular R(z)?, r

for

I z - Zo I < 0, and we have

lim R(z) 2. R{zo). r-z

Let K = { z. ell z - zol :::. p} C fl.

o

harmonic on K.

Suppose that, for

o

Let h be continuous on K and

Iz-zol = p,

- log R(,,) :::. h{z) .

(I)

We have to show that (I) holds for converges for

17o-zo I < p.

Since the series (H)

I,,-zol = p, Iwl < R(z), we have

If)z)1

l/v

~

1 R{z)

h{z) !>. e

for

V-DO

On the other hand, R{z) is bounded below on K, hence (H) converges uniformly on a set of the form

K

X {

I wi!>.

1')},

v z • K, where

1')

> O. In particular,

V v = 0, I, •••

,

M is a constant.

Hence by the remark following Proposition 7, chapter 3, we have

lim

a)r)

5.

0,

V-DO

where a)r) =

sup

I z- z o l5. r

Hence, for € > 0,

v> v o(E).

{.!..loglf(z)l-h(z)lJ. v v

I f)z) Ii/v

!>. eh(z)+s

O
uniformly for

I z-zo I :::. rand

Hence the series (H) converges uniformly in the neighborhood

of (z,w) for

I z-zol < p , Iwl < e -h(z) -t •

I z-zo I < P , which proves (I) for these

z.

Hence

R(z)?e

-h(z)

for

58

We now tum to the proof of ThID rem 3 in the caSe n = 1. may suppose that 'I'

0

Ene

4:

and that

rp{O)

= O.

is holomorphic in the neighborhood of z = O.

We have to show that Let

lR > '1 > 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 <

o.

Izl < 0

We

(Lemma 1).

Let Wo

E (\; ,

Let

The above series converges uniformly in the neighborhood of (z,w) for

Iw-wol< l
since f is holomorphic ina neigh-

borhood of such a {z, w} (note that

I w 0 I < R-'1, so that

I rp{z} - w 0 1< R}.

Hence, the Hartogs radius associated with such a series, R{z) ~ Irp{z) - w 0 I.

fies the inequality R{zo) > Irp(zo) - wol is semicontinuous, with

R{z), satis-

On the other hand, if

for some zo' Izol < 0 , we would have, since R R(z) > 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
Proof.

By Proposition 3 and

Lemma 2. u(z. w 0) is a subharmonic function of z for p < 0 - I z I.

1o

2"

o

u(z:,.re irp )drp=-

o

T) < r < R-T). we

1

2"

1 loglrp(z:,)-re irpl drp=2"10g~

for

0

since

I al < r.

0

We integrate with respect to rp from 0 to 2".

I"

1z:,1 < 0

°a

u(z + pel .w )da

= rei'P in this inequality! for fixed r.

have (II) for 0 ~ rp ~ 2". Now.

Hence.

we have

(II)

We set w

in I z I < O.

1
and

1o

2"

°

logla_relrpldrp = 2'1flogr for

[To prove this latter formula. it is enough to remark that

logla_reirpl =loglae-i
I z I < r, so that we can use chapter 3, Proposition 3.

Corollary 1. )

1

2"

This gives

u(z. re i'l' )d'l' = 2" log -1 =

1 2" o r "

1 1o · 2"

d
2"

i9 re i'l' ida u(z+pe.

0

This and (II) imply, since u is continuous, that

f"

1 U(Z'W)=-2 o " 0

°a

u(z+pe l ,w)d8 0

By Corollary 7 to Proposition 4 in chapter 3, this proves the result. Lemma 3. T)

< Iwl <

Since log I rp(z) - w I is harmonic, it is a real analytic

function, hence

Ceo; hence so is

IjJ (z, w) = I 'I'(z) /2 - w
o ~ ~ ~ 2.....

is harmonic in z for all w,

then rp or -; is a holomorphic function of z.

P-T),

Proof.

If logl
Hence W

w~(z)

w"i

=

/'P(z)_wI 2 = exp(210gl
~(.y(z,

rei~

,T)

-wi - .y{z, wi) is Coo.

< r<

R-T),

Taking

w= r. w= ir, we find that q>+q; and q>-"i, hence also rp, are Coo.

60

fJ,p/a-z

To prove our result, we have now to show that either a<j/az

is identically O.

al

o=

or

Now -- _

azaz log (q>(z)-w)( ",(z) - w) = (",-w)

-1

al a a iat- (",-w) - l Tz-' -Fz

a - _ (;;;" _ w)- l J!. a- . .....!! a- , + (- -)- 1 ...!!....!L l

'" - w

az O z ' "

for all w in a nonempty open set •

8z

a!"

Multiplying by (",_wflcq;-_w)l and

equating powers of wand ;;; , we find that

The first relatio:l implies that

: : vanishes on a nonempty open

set.

Hence, by the principle of analytic continuation, either

~

( : : ) is identically 0, and the lemma is proved. Corollary.

Under the conditions of Theorem 3, if n

fJ,p

Or

8Z

= I,

then

either '" or ;; is holomorphic. Proof of Theorem 3 when n Then, for

11;1, 1w 1

= 1.

We write

w

= w+ z,

1;= z.

1

sufficiently small, the set {(I;.w) w = ",(I;) + I;}

satisfies the hypotheses of Theorem 3, n = 1 with respect to the function f(I;, w-I;). Hence, by the above corollary, in a sufficiently small neighborhood of then

1;=0,

",(9 or ",(I;) + I is holomorphic. If '" is not holomorphic,

q;(O and q;(~)+ ~ are holomorphic, hence so is t, which is absurd.

Hence '" is holomorphic at 0, and the result is proved. Proof of Theorem 3 in the general case. a sufficiently small polydisc with center a, shall prove that the function convex set D

Let a. rl and P be

perl.

Let z

o

• p.

We

>. - ",(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
Iq:o(z)I<" for

a

=0,


Izl
Then f is

{(Z,W)E cn+lllzl
expand in a Laurent series: f(z,w) 11= -00

v < 0 so that fv ¢ 0 on o 0

Since f is singular at (0,0), there is p

= {I z I <

o}.

Let S

= {z <

pi fv (z) o

z is holomorphic on the set

o

-I

0 }.

The function

• S

{(lI.,w)Ill. • D, wi q:o(lI.zol)

and is moreover

singular at some point (lI., w 0) for each lI. (since otherwise f

Vo

(fLz) 0

=0

for fL in a neighborhood of Some lI., hence f

by the principle of analytic continuation, a contradiction).

Vo

=0

(z) 0

Hence

g(>.., w) is singular precisely on the set w = q:o{ll.zo)' and the result follows from the case n = i

proved above.

Thai q:o is holomorphic on

n follows now from chapter 3,

Lemma 3. Proposition 5. D = {w. a:llwl < R}

Let 0 be a connected set in C n , and let K be a compact subset of D.

Let

A C 0 XK and suppose that there exists p> 0 so that, for all z

is a finite set containing at most p elements.

E

0,

Suppose that there is a

holomorphic function f on OX D-A which is singular at every point of A. so that

Then, there are holomorphic functions
62

A = {(z. w) • rl X D I w in particular.

runs over

elements.

+
Let

n.

and let U be the set of z,

=A

a

Let €. <

•

I

2:

n

so that

.f>, Z

contains

q

Let a < U and

max Iw. - w.l. 1 J

i/j

It is enough to prove

f. r6

that. if z is near a. A {\ {w, cllw-w.1 <£}

z

Since A z

}

q be the maximum number of elements in A z when

We claim that U is open.

{w l' •••• w } q

0

A is an analytic set.

Proof. z

q

1

for i = l ••••• q.

is. by assumption. the set of singularities of the form

(z. w) of f. this follows at once from Theorem 2. For z, U. let A g(z)

= i

z

= {wl(z) ••••• w (z)}. and let q

n <j

(w.(z) - w.(z»2 . 1 J To prove this. let a < U and

We claim that g is holomorphic on U. Let

E<

1

"2

z

be near a.

max Iw.(a)-w.(a)l. and for each i and z i =j 1 J near a. let w.(z} be that element of A with Iw.(z)-w.(a) 1 < £ • For z

near a. we can apply Theorem 3 to wi(z); hence wi(z} is holo-

1

Z

1

1

morphic in z for z near a. and it follows that g is holomorphic. Let a v '

U. a v - a, (au)

r'I

n. Since. by assumption wi(a) < K.

there is a subsequence {vk } so that Since

A

wi(a v ) - Wi < K. i

= 1.2 •...• q.

is closed (see remark after Definition I). (a. Wi) < A.

i = 1 ••• , • q.

Since a ( U. we conclude that w. = w. for some i'; j. 1 J

It follows. since .. aeh w;
Vk

) - 0 as k - co.

Since this is true for any

sequence a v - a. it follows that g(a) - 0 as Rado's theorem (Theorem 1). the function h on

v - co.

n

Hence. by

defined by

63

g(Z} ,

Z,

U

h(z} = {

o n.

is holomorphic on

Moreover,

,

n-u= {zEnl h(z)=O}, and so is an

analytic set in n. For z ( U, define q

IT

pl(Z, w) =

,

(w - w.(z})

i= I

where I"'(z} = (-I) k

~ 2.....

k

I ·· ~ ' 1< ... < 'k~q

. w. (z} ••• w. (z) ,sthek-thele'I 'k

mentary symmetryc function in the wi(z). bounded on

U.

Since wi(z} E K, I"k is

Further, the ~ are holomorphic on U (see proof that

g is holomorphic given above).

Hence, by PropositlOn 2, there are

holomorphic functions I"k on n, whose restrictions to U are the

'PI.

above. Let p(z, w} = w P{z,w}

00,

lj

'1'1 (z}w

t-

then w, K.

Since U i.s dense in

n,

q-l

+ .••

Clearly, for

Z

+ I" q(Z).

If (z, w) ,

E C", A z = {w

we conclude that A C {(z, w)

E

E

n X C and

([;Ip(z,w) = A}.

nx([; I P(z, w} = OJ.

Conversely, if (z, w) E n X([; and p(z, w} = 0, then w ( KeD and if z

v

-z,

Z

v

E

Dr there are

W

v

E

ct, P{z,w )=0 so that w -w. Since v V v

(z ,w ) E A C A and A is closed, (z,w) v v Zv

E

A, So that

A = {(z,w) E ax clp(z,w} = oj . Theorem 4.

Let W be an open set in ([n+1

and A C W a sub-

set such that (a) For all a E W there is a neighborhood n X D , nCCn,D= {!w-an+I!O

suchthat

{w ( D! (z, w) • II) contains at most p points for any z, n.

64 (b)

There is f holomorphic on W - A which is singular at every

point of A. Then A is an analytic subset of W. Proof.

n

Let a. Wand

{WE DI {a'.w}. A}

X D be as in (a).

= {wl •.••• w}. a'= {al ••••• a n }. a= (al ••••• a ntl ).

Let O
{a'}x{wllw-antll=r}()A=~.

E is small enough. the set Iz -a'i $ L not meet A.

Consider

If P= {z
Then. if

r-f $ Iw-antl ' $ rH

does


it follows that the projection on Do of A () P X Do is contained in the compact set {w II w-antll $ r} holomorphic functions A

n

of D.


P X Do = {{z. w} • P XD

o

By Proposition 5. there are

q~P.

on P

with

I wqt
Clearly. then. A is an analytic set. For the results on analytic sets. see [16].[20].

Hartogs proved

Theorem 3 in [14J. A form of Rado's theorem occurs in [26].

Its usefulness in com-

plex analysis was recognized by H. Behnke and K. Stein. We have assumed in Proposition 5 that the number of points in the set A

z

is bounded as a function of z.

sis is essential is unknown.

The extent to which this hypothe-

5

A UTOMORPHISMS OF BOUNDED DOMAINS

Let 0 be an open set in {;n and f

= {f l , ••• , fm):O

We say that f is holomorphic if each fj is, Definition 1.

I So j

~

- {;m a map.

m.

Let 0 be a connected open set (domain) in {;n.

A holomorphic map f:O - 0

is called an (analytic) automorphism if

there is a holomorphic map g:O -0

with gof = id., fog = id.

lid is

the identity map of 0.) Remark.

A theorem of Osgood, which we shall prove at the end of this

chapter, implies that a bijective holomorphic map of 0

onto itself is

an automorphism. If D is a bounded domain in (;n, we denote by Aut(D) the set of

automorphisms of D. maps.

Aut(D) is a group relative to composition of maps

We introduce a topology on Aut{D) as follows. Let KeD be compact,

the set (er

E

U and open set contained in D.

Then,

Aut(D) I erIK) C U} runs over a basis of open sets for the

topology of Aut(D) when K runs over the compact subsets of D, U over the open sets. the sets

(er

E

This topology has a countable basis

Aut{D) I er{K )

C U} whe re

"'"

able base of open sets in D and {K,,}

{U}

formed by is a count-

'" '" = I, Z, ••.

= 1, Z,...

is a sequence of

o

compact sets such that the {K.,J

form a basis of open sets for D}.

65

66 (.-) C Aut(D) converges if and only if (Tv converges

A sequence

uniformly on compact subsets of D to an element over.

(0-,


(Aut(D).

More-

Aut(D) is Hausdorff. and is a topological group [the map

T) ...... 0' OT- 1

is continuous].

Proposition 1 (H. Cartan).

Let D be a bounded domain in en

and f:D - D a holomorphic map. with ita) = a.

Suppose that there exists a. D

Let the expansion of f in a series of homogeneous poly-

nomials about a be of the form

f(z) = z where P k

is an n-tuple of homogeneous polynomials of degree k.

f{z)

Then Proof.

=

z.

We may suppose that a = O.

tained in the polydisc bounded).

+ P 2{z-a) + ..•

If F

{z II z.1 < J

R}

Suppose that D is con-

(such an R> 0 exists since Dis

is a holomorphic map of D into itself.

F=(F 1 ..... F). and F(z)= La ZCl n

Cl

(a

(a 1 ••••• an ) • en) Cl

Cl

is

Cl

the Taylor expansion about O. then. by Cauchy's inequalities (chapter 1, Proposition 3), we have

{ZE

cnllz.1 J

la",!

~

Rp-IClI ,

p> 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
foranyr
j =i

i. e.

J

(I) Consider now. for a given j. the sequence z" wher.e we haTe 1 -

~

1

= (0 ••.•• 1- v ..... 0)

in the j-th place and 0 elsewhere.

By the

remar'k after Definition 3. every limit point of fez ) in Q;n is on II

Smce f(z,,) = (1 -

~ )(a 1 .••••• a

latter point is in

aD.

J

.) - (a 1 .••••• a .). we conclude that this J nJ

nJ

i. e ••

max k=1 ••••• n Let k(1) be so that j . Z••••• n.

for j= i •...• n.

la k (1).1 1 = 1.

Let k(Z) be so that

By (I) above. a k (1).j = 0 for

Iak(z).zl

(smce ak(t).Z = 0) and ~(2).j = 0 for j so that

Iak(j).j I =

aD.

I

= 1.

Then k(Z)

2 by (I).

I

k(i)

Thus if k(j) is

1. then (k(t) ••••• ken)) is a permutation of

(1. ••••• n) and ak(j). i

=0

for i

I

j.

If P is the inverse permutation

70 of the one just constructed, we have fk{z} = ak,p{k}zp{k} , 1ak,p{k} 1 = 1.

q. e. d.

It is a classical theorem, due to Riemann, that any two simply connected domains in

e, fe,

are analytically equivalent.

Proposition

3 enables one to show that the situation is very different for domains

PropositlOn 4 j=l,2}

(Poincar~).

and B={ZEC 21

Let P = {(zi ,z2)

IZII 2 +Izli 2
E

e 2 11 Zj 1< I,

Then there is no analytic

isomorphism of Ponto B. Proof.

Since there is an automorphism of P

taking any point

of Ponto 0, it suffices to show that there is no iSOTTlorphism f: B - P 0--

with f{O) = O.

foo-o£-l

Suppose that there is.

Then, the map

is an isomorphism of Aut{B) onto Aut{P) as topological

Hence, in particular, if Go denotes the connected component

groups.

of e of the topulogical group G, Aut{B)o and Aut{P}o are isomorphic. Moreover, the above map induces an isomorphism of the subgroup G o£ Aut{B} leaving 0 fixed onto the subgroup H of Aut{P} leaving 0 fixed.

Hence Go is isomorphic to Ho'

that any

0- E

It follows from Proposition 3

H which is sufficiently close to e is of the form

Ho contains only clements of the form (1) above, and so is abelian.

On the other hand, G

o

contains all elements

where A is a 2 X 2 unitary matrix.

T

of the form

Z -

Az

Hence the 2 X 2 unitary group

71

U(2)

C

Go as a subgroup.

not isomorphic to

Hence G

cannot be abelian so that G

o

0

is

Ho'

We shall now prove some theorems, due to Remmert and Stein, which shows that the situation in n > I

variables is really complicated.

We begin with a lemma. Lemma I.

Let

n be a connected open set in (;n and £1' ••• , fm

n.

holomorphic functions on m

L If.{z) I j=1

Suppose that 2

n.

is constant on

J

Then the f. are all constant. J Proof.

n.

If rp is holomorphic on

we have. for k

= 1 ••••• n.

rprp

2: If.{z) 12 = const .• J

Hence. if

L

o

we have 2

If.{z)1 J

af.

LI~I

2

ilL so that

---1.. _ 0 for j = I ••••• m. k= I ••••• n. a~

Theorem I (Remmert-Stein). let n

= n l + n 2•

n 1 • n 2 > O.

open set U = U 1 X U 2' U j

un

D

Let D be a domain in

ce n •

and

Suppose that there is a point a < aD and an

C ce

n· J • containing

a such that n. _ 1 X OZ' OJ open and connected in C J. with 02" U 2

=°

Then. for any m L I. there is no proper holomorphic map of D m 2 the ball B {(wl ..... w ) ( em I Iw.1 < I}. m m j =1 J

f

Uz

.

into

2:

(Note that the condition is. in particular. satisfied if and is bounded; we may take U

= (;n.)

° = DI X D2

7l

Proof.

Let f = (fl' •••• fm) be a proper ho10morphic map of n1

into Bm.

We shall write z = (I;,. w). 1;,. (;

and suppose that "'v -

w.

°

nZ

• w. (;

•

Let..,v. Oz

(aDz)" U z • For j = 1 ••••• m. the functions

I;,"'f.{z • ..,) is aholomorphic function '1'. on 0 1 with 2::1'1'. I Z < 1. J v J.v J. v Hence. by Montel's theorem (chapter 1. Proposition 6). there is a subsequence {v k } so that 'P. - 'P. uniformly on compact subsets of 0 1 • J, v k J Now. for any 1;,< 0 1 , {{I;,.",

)} has no limit point in D. Since f is "k proper. f{I;, • .., ) = ('1'1 (1;,) ••••• 'P (1;,» has no limit point in B "k '''k m. "k m Z m l Since 2::1'P j .,,(QI < on D 1 .weconcludethat ~I'I'/I;,)I

lim.@-

k-oo

~ l'Pj."k(l;.)1

'I' j = const..

z

for all 1;,. D 1 •

Hence by Lemma 1.

j = 1 ••••• In.

Now. by Weierstrass' theorem (chapter 1. Proposition 5). if I;, = (1;,1 ••••• 1;,

nl

).

af.(I;, • .., k) J

"

(p = 1 ••••• n 1 ). af.(I;, • ..,) J

Hence. for p= 1 ••••• n 1 •

a I;, p

tends to 0 if .., tends to a point

Hence. for fixed 1;,. D 1 • the function

..,-

{

af.(I;. • ..,)/81;. J p

o

if..,.

otherwise

is ho10Inorphic on U Z (Rad~'s theorem: chapter 4. TheoreIn 1). Uz

by assumption.

- 15Z f

~. this implies that

af. _J_ '" 0 on D 1XD Z ' al;.p hence CJf./a I;. J

p

'"

0 on D.

Since.

p= 1 •••.• n 1 ;

p = 1 •.••• n 1 • Hence the map f is constant

73

on any connected component of the set Dl (wo ) = Clearly.

{w = wO •

W

Dl (wo ) is not relatively compact in D. so that

O

E

D Z }.

cannot be

proper. Let D be a domain in
Theorem Z. n l • n Z > O.

Suppose that there is a point a. aD and an open connected

n· J. containing a n. with D j open and connected in (; J and

C «;

set U = U l X U Z ' Uj

such that U 1"\ D = Dl X D Z

D Z (\

UZ'l U Z•

Let

W=WIX ••• XW m be a domain in em where Wl ••••• W m are bounded domains in (;. WE

(We denote a point in ([;n

by

(z. w).

nl Z E

G:

•

G: n z.)

Let £ = (£1 ••••• fm) be a proper bolomorphic map of D into W. Then. for at least one j. 1

~

i. e .. if z = (zl' ...• z ). n1

af./az

Proof.

fj is locally independent of z.

j ::;. m.

J

p

= 0 for p = 1 •••• ,n l .

We shall need the following generalization of Rado's

theorem.

( *)

Let (
be a matrix of holomorphic

functions on DC U, where D and U are connected open subsets of C n • and U _

D 'I ¢. I

Suppose that

k

n 2: /<1'

p(z) '"

11=1 fl.=1 for any I;,

E


tained in U. by

Let

ZE

D. z-I;,

Then. for some v o' 1:So v 0:5.1 • we have

(aD) 1"'\ U.

Proof of (*).

(z)/Z - 0 as fl.v

fl.Vo

'" 0

Let a

£

• fl. = 1 •.•. k.

(aD) 1"\ U and P

a. D "P and

r. E

P -

D.

a polydisc about a conand let VC (; be defined

74

and let s(>..) = 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 {
converges uniformly

on compact subsets of 0 1 to a function 'l'j (Montel's theorem; the Wj

75 are bounded). in W.

Then

Since f is proper. no limit point of fez. w v ) can be

Hence. for all z. D I •

U E j = D t.

that E j

aw.

('I't(z) •...• 'I'm(z»,

Since each E j

Set

is obviously closed in D I' it follows

has a nonempty interior V for some j. I:>' j

:>. m.

Since 'I' j

maps V into 8W. and a nonconstant holomorphic function is an open J map into C. it follows that '1'. is constant. rNote that the index j J aq>. depends. ~priori. on the sequences {w).{wv }.] Hence 0 k v

if-:;

On D l'

v

= I •.••• n l

(z = (zl ••••• z

for any sequence {w)

In any case. it follows that

tending to a point of (aDZ) n U Z ' a subsequence m

of j

tends to zero.

».

nl

if

=1

n1

8f

Z

~ I~

v= I

(z.w

Zv

v

)1

Hence

n

IT

a:at.

nl

~

I

j=1 v=1

Z (z.w)1

v

tends to 0 as w tends to a point of (8D Z)

n

U Z•

It follows from (*)

that we have:

(**)

For each

Z E

D l • there is j = j(z) nl

p.{z, w) = J

Since the set

8f.

:Llaf (z.w)1 v =1

such that

2 = 0

for all w ( D2 •

v

F j = {(z. w). DI X D2 1 p/z. w) = o}

is closed and

UFj = Dl X D z• at least one F j has an interior point. llf.

~(z, w) :; 0 on D,

v = I, •••• n l •

v

for this value of j

(principle of analytic continuation).

Clearly

76

Let

Lemma Z.

n

be an open set in (l;n, n::: Z.

n

no proper holomorphic map of Proof.

Let a.

n

into a domain

and A = f- 1f(a).

-1 g(z) = (f(z) - f(a)) ; g is holomorphic on

we

Then there is

<1:.

Then A is compact.

n - A.

Let

B be the closed

ball about 0 with the smallest radius containing A (in general, B Then, there exists radius 1.

zo' (6B)

n

A.

I z1 1

Hence, if I zjl set.

:::.£ ,

¢.

n).

We may suppose that B has

By choosing a suitable orthonormal basis of ([n, we may

suppose that Zo = (0, ••• ,0,1). the set

Let

= 6,

Zz

= 0,

Then if

= 0,

zn_1

zn

6> 0 is sufficiently small,

=I

is contained in

n-

A.

e> 0 is small enough, so is the set 6-£ :::. I zll 5. 6+e , j = Z, ••• ,n-l, I zn -115.£ , so that g is holomorphic on this

Moreover, g is holomorphic on the set

j = Z, ••• ,n-l, zn = 1 +",

,,>

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
C
a proper holomorphic map of D into W.

f.( Z ('))' j J p J

n-

whe re

Let f = (f l , f 2 ) be

Then f. is of the form J

p is a permutation of (I, 2).

Proof. independent of zl' one of z2'

It is sufficient to prove that neither fj

is independent of zi and of zz.

But if, for instance, fl

is constant,

fZ'D - W 2 would be a proper holomorphic map, contradicting Lemma 2.

77 Theorem 3 applies in particular to analytic isomorphisms of

D onto W.

The most general result that follows easily from the

reasoning given in Theorem Z is the following. nj Proposition I.

Let Dj' Wj be bounded domains in {;

j = 1, ... , k, and suppose the following about W/

aw.

contains no positive dimensional analytic set in an open set J in {;nj. Let f = (f l , ••• , f k ) be a proper holomorphic map of D

= Dl

n. C J.

X ••• X Dk into WI X ••• X W k ; here fj

is a map of D

".. nj Denote a point z. D by (zl' •.• ' zk)' Zj' ....

a permutation p of the set

into

Then, there is

{I, ... , k} and proper maps

such that

f = gl X ••• X ~ • The proof uses the following generalization of Lemma 2. If

nC

Cn

and

we

proper holomorphic map of

{;m,

n

m < n , are open sets, there is no

into W.

For a proof of this latter fact, see [ZO J, chapter III, Proposition 10 and its corollaries, and chapter VII, Propositions I and Z. Further, in the case of automorphisms, H. Cartan has proved the following generalization of the proposition above Proposition II (H. Cartan). n

= n l + n Z'

Let D

= Dl

n l , n 2 > 0, be a bounded domain.

(see [8]).

X D Z' Dj

C

n· C J ,

Then, any f. Aut (D)

which belongs to the connected component of the identity in Aut (D) is of the form

78 where f. ( Aut (D.). J J We shall prove now a fundamental thea rem, due again to H. Cartan, and give some of its applications. Theorem 4 (H. Cartan). Let

{fv}

that

fv

Let D be a bounded domain in «;n.

C Aut(D) be a sequence of automorphisms of D. Suppose converges, uniformly on compact subsets of D, to a holo-

morphic map f:D - «;n.

Then, the following three properties are

equivalent.

(i) f. Aut (D) (ii) £(D)

1.

aD

at

(iii) There exists a

E

D such that the jacobian matrix (~(a)), j

f

= (£l

J

•

a

•

J

fn)' has a nonzero determinant.

We shall need two preliminary propositions. If f:n - «;m

is a holomorphic map, we denote by (df)a' a ( n,

the linear map of (;n into

a;m defined by

where n

w

Lemma 3. Suppose that

j

=

L

vk

k=l

a£. a::k

(a) ,

Let n C (;n and f:n - «;n be a holomorphic map.

dct(df)a

i

0 for some a

neighborhoods U of a and V of f(a)

E

r!.

Then there exists

such that feU)

C

V and fl U

is an analytic isomorphism onto V. This is a special case of the implicit function theorem which we do not prove.

For a proof, see, for example, [21, chapter 1).

79 Proposition 5.

Let {'I' } be a sequence of continuous open v n n mappings of rl C C into If: • Suppose that 'I' converges, uniformly v n on compact subsets of n to a map 'I':n _ If: • Suppose that, fo r some a, n, a ,,-l,,(a).

is an isolated point of

Then, for any neighborhood U of a, there is avo such that

Proof.

Suppose that this is false.

By passing to a subsequence,

we may suppose that U is such that

U

is compact, ",(a)

i ",)U) ,

v = 1,2, •.• ,

C("\",-l",(a)= {a}. Then 'I'UH.;) is compact and 'I'(a) hood V of q>(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.

Let r' be the subgroup generated by {'1 1 ••••• 'I N}' -1

'I"

,h i •...• '1",}

for

"

If r'

f

Let r'(K) =

U

'1'(K). r"(K) =

'1',r'

= r' v

Since

i!::.i!::.N. r' is the set of products y .••• '1. 'i 'p

r, let r" = r - r'.

Then, since r

can be

r", and

U

'1(K)

= D,

W

'I "(K).

'1',r"

we have

'1,r r'(K) v r"(K) = D.

Further, r'(K)

1"\

r"(K) =~.

x,y' K. 'I"

r'. 'I"' r". we would have

x,y' K, 'I'

h i ; •..• '1 N },

diction.

say 'I = 'Ii'

In fact. if '1'x = '1"y.

'1Y = x. 'I = 'I'

'I"

Since

Then 'I" = 'I ''Ii • r'. a contra-

Hence r'(K), r"(K) are disjoint.

Moreover, the family {'1(K)}

Y'

r

is locally finite (i. e •• any point

of D has a neighborhood U such that {y E r

I y(K)"

it suffices to take for U any compact neighborhood. rn(K) are closed in D.

U jI'~}

is finite;

Hence r'(K) and

Since D is connected and r '(K) jI'~. this

implies that r"(K) = ~. i. e., r" = ~. so that r' = r proved.

-i

and the result is

86 We proceed now to prove the theorem of Osgood referred to at the beginning of this chapter.

We shall need the rank theorem (which is

stronger than Lemma 3).

For a proof, sec e.g., [21, chapter 1].

The rank theorem.

Let 0 be an open set in a;n and f:O-

a holomorphic map. an integer k

~

Suppose that the rank of the linear map (df)a is

n independent of a

E

O.

Then, for any a

exist neighborhoods U of a, V of {(a), polydiscs P Q about

0 in a;m

and v: V -

n

E

n,

there

about 0 in cr;n,

respectively, and analytic isomorphisms u, P - U

Q so that the map v. f. u: P -

(Zl' ••• , zn) ~ (zl'· •• zk' 0, ••• ,0). a.

ct m

Q is given by

In particular, if k

< n, no point

is isolated in f-If{a). Theorem 5.

Let

n

injective holomorphic map.

be an open set in

ct n and f:O - C n an

Then f is a homeonlOrphism of 0 onto

an open set 0' C a;n and the inverse map f- I :0' Proof.

n

We may suppose that

n

is connected.

is holomorphic. We first assert

that there exists a (fl such that {df)a has rank n (so that Suppose that this is false, and let k = max rank{df) < n. a (fl a Let Zo

E

fl be such that

= k.

rank {df)x

Then clearly, there is a

o

neighborhood U of Xo such that rank{df)x ~ k (hence = k since k is the maximum rank) for x

E

U.

in f-If{xo )' hence there is dieting our assumption that Let A

By the rank theorem, Xo cannot be isolated xl ( U with f(x l )

= f{x o ) ,

Xo


contra-

is injective.

= {x E n I det{df)x = OJ.

Since

x"""'" det{df)x is obviously

holomorphic, and, by what we have seen above, A


it follows that

87 A is an analytic set in A

= _,

(!,

and

(! -

the theorem follows from Lemma 3. Let a

E

0, and let P be a small polydisc about a,

SP is compact, hence so is f(ap) = K. f(a) , K. U

Ii we prove that

A is dense in O.

p.

O.

Then

Since a , ap and f is injective,

Let V be a small polydisc about f(a),

= f- I (V)"

PC

V " K = _, and let

We assert that the map f: U - V is proper.

In fact,

if C is compact in V, f-I(C) cannot be adherent to ap (since there is a neighborhood N of ap with feN) '"' V compact in P.

Let W = feU - A).

Let cp(z) = det«df) ), z ul cp(z)

E

= O}.

Let

Let g: W - U - A be the inverse of E

z

={z

U" A

hence is relatively

By Lemma 3, f: U - A .... W is an analytic iso-

morphism (and W is open in C n ). flu - A.

= _),

It follows that f- I (C)" D' is compact.

U.

",(x)

cp is holomorphic on U, and

=det«dg)x)'

x

W.

E

Then'" is

helomo rphic on W, and, since f. g = identity on W, we have cp(g(x».ljI(x) = I, Let a g(x ) v

E

x < W •

(aw) " V and xvE W, Xv - a, v ....

= f-l(x v )

is in

a compact subset of

00.

Then

U (since f: U - V is proper).

Further, every limit point of g(x) lies on A n U.

Hence, since cp is

holomorphic on U and cp IA '"' U = 0, it follows that cp(g(x» - 0 as

v-

CD.

Hence, the function 1/", on W tends to 0 as we tend to a point

of (aW) '"' V.

By Rado1s theorem (chapter 4, Theorem I), the function 1/ljI(X) 'lex) = {

o

XEW, ,x< V-W

is holomorphic on V and B = V - W = {x E V I 'lex) =

subset of V, B g

t

= (gl' •••• gn)'

V.

Hence W is dense in V.

OJ

is an analytic

Further, if

the gj are bounded on W since U C P

is bounded.

88 Hence, by chapter 4, Proposition 2 (Riemann's continuation theorem), there is a holomorphic map G: V -

U en such that GIW = g. Then

foG is the identity on W since fOGI W dense).

= f. g = identity

(and W is

This proves that (f I U)-l = G is holomorphic, and the theorem

follows The basic theorems of Cartan are in (5] (Propositions 1 and 2). The theorem on limits of automorphisms in (6] and the result, stated without proof in this chapter, on the automorphisms of a product will be found in (8). For Theorem 5 about the inverse of an injective holomorphic map of

n C (;n into (;n, see Osgood (23] and the references given there. The results of Remmert-Stein will be found in (28].

They prove

Theorem 2 (see also Proposition I) only in the case of products of two plane domains, since RadO's theorem applies directly only to this case. Several references to related results will be found in [1].

Chapter (, ANALYTIC CONTINUATION: ENVELOPES OF HOLOMORPHY

We have seen, in chapter Z, that the domain of existence of a holornorphic function in a domain in cr;n can be constructed as a domain over cr;n.

We shall now develop analogues of these results for

a family of holomorphic functions. Let S be a set. we define the sheaf

In analogy with the considerations of chapter Z,

f9 (S)

of S-germs of holomorphic functions on en

as follows. Let U be an open set in cr;n and {fs}s.S a family of holomorphic functions on U, indexed by S. For a. cr;n and (U, {f }),

s

(V,{gs}) with a. U, a. V, we say that these two pairs are equivalent if there exists a neighborhood W of a, W C U (V such that, for all S

E S, fs IW = gs Iw.

An equivalence class with respect to this relation

is called an S-germ of holomorphic functions at a.

We denote the set

=U

(9 (S). We have

of 5-germs at a by

(9 (5). We set

(P(5)

a

a natural projection p = P5: g.

C9 (5) --- en

(9 (S). We define a topology on (J) a

a.cr;n

a

defined by peg) = a if (5) as follows (see chapter Z

for the case when 5 reduces to a single element). and let (U, {f }) be a representant of .& • s a

Let.&

defined by {fs } at bE U, and let N{U, {f }) =

s

Let .& b

E

(9 (5) a

be the S-germ

U'&b' b.U

N(U, {fs }) form a fundamental system of neighborhoods of

89

a

The sets

la'

We

90 prove, as in chapter 2, the following: The map p: (9(5) ~ CC n is continuous and is a

Proposition 1.

local homeomorphisITl of

(0(5) onto CC n • Further, (9(5) is a

Hausdorff topological space.

The triple (Q(S), p, CC n ) is an unraITli-

fied domain over ([n. Let p: X ~ <en, p': X' ~ <en be dOITlains over ([n.

A continuous

map u: X ~ X' is called a local isoITlorphisITl (or local analytic iso

I

morphism) if every point a ( X has a neighborhood U such that u U is a homeomorphism onto an open set U' C X' and ul U and (ul U)-l are holoITlorphic (on U and U'

respectively).

If, in addition,

U

is a

homeomorphism of X onto X', we say that u is an isomorphism (or analytic isomorphism). If u: X - X' is a continuous map such that p'o u = p, then u is

automatically a local isomorphism. Note that the analogue of chapter 5, Lemma 3 holds for holomorphic maps

n - n'.

Definition 1. 5

C

/C.(n).

Let p : n ~ en be a connected domain and o

Let PIX - en be a connected domain and q>:n - X a con-

tinuous map with pol" = po.

We say that PIX - en, q>: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'

=
It is sufficient to require that u o
p

-1

If the functions

{Ff } f<S

separate the points of

pta) for a point a <
of the other requirements of (*). The S-envelope of holomorphy, if it exists, is unique up to

Remark.

"isomorphism".

In fact, let p:X - I!;n,
q>':n - X' be two S-envelopes of holomorphy.

Then, by (*) of

Definition 2.: there are holomorphic maps u: X' - X, v: X - X'

that

p

= plov,

p' = poU, r;p = UClIjP'. #pI = V.f/'.

Then

UoVO'

so that uo v is the identity on q>(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