wolniewicz introduction to several complex variablewols

8/12/2019 Wolniewicz Introduction to Several Complex VariableWols

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Introduction to Several Complex Variables

T. M. Wolniewicz

February 5, 1998

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Contents1 Preliminaries 1

2 Complex derivatives 3

2.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.2 Rules of complex differentiation . . . . . . . . . . . . . . . . . 3

3 Differential forms 6

3.1 Real forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3.2 Complex forms . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.3 Operatorsand . . . . . . . . . . . . . . . . . . . . . . . . 9

4 Holomorphic functions 10

4.1 One variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4.2 Several variables . . . . . . . . . . . . . . . . . . . . . . . . . 11

4.3 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4.4 Cauchy formula in the polydisk . . . . . . . . . . . . . . . . . 14

4.5 Holomorphic mappings . . . . . . . . . . . . . . . . . . . . . . 19

5 Harmonic and subharmonic functions 25

5.1 Harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . 25

5.2 Subharmonic functions . . . . . . . . . . . . . . . . . . . . . . 26

5.3 Pluriharmonic and plurisubharmonic functions . . . . . . . . . 33

5.4 Smooth approximation . . . . . . . . . . . . . . . . . . . . . . 38

6 Some applications of subharmonicity to SCV 41

6.1 The Bergman space . . . . . . . . . . . . . . . . . . . . . . . . 41

6.2 Hartogs theorem on separate analyticity . . . . . . . . . . . . 42

7 Pseudoconvexity and domains of holomorphy 46

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

7.2 Smoothly bounded domains . . . . . . . . . . . . . . . . . . . 46

7.3 Geometric convexity . . . . . . . . . . . . . . . . . . . . . . . 49

7.4 Pseudoconvexity . . . . . . . . . . . . . . . . . . . . . . . . . 55

7.5 Domains of holomorphy . . . . . . . . . . . . . . . . . . . . . 66

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8 The problem 728.1 The problem and its consequences . . . . . . . . . . . . . . . . 72

8.2 Solution for compactly supported forms . . . . . . . . . . . . . 778.3 Elements of Hormanders solution for smooth forms . . . . . . 80

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1 Preliminaries 1

1 PreliminariesFirst we introduce some notational conventions.

Ifw, z Cn, w= (w1, . . . , wn), z= (z1, . . . , z n) then

w|z=

wj zj ,

|z|=

z|z= (

|zj |2)

1/2.

A multi-index is any n-tuple = (1, . . . , n) with j N (we include 0 inN). If is a multi-index then

! =1! . . . n!, ||= 1+ +n, z

=z11 . . . z

nn .

Exercise.

(z+w) =

+=

!

!!zw

where,are multi-indices and the sum +is taken in the ordinary vectorsense.

Exercise. nj=1

ajk

=||=k

k!

!a

where a= (a1, . . . , an).

Ifr= (r1, . . . , rn),rj >0 then the polydisk centered atw with polyradiusris

Dn(w, r) ={z Cn :|zj wj|< rj, j = 1, . . . , n}.

A ball centered at w with radius is

B(w, ) ={z Cn :|z w|< }.

Ifz= (z1, . . . , z n) then z = (z1, . . . , z n1) so that z= (z

, zn).Whenever it is convenient we will identify Cn with R2n. Thus a vectorw =

(w1, . . . , wn) = (a1+ ib1, . . . , an + ibn) is the same as (a1, . . . , an, b1, . . . , bn)R.To get used to this identification lets realize how multiplication by i workson this R2n. We have

i(a1, . . . , an, b1, . . . , bn)R =iw = (iw1, . . . , i wn)= (b1+ia1, . . . , bn+ian) = (b1, . . . , bn, a1, . . . , an)R.


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2 1 PRELIMINARIES

IfD, are subsets ofRm

then we will write D ifD int andDis compact. Thus for instance D Rm means simply that D is bounded.Whenever we say that is a domain we mean that it is open and con-

nected.Bywe will always denote the Lebesgue measure of the underlying space,

will be reserved for surface measures.Uniform convergence on compact subsets of a given set plays a very im-

portant role in function theory of several complex variables and it is usuallyreferred to as normal convergence. We will always use this term in thismeaning.


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3

2 Complex derivatives2.1 Definitions

Let Cn be open and f: C be differentiable in the real sense at apoint P . We define complex derivatives off at P as

f

zj(P) =

1

2

f

xj(P) i

f

yj(P)

(2.1.1)

f

zj(P) =

1

2f

xj(P) +i

f

yj(P)

. (2.1.2)

These definitions are extremely useful but they are purely formal and do notcarry any particular geometrical meaning.

As a consequence of (2.1.1) and (2.1.2) we get

f

xj(P) =

f

zj(P) +

f

zj(P)

f

yj(P) = i

f

zj(P)

f

zj(P)

.

2.2 Rules of complex differentiation

zj(f g) =

k

f

zk

gkzj

+k

f

zk

gkzj

and similarly for zj

. We also have

f

zj

=

f

zj.

We can express the action of the real global derivative of a functionf ona vector w Cn as

DRf(P)(w) =n

j=1

f

zj(P)wj+

nj=1

f

zj(P)wj (2.2.1)


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4 2 COMPLEX DERIVATIVES

To prove this let w= (aj+ ibj)j. Then

DRf(P)(w) = f

xj(P)aj+

fyj

(P)bj

= f

zj+

f

zj

aj+

i

f

zj

f

zj

bj

= f

zj(aj+ ibj) +

fzj

(aj ibj)

= f

zjwj+

fzj

wj

We define higher order complex derivatives in the obvious way. Of coursethese derivatives do not depend on the order of differentiation provided thatf is smooth enough. Sometimes we will use shorthand notation like

||+||f

zz =

1

z11

n

znn

1

z11

n

znnf.

Observe also that

zj

zj=

1

4

xj i

yj

xj+i

yj

=

1

4

2

x2j+

2

y 2j

hence we get thatn

j=1

2f

zjzj=

1

4f. (2.2.2)

To get a bit more feeling about complex derivatives let us derive thecomplex form of the Taylor expansion.

If Rm and f: Rp is of class Ck in then

f(x0+h) =k

j=01

j!DjR

f(x0)(hj) +o(hk)

=k

j=0

1

j!

||=j

||!

!

||f

x (x0)h

+o(hk)

The reason for the factor |!|/! is that to each ||/x correspond|!|/!derivatives which have a different ordering of variables.


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2.2 Rules of complex differentiation 5

If we consider the differential operatorml=1

xl(x0)hl (2.2.3)

corresponding to the global derivative evaluated on the vector h then thecomposition ofj of such operators gives

ml=1

xl(x0)hl

j=

||=j

||!

!

||

x(x0)h

which is (modulo 1/j!) exactly the j-th term of the Taylor expansion.Now let us pass to Cn R2n substitutingP forx0and w for h. It directly

follows from (2.2.1) that in this case operator (2.2.3) can be written as

nl=1

zl(P)wl+

nl=1

zl(P)wl

hence the j-th term of the Taylor expansion will be:

1

j!

n

l=1

zl(P)wl+

n

l=1

zl(P)wl

j=

1

j! ||+||=j(|| + ||)!

!!

||+||

zzww.

Thus we get Taylors formula

f(P+w) =

||+||=j

1

!!

||+||

zzf(P)ww +o(|w|k). (2.2.4)

Now let us use Tylors formula to derive another useful identity

ni,j=1

2f

zizj(P)wiwj =

2f

Rw2+

2f

R(iw)2 (2.2.5)

On the right-hand side of (2.2.5) we have the sum of the second orderterms in the Taylor expansion off(P+ w) andf(P+ iw) respectively. Whenwe apply (2.2.4) and sum the corresponding terms we get cancellation ofterms where||= 2 or||= 2 and only the terms with ||= || = 1 remainproducing the left-hand side of (2.2.5).


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6 3 DIFFERENTIAL FORMS

3 Differential forms3.1 Real forms

Let Rm be open. We say that is a p-form in if

(x) =

1j1


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3.2 Complex forms 7

Any function fcan be considered to be a 0-form and then

df= f

xldxl.

3.2 Complex forms

Define

dzj = dxj+ dyj

dzj = dxj dyj

then

dxj = 1

2(dzj+ dzj)

dyj = 1

2i(dzj dzj).

As a consequence we get that every form on Cn R2n can be writtenin a unique way as a sum of terms of the form

aj1,...,jp,l1,...,lqdzj1 . . . dzjp dzl1 . . . dzlq (3.2.1)

The quickest way to see the uniqueness is to observe that forms like (3.2.1)with coefficient 1 span the space of forms and there is the sane number ofthen as of those composed by dxj and dyj. Since the latter form a basis ofthe space of forms the former must also be a basis.

A form which is a sum of components like (3.2.1) with the same p andqis called a form of type (p, q) or shortly a (p, q)-form. Of course forms ofdifferent types cannot be equal unless they are both zero.

Let : [a, b] C R2 be a C1-curve then according to the standarddefinition

f dz= b

af (t)(t)dt. (3.2.2)

But if we look at f dzas at a differential form then

f dz =

(f1+if2)(dx +idy)


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8 3 DIFFERENTIAL FORMS

= (f1dx f2dy) +i (f2dx +f1dy)=

ba

(f1 )1 (f2 )

2dt +i

ba

(f2 )1+ (f1 )

2dt

= ba

((f1+if2) )(1+i

2)dt

= ba

(f )dt.

It turns out that the seemingly completely arbitrary definition (3.2.2)given in the course of one complex variable is a simple reformulation of the

standard curvilinear integral.

Lemma 3.2.1 (Complex form of Greens theorem) LetD C beC1-bounded and letf be aC1-function in a neighbourhood ofD then

1

2i

D

f()d= 1

D

f

z()d().

Proof. Recall Greens theorem:

D(g

1dx +g2dy) = D g2x g1y d(x, y). (3.2.3)Then

Df d =

D

(f1+if2)(dx +idy)

=D

(f1dx f2dy) +iD

(f2dx +f1dy)

=D

f2x

f1

y

d+i

D

f1x

f2

y

d

= i D

f1x

+ f2x

+if1y

+i f2y

d= 2i

D

f

d.


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3.3 Operatorsand 9

3.3 Operators and

Iff: C is of class C1 then

df= f

zjdzj+

fzj

dzj.

Indeed

df= f

xjdxj+

fyj

dyj = f

zj(dxj+ idyj) +

fzj

(dxj idyj)

(this is the same formal calculation as in the proof of (2.2.1)). It is quitenatural to split the operator dinto two parts denoted and acting:

f = f

zjdzj ,

f = f

zjdzj .

These definitions are extended to arbitrary forms by action on the coefficients.If is a (p, q)-form then is a (p + 1, q)-form and is a (p, q+ 1)-form.We have

dd = (+ )(+ ) = ++ + .

Since is a (p+ 2, q)-form, and are (p+ 1, q+ 1)-forms and is a (p, q+ 2)-form and dd = 0 we get that = 0 = and = ().


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10 4 HOLOMORPHIC FUNCTIONS

4 Holomorphic functions4.1 One variable

Let C be open. A function f: C is called holomorphic if it isdifferentiable in in the complex sense. Its derivative is denoted byf ordfdz

. If we only assume that f is R-differentiable then it will be holomorphicif and only if it satisfies the Cauchy-Riemann equations

Ref

x =

Imf

y ,

Ref

y =

Imf

x .

These can be reformulated to give:

0 = 1

2

Ref

x

Imf

y

+

1

2i

Ref

y +

Imf

x

= 1

2

Ref

x +i

Imf

x

+

1

2i

Ref

y +i

Imf

y

= 1

2

f

x+i

f

y

= f

z.

Thus we obtained an equivalent formulation of the Cauchy-Riemann equa-tions:

f

z = 0.

Next observe that if fz

= 0 then fx

= fz

+ fz

= fz

. But fx

= dfdz

so weobtain that for holomorphic functions

f

z =

df

dz. (4.1.1)

In other words whenever the right-hand side of (4.1.1) makes sense we getthe equality (4.1.1).

Another way to state that fz

= 0 is that f = 0. This is the mostcompact form of Cauchy-Riemann equations.


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4.2 Several variables 11

4.2 Several variablesNow consider Cn. We want to define complex differentiability of afunction f: C. The most natural way is to follow the definition for thereal case and say that f isC-differentiable at a point P if there exists alinear mapping A:Cn C such that

limh0

f(P+h) f(P) Ah

|h| = 0. (4.2.1)

Suppose this condition is fulfilled. ThanAcan be considered to be a linear

mapping from R2n

to R2

and (4.2.1) implies thatfis differentiable in the realsense, A being its global derivative at P. If in turn we start with assumingthat f is R-differentiable at P then f needs not be C-differentiable sinceDRf(P) is not necessarily C-linear. But if (and only then) this derivativehappens to be C-linear then we do get C-differentiability. Since an R-linearmapping A is C-linear iffA(iw) = iAw we get that f is C-differentiable atP iff it is R-differentiable and

DRf(P)(iw) =iDRf(P)(w). (4.2.2)

So if we want C-differentiability offwe must verify (4.2.2) and it is enough

to do this on some basis ofCn. If we take the basis

ej = (0, . . . , 1, . . . 0)j

Cn

then

ej = (0, . . . , 1, . . . , 0, 0, . . . , 0)Rj

iej = (0, . . . , 0, 0, . . . , 1, . . . , 0)Rn+j

In particularDRf(P)(ej) =

f

xj

while

DRf(P)(iej) = f

yj.


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Thus we get the system of equationsf

yj=i

f

xj

or equivalently

Ref

xj=

Imf

yjRef

yj=

Imf

xj.

These are the Cauchy-Riemann equations in several variables and as we haveseen, together with R-differentiability they are equivalent to the fact thatf isC-differentiable. Other equivalent formulations of the Cauchy-Riemannequations are:

f

zj= 0, j = 1, . . . , n

orf= 0.

Definition 4.2.1 A function f: C is holomorphic in if it is C-differentiable in or equivalently if it isR-differentiable and satisfies theCauchy-Riemann equations in . The collection of all holomorphic func-tions on will be denoted byH().

It is easily seen that H() is a linear space.Iff is holomorphic in Cn then for every z and j = 1, . . . , nthe

functionfz,j() =f(z1, . . . , , . . . , z n)

j

is holomorphic in the open set z,j ={ C: (z1, . . . , , . . . , z n) }. Thisis completely obvious since

fz,j

() = fzj

(z1, . . . , , . . . , z n) = 0.

It is not at all obvious (but true) that the opposite holds as well i.e., ifall fz,j -s are holomorphic than so is f. (The proof of this will be given inSection 6.1.)


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4.3 Power series 13

We will introduce a temporary definition and say that a functionf: C is coordinatewise holomorphic if all fz,j -s are holomorphic.Of course iff is C1 and coordinatewise holomorphic then it is holomor-

phic.

4.3 Power series

A power series in Cn is a series

a(z P)

(4.3.1)

where ranges over all n-position multi-indices.We will only consider absolute convergence of power series; then the sum

does not depend on the order of summation. Whenever we say that a powerseries is convergent we mean that it is absolutely convergent.

Theorem 4.3.1 If the power series (4.3.1) converges at a point z0 Cn

then it converges normally on the polydisk

{z Cn :|zj Pj|< |z0,j Pj|}.

Proof. Exercise.

The above theorem tells us that if a power series converges pointwise insome open polydisk then it converges normally.

Definition 4.3.2 A functionf: C is analytic if for everyP thereexists a neighbourhoodUP ofP insuch thatfcan be represented inUP asa sum of a power series

a(z P)

.

Every analytic function is of class C and holomorphic. If

f=

a(z P)

then||f

z (P) =!a. (4.3.2)


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4.4 Cauchy formula in the polydiskTheorem 4.4.1 (Cauchy formula) Let Cn be open andf a coordi-natewise holomorphic function in. LetP and r= (r1, . . . , rn) be suchthatDn(P, r) . Then for everyzDn(P, r)

f(z) = 1

(2i)n

|nPn|=rn

|1P1|=r1

f()

(1 z1) . . . (n zn)d1 . . . d n

(4.4.1)

Proof. We proceed by induction. For n = 1, (4.4.1) is the usual one-di-mensional Cauchy formula. Assume (4.4.1) is true for n 1. Then for everyz with |zj| rj , j = 1, . . . , n 1 the function fz,n is holomorphic in aneighborhood ofD(Pn, rn) hence

f(z, zn) = 1

2i

|nPn|=rn

fz,n(n)

n zndn (4.4.2)

Next consider the function z fz,n(n) = f(z, n) where n D(Pn, rn)

is fixed. This function is coordinatewise holomorphic in a neighborhood of

Dn1(P, r) hence, by the inductive hypothesis

f(z, n)

= 1

(2i)n1

|n1Pn1|=rn1

|1P1|=r1

f()

(1 z1) . . . (n1 zn1)d1 . . . d n1

Inserting this into (4.4.2) we get (4.4.1).

The integral on the right-hand side of (4.4.1) has to be understood as

an iterated integral. One cannot readily use Fubinis theorem at this pointwithout some additional assumptions on f.

Corollary 4.4.2 Iffis coordinatewise holomorphic and locally bounded thenf is analytic, in particular also holomorphic.


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Proof. Since holomorphic functions are continuous they are also locallybounded.

Corollary 4.4.4 Iff is holomorphic then

|+|f

zz = 0

if= 0. In particularf= 0.

Proof. Since fzj = 0 for every j, every other derivative will also vanish.The second part follows from (2.2.2)

Corollary 4.4.5 If Cn is open and connected, f, g H() and{z :f(z) =g(z)} has non-empty interior thenf=g on.

Proof. Let h = (f g). Let V = {z : ||hz

(z) = 0 for all }.Then Vis closed in and non-empty, but it is also open because ifP Vthen the series expansion ofh about Phas zero coefficients so his zero in a

neighbourhood ofP.

Corollary 4.4.6 If Cn is open and connected and f H() is such

that at some pointz0, for all, ||fz

(z0) thenf0 in.

Proof. Apply the proof of the previous corollary.

Definition 4.4.7 We say that is a homogeneous polynomial of degrees if

(z) =||=s

az.

This definition can be equivalently stated as (lz) = ls(z), for z Cn,l C.


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4.4 Cauchy formula in the polydisk 17

If f is a holomorphic function in a neighbourhood of 0 then f can bewritten as the Taylor series

f(z) =j=0

1

j!DjC

f(0)(zj),

the same effect can be obtained by grouping the power series

f(z) =

az =

j=0

||=j

az

.

Both of these approaches give us the same expansion

f(z) =j=0

j(z)

where j are homogeneous polynomials of degree j.Another approach to such an expansion is by considering the function

C l f(lz). Then

f(lz) =j=0

ljj(z).

This is nothing else but the Taylor expansion of the function of one variableand j(z) serve as coefficients. This shows at once that the homogeneousexpansion is unique and

j(z) = 1

j!

dj

dljf(lz)

l=0

(4.4.3)

j(z) = 1

2

20

f(zei)eijd (4.4.4)

We will later need one technical lemma about homogeneous polynomials:

Lemma 4.4.8 If is a homogeneous polynomial of degrees then

(z+o(z)) =(z) +o(|z|s).

The proof is quite easy, it is enough to consider of the form z where||= s, we leave the details to the reader.


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Theorem 4.4.9 (Cauchy inequalities) If f is coordinatewise holomor-phic in and|f| Min a polydiskDn

(P, r) then for every multi-index ||fz

M !r.Proof. We have to estimate the -th coefficient in the series expansion off about Pwhich is equal to

a= 1

(2i)n |nPn|=rn

|1P1|=r1b()f()d1 . . . d n.

So |a| supDn(P,r) |b()| supDn(P,r) |f()|r1 . . . rn. But

|b()|r1 . . . rn= 1

| P1|1+1. . .

1

|n Pn|n+1r1 . . . rn = r

11 . . . r

nn = r

and sup |f| M.

Corollary 4.4.10 If f and r are as in Theorem 4.4.9 and r = r/2 :=(r1/2, . . . rn/2) then

zDn(P,r)

||fz (z) M !2||r.

Corollary 4.4.11 Iffn H() andfnf normally on thenfH()

and for every multi-index, ||fnz

||fz

normally on.

Proof. It follows from Cor. 4.4.10 that fnzj

converge normally on . Sincefnzj

= 0 we get that also fnxj

and fnyj

converge normally. It follows that f is

of classC1 and fnxj

fxj

and fnyj

fyj

. In particular fnzj

fzj

so f= 0

and fH(). Convergence of higher order derivatives follows at once fromCor. 4.4.10 applied to fn f.


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Now suppose m= n, then all matrices are square and using the fact that

det

A BB A

= |det(A +iB)|2

we get thatJR =|JC|2.

Definition 4.5.1 Let1, 2 Cn be open and: 1 2. is said to be

biholomorphic if is a holomorphic bijection of1 onto 2 and1 is also

holomorphic.

It is a highly nontrivial fact that every holomorphic bijection is automaticallybiholomorphic.

Corollary 4.5.2 If : 1 2 is biholomorphic and f: 2 C is inte-grable then

2f d=

1

(f )|JC|2 d.

Theorem 4.5.3 (H. Cartan) Suppose

a) is a bounded domain inC

n

,b) : is a holomorphic mapping,

c) for someP , (P) =P andDC(P) =I.

Then(z) =z for allz .

Proof. Without loss of generality we may assume that P= 0. Then thereexists r1 > 0, r2


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4.5 Holomorphic mappings 21

then =

j=0

Fj(z).

The terms of this expansion correspond to the derivatives of at 0 soF0(z) =(0) = 0, F1(z) =DC(0)(z) =zand we have

(z) =z+j=2

Fj(z).

It is our aim to show that Fj = 0 forj 2. Suppose this is not true and letm j be the first index for which Fj 0. Then

(z) =z+Fm(z) +o(|z|m).

Let s be the s-th iteration of i.e., s = . . . s times

then

s(z) =z+sFm(z) +o(|z|m) (4.5.1)

We prove (4.5.1), again by induction, this time on s. For s= 1 the claim isobvious and fors >1,

s(z) = (s1(z)) =

z+ (s 1)Fm(z) +o(|z|m)

= z+ (s 1)Fm(z) +o(|z|m)

+Fm

z+ (s 1)Fm(z) +o(|z|m)

+o

|s1(z)|m

= z+ (s 1)Fm(z) +Fm

z+o(|z|)

+o(|z|m)

= z+ (s 1)Fm(z) +Fm(z) +o(|z|m) +o(|z|m)

= z+sFm(z) +o(|z|m)

It follows, by (4.4.4), that

sFm(z) = 1

2

20

s(zei)eimd.

Each s

maps into hence |s

(z)|< r2 and so we have|sFm(z)| r2 for all s 1, |z|< r1

hence Fm(z) 0 on B(0, r1) and so must be identically zero, which is acontradiction.


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Definition 4.5.4 A setE Cn

is called circular if lzE wheneverzEand l C, |l|= 1.

Theorem 4.5.5 (H. Cartan) Suppose

a) 1, 2 are circular domains inCn with0 1 and0 2,

b) is a biholomorphic map of1 onto 2 with(0) = 0,

c) 1 is bounded.

Thenf is a linear.

Proof. Let, for [0, 2]

(z) : = 1(ei(eiz)).

is well defined since 1, 2 are circular and (1) = 1. If we computethe derivative of at 0 we get

D(0) =

D(1)(0)

eiD(0)

eiI

= I .

Since (0) = 0, we get by Theorem 4.5.3 that =Iso we have

z= 1(ei(eiz))

hence

ei(z) = (eiz).

But if

=j=0

Fj

is the homogeneous expansion of then, by (4.4.4)

Fm(z) = 1

2

20

(eiz)eimd = (z) 1

2

20

ei(m1)d= 0

ifm= 1 and thus = F1 so is linear.


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4.5 Holomorphic mappings 23

Theorem 4.5.6 LetU C be the unit disk. Ifa U and

Ba() : = a

1 a

then Ba is a biholomorphic mapping of U onto itself and Ba(a) = 0,Ba(0) =a.

Proof. We have Ba Ba() =so we only have to show that |Ba()|< 1for Uwhich is a trivial calculation.

The function Ba

() is called the Blaschke factorat a.

Corollary 4.5.7 LetD= Dn(z, r). Ifa D then

a(w) =

r1

w1 a11 a1w1/r21

, . . . , rnwn an

1 anwn/r2n

is a biholomorphic mapping ofD onto itself witha(a) = 0, a(0) =a.

Proof. If r= (1, . . . , 1) then

a(w) =Ba(w) : = (Ba1(w1), . . . , Ban(wn))

so a acts into the polydisk and a a= I hence a is a biholomorphismofD onto D with a(a) = 0, a(0) =a.

In the general case we first map D onto the unit polydisk by the linearmap (w1, . . . , wn)(w1/r1, . . . , wn/rn), then send the image ofaonto 0 bya suitable Bb and finally map the unit disc onto D by the linear mapping(w1, . . . , wn) (r1w1, . . . , rnwn). This shows that a is a composition ofthree biholomorphic mappings hence is biholomorphic.

We have just proved that a polydisk is an example of a homogeneousdomain i.e., a domain in which any point can be mapped onto another bya biholomorphic transformation of the domain. (For the polydisk just takeb a.) A ball is another example of such a domain, see [R] sec. 2.2.

Corollary 4.5.8 Ifn >1 then there is no biholomorphic mapping of a poly-disk onto a ball inCn.


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Proof.Without loss of generality we may assume that both the ballBandthe polydisk D are centered at 0. If there were a biholomorphic mapping

ofD ontoBthen we would take a= 1(0) and define

= a.

Then (0) = (a(a)) = (a) = 0 and is also a biholomorphic mapofD ontoB, hence by Theorem 4.5.5 is linear, which is clearly impossiblesince a linear image of a polydisk is a rotated polydisk.

This corollary shows that there is no extension of the Riemann mapping

theorem to several variables.


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25

5 Harmonic and subharmonic functions5.1 Harmonic functions

Even though harmonic and subharmonic functions are a real variable topicthey are one of the most important tools of the theory of several complexvariables.

Let Rm.

Definition 5.1.1 A functionh: R is harmonic ifh is of classC2 inandh= 0.

It can be shown that every harmonic function is of class C. We will oftendeal with complex-valued harmonic functions which are of course definedexactly as the real-valued ones. We have seen that holomorphic functions,hence also their real and imaginary parts are harmonic. If C is simplyconnected then every real-valued harmonic function on is a real part of aholomorphic function. In Cn, (n > 1), no matter what is like, this is nolonger true.

Example. Let h:C2 R, h(z1, z2) = |z1|2 |z2|

2 = z1z1 z2z2. Then2h

z1z1= 1 ,

2hz2z2

= 1 hence h = 0 so h is harmonic. But ifh = Refwith fholomorphic then h would also be coordinatewise harmonic which itis not.

One of the most important properties of harmonic functions is the meanvalue property:

Theorem 5.1.2 (Mean value property) Leth: R be harmonic andx0 , r >0 such thatB(x0, r) . Then

h(x0) = 1

(B(x0, r))

B(x0,r)

h(x) d(x)

h(x0) = 1

(B(x0, r)) B(x0,r)h(x) d(x)

where is the surface measure on the sphereB(x0, r).

Corollary 5.1.3 (Maximum principle) If Rm is connected then ev-ery harmonic functionh: Rwhich attains its maximum in is constant.


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26 5 HARMONIC AND SUBHARMONIC FUNCTIONS

Proof. The set {x :h(x) = maxh} is non-empty and closed in , butit follows from the mean value property that it is also open hence it must be

equal to .

Corollary 5.1.4 If h1, h2: C are harmonic in a bounded set andcontinuous on andh1=h2 on thenh1 = h2 on.

Proof. Let h = Re(h1 h2) then h is harmonic and h = 0 on henceh 0 on . We repeat the same with (h) and get (h) 0 thus h = 0.Next do the same with the imaginary part.

Theorem 5.1.5 (Solution to the Dirichlet problem in a ball)Iff: B(x0, r) R is continuous then there exists a unique continuous func-tionh: B(x0, r) R which is harmonic inB(x0, r) andh|B(x0,r)= f.

5.2 Subharmonic functions

Definition 5.2.1 A functionf: R {}is called subharmonic if

a) f is upper semicontinuous, that is{x :f(x)< t} is open for every

real numbert.b) For every ballB and every continuous functionh: B R which

is harmonic inB andh f onB we haveh f onB.

A functionfis called superharmonic if(f) is subharmonic.

One of the reasons to consider semicontinuous rather then continuousfunctions is that decreasing pointwise limits preserve subharmonicity but donot preserve continuity. This is also one of the reasons why value{} isadmitted. Another reason for this is that log |f| is subharmonic whenever fis holomorphic and this function is () at the zeroes off.

Lemma 5.2.2 (Integration in spherical coordinates) Letfbe an inte-grable function on the ballB(0, R) inRm. Then

B(0,R)f(x) d(x) =

R0

rm1B(0,1)

f(r) d() dr


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5.2 Subharmonic functions 27

The proof is a simple observation that when one integrates the functionon the left using classical spherical coordinates then integrating first in theangle variables we get the integral over the unit sphere, the remaining ra-dius variable gives us the second integral.

Theorem 5.2.3 Let Rm be open and f: R {} be uppersemicontinuous. Then the following are equivalent:

1) f is subharmonic in.

2) Ifx andr >0 are such thatB(x, r) then

f(x) 1(B(x, r))

B(x,r)

f(t) d(t). (5.2.1)

2a) xr00


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(1

2) For M R let fM = max{f, M}. Then fM is upper semicon-tinuous and R-valued. Let x, r satisfy the hypothesis of 2) and hM be thesolution to the Dirichlet problem in B(x, r) with boundary values fM, i.e.,hMis continuous in B(x, r), harmonic in B(x, r) and hM=fM on B(x, r).It follows that hMfM on B(x, r), in particular hM(x) fM(x). But then

fM(x) hM(x) = 1

(B(x, r))

B(x,r)

hMd = 1

(B(x, r))

B(x,r)

fMd

and B

fm dM()

B

f d. (5.2.3)

In (5.2.3) we used the fact that there exists an M0 Rsuch thatfM0 onB(x, r). Then forM < M0we haveM0 fM 0 and (M0 fM) (M0 f).Hence we can use Lebesgues monotone convergence theorem to get (5.2.3).We end by observing that fM(x) f(x) as M().

(3a 4). Let h be as in 4). Let G be a component of int K. Then(fh) satisfies 3a) in G and (fh) 0 on G (G K). LetM =supG(f(x) h(x)) and letV ={x G : f(x) h(x) M}. ThenV is closedin G and it follows from 3a) that it is also open. Hence V =f or V =G.In either case we get that M 0 so f h on G. Since the choice of thecomponent was arbitrary we get fh on int Kthus on K.

Corollary 5.2.4 Every function which is both super and subharmonic is har-monic.

Proof. Let f be such a function in . Then first of allf is continuoussince it is both lower and upper simicontinuous. Let B be a ball andleth be the solution of the Dirichlet problem in B with dataf. Then by thedefinition f h in B. But (f) is also subharmonic and (f) = (h) onB hence (f) (h) in B and we get f=h in B sof is harmonic in B.

Corollary 5.2.5 Iff: C is continuous and

xr00


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Proof. It is enough to consider real valued f. By condition 3a) of Theo-rem 5.2.3 fand (f) are subharmonic, hence f is harmonic.

Exercise. Let f: C be such that for every K there exists anrK>0 such that

xK0 j0.


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Proof. Fix a compact set K . We may find an open set 0 suchthat K 0 , then fj are bounded above, by say M, on 0. Let

vj =M fj . Thenvj 0 on 0 and limj vj(x) C1: =M C forx .We are to show that vj(x) C1 forx K,j > j0. We may assume thatC1 > 0 otherwise there is nothing to prove. Choose r > 0 so that for everyx K, B(x, 2r) 0. By subharmonicity offj we have, for x K and0< 2r,

fj(x) 1

(B(x, ))

B(x,)

fjd

hence

vj(x)

1

(B(x, )) B(x,) vjd.By Fatous lemma we also have

limj0

1(B(x, r))

B(x,r)

vjd

1

(B(x, r))

B(x,r)

limj0

vjdC1.

Fix >0, then for each x Kthere is a j (x) such that for j > j(x)

1

(B(x, r))

B(x,r)

vjdC1

2.

Now let 0< < r be such that rr+

mC1

2

C1

and y B(x, ) then B(x, r) B(y, r+) and for j > j(x)

vj(y) 1

(B(y, r+))

B(y,r+)

vjd

r

r+

m 1(B(x, r))

B(x,r)

vjd

r

r+m

C1

2 C1 .Hence we have shown that for each x K there exists a j(x) such that ifj > j(x) then for yB(x, ), vj(y) C1 . Since Kmay be covered by afinal number of balls B(xk, ) we can take j0= max{j(xk)}.


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Proposition 5.2.9 If Cn

, f H() then log |f| is continuous andsubharmonic in and harmonic on the set where it is not().

Proof. Let x be such that f(x) = 0. Take r0 > 0 such thatf(B(x, r0)) C\L where L is a ray starting from the origin. Let bea holomorphic branch of logarithm on C\L then f is holomorphic inB(x, r0) so log |f| = Re( f) is harmonic. Thus log |f| is harmonic in

1 = \

x: log|f(x)| =

. Hence we get condition 2a) of Theorem 5.2.3

for x 1 and this condition is obviously fulfilled if log|f(x)|= .

Theorem 5.2.10 (Jensens inequality) Let be a probability measure onA, f: A R a measurable function and :R R a continuous convexfunction such thatf and fare integrable onA. Then

Af d

A

f d.

Remark. It is also possible to formulate Jensens inequality when f andA f dcan take the value (). We will use such a version below.

Corollary 5.2.11 Iffis subharmonic in and:R{} R{}

is convex and increasing then f is subharmonic.

Proof.

( f)(x) 1

(B(x, r))

B(x,r)

f d

1

(B(x, r))

B(x,r)

f d

The first inequality follows from the fact that is increasing and f issubharmonic, the second from Jensens inequality applied to the measured = 1

(B(x,r))d on B(x, r). Thus we have verified condition 3) of Theo-

rem 5.2.3 for f.

Corollary 5.2.12 IffH() then log+ |f|, |f|p (p >0) are subharmonic.

Proof. Functions t t+, t ept are convex and increasing and composedwith log |f|give log+ |f| and |f|p.


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Definition 5.2.13 Forf defined onB(x, r0) Rm

and0< r < r0 let

f,x(r) := 1

m1

B(0,1)

f(x +r) d() = 1

(B(x, r))

B(x,r)

f() d()

wherem1 = (B(0, 1)).

Lemma 5.2.14 Letf: R {} be continuous and subharmonic in andB(x, r0) . Thenf,x(r) is an increasing function of r.

Proof. Assume f = () on and consider r1 < r2 < r0. Let h be a

continuous function on B(x, r2) with h = f on B(x, r2). Then h f onB(x, r2) and, by the mean value property for h we have

f,x(r2) = h(x) = 1

(B(x, r1))

B(x,r1)

h d

1

(B(x, r1))

B(x,r1)

f d=f,x(r1).

If we allow that f attains the value () then we must first consider fM: =max{f, M}for which the lemma has already been proved and then let M().

Corollary 5.2.15 LetfH(BCm(0, 1)) and

Nf(r) : = 1

2N1

B(0,1)

log+|f(r, )| d()

Mpf(r) : = 1

2N1

B(0,1)

|f(r)|p d() 1p .

ThenNf andMpf are increasing functions of r.

Lemma 5.2.16 LetfC() then limr0 f,x(r) =f(x) and if in additionfC2() then

df,xdr

(r) = 1

m1rm1

B(x,r)

f d.


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5.3 Pluriharmonic and plurisubharmonic functions 33

Proof.The first part is obvious. For the second we have

m1d

dr =

d

dr

B(0,1)

f(x0+r) d()

=B(0,1)

fxi

(x0+r)i d()

=B(x,r)

fxi

()1

r(i xi)

1rm1

d()

= 1

rm1

B(x,r)

f

n() d()

= 1

rm1 B(x,r) f() d()where n is the unit normal vector and we use a version of Greens theorem.

Theorem 5.2.17 Let f: R be C2 in . Then f is subharmonic ifff0 on.

Proof.

f0 xr0r


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Exercise.Every pluriharmonic function is harmonic.

Observe that iff is pluriharmonic then so are f, Ref, Imf.

Theorem 5.3.2 f is pluriharmonic in iff

i,j2f

zizj(z) = 0 on . (5.3.1)

Proof. We have

2

f(z+w)

=

nj=1

f

zj(z+w)

(z+w)j

+n

j=1

f

zj(z+w)

(z+w)j

=

nj=1

f

zj(z+w)wj

=

nj=1

ni=1

zi

fzj

(z+w)

(z+wi

wj

+n

i=1

zi

fzj

(z+w)

(z+w)i

wj

= nj=1

ni=1

2fzizj

(z+w)wiwj

(5.3.2)

Evaluating (5.3.2) at = 0. We get

1

4f(z+w)

=0

=n

i,j=1

2f

zizj(z)wiwj .

Thus (5.3.1) implies thatf is pluriharmonic. It is enough to prove the reverseimplication for real f. Then we have

2

fzizj

= 2fzizj

so the matrixH=

2fzizj

i,j

is hermitian andHw|w = 0 for everyw Cn.

It follows that H= 0 so all its entries are zero.


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Corollary 5.3.9 Iff is plush thenf is subharmonic.Proof. We will show that condition 2) of Theorem 5.2.3 is satisfied for f.We are to show that

f(z) 1

(B(z, r))

B(z,r)

f(w) d(w).

But observe that w z+ ei(w z) is a measure-preserving transformationonB(z, r) (it is a rotation about the center of the ball). Hence for each fixed we have

B(z,r)f(w) d(w) =

B(z,r)f(z+ei(w z)) d(w) (5.3.4)

Integrating both sides of (5.3.4) in over [0, 2] we get

2B(z,r)

f(w) d(w) = 2

0

B(z,r)

f(z+ei(w z)) d(w) d

=B(z,r)

20

f(z+ei(w z)) d d(w)

Using (5.3.3) with r= 1 we get

f(z) 1

2

20

f(z+ei(w z)) d.

Hence B(z,r)

f(w) d(w) = 1

2

B(z,r)

20

f(z+ei(w z)) dd(w)

B(z,r)

f(z) d(w) =f(z)

Remark. The trick used in this proof based on circular invariance of the ballis very commonly used in SCV. The same argument proves that whenever Dis circular with respect toz, i.e.,wDC(||= 1 z+ w D) andf isplush in D we have

f(z) D f() d()Theorem 5.3.10 Letf: R beC2 in. Thenf is plush in iff

zwCnn

i,j=1

2f

zizj(z)wiwj 0 (5.3.5)


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5.3 Pluriharmonic and plurisubharmonic functions 37

Proof. We know that f(z+w) is subharmonic iff f(z+w) 0.But

1

4f(z+w)|=0=

ni,j=1

2f

zizj(z)wiwj

and the claim follows.

Definition 5.3.11 A C2 function f: R is called strictly plurisubhar-monic if

zwCw=0

n

i,j=12f

zizj(z)wiwj >0.

The matrix 2fzizj

(z)i,j

is called the complex Hessian off at z. Thus we

can say that a function is strictly plush iff its Hessian is positive definite on.

Proposition 5.3.12 Iff : R is strictly plush andK then thereare constantsC, c >0 such that

C|w|2 n

i,j=12f

zizj

(P)wiwj c|w|2 for PK, w Cn.

Proof. Consider the function

Cn (P, w)

ni,j=1

2f

zizj(P)wiwj.

Obviously is continuous and (P, w)> 0 ifw= 0. Let

c : = min{(P, w) :P K, |w| = 1}> 0

C : = max{(P, w) :PK, |w| = 1}<

Since (P, w) =|w|2

P, w|w|

we get

C|w|2 (P, w) c|w|2.


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5.4 Smooth approximationLemma 5.4.1 Let be a C function onRm which is non-negative, sup-ported on the unit ball and

Rm

= 1. Let(x) =m(x/). Iff: R

is locally integrable then the functionf(x) =Rm

f(y)(x y) d(y) is ofclassC on = {x : (x) > } and iff is continuous thenf funiformly on compact subsets of.

The family {}>0 is an example of a so called approximating identity.The integral defining f is called the convolution off and and denotedf. Convolutions with smooth approximating identities are very oftenused to approximate a given function by smooth functions.Proof. Since integrals defininff are taken only over balls B(x, ) ,we have that f is integrable on thes balls so the integrals are well defined.The fact that f is C

follows at once by differentiating integral on theparameter.

Since

= 1 we get

f(x) =f(x)

(y) d(y) = f(x)

(xy) d(y) =

f(x)(xy) d(y).

Now observe that is supported on B(0, ) so

|f(x) f(x)| = (f(x) f(y))(x y) d(y)

B(x,)

|f(x) f(y)|(x y) d(y)

IfK is compact then for small we have K . For such letK ={x: dist(x, K) }thenK andK is compact hence if we fix an >0 then there is an 0 < such that for x, yK

satisfying||x y|| < 0we have

|f(x) f(y)| < . (5.4.1)

If we now use (5.4.1) for xK, < 0 then for y B(x, ) we get |f(x)

f(y)| < so |f(x) f(x)|< .

Lemma 5.4.2 Let be as before and in addition radial (i.e., (x) =(||x||)). Iff is a subharmonic function on thenf is also subharmonicand for1> 2 > 0 we havef1 f2 f.


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5.4 Smooth approximation 39

Proof.We have

f(x) =B(0,)

f(x y)(y) d(y)

=

0rm1

B(0,1)

f(x r)(r) d() dr

=

0rm1m(r/)

B(0,1)

f(x r) d() dr

=

0rm1m(r/)f,x(r) dr

= 1

0

tm1(t)f,x(t)dt.

Since f,x(r) f(x) and1

0 tm1(t) dt =

B(0,1) (y) d(y) = 1 we get

that f(x) f(x) and since f,x(t) decreases together with we get thatthe whole integral decreases.

Now we establish subharmonicity off. Let x and r >0. ThenB(x,r)

f(x) d(x) =B(x,r)

Rm

f(x y)(y)d(y) d(x)

=Rm

B(x,r)

f(x y) d(x) (y) d(y)

= Rm

B(xy,r)

f(w) d(w) (y) d(y)

Rm

(B(x y, r))f(x y)(y) d(y)

= (B(x, r))f(x)

Remark. If Cn andfis plush in then we get that under assumptionsof Lemma 5.4.2 f is also plush, we simply have to repeat the above proofintegrating over circles instead of balls.

Lemma 5.4.3 If1 Cn, 2 C

m are open, : 1 2 is holomorphicandf: 2 C is of classC

2 then

ni,j=1

2

zizj(f )(P)wiwj =

mk,l=1

2f

kl((P))vkvl


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wherev : =

ni=1

zi(P)wi= D(P)(w)

Proof. It is an exercise in complex differentiation, one has to remember

that if is a holomorphic function then zk

= 0 and zk

=zk

= 0.

Theorem 5.4.4 If1 Cn, 2 C

n are open,: 1 2 is holomorphicandf is a continuous plush function in2 thenf is plush in1. If inaddition is biholomorphic andf is strictly plush thenf is also strictly

plush.

Proof. Assume thatfis C2. Then it immediately follows form Lemma 5.4.3thatfis plush. If is biholomorphic thenD(P) is invertible hence ifw = 0thenv =D(P)(w) = 0,and we get that iffis strictly plush then so is f.Iff is not C2 then we consider approximating functionsf, which are plush,C andf f as 0. It follows that f are plush andf f hence f is plush.


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41

6 Some applications of subharmonicity toSCV

6.1 The Bergman space

As we have seen in Section 4.4, H() can be considered in a natural way as acomplete metric linear space. Unfortunately this space cannot be normed sowe cannot use the tools developed for Banach spaces. It is however possibleto introduce a large Hilbert space of holomorphic functions and then use therich theory of Hilbert spaces.

Definition 6.1.1 Let Cn be open and

L2H() =L2(, d) H().

L2H()equipped with the scalar product fromL2(, d)is called the Bergmanspace on.

Theorem 6.1.2 If Cn is open thenL2H() is a Hilbert space.

Lemma 6.1.3 LetfH(B(z, r)) then

supB(z,r/2)

|f| 2n

1

(B(z, r)))

B(z,r)

|f(w)|2 d(w)

12

Proof. Let B(z,r/2). Since |f|2 is subharmonic in B(z, r) andB(,r/2) B(z, r) we have

|f()|2 1

(B(,r/2))

B(,r/2)

|f(w)|2 d(w)

1(B(,r/2))

B(z,r)

|f(w)|2 d(w)

= 22n

(B(z, r))

B(z,r)

|f(w)|2 d(w).


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42 6 SOME APPLICATIONS OF SUBHARMONICITY TO SCV

Proof of the theorem.We only have to prove that L

2

H() is closed inL2() so let fn L

2H(), fnL2

f L2(). Then {fn} forms a Cauchysequence in L2(). Fix x and r > 0 such that B(z, r) then ittrivially follows from above lemma that {fn} form a Cauchy sequence onB(z,r/2) in the topology of uniform convergence so fn converge uniformlyon B(z,r/2), of course to f, hence f is holomorphic in B(z,r/2).

Proposition 6.1.4 Let: 1 2 be biholomorphic then the operator

T: = (f )JC

is an isometry ofL2H(2) onto L2H(1).

Proof. Exercise.

Of course any two non-zero L2Hspaces are isometric since they are separa-ble Hilbert spaces but one usually needs an explicit and natural isomorphismso this proposition is quite useful.

Let Cn be open and: R a positive, continuous function thendefine

L2

H(, ) ={fH() : |f|2 d


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6.2 Hartogs theorem on separate analyticity 43

Recall that according to Cor. 4.4.2 every locally bounded and coordinate-wise holomorphic function is holomorphic.We will proceed by induction and assume that Theorem 6.2.1 has already

been proved for dimensions smaller then n (for n = 1 there is nothing toprove).

Lemma 6.2.2 Let f be coordinatewise holomorphic in and let D =nj=1 Dj , where Dj C are discs. Then there exist discs Uj Dj

withUn=Dn such thatf is bounded inU=n

j=1 Uj, hence holomorphic inU.

Proof. Denote by D =n1j=1Dj . For M NletEM=

z D

:|f(z, zn)| M when znDn

.

By inductive hypothesis for a fixed zn the functionz f(z, zn) is holomor-

phic, in particular continuous so EM is closed. We also have

M=1 EM=D

hence, by Baires category theorem there is an M such that EM has non-empty interior and we may find U =

n1j=1Uj EM. Then U =U

Dn isthe desired polydisk (f is bounded by M inU).

This lemma does not yet prove our theorem as we have no control overthe polydisk U.

Lemma 6.2.3 Letfbe a complex valued function in a neighbourhood of thepolydiskD= {z:|zj z

0j | R, j= 1, . . . , n}, assume thatf is holomorphic

inz ifzn is fixed and thatfis holomorphic in a neighbourhood of

U={z:|zj z0j | r, j= 1, . . . , n 1 , |zn z

0n| R}

for somer >0. Thenf is holomorphic inD.

Proof. We may assume that z0 = 0. Choose R1 and R2 with 0 < R1 0}

= 0 on .

We will often use directional derivatives as they seem to be most intuitive.It is important to remember that ifP, w Rm then the directional derivativeoffin the direction ofw may be obtained in the following way: let be anysmooth curve such that (0) =P, (0) =w, then

f

w(P) =

d

dt(f )

t=0

.

We also have expressions

f

w(P) = Df(P)(w)

fw

(P) =mj=1

fw

(P)wj.

Definition 7.2.2 A function satisfying conditions of Theorem 7.2.1 iscalled a defining function for.

It is obvious that if Ck(Rm) is such that = 0 on the set{x: (x) = 0} then each component of the set {x: (x)< 0} is aCk-boundeddomain.

Proposition 7.2.3 Let Rm be aCk-bounded domain(k 2) and letbe a defining function for. LetV Rm be open, V = f . Suppose: V R is a function of classCj , (2 j k) such that= 0 onV .Then there is aC(j1)-function onV such that= . We have then

= on V (7.2.1)


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48 7 PSEUDOCONVEXITY AND DOMAINS OF HOLOMORPHY

and if w, v Rn

are tangent to at P (i.e., (P)|w =(P)|v = 0) then2

vw=(P)

2

vw. (7.2.2)

Sketch of the proof. One has to define onV\ by= /. Usinga suitable change of variables we see that extends to a C(j1)-function onV. Then

= () =+.

On this equals to hence

w (x) =

w (x) for w Rm

, x . (7.2.3)

Let v TP. If we take a curve in defining this vector then we have(7.2.3) satisfied all the time on this curve and differentiation of

w along it

produces

v

w =

v

w+

2

vw.

If in addition w TP then the first term of the right-hand side vanishes.

Definition 7.2.4 For x Rm let (x) = dist(x, ), and for A Rm,

(A) = infxA (x). Define also

(x) =

(x) x (x) x

Theorem 7.2.5 If isCk-bounded(k 2) then is of classCk in some

neighbourhood of and= 0 on.

Ifk = 1 then need not be differentiable, also even ifk 2 need not

be differentiable in all of .Hints for the proof of this theorem are given in [KR], Exercise 4 in Chap-

ter 3.

Sinceneed not be of classC

2

in all of we cannot take it as a definingfunction but it is trivial to find a defining function which is equal to on a neighbourhood of . Just take any defining function and a C-function 0 such that = 1 on some neighbourhood of and supp()is contained in the set where is smooth. Then : = + (1 )

has all

desired properties.


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7.3 Geometric convexity 49

7.3 Geometric convexityThe purpose of this section is to present some characterizations of convexityin order to explain, later on, what is the origin of certain definitions used inthe theory of SCV. We also introduce some important definitions.

Fix an open connected set Rm.

Definition 7.3.1 Let: R. We say that is convex in if whenever[x, y] andt (0, 1) we have(tx + (1 t)y) t(x) + ( 1 t)(y), andthat it is strictly convex if the inequality is strict.

Proposition 7.3.2 If: R is of classC2 then it is convex iff

PwRmm

i,j=1

2

xixj(P)wiwj 0

and strictly convex iff

P wRmw=0

mi,j=1

2

xixj(P)wiwj >0.

Proof. Both parts become obvious if one realizes that is (strictly) convexiff every function g(t) = (P+tw) is (strictly) convex in a neighbourhood

of zero and that in turn can be expressed as g 0 ( g >0). But

g(0) = 2

w2 =

mi,j=1

2

xixj(P)wiwj. (7.3.1)

so our conditions seem to say the right thing but only at 0, still since we canvary the point Pwe can get that they imply g being (strictly) positive in aneighbourhood of zero.

Corollary 7.3.3 EveryC2 (strictly) convex is (strictly) subharmonic.

Proof. It follows from the above proposition that the real Hessian (i.e., thematrix of the second derivative) of a (strictly) convex function f is (strictly)positive definite hence has (strictly) positive trace so fis (strictly) positive.Now use Theorem 5.2.17.


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Proof.We will only show that condition (7.3.2) is necessary.Let P and H+P be the hypher-plane separating P from . Let

w H. Then t (P+tw) 0 and (P) = 0. It follows that the functiong(t) := (P+tw) has a minimum at 0 so its derivative vanishes and thusw

(P) = 0, sow Tp. HenceHTp() and considering the dimensionswe see that they must be equal. Considering the minimum ofg again we seethat g(0) 0 so

2w2

(P) 0. This condition holds for everyw Hso forevery w Tp.

Definition 7.3.8 SupposeP is such that is of classC2

in a neigh-bourhood of this point. P is said to be a point of strict convexity if

wTPw=0

mi,j=1

2

xixj(P)wiwj >0

and a point of strict concavity if the strict reverse inequality holds.

Definition 7.3.9 A C2-bounded domain is called strictly convex if all itsboundary points are points of strict convexity.

Proposition 7.3.10 A strictly convex set is convex.

Obviously this proposition follows at once from Theorem 7.3.7 but fromthe implication which we did not prove. Since we will use this proposition inthe proof of Theorem 7.3.13 we give an independent proof.

Proof. Let

S={(x, y) : [x, y] }

thenSis a non-empty open subset of . Let (xj , yj) , (xj, yj)

(x0, y0). Suppose (x0, y0)Sthen [x0, y0] but certainly [x0, y0] soletP]x0, y0[,w = x0y0andg(t) : =(P+tw). Theng(0) =(P) = 0and g(t) 0 in a neighbourhood of 0 since [x0, y0] and 0 on . Itfollows that g has a local maximum at 0 so g(0) = 0 and g(0) 0. But

g(0) = w

(P) sow Tp andg(0) =

2w2

(P)> 0 which is a contradiction.


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Definition 7.3.11 LetFbe a family of real or complex valued functions on. IfK then we define theF-convex hull ofK to be

KF=

x :|f(x)| suptK

|f(t)| fF

.

is said to beF-convex (or convex with respect to F) if

K KF.

Example. IfA() is the family of all affine functions on then KA() =

conv(K) . Indeed

KA=A

{x: (x) maxK

}

.

The first set is an intersection of closed half-spaces containing Khence con-tains conv(K). It also follows from the separation theorem for convex setsthat it is the smallest closed convex set containing K.

Example. KC()= K .

Definition 7.3.12 A functionl: R is said to be an exhaustion functionfor if

tR{x :l(x)< t} Example. l(x) = max{( log ), x}is always an exhaustion function for.

Theorem 7.3.13 The following conditions are equivalent

a) is convex.

b) is convex with respect to the family of all affine functions on.

c) ( log ) is a convex function on.

d) has a convex exhaustion function.

e) has aC convex exhaustion function.

f) =jj wherej j+1 andj are strictly convex.


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g) For every affine functionh: [a, b]

h([a, b])

=

{h(a), h(b)}

.

h) If {I} is a family of intervals in such that

I then I whereIdenotes the set of endpoints ofI.

Proof.

g) c) f)

h) a) b)

d) e)

-

-

6

?

?

6

-

H

H

H

H

H

HY

(a b) Let A() be the family of all affine functions on then, as wehave seen KA() = conv(K) . Now if is convex and K then

K and is compact so conv(K) is compact and of course conv(K).We also have conv(K) = conv(K) thus KA().

(b a)Take K={x, y} then by b), [x, y] = conv(K) = KF so is convex.

(a c) Suppose ( log ) is not convex. Then there are pointsx, y and a t (0, 1) such that

log

(tx + (1 t)y)

< t log (x) + (1 t)log (y)

hence

(tx+ (1 t)y)< et log (x)+(1t) log (y) t(x) + (1 t)(y).

Let P = tx+ (1 t)y and let P Rm \ be such that ||P P|| 0 such thata < (x),b < (y) andta + (1 t)b= ||P P||. Finally let w= (P P)/||P P||, x =x+ aw,


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y

=y +bw. Then x

, y

and P

=tx

+ (1 t)y

. Since is convex weget P which is contradiction.

(c d)Define l(x) = max{ log (x), ||x||}.

(d e f) These proofs are identical to the proofs of Theorem 7.4.8((3)(4)(5)) after substituting convex for plush. Since these proofs arequite complicated and this is, after all, a course in SCV we refer the readerthere.

(f a)Since each of j is convex is also convex.

(c g)Ift [0, 1] then

log

h(ta+ (1 t)b)

= log (th(a) + (1 t)h(b))

t log (h(a)) + (1 t)log (h(b))

min

log (h(a)), log (h(b))

= log

{h(a), h(b)}

and the claim follows.

(g h)What g) really says is that the distance from an interval in tothe boundary of is attained at the endpoints of that interval so for eachinterval I we have (I) =(I). The fact that K is equivalentto the fact that K is bounded and (K) > 0. Now let {I} satisfy thehypothesis of h). Then

0< (

I) = inf

(I) = inf

(I) =(

I)

and since

I is bounded we get that

I is bounded so

I.

(h a)LetS={(z, y) : [x, y] }

then Sis non-empty open subset of . Now let (xj , yj) S, (xj, yj)(x0, y0) and let Ij = [xj, yj], j 0. Then

j1 Ij so

j1 Ij but it is easily seen that I0

j1 Ij so I0 and thus(x0, y0) Swhich shows that Sis also closed and as is connected weget that S= so is convex.


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7.4 Pseudoconvexity 55

7.4 PseudoconvexityWe will now consider an extension of the notion of convexity to one whichis more suitable for the study of holomorphic functions. One of the mainobjectives is that the class we define be closed under biholomorphic trans-formations. The class of all convex domains does not satisfy this conditionbut it can be a starting point. Suppose : 1 2 is biholomorphic and 2is convex. What can be said about 1? One approach is to use exhaustionfunctions. If l: 2 R is a convex exhaustion function then l is anexhaustion function for 1 but it need not be convex, however it must beplush (as l is plush and this is preserved). Another approach is to look at

C2-bounded domains and assume that is also aC2-diffeomorphism definedon a neighbourhood of 1 onto a neighbourhood of 2. In this case ifis adefining function for 2 then is a defining function for 1. satisfiescondition (7.3.2) which can be stated as

2v2

(Q) 0 forQ 2,v TQ2.This condition need not be preserved by holomorphic mappings but we canderive one that will be. Namely ifv, iv TQ2 then

ni,j=1

2

zizj(Q)vivj =

2

v 2(Q) +

2

(iv)2(Q) 0

but ifQ = (P) and v= D(P)(w) then , by Lemma 5.4.3

ni,j=1

2 zizj

(P)wiwj =n

i,j=1

2zizj

(Q)vivj.

Since (v TQ2) (w TP1) and D(P)(iw) = iv we see thatcondition

Q2vv, ivTQ2 2

v2(Q) +

2

(iv)2(Q) 0 (7.4.1)

is preserved.

Definition 7.4.1 If is a C1 bounded domain then the complex tangent

space to atP is defined asTCP ={w TP :iw TP}

Proposition 7.4.2 Let be a defining function for then w TCP iffnj=1

zj

(P)wj = 0.


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Proof. Let H= w:nj=1 zj wj = 0. Then

zjwj =

1

2

xjaj+

yjbj

+

1

2i

xjbj

yjaj

.

Hence w H n

j=1xj

aj +n

j=1yj

bj = 0 so w TP. Now we have

H TP, and H is closed under multiplication by i so H TCP, but

the complex dimension ofH is (n 1) so H=TCP.

We can now state our first definition

Definition 7.4.3 A C2 bounded domain is called Levi pseudoconvex if

ni,j=1

2

zizj(P)wiwj 0

wheneverP andn

j=0zj

(P)wj = 0, where is aC2 defining function

for.

Definition 7.4.4 The form

ni,j=1

2zizj

(P)wiwj

is called the Levi form of.

Some as in the case of convexity we define a strictly pseudoconvex domain:

Definition 7.4.5 A C2 bounded domain is strictly pseudoconvex if

n

i,j=12

zizj

(P)wiwj >0

wheneverP, w= 0,n

j=0zj

(P)wj = 0.

It is easily seen that C2 bounded convex domains are pseudoconvex andstrictly convex domains are strictly pseudoconvex.


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Remark.It follows from Proposition 7.2.3 and Theorem 7.2.5 that the def-initions of Levi pseudoconvexity and strict pseudoconvexity do not depend

on the choice of the defining function. It is in fact enough to check condition(7.4.1) for as only the behaviour close to is important. It also followsfrom considerations at the beginning of this section that Levi pseudocon-vexity and strict pseudoconvexity is preserved by biholomorphic mappingswhich smoothly extend onto a neighbourhood of the closure of the domain.

It should be apparent by now that if we want conditions which are pre-served by biholomorphic mappings then convex functions must be substitutedby plush functions, so the next definition should not be surprising.

Definition 7.4.6 A domain Cn is called Hartogs pseudoconvex if thefunction( log ) is plush in.

We will now state a theorem which will introduce some order to this newtopic. Before that we will need one more definition:

Definition 7.4.7 LetU be the unit disk inC. An analytic disk inCn is anon-constant holomorphic mapping: U Cn. If extends continuouslyto Uthen we call it a closed analytic disk and denote = (U) to be theboundary of the analytic disk.

We will often identify an analytic disc with its image, so depending on thecontext can be either a function or a set.

Theorem 7.4.8 The following conditions are equivalent

1) is convex with respect to the family of all plurisubharmonic functionsin.

2) is Hartogs pseudoconvex.

3) has a continuous plush exhaustion function.

4) has aC strictly plush exhaustion function.

5) =jj wherej j+1 andj are strictly pseudoconvex.

6) If is a closed analytic disk then() =().


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7) Let () be a family of closed analytic disks in. If then.Under the additional assumption that isC2-bounded all the above are equiv-alent to

8) is Levi pseudoconvex.

Proof.

6) 7) 1)

2)

8)

3)

5) 4)5)&8)

-

6

?

6

?

: X

X

X

X

X

Xz

1) 7) Let P() denote the family of plush functions in . Let be aclosed analytic disk and fP() then f is subharmonic, hence

|t|


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hence (j) 1

j |w|> 0 so j . Next let >0 be such that ()>() +. Then for P , z we getP+ 1 1j

w

z () + 1 1

j

|w|

1

j() +.

Hence j(j) soj(j) and by the assumption

jj .

But P0+ (1 1j

)w j and P0+ (1 1j

)w= (z0 1j

w) z0 which isa contradiction.

6) 2)We have to show that for a fixed z and smallw Cn we have

log (z) 12

20

log (z+wei) d.

To shorten the notation we will denote

() : = log (z+w) C, || 1.

Hence we are to show that

(0) 1

2

20

(ei) d.

We will use the fact that every real, continuous function on Ucan be uni-formly approximated by real parts of holomorphic polynomials. (This isa restatement of the fact that real trigonometric polynomials are dense in2-periodic, continuous, real functions).

So let pbe a holomorphic polynomial in C , such that

|Rep() ()|< for U.

Starting with a smaller and adding a constant to p we may assume thatRep > on U.

Take a point b Cn with |b| 1 and define

() :=z+w+bep() for U

then is a closed analytic disk. We will show that . Assume for amoment that this has been shown, then we finish as follows:

(0) =z+bep(0) .


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Since b was arbitrary with |b| 1 we get that B(z, |e(p(0)

|) , orequivalently (z) |ep(0)| = eRep(0). Since (0) = log (z) we get

(0) Rep(0). But then

(0) Rep(0) = 1

2

20

Rep(ei) d + 2

0(ei) d.

Take 0 to get the desired inequality.It remains to show that . It will be enough to show that ,

so let U then

|z+w (z+w+bep()

)|= |b||ep()

| eRep()

< e()

=(z+w).

Hence () =z+w+bep() if||= 1.

2) 3)Obvious.

3) 1)Let l be a plush exhaustion function for and let K. Thenfor some t R, K {l < t}. If z KP() then l(z) supKl t soz {l < t+} .

2) 8)We assume now that is C2 bounded. We know that is C2

in a neighborhood of and ( log()) is plush in . It follows that

2zizj

( log())(z)wiwj 0

for z in an -neighborhood of and w Cn. But

zjlog() =

1

zj

so2

zizj log(

) =

1

()2zi

zi +

1

zizj .

Hence

0i,j

2

zizj( log())wiwj =

1

()2

i

zi

wi

2

1

i,j

2zizj

wiwj.


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It follows that there exists an r >0 such that for 0 < ||< r, ()and (0) . In particular has a maximum at 0. Next we consider

||2 (())

= ( )(0) + 2Re

( )

=0

+1

2

2

2( )

=0

2

+ 2

( )

=0

|2| +o(||2)

= Re

2

2( )|=0

2

+ 2zizj

(z+a)i(0)j(0)||

2 +o(||2)

Re

2

2( )|=0

2

+o(||2).

The last inequality holds because (0) TCz+a and the Levi form of is

positive definite on this space. The inequality

||2 Re(2) +o(||2)

cannot hold on any neighbourhood of zero, it is false if = 0 and if = 0

than we can always make the right-hand side positive. We have thus provedthat ( log ) is plush close to .

Now we will show that if has any continuous exhaustion function lwhich is plush close to then we also have an exhaustion function whichis plush in all of .

Let C > 1 be such that l is plush on the set {z : l(z) > C}. Let:R R be convex, increasing, = 0 on (, 1], (x) = x for x > 1.Define

(z) =(l(z) 2C) + 2C.

According to Cor. 5.2.11 is plush when l > Cbut ifl(z) C+ (C1) then

(z) = 2Cso is plush on all of . We also have =l on {z:l(z)> 2C+1}so is an exhaustion function.

3) 4) Let l be a continuous plush exhaustion function for . Let t ={z : l(z) < t} and = {z : (z) > }. On define l =(l|3) , where are the functions from Lemma 5.4.1.


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It follows that j is plush on j+2

and strictly plush on the subset ofj+2 where j j+ 1> 1/2 (because (j j+ 1)> 0 on this set), we

also have that j = 0 on the set where j j+ 1 1/2. Now let us getsome information on these sets. We have

j lj l on j+2

and l (j 3/2) on \ (j3/2) so

j j+ 1 1/2 on j+2 \ (j3/2).

Next j l + 1/2 on j

soj (j 2) + 1/2 on

j2

hencej j+ 1 1/2 on

j2.

It follows that j is 1 and strictly plush on j+2 \ j3/2 and zero on

j2.

So now finally we can construct our exhaustion function as

:= 0+

k=1

akk

where ak are defined as follows: suppose for k < j we have defined ak >0 sothat 0+

j1k=1 akk isl and strictly plush on

j1. (This is satisfied for

j = 1). If we takeaj j thenajj j > lon j\

j1, so 0 +

jk=1 akk

l on j . Since j is strictly plush on a neighborhood of j+1

\ j1, usingLemma 5.3.12 we can take aj so large that the sum 0 +

jk=1 akk becomes

strictly plush on j \ j1 and it remains strictly plush in j1 because theshorter sum was such and ajj is plush there.

The last thing to observe is that the sum did not change on j2 as jis zero there. Hence the series 0+k=1 akk converges to a C, strictlyplush function satisfying l. This last inequality guarantees that isan exhaustion function for .

(4 5)The proof of this implication would carry us a little to far from thegeneral subject so we just give an idea. Since we have a C, strictly plush


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exhaustion function l then it is tempting to define j ={z :l(z)< tj},where tj as j . Since the complex hessian of l is strictly positivedefinite on it would seem that j = l tj is a defining function for jwhich makes j strictly pseudoconvex. This would be true if we knew thatj does not vanish on the set {j = 0} but there is no guarantee that thismay be achieved. Fortunately, using a lemma of Morse it is possible to find anew exhaustion function l which differs froml by a linear term and has onlyisolated critical points. Then l is alsoC and strictly plush and now we canchoose tj so that l = 0 on {l =tj} which shows that j ={l < tj}are strictly pseudoconvex.

5)&8) 2)Since j are strictly pseudoconvex, they are Levi pseudocon-vex hence by 8) 2) we see that ( log j) is plush. But j so( log j ) ( log ) and, by Proposition 5.3.5, ( log ) is plush.

Definition 7.4.9 A domain is called pseudoconvex if it satisfies any of theequivalent conditions of the previous theorem.

Corollary 7.4.10 Pseudoconvexity is a local property i.e., if everyP has a neighbourhoodV such thatV is pseudoconvex then is pseudo-

convex.

Proof. Let P, Vbe as above. There is a neighbourhood V1V ofP suchthat

V |V1= |V1.

(For instanceV1 = B(P, V(P)/2).) It follows that log is plush inV1 .Repeating this for every point of we see that log is plush close to theboundary of and, as shown in the proof of the previous theorem (8) 3)),this implies pseudoconvexity of .

Corollary 7.4.11 Pseudoconvexity is preserved by biholomorphic mappings.

Proof. It has already been remarked before that biholomorphic mappingspreserve plush exhaustion functions.


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Remark.The proof of 8) 3) in Theorem 7.4.8shows that if is pseudo-convex and is an analytic disk which has a common point with

then (hence the situation shown on Figure 3 is impossible); this isthe same with convex domains and open intervals. Hence a domain whichhas a strictly concave point at the boundary cannot be pseudoconvex (suchan analytic disc can be found in the tangent plane).

Figure 3.

7.5 Domains of holomorphy

As mentioned before there exist pairs of domain (1, 2) inCn such that

1 2 & fH(1)fH(2) f

1=f .

As an example one may take 1 = B(0, 1) \ B(0, 1/2), 2 = B(0, 1) (laterwe will prove a general result from which it will follow that this is a correctexample). It is therefore natural to define a class of maximal domains.

Definition 7.5.1 Cn is a domain of holomorphy if for any two domains1, 2 with2 connected, 2 , 1 2 there is anf H() such

that there is no fH(2) withf= f on1.

This definition is somewhat obscure so consult Figure 4. The complicationcomes from the fact that it may happen that 2 is not connected andwe do not want to require that an extension from one component of thisintersection agrees with fon other components. Think of the complex plane


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7.5 Domains of holomorphy 67

with a ray removed and of extending a holomorphic branch of logarithmfrom this set. This function can be extended only in the way described inthe previous definition.

Figure 4.

We will now present several equivalent formulations of this definition.

Theorem 7.5.2 Let be a domain inCn, then the following are equivalent:

1) K(K) =(K) (whereK: = KH()).

2) is holomorphically convex (i.e., H()-convex).

3) fH()1,2 if2 , 2 connected, 1 2 then there is no

fH(2) withf=f on1.

4) is a domain of holomorphy.

5) zr>(z)fH() such that f does not extend from B(z, (z)) toB(z, r).

6) IfX is infinite with no cluster points in then there is anf H() which is unbounded onX.


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Proof.Within the proof we will let

Kstand for

K. We will introduce anauxiliary condition

a) AssumeK1 K2 . . . be such that

jKj = . For eachj let

j \ Kj andj C. Then there exist functionshj H() suchthathj(j) =j and |hj| 2

j onKj.

Then the proof will run as shown below:

5) 4) 3)

1) 2) 6)--

6

?

Q

Q

Q

Q

Q

Qs

1) 2)It is only necessary to show that Kis bounded but Kconv(K),and this larger set is bounded.2) a)Fix j. Since j \ Kj there is an h H() such that |h()|>supKj|h|. Multiplyingh by a suitable constant we may assume that h() = 1and then c : = supKj|h|< 1. Next observe that h

m(j) = 1 and supKj|hm|=

cm and take m so large that cm 2j/|j|. It follows that hj: = jhmsatisfies the claim.

2) 3)Suppose 3) does not hold then

fH()1,2fH(2)f|1=f|1 . (7.5.1)

Let {wj}be a dense sequence in and for each j let Bj: =B(wj, (wj)).Let {Kj} be a sequence of subsets of satisfying the assumptions of a). Since

Kj we have Bj\ Kj =f (consult the definition ofBj) hence we may

find j Bj \ Kj . Now take hj H(), |hj| 2j on Kj, hj(j) = 1 and

define

f=j=1

(1 hj)j.

Since

j=1j|hj|converges normally on , we have that the product definingf converges normally to a function which is not identically zero hence f is


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offat z0 is convergent in z0+ D which implies that this series is convergentin B(z0, r). This means that everyf H() can be extended to B(z0, r),taking 5) into account we get that r (z0) so (K) (K) and thusthey must be equal.

2) 6) Suppose X is as in 6) and Kj are as in a). Let l1 = 1, and take

1 X\ K1 (K1 so such a point exists). Then there is an l2 > l1such that 1 Kl2 and we pick a 2 X\ Kl2 and so on. We can now findfunctions hj H() such that |hj | 2

j on Klj and

|hj(j)| j +j1

l=1|hl(j)| + 1.Then

hj converges normally in to a holomorphic function f and

|f(j)| =

l=1

hl(j)

|hj(j)|

j1l=1

|hl(j)|

l=j+1

|hl(j)|

j+ 1

l=j+1

2l

jHence f is unbounded on X.

6) 2)IfK then everyfH() is bounded onKsofis boundedon Kbut, by 6), this implies that every infinite subset of Khas a clusterpoint in so K.

Corollary 7.5.3 is a domain of holomorphy iff for everyP there isanfH() such that limP,|f()| = +.

Proof. Take any sequence zj P, zj and apply 6) to X={zj}j=1.

Corollary 7.5.4 A biholomorphic image of a domain of holomorphy is adomain of holomorphy.

Corollary 7.5.5 Every convex domain is a domain of holomorphy.


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7.5 Domains of holomorphy 71

Proof. IfK then

Kconv(K) .

Corollary 7.5.6 Every domain of holomorphy is pseudoconvex.

Proof. Iff H() then |f| P() so KP() K and thus ifK

then K so also Kp() and hence is pseudoconvex.

Remark. The question whether the reverse implication is true is calledthe Levi problem. We will show later that it has a positive solution. See

Section 8.1 for some consequences this fact has for domains of holomorphy.Remark. As we mentioned in the introduction every domain in C is adomain of holomorphy (exercise) hence pseudoconvex.


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72 8 THE PROBLEM

8 The problem8.1 The problem and its consequences

Recall that we have introduced an operator acting on complex differentialforms. So far it was only used to write a condensed form of Cauchy-Riemannequations but now we will show how it can be put to much more use.

Definition 8.1.1 A differential form is said to be-closed if = 0 and-exact if there exists a form such that = .

Since 2 = 0, a -exact form must be -closed. The question whetherevery-closed form is -exact is called the -problem. A similar problem canbe posed for real forms and the operator d and then Poincare Lemma saysthat in a starlike domain (or more generally contractible to a point) everyd-closed form is d-exact; in completely arbitrary domains this need not betrue.

We will begin by stating some theorems about solutions to the problemand listing their consequences.

Theorem 8.1.2 If is aC1, compactly supported, -closed, (0, 1)-form on

Cn (n 2) then there exists aC1, compactly supported functionu such thatu = andu 0 on the unbounded component ofCn \ (supp ).

Corollary 8.1.3 (Hartogs extension phenomenon) Let Cn (n 2)be a domain and letK be compact and such that\ K is connected.Supposef H( \ K) then there exists anF H() such thatF = f on \ K.

Proof. Let Cc (Cn) be such that = 1 on K and supp then

(1 )fcan be considered to be in C(). Let

=

((1 )f) in 0 in Cn \ .

Since f is holomorphic in \ Kand (1 ) = 1 in a neighbourhood ofwe get that ((1 )f) = 0 in a neighbourhood of so is a (0, 1)-form


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8.1 The problem and its consequences 73

with C

c (Cn

) coefficients and of course is -closed. Letu be the solutionto u = given by the previous theorem and define

F: = (1 )f u in

thenF = u = 0,

soFis a holomorphic function in . We know that = 0 on the unboundedcomponent ofCn \supp and this set has a non-empty intersection withevery neighbourhood of , in particular F = f on some open subset of \ Kand since \ Kis connected F =fon all of \ K.

A few words of comment are in order. One might be surprised why take of the form g = and then seek solution to u = , when u = g seems ahandy answer. The point is that uhas an additional property of being zeroin a neighbourhood of, thus it does not change g on the set where g isalready holomorphic, it changes it however and makes holomorphic on theother part of .

Corollary 8.1.4 If Cn (n 2) is open andf H() thenfhas noisolated zeroes.

Proof. Supposez0 is the only zero off inB (z0, ) then 1/f is holomorphicinB(z0, ) \ {z0} so, by Hartogs extension phenomenon, can be extended toB(z0, ) which is impossible since limzz0 |1/f| = +.

The above proof can be used to show much more, namely that no com-ponent of the zero set of f may be compact, in other words every suchcomponent must extend to the boundary of . This is again completelydifferent from the situation we have in C .

Theorem 8.1.5 If Cn is pseudoconvex and is a C, -closed,(p, q+ 1)-form on then there exists aC, (p, q)-form such that = .

If q= 0 then every other form satisfying = also has to be of classC.

Corollary 8.1.6 Let Cn be pseudoconvex andD = {zn = 0}. ThenD can be considered as an open subset ofCn1 and iff H(D) then thereis anF H() such thatF|D =f.


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74 8 THE PROBLEM

Proof.Consult Figure 5 for illustration.

Figure 5.

Let be the projection onto Cn1 defined by (z, zn) = z. Let B =

1(Cn1 \ D) and let C() be equal to 1 on a -neighbourhoodE ofD and to 0 on \ B. We will try to find F in the form

F = (f )+znv

where v remains to be defined. The fact that f is not defined outside Bdoes not matter since was taken such that (f ) can be extended to by putting 0 on \ B.

Applying to the above equality we get

F = (f ) +znv

hence ifF is to be holomorphic v must satisfy

znv = (f )

or equivalentlyv =

(f )

zn. (8.1.1)

Observe that the right-hand side of (8.1.1) is a well defined, C, (0, 1)-formin since vanishes on Eso division by zn does not introduce problems.We also have


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8.1 The problem and its consequences 75

f

zn

= (f )

zn

and /zn = 0 o n E and (/zn) = ()/zn = 0 o ff E. (We haverepeatedly used the fact thatvanishes on holomorphic functions.) It followsthat (8.1.1) can be solved so F can be found in the desired form and thenobviouslyF|D =f.

Once again we draw attention to the technique that was used. It wasno problem to extend f to B, f does just that, then by taking (f )

we got a function which was fine on E but was not holomorphic, so weused the solution to the -problem to find an extra term that would make(f )holomorphic without changing it onD , this makes the proof similarin principle to the proof of Cor. 8.1.3.

Exercise. Modify the arguments of the above proof to show that if Cn

is such that for every C, -closed, (0, q)-form in there exists a (0, q+1)-form with = then has the extension property of Cor. 8.1.6 for all-closed, C, (0, q)-forms on D.

Corollary 8.1.7 (Solution of the Levi problem) Every pseudoconvexset is a domain of holomorphy.

Proof. We proceed by induction. Ifn = 1 then every open set is a domainof holomorphy so there is nothing to prove. Assume the corollary is true forn 1. Fixz0 . We will want to satisfy condition 5) of Theorem 7.5.2.

Take P B(z0, (z0)), by changing coordinates we may assumethat z0 = 0 and P D = {zn = 0}. Since D = on D we see that( log D) is plush so D, or rather each of its components, is pseudoconvexand hence, by the inductive hypothesis, is a domain of holomorphy.

It follows that anfH(D) can be found which cannot be extended fromBCn1(0, (0)) onto any larger ball. Now use the previous corollary to pro-duce an extension F offonto . IfFcould be extended fromBCn(0, (0))onto a larger ball then so could be f hence Fis the function we wanted tofind.


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76 8 THE PROBLEM

Exercise.Modify the arguments of the above proof to show that if C

n

is such that for everyC, -closed, (0, q)-form in there exists a (0, q+1)-form with = then is a domain of holomorphy. (Use the previousexercise.)

This exercise, together with Theorem 8.1.5, shows that the fact that -p

wolniewicz introduction to several complex variablewols

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