Differential Forms Orthogonal to Holomorphic Functions or Forms, and Their Properties

HISTORICAL SURVEY AND OPEN PROBLEMS

88

7) Are the orthogonal polynomials P(w, z) E O(aD) dense in O(aD )? A positive answer to this question would enable us to state the analogue of the Hartogs-Riesz theorem of §12 in a way similar to the classical theorem: one would be considering measures orthogonal to the polynomials in O(aD ) described in §5.

8) Elucidate for which domains the second summand in Theorem 7.1 is necessary. The results of §§6-8 solve the problem only in certain cases (the case of strictly pseudoconvex domains from which a finite number of strictly pseudoconvex domains have been removed; the case of domains not covered by Theorem 8.1, when n = 2).

9) Characterize the class of domains for which the Cauchy-Fantappie formula is the general form of the integral representations for holomorphic functions (analogues of the Cauchy integral). The theorems of §13 show that for n = 2 this class contains the strictly pseudoconvex domains and consists only of domains of holomorphy. This problem is related to the next two problems. 10) Is the Cauchy-Fantappie formula invariant under biholomorphic maps?

11) Find the general form of the integral representations of holomorphic functions for an arbitrary bounded domain with smooth boundary.

12) Prove approximation theorems for forms in Z(,_ I ), similar to the well-known results of Vitushkin [1] and Mergelyan [1].

13) Extend the definition of the product of distributions of class A in R2r-' to arbitrary distributions. 14) The representations of distributions in §14 are closely related to their Fourier transforms when n = I (see Bremermann [11, Chapter III). Is there a similar convenient connection when n > I? A positive answer to this question would lend further support to our view that forms in B,_1 (or Z(n, n _ I)) are good analogues of the holomorphic functions of one variable. Note. After this book went to press, solutions have emerged for some of the problems listed above. Aronov and Kytmanov [1) have obtained an affirmative answer to question 2) for functions of class C'(D). Also, Kytmanov [3] has

settled question 3) positively for forms in ,q)(D). In another paper [5] by the same author, problem 13) is solved in arbitrary dimension.

CHAPTER V

INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED DIFFERENTIAL FORMS AND HOLOMORPHIC FUNCTIONS

§16. A characteristic property of a-closed forms and forms of class B

In this section, we answer question 3) in the " Brief historical survey and ' open problems" section. THEOREM 16.1. Suppose a (D) E C 2" and y E C( P Q)(D ), X > 0. Then y is a a-closed in D if and only if, for all z E D,

Y(z) = Ip.9(D, Y) - Blp.a-i(D, Y)

(16.1)

If the coefficients of y are harmonic in D, then the conditions on aD and the behavior of y on aD can be relaxed. THEOREM 16.2. Let D be a bounded admissible domain with piecewise smooth boundary, and suppose y E C(p,q)(D) has coefficients harmonic in D. Then Y is 8-closed in D if and only if (16.1) holds.

10, and I.,,

Note that, if q = 0, then I p, q 9(D, y) has coefficients which are harmonic in D. Hence from Theorem 16.2 we deduce COROLLARY 16.3. Suppose D is as in Theorem 16.2, and -y E q,O)(D). Then,

for z E D, -y:= Ip,o(D, -y),

if and only if the coefficients of y are holomorphic in D. 89

(16.2)

90

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

For the proof, we need to use the double form

z) =

((2n1Ti)n'

karU(I'

I

[I, k]dz,

k)

(a part of UP.Q), as well as the following lemma. LEMMA 16.4. Let D be a bounded admissible domain in C", n > 1, and suppose

f E C '(D) is harmonic in D. Then f can be approximated in the topology of zk (Z-2", zk (4 D,

CA(D) by linear combinations of fractions of the form I

PROOF. Since D is admissible and bounded, the space of functions harmonic

in a neighborhood of D is dense (in the topology of CA(D)) in the space of functions of class CA(D) harmonic in D (see Weinstock [3]). Hence we need only prove the lemma for functions harmonic in a neighborhood of D.

Now let f be harmonic in the closure of a domain G D D with smooth boundary. Then, by Green's formula, for z E G

.t(Z) _ (2n 12)Q2n

a(n) I [g,z ) a

an

Gt

Z)I

do, (16.3)

z) = I - z I22n; also, where Q2" is the area of the unit sphere in C", and a f(')/an denotes the derivative of fin the direction of the outward normal to aG at , and do the surface element of aG. By replacing the integral in (16.3) z)/an by difference quotients, by the approximating Riemann sums and we obtain the assertion of the lemma. PROOF OF THEOREMS 16.1 AND 16.2. The necessity follows immediately from

the Martinelli-Bochner-Koppelman formula (1.1). Let us prove the sufficiency. If the form

2'

A dfJ

J

I

satisfies (16.1), then each of the forms

d

Yf

I

also satisfies (16.1), Indeed,

IP,9`Y) = I D

I'

A If

df, n

n (l)'wq(, Z)

A dZk

k

= f 11 2' (,j()df, A wq A aD

A a(J)d [J] A dz,)

§16.5-CLOSED FORMS AND FORMS OF CLASS B

_ I I fo _

A w4 /

91

A dzj

A dzj. J

Similarly,

Y) _ 2 aloqJD1 Y) A dzJ.

J Hence it is enough to consider forms of the type (0, q)

y(z) = 2'yr(z)dY,. I

Suppose now that the conditions of Theorem 16.2 are satisfied. If (16.1) holds for y, then y1(z) = II,q(D, ay) = 0 for z E D by (1.1). But the coefficients of y, are harmonic outside D, and tend to 0 as I z I -+ oo; also, being convolutions of compactly supported bounded measurable functions with the locally integrable functions I Ik, they are continuous in C". Hence by the uniqueness theorem for harmonic functions we have y, = 0 on C". Writing out this identity, we see that, for every multi-index I'(i,< i 2 < . < '0,

ay, D 11

p-

k4Z Il

aJk

-

A

-

ag m6ZI

L

d-

A d = 0.

(16.4)

By Lemma 16.4, the y, ( ) can be approximated in C' (D) by linear combinations of the g(', zs), zs (4 D, s = 1, 2,.... Hence (16.4) yields D 1,

7 k6-11

rr-.

aak

A 7 a(I,m)-a df [I,m}Add--0.

(16.5)

m(41

Summing over I in (16.5), we get

fDh1d= 0, i.e. 8y = 0.

Suppose next that we are in the situation of Theorem 16.1. The coefficients of the form y2 = Io, q(D, y) are harmonic in D, and y2 E Coo, q)(D) (see Corollary 2.4). Further, a -y2= a y by (16.1), and hence

,02, JD, aye) = 10,q(D, ay) = 0. Thus, (16.1) holds for y2, and it remains to apply Theorem 16.2. We have just considered the properties of forms representable by (16.1), i.e. the Martinelh-Bochner-Koppelman formula without the second summand. It is

92

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

interesting to discuss the class of forms to which the first summand belongs. If we consider (1.1) with p = n and q = n - 1, we see that the first summand has the form

In1n-1(D,Y) = f

o

It is clear that this form is of class B,,,,,-1(D). Forms of this important class have already been encountered before in this book (see Theorems 2.12, 7.1 and 14.1-14.3). Let us present some new properties of this class of forms. They can be written as

Y = YG(Z) _ 2 (-1)k-I aGdi[k] A dz, k=I

with G harmonic in D. THEOREM 16.5 (UNIQUENESS). Let aD E C', and suppose G E C'(D) is harmonic in D. If the restriction of YG to aD is zero, then YG = 0. PROOF. n

O=f GYG(z)= f D

D

f

aG

D k=I

azk

2

diAdz.

Hence aG/azk = 0, k = 1, ... , n, i.e. G is antiholomorphic in D, and so YG = 0. We remark that Theorem 16.5 remains valid even for unbounded domains, if we assume G = 0(1 z 12-2n) as I z I -> oo. THEOREM 16.6. If n > 1 and aD is connected, and y E C(n,n-1) is such that

Y(z) = In,n-I(D, Y),

z E D,

(16.6)

then y =- 0. PROOF. Consider the functions

2)I

- (n (27ri)

f

aD

z)

G+(z), j l G-(z ),

z E D, z V- D.

These functions are harmonic (outside aD). It follows from the properties of the potential of a simple layer that G' = G- on aD. By Theorem 2.5, YM IaD=YG+IaD-YG-IaD1

(16.7)

so that (16.6) implies YG-IaD = 0. By Theorem 16.5, YG-= 0 outside D, i.e. G- is

antiholomorphic outside D and vanishes at infinity; hence G-= 0. Hence G+ = 0 as well. Now the assertion of the theorem follows from (16.7).

93

§16. a-CLOSED FORMS AND FORMS OF CLASS B

This theorem shows that there are no nontrivial forms representable in D by (16.6) (i.e. by (1.1) with only the first summand). The next result describes forms representable by (16.6) outside D.

THEOREM 16.7. Let a(D) be connected and of class C, and suppose G CG 1 (CD) is harmonic in CD, with G = 0(1 z 12-2") as I z I -* oo. Then

Z a D, YG(z) = -In,"-1(D, YG), if and only if G can be continued antiholomorphically into D. PROOF. Necessity. By considerations similar to those in the proof of Theorem 16.6, we see that G- is antiholomorphic in D and that Yc = Yc- Hence G is antiholomorphic in CD and tends to zero as I z I -* oo, so that G = G-. Sufficiency. Suppose G can be continued antiholomorphically into D: denote the continuation by G,. Then G, E C'(8D) and G, is antiholomorphic in D; hence G, E C'(D) (Shabat [1], p. 334). Let

n- - 2)! (2,.Ti

)n

Z) =

aD

G'(z), G-(z),

z E D, z (4 D;

then 1'c' Ian - Yc-Ian = -Yc laD = 1'c, Ian - Yc laD'

so that

(-fG+ -yc,) ID = (YG7G) IDLet D = {z E C": p(z) < 0), where p E C'(C") and grad p 18D 11 G

0. Set

a

}" aG ap grad pr-1 ; k=1 34 azk

NG is defined near aD. It is easy to see that NG IaDa'ff = YG Ian'

where as is the surface element of W. Now we see from (16.8) that NG+_G, = NG--G for z E 1D, and G+ -G, _ G - - G on 1D. It follows from this and the complex form of Green's theorem that the function G+ (z)

G,(z),

,.G-(z) - G(z), is harmonic in C" with

0; hence

z Ei

z (4 D,

= 0. Consequently G+ (z)

= G, (z) and G -(z) = G(z). Hence YG+ = 0 and YG- = YG

94

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

§17. Holomorphy of continuous functions representable by the Martinelli-Bochner integral; criteria for the holomorphy of integrals of the Martinelli-Bochner type

If p = 0, the kernel in (16.2) becomes the Martinelli-Bochner kernel UO.OIJ. Z) _ -Un.n-1(Zr

0 - `n - 1)!

nk

Zk)d [kJ A d

I-zI

(2irz)"

nd the formula itself reduces to the Martinelli-Bochner formula for holomorphic functions:

rcZ) = f o

ZED,

z),

(17.1)

which is valid if D is a bounded domain with 8D piecewise smooth, and f E AC(D). The Martinelli-Bochner kernel z) is not holomorphic in z for n > I. Nevertheless, the following sharpening of Corollary 16.3 holds. THEOREM 17.1. If D is a bounded domain with piecewise smooth boundary, and

the representation (17.1) is valid for an f E CA(D), then f E A(D).

We may drop the condition f E C'(D), but we must then impose additional conditions on 8D: THEOREM 17.2. If f E C(D) is representable in a bounded domain D by the Martinelli-Bochner integral (17.1), then j is holomorphic in D in each of the following cases:

a) 8D (a C2; b) n = 2, and 8D is connected and of class

If f E AC(D), then

iDo,0(r, Z, = o,

Z

D.

(17.2)

As it turns out, (17.2) completely characterizes the traces on 8D of functions holomorphic in D: COROLLARY 17.3. In each of the two cases considered in Theorem 17.2, the condition (17.2) holds if and only if for f E C(8D) there exists F E AC(D) such that F lao = f. .

Let Pk,l be spherical (i.e. homogeneous and harmonic) polynomials forming an orthonormal basis for LZ(8B(0,1)), where 8B(0,1) is the unit sphere in C". Here,1 is the degree of the polynomial, and k its index among the polynomials

95

§17. HOLOMORPHY OF CONTINUOUS FUNCTIONS

of degree I occurring in the basis, so that 1 = 0, 1 , 2 ... , and k = 1, ... , a(1), where 2(n + k - 1)(2n + k - 3)!

k! (2n_2)! COROLLARY 17.4. Suppose D is bounded, and 8D is connected and of class C2.

Then an f E C(8D) is the trace on 8D of a function from AC(D) if and only if "

aPk,r f(f) I (_1)j_' aD j=1

k = 19 21 ...,x(1). For n = 1, this corollary is the well-known classical criterion, namely

1= 0,1,2,...;

orthogonality to all the monomials i",1= 0, 1, 21 ... .

We shall not present the proof of the rather easy Theorem 17.1 here, but refer the reader to Aronov and Kytmanov [11 (see also Corollary 16.3), or to the book by Aizenberg and Yuzhakov [1], §1. We proceed to the proof of Theorem 17.2, which is more difficult.

Let f E C(aD ), aD E C'; consider the following integral of MartinelliBochner type:

f()Uo,o(z)f f+(z), f (z),

z E D, z (4 D.

aDD

Consider lim Z, Z, t E a D [ f + (z) - f -(z')], where z, z' -- in the same manner as in (2.9). By Theorem 2.6, this limit is equal to f(), so that Theorem 17.2

and Corollary 17.3 are equivalent. Therefore it suffices to prove Corollary 17.3.

We introduce some notation. Let z0 E 3D, z+ E D and z - E CD, with the last two lying on the normal to aD at z ° and I z

=

P

- ZEaD inf I'-zI, inf 1-zI, zEaD

z

z

z0

.

Define

QED,

then D = (z E C": p(a) < 0). If 8D E C2, then (see Volkov [1], §2) the following assertions hold: a) There exists a neighborhood U of 8D such that p E C2(U). 1 in U.

b) I grad p C)

'P (Z+ k

and z$EU.

ap Z'

(z

ap

a(Zk)

(zo),

k = 19 2,... n,

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

96

Let

ap aF±

NF -k=I" I

a_

aZk

.

k

LEMMA 17.5. Suppose aD E C 2 and F E C(aD ); then lim Z+, Z

(NF (z+) - NF (z-)) = 0

uniformly in z °. If NF can be extended continuously to D, then NF can be continuously extended to CD, and conversely.

PROOF. Let F E C(aD). It is clear that the second part of the lemma follows from the first. Since NF = NF+ c for any constant C, we may assume F(z °) = 0. Observe that

df [k] A d IaD =(2i

)n(_I)k-I

a

, k

where do is the surface element of W. Therefore

z) Ian -

(n-i)!

f_kI'

I

,k_ Iznda

hence

NF (2+) - Ni (z-)

(n 0! J

F(J) ko)

D

+n n' aD f '

k21 a a k=1

J ak

Z+

I2"

I_z_I2n] J

Zk F+m=1

C10

(aplafm) (fm - Zm

I r . Z+2n+2

Ik_1

Zk )2m=1

(ap/am)(m

- Zm))

z-12n+2

Let us denote the first integral above by I,, and the second by I2. By performing a unitary transformation and a translation, we may take z° to

0, and the tangent plane to aD at z ° to the plane a = (w E C": Im w" = 01. Then aD will be defined in a neighborhood of the origin by the system of equations

'p='w, where

"=u"+h (w), ),'w=(wl,...,w,_1),and w=CM

be of class C2 in a neighborhood W of 0 in a, and z ± will be of the form (01000909 iyn ).

97

§ 17. HOLOMORPHY OF CONTINUOUS FUNCTIONS

aD is a Lyapunov surface with Holder exponent 1. Hence

w E W,

14'(w)I < cIw12,

a

_


_ =1

a

1 2 ...,n

±HcIwl, yj

29. ..n- 1 (17.3)

where zj = xj + iyj (see Vladimirov [5], §22.4; these estimates are derived there for R3, but the proofs easily carry over to R", n > 3). Hence ap

aP

aZk

c1 1wI,

w E W,

cilwl,

azk

(17.4)

for k = 1,. .. , n - 1, since

ap

_

az;

ap

ap

ap

aZ,

ayn'

ay"

wEW.


Here the constants are independent of the point being considered. Furthermore,

Ic31w1,

w E W.

Fix an e > 0. Choose a ball B in the plane a with center 0, and an a > 0, such that I) B C w,

EB,

2) 1

3) B X [-a, a] C U,

+cI wI2)6d
oandwEB.

, we have I (w) - z IZ = I w +(Y - 41(w))2. Hence 1

1

I -z12

I w-

1

-

(2iGYn

- lpZ)/I w - z

12'

but

(w-z12

cIwlZ(2lynl+clwlZ) < c(21yn lwl2+yn

l+C iwl2)< a<1.

Consequently, 1

1 - (2py-gv2)/w-z 1 2

(2

= k=b

=1+

2)k 1

- z 1ik

Iw-zI

Z

S 1(w Z)91

V. INTEGRAL PROPERTIES CHARACTERIZING 3-CLOSED FORMS

98

where gl(w, z) is bounded uniformly for z E B and I yn I- a. Therefore __

1

IZI2n

1

IW-ZI2n(

-

1

1+20yn-4'2/ gz) IW-ZI2

(17.6)

2¢yn-02

I

1+ Iw-z12

Iw-zI2n-2

(17.7)

93

and g2 and g3 are uniformly bounded for w E B and I yn I < a.

Let us estimate the integral I, over the portion of 8D corresponding to the ball, i.e. the integral I. By (17.6) and (17.3), we have

_

1

z I2n

I

n+

c 2 12 yn

1

Z+ I2n

I

w - z+ I

+clwl Igzl --< c4

n

(I Iw I2I+ y,

J

Further, do < d, dS, where dS is the surface element of the plane a; hence II BI

Ec5 fB (IWI2+Ynl

» ) IdS

I

ECs

(

I y.+ I dS

2

n

+c ('B

dS

(IWI2+'Y.2)

The first integral above is a Poisson integral for a half-space (see, for example, Stein [2], Chapter III, §2.1) and is independent of y,,+, while the second integral is continuous at 0. Hence I IB I s me, where m is independent of F and the point considered.

The integral 1, over the remaining portion of aD can be obviously made arbitrarily small as z -+ 0. It is easy to see that the form of 12 is unchanged by the unitary transformation and translation. Hence

I

k=l

aP p10)Zk

(f -Zm)

m=1

aZk

I (Jm-Zm) M=1 a.m Op

nm=1

aP m-

am

(

n-Zn)+j 3P (un+ 2+Yn -24yn) an

§ 17. HOLOMORPHY OF CONTINUOUS FUNCTIONS

99

The integral I2 over the part of aD corresponding to B can be split into three integrals

12 - - tin! " B :tin! n

B

X

(ap/a)2py, + In-1 M= I

F(e(w))

(aPlafmXmYn

da

I

F(e(w)) 2

n)(un + ,2 +

2pyn I + 7.m='1

yn

(aP/afmXm(L - Zn )

I w - z+ (2n+4 n

a

i+2

(-a

ds'

=

X

n-I

Z

X)

+

a

2

as,

and each of these can be estimated in the same way as J B' using (17.3)-(17.5) and (17.7). LEMMA 17.6. Let D C C" be a domain with C' boundary, and let F E C(aD ). 10

Set

Fdrg A d' [k, m] A d' aD

where

H,. (z),

z E D,

IH;(z),

z Qt D,

g(',z)=IzI2-2n. Then

(H,(z) -HF(z)) =0,

lim z+,

z °. If HF (z) can be extended continuously to D, then Hi (z) can be extended continuously to CD, and conversely (z :L- and z ° are as in Lemma 17.5).

PROOF. We essentially repeat the proof of Theorem 2.6. Since HF = HF+ c we may suppose F(z °) = 0. As in Lemma 17.5, we make a unitary transformation and translation so that aD may be defined in a neighborhood of the origin by the system of equations T

w,

Jn= Un+l (w)

(we preserve the notation of Lemma 17.5). In addition, 4(w) is of class C' in a neighborhood w of 0 in the plane a, and t(w) = Ofl w 1). Fix an F > 0 and choose a ball B in a with center 0 such that

1)BCw, 2) IFU(w))I<eforw E B,

3) 1 w - zI
100

V. INTEGRAL PROPERTIES CHARACTERIZING 5-CLOSED FORMS

Condition 3) is guaranteed by the relations

Iw - 0w)I = WWA = 0(I w 1),

Iw - z-I
Let B' = ( E 8D: ity

_ (w), w E B). Because of 3) above and the inequalI we have IZI2nI

Z+ 12n

Z+I-Iz II2n-1

-

I'kI IJ

I

Iw-z+I2n

I

(17.8)

Similarly, Zk

Zk

IS -Z+I2n

IZ-I2n

Iz+I

(17.9)

Iw-z+I2n

Finally,

da"d,dS.

(17.10)

Now condition 2) on B and (17.8)-(17.10) yield

f

d

z+) A df [k, m] A dg s ed2f

(Yn Ids

B /(Iw12+y, 2)

A d]]

< ed3.

Here d3 is independent of z° and F. Thus the first part of the lemma is proved, and the second part follows from the first, as in Lemma 17.5. PROOF OF COROLLARY 17.3. Suppose f as in Corollary 17.3, i.e. f - n 0. Then,

by the jump formula for integrals of Martinelli-Bochner type (Theorem 2.6), we see that f+ can be extended continuously to D and its restriction to aD coincides with f. Consider now a function p harmonic in some neighborhood of D. There exists a domain G with smooth boundary such that D C G and p is harmonic on G. Then, by Green's formula (16.3),

T(z) _

a-

[g(,z)W

(n ,aor

q G) agan' z) ]de,

z E G.

101

§ 17. HOLOMORPHY OF CONTINUOUS FUNCTIONS

Replacing the integral on the right by approximating Riemann sums, and the derivatives by difference quotients, we see that cp can be approximated uniformly in some neighborhood U(D) of D by linear combinations of fractions of the form zk), zk U(D), k = 1, 2,.... Since

Vo.o( z) = - ((21ri))'

k (-1)'-d [k] A

it follows from (17.2) that

f

f()

an

(_1)k_1d[k]Ad=O k

n

(».ii)

k=1

for all p harmonic in a neighborhood of D. Consider now a s e q u e n c e of domains Dm, m = 1, 2, ... , with smooth boundaries, such that D. C Dm+ 1, U 7 D. = D, and

a=

lim

m-. oo

aD,,,

a

(17.12)

aD

for all continuous (2n - 1)-forms a on D. By Stokes' formula I)n

f f a

D-

1)

k-1

Tdk [k] A d

k=1

fa

fpuf + m

A d. The integral on the left tends

where µf+ = 2k ,

to zero by virtue of (17.11) and (17.12), so that

lim f q?µf+ = 0

(17.13)

aDm

for all functions T harmonic in a neighborhood of D. We now want to show that (17.13) holds for all q) continuous on D. Using Lemma 17.5 we finish the proof of case a) of Corollary 17.3. Choose Dm = (z E C": p(z) < -1/m}. Then clearly (17.12) holds. It is easy to verify that Adf 13DM = (2.)fl(

1)n±k-t a WdGm

where dam is the surface element of aDrn. Hence JL f+ IaD, = (_2i)'1N7 darn.

Since Nf = 0, Nf can be extended continuously to D by Lemma 17.5, and vanishes on aD. Therefore (17.13) holds for all p in C(D). Choosing f + as ca,

102

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

we then get by Stokes' theorem

I

= (-2i )nf

f 1+ it anm

dj

BDm k=1

2

dV-*O

-

ask

as m -+ oo. Hence f+ is holomorphic in D. We have already shown that the restriction off`' to 8D coincides with f. We now consider case b) of Corollary 17.3. Let n = 2, and suppose 8(D) is

connected and of class C'. We have proved that if J E C(D) satisfies (17.2), then (17.11) holds for f. Consider the function

z,) + bG2 - Z2)

-

b(f, -,F,) - a(

Z 12

I

in D, =

b(f, z1) - a(fZ - i2) = 0). Let V be a ball containing D. Then, for each z E V, we can choose a and b such that This is harmonic in

V C D,. As is easily verified,

dg n d = 1

a'

a2

1

n

Thus, by (17.11),

ffdgAd=0,

zEV.

But since aD is connected, the above integral vanishes for all z E CD. We can now conclude the proof of Corollary 17.3 with the aid of Lemma 17.6. Indeed, we have shown that Hi 0; therefore, by Lemma 17.6, H7 can be extended continuously to D and vanishes on aD. Being harmonic in D, Hf therefore vanishes identically in D. Further, a U

z

=

1

4 a z2

a d 2 az2 g A

az,

dg

Thus

ZEn,;=i,z, i.e. f + is holomorphic in D.

PROOF OF COROLLARY 17.4. Since f ' is harmonic in CD (and CD is connected), we will have f(z) = 0 for all z E CD if we show that f(z) = 0

103

§17. HOLOMORPHY OF CONTINUOUS FUNCTIONS

outside a ball B containing D. For z a B,

z) is harmonic in B with respect

to , and hence can be represented as a series in the basis of polynomials converging uniformly in B. Hence the equality in Corollary 17.4 implies (17.2). Conversely, if f extends holomorphically into D as a function of class AC(D), then the above-mentioned equality follows from the fact that the integrands are a-closed forms.

To obtain another application of Corollary 17.3, we need the following notion. Suppose D is a bounded domain with aD E C', and let f E C(aD). We shall say that f is holomorphically continuable in dimension one if, for each complex line 1, corresponding to the restriction f, = f 13n11 there exists F, E C(D fl l ), holomorphic at interior (relative to 1) points of D fl 1, such that

f=F, onaDfll. COROLLARY 17.7. If D is bounded and aD E C2, and if f E C(aD) is holomorphically continuable in dimension one, then f is holomorphically continuable into D.

This is a consequence of Corollary 17.3 and the following lemma: LEMMA 17.8. If a function is holomorphically oontinuable in dimension one, then (17.2) holds for z E CD.

PROOF. Suppose z lies in the unbounded component of CD. By performing an orthogonal linear transformation and translation in C", we may ensure that z = 0 and that the plane {z: z1 = 0) does not meet D. This can be done for z far enough from D. Thus U0,0Q, z) becomes 0). We now make the change of variables = v;1, j = v1 v,' for j = 2,.... Then 1

vt2

0

0

V2

0 v2I

vn

do

I

0

v1

vt2

Also,

do.

0) is mapped biholomorphically onto C"\{v: v1 = 0).

Observe that n (-1)k-If

kd$

k=1

[k]

n

akdo[k] k=1

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

104

where

v;, k= 1, I

Ak= 10,

k> 1.

Hence v

n

[1]n + dv[1] v.121 = dvI A ACV). n + v /1 t + v2 12

Here, 'v = (v2, ... ,v). Under our change of variables, the lines (: i = t, 2 = c2t,..., = go ov r to the lines {v: v2 = c2,...,v = and the boundary of D to a certai cycle F. If f is holomorphically continuable in dimension one, then, after th change of variables, the new function f(v) will be continuable in dimension one along all the lines (v: v2 = C2,... ,v = Cn). Furthermore,

f

frf(v)

vv1

A X('v) =

f '(o)nrf (U) vvl

where 17 is the projection v - 'v. Note that, by Sard's theorem, the ¶r-'('v) fl 1' are smooth curves for almost all 'v E 7r(F). The origin lies outside the domain bounded by r; hence the inside integral above vanishes for almost all 'v, i.e.

(17.2) holds at points z sufficiently far from D. Since an integral of the Martinelli-Bochner type is harmonic, it follows that (17.2) holds on the unbounded component of CD. Suppose now that z belongs to a bounded component Q of CD, and suppose z = 0 (we can always achieve this by performing a translation). Consider a cone 11E with vertex 0, formed by complex lines through 0, where e denotes the area of the portion of 3Q lying inside He. By the same change of variables as

before, this cone goes over into the cylinder formed by the lines (v: v2 = c2, ... , v = And H, can be chosen so that: 1) as e --> 0, the area of 3(Q fl II,) fl 3D tends to 0, therefore

f o wU0.a,0) = li m

foa(Qnn,)f()U0.0(,0),

105

§17. HOLOMORPHY OF CONTINUOUS FUNCTIONS

and 2) aD \a(Q fl r1,) maps to a compact set I', under the change of variables. Then it can be shown, as before, that

ff(v) v re

AX'v

C

)

= 0.

1

Hence (17.2) also holds for z E Q.

Corollary 17.3 is closely related to the criterion for the holomorphy of integrals of Martinelli-Bochner type. THEOREM 17.9. Suppose n > 1, and D is bounded with aD connected and of

class C'; and let F E C1(0). If F+ is holomorphic in D, then F+ extends continuously to D and its restriction to aD coincides with F.

THEOREM 17.10. Suppose n > 1, and D is a bounded domain with aD connected and of class C 2. Let F E C(aD ). If F+ is holomorphic in D and extends continuously to D, then its restriction to aD coincides with F.

The content of the above two theorems is that, in contrast to the onedimensional case (integrals of Cauchy type), an integral of Martinelli-Bochner type is holomorphic in the case n > 1 if and only if it is a Martinelli-Bochner integral, i.e. F+ = F on aD (or equivalently, F 0). The condition that aD be connected cannot be dropped, as the example of the spherical shell D = (z: 1 < I z I < 2) and the function

_

1 1= 2,

1,

F E C'(aD), but F+ = 1, i.e. F+ does not coincide with F on aB(0,1) C 3D. For the proof of Theorems 17.9 and 17.10, we need the following assertion on the saltus of derivatives of an integral of Martinelli-Bochner type (Aronov [3l).

LEA 17.11. Let D be a bounded domain with 3D E Cl: D = {z E C": p(z) < 0}, where p E C'(C") and grade IaD =# O. If F E 0(8D), then Z

aF4- (z +

lim o

(

aF-(z-)

ask

azk

1

-

aF(Z°) aZk

I

aF(Z°)

Pk m=1

aZm

Pm+

uniformly in a°. Here z° E 8D, Z": lie on the normal to 8D at a°, z+ E D, a-(4D, I z+ -z°1=1 z--r° I ,and

Pk(z°) =

a a

ko)

Pk - Pk

106

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

PROOF OF THEOREM 17.9. We extend F to a function F of class C' in a neighborhood of D. For F, the Martinelli-Bochner formula for smooth functions (the special case p = q = 0 of (1.1)) is valid:

f

f

U0,0( z)

8D

f

A

D

z) (17.14)

z).

A

D

The second integral in (17.14) is the convolution of a bounded compactly supported measurable function with the locally integrable functions I fk - Zk I - Z I-In , and hence is continuous in C". Consequently the first integral, i.e. F+ , extends continuously to D. Now F+ is holomorphic in D, and therefore I

aF+/aik = 0; hence it follows from Lemma 17.11 that the aF-/aik extend continuously to CD and their restriction to aD has the form aF -

aF

aZk

aZk

aF

" m=1

(17.15)

1,2,...,n.

k

pm,

Zm

Multiplying (17.15) by Pk and summing over k, we then get "

k=1

aF -

- Pk =

aZk

"

2

aF

k=1 aZ k

" " aF Y Pk Pk I al_ pm m

Pk

k=1

m=1

since n

2 PkPk = 1. k=1

Next, consider the form n

µF-

()k -I ai- dz[k] n aF.

k=1

k

Recall that

dz[k] A Lil 13D =(2i

)n(_I)n+k-

lPkda

where da is the surface element of D. Hence

AF-6

-

)n y

V

da = 0.

k=1

By Stokes' formula, we deduce that OF-

0 = J DF N'F-= k=

azk

2

dv;

Q,

§ 18. TRACES ON THE SHILOV BOUNDARY

107

this is permissible, since F - = 0(1 z 11-2" ), and aF -/aZk = 0(l z I2-2n ) as zI

oo. (17.16) means that F - is holomorphic outside D. Since aD is

connected, Hartogs' theorem now shows that F- extends holomorphically into D. However, F - - 0 as I z I--* oo ; hence F - = 0. By Theorem 2.6, F+ extends continuously to D and F+ 18D = F. PROOF OF THEOREM 17.10. Let z o and z ± be as in Lemma 17.11. Apply Lemma 17.5. In our case NF = 0, and so NF extends continuously to CD and Ni 6 = 0. The proof proceeds from this point exactly in the same way as that of Theorem 17.9. Note that neither of the Theorems 17.9 or 17.10 contains the other. They can

be regarded as results concerning the possibility of extending F from aD holomorphically into D. §18. The traces of holomorphic functions on the Shilov boundary of a circular domain

Let D be a bounded strongly starlike circular domain in C", i.e. a domain invariant under all rotations z -- a Fez, 0 < 0 < 217, such that XD = (Az: z e D) C D for any A, 0 < A < 1. Let S denote the Shilov boundary of D, i.e. the smallest closed set lying in D and such that

sup{If(z)I : z ED} = sup{If(z)I : z E S} for all f EAC(D). If D is strictly pseudoconvex, then S = aD; in general S is a subset of aD. Before proceeding to the description of the traces on S of functions from A C(D ), we give a characterization of polynomials orthogonal to A C(D ). Let it

be a nonnegative measure on S, invariant under the rotations z -+ eiez, 0 < 0 < 27r, and massive on S, i.e. S \ E = S for every E C S with µ(E) = 0. Denote by [A C(D)] the space of functions 9p E C(S) such that f S cp fdµ = 0 for all f E A C(D). If D is the unit ball and µ = w(i, z), then [A D)]µ = O(D). Let P 2(µ) be the space of polynomials in z,,.- , ", with the inner product

(f,g) = ffd&. If the multi-indices I and J are such that 11111 µ under rotations we have (Z19 Zj) =

II J II, then by the invariance of

f

Z1i.rdµ = s

= e i801 111 -

Of 11111),

(ZI

)

Zj

).

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

108

Hence (z', z') = 0; thus, if f and g are any homogeneous polynomials of different degrees, then

(f,g)=0.

(18.1)

If follows that we can choose a complete orthonormal system of polynomials ((pl } in P2(µ), where I is the degree of the homogeneous polynomial, and k its index among those of degree 1. For multi-indices I and J with 11111 - II J II, we

put P1.J(z) = :Ea1JkcIIJ11-11111(=), k

where a1Jk - SZ Z q J11-11111 des.

LEMMA 18.1. A polynomial P(z, i) lies in [AC(D)]µ if and only if it is a linear combination of polynomials of the form

z'z',

11111 > IIJII,

z'Z'-P1J(z),

IIIII - IIJII.

(18.2)

(18.3)

PROOF. Sufficiency. D is strongly star-like, so functions from AC(D) can be approximated uniformly on D by polynomials. Since (q4) is complete, we need only prove that, for a polynomial P(z, Z) of the form (18.2) or (18.3) and for all k and 1,

f P(z, Z)Tldµ = 0.

(18.4)

S

If P(z, i) is of the form ( 8.2), then by (18.1)

f P(z, z)cpjd, = (z'q),, z') = 0, S

since z'(pk is a homogeneous polynomial of degree 11 1 II + 1 > 11 J II. Next, if P(z, Z) is of the form (18.3) and I II J II - 11111, we have (using (18.1) again) S f/(Z, Z)q d.& = (alai, z

J) - ((pi, PPJ(Z)) = 0.

Finally, if I = II J II - 11111, then (18.4) holds by the choice of the alik. belong to [AC(D)]µ ; then Necessity. Let P(z, Z) = El,,

b1Jz'z' +

P(z, Z)

11th. HJ I

+

I

11111--11!11

§18. TRACES ON THE SHILOV BOUNDARY

109

i.e. P(z, Z-) = P1(z, i) + P(z), where P1 is a linear combination of polynomials

of the form (18.2) or (18.3), and P(z) is a holomorphic polynomial. Since P(z, i) E [ A C(D )] µ by assumption, and P1(z, i) E [ A C(D )] µ by what we have already proved, we see that P (z) E [A C(D )] µ . Hence

f1P(z)I2c1µ = js P(z)' P(z)dµ = 0. Since u is a nonnegative and massive on S, it follows that P(z) = 0 on S, and hence that P(z) = 0 on D (S is the Shilov boundary). Thus P(z, i) = P1(z, z ). Q.E.D. If D is n-circular, i.e. invariant under all n-rotations z -> (eie,z1, ... , eienz ), 0 < 01, ... , 0 < 21r, and if u is also invariant under n-rotations, then as a

complete orthonormal system we an choose {ajz'}, consisting of monomials

with al = (fSZJiJdµ)-'

.

In this case Lemma 18.1 becomes

LEMMA 18.2. The polynomial P(z, z) lies in [AC(D)Iµ if and only if it is a linear combination of polynomials of the form

z,ij,

with 'k > jk jor some k, 2

ZIZI+J

al+J

(18.5)

ek1z

where ek is the multi-index with 1 at the kth place an 0 elsewhere.

Lemmas 18.1 and 18.2 are analogues of Lemma 5.7. When D is the unit ball, w = I and ,4 = w (z, z ), Lemma 18.2 and Lemma 5.7 are identical.

For the proof of the main result, we need the Poisson kernel. First we construct the Szegd kernel h (r, z) of the circular domain D by the formula

h(f, z) =

I q)k(Z) (Pk

k,!

z Ei D, E

I

It can be shown that: 1) h (f, z) is holomorphic in z on D and continuous in 2) h (f, z) = h (z , ) on D X D; and

on D;

3) forallfEAC(D),

f(Z) = ff()h(, z)

dµt.

The Poisson kernel can be constructed from the Szegb kernel:

z)

Ih(z)I2 h(i, z)

(18.6)

110

V. INTEGRAL PROPERTIES CHARACTERIZING 2-CLOSED FORMS

For z E D we have h(a, z) > 0, since (18.6) and 2) above imply

h(z, z) = fh(z, )h(f, z)dµt = fs Ih(, z)12

dµ > 0.

Further, if f E AC(D), then

fez)

h(z

z)f(z)h(z>

z) - h(i, z) f f(flh(i, )h(, z)dµt

z)dµt

=

(18.7)

THEOREM 18.3. Let D be a bounded strongly star-like circular domain in C"

such that every point of its Shilov boundary S is a peak point for the algebra AC(D), and let µ be a nonnegative massive measure on S invariant under the rotations z -> e'BZ, 0 < 0 < 2r. Then a function qD E C(S) is of the form f IS for some f E AC(D) if and only if

=0

(18.8)

fSTPdj%

for every polynomial P of the form (18.2) or (18.3).

If D is n-circular and u is invariant under all

the n-rotations z -30 (eia, z,, ... , e`0nzn ), 0 < 0s, ... , On < 21T, then we may replace (18.2) and (18.3) above by (18.5). PROOF. The necessity follows from Lemma 18.1.

Sufficiency. Since each point of S is a peak-point for AC(D ), we have for each z 0 E S and any neighborhood W of z 0

P(', z)dµq = 0.

lim z-'z zED

(18.9)

Sew

The argument for this can be found in Koran [1]. Further, since P(', z) > 0 and

fP(',z)d,.i.-l,

z ED,

S

it follows from (18.9) (see, for example, Hoffman [11, p. 18) that if p E C(S), then the function

Z fp()P(, z)dµr,

z E D,

§19. COMPUTATION OF MARTINELLI-BOCHNER INTEGRAL

111

is continuous on D U S. Since S is the Shilov boundary of D, {f(rz)) therefore

converges uniformly on D as r -4 1; hence f extends continuously to D. It remains to show that, if p satisfies (18.8), then f E A(D). From (18.7), we see

that, for any g E A C(D) and any differential operator T with constant coefficients,

fg(')7(P(, z))dµ = 0. It follows in particular that

z)) L=o E[AC(b)] From the form of the Poisson and Szego kernels, we see that the function on the left above is a polynomial in and . By Lemma 18.1, it is a linear

combination of polynomials of the form (18.2) and (18.3). This and the hypothesis of the theorem imply that

f (pa) [TzmPal z))] 1z=0 dji = 0, i.e. all the coefficients of the form

af(z)

f

z)dp

(18.10)

and all their derivatives vanish at z = 0. Since the Poisson kernel is realanalytic in z, the form (18.10) has real-analytic coefficients; hence 5f = 0 on D, i.e. f E A(D). §19. Computation of an integral of Martinell i-Bochner type for the case of the ball

We consider on the unit sphere aB(0,1) the space L2(aB(0,1)) with the usual inner product

(f, g) =

aB(o, 1)

fds,

where do is the surface element of the sphere. We shall identify this space with

the space G 2(B(0,1)) of harmonic functions f in B(0,1) represented by a Poisson integral from L2(aB(0,1)). Let 6Y,,t be the finite-dimensional vector space consisting of homogeneous harmonic polynomials of the form Ps,t . aIJZ J-J II111=s IIJII=t

112

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

is dense in L2(aB(0, 1)). As we know, (17.1) is valid The union of all the for f E G2(B(0, i)) n A(B(0,1)):

z e B(0, 1).

z) =f(z),

(Mf )(z) = J 85(0.1)

LEMMA 19.1 Let P,,, E 9,,t. Then the Martinelli-Bochner type integral of PS r is of the form MPS,t

n+s-1

_

n+s+t-

This follows from the fact that the harmonic extension of k P,,, from aB(O, 1) into B(0, 1) is given by 1

a

- I fl2

kPst+n+s+t-1 afkPs,t, and from the easily verified formula U0,0G, Z) IaB(0,1) = 1

here

I - IZ

z)da;

12

z) is the Poisson kernel for the ball: 1p-IZ 2

PQ,Z)_(n - 1)! 'J -ZI2n COROLLARY 19.2. If

Qs t =

2

2

aIJZIZJ

II711=s IIJ11=t then

MQ5,, = I R.,P, t-P P.o

where the R,._P. t-P are the harmonic polynomials defined by

_ R' P11-17

s+p+n-1

p!(s+t+n-p- 1)!

X'f

(_i)i(s+t-j-2 p+n-2)!IzI2jAj

A being the Laplace operator:

DR=

n

a2R

i=1 aziazi

.

PQs,t'

§19. COMPUTATION OF MARTINELI-BOCHNER INTEGRAL

113

This corollary follows from Lemma 19.1 and the Gauss representation of an

arbitrary polynomial on the sphere as a sum of homogeneous harmonic polynomials (see, for example, Sobolev [IJ, Chapter 13). We introduce the differential operators b and b by setting

bz-, 6=±ik n

n

a

k=I kaZk

a

azk

k=1

THEOREM 19.3. Suppose f E G2(B(0, I)). Then the Martinelli-Bochner type integral off has the form [n

(19.1)

kYk, k=1

where

II - IZ'I-n

IJ'n-2J

fo

(J )dI i,

a

ICI"-I

k

Iz) nJkl

'Yk

= 1,...,n.

a

PROOF. Since f E G2(B(0, I)), in the interior of B(0,1) an expansion f(0 = 2 P"'(0,

(19.2)

s, t>0

converging uniformly on compact subsets of B(0,1), in particular on each sphere aB(O, r), 0 < r < 1. From the definition of the class G2(B(0, 1)), it is not hard to show that (19.2) is also valid on aB(0,1), but in the sense of convergence in L2(aB(0,1)). By Lemma 19.1, we deduce that, in the interior of B(0, 1),

n+s-1

Mf _

t O n+s+tI P

S

lPs`-f-

o n+s+t -

S

To prove the first part of (19.1), we observe that IJIn-2+t+sPs

=IZI1-n

G

Y (Z)

s, two o

I

Iz I t+S S

B¢

)

f'ZI

o

_

n+s+t I

S

O

z

I

__

1PS'`(Izj) n

l 1

FU n+s+t- IP S,

apSt

n+s+t - 1 k=IZk aik

t

S1 0

n+s+t-

1P"t'

114

V. INTEGRAL PROPERTIES CHARACTERIZING 5-CLOSED FORMS

The proof of the second part of (19.1) is similar. The above theorem enables us to compute Mf for any f E G2(B(0,1)). The functions V/ and 'Pk are harmonic in B(O, 1). To compute them, it is convenient

to change over to polar coordinates. If f is polyharmonic or real-analytic in B(O, 1), then it may be convenient to compute Mf by using the Almani representation of f (see Sobolev [1], Chapter 14): {_ 2k

=

fkIZI

>

with the fk harmonic in B(O,1). We present some examples to illustrate the applications of Theorem 19.3. EXAMPLE 1. Let f E G2(B(O, 1)), and

Tl

_ (n - 2)! f (2ir)n

dat

z E B(0,1) ,

- Z I2n-2

1)

3B(0

I

the potential of a simple layer. Then T f=IZI1-n

lz' pIJln-2f(J)dIJI. y

r

J0

In particular, if Tf = F, then f = (n - 1)F+ bF + bF. Consider the integral equation

q) E G2(B(0,1))

f + ATf = (p,

(19.3)

on aB(0,1). Then its solution is of the form

f= , -

AIZI1-n-A

IAI < n - 1.

f1=I

(19.4)

EXAMPLE 2. Let f E L2(aB(O, l)), and

Sf = 2f

z),

z E aB(O, 1),

3B(0,1)

be the singular integral of the Martinelli-Bochner type. Consider the singular equation

f+ XSf= q,

cp E L2(aB(0,1)).

(19.5)

If T has an expansion T = 25-0 PpS.gS in B(0,1), where p and q are nonnegative integers, then

f=

p+q (1 - X)q + (I + A)p

2Aq(P + q)(n - 1) [(1 - X)q + (1 + A)p]2

IzI-" 0

(19.6) IAI

< 1,

- (n - 1)(1 + A)(P + q) (1 - X)q + (1 + X)p

§19. COMPUTATION OF MARTINELLI-BOCHNER INTEGRAL

115

If we know only the values of p on aB(0,1), we must replace p in (19.4) and (19.6) by its Poisson integral. Then we obtain integral representations for the solutions of (19.3) and (19.5), i.e. resolvents of the operators T and S. To compute integrals similar to Mf in more general domains than the ball, we may use the ideas of §5. We will have to require that (5.8) holds, or that the polynomials P(z, w) are dense in L2(aD). We indicate other applications of Lemma 19.1. The following theorem is an immediate consequence of the lemma.

THEOREM 19.4. For n > I the operator M maps L2(aB(O, I)) into itself, is bounded, and has norm 1. All positive rational numbers in [0, 1] are eigenvalues of infinite multiplicity for M. The spectrum of M is [0, 1].

Denote by PrH the orthogonal projection of G2(B(0,1)) into the Hardy space H2(B(0,1)) consisting of elements of A(B(O,1)) satisfying the condition that JaB(O, r) If I2 do is uniformly bounded for 0 < r < 1. This operator is called Szego's operator, and has the integral representation (-(1)k-1(JkdS

(PrHf)(Z)

f

kB(a 1)f(J)

[k] A d \f, z))n

l

For n = 1, integrals of Martinelli-Bochner type are integrals of Cauchy type, and M = PrH. For n > 1, the kernel U00 0 is not holomorphic, and M and PrH are not the same, as the following example shows. EXAMPLE 3. Let f = ak. Then f' is not holomorphic in B(0,1). Indeed,

aakf+(Z)

(n - 1)! (21ri )n

1 m-1

n

L,

1

(- )

I

Zm)(Jk - Zk) dy rm A d J J ` ]

p __ z I2n+2

(^1)k-1

I

IZ

I2n

df [k] A dr

Since

- (n - I)l

(p

U0,0(r, Z) I3B(0,1) =

24rn

[[n ZI2n

do'

m=1

we have a

f

(z-0 aZk +

(n

- 1)1 21rn

and the integral vanishes only if n = 1.

(n

- I) ° 8B(0,1)

I

kI2 da,

.

116

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

However, the following theorem is true: THEOREM 19.5. For every f E L2(aB(0,1)),

lim Mkf = PrH f,

(19.7)

k -oo

where Mk denotes the kth iterate of M.

PROOF. For each homogeneous harmonic Polynomial P,,t, (19.7) follows immediately from Lemma 19.1, since PrH Ps,t =

fP,1,

if t=

to,

if t > p,

But such polynomials are dense in G 2(B(0,1)), and the family of iterates of

M is uniformly bounded in norm; hence (see Dunford and Schwartz [1], Chapter II, §3, Theorem 6) (19.7) holds on all of L2(B(0,1)). §20. Differential boundary conditions for the holomorphy of functions

Let D = (z E C": p(z) < 0) be a bounded domain with aD E C'. We consider the following problem: suppose we are given a function f E C'(D) harmonic in D, and a continuous vector field w = (w, (z),. . . , wn(z )) on aD, with

(w,gradp)=

W 8z

'

Z E an,

(20.1)

i.e. the vector w never lies in the complex tangent plane TTD(z) at z to aD for any z E aD. Then does the condition " k = =1

al = 0, -wk azk

z E aD,

(20.2)

imply that f is holomorphic in D? This problem is an analogue of the oblique derivative problem for real-valued harmonic functions. If we do not require (20.1), then there does exist an f e A(D) satisfying (20.2), e.g. D = B(0,1) C

C2,w=(0,1)andf=z,. When w = grad p, (20.2) becomes the Neumann normal a-condition for functions harmonic in D. and the answer to our question is affirmative: THEOREM 20.1. Let D be a bounded domain in C", and let f E C(D) be harmonic in D. Then f E A(D) if and only if the a f/azk, k = 1,. .. , n, can be extended continuously to D and satisfy the Neumann normal a-condition n

k =1

af

azk

ap

azk

-

01

z E aD.

(20.3)

117

§20. DIFFERENTIAL BOUNDARY CONDITIONS FOR HOLOMORPHY

PROOF. If f e A(D), then all a f/aik = 0, and (20.3) is trivial. Conversely, if 8 fl azk E C(D), k = 1, ... , n, and (20.3), holds, then n

of

k= , azk

ap

azk

where do is the surface element of aD. As in the proof of Theorem 17.3, we now conclude easily that f E A(D ).

Let us reformulate our problem differently. Decompose w into its normal and tangential components to aD at a point z E aD:

w(z) = a(z)grad p+ T(z); here a never vanishes on aD, and T(z) is orthogonal to grad p,

i.e.

(grad p, T)=0 on 8D. Then (20.2) becomes

of aP

of , ?k(z) a(Z)

k=I azk

k_, ask aZk Consider again the form n

kI of d; [ k ] A dz-.

k=1

Using (20.4), we can write it on 8D as

µI JaD = I ak,m(z)djn dz[k, m] A dz IaD,

(20.5)

k>m

where the ak,m E C(8D). Then the new formulation of our problem is: suppose f E C'(D) is harmonic in D, and (20.5) holds; does it follow that f is holomorphic in D? (20.5) reduces to the Neumann normal 8-condition when all the ak,M are zero. Condition (20.5) can also be written in integral form:

AZ) = fan with

z) + (-1)n 2 d(gak.m) A

[k, m] n dal = 0

k>m

z) = -(n - 2)!(21ri)-" I

z 12-Z". It is then clear that the problem

of the holomorphy of the functions defined by a Martinelli-Bochner integral (see Theorems 17.1 and 17.2) is a special case of the present problem (the case

[1km=4 We now present the solution to our problem for certain classes of domains D and functions THEOREM 20.2. If n = 2, and 8D is connected and of class C', and a harmonic function f E C'(D) satisfies the condition

µJ1an=adfA where a is a constant, then f E A(D).

(20.6)

118

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

PROOF. If we show that, for each Ps,1 E 6Ys,t (see § 19 for the notation),

S,t=0,1,2,...,

f

2

an

I ) k- I OPS,t d

k=1

[k] A d = 01

(20.7)

holds, then our assertion will follow from Corollary 17.4. By (20.6) and Stokes' theorem, Z

(-l)'d[k]Ad=fakDPs,1µf k

=d= '3D'"

,Ad. aD

If t = 0, i.e. Ps,, is a holomorphic polynomial, (20.7) holds. Let t > 0. We transform dPs,t A d' into the form

nd _

aPd-

,

aP

Ad

.

This can be achieved by setting

- 4 dr

P

ar,

2

(formal integration with respect to 2 ). Then a2Ps,t

aP

a2Ps,t

=

d _ aps,t 2

aft

so that AP =0. Also, P E 33+,,t_ 1 Thus 2

aD

f(U)

(_l)k_td[k]Ad

k=1

4k

2

an

k-(_l)hff.d

k=1

Now we apply induction on t, using the fact that (20.7) holds for t = 0, and conclude from (20.8) that (20.7) is valid for all t. THEOREM 20.3. Let D = B(0,1) and suppose that the lie in AC(D). Then, if f E CA(D) is harmonic in D and satisfies (20.5), it follows that f E A(D). PROOF. Consider first the case µ113a(0,1> =

where Qi.o E 'Yio

n df [k,m] A d IaB(o,1)'

(20.9)

§20. DIFFERENTIAL BOUNDARY CONDITIONS FOR HOLOMORPHY

119

Writing out (20.9) for the case of the ball, we have n

2A Jj - ±Q1,0(J)

E M(0, 1). (20.10)

kJ,

kkJm

a

J

We shall look for the solution of (20.10) in the form of a series

f = I Ps,r+ s,ta0

where Ps.t E s,t Then (20.10) yields (20.11)

s.ra0

3m

aJk

0

We now extend the function on the right above harmonically into B(0,1) using Gauss' formula (see, for example, Sobolev [1], Chapter 11): if Pk is any homogeneous polynomial of degree k, then its harmonic extension Pr from M(0, 1) into B(0,1) is given by

Pr = 2 Zk-2s where

k-2s+n-1

Zk-2s

+

s1)!JG

j(k-j-2s+n-2)! j! z 2j 1+sPk.

(-1)

(20.12) In our case

StaP" -k

SmaP

Pk = Q1,0

a

p

a ym

J

k

and k = I + s + t. Hence (20.11) leads to the system of equations yyt2j

tPt r =

aPs-ry-l,r+, Q1,0

+ 2 bjm2jA1+, Qr o (

j

l

2

t', -

_Ps-lp-l,t+I

aJk

aJm

aPs-1,r+2

aPs-r,r+2

ask

aim

aPs-1,tl±l+1

Q10(

m_

k1

kl l+

aPs

aim

+ .. .

k

(20.13)

where the a3, 1,,...,cj are constants. For s = 0, we have tPO,r = 0 for any t by (20.13); hence Po,, = 0 for t 0. Next we have tP,,r = 0,...,tPs.r = 0, so that the series for f consists only of holomorphic summands, i.e. f is holomorphic in B(0,1).

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

12 0

If the ak,m are arbitrary holomorphic functions, we can decompose them into sums of homogeneous polynomials and apply the earlier considerations. Then we get a system of equations of the type of (20.13), with each equation involving only finitely many summands. In the case of B(0, r), the Neumann normal a-condition has the form _ of In zka_ =0 zk

on aB (O, r).

k= 1

Suppose now that this condition is satisfied everywhere in B(0, r). Then it turns out that the requirement that f be harmonic can be dropped, and f will still be holomorphic in B(0, r) as before: THEOREM 20.4. Let f E C'(B(O, r)) fl C°°(0), and suppose that 91

ik

of aZ-k

=0

z EB (O, r )

(20.14)

k

Then f is holomorphic in B(0, r). If f E C2 (B(, r)) fl C°°(0) and a 2f

n ZkZm azkazm

09

z E B(0, r),

(20.15)

k, m=,

then f is pluriharmonic in B(0, r). PROOF. Observe that (20.14) implies that f is holomorphic on each complex line I through 0, and (20.15) implies that f is harmonic on each such line. Thus we need only appeal to a result of Forelli [1], which asserts that an f which is harmonic on each complex line 1 through 0 and C°° at 0 is pluriharmonic in B(0, r).

Rudin [1] has shown that, if a function f in B(0, r) satisfies the Laplace equation A f = 0 and the Laplace-Beltrami equation d' f = 0, where a2

n

A

A k I

m=

ZkZmaZkaZm -

then f is pluriharmonic in B(0, r). Theorem 20.4 shows that, if f E C2 (B) ft C°°(0), then the single equation (20.15) is sufficient. The single condition A,f = 0 is not sufficient. For example, let

f-

(IzI2

Iz2I)

+11)I IZlzk.

ko (n'+ k

(20.16)

Then f is real-analytic in B(0,1), since the radius of convergence of (20.16) is 1. It is easy to verify that A' f = 0, but f is not pluriharmonic, since

(IZ112 _Iz212) ; (j+ 1)! (j + l)(n + 1)! k.m

ZkZm

aZ a

(n +j +

IZI2j.

§20. DIFFERENTIAL BOUNDARY CONDITIONS FOR HOLOMORPHY

121

If f E Ck(B(0, r)), Theorem 20.4 is not true; e.g. let

- zj=1 I zj

f(Z) Since

I z j I4


Izi

I4

4

we see that, by choosing a multi-index I with

11 I 11

sufficiently large, we can have f E Ck(B(0, r)) for any prescribed k. Clearly f is not holomorphic in B(0, r), although 7; zka f/aik = 0. Another theorem related to the above theorems by the method of proof is the following:

THEOREM 20.5. Let Pk be an arbitrary homogeneous polynomial of degree k > 2. Then the restriction of Pk to 8B(0, r) can be extended holomorphically into B(0, r) if and only if ek Pk = 0 on 0B(0, r), where

(n+k-2l--3)!(2l-1)!!2,1 (n+k-2)!(l+1)! !0

n

k

2 Zj j_1

j

PROOF. Let Pk = Pm, t, i.e. a1JZIZ-J .

Pk 11111=m II.JII=t

We extend Pk from aB(0, r) into B(0, r) by Gauss' formula; the extension, call it Pr, can be written as Pr = 7sz0 Zk _ 2s, where Zk _ 2s is defined by (20.12). P. is holomorphic if and only if

apr

n

2 zj azj_ = 0

j=1

in B(0, r) (see Theorem 20.4). But n

j= 1

ij aZaZ-2s - (t - S)Zk-2s j k

Then

2 j=1

i 8Pr a1i

1)(k-j-2s+n-2)!A+sP

I

S!j! (k + n

j, S

s

AP(-1)(k-2s+n-1)(k-l-s+n-2)!(t-s) !>0

s:kt - s): kk -r n - s - i):

j+s=1

(-1)5(k-2s+n-1)(k-l--s+n-2)!(t-s)

(_')/Q!Pk I;;,__0

s=0

s:kt - sJ:kK -rn - s - iJ:

The following lemma now concludes the proof of the theorem.

k

12 2

V. INTEGRAL PROPERTIES CHARACTERIZING a-CLOSED FORMS

LEMMA 20.6. If k > 21, then

5=o

(-1)3(k---2s+n- 1)(k-l-s+n-2)! s!(1-s)!(k+n-s- 1)!

JO,

ifl > 0,

ii, ifl=0,

and

(-1)3(k - 2s + n - 1)(k-l-s+n-2)!s

Sl.(I-s)!(k+n-s- 1)!

(-1)'(n+k- 21 - 1)! (21 - 2)! (n + k - 2)!1! (1 - 1)!

ifl=0. PROOF. Let us compute for instance the first sum, which we denote by S. We have

S_ (-1)I (k-21+n- 1)(k-21+n-2)! 1! (k + n - I - 1)!

+(-1)1-1(k-2I+n+1)(k-21+n-1)!+ (_1)'(k-21+n-1)![' k-21+n+ Il 1+ k+n-1 + (1- 1)! (k+n-l- 1. 1

(-l)'-'(k-21+n-1)!(1_ 1)(k+n-21)+ 1! (k + n - 1)! 1)'-'(k - 21 + n)! (I - 1) 1! (k + n - 1)!

+

(_1)1+2

(k - 21 + n + 3)(k - 21 + n)! (I - 2)!2!(k + n - 1 + 1)!

+...

_ (-1)'2(k-2l+n+ 1)!(1- 2) + .. . 2!1(1- 2)!(k + n - 1 + 1)!

(k-l+n-2)! (1- 1)!l(k + n - 2)!

+

(k+n- 1)(k-l+n-2)! 1!(k + n - 1)!

The second sum can be computed in the same way.

§20. DIFFERENTIAL BOUNDARY CONDITIONS FOR HOLOMORPHY

123

COROLLARY 20.7. Let P = 20 Pk be an arbitrary polynomial, where the Pk are homogeneous polynomials of degree k. Then P can be extended holomorphically from aB(0, r) into B(0, r) if and only if

k=0

ek pk IaB(0, r)

0.

We note that the condition EkPk = O on 8B(0, r) may be replaced by the condition ek Pk = 0 in B, where

_ k

n

_

a

2 Zi a

j=1

-

2 1;190

(n+k-21-3)!(2l- 1)!!21 1

21+2 !+ IZI

CHAPTER VI

FORMS ORTHOGONAL TO HOLOMORPHIC FORMS. WEIGHTED FORMULA FOR SOLVING THE a-EQUATION, AND APPLICATIONS §21. Forms orthogonal to holomorphic forms

In this section, we sharpen the results of §6. First some ancillary results: LEMMA 21.1. Let D be a bounded strictly pseudoconvex domain in C". Then there exist positive constants r, 8 and c such that

I) For any

with I p(') I< r in

8), we can perform a smooth change of

variables T = T(Z)with the properties T2 = Im z); 1) T1 = P(z) -

3)1/c >I aT/aZ I > c; z E

8). II) For any z such that I p(z) I < r in B(z, 8). we can perform a smooth change of variables T = r() with the properties 1) T1 = p(J) P(Z); T2 = Im `W, Z);

EB(z,6); 3) 1 /C > 18 T/af I> C;

E B(z, 8).

z) was introduced in §4.2'. The function PROOF. We need one more property of Khenkin's barrier function not noted in §4.2°: there exist r > 0 and a nowhere-vanishing continuous function h on (z: I p(z) I < r) such that, for z e Vwith I p(z) I < r, d14 (', z) lz=r = d3' I

z) I= hdp

(see ovrelid [1]). From this we conclude that the maps

Ti = p(z), T2 = Im'DG, Z),

and 125'

Ti = pM =

lr2Im(,z),

126

VI. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

and grad p(z)

have rank 2 at points where z =

0. The lemma follows

easily from this. LEMMA 21.2. Let

(T1 + e)adT, "' dT2n 1a,k,f(E) -

ITI
I +IT2I +E+ ITI2)kITI!'

( I T,

where a, k and/ I are real and a > -1. Then a) Ia,k,2n-I(e) = O(1) E1-k+a; k - a > 1; b) 1a,k,2n-3-2m(E) = O(l)e-k+a+m+5/2;

k-a-m

> 5/2; m > 0.

PROOF. In the integral defining 1.,k,11 we introduce spherical coordinates

and perform the change of variables

T, = r cos q) 1, T2 = r sin qp I cos (p2, ... ,

sj = cos q)j, j = 1, 2. Then

(rs1 + E)ar2n-1-1drdslds2

1a,k.I(E) = 0(1)

l HI sI I +s2(1 - sI )1/2) + e + r2,

°<

IS1I < I

k

rs, +E>O

We use the inequality I S1 I +s2(1 - s1)I/2 4(i s1 I +s2) for I s1 I< I and 0 < s2 < 1. Settings, = su and s2 = s(1 - I u I), we have

(rsu + e)ar2n-'-(sdrdsdu 0
(rs + e + r2)k

lul
rsu+E>O

For I = n - 1, we go over to the coordinates (v, s, u), where v = e-Irs. Since

rs+e+r2>rs+E, we get 1 .k.2n-1(E) = 0(1)

El-k+a

I

J0

dv

f (vu + 1) adu + j 4- 1

dv

1

k

(v + I) J-l

1

(v + 1)

k

j

i

(VU + 1)adu

.

Assertion a) follows. In the case b), we have 22m

1a.k,2n-3-2m(E) = O(1) j r dr j

rdS

2

(rs + e s = O(1) .0(2r2mdr (2

.

rds (rs+e+r2)k-a-I

0

=

O(1)I2

(re + r

r2)k I

I

a

(rsu + e) rsdu

£(rs)

= 0(1)

J0

dr )2(k-a-m-2)

Assertion b) now follows by direct integration.

r2mdr

/'2

(e +

r2)k+a-2

§21. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

127

LEmmn 21.3. Let D C R" be a bounded domain with smooth boundary and let f E C '(D). S u p p o s e I grad f(z) (= O(1) I P(z) I' Co,aGD).

for 0 < X < 1. Then f E

This assertion is a slight generalization of a result of Hardy and Littlewood (see, for example, Goluzin [1], Chapter IX, §5, Theorem 3).

We now improve the estimate for the solution of the a-problem given by Theorem 4.4. THEOREM 21.4. Let D be a bounded strictly pseudoconvex domain in C" with boundary of class C m+ 2, and let y E Z p; 9)(D ), m > 0, A > 0. Then there exists

a form a satisfying as =yin D, with a E C(p,q± lj 2(D ), C"'+1,a-1/2(D)

a E (p9-1)

0 < X ' < J1, if 0 < 71

2;

2

1.

(21.1) '

2

PROOF. We proceed as in Theorem 4.4. We need only show that the form

A u'q-, A Dp,nr-p(aZ, 3 ) (V\D),

satisfies (21.1). Lemmas 1.4 and 4.1 show that it is enough to verify that (21.1) is true for the function

H(Z) _ f (V\D)r

4Dr+

Z 1-r (,z)I-zl(" Z) I

1>d n

(21.2)

C'"-1,11(C") and h = 0 on D. If m = 0, we may suppose that with h E h E C(V\ D) and I for this, we have to extend y to I= 0(1) all of C" by means of Theorem 3.2 and Supplement 3.3 of Malgrange [11,

Chapter 1. For arbitrary m, by working out the argument in Lemma 4.2 more carefully, we see that 0(1) I Im1 . We shall carry out the rest of the proof in the case X > Z. The case A < Z can be handled similarly.

Differentiating under the integral sign and using the argument in Lemma 4.2, we conclude that, for z E D, 13jIJ

0(1) f V\D)s

1

(

Z)

CItJZ

12n-3+j'

Choose r and S according to Lemma 21.1 and split the integrals over V \ D

into integrals over (V\D)\B(z, S) and (V\D) Q B(z, S). The first integrals

128

VI. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

can be estimated above by a constant independent of z. Thus 2

%=0

=o(i)l 1 +

fl(V\D)f1B(z.S)]; I(W' Z)I3-,IJ -

ZI2n-3+j

2

j=0

p -'dvt °[(V\D)f1B(z.6)]t

rC1(P(0

+IIm4D(J.

P(Z)) + C2IJ - ZIZ

Z)I)3-JIJ

-

ZI2n-3+j

Here, we have used inequality 4 from §4.2°. Now make the change of variables

occurring in part II of Lemma 21.1. Using the definition of the Ia,k,1, the obvious inequalities

IA-1,2,2"-2\ 0 =

0(1)Ix-1,12,2"-1(e)'

4-1,1,2n-I(e) = 0(1)Ia-1,12,2n-1(e) and Lemma 21.2, we get for 11111 + IIJ II = m + 2 2

ID''JH(Z)l = 0(l)l 1 +2 4_1,3_J,2fl_3+f(IP(z)I) j=o

= O(1)11 + Ia-i.3.zn-s0P(z)D + 0(1)IP(z)I

A-12

Hence we conclude by Lemma 21.3 that all derivatives of H of order m + 1 lie in C °"X -1 /2(D ). Thus H satisfies (21.1).

We next give the sharpened version of Theorem 6.1.

THEOREM 21.5. Let D be a bounded strictly pseudoconvex domain in C", n > 1, with aD E C"`+2, m > 0, and let a E C(P,n_1)(aD), A > 0. Then a E

A ,n p(D) if and only if there exists a form y satisfying (21.1) such that

--a.

PROOF. The sufficiency was proved in §6. We prove necessity. By Theorem 2.12,

a=a,1an-az1an9 where a1 E Z(_ 1)(D) and a2 E Z_ 1)(CD). By Theorem 21.4, a1 = ay1 and y1 satisfies (21.1). Extend a2 to a form a2 E cp:n_1(C ). By (1.1), for

129

§21. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

z E (V n G )\ D we have 2

a2 = a2 = Ip,n-1(G, a2) - II2 ,n_ 1(G, aa2) - alp,n_2(G, a2

).

(21.4)

Here G is any domain of holomorphy with boundary of class C,+2 such that D C G. Using Lemma 1.3 and the fact that aa2 = aa2 = 0 on CD, we get for

zEG\D Ip.n_1(G, aa2)() = I p.n_1)(D, aa2)(')

-a D aa2 A µo -

4

aa2 A WW,0(u,', z). (21.5)

Here µo is the form defined in Lemma 1.3, for the choice

wI =t=(f -z)/l -z12 and The coefficients of WP.o(u, , z) are holomorphic in z, and

a2Ian-02 Ian-a1 Ian - a E so that the last integral on the right side of (21.5) vanishes by Stokes' theorem. The integral

72(r) =

aa2 A µo DZ

and its derivatives can be estimated as in the proof of Theorem 21.4 to conclude that y2 satisfies (21.1) with G \ D in the place of D. Extend y2 to all of G with the same smoothness, and denote the extension also by y2. Next, since

Ip,n- I(G, a2) E Z(1P,,,_,)(G), we have ay3 = Ip,n_ t(G, a2), where y3 E Then it follows from (21.3)-(21.5) that a = a y IaD, where y = yl C(pp, n _ 2)(G).

12 lp,n _ 2(G,

a2). Corollary 2.4 now finishes the proof. If we use Theorem 21.5 in place of Theorem 6.1 in the proofs of Theorems 13.2 and 13.4, then we can weaken their hypotheses: instead of the condition µ Z E C2» PD), A > 0, we need only require µ z E C20 - 1(aD ), A> 2 . We state the next theorem without proof. + Y2 - Y3 +

THEOREM 21.6. Let D be a bounded strictly pseudoconvex domain in C", and

suppose a is a (p, n - 1) form on aD with coefficients in L'(aD ). Then a E A n , (D) if and only if there exists $ E Zp,,,_ 1(D) such that lim E-o {p(Z)=-E)

$(z)Ap(z)f = a(z) A (p(z) aD

(21.6)

for any p E q,O)(D). Also, N(Z) A 5P(z) +IP(Z)I'/2I$(z)I dvZ = O(1)jjajjL'. (21.7) (p(z)=-E)

130

VI. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

Here, as before, p is a function defining D; by the absolute value of a form we mean the sum of the absolute values of its coefficients, and II II V denotes

the L' norm of the absolute value of the form. (21.6) says that, in some (generalized) sense, the restriction of P to aD coincides with a. (21.7) says that the tangential part of P has the same properties as a. For forms of type (p, n - 1), the a-problem is solvable not only for domains of holomorphy but also for domains D with aD E C2 for which the restriction of the Levi form (cf. §4.2°) to the complex tangent space to aD at a point has at least one positive eigenvalue for each E 3D. In the C°° case, this was done by Kohn and Rossi [1]. By combining the methods of proof of Fischer and Lieb [I ] with those of Theorem 21.4, it is possible to show that Theorem 21.4 is

valid for q = n - 1 and for domains of the above kind. After making the corresponding changes in the proof of Theorem 6.2, we thus obtain the following improved version:

'

THEOREM 21.7. Let D = S2\ U Sz;, where Sa is a strictly pseudoconvex domain

with al E Crn+2, m > 1, and the Ii C Sz are domains with boundaries of class Crn+2 such that S z; n SZ, =0 f o r j 1, j, I = 1, ... , k, and the restriction of the Levi form to the complex tangent space to afli at ' has at least one positive eigenvalue at each ' E ale , j = 1, ... , k. Let a E Cep: n_ 1)(8D ). Then a E An p(D) if and only if there exists a form it such that it IaD = a and

a E Z(p,n'' 1) '/2(D ),

j

a E Ztp;n'= 2(D),

0 < X' < X , if 0 < A < 1/2;

1/2 < A' < A, if 1/2 < A < 1.

§22. Generalization of Theorem 8.1 THEOREM 22.1. Let D be a bounded domain in C n with boundary of class C2, and suppose that for each a E A 1(D) n C(n,n _ 1)(D) there is an a e Z(1,, n _ 1)(D) such that a 18D= a. Then D = SZ \ U 0i, where the SZi a Si are domains such that ii, n Ii, =0 for j 1, j, 1 = 1,. . ,- , k, and the restriction of the Levi form of a 2i to the complex tangent space to a SZj at ' has at least one nonnegative eigenvalue at each E a Sz;, j = 1, ... , k.

Theorems 21.7 and 22.1 give similar necessary and sufficient conditions on the bounded components of the complement of D in order that all forms in have a-closed extensions into D. However, the most interesting part of the boundary, namely that of the unbounded component of the complement, remains unexplored.

131

§22. GENERALIZATION OF THEOREM 8.1

PROOF. Suppose that, for some j, SZi does not satisfy the conclusion of the theorem, i.e. there exists ° E 80, at which all the eigenvalues of the Levi form

are negative. Then there exists a neighborhood U of ° such that for any E U fl S j the function FE(Z)

ap

11

(J )(z - J!) + r

a 8 Pm

1

(J)(z, - 0(Zm - Jm)

vanishes in u fl CStj if and only if z = E aRj U U (see, for example, Gunning and Rossi [1], Chapter IX, §B). A similar assertion was used in the first part of the proof of Theorem 8.1. It follows that there exist 1 E U fl oil z° E U fl C9i and r > 0 such that the function f(z) = 1/Fti(z) satisfies

and CSZj fl

r),

z E CStj fl

11(20)1>Jf(z)1,

(22.1)

r) has piecewise smooth boundary. The form a

U°.° , z0),

E a0,,

0,

E aD\ auj,

lies in A' (5) fl C( _1)(D); hence there exists a E Z("",- )(D) such that a I8D = a. As in the proof of Theorem 8.1, we see that U0,0(t, z°) = ay in 52., where

-'U)

- f J)WU0,0(w, Z°)U,,n-2(w,

+ f &(w) A Un,n-2(w, J )

By Corollary 2.4, y E C( 2)(S2f). By the Martinelli -Bochner formula (i.e. (1.1) for p = q = 0) and Stokes' theorem, for any g E A Csaj n B(r°, r)) we have

g(z°) = =

a

Z°)

8(r°, r))

fr

(Ca,naB(U°, r))1

Here IF = 8uj n 8B(3°, r). It follows that, for some constant c,

lg(z°)) < C

max tECUjnaB(('°, r)

lg(z)l.

(22.2)

But it is clear from (22.1) that if g(z) = f m(z) with m sufficiently large, then (22.2) cannot hold. This contradiction proves the theorem.

132

VI. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

§23. Weighted formula for solving the a-problem in strictly convex domains and zeros of functions of the Nevanlinna-Dzhrbashyan class

To obtain certain precise results in strictly pseudoconvex domains, it became necessary to have the solutions of the a-equation with a right-hand side that grows near the boundary expressed by a formula. For simplicity, we present the formula here only for a strictly convex domain.

Let D be a strictly convex domain in C", i.e. a domain of the form D = (z E U, p (z) < 0), where U is a neighborhood of D, p E C 2 (U ), and the second differential (i.e. Hessian) of p is positive definite at all ' E U. Hence grad pIaD 0. For a strictly convex domain, the Khenkin barrier function (see §4.2°) may be introduced as follows:

PM = (P1,...,P) =

ap a

1

z) -

We set 1$(i, z) = of p, we have

,

...,

ap

Z) = (P, - Z).

By Taylor's theorem and the strict convexity zI2

for z,

2 Re z) + c, 1 p(z) > E D', D C D' C U, with a constant c1 > 0 depending only on Y.

Therefore 2 Re (D

z) -> p

2 Re

z) >

P(Z) + Cjj _ Z12,

p(z) + c, 1 - zj

2

(23.1)

Hence I41(3', Z)I =

{- P(z)I - Pq) - P(Z)

Z)k

+IzM $q, Z)I+I

(23.2)

-

0(1) depend only on D' and not on z, For a (p, q)-form 7; we define (Gp",qy)(z)

ZIZ

=

E Y.

`IP,4-I(D, akGI Z)7(0)(Z)

f 5takGl z) A y(J) A tJ'q

A p!

(n- p)! Dp,n-p(aZ+ (23.4)

§23. WEIGHTED FORMULA FOR SOLVING THE a-PROBLEM

Here, µ9

133

z 12 and u =

is given by Lemma 1.4, with w = t = ( -Z )/I z), and ak is defined by Z)

ak(

2n k

PG) Z)

Z)

L

k 2,

Z)

j=0

Z)

k

(23.5)

We have Z)

_ -k2nak-IM z)

2n- I

Z)

(W, zPP(0 - P(WeW, Z)

$Z( i)

4bq, =)1

(23.6)

From Lemma 4.1 and (23.6),.and the fact that .t , is a linear combination of forms of the type (1.10), we see that the kernels of the second term on the right in (23.4) have no singularitites. Also, ak contains a factor hence the

operator Gpq can be applied to forms y such that pkE Lt p,q)(D). THEOREM 23.1. Let q > 0, and suppose y E Z(' p,q)(D) and 0 > 0 are such that

pay E L'(p, JD) and pe-'i2ap n y E L(p,q+I)(D). Then for all k > 0 + 1 the forms I3k = Gp, qy satisfy aflk =yin D, and PP -'13k E L p q _ 1(D ).

PROOF. Assume first that y E Zp, q(D ). Substitute z)y(') in the Martinelli-Bochner-Koppelman formula (1.1). Since akG, z) = 0 for E aD, ak is holomorphic in z on D and ay = 0, we get

y(z) = -Ip,q(D,

z) A Y) - aIp,q-,(D,

z)y).

(23.7)

We transform the first term on the right above by Lemma 1.4 to

IP,9(D' ataklJl Z) A -y) =

=

ID;«kG,

8Z

Z) A

z) A

1DklJ

z) A Y\J 1 A Up,9Gl Z)

A (kjtiqI + afµa) A A µq-1 A

(_i)f d[ak(, z) A

P, (n'_ P),

p Au' A

DP.n-p(az,

p! (n-p)!

Dp

As already observed, the integrals have no singularities, so it is legitimate to bring O. outside the integral sign. Now, since arClk(t, z) = 0 for E aD and k > 1, we see by Stokes' theorem that the second integral above vanishes.

134

VI. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

Hence (23.7) yields y = aGp Qy. In the general case, we apply this equality to the form y,(r) = y(rz) E ZP,q(D) and pass to the limit as r -> 1.

We proceed to estimate the solution thus obtained. From (23.6) and the definition of Gp Q, we have fDz

lPe-i(Z)I (Gpkqy)(z)dvZ = 0(1) JD;(flIdvfDt

Ip°(z)I

UP.9-11J, Z)I IaklJr Z)I

Jr

+lak-IIJ

z) 12P11

I

z) Izn+ 1

I()I lk4D(J Z)I IiI}dv_ 1 4,j /yQZl Z) 12n

PB-1IZ)I

+O(l)fD (Y(J)n

I

l

lak-IIJ1Z)I

-

Izn+l lltlq-1jdVz-

Here, as before, dv denotes the volume element, and the absolute value of a form is the sum of the absolute values of its coefficients. Next, using (1.4), Lemma 4.1, (23.5), (23.2), (23.3), the relation q9 Z) I = I

- z ) I = 0(l) 1 - ZI 3,

9

and the definition of µ9_, (see Lemma 1.4), we get

fD 10'-'(Z)l i(Gpkq-Y)(Z)IdVz =Dn00) f I()I Ipk()IdvD I I°(z)I t

Z)

Ik

Z I2n-1 I

Z

+ I $(Jq Z) (k+21

+0(1) t fD IA ()I

I

Z

I

I''

Hence, by the same argument as in the proof of Theorem 21.4,

fDIPP-'(Z)l (G,'qy)(z)dvz t = 0(1){ 1 +1Ds I)I IPk( )- I dvS[

+fp jA ()I t

An application of Lemma 21.2 finishes the proof of the theorem.

) (dvi I2n-3

§23. WEIGHTED FORMULA FOR SOLVING THE a-PROBLEM

135

We state without proof an application of Theorem 23.1.

Let D = {z: p(z) < 0} be a bounded domain in C" with C2 boundary. We denote by Na(D) (a > 0) the class of holomorphic functions F in D such that

fDI()I

°`-' 1n+IF(Z)IdQ2jz)

< o0

(the Nevanlinna-Dzhrbashyan class). Here dak is the k-dimensional LebesgueHausdorff measure, and ln+ I = max{In t, 0}, t > 0.

THEOREM 23.2. Let D be a strictly convex domain in C". Then an (n - 1)dimensional analytic subset M of D is the set of zeros of a function from N.(D ), a > 0, if and only if

f

M

IP(Z)Ia+Ida2n-2(Z)

- >21'.,f I< 001, f Mj

where the Mi are the irreducible components of M, and yj is the multiplicity of Mj in M.

CHAPTER VII

REPRESENTATION AND MULITPLICATION OF DISTRIBUTIONS IN HIGHER DIMENSIONS

§24. Harmonic representation of distributions

Recently, there has been increased interest in the problem of multiplication

of distributions, and there has been a lot of work on the problem. In the International Conference on Generalized Functions held in Moscow (Novem-

ber 1980), different approaches to solving this,problem were discussed in several papers. Here we present a method based on representing the distributions by harmonic functions.

Let W" be the space of the variables (x,,.. . , x,,, y) = (x, y), and let R+ ' = {(x, y) E W+', y > 0), then R" = ((x, y): y = 0). We shall say that a function f defined in W++' has finite order of growth as y -+ + 0, if for every compact set Min R" there exist c > 0 and m > 0 such that If(x, y) I< cy -I for x E M and 0 < y < 1. Let H denote the set of harmonic functions f in W++' having finite order of growth as y --+ + 0, and K the set of harmonic functions

in R+ ' which can be extended continuously to R' U R" and vanish on W. Then K C H. Let K(x, y) be the Poisson kernel for the half-space R+ '

K(x, y) =

:

r(((n+

>>>z 2)

I X IZ

+ y2)cn+>>,z

.

If T E E'(R" ), then the harmonic function T*(x, y) = T K(x - t, y) is called the Poisson representation of T.

TlHouHt 24.1. If T E E', then T* E H and represents T in the following sense:

lim f

F-.f0 y =E T''`(x, Y)4)(X)dx = (T9 (P), 137

9P E Oo(R").

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

138

Here T* extends continuously to R+ ' U (R" \supp T), and vanishes on R" \supp T, and T *(x, e) = 4(1 x I -"-') as I x I -+ oo. The Poisson representation of a derivative of T is the corresponding derivative of the Poisson representation of T. This theorem generalizes Theorem 14.1, since (see § 14) C" = R2 n

aT(Z) ly=, - aT(Z) ly=-, = T*(x1,...,x2,,_1, E). The proof of Theorem 24.1 is the same as that of. Theorem 14.1. We need only show that T* E H. But since T E E', it is bounded with respect to some seminorm sup

I

xEK,IIIII<m

K

supp T,

and so

I (T, K(x - t, y)) I k cjjK(x - t, y)11 < c, y-s,

x E K.

THEOREM 24.2. For each T E D'(R" ), there exists an f E H which extends continuously by 0 to R+' U (R" \supp T) and which is a harmonic representation of T, i.e. lim E-i

+O

y=E

f(x, y)gv(x)dx = (T, q)),

qp E 6 (R").

(24.1)

If f E H represents 0, then f E K. Every f E H represents some Tf E 6D ' via (24.1). Thus 61 '

H/K.

THEOREM 24.2 generalizes Theorem 14.2 to the case of an arbitrary R" and an arbitrary T E 6D'. Thus, for distributions defined on hyperplanes, it is more convenient to construct their representations by means of the Poisson kernel rather than the Martinelli-Bochner kernel. The latter can be used for distributions defined on smooth hypersurfaces in C". The first part of Theorem 24.2 has in fact been proved already (Theorem 14.2). We proceed to the proof of the second part. LEMMA 24.3. Let f E H. Then, for any bounded domain a C R", there exist an index p and an F E H which can be extended continuously to R++' U SZ, such that

f = a 2pF/ay 2P.

For n = 1, as ' ' ar lemma for holomorphic functions is due to Tillmann [1]. PROOF. Let 0 be the La lacian in R"', and A that in W. Given f harmonic in R + ' , consider

Fj(x, y) = fF(x, 7J)dn + fi(x).

139

§24. HARMONIC REPRESENTATION

Then o F,(x, y) = a f (x, l)/ay + 04(x ). Choose T E C°° such that AT = -a f (x, 1)18y in R" (this is possible since the Poisson equation can be solved in

W). Then AF, = 0, and aF, /a y = f. If f E H, then there exist c > 0 and m > 0 such that cy_m,

lf(x, A I

x E S2, 0
1.

Hence F, E H and IF1(x,

y)I
x E2,0
At some step we can find an F2 E H which is bounded for x E SZ and 0 < y < I and such that asF2/ays = f. The next harmonic primitive F E H of f will then be continuously extendable to R"+' U 2, since

xES1,0
IF(x,yI)-F(x) y2)I
Using this lemma, it is easy to show that the limit in (24.1) exists for any f E H and T E G1. Indeed,

j a2 pF(x, fy-E_ I(x, y)ro(X)ax = y-e 8y

_ (_l)"f _ ApF(z, Y)9)(x) dx -E

_ (-1)p

F(x, y=e

a

+U

F(x,0)APp(x)dx. JR"

Consequently, since GD is weakly dense in 6D' (Shilov [11, Chapter II, §9.3), the above limit defines a certain distribution Tf e 6D.

Suppose now that, for some f E H, the limit in (24.1) is 0 for all p E 6D. Then, by Lemma 24.3, for each bounded domain SZ C R" we have an F E H

and an index p such that a 2PF/a y 2P = f and F extends continuously to R+ ' U I. Since f represents 0, &PF(x, 0) = 0 in SI in the sense of distributions. Hence it follows from the ellipticity of A' in 9 (see, for example, Schapira [1], Theorem A.22) that F(x, 0) is real analytic in I. Hence F extends in a C°° way to SZ (see Brelot [1], Chapter I, §8). Then

f(x, y) _ (-l)'4PF(x, y --> +0.

LEhnKn 24.4. Let f be a polyharmonic function of order m (i.e. 0'"f = 0) defined in W++'. Then

f(x, .v)

f0(x, y) +vfl(x, y) + ... +ym-'fm-1(x, y),

(z4.2)

140

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

where the f , j = 0, 11... , m - 1, are harmonic functions in R+ ; if f has finite order of growth as y - + 0, then the f lie in H, and if f E C°°(7+ ' U I) for an open I C R", then the f are C°° in R"+' U f2. 1

The proof is by induction on m. If m = 1, (24.2) obviously holds. Let e f = 0, m > 1. Then we must have f = 20 -' yif, which would give us m-2

Of

Ij=0yj (j+2)(j+ l)f+2+2(j+ 1) aJj+ay1

By the induction hypothesis, since 0 f is polyharmonic of order m - 1, m-2

Af = I y'T,, j=o

so that we obtain the equations

4pj = (j + 2)(j + 1)fj+2+ 2(j + I

afj+

j=0,...,m-2,

ay

for the j; in particular,

Tm-2 - Z(YIi - ])

afm-1

ay

As in Lemma 24.3, we seek the solution of this equation in the form of the integral

fm_i(X, Y) = 2(m1

1)

f yTm-i(x, n)dq + (x)

Since

AfM-'

I

2(m - 1) we find

(X, 1) +

0

8y

from this equation. Next, from the equation rn-3

a!m-Z rn- 1

ay

we determine frn - 2 in a similar way, etc.

It is clear from the proof above that the f lie in H if f is of finite order growth, and that f E C°°(cl U IV ++') for all j if f E C00(2 U R+ ').

of

COROLLARY 24.4. If 0'"f = 0 and f is of finite order of growth as y --> +0, then f(x, y) -- Tf E 6D' in the sense of (24.1) as y -), + 0.

Thus, polyharmonic functions also represent distributions in 6D', and in this case Tf = Tfo

§25. MULTIPLICATION OF DISTRIBUTIONS

141

§25. Multiplication of distributions and its properties

Using the harmonic representations, we can define the product of two distributions by the method of Ivanov [1], [2]. Let H* be the algebra of functions generated by H, and K* C H* the subset consisting of functions representing 0 in the sense of (24.1). We shall call the quotient H*/K* =6D* the space of hyperdistributions, and denote by f* the image of f e H* in G *,

i.e.f*=f+K*. The map A : 6D' -> 6D * defined as Tf -* f * (f a harmonic representation of T) is an imbedding, since K C K*. Hence we shall suppose GD' C 6D*. Let T. S E

6D', and let f and g be their harmonic representations. Then by the product T o S of T and S we shall mean the hyperdistribution (fg)* = fg + K* E 6D*. In the sequel, we shall be interested in the question as to when T o S is a distribution, in the sense that the limit (24.1) exists for fg and an arbitrary ( E 6D. Note that if this limit exists for fg, then it exists for any function in fg + K*, and coincides with the limit for fg. We give some properties of this product. 1. S o T is well-defined, i.e. the class fg + K* is independent of the choice of the representatives f and g.

PROOF. It suffices to show that, if f E K and g E H, then fg E K*. Let p E 6D. It is easy to show that f = yf, in a neighborhood of the support of p, where f, is a real-analytic function. Hence ftp E 6 for fixed y, and ftp -+ 0 in GD as y --> + 0, while g tends to S in 6D' as y - + 0; consequently, by a lemma in Shilov [11 (Chapter II, § 9.3),

f(x,e)p(x)g(x,e)dx= (S,0)=0,

lim

e-++0

R"

i.e. fg E K*.

2. supp(T o S) C supp T fl supp S. In particular, T o S = 0 if supp T n supp S = 0 . Here, the support of a hyperdistribution is defined as follows. We say that f * E 6D* vanishes in an open set f in R" if, for any qv E GD with support in 0, lim+o e-.+

y=e

f(x,y)p(x)dx=0,

fef*.

Now the support supp f* of f* is by definition the complement of the largest open set in which f* vanishes. It is easy to verify by means of a partition of unity that this notion is well-defined. PROOF OF PROPERTY 2. Let qv E GD with supp qp C R" \(supp T fl

supp s) _

0. We may suppose that, for example, supp q) fl supp T = 0 (by means of a

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

142

partition of unity it is always possible to pass to functions 9) with arbitrarily small support). Then f can be written in the form yf 1 in a neighborhood of supp qn, and the rest of the proof proceeds as for Property 1.

3. If T E 6D' is defined by a function T(x) E L p(loc), and S by S(x) in LQ(loc), 1/p + l/q = 1, then To S = T(x)L(x) E L'(loc). PROOF. It can be shown that, corresponding to S(x), there exists u(x, y) E H such that u(x, y) -* S(x) almost everywhere as y - + 0, and for each compact set M

IIu(x, Y) - S(x)II as y -4 + 0, where q

IIrIIM

=

f

-M+ 0

If IfIgdxJ

q> 1.

To do this, we must use:

(a) Theorem 1 of Stein [2], (Chapter III, §2.1), asserting that, for every f e LQ(R" ), its Poisson integral defines a harmonic function J(x, y) such that f(x, y) converges to f(x) almost everywhere and 11 f(x, y) - f(x)II - 0 as y --- + 0; and (b) the procedure in Bremermann [11, §5.9, replacing power-series expansions by expansions in harmonic polynomials. Since S(x) E L9(loc) and T(x) E LP(loc), we will have S(x)T(x) E L' (loc). Let u(x, y), v(x, y) and w(x, y) be the respective harmonic extensions to R+ ' with the properties mentioned above. Then u, v and w are harmonic represen-

tations of T(x), S(x) and T(x)S(x). Furthermore uv - w E K*. Indeed, let 4p E 61)1 with supp qQ C M. Then

(uv - w)yv dx

c fMjuv

- wIdx < c(Ijuv

fRfl

-

+11w -

but II uv - ST 1) M --*0 as y - +0, since II u - T II°f and 11v- S II %f tend too. 4. The derivatives of hyperdistributions are defined as follows: a

f*

a

_1

_ of

This definition makes sense, since af/axk E K* for f E K*. For the derivatives of the product of two distributions, we have the Leibniz rule a

aXk

(To S =

aT aXk

oS+ To as

axk

143

§25. MULTIPLICATION OF DISTRIBUTIONS

This follows immediately from the fact that if f represents T, then a f/axk represents aT/axk. 5. The product of distributions is compatible with multiplication by multipliers,

i.e.ifaEC°°(R") andTE6',then a°T=aT. PROOF. Since a E C°°, its harmonic representation g(x, y) E H extends continuously to R+ ' U R" (solution of the Dirichlet problem for the half-space). Hence g E C°°(R. ++' U R") (Brelot [1], Chapter I, §8), so that p(x)g(x, y) -+

(p(x)a(x) in '! as y -* +0, for any T E'D. On the other hand, f(x, y) -> T in 6D' as y --> + 0, so that, by the lemma already mentioned (Shilov [ 1 ], Chapter II, §9.3),

j

Y

_ _e

f(x,y)p(x)g(x,y)dx -' (T, a(p) - (aT, 97) e- +0

6. If f E H is a homogeneous function of degree p, then its boundary value as y -+ + 0, i.e. Tf , is also homogeneous of degree p.

limo f_,)p(x)dx TJ, (P(tX))- elimo f f(x,y)p(tx)dx= Indeed,

y-et

\t

= tp+n (limo

If T E d' is homogeneous of degree p, then it is not hard to see that T*(x, y) is a homogeneous function of degree p, since

T*( tx t y) y)

(ii, K (u -- tx ,

)) =

t

n

Tu K

t

- x ,Y

= tp (Tu, K(u - x, y)) = tPT*(x, y). By means of the general procedure for constructing a harmonic representation, it can be shown that, if T E 6 ' is homogeneous of degree p, then T admits a harmonic representation which is homogeneous of degree p. It follows that, if T, S E 6D' are homogeneous of degrees p and q respectively, and T o S E 6D', then T o S is homogeneous of degree p + q. 7. Let x' _ (x l , ... , x m ), x" _ (Xm + 1, ... , x n ), and T E GL'(Rm ), S E 6D "(W - m ). Let f(x', y) and g(x ", y) be harmonic representations of T and S in R++' and R+ m+ 1 respectively. Then

(f +Kj-(g+Kx..) Ch+K* where h is a harmonic representation of the tensor product T X S. Thus multiplication of distributions is compatible with the formation of products.

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

144

PROOF. It suffices to show that the boundary value of f (x', y)g(x", y) in the sense of (24.1) is T X S. Consider the integral

fg(x", y)cp(x)dx" = (x', E),

(p e 6D.

y=E

We have (x', e) E 6 (Rm) and 4 (x', e) -> (S, cp(x)) = J(x') in 6D(Rm) as J(x') as c -- > + 0 (Lemma 24.3), and

er -* + 0. Since J(x', e)

DX, (x', e) = fg(x", y)DX-p(x)dx we have, by the same lemma, D x' (x, e) o D,' (Ji(x')) as e -4 +0. Now we apply the lemma from Shilov [1] (Chapter II, §9.3) once again and conclude

f

f(x',

_

y)s(x,,,

.v)'T(X)dx

E

=

f

_ .f(X', F

y)aX'F+0 -- (Tx,, Ox'))

= (Tx,, (Sx, T(x))) = (T X S, T(x)). Thus the definition of product which we have given has several natural properties; in particular, we will not have counterexamples of the Itano kind with our definition. §26. Examples of products of distributions Consider the vector space WI of functions, generated by functions of the form

Ax, y) =

Pk(x y)

(lxi2+y2)

k-m+(n+1)/2'

k> ,0 , m>

1,

(26.1)

where Pk(x, y) is a homogeneous polynomial of degree k and 1mPk = 0. These functions are polyharmonic of order m; actually they are the Kelvin transforms of the Pk(x, y). We first elucidate what kind of distributions they represent.

LEM&A 26.1. Suppose f is a function of the form (26.1) lying in

k+1-2m>0.Then

I

Pk(X, Y)xadx (I

=0

I

for all monomials x° with Ilall < k + 1 - 2m. PROOF. Let

AY) =1X,=1

PJx9 Y)xadx I2 +

2)k-m+(n+ 1)/2 '

y

Hall
T1,

and

145

§26. EXAMPLES OF PRODUCTS

We shall show that the limit

lim I(y)-= c

y-.+0

exists and is finite. Indeed, by Lemma 24.4, we can write fin the form

f=fo+yfi+...+ym_hfm_i; then m-1

yj I(y) _ j=0 7 y-lim +o Y--+ +o

f(X,y)XadX.

lim

Ixl< l

According to Lemma 24.4, the fj are C' on W++ ' U (Rn \ (0)). We now show that the limit

f(X,y)XadX

lim y-.+o exists.

By Lemma 24.3, there exists F (x, y) E H extending continuously to W++' U fx E Rn : I x I< 2) and such that a2p

ay 2p

F (x, y) = M X7 y)

Then f (x, y) = (- l) pOp1 (x, y), and iXi

aXk F(x

y)q)(x)dx

f

F.(X, y)

_-

ax + (-i)f

F(p dx

Thus

i

x

f(x, y)xadx= (_lYf 1

(xI

F(x, y)APxadx+ (x =1

1

where D is a differential operator, and do the surface element of the sphere.

We need only observe now that the F, like the fj, are C°° on R+ ' U ({ x I < 2)\{O)) (see the proof of Lemma 24.3). Thus the limit

lim AY) = c

Y--* +0

exists. We now transform the variables in I(y) by setting x = rE, rllall+n-l+k-s I I(y) _ ys _ Pk()adc 0 (r2 + y2) k-m+ (n+1)/2 dr

-1

1. Then

146

V11. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

where Pk(x, y) _ 2saa y5Pk(x). If we further set r = yt, then =Y11all -k - l+2m (

1(y)

tllall+n-1+k-sd1

1/y

(1 +

J

Since I(y) has a limit as y

J = lim I(y) = f Y

0

I£1=1 Pk(S)Sada.

(26.2)

+0, we necessarily obtain from (26.2) that tllall+n-(+k-s

op

t2)k-m+(n+1)/2

sa0 (1 +

0

tt

f

t2)k-m+(n+])/2

dt f

0.

ICI=1

The improper integral above is convergent, since II a II < k + I - 2m. It is easy to see that

Pk(x, y)x adx R"

(' x 12 + y2)k-m+(n+1)/2

= yllall-k-l-2mJ = 0.

LEMMA 26.2. Suppose an f of the form (26.1) lies in V and k + 1 - 2m' 0. Then

1=p.v. f

Pk(x, n

Hall =k+I - 2m,

k)M+q/2=ca,

R" (I X'2 +. y2)

where

-

,,n/2r(s

2s-1(X)Xa)

1/2)Ak+l-s-m(Pk

Ca = sat 22(k+l-s-m)(k + 1

ll

- s - m)!F(k - m + (n + 1)/2)

(26.3)

PROOF. Consider the integral

Pk(x, y)x"dx

I(Y) -'1xlc1 (Ix I2 I y2)k-m+(n+1)/2 Transforming to polar coordinates as in Lemma 26.1, we get

I(y) =

f

l/y

saO 0

t2k+n-s-2mdt

2 k m+(n+l)/2

(1 + t)

f

t tada

Pk(s)s.

ICI=1

The polynmials Pk have degrees k - s; hence the Pkj" have degrees 2k + 1 - 2m - s. Thus, if s is even, the degree of the homogeneous polynomial Pkta is odd, and so Pk6ada = 0.

41 = I

Then,

I(Y)

fl/Y

-

0

dt f III=IP2s- 1()ada.

t 2k+n+1-2s-2m

(1 +

t2)k-m+(n+l)/2

§26. EXAMPLES OF PRODUCTS

147

Clearly I (y) -* I as y -* + 0, so that 00

t2k+n+ 1-2s-2mdt 12)k-m+(n+ 1)/2

+

fo

.

2S-1

I,I _ 1

ada.

k

But

f

0

2k+n+1-2s-2mdt

00

1+ t2

k-m+(n+ 1)/2

I

n

2

2

2

and

IEI=1

Pk

to )

2iI n/2y+ 1-s-m

2s-1 a

S da = 2(k+l -s-m) (k+1-s 2

- m)!I'(k+1-s - m + n/2)

(This follows from the Gauss representation of homogeneous polynomials; see Sobolev [1J, Chapter 11). Hence I = ca. Observe that if Pk expands only in even powers of y, then ca = 0 for all a such that II a II < k + I - 2m. Let h kX)

=

Qk(X) I x 12k-2m+n+ 1 '

where Qk is an arbitrary homogeneous polynomial of degree k in W. We denote by h the distribution obtained by regularizing h (x ), i.e.

h, .p1 = f.p. f = P.V.

Qk(X)V(x)dX 2k-2m+n+l

Qk(x)((P(x)

fr

1Ilali
if k + 1 - 2m > 0, where the abbreviations "f.p." and " p.v." signify "(Hadamard) finite part" and "(Cauchy) principal value".

If k+ 1 -2m<0,then (ii, qg)=

q) E 6D.

In other words, h is the Hadamard finite part of the integral

f h(x)qp(x)dx.

148

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

LEMMA 26.3. Let Q be any polynomial in Rn. Then

Q(x, Y) _ I (-1)ry2OIQ(x) (21)! I>0

is a harmonic extension of Q to all of R 4 This follows from direct computation. If

h(x) =

Qk(x, Y) (I x 12 + y2)k-m+(n+I)/2'

then we denote by h(x, y) the function of the form Qk(x, Y)

h(x, Y) =

y2)k-m+(n+1)/2

(I x 12 +

inTZ. THEOREM 26.4. The distributions h and T, are identical.

This follows immediately from Lemmas 26.1-26.3 and the definitions of h and Let W't 1 (respectively 12) denote the subset of WI consisting of functions of the form (26.1) with Pk expanding only in even (odd) powers of y. It is easy to verify that span(TZ I + WI 2) = WI .

Theorem 26.4 implies that the boundary values of functions f in WI I in the sense of (24.1) are the distributions (f(x,0))^. THEOREM 26.5. If f E SWI21 then

G Tf

t

'

ifk+ 1 -m>0,

Ha1l=k+I-2m

-

0,

ifk+ 1 -m<0,

where the c" are defined by (26.3) and S(") = a"S/8x", S being the Dirac 8 -function.

This is immediate from Lemmas 26.1 and 26.2. We have thus'a complete description of the class of distributions representable by the functions in V. EXAMPLE 1. Let f(x, y) = Y/(I x 12 + y2)(n+I)/2. Then 7 = 1T (n+I)/28/I'((n + 1)/2).

§26. EXAMPLES OF PRODUCTS

149

We now consider the question: when is the product of two functions in J31 again im V? Let Pk(x, y) A g = (I x +Q1(x, f= (I 12 +y2)k-m+(n+1)/2' y2)1-5+(m+1)/2 x

12

then

PkQ1

fg

(I

x

12 + y2)I+k+(n+1)/2-(m+s-(n+1)/2)

Hence we see that fgEW if and only ifnisodd(n+2p- 1),m+s.>-p, and

pm+s-p(PkQ,) = 0. Thus the class of such functions f and g is fairly large.

THEOREM 26.6. Suppose f, g and fg lie in V. Then:

a) if f, g E TI , then fg E 'fi't 1 and (f (x, 0)) ^ a (g(x, 0)) ^ _ (fg(x, 0))^;

b)iff,gEan2,then TfoTg=O; c)iffE11P11and gETZ2,then fgE D12and (-1)k-2p-2m {(x' O) o J

C. &(a)

=

IIall=tt+l-2s

2

c 1

II.811=1+k+2p-2m-2s- 1

For n = I, identities of this type have been obtained by Ivanov [3]. EXAMPLE 2. Let f = (p - 1)!y/irp(1 x 12 + y2)P. Then Tf = S. Let h(x) Qk(x)/I x 12k with AQk = 0; then

a= (-I)k 2

(p - 1)!Z (QkXa) 6(a).

I1all=k

k! (k + p - 1)!

Ifh=xj/Ix12,then a

-8 = -

8.

I

For p = n = 1, this identity reduces to the familiar identity P(1 /x) .6 = -82.

SUPPLEMENT TO THE BRIEF HISTORICAL SURVEY

§ 16 contains the solution to Problem 3 (see "Brief historical survey and open problems" which follows Chapter IV). All the results of this section are due to Kytmanov [2]-[4].. A special case of Problem 3 (viz., Problem 2) is discussed in

§ 17. Theorem 17.1 was proved for the ball in C" by Aronov [2], and in the general case by Aronov and Kytmanov [1]. Theorem 17.2 and Corollaries 17.3 and 17.4 were obtained by Aizenberg and Kytmanov [ 11, [2]. Corollary 17.7 was obtained for the ball by Agranovskii and Val'skii [11, and for arbitrary domains by Stout [ 11. Theorems 17.9 and 17.10 were proved by Kytmanov [ 1 ]. The results of § 18 are due to Aizenberg and Dautov [ 1 ]. All the assertions of

§19 except for Theorems 19.4 and 19.5 are due to Kytmanov [8]. Theorems 19.4 and 19.5 were proved by Romanov [ I]; Theorem 19.5 is in fact true for arbitrary domains with smooth boundary in C" (Romanov [2]). In §20 differential criteria for the existence of holomorphic extensions of functions are presented, which are criteria different from the tangential Cauchy-Riemann equations. Theorem 20.1 was proved for domains D in Kahler manifolds with OD E C°° and f E C°°(D) by Folland and Kohn [1]; in the form stated here, it is contained in Aronov and Kytmanov [1]. Theorem 20.4 is a reformulation of a result of Forelli [1]. The remaining results of §20 are due to Kytmanov [7]. The strengthening of Theorem 6.1 (Theorem 21.5) was obtained by Dautov (see Aizenberg and Yuzhakov [1], Theorem 26.1). The results of §22 are due to Dautov. Weighted formulas for solving the a-problem and weighted estimates (Theorem 23.1) were proved by Dautov and Khenkin [ 11, [2]. Theorem 21.6 is due to Khenkin [5]. For a = 0, Theorem 23.2 (for functions of the Nevanlinna class) was proved by KhenkiJ41-[6] and Skoda [ 11, [2]; for a > 0 (functions of the Nevanlinna-Dzhrbashy class) it was proved by Dautov and Khenkin [1], [2].

In these papers, there is also a generalization of Theorem 23.2 to strictly pseudoconvex domains in manifolds. 151

152

HISTORICAL SURVEY

Chapter VII is devoted to the solution of Problem 13. The results of this chapter are due to Kytmanov [5], [6]. We note that, for n = 1, Theorem 24.1 is proved in Bremermann [ 1 ] _(p. 49). That part of Theorem 24.2 which proves that the boundary value of a harmonic function of class H is a distribution can be deduced from a theorem of Roitberg [1] which states that, if L is an elliptic operator of order 2m in a bounded domain 2 C R" with sufficiently smooth

boundary, then every solution f of the equation Lf = 0 with finite-order singularity on aft defines a certain distribution on M. Theorem 24.2 gives a method for constructing this distribution. A different method of multiplying higher-dimensional distribution by means of analytic representations has been proposed by Ivanov and Verzhbalovich [1]. We rote that, with this method as with ours, it would not be possible to have counterexamples like those of Itano M.

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SUBJECT INDEX a-problem, 31, 86

Formula, of Martinelli-Bochner for exterior differential forms, ((1.1)), 5, 7,..., 87 Functions orthogonal to holomorphic functions, 37

Cousin data, 58, 59 Derivatives of distributions, 82, 84 Determinant, Levi's, 42 Distributions of class A, 79, 81, 82, 88 Domain, admissible, 62 linearly convex, 41, 68, 74, 75 in the sense of Martineau,

Kernel, reproducing, 72-74 , Cauchy, 70, 71 , Cauchy-Fantappi6, 7, 73 , of Martinelli-Bochner, 71, 77

43

Problem, first Cousin, for forms, 58 of Mittag-Leffler for forms, 57 Riemann's additive, 18 for forms, 26 Product of distributions, 2, 76, 82, 88

, with smooth boundary, 1, 4, 18, 22, 26, ... , 86, 88 , with boundary of class C"n,\, 4, 19, 20,241..., 73, 74 , strictly pseudoconvex, 2, 31,..., 87, 88

Representation, integral, of holomorphic functions, 2, 71, 88

Envelope of holomorphy of a compact set,

Martinelli-Bochner, of a distribution, 77, 78 analytic, of a distribution, 76, 84

47

Equations, tangential Cauchy-Riemann, 30, 85

Solution, weak, of the tangential Cauchy-

Form, a-closed, 2, 27, 46,..., 87 , o-exact, 2, 39,..., 74 , double, 6, 7

Riemann equations, 30, 69, 70

- , of the first Cousin problem, 58, 59

Levi, 2, 31, 49 orthogonal to holomorphic forms or functions, 1, 2, ... , 71, 72 Formula, of Cauchy-Fantappie, ((13.2)), 38,

Theorem, of F. and M. Riess, 68-70, 71 Runge's, 53, 86 , for forms (Theorem 9.1), 53,

...,86

72,73,86-88

163

INDEX OF SYMBOLS A, 82

Hl, 67

lzI, 3

ACC(S2), 4

, 3

Ao (aD), 39

H(K)7 47 int M, 3

An (D), 45

11(P,,)(D, 'Y)(z), 8

ar, 56 an , 79

AP(K), 4 A(fl), 4 BA, 82 BA, 82 Bp,n-1, 27 B(z°, r), 3 Cn, 3

I(P,e)(D,'Y)(z), 8

7±, 20

11111, 3

'Y+, 26 6, 79

L (P); 42

M, 3 Mr, 3

6+, 79 &-, 81 °k, 81 0o, 81

MCN, 3 O(aD), 37 01(3D), 68 02(aD), 68

Cm'>`(F), 3 Cm,q (P) `F), 3

P0, 11

P(x-1), 81 P(xnI xI -2n), 81

C (P) ,q (11)' 4

diam M, 3 dz, 3

P(S, z), 31

dzj, 3 dz[4, 3

r(z), 62 supp a, 4

aa, 4

t(S, z), 3

D, 43

Up,q(S,z), 7 Vq, 9

Dijj 3 )em), 5

g*f, 56

Z p,q (1l)' 4

i;,11 µ, 11 Q

Pr(z), 56 P(z, M), 3 cr(I), 9 a(I, k), 9

>', 3 ±1 18

Wp,q(w, c, z), 7

I'(S,z), 31

z, 3

W (w , S, z), 7

ABCDEFGHIJ -EB-89876543 165