Differential Forms Orthogonal to Holomorphic Functions or Forms, and Their Properties (Translations of Mathematical Monographs)

Ia,k,t(e)

= 0(1) 1

(rsl

O
+ e)"'r2n-l-ldrdslds2

.

I+ s2 {1 -sl2)1/2) +e+r 2]k

We use the inequality ls 11 +s 2(1-sf) 112
Ia,k.t(e)

= 0(1) 1

(rsu

O
+ e)"'r2n-l-tsdrdsdu (rs+e+r2)k

rsu+e>O

For I= n - 1, we go over to the coordinates ( v, s, u ), where v rs + e + r 2 > rs + e, we get la,k,2n-l(e) 0{1)

= e- 1rs. Since

=

J'

dv j'(vu+1)"'du+j4 e-• dv (vu+1)"'du]. o (v+ 1)k -t t (v+ 1)k -v-•

·el-k+a[1'

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

Ia,k.ln-J-2m(e)

= 0(1) 1 2r 2mdr1 2 0

0

(rs

rds

+ e + r 2)

k

t

(rsu

+ e)"'rsdu

-e(rs)- 1

= 0(1)12r2mdr12 . rds = 0(1)12 r2mdr o o (rs+e+r2)k-a-l o {e+r2)k+a-2 dr

= 0(1)12 0

(

{e + r )2(k-a-m-2).

Assertion b) now follows by direct integration.

127

§21. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

LEMMA 21.3. Let D C Rn be a bounded domain with smooth boundary and let fE C 1(D). Suppose lgradf(z)l= 0(1)lp(z)l"- 1 for O
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.

en

THEOREM 21.4. Let D be a bounded strictly pseudoconvex domain in with boundary of class cm+ 2, and let y E zr;·.~)(D), m:;;;. 0, A> 0. Then there exists y in D, with a form a satisfying a

a=

a E cm,A+I/2(f5) (p,q-1) •

0 < A' < A, if 0 < A .so;; t;

a E cm+I,A-I/2(f5) (p,q-1) •

t < A' <

(21.1)

A, if t < A .so;; 1.

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

1

a·Y(r)

(V\D),

1\

1'~-~

1\ DP

'

nrp(oz, or}

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)=1 ,

h(r)q>(r,z)(fk-zk)

dfl\dr

«<>'+'(r,z)lr-zi2Cn-r-t>

(21.2) '

=

with hE cm-t,>.(Cn) and h 0 on D. If m = 0, we may suppose that hE C(V\D) and lh.. We shall carry out the rest of the proof in the case A> f. The case A< t can be handled similarly. Differentiating under the integral sign and using the argument in Lemma 4.2, we conclude that, for zED, •

,f- 1(r)dv

ID 1·JH(z)l = 0(1) j=O ~ 1 r (V\D>,I «~>(r, z) 13 -j I r- z l2 n-3+j. 2

•

IIIII + 11111 = m + 2. Chooser and 8 according to Lemma 21.1 and split the integrals over V\D into integrals over (V\D)\B(z, 8) and (V\D) (,l B(z, 8). The first integrals

128

VI. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

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

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,t• the obvious inequalities

IA-t,2,2n-2(e) = 0(1)/,__t,J1,2n-t(e), IA-t,t,2n-t(e) = 0(1)l,__t,J1,2n-t(e) and Lemma 21.2, we get for II I

II + II J II

=m + 2

iD'·JH(z ll = 0(1)[ I + i i h-1>-J>•-Hi(ip(z )I)]

= 0(1)( 1 + l>.-t,3,2n-3(1P(z)l) + l>.-t,t4,2n-t(IP(z)l)] A-t! =0(1)lp(z)l •. Hence we conclude by Lemma 21.3 that all derivatives of H of order m lie in co,A-t/ 2(D). Thus H satisfies (21.1). We next give the sharpened version of Theorem 6.1.

+1

THEOREM 21.5. Let D be a bounded strictly pseudoconvex domain in C", n > 1, ~ith 3D E cm+z, m;;;. 0, and let a E C(;:~-t)(3D), ~ > 0. Then a E ~;_p(D) if and only if there exists a form y satisfying (21.1) such that ay lao:::; a .

• PRooF. The sufficiency was proved in §6. We prove necessity. By Theorem 2.12, (21.3) where at E Z{';,·.~-t>(D) and a 2 E Z{';,·.~-t>(CD). By Theorem 21.4, a 1 = ayt and Yt satisfies {21.1). Extend a 2 to a form a2 E C(;:~:.._t)(C"). By (1.1), for

129

§21. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

z E (V n G)\D we have a2

= li 2 = I~.n- 1 ( G, liz) -

I},n- 1( G, aliz} - ai},n-z( G, liz). (21.4)

Here G is any domain of holomorphy with boundary of class cm+Z such that D <£G. Using Lemma 1.3 and the fact that aliz = aaz = 0 on CD, we get for z E G\D

I}.n- 1( G, aliz}(r}

= lc~.n-•>( D, aliz}(r)

-arf aliz

t\

l'o-

D,

f aliz

t\

Uj,,o(u,

D,

r. z ).

(21.5)

Here l'o is the form defined in Lemma 1.3, for the choice

w1 =t=(f-z)/lr-zlz and wz=u=P(r,z)/~(r,z). The coefficients of Uj,,o( u,

r. z) are holomorphic in z, and

liz lao= azlao

= a• lao- a E A;_P(D),

so that the last integral on the right side of (21.5) vanishes by Stokes' theorem. The integral

and its derivatives can be estimated as in the proof of Theorem 21.4 to conclude that "Yz satisfies (21.1) with G\Din the place of D. Extend Yz to all of G with the same smoothness, and denote the extension also by "Yz· Next, since l~.n-l(G, liz) E Z{;,n-I)(G), we have ay3 = l~,n-l(G!.. liz), where y3 E Cc';.n-z)(G). Then it follows from (21.3)-(21.5) that a= ay lao• where y = y1 + "Yz- y3 + Ii.n-z(G, liz)· 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 l'z E Cd~"- 1 (aD), ~ > 0, we need only require l'z E Cf~~ 1(aD), ~ > t. We state the next theorem without proof.

en,

21.6. Let D be a bounded strictly pseudoconvex domain in and 1 with coefficients in L (aD). Then a E suppose a is a (p, n- I)-form on A;_ 1(D) if and only if there exists P E zp,n- 1(D) suc'h that THEOREM

aD

lim

J

e.... o (p(z)=-e}

fJ(z)

t\

cp(z) ~

1aoa(z)

t\

cp(z)

(21.6)

for any cp E Ccn-p,O)(D). Also,

f

(p(z)=-e}

(IP(z) A ap(z)l +lp(z)l 11ziP(z)l)dvz = O(I)IIaiiL·· (21.7)

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 L' denotes the L 1 norm of the absolute value of the form. (21.6) says that, in some (generalized) sense, the restriction of fJ to aD coincides with a. (21.7) says that the tangential part of fJ has the same properties as a. For forms of type ( p, n - I), the a-problem is solvable not only for domains of holomorphy but also for domains D with aD E C 2 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 aD. In the coo case, this was done by Kohn and Rossi [I]. 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 - I 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:

r

r

THEOREM 21.7. Let D = 0\ U~ OJ, where 0 is a strictly pseudoconvex domain with aO E cm+ 2 , m;;;;. I, and the OJ <:£ 0 are domains with boundaries of class cm+ 2 such that SlJ n Sl 1 = 0 for j =I= 1, j, I= I, ... ,k, and the restriction of the Levi form to the complex tangent space to at has at least one positive E aOJ, j = I, ... ,k. Let a E C{;:~-t>(aD). Then a E eigenvalue at each A;_P(D) if and only if there exists a form a such that a a and

aoj

r

r

laD=

O<"A'<"A,ifO<"A..;, lj2; lj2 < "A' < "A, if lj2 < "A .,;;;, I. §22. Generalization of Theorem 8.1

22.1. Let D be a bounded domain in en with boundary of class C 2, and suppose that for each a E A J.. (D) ~ _Cc_":,n-l)( D) there is an a E Z/n,n-t)( D) THEOREM

laD =

=

u oj'

oj

such that a a. Then D 0\ I where the <& 0 are domains such that OJ n = 0 for j =I= l,j, I= I, ... ,k, and the restriction of the Levi form of to the complex tangent space to at has at least one nonnegative eigenvalue at each E aOJ' j = I, ... ,k.

aoj

n/

r

aoj

r

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 A;_p(D) 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, nj does not satisfy the conclusion of the theorem, i.e. there exists ro E ()ijj at which all the eigenvalues of the Levi form are negative. Then there exists a neighborhood u of ro such that for any r E U n Slj the function

Fr(z) =

"

I

f=l

a;,

az-(O(z,- r,)

"

a2p

+ I

(}z (}z (r}(z,- r,)(zm- rm)

l,m=l

f

I

m

vanishes in U n COj if and only if z = r E ()ijj 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 r 1 E U n Oj, z 0 E U n CSlj and r > 0 such that the functionf(z) = 1/Fr•(z) satisfies

z E cnj n aB(r 0 , r),

lf(z 0 )l>l/(z)l, and CSlj

{22.1)

n B(r 0 , r) has piecewise smooth boundary. The form

a{ Uo,o(r, z 0,

0 ),

r E ()ijj• E
r

lies in A J. (D) n <10:.n-l)( D); hence there exi;ts ci E Z(111,n-1)( D) s~ch that a lao= a. As in the proof of Theorem 8.1, we see that Uo.o(r, z 0 ) =()yin nj, where

By Corollary 2.4, y E G~.n- 2>(0j). By the Martinelli-Bochner formula (i.e. ( 1.1) for p = q = 0) and Stokes' theorem, for any g E A ( COj n B(f 0 , r}) we have

= 1g(r) 1\ y(r) r Here f =

+[

(CSlinaB(!' 0 , r))r

anj n aB
max

lg(z)l.

(22.2)

!'ECSlpaB(r 0 , r)

But it is clear from (22.1) that if g(z) =/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 fonnula for solving the a-problem in stricdy 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 Uis a neighborhood of i5, p E C 2(U), and the second differential (i.e. Hessian) of p is positive definite at all ~ E U. Hence grad P =I= 0. For a strictly convex domain, the Khenkin barrier function (see §4.2°) may be introduced as follows:

lao

<1>(~,z)= (P.~-z).

We set ci»(r, z) = {r, z)- p(n, By Taylor's theorem and the strict convexity of p, we have 2

for z, ~ E Therefore

p(z);;;;.. p(r)- 2Re <1>(~. z) + e 1 1~- zl D', D <& D' <& U, with a constant e1 > 0 depending only on D'. 2Re(~,

2

z);;;;.. p(r)- p(z) + e 1 1~- zl,

2Re ci»(~. z);;;;.. -p(r)- p(z) + e 1 1~- z(

(23.1)

Hence

I(~,z)l= O(I)Ici»{r,z)l,

(23.2)

jp(r)- p(z)l- p(r)- p(z) +lim (~. z)l +I~- zl 2 = O(l)j(r, z)j. (23.3) The constants in 0(1) depend only on D' and not on z, ~ ED'. For a (p, q)-form y; we define

( G:.qY )(z) =-Ii.q- 1(D, ak(~, z)y(r))(z)

-f.o,a,ak(~, z) 1\ y(r) 1\ P.~-l 1\ p.

1(

n

1_

)' Dp,n-p(az, ar). p . (23.4)

133

§23. WEIGHTED FORMULA FOR SOLVING THE a-PROBLEM

Here, 1-'~- I is given by Lemma 1.4, with w P(n/<~~(r, z ), and ak is defined by

ak(r,z}

= t =
=[ 1 _(<~~(r, _ z} )2nlk ={- _p(r)

<~~(r. z >



z 12 and u

=

)k[2n-I ~ ( <~~(r, z} )J]k -=--..>..::...:..._-<-

j=o

<~~(r. z)

(23.5} We have

arak(r. z)

::~: ::

= -k2na,_,(;, z)(

r-·1

<)(!. z)ilp(~'~:x)ii,o)(l'. z)

l·

(23.6)

From Lemma 4.1 and (23.6), .and the fact that 1-'~- I 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 pk(n; hence the operator G;.q can be applied to forms y such that pk-I(nY 0, and suppose y E z(Ip,q)(D) and 8 > 0 are such that 8 p y E L:p.qJ(D) and p8 -~1 2 ap 1\ y E L:p.q+Il(D). Then for all k;;;. 8 1 the

forms {3k

= G:.qY satisfy apk =yin D, and p8 -If3k E L~.q-I(D).

+

PRooF. Assume first that y E Z~.q(i)). Substitute ak(r, z)y(n in the Martinelli-Bochner-Koppelman formula (1.1). Since ak(r, z) 0 for rEaD, ak is holomorphic in z on D and = 0, we get

=

ay

y(z) = -I;,q( D, a.cak(r, z) 1\ y) - ai;,q-I(D, ak(r, z }y ).

(23.7)

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

1;,q( D, a.Cak(r, z) 1\ y)

= fo,arak(r, z) 1\ y(r} 1\ ~,q(r, z)

= f o,a.Cak(r, Z) 1\ y(r) 1\ (azl'~- I + arl-'!) 1\ p.1 ( n 1_ p ) 1. Dp,n-p( az, ar) = azfo,arak(r, z) 1\ y(r} 1\ I'~- I 1\ p.I( n ~ p }I. Dp,n-p{az, ar) + (-IY+q-I f

d[arak(r, z) 1\ y(r) (\I'! 1\

o,

I ( 1_ • )' Dp,n-p(az,

p. n

p .

an].

As already observed, the integrals have no singularities, so it is legitimate to bring outside the integral sign. Now, since arak(r, z) = 0 for rEaD and k > I, we see by Stokes' theorem that the second integral above vanishes.

az

134

VI. FORMS ORTHOGONAL TO HOLOMORPHIC FORMS

Hence (23.7) yields y = aG:.q'f· In the general case, we apply this equality to the form Yr(z) = y(rz) E Z~.q(D) and pass to the limit as r ..... I. We proceed to estimate the solution thus obtained. From (23.6) and the definition of G;,q, we have

J. ll- 1(z)lj(G;,qr)(z)jdvz = O(l)J. lr(OidvrJ. IP11 - 1(z)l D, Dr D,

· {lup,q- 1 (~. z)llak(~. z)l I «P(~ z) 12n-l } +lak-1(~. z)JI «il(~: z) l 2 n+ 1 1P(01Iar«P(~. z)jjp~-d dvz + 0(1) fvr lr
l~t~-ddvz.

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

lar«P(~. z )j =I (aP(O. ~- z )I= 0(1)1~- zl, and the definition of Jl~-l (see Lemma

1.4), we get

J. ll- 1(z)lj(G;,qr)(z)jdvz = O(l)J. lr(OIIl(OidvrJ. IP11 - 1(z)J D, Dr D, X {

«il(~, Z) lk; ~ _ z 12n-l + I «il(~, z) lk+! I~_ z l2n-4 }Jp'-l(z )ldvz

1

f.

II k I( I J. ll-l(z)ldvz I ) Aap(O +0(1) Dry(~ P- 0dvr D, l«il(~,z)lk+ll~-zl2n-3' Hence, by the same argument as in the proof of Theorem 21.4,

J. IPIJ-l( Z )lj( G:.q'f ){ Z )jdvz D,

= 0(1){1 + J.Dr lr(OIIl
§23. WEIGHTED FORMULA FOR SOLVING TIIE a-PROBLEM

135

We state without proof an application of Theorem 23.1. Let D = {z: p(z) < 0} be a bounded domain in en with C 2 boundary. We denote by Na.( D) (a > 0) the class of hoiomorphic functions Fin D such that

j0 lp(z}la.-Jin+IF(z}lda2 n(z} < oo (the Nevanlinna-Dzhrbashyan class). Here dak is the k-dimensionai LebesgueHausdorff measure, and In+ t = max{ln t,O}, t > 0.

en.

THEOREM 23.2. Let D be a strictly convex domain in Then an (n- I)dimensional analytic subset M of Dis the set of zeros of a function from Na.(D), a ;;;;, 0, if and only if

f.

M

lp(z)j"+ 1da2 n_ 2 (z}

= ~Yj/. j

lp(z)j"+Jda2 n- 2 (z} < oo,

~

where the Mi are the irreducible components of M, and yi is the multiplicity of~ inM.

CHAPTER VII

REPRESENTATION AND MULITPLICATION OF DISTRIBUTIONS IN IDGHER DIMENSIONS

§24. Hannonic 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 (November 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 R"+ 1 be the space of the variables (x 1, ••• ,x", y) = (x, y), and let R~+l,; {(x, y) E R"+ 1, y > 0}, then R" {(x, y): y 0}. We shall say that a function f defined in R~+ 1 has finite order of growth as y -+ + 0, if for every compact set Min R" there exist c > 0 and m > 0 such that lf(x, y) I< cy-m for x E M and 0 < y E;; 1. Let H denote the set of harmonic functions f in R~+ 1 having finite order of growth as y-+ +0, and K the set of harmonic functions in R~+l which can be extended continuously to R~+l U R" and vanish on R". Then K c H. Let K(x, y) be the Poisson kernel for the half-space R~+l:

=

K

_ f((n

(x, y)-

+ 1)/2)

7T(n+l)/2

=

y • {lxl2

+ y2)(n+l)f2

If T E E'(R"), then the harmonic function T*(x, y) = T,K(x- t, y) is called the Poisson representation of T. THEOREM 24.1. If T E E', then T* E H and represents T in the following

sense:

e~O £=eT*(x, y)rp(x)dx = (T, fP ), 137

138

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

Here T* extends continuously to R~+ 1 U (R"\supp T), and vanishes on R"\supp T, and T*(x, E) = 0(1 x 1-n-l) as Ix 1-+ oo. The Poisson representation of a derivative of T is the corresponding derivative of the Poisson representation ofT. This theorem generalizes Theorem 14.1, since (see §14) C"

= R2 "

«r(z) ly=e- «r(z) ly=-e = T*(xl, ... ,x2,-l, 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 TEE', it is bounded with respect to some seminorm

IICJ>II;=

ID 1cp(x)!,

sup

K2) suppT,

xEK,IIlli""m

and so

xEK. THEOREM 24.2. For each T E 6D'(R"), there exists an f E H which extends continuously by 0 to R~+l U {R"\supp T} and which is a harmonic representation ofT, i.e.

J

lim f(x,y)cp(x)dx= (T,cp), e... +0 y=e

(24.1)

Iff E H represents 0, then f E K. Every f E H represents some 1j E 6D' via (24.1). Thus 6D' ~ HjK. 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 frrst 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 0 C R", there exist an index p and an F E H which can be extended continuously to R~+ 1 U 0, such that f= a2P£fay2P. For n = 1, a similar lemma for holomorphic functions is due to Tillmann [1]. PRooF. Lett:. be the Laplacian in R"+ 1, and~ that in R". Given/harmonic in R~+ 1 , consider

1

F (x, y) =

JY F(x, 1J)d7J + cp(x). I

139

§24. HARMONIC REPRESENTATION

Then l:lF1(x, y) = 'iJf(x, 1)/'dy + .iict>(x). Choose q> E coo such that .iiq> = -'iJf(x, 1)/'dy in Rn (this is possible since the Poisson equation can be solved in Rn). Then l:lF1 0, and 'iJF1j'iJy =f. Iff E H, then there exist c > 0 and m > 0 such that

=

Hence F 1 E Hand

IF1(x,y)loEO;c 1y-m+l,

X

EO,O
At some step we can find an F2 E H which is bounded for x E 0 and 0 < y oEO; 1 and such that 'iJ 3Flf'iJ ys =f. The next harmonic primitive F E Hoff

will then be continuously extendable toRn+ 1 U

f

n, since

Using this lemma, it is easy to show that the limit in (24.1) exists for any E H and q> E 6j). Indeed,

fy=e f(x, y )q>(x )dx = 1y=e -ay

(j2P 2

F{x, y )q>(x) C/x

p

= (-1)P £=e.iiPF(x, y )q>(x) dx = (-1)P1

y=~t

F(x, y).iiPq>dx- (-1)P e .... +O

1F(x,O).iiPq>(x) dx. R"

Consequently, since 6j) is weakly dense in "D' (Shilov [1], Chapter II, §9.3), the above limit dermes a certain distribution 1i E GD. Suppose now that, for some f E H, the limit in (24.1) is 0 for all q> E "D. Then, by Lemma 24.3, for each bounded domain D C Rn we have an F E H and an index p such that 'iJ 2PFj'iJy 2P = f and F extends continuously to R~+ 1 u n. Since f represents 0, .iiPF( x, 0) = 0 in 0 in the sense of distributions. Hence it follows from the ellipticity of .iiP in 0 (see, for example, Schapira [1], Theorem A.22) that F(x, 0) is real analytic in 0. Hence F extends in a coo way to 0 (see Brelot [1], Chapter I, §8). Then

f(x, y) = (-1)P.iiPF(x, y) =:(-1)P.iiPF(x,O) = 0, asy-

+0.

LEMMA 24.4. Let f be a po/yharmonic function of order m (i.e. l:lmf= 0) defined in R~+ 1 • Then

f(x, y)

=fo(x, y) + Yf1(x, y) + · · · +ym-lfm- (x, y), 1

{24.2)

140

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

h·

where the j = 0, I, ... , m - I, are harmonic functions in R"t+ 1 ; iff has finite order of growth as y ..... +0, then the h lie in H, and iff E C 00 (R"t+ 1 U 0) for an open 0 c Rn' then the h are ceo in Rn+ I u n.

The proof is by induction on m. If m = I, (24.2) obviously holds. Let N"f = 0, m > 1. Then we must have f = IQ'- 1yih· which would give us m-2 [ A/= -~ yi J=O

u + 2)(j + l)h+2 + 2(j + 1) 'd/;r·y ] .

By the induction hypothesis, since Af is polyharmonic of order m - I, m-2 A/= I yicpj, j=O

so that we obtain the equations cpj = (j

+ 2)(j + l)h+2 + 2(j + 1) a~;·'

j

= 0, .. . ,m- 2,

for the h; in particular,

_ 2(

cpm-2-

) 'dfm-1

m- 1

a:Y""

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

J.y 1 fm-l(x, y) = 2(m _I) 1 cpm_ 2(x, Tl)dTI

+ !f(x).

Since

Afm-1 = 2(m 1_ 1)

a~~- 2 (x, 1) + a!f(x) = 0,

we find t/1 from this equation. Next, from the equation cpm-3

= (m- 1)(m- 2)/m-1 + 2(m- 2) a~n; 2

we determinefm- 2 in a similar way, etc. It is clear from the proof above that the~ lie in H iff is of finite order of growth, and that~ E C 00 (0 u R~+l) for allj if IE C 00 (0 u R~+ 1 ). CoROLLARY

f(x, y)

-+1j E

24.4. If Amf = 0 and f is of finite order of growth as y 6j)' in the sense of (24.1) as y ..... +0.

-+

+0, then

Thus, polyharmonic functions also represent distributions in 6D', and in this case 1j 1j0 •

=

§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* jK* = GD* the space of hyperdistributions, and denote by f* the image off E H* in GD*, i.e. f* = f + K*. The map A: Gj)' --+ Gj)* defined as 1j--+ /* (fa harmonic representation ofT) is an imbedding, since K C K*. Hence we shall suppose GD' C GD*. Let T, S E GD', and let f and g be their harmonic representations. Then by the product To S of T and S we shall mean the hyperdistribution ( fg )* = fg + K* E GD*. 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 cp E GD. 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. SoT is well-defined, i.e. the class fg the representatives f and g.

+ K*

is independent of the choice of

PRooF. It suffices to show that, iff E K and g E H, then fg E K*. Let cp E GD. It is easy to show that f = yf1 in a neighborhood of the support of cp, where / 1 is a real-analytic function. Hence fcp E 6j) for fixed y, and fcp --+ 0 in 6j) as y --+ +0, while g tends to Sin Gj)' as y -+ +0; consequently, by a lemma in Shilov [I] (Chapter II,§ 9.3), lim

e~

+0

J.

R"

f(x, t)cp(x)g(x, e)dx:;:: (S,O)= 0,

i.e.fg E K*.

=

2. supp(T o S) c supp T n supp S. In particular, To S 0 if supp T n suppS 0. Here, the support of a hyperdistribution is defined as follows. We say that f* E Gj)* vanishes in an open set gin Rn if, for any cp E 6j) with support inn,

=

lim

e~+o

J

y=e

f(x,y)cp(x)dx=O,

/Ef*.

Now the support supp f* off* 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 cp E 6j) with supp cp C Rn\(supp T n supp S) = 0. We may suppose that, for example, supp cp n supp T = 0 (by means of a

142

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

partition of unity it is always possible to pass to functions cp with arbitrarily small support). Then f can be written in the form y/1 in a neighborhood of supp cp, and the rest of the proof proceeds as for Property 1. 3. If T E Gj)' is defined by a function T(x) E LP(loc), and S by S(x) in U(1oc), 1/p + ljq 1, then To S T(x)L(x) E L 1(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 asy-+ +0, and for each compact setM

llu(x, Y)- S(x )II~-+ 0 as y

--+

+ 0, where 1/q

11/ll:!t = ( JM l!lq dx )

q ~ 1.

,

To do this, we must use: (a) Theorem 1 of Stein [2], (Chapter III, §2.1), asserting that, for every f E U(R"), its Poisson integral defines a harmonic function j(x, y) such that j(x, y) converges to f(x) almost everywhere and llj(x, y)- /(x)llfr.-+ 0 as y--+ +0; and (b) the procedure in Bremermann [1], §5.9, replacing power-series expansions by expansions in harmonic polynomials. Since S(x) E U(loc) and T(x) E LP(loc), we will have S(x)T(x) E L 1(loc). Let u( x, y ), v( x, y) and w( x, y) be the respective harmonic extensions to R++ 1 with the properties mentioned above. Then u, v and w are harmonic representations of T(x), S(x) and T(x)S(x). Furthermore uv- wE K*. Indeed, let cp E Gj) with supp cp C M. Then

l!r

(uv-

w)cp dxii!O; c J)uv- wldx < c(lluv-

STII~ +llw- STII~ ),

but lluv- STII~-+ 0 asy-+ +0, since llu- Til~ and llv- SIIX, tend to 0. 4. The derivatives of hyperdistributions are defined as follows:

_a r = (_i_t)* =X+ K:. axk

axk

axk

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

_i_(ToS)= axk

aT axk

oS+

To

as.

axk

§25. MULTIPLICATION OF DISTRIBUTIONS

143

This follows immediately from the fact that if I represents T, then ofjoxk represents oTjoxk. 5. The product of distributions is compatible with multiplication by multipliers, i.e. if a E C""(Rn) and T E "D', then a o T =aT. PRooF. Since a E C"", its harmonic representation g(x, y) E H extends continuously toR~+ 1 U Rn (solution of the Dirichlet problem for the half-space). Hence g E C""(R~+ 1 u Rn) (Brelot [1], Chapter I, §8), so that q>(x)g(x, y) ..... q>(x)a(x) in 6j) as y ..... +0, for any q> E "D. On the other hand, f(x, y) -+ Tin "D' asy-+ +0, so that, by the lemma already mentioned (Shilov [1], Chapter II, §9.3),

~=/(x, y )q>(x)g(x, y )dx e=-:o (T, aq>) =

(aT, q> ).

6. Iff E His a homogeneous function of degree p, then its boundary value as y -+ +0, i.e. Tj. is also homogeneous of degree p. Indeed, (Tj,q>(tx))= lim

e-+0

=

t

J

y=e

f(x,y)q>{tx)dx=

p~n e-+0 lim J f(x, y=et

lim~~ !(~.l)q>(x)dx t y=et t t

e-+0

y)q>(x)dx =

t

p~n (1j, q> ).

If T E 0~n 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, ty) =

(T.,, K(u-

tx, ty))= tin ( Tu, K(

= tP(T.,, K(u-

X,

!j-- x, y))

y))= tPT*(x, y).

By means of the general procedure for constructing a harmonic representation, it can be shown that, if T E 6J)' is homogeneous of degree p, then T admits a harmonic representation which is homogeneous of degree p. It follows that, if T, S E "D' are homogeneous of degrees p and q respectively, and To S E "D', then To Sis homogeneous of degree p + q. 7. Let x' = (x 1, ••• ,xm), x" = (xm+l•· .. ,xn), and T E 6D'(Rm), S E "D'(Rn-m). Let f(x', y) and g(x", y) be harmonic representations ofT and Sin R~+ 1 and R~-m+ 1 respectively. Then

(! + K x') · ( g + K x")

C h

+ 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.

144

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

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

~=Eg(x", y)q>(x)dx"

= o/(x', e),

'P E 6D.

We have 1f(x', e) E 6D(Rm) and o/(x', e) ..... ( S, q>(x)) = o/(x') in 6D(Rm) as e ..... +0. Since o/(x', e) ::4 o/(x') as e ..... +0 (Lemma 24.3), and D;o/(x',

e)= ~=Eg(x",

y)D;q>(x)dx",

we have, by the same lemma, D;o/(x, e) ::4 D;(o/(x')) as e ..... +0. Now we apply the lemma from Shilov [1] (Chapter II, §9.3) once again and conclude

~=/(x', y)g(x", y)q>(x)dx = ~=/(x', y)o/(x', y)dx' E-=-:c, (Tx'• o/(x')) = (Tx.,(sx"•q>(x)))= (TXS,q>(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 ID1 of functions, generated by functions of the form

f

Pk(x,

)

( X,

Y = {

lxl

2

+y

y)

2 }k-m+(n+l)/2'

k~O,m~1.

(26.1)

where Pk(x, y) is a homogeneous polynomial of degree k and ll"'Pk = 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.

LEMMA 26.1. Suppose f is a function of the form (26.1) lying in ID'l, and k + 1 - 2m > 0. Then

1 {I R"

Pk(x,y)x 4 dx X 12

+ y2}k-m+(n+l)j2 =

for all monomials X 4 with II all< k PROOF.

0

+ 1-2m.

Let

I(y) = [

Pk(x, Y )xadx

lxl
{I X 12 + y2t-m+(n+l)/2'

II all< k

+ 1-2m.

145

§26. EXAMPLES OF PRODUCTS

We shall show that the limit lim /(y) = c

y-+0

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

f= fo + Y/1 + · · · +ym-lfm-1; then m-1

lim l(y) = ~ y- +0

According to Lemma 24.4, the~ are that the limit lim

1lxl< ~(x, y )xa dx.

lim yi

j=O y-+ +0

1

I

coo on R~+l

y-+0 lxl
U(Rn\{0}). We now show

~(x, y)xadx

exists. By Lemma 24.3; there exists .fj(x, y) E H extending continuously to R~+l U {x ERn: I xI< 2} and such that

a2p , .fj(x, y) =~(x, y). ay p

-

2

Then~(x, y) = (-l)P~P.fj(x, y), and

1lxl(x)dx Xk

= -1

lxt
.fj(x, y) aa"' dx

+ (-l)k-11

lxt=•

xk

Fj'Pdx

[I].

Thus

1lxl""l ~(x, y)x dX = (-IY1lxl
0

where Dis a differential operator, and da the surface element of the sphere. We need only observe now that the .fj, like the ~. are coo on R~+l u ({I xI< 2}\{0}) (see the proof of Lemma 24.3). Thus the limit lim /(y)=c

y-+o

exists. We now transform the variables in J(y) by setting x =

1

I(y) = ~ ys s>O

1

0 (r2

rllall+n-l+k-s

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

dr

1

1~1=1

r~, I~ I=

Pt( ~)~ 0da

1. Then

146

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

= }";s>O y•Pt(x ). If we further set r =yt, then

where Pk(x, y)

I(y) = yliall-k-l+2m1

1/ Y

0

I llall+n-J +k-sdt

~

(1 + t2)k-m+(n+l)f2

s;;.O

1

1~1=1

Pta)~ada. (26.2)

Since I(y) has a limit asy--+ +0, we necessarily obtain from (26.2) that J

= y-+0 lim I(y) = 1 0

00

1nall+n-l+k-s

~

(1 + t2)k-m+(n+l)f2

s;;.O

dt

1

1~1=1

Pt(~)~ada

= 0.

The improper integral above is convergent, since II a II < k + 1 - 2m. It is easy to see that

f.

R"

LEMMA

Pk(x, y)xadx

(I X 12 + y2t-m+(n+l)/2

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

26.2. Suppose an f of the form (26.1) lies in 9R and k + 1 -2m ;a. 0.

Then I - pv -

f.

•• R"

Pk(x, y )xadx

(I X 12 +y2t-m+(n+l)j2 = ca•

II a II

=k + 1 -

2m,

where ca

= s~J

w"l2f(s- l/2)ak+l-•-m(pf•-l(x)xa} 22(k+l-s-m)(k + 1 - s- m)!f(k- m + (n + 1)/2).

(26.3)

PRooF. Consider the integral

() 1

Pk(x, y )xadx .

I y = . . lxl
Transforming to polar coordinates as in Lemma 26.1, we get

I(y)

= ,r;;.O ~ 1 0

Jjy

(1 + t 2)k-m+(n+l)/2 11£1= I P[(~)~adCJ. 12k+n-s-2mdt

The polynmials Pt have degrees k - s; hence the Pt~a have degrees 2k + 1 -2m- s. Thus, if sis even, the degree of the homogeneous polynomial Pt~a. is odd, and so

Then,

147

§26. EXAMPLES OF PRODUCTS

Clearly I(y)- I asy- +0, so that

But 1zk+n+ l-2s-2mdt

00

fo

(I

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

1 ( 2

n

I) 2 '

=-B k-s-m+I+- s-2'

and

r

pfs-l~ada

JIEJ=I

2'1Tn/2'f:.,k+ 1-s-m( pk2s-l(~)~a) = -;;-:-:-::----:-;------~___::.-..:...:...:_:__::....._ ____

22(k+l

s

m>(k +I-s- m)!f(k +I-s-m+ n/2).

(This follows from the Gauss representation of homogeneous polynomials; see Sobolev [1], 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 all.;;; k + 1-2m. Let

h(x)-

Qk(x)

- 1X 12k-2m+n+ I

'

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

1 Qk(x)4p(x)dx IX12k-2m+n+l


R"

= p.v.1R" Qk(x)(qJ(x)- ~llall"'k+l- 2 m(l/a!)(a~;axa)(O)xa)dx 1X 12k-2m+n+ I

if k + I - 2m ;;a. 0, where the abbreviations "f.p." and "p.v." signify "(Hadamard) finite part" and "(Cauchy) principal value". If k + 1 - 2m < 0, then

qJ

E

GD.

In other words, his the Hadamard finite part of the integral

1h(x)qJ(x)dx. RN

148

VII. REPRESENTATION AND MULTIPLICATION OF DISTRIBUTIONS

LEMMA

26.3. Let Q be any polynomial in Rn. Then

Q(x, y) = ~ (-I}'y2'~Q(x} f;;.O (2/). is a harmonic extension of Q to all of Rn+ 1

This follows from direct computation. If

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

li(x

)-

Qk(x, y)

'Y - (

2

lxl + Y

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

in IDl. THEOREM

26.4. The distributions h and Th- are identical.

This follows immediately from Lemmas 26.1-26.3 and the definitions of h and T;. Let 9R 1 (respectively IDl 2) denote the subset of 9R 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( 9R 1 + 9R 2) = 9R. Theorem 26.4 implies that the boundary values of functions f in 9R 1 in the sense of (24.1) are the distributions (f(x, 0)) A.

THEOREM

26.5. Iff E IDl 2 , then

1j = {

(-l}k+l+2m

I

C~ 8(a), llall=k+l-2m a.

Ca

m ;a. 0,

ifk +1-m< 0,

0, where the 8-function.

if k + 1 -

are defined by (26.3) and

8(a)

= 0°8jox 0 , 8 being the Dirac

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 9R. ExAMPLE 1. Letf(x, y) = y/(1 x 12 + y 2)l 2. Then

1j = w
§26. EXAMPLES OF PRODUCTS

149

We now consider the question: when is the product of two functions in [Q again im \JR? Let Pk(x, y)

then fg = (jxj 2 + y 2)i+k+(n+l)/2-(m+s-(n+l)/2)

•

Hence we see thatfg E [Q if and only if n is odd (n + 2p- 1), m + s;;;. p, and 6_m+s-p( PkQ1) = 0. Thus the class of such functions f and g is fairly large.

26.6. Suppose f, g and fg lie in \JR. Then: a) iff, g E \JR 1, thenfg E \JR 1 and(f(x,O))A o (g(x,O))A = (fg(x,O))A; b) iff, g E \JR 2 , then 1j o Tg = 0; c) iff E \JR 1 and g E \JR 2 , thenfg E \JR 2 and

THEOREM

(-1)k-2p-2mj(x,O)o

~

:~8=

llall=l+l-2s ·

~

~8