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2) from the theory of functions of a single complex variable. To prove this assertion, we shall exhibit a single counterexample in C2 which is a slight modification of the example presented by Bochner and Martin ([31, p. 91). Let us show that every function f(z) that is holomorphic in the semihollow sphere
Q:
[Z=(ZI'
Z2):{o, is a subharmonic function, the function uP (z) (for
p> I) is also subharmonic.
This assertion follows from Holder's inequality
uP(z)
< [ ;"
l
< 2~
u(z+re IS )d8]P
i
uP (z+re'S)d6
and the test of section 9.4. If u (z) is a. subharmonic function, the function eU (Z) is also subharmonic. This assertion follows from the inequality between the geometric and arithmetic means: e J p In /
dx ..... ./
f pi dx.
1'f
f
P d x= 1•
...... 0 Po?
I V' .... 0
(see [101, p. 167) and section 9.4. Specifically. eU(Zl<ex p
!;" l
u(z+re iQ )d8
1< 2~
< i" J exp (u (z+ re lS»dO. o
The converse is not true (see section 9.15).
12. Jensen's inequality Suppose that f(z) is hoiomorohic in G. Then. the functions In 1f(z) I. In+ ! f(z) 1 and I f(z) IP (p ~ 0) are su/Jharmonic in G. Proof: Since
I/(zW=ePlnl/(Zll. In+l/(z)l=max(O. Inl/(z)l>.
66
PLURISUBHARMONIC FUNCTIONS AND PSEUDOCONVEX DOMAINS
it will, by virtue of section 9.11, be sufficient to prove the above assertion for the function In I/(z)l. But this function is subharmonic according to the test of section 9.4 since it does not take the value Xl; it is upper-semicontinuous in a; and, it satisfies inequality (8) for all sufficiently small r -< ro (z) for z Ea. The last assertion follows from the following considerations: Either In If (z)1 = - 00 and inequality (8) is trivally satisfied or Inl/(z)I=Reln/(z) >-ex:; is a harmonic function in some neighborhood of the point z and inequality (8) is satisfied by virtue of Eq. (5). Thus, if a function 1 (z) is holomorphic in S (0. r), Jensen's inequality
+
2_
Inl/(z)l<j P(z. re I8 )lnl/(re I8 )ld8 o
follows from inequality (7).
13. The approximation oj subharmonic junctions Por u(z) to be a subharmonic function in a domain G, it is necessary and sufficient that u (z) be the limit of a decreasing sequence of subharmonic functions Ua(Z). for a = 1.2•...• in the class C(CD)(a:,), where the domains Gil are Ga = G. such that Ga C Ga + 1 and
U a
Proof: Sufficiency of the above condition follows from section 9.6, Let us prove the necessity. If u (z) = - 00, the required sequence is u. (z) = - a. If u (z) ¢ - 00, u (z) is locally summable in a (see section 9.9). Suppose that a function w(Izl) defined in D possesses the properties:O<w(lzl) w(jzl)=O forlzl~ I; I
2"11:
f w(p)pdp= 1
(16)
o
(see section 3.1). We construct a sequence of mean functions: u.(z)=
f u lz + :' )w(lz'l)dx' dy' = = f (z')0l(,;lz-z'I>:x dX'flY'= = / rw(n [Iu (z +~ej~)dO Jdr = 2
U
(17)
I
=27./Jl~·' "
z; Il)rw(r)dr.
a=1.2 • ....
Since the function u (z) is subharmonic in a. inequalities (17) and the properties of the function 1D{lz/) imply, by virtue of the test of
SUBHARMONIC FUNCTIONS
section 9.4, that the functions differentiable in the open sets
o.=[z:
u. (z)
67
are subharmonic and infinitely
Aa(Z»~. zEO].
where O.CO.-IlandUO.=O .
•
As our set 0., we take that component o. to which the fixed point of the domain 0 belongs. Let us show that the monotonically decreasing sequence of functions [u.! converges to u(z) at every point zEo. If zEO, then S(z. p)eO for some p > O. But then, there exists a number N>- 1 such that S(z. p)@O.for all a.>-N. Since the function J (rjll, z; u) decreases monotonically with increasing a. (see section 9.8), it follows from the last term of Eq. (17) that the sequence u. (z). for a. ~ N, decreases monotonically. On the other hand. the test of section 2.2 tells us that. for arbitrary positive e. there exists a positive number 8 (where (8 ~ p) sucb that u (z') < u (z) I! for all Iz' - z I < 8. From this and from (16 and 17). it follows that u. (z) < u (z) +e for all sufficiently large «. This. together with tbe inequality u (z) ~ u. (z), proves that lim u.. (z) = a (z) for aU z EO, which completes the proof.
+
--:toco
14. A property of positivity of subharmonic junctions If u (z) equality.
~
-
00
is a subharmonic function in a domain G, it satisfies the in-
02U
(z)
4 --_- = tJ.a (x. y)
oz oz
>- o.
zEO.
(18)
Conversely, if u E D·(o) and satisfies inequality (18) in G. then u(z) is a measurable function and there exists a unique subharmonic function in G that coincides with u(z) almost everywhere in G.
Remark: Differentiability and positivity in inequality (18) must be understood in the sense of generalized functions: Au >- 0 means that
f a tJ. 'f' dx dy >- 0 for arbitrary 'f' (z):? 0 in D (0) (see section 3.4).
Proof: Suppose that a (z) is a subbarmonic function in O. Since u (z) '¢ - 00, u (z) must, because of section 9.9, be locally summable In O. Hence, u (z) defines a continuous linear functional u over D (0)
f
by the formula (a. 'f')::::: u'f'dx dy (see section 3.2); that is, u ED' (0). Inequality (8) for the function u (z) can be rewritten in terms of generalized functions in the form of the convolution (see section 3.3)
68
PLURISUBHARMONIC FUNCTIONS AND PSEUOOCONVEX DOMAINS
(19) where the generalized function p., is defined by the equation 2.
2~
(p." 'f') =
J
'f'ED(O),
'f' (re li ) da,
o
But since
(fl';:;-~. 'f) = ;2 [21"l'f(retU)da-'f'(O)]-ill'f'(O) for arbitrary 'f' ED (0). it follows that flr -
~
1
~
-,-.- - 4" Ilu.
as
r-+ O.
(20)
Using the continuity of the convolution (see section 3.3), we derive inequality (18) from (19) and (20):
Conversely, suppose that Il (z) ED' (0) and satisfies inequality (18) in the domain O. Then, IlU defines a positive measure in 0 (see section 3.4). We denote this measure by /1.. Thus, Il satisfies Poisson's equation 1l1l=P.. We denote by E(z)=(I/:.1,,)lnlzl thefundamental solution of this equation, IlE = 0, and we denote by p.O' the restriction of the measure (l to an arbi~rary subdomain 0' c: O. Then, u (z) can be represented by Riesz' formula in A' (see section 3.10): u=E*p.o'+Vo',
(21)
where Vo' (z) is a harmonic function in the region 0'. Furthermore, since the function !Ilnlz- z'lldxdy
o·
is continuous with respect to Zl in 0' and since the measure P.o' defines a continuous linear functional over the space of functions that are continuous in 0', the iterated integral
exists and is finite. Then, we conclude from Fubini's theorem (see
69
SUBHARMONIC FUNCTIONS
section 2.9) that, for almost all z Ea', the integral
f
2~
Inlz-z'l dp.(z')=E*p.a·,
0'
exists and is finite. This integral is a summable function on 0'. From this and from (21), it follows that the generalized function u is locally summable in A' and can, for almost all z Ea', be represented in the form u (z) =
2~
I
In jz - z'ldp. (z')
+ Ua·(z).
(22)
0'
Let us show now that the right member of Eq. (22) is a subharmonic function in the domain O~ We denote it by 1'1 (z). Since, for every z Ea'. the sequence of continuous functions In (Iz - z' I E) approaches In Iz - z'l monotonically from above as E ~ 0, it follows on the basis of Levi's theorem (see section 2.7) that
+
1'J(z)= lim u.(z)+Uo·(z).
+
(23)
'-+-10
where u,(z)=
ill Iln(lz-z'I+E)dp.(Z')' a'
The functions u, (z) are continuous in measure Jl. > 0 and the function
a'.
Furthermore, since the
In(lzl +e)= sup Inlz+CI I CI'"
is subharmonic (see sections 9.6 and 9.12). we conclude, on the basis of the test of section 9.4, that the function u, (z) is subharmonic in A' and that it decreases monotonically as E -+ + o. Remembering that U a · (z) is a harmonic function in a', we conclude from equation (23) (see section 9.6) that the function 1'J(z) is subharmonic in a'. Thus, the function u(z) coincides (almost everywhere) in an arbitrary subdomain Q'~O with a subharmonic function. This means that u (z) coincides almost everywhere in a with a subharmonic function. From section 9.10, this subharmonic function is unique, which completes the proof.
15. Logarithmically subharmonic junctions . A nonnegative function u (z) is said to be logarithmically subharmonic in a domain a if the function In u(z) is subharmonic in O.
70
PLURISUBHARMONIC FUNCTIONS AND PSEUDOCONVEX DOMAINS
On the basis of section 9.11, the function u (z) Is also subharmonlc. Examples of logarithmically subharmonic functions are the absolute values of holomorphic functions (see section 9.12). THEOREM. For a nonnegative function u{z)to be logarithmically subharmonic in a domain G, it is necessary and sufficient that the function u{z) I e az I be subharmonic in G for all complex a. PrOOf: The necessity of the condition is obvious since the function In u (z)+ Re(az)is sliliharmonicfor all a and, by virtue of section 9.11,
the function u (z)leaZI is also subharmonic. To prove the sufficiency, note that the function In u (z) < is upper-semicontinuous in a since
+
00
and
lim u (z) == u (zO) == 0 = lim u (z) ......... HIt
and hence the nonnegative function u (z) is continuous at points zO at which u (zO) = O. We introduce the decreasing sequence u.(z)=
f u(z+:' ) 1D(lz'\)dx'dy'++.
«== l. 2••.•
of positive infinitely differentiable subharmonic functions in the corresponding domains a... where a.cad l and Ua.=a (see
.
section 9.13). It follows from the equation u. (z)
=
leaz I =
f u( z+ z: )1/ (z+:') 1111 (lz'l> e
-Re
(II:') dx' d.v'+~1 eazl
and the hypothesis of the theorem that the function u. (z) I eaz I is subharmonic in a. for all «. Then, by virtue of section 9.14, we have
+
for all b and e, where a = b Ie. From this, taking the minimum over b and e and using the fact that u. ~ 1 / II is positive, we obtain the inequality
ou,. a.b.u.- (ax
)1 - (dY ou,. )1 ~O.
On the other hand, we have 1 [ u. b.u. - ( h iJu. 11 In u. = u:
)1 - (T, ou. )'] .
SUBHARMONIC FUNCTIONS
71
Therefore. fJ.ln u. ~ 0 in 0 •• From this. we conclude on the basis of the test in section 9.14 that the function In u. (z) is subharmonic in 0 •• Since u. -+ u, monotonically decreasing (see section 9.13), as 11 00 , it follows that In u. _In u. also decreasing monotonically. Therefore, the function In u is subharmonic in a (see section 9.6). which completes the proof.
16. The trace of a subharmonic junction on a Jordan cunJe Suppose that a function u (z) is subharmonic in a domain G and that L = . [z: z' = z , 0 ~ t ~ I] is a Jordan curve lying entirely in G. Then. lim
u [z (t») = u [z (0»).
1-+0. '".0
(24)
(See Oka [75] and Rothstein [66].) Proof: Without loss of generality. we may assume that z (0) = O. Let us denote the left member of Eq. (24) by a. Since the function u [z (I)] is upper-semicontinuous for t E[0. I) (see section 9.1). we have a
-< lim u [z (I») = 10+0
II
(0)
< + 00.
Let us suppose that a lies on the circle I z.1 " r and that the curve z = z (I). for 0 -< I -< 10• lies entirely inside the closed circle I z I = r. Suppose that p < r. We denote by Lp that portion of the curve L that lies between the points z (to) and z (I p)' where z (I p) is the last point of intersection of L with the circle Iz I= p (as we move along L to the point 0 [see Fig. 11]). We denote by D, the circle lei < r deleted by the curve L,. Since L is a Jordan curve. D is a simply connected domain and all points l o its boundary dD, are accessible from D (see section 1.5). Here. points of curve L' are double points. We denote by d'D, thJ Fi8. II boundary of D, with the multiplicity of its points taken into consideration 1see secfion 7.3].
72
PLURISUBHARMONIC FUNCTIONS AND PSEUOOCONVEX DOMAINS
Let us consider the function z =
that maps the open circle D, in such a way that cop (U) = 0 and wp (0) > O. According to Caratheodory's theorem (see W.
(e)
Ie 1< 1 single-s~eetedly and conformally section 7.4), i)'Op'
w.
on~
maps the circle IeI-< I in a one-to-one manner onto
lei < 1. c is holomorphic in IeI < 1 and is con-
Here, the function "\(C) is continuous in
Thus, the function
"'p (C) /
tinuous and nonzero in
ICI = I.
Therefore. the function In
is harmonic in ICI < I and continuous in ICI (5) to it, we obtain In
I'\(;) I
< 1. If we apply formula
Iw; (0) I=
2.
2.
r
f '' I
="2 1
(/8) dO="2 1It ;, eli
In
"
-p--
1>
In I wp(e IB ) Id6.
(26)
We denote bv Ip the preimage of the curve Lp C d'Dp under the mapping z = Wp (e) and we denote by 21ta p the measure of the set Ip (on the circle ICI= I). Obviously,
I wp (e) I > p.
eEIp; IlI)p (C) 1-< r -< I. IeI = I.
From Eq. (26) and the above inequalities, we obtain the inequality In
III); (0) I> a
p In
(27)
p.
On the other hand, since the function Olp (e) does not assume the value p in the circle lei < I, it follows, on the basis of Koebe's theorem (see section 7.4) that Iw' (0) I -< 4p. From this and from inequality (27), it follows that P I-a 2 1 P
P9'4"'
Since a p -< I, this inequality means that a p -+ I as p -+ + o. The function u [wp (e)) is subharmonic in the open circle I e I < I (see section 10.11) and upper-semicontinuous in the closed circlel 1: 1-
PLURISUBHARMONIC FUNCTIONS
75
plurisubharmonic. Examples of logarithmically plurisubharmonic functions are the absolute values of holomorphic functions. THEOREM. For a function u(z) .? 0 to be logarithmically plurisubharmonic, it is necessary and sufficient that the function u(z) I e az I be plurisubharmonic (or all complex vectors a.
This follows from the theorem of section 9.15. We conclude from this theorem that the property of logarithmic plurisubharmonicity is conserved under addition and multiplication by a positive number. Therefore, the function V(z)=
f uP(z; ~)dp.m,
Is logarithmically plurisubharmonic if the function u (z; 0 is logarithmically plurisubharmonic with respect to z for every ~ and uP [s summable with respect to the measure p. 0 (under the assumption that V (z) is upper-semicontinuous).
>-
6. Jensen's inequality If a function u (z) is plurisubharmonic in then, for all z ES (zo. r), 21t
U (z)
-'O!l1
•
U o (~)
=
(re iO ) •
lim u. (re/~). «-++.:0
Then, on the baSis of (33) and (35), we have. for z E:~ (0, u.(rel~)
-00.
It then follows from (37) and
° from section 2.8 that 0
E. _ 0, monotonically decreasing. as (,t 00. We choose for N = N (E. rO) the smallest integer such thatM n (r. rO) eN - N. On the other hand, if 21t
2n:
f ... f uu(O)dO=-oo. °
°
then 2ft
1).
=f
2.:
... f u: (6) dry -
°
-
00.
a-
00.
°
monotonically decreaSing, as a _ 00. By using inequalities (36), (37), and (32) and Eq. (31), we obtain, for all z ES (0. -rO), 2.
u.(z)'<M
2.
f ... f Pn(z. re/~)dO+
°
2.
+f o
0
21l
... JPn{r.
e i O)[u:(6)-Mjd6- 1;
I
02.
f
w(p)p2n- 1 dp= 1
I)
(see section 3.1). We define the required sequence of functions u. (z) for 11 = I. 2.... , as in (17): u.(z)=
f
u
(z + ~ )w( Iz'l)dx' dy' =
=
.r
=
f r2n-
u (z')w (a. iz - z'l )112n dx' dy' =
I
o
Iw
(r) [
f u (z + ~ a) daJ dr f J(~.
=
1'1= 1
1
z; u)r2n - 1w(r)dr.
=02n
o
The monotonic decrease of the functions u. (z) follows from the theorem in section 10.8. Remark: If a function u (z) is logarithmically plurisubharmonic, the sequence u (z) constructed in the theorem consists of logarithmically plurisubharmonic functions. This follows from the theorem in section 10.5 (cf. also the proof of the theorem in section 9.15).
80
PLURISUBHARMONIC FUNCTIONS AND PSEUDOCONVEX DOMAINS
10. The property ojpositivity jar Plurisubharmonic junctions If u(z)
~
-00
is a plwisubharmonic function in a domain a,the Hermitian form ~ d 2a ~--_- ajak=(!f(zj u)a. a)
J•• dZJ oz.
is positive in Gl that is, for an arbitrary nonnegative function arbitrary vectors a,
cp
€
D (0) and,
~! u(z)--_-dxdyajak"';pO. d2'1'Cz) -
~
l.k
dZjOZll
Conversely, if u € D*(O) is such that the Hermitian form (H(z; u)a,li) is positive in a, there exists a unique function that is plurisubharmonic in G and that coincides with u(z) almost everywhere (cf. section 9.14). Proof: If U (z) =1= - 00 is a plurisubharmonic function in a, there
exists, by virtue of the theorem in section 10.9, a sequence of plurisubharmonic functions Ua (z), for II = 1. 2, ...• in the class C(OO) (Oa)'
where 0ac.0a+ 1 and UOa=O such that ua(z)_u(z), de-
creasing monotonically. Therefore. if we use the definition of plurisubharmonic functions and the theorem in section 9.14, we conclude that d2ua (z+Aa) (!f(z' ")a. a)= • a d). dJ..
I
.......
A=O -:P'
0
•
zE a•.
Since "a(z)-u(z)in the sense ofD'(O), if we take the limit as a_oo in the inequality that we have just obtained and make use of the continuity of the differentiation operator in D' (a) (see section 3.3), we see that the form (H (z; u) a. li) is positive in O. Conversely, suppose that u ED' (a) and that the form (H (z; u) a. a) Is positive in O. Then, Au> 0 in the sense of D' (a). In accordance with the theorem in section 9.14 (more preCisely, in accordance with the analogous theorem for the 2n-dimensional case), the function u (z) coincides almost everywhere in a with the unique subharmonic function v(z). It then becomes clear that (H(z; v)a. Suppose that the function w( Izl) satisfies the conditions of section 10.9. We construct a sequence of mean functions
a»O.
V.(z)=
/v(Z-T- :')~(IZ'I)dx'dY'=
=
f
(40)
v(z')w( O. q. e. d.
12. Analytic surfaces A set FcC" is called a 2k-dimensional analytic surface if it is defined in some neighborhood of every point zO EF by equations of the form
where the functions matrix
zJ (zO; A)
are holomorphic and the rank of the
(j= 1. 2••..• n.
is equal to k
s= 1. 2•.•• , Ie)
in the corresponding neighborhood of the point
),=0, zO=z(zO; 0).
It follows from this definition that, in a neighborhood of a point zO EF, a 2k-dimensional analytic surface F is defined by the equations
Is (zO; z) = O.
s= 1.2 ..•.• n-k.
where the functions Is (ZO; z) are holomorphic and the rank of the matrix of the 01.1 oZJ is equal to n-k in that neighborhood (cf. 7.2). In accordance with the definitions of section 1.4, an analytic surface is a surface of class C 0, for j > 1,2, .... n,itisnecessaryandsufficientthatthefunctionu(z) = R(jZll, ... , I Zn i> be plurisubharmonic in the multiple-circular domain
Proof: Either the function R (r) is continuous in B or R (r) (see section 11.1). Furthermore, taking
we obtain ~=..!.- a2~ +_1_~. iizj iiz,
4
ar,
4'j
ii"
a·u_ aZj aZIl
(} =1= k).
It follows from this that
=~~ei(l'll-") 4
a,) a'll
=-
00
88
PLURISUBHARMONIC FUNCTIONS AND PSEUDOCONVEX DOMAINS
If we define a = (r11 a.lcos «(jll -
~j)' •..• r nl an Icos (CjIn -
~=(rllallsin(~l- 61),
....
an»'
rnl a nl sin (CjIn - 6n
».
we obtain -
1
1
(H(z; u)a. a)=4"(H(lnr; R)a. a)+"4(H(lnr; R)~. ~).
from which the assertion made follows.
5. Functions that are plurisubharmonic in multiple-circular domains For a {unction u (z) = R to be plurisubharmonic in a multiplecircular domain G. it is necessary and sufficient that the {unction R (r) possess the following properties: (1) R (r) < + 00 and is upper-semicontinuous on the set j = 1. 2..... n; z EOJ;
2) R(r? ...• rj • ...• r~) increases with respect to each variable rj( with the others fixed> in the interval [O.R~) whenever the interval er? ...• rj' ...• r~). where 0 S rj < RJ. is contained in B; (3) ~(r) is convex with respect to 0 for all z in O'eO. From this, we conclude, on the tlasis of formula (3) of section 1.3, that
.
.
-In ~o (z) = sup [ - In flo (z)]
is a plurisubharmonic function in a (see section 10.3). Thus, pseudoconvex domain, which completes the proof.
a
is a
95
PSEUDOCONVEX DOMAINS
ocC R and DcCm are pseudoconvex, the domain is also pseudoconvex. Proof: Since a x D = (a x Cm) n(C n X D), it will, by virtue of the preceding results, be sufficient to show that the domains a x cm and cn X Dare pseudoconvex in cn + m .• But this assertion follows from the relation If the domains
a x Dcc n + m
-In 1:.0 (z) = - In I:.oxcm (z. U7).
according to which the function In 1:.0 x em (z. fII) is plurisubharmonic in (j X Cm j that is, the domain a X Cm is pseudoconvex, which completes the proof. The union of an increasing sequence of pseudoconvex domains is a pseudoconvex domain. Proof: Suppose that 0.co.+ 1 and o=UO•. Then, on the basis of
•
(2) (see section 1.3),
-In /lo. (z) ~ -In AOHI (z) - -In Ao (z)
as
11_ 00
is an arbitrary subdomain o'eo. The functions In 1:.0 (z) are plurisubharmonic in a'. Therefore, on the basis of section 10.3, the function In 1:.0 (z) is plurisubharmonic in a', Since a'is an arbitrary compact subdomain of a, this function is necessarily plurisubharmonic in a, which completes the proof.
3. The weak continuity principle We shall say that the weak continuity principle applies to a domain a if the following assertion holds: Let (S.I denote a sequence of domains that lie, together with their boundaries as., on the twodimensional analytic surfaces F. and suppose that
S.u oS.co.
lim S. = So. .~CD
lim
_-+00
Then, if So is bounded, SocO (see Fig. 15). /
,. f". /,
/ / /
rill
~\~v. I 1/
11/
, II
Fil;.15
as. =
TocO.
96
PLURISUBHARMONIC FUNCTIONS AND PSEUDOCONVEX DOMAINS
Clearly, the weak continuity principle holds for every domain in CI. LEMMA (Bremermann [13]). If the weak continuity principle applies to a domain G, the following equations hold:
inf
zESUoS
1x.(z)l~a,o(z)==
lof
zEoS
Ix. (z)I~II'o(z), lal= 1;
lof 1x.(z)l~o(z)= lof 1x.(z)l~o(z), zESUoS zEoS
(45) (46)
where S is an arbitrary domain lying on an arbitrary two-dimensional analytic surface z = zO + Ab such that S U as c G and X (z) is an arbitrary function that is holomorphic and nonvanishing in the domain G. Proof: SinceoSeO and y.(zh~'O in 0, the number m= lof ~a o(z)Ix.(z)l> 0
zEoS
'
(see section 1.3) for every vector a such that Ia 1= 1. From this and the fact that the set oS is closed, it follows that the sets
are such that T.f::O for arbitrary ex < m. Suppose that the vectors a and b are linearly independent. Then, for ex E[0, m), the two-dimensional set F.=[z:z=zo+Ab+aCl X(zo+M) -t,
AEO.....].
where (see section 1.3)
is an analytic surface because (see section 10.12)
~: =b+a~ ~ o:~ x.-I(zo+Ab)b.~O. 1< • ..;n
Furthermore, for each
S. =[z: z =
:x
in the interval indicated, the set
z' +aCl
x(z')
-I,
z' ES]
is a domain that lies together with its boundary as. = T. on F •• However, by hypothesis, S(O)=SeO. Therefore, there exists a number ClO in (0, m) such that S.f::Q for all Cl E[0,(10). Thus, since oS.=T.cO for arbitrary ex < m, we conclude from the assumption that the weak continuity prinCiple holds that S.tElO for arbitrary :z < m. This means that
PSEUDOCONVEX DOMAINS
If we take the limit as
11_ m -
97
0 in this inequality, we obtain
6/1.0(z)~mlx(z)rl.
zES.
from which Eq. (45) follows. In particular, we have shown that Eq. (45) holds for an arbitrary domain in CI. In this case, it reduces to Eq. (46). Let us assume now that the vectors a and b are linearly dependent. Without loss of generality, we may assume that a = b. In this case, if we use the formula 6/1.0 (ZO
+ aA)=~Oz" .. (k)
(see section 1.3), Eq. (45) reduces to Eq. (46) for n = I (in the A-plane) with 0 replaced by Oz'. /I' X(z) by X(zO + ).a), and Sand iJS by the hulls of these sets in the A-plane. But, for n = I, Eq. (46) is already proven. Therefore, Eq. (45) is established for all a such that lal == 1. By use of formula (4) (see section 1.3), we can obtain Eq. (46) from Eq. (45). Specifically, in!
in.! IX(z)l~a.O(z)= in!
lal=1 zE.dS
= in!
in!
IX(z)l~a.o(z)=
lal=1 zESUJS
in!
Ix (z)1 ~a.O (z) =
zE.dS lal=1
=
in! Ix (z)I.la. aCz) =
in! Z;:SUdS
in! 1x'(z)I.lo(z)= zEdS
lal=1
in!
Ix(z)l~o(z). q. e.
d.
z(SUdS
4. A condition for pseudoconvexity, I For a domain G to be pseudoconvex, it is necessary and sufficient that a weak continuity principle apply to it. Proof of the SUfficiency: Since a weak continuity principle applies to the region a, Eqs. (45) are valid for it. In the proof of the
necessity of the second condition for pseudoconvexity (see section 12.5), it will be established that Eqs. (45) imply that the func-
tion In /la. 0 (z) is plurisubharmonic in a for all a such that Ia I = I. The function In /l... o (z) is uniformly bounded above for all a such that lal=1 and all ZEO'~O. We then conclude on the basis of formula (4) in section 1.3, that the function - In flo (z) = sup [- In ~a. 0 (z») lal=1
is plurisubharmonic in a (see section 10.3). This means that the domain a is pseudoconvex.
98
PLURISUBHARMONIC FUNCTIONS AND PSEUDOCONVEX DOMAINS
Proof of the necessity: Suppose that a is a pseudoconvex domain. Consider a sequence of domains S.. for IX = I, 2, , .. , that lie together with their boundaries as. on the two-dimensional analytic surfaces F •• Suppose thatS.uaS.eO. that lim S. =S'l is bounded. and that lim as. = Tue O. Since the function In ~a (z)is plurisubharmonic, we have, on the basis of the maximum theorem (see section 10.14) sup [-In~o(z»)= sup (-In~a(z»). zE. iJs. zE. s. UiJs.
If we take the limit as at -+ 00 in this equation and use the continuity of the function In 10 (z) (see section 1.3) and the boundedness of the sets So and Tn, we obtain the equation sup ( - In ~o (z») = sup (-In 110 (z»). zE T, zES,U T.
It follows from this equation that ~a(So)>-l1o(To). But To=O and hence l1a(Sn) l1a (To) > o. In view of the boundedness of the set So. this inequality implies that So~O, which completes the proof. COROLLARY. Every domain in Cl is pseudoconvex. We note in passing that every domain in Rl is convex.
>-
5. A condition jor pseudoconvexity, II For a domain G to be pseudoconvex. it is necessary and sufficient that the function -In Ao.. G (z) be plurisubharmonic in G for all a such that Ia I = 1.
The sufficiency of the condition was proven when the preceding condition for pseudoconvexity was proven (see section 12.4). To prove the necessity. note that since the domain a is pseudoconvex, it follows from the necessity of the preceding condition for pseudoconvexity that a weak continuity principle applies for a and hence Eqs. (45) are valid. Let us show that in this case, the function In ~". a (z) is plurisubharmonic in a for arbitrary a such that lal = I. Suppose that this is not the case. Then, there exist a vector a such that lal=1 and an analytic plane z=zo+l.b, where Ibl=I, such that the function V (A) = - In fl",o (zu+ I.b) is not subharmonic in OZ" b (see section 10.1). Therefore, there exists a function h (A.) that is harmonic in the open circle [A: IAI < r)g;Ozo,b and continuous in the closed circle IAI r such that
o. But the function V (A) < + 00 and is upper-semicontinuous in the circle IA1- min (tJ.o(To). pl. It follows from this inequality and from the inor; clusion Toea that tJ.o(So) > o. But So is a C set. Therefore, Suc:O. Thus, a weak continuity principle applies to the domain a and, hence, a is a pseudoconvex domain. Suppose now that a is an unbounded domain. We introduce the open bounded sets OR = On U (0. R) (see Fig. 16). On the basis of formula (3),of section 1.3, we have -)0
- hdoR(z)= lIIm > 0 for all z EdO. Consequently, there exists a neighborhood V (ZO) such that -In tJ.OR(z) = -In (R -Izl).
z EV (ZO) nV (0. R). Therefore, the function -In ~aR (z) is plurisubharmonic in V (ZO) n U (0. R). Thus, the function -In tJ.OR(Z) is plurisubharmonic in the boundary strip of the open bounden set OR' From what has been proven, each component of this set is a pseudoconvex domain. But OR increases monotonically and
U 0R= O.
Therefore, the domain
R>O
a '1.
is a pseudoconvex (see section 12.2), which completes the proof.
A condition jar pseudoconvexity, IV Each component of the open set
O=(z:
V(z) 2r, and O=[z:
IXII < yr2-(lz2/-R)2, R-r < /Z2/ o.
Furthermore. it is clear that the family of functions (- In a (x - X O). ;cOEdB) Is uniformly bounded above on every subdomain B'EB. Therefore, on the basis of section 11.3, Eq. (58) implies that the function - In 6. B (x) is convex in B. Proof of the sufficiency: Suppose that the function - In tJ. B (x) is convex in B. Let us show that the weak continuity principle applies to
111
CONVEX DOMAINS
the domain B. Let S., where 11= 1. 2.... , be a sequence of intervals such that UaS.CiiB., IimS.=S~ is bounded, and IimoS.=TocB. For the convex domain-ln6 B (x), the maximum theorem holds: JC
E: Sa'
which is equivalent to inequality (56). Just as in the proof of the necessity of the first condition for convexity (see section 13.2), it follows from this that the weak continuity principle applies to the domain B. From the first condition for convexity, the domain B is convex. This completes the proof. COROLLARY. Every convex domain G c en is pseudoconvex. Proof: Since the function - In 60 (x. )/) is convex in Q, it is also plurisubharmonic in Q (see section 11.3).
4. Properties of convex domains The interior of the intersection of convex domains is a convex domain (see section 12.2). The union of an increasing sequence of convex domains is a convex domain (see section 12.2). For a domain B to be convex, it is necessary and sufficient that there exist a function that is convex in B and that approaches + 00 everywhere on iJB (see section 12.8). If V (x) is a convex function in U (8) the domain B=[x:
V(x)O
plete logarithmically convex domains. - I.
+ ...
But, for a,>-O. the function alln IZII +an In IZnl-ln A«is plurisubharmonic in C· (see section 10.5). Therefore, (see section 12.9). the domain [z : IZ"I < A.l is pseudoconvex. But then the domain OR' being the interior of the intersection of pseudoconvex domains. is pseudoconvex (see section 12.2). Suppose now that 0 is a logarithmically convex domain containing no points of the manifold Zl ... Zn = O. Since the corresponding Q is convex, there exists a function V (~) that is convex in Q and that approaches +"" everywhere oniJQ (see section 13.4). Therefore. the function V (In Iz.l .... In Iz.1> is plurisubharmonic in 0 (see section 11.4). Furthermore. since the mapping ~ = In r is a homeomorphism of(O. 00) onto (-00. (0), the function V(ln Izl) approaches +00 everywbere on dO (cf. lemma in section 12.11). Therefore, (see section 12.8), the domain 0 is pseudoconvex, which completes the proof. In concluSion, let us take a look at complete logarithmically convex domains in the space (;2. Such a domain can be represented in one of the following two forms: [z:
Iz.l
[z: Iz?1
< RI (I Z 2/)' IZ21 < R'J. < R2 (l z l/)' IZII < RJ.
wbere the function R2 is the inverse of the function R, (see Fig. 26).
115
CONVEX DOMAINS
The functions RI (r) and R2 (r) are lower-semicontinuous and decreasing functions in (0. R')and (0. R). relIz' spectively. F'or a complete multiple-circular domain G to
R'
, --4-----, I I
be logarithmically convex, it is necessary and sufficient that the function Rl (r) possess the following rroperties: (1) It is lower-semicontinuous and decreasing in [0, It); (2) the function -Rl (r) is convex with respect to In r in (0, It).
It follows from these properties that RI (r) is a continuous function in (0. R') and
I
R,(izzi)
~--~I------~--~~ 0
Fis·26
that R=R1(0)= lim Rdr). r-++O The function R2 (r) possesses analogous properties.
CHAPTER III
Domains and Envelopes of Holomorphy In this chapter. we shall expound the general theory of (singlesheeted) domains of holomorphy and envelopes of holomorphy. We shall take up the following four types of properties of domains of holomorphy: (1) holomorphic convexity. (2) principles of continuity. (3) local pseudoconvexity. and (4) global pseudoconvexity. Although all these properties are closely interrelated. theycharacterize domains of holomorphy in quite different terms. In particular. according to the fundamental theorem of Oka t the class of (singlesheeted) domains of holomorphy coincides with the class of pseudoconvex domains. Therefore. study of domains of holomorphy amounts to a study of geometric objects. namely. the pseudoconvex domains examined in the preceding chapter. Before embarking on an expOSition of the general theory of domains of holomorphy t let us examine in greater detail functions that are holomorphic in multiple-circular domains and Hartogs domains.
14, MULTIPLE-CIRCULAR DOMAINS AND POWER SERIES
1. Holomorphic extension o/multiple-circular domains Suppose that a multiple-circular domain 0 contains its center (see section 7.5). Let f (z) be a function that is holomorphic Let us take an arbitrary multiple-circular subdomainO' a 0 that contains the center O. We define a= 0 in O.
R- sup Izl. aEO'
b.o(O')==lJ>O.
r=l+ 2R'I ,
Let us construct the function (see Cartan [118]) 1_ ( ) __ If z - (2"j)/I
f IAd='
•••
r
w
P"I=' 116
tfA,z, ... ,. A"z,,) (A _ I)'
). d ,
(1)
117
MULTIPLE-CIRCULAR DOMAINS AND POWER SERIES
When z E a'. the points (AJZI' .... AnZn) are strictly contained in IA}I =r. where j = 1. 2..... n. since
a for
~lzl(r-l)~-;-.
Therefore. the function !p(z) is holomorphic ina' and I/(A.Z ...... Anzn)l-< M for all zEO'. where IA,I =r. for j= 1. 2..... n. Furthermore. since 0 E 0'. the function I (z) can be expanded in a Taylor series I(z)=
~
aa = -;DOl (0) Q •
aa z ••
(2)
•'1>0
in some polydisk S (0. p). If we substitute this series into formula
(1). we see that !P (z) =
I
(z).
On the basis of the holomorphic continuation theorem (see section 6.1). we conclude that !p(z)=/(z) in the domain a'. that is. that
From this. we get the expansion
I
(z) =
l:"" !Po (z).
(3)
zEO'.
«:=;:O
where
.f ..
'I Zn H z~ I
f(z')
z' a+1
dz'.
(4)
The functions !Po (z) are holomorphic in A' and the series (3) converges absolutely and uniformly in 0'. Let us show that/.(z)=a.inO'. It follows from (4) and Cauchy's theorem that these equations are valid in But then. the equations !Po (z) =
a.zo.
which are
0'. This proves our assertion.
s(o. f). valid in S (0. f). remain
valid in
118
DOMAINS AND ENVEL..OPES OF HOL..OMORPHY
Thus, the series in (3) is independent of r (and hence of 0') and coincides with the power series (2). Remembering that 0' is an arbitrary compact subdomain of 0, we can, by use of Abel's theorem, obtain the following result (see Hartogs [6], Cartan [118]): If a function f(z) is holomorphic in a multiple-circular domain G that contains its center a; 0, it is holomorphic (and, consequently, single-valued) in the smallest complete multiple-circular domain TrW) containing the given domain G,
and it can be expanded in the absolutely convergent power series
Thus, "!teO) is a hoi om orphic extension of the domain
(2)
o.
in TrW).
2. Domains oj absolute convergence oj power series We shall say that a multiple-circular domain a is a domain of absolute convergence of a power series if there exists a power series that converges absolutely in the domain a but does not converge (absolutely) in any larger domain. Thus, the preceding theorem reduces domainS of absolute convergence of power series to complete multiple-circular domains. We shall see below that not every complete multiple-circular domain is a domain of absolute convergence of any power series. The question arises as to how domains of absolute convergence of power series may be characterized. The answer to this question for boWlded domains was already known to Hartogs [6]. Here, following the work of Ayzenberg and Mityagin [62], we shall give a simple solution of this problem. We introduce the sequence of numbers (see Eq. (59) in section 13.6)
LEMMA 1. If the power series (5)
converges absolutely in a closed bounded complete multiple-circular domain G, then where
M=ma~1 ~ zEO
Proof:
For an arbitrary polydisk
S(O,
r) c
101;;.0
a,
aoz o'.
where
rj
= Ihjl for
hE a, the following inequality of Cauchy (see section 4.4) is valid:
MULTIPLE-CIRCULAR DOMAINS AND POWER SERIES
119
Since this inequality holds for all bE a, we have
la.1
(a)
m:O
m,
0
also converges absolutely in S (0. I). which is impossible. Therefore. A. (0')
But the inclusion
-< A. (0)
for all a.
a implies the reverse inequalities. Therefore, follows from this that 0'=0"=0. which is impossible. Consequently, a is a domain of absolute convergence of the series (10). which completes the proof. 0':::>
A.(O')=A.(O). It
15. HARTOGS' DOMAINS AND SERIES In sections 6. 8 and 14, (multiple) power-series expansions of functions constituted our basic tool for holomorphic expansions. In this section. we shall introduce another method of holomorphic continuation. namely. the method of expanding functions in power series in a single variable (Hartogs' series). In contrast with the preceding method. this method can be applied only in the case of several complex variables (n>- 2). The results that we shall present in connection with Hartogs' series are of independent interest but they also have a number of applications in the theory of several complex variables. for example. in the proof of the fundamental theorem of Hartogs (see section 4.2) and in the study of semi tubular domains.
1. Expansion in Hartogs series Suppose that a Hartogs domain a (see section 7.5) contains points of its plane of symmetry zl=a 1• For simplicity,we shall assume that a 1 = O. Suppose that a function f (z) is holomorphic in the domain O. Let 0' be an arhitrary domain that is compact in a and that contains points of the form (0. z). Without loss of generality. we may assume that 0' is a Hartogs domain. We define Ao (0') = "IJ.
r=1+2k'
(12)
123
HARTOGS' DOMAINS AND SERIES
Let us construct the function (see Cartan [118]) (13) When z EO', the points P·d = r are strictly contained in 0 for all (AIZi' since, on the basis of (12),
z)
I(AIZ •• z)_(elarg1,z •.
i)1 ~ IAz.-elarg1'zd ~ .
'I
~ IZII(IAd -1)~2"
Therefore, the function cp(z) is holomorphic in Ci' (see section 4.6) and II (A.ZI' %)1 ~ M for IAII = r. and Z E0' • _ Furthermore, since there exists a point (0. zO), belonging to 0', the function I (z) can be expanded in an absolutely convergent power series I(z)= ~ aozi'(z-zo)".
1 a. = at DOl (0.
-
ZO)
lel>O
in some polydisk 5«0. Eq. (13), we see that
;0). p). If
we substitute this series into
=
z
~ ao zj'(z- O)7i=/(z) 101:>0
for
Therefore, OD the basis of the holomorphic continuation theorem (see section 6.1). we conclude that cp(z)=/(z) in the domain a'; that is.
zEO'. From this we get the expansion
I
""
(z) = ~ CPo (z).
_=0
zEO'.
(14)
124
DOMAINS AND ENVELOPES OF HOLOMORPHY
where CPo
(z) =
I :bti
r ' I
•
(AI ZI.
dA, = Z) ).a+1
I A,l -r
I
I
(z)=_ CI
21ti
f
• (Z). zd.
(15)
I
f(Z;. i) 10+1 ,Zl
1
dz l •
0:= 0, 1. ••••
r/zd-lzd
(16)
The series (14) converges absolutely and uniformly in 0'. According to Cauchy's theorem. the integral (16) defining the function I. is independent of r. Therefore. we may take the limit in formula (16) as r _ 1 + 0.• We then obtain 1 I.(z) = 2" 1tl
f
J(z;. i) ,.+1
I zI\-/ z;1
ZI
,
dz l •
0:==0, 1. ••••
(17)
The representation (17) is valid for z EO~ But the right member in (17) is independent of 0' c= O. Therefore. the functions I.(z) are defined throughout the entire domain o. Furthermore. on the basis of Cauchy's theorem. these functions are independent of ZI when ZI varies in some annulus r (z) < 1 zll < R (Z) (or in the Circle I zll < R (i» lying entirely in the domain a (see Fig. 27).
In particular. if (0. ilEa. by taking the limit in (17) as we obtain 1 .1(0. I. (z) =--;;:D
z).
if
[z:
IZII < R (z»)c:: a,
ZI_O.
(18)
Let us show that the functions I.(z) are holomorphic in a. It follows from (15) that the functions CPo (z) are hoI om orphic in a' and that I.(z) =';>,(z)zl·. Therefore. the functions I.(z) are _holomorphic in a' if ZI 4= o. en the other hand. if the point (0, z) Ea'. the functions f. (z) will. by virtue of what we have proven. be
125
HARTOGS' DOMAINS AND SERIES
independent of ZI in some neighborhood of that point. Thus, the in the functions f. (z) are also holomorphic at the pOints (0. domain a'. Since A' is an arbitrary subdomain that is compact in a, this means that the f.(z) are holomorphic in O. Summing up what we have said, we have the following result (see Hartogs [6], Cartan [118]).
z)
Every function f(z} that is holomorphic in a Hartogs domain G that contains points of its plane of symmetry Zl = 0, can be expanded in an absolutely and uniformly convergent Hartogs series 00
f(z)= ~zU.(z).
(19)
Cl=U
in that domain. Here, the functions fa (z) are holomorphic in G and are independent of Zl in every component (annulus or circle) of the open set
2. Holomorphic extension of Hartogs domains We denote by B the projection of the domain a onto the plane of symmetry ZI = O. Obviously, 8 is a domain in en -I. Let zo=( zY. ZO) Ea. Then, zOEB. Let a., be any subdomain of the domain a that contains the point ZO and that can be represented in one of the following forms (see Fig. 27):
a., = a., =
< I ztl < RoO (z). ZEB').
[z:
roO (z)
[z:
I zll < Rz.(i).
zEB'].
where 8' is a subdomain of the domain B that depends on zO. On the basis of section 2.2, the functions r z. (z)and R•• (z) must be upper- and lower-semicontinuous, respectively, in B'. On the basis of the results of the preceding section, in the expansion (19), the functions f. (z) in the domain Oz' are independent of Zl. Therefore, the functions f.(Z)=f.(Z), being holomorphic in O. are also holomorphic in B'. Thus, the expansion (19) in the domain oz' can be written as follows: 00
f (z) = ~ zil. (z).
(20)
•• u
According to Abel's theorem, the series (20) converges absolutely and uniformly in the smallest complete Hartogs domain (see Fig. 27)
'It(O••)=[z: Zl=/'Z;. z=z,. z'EO••. P,I< 1]= = [z: I z./ < R:' (.i), .; EB'].
126
DOMAINS AND ENVELOPES OF HOLOMORPHY
containing the domain Oz, and thus defines a holomorphic function in 1t (OzO). Thus. every function f (z) that is holomorphic in a Hartogs domain 0 containing points of its plane of symmetry Zl = 0 is holomorphic in every complete Hartogs domain IT (for (zO € 0» and is represented in it by the series (20).
Therefore, by means of the series (20), the function / (z) can be holomorphically continued to every point of the smallest complete Hartogs domain 1t (0) containing the given domain 0 (see Fig. 27): 1t(0)=
U 1t (Oz,) = z'EO
= [z:
Zl =
Az;. z=z. z'EO. 11.1- will be holomorphic (and single-valued). Here. the series (20) defines a function ! (z) that is holomorphic in the corresponding covering domain (over 1t(0». The exposition makes it possible to construct examples of single-sheeted Hartogs domains with nons inglesheeted envelopes of holomorphy (see section 20.2,.
3. Hartogs' theorem Here, we shall confine ourselves to a consideration of the simplest case. that in which the function! (z) is single-valued in It (0) , that is. to the case of complete Hartogs domains. Suppose that the function! (z) is holomorphic in a complete Hartogs domain
O=(z: IZl_1 < R(Z>. zEB). where the domain 8 is the projection 0 onto the plane of symmetry Zl = 0 and the function R (Z) is lower-semicontinuous in 8 (see section 2.2). (8 coincides with the intersection of the domain 0 and the analytiC plane Zl = 0.) On the basis of the results of sections 15.1 and 15.2, the function! (z) can be represented in 0 by a Hartogs series go
!
(z)
= ~ z~!o (z). .-0
f. (z) = ~ DO, (0.
z).
which converges absolutely and uniformly in the domain functions ,. (z) are holomorphic in B.
(21)
o. The
127
HARTOGS' DOMAINS AND SERIES
z
For each EB. we denote the radius of convergence of the series (21) by R~ (z). ~ince !he series (21) converges for I zil R (z). it follows that R~ (z) R (z). Furthermore. on the basis of the Cauchy-
-
Hadamard theorem. we have (22)
The function R~ (z) does not need to be lower-semicontinuous. If it is not. we set RJ (i) =
lim R~ (if). i' ....&
(23)
Obviously. RJ(z)0.
turns out that the series (21) converges absolutely and uniformly in the domain 01' We shall prove this statement in the following section. As a preliminary. we shall prove the following. less general. theorem, belonging to Hartogs (6). It
--
8
o
1'" ........
-
~
.I -
1
Fit;. 28
If the Hartogs series (20) converges absolutely and uniformly in a closed polycircle S Rl < 'I for arbitrary fixed ~ in S(~O :;:-), this series converges absolutely and uniformly in the polycircle S 0 (see section 2.3). Therefore. on the basis of (22).
R;
-:J-1m
. - . He
'VII • (-)1 Z =
I
./
1
-R" - -.. - R j (z)
f (Z)
./ I
. 0 [see section 2.3].) According to Hartogs' theorem (see section 15.3). the series (21) converges absolutely and uniformly in the polydisk S(o. R,} X S(zo. i). Since an arbitrary subdomain 0, that is compact in 0' can be covered by a finite number of such polydisks, the series (21) converges absolutely and uniformly in 0'. This means that the series (21) converges absolutely and uniformly in 0, (see section 1.6). It follows from the absolute and uniform convergence of the series (21) in the domain 0 that. for an arbitrary subdomain B' reB. there exist numbers M and r, such that IfQ(i)lr~-<M.
zEB'.
a=1. 2 ..•..
From this we conclude that the set { ~ In If. (z) I } of functions that are plurisubharmonic in B is uniformly bounded from above on B'. But then. the function -In RI (z) is plurisubharmonic in B (see section 10.3). which completes the proof. Proof: The function -lnR,(i) belongs to the class FB of Hartogs' functions (see section 10.15). COROLLARY. Every function that is holomorphic in a complete Hartogs domain G is holomorphic (and single-valued) in the domain Gr.
5. Logarithmically plurisubharmonic envelopes Consider the complete Hartogs domain 0= [z:
I z,1 < R (i). i EB)
.
We shall refer to a complete Hartogs domain of the form
o"=[z: Iz,l<ev(Z). zEBl. where V (z) is the least plurisuperharmonic majorant of the function InR(z) in the domain B. as the logarithmically plurisuperharmonic envelope of the complete Hartogs domain O. Since the functions R (i) is lower-semicontinuous (see section 2.2) and positive in B. the function In R (z)< 00 and is lower-semicontinuous in B. Therefore. this function has a least plurisuperharmonic majorant V (z) (see section 10.4). Therefore. the domain o· is pseudoconvex (see section 12.13). Clearly. 0*::;)0. The logarithmically plurisuperharmonic envelope of the complete Hartogs domain is a holomorphic extension of that domain.
130
DOMAINS AND ENVELOPES OF HOLOMORPHV
Proof: Let f EH o. Then. the function / larger domain (see section 15.4)
(z)
is holomorphic in the
where V,(z)= -
lim
i' -+z
1 lim -lnl/.(z')1 11-+ +00
(l
is a plurisuperharmonic function in the domain B that satisfies the inequality v,(z) >-In R(z). From this and from the definition of least plurisuperharmonic majorant (see sections 10.4 and 9.7). we get the inequality (26) Inequality (26) means that O'cO,. Thus. every function/(z) that is holomorphic in 0 is holomorphic in 0'. which completes the proof.
6. Hartogs-Laurent series Let us now consider a Hartogs domain that does not contain a single point of its plane of symmetry (see section 12.13) 0= [z:
r
(z) < 1zll < R (z).
zE BI
(see Fig. 18). Every function f(z) that is holomorphic in a Hartogs domain G is holomorphic in the domain
and. throughout this domain, can be expanded in an absolutely and uniformly convergent Hartogs-Laurent series 00
/ (z) =
~ z~/. (;).
(27)
• __ co
Here, the functions fa ('2) are holomorphic in B and
- = f • (z)
1
2";
J1(Z;.,•• z) dz.., I
z.
1
Cl
=
o. ± 1.
..••
(28)
where the integration is carried out over an arbitrary closed piecewise-smooth contour lying entirely in the annulus
HARTOGS' DOMAINS AND SERIES
and containing the point Zl
=
131
O. F'inally.
z
Proof: For every EB. let us expand the function! (z) in a series of the form (27). Bere. the functions !. 1 and a is an arbitrary complex number. also belong to Ko • For example, the set Ho of all functions that are holomorphic in 0 constitutes a class. The set P of all polynomials constitutes a class. Obviously. Ho contains every class Ko. THEOREM (Canao-fiolleo [7]). Suppose that the maximum principle holds for a point zO E G and a set A c: G with respect to the moduli of functions of a class KG, that is, that
If (zO)1
< sup II (z)l· .tEA
IEKo·
(33)
135
HOLOMORPHIC CONVEXITY
Then, every (unction ( € KG can be holomorphically continued into the polycircle S(zo,rll, where r = 8G(A), and (or all p < r, the maximum principle sup
6ES (zO. pI)
where Ap
=
II (z>l '"6EA, sup If (z)1
(34)
U S(z', pI) (see Fig. 30), holds. ilEA
If the open set S (zO, r/) n0 is a domain. then. on the basis of the holomorphic continuation theorem (see section 6.1), the continued function I (z) is single-valued in S (ZO, rl) UO. Remark:
Proo(:
Since Ap @
a and p < r.
if we de-
fine
we obtain I/(z)1 <MI . p'
zES(z', pI).
z'EA.
From this and from Cauchy's inequality (see section 4.4). we deduce
I~,
DO/(z')I<MI.pp-'o,.
z'EA.
(35)
Since D'/EKo. by using inequality (33). we obtain, on the basis of the above inequality.
l ~D'/(ZO)1 a' """~MI. F P -\01 •
(36)
It follows from this inequality that the series
I (z) = ~
!, D'I (zO) (z -
zO)"
\01:>0
converges uniformly in the polydisk S(zo. pI) for arbitrary p < r. This means that the function/(z)is holomorphic inS(zO. r/). To prove inequality (34). it will be sufficient to show that (37)
for arbitrary PI < pI. Suppose that inequality (37) does not hold for some PI < P. that is. that
1 36
DOMAINS AND ENVELOPES OF HOLOMORPHY
It would then follow that the function 'fIp (Z) = [
~ (z)
]P •
/' p
where P is a positive integer, satisfies the equation (38)
On the other hand. cpp EKo. M,p. p = I. and on the basis of inequality
(36), which we have proven, we have
Therefore, the function 'fp (z) satisfies the following inequality in the polydisk S (ZO, PI/): l'fp(z)!""
~ ;1
ID"'fp(zO)/lz-
zOI"""
10 1:>0
~ ~ ~ ~(J'.!.)101=(1 al p
......
_J'.!.)-" . p
1"1:>0
This inequality. together with Eq. (38). yields the inequality
for arbitrary integral P> O. But this inequality is impossible for sufficiently large P since, by hypothesiS. a. > 1. This contradiction proves inequality (36) for arbitrary PI < P and this completes the proof. This theorem has the COROLLARY. Suppose that the maximum principle holds for sets S c:: G and T c:: G with respect to the moduli of functions of a class KG. Then. every function in KG can be holomorphically continued in Sr (possibly nonsingle-valuedly Sr U 0), where r = BG (T). Also, for arbitrary p < r. the maximum principle is valid for the sets Sp and T p with respect to the moduli of functions of the class KG·
2. K-convex domains In connection with the results obtained in the preceding section, we may introduce the concept of a K-convex domain. A domain a is said to be K-convex (convex with respect to the class of functions Ko) if. for an arbitrary set Ac:Q, the set
137
HOLOMORPHIC CONVEXITV
FA=-
n [z:
IEKo
I/(z)l<sup/!(z)l· zEO] zEA
is compact in O. In other words. a domain 0 is said to be K-convex if. for every set AgO. there exists a set FA such that AcF AC:O (see Fig. 31) and for every point ZO EO" FA. there exists a function f EKo such that
suplf(z)1 'EA
< I/(zO)I·
(39)
If a domain 0 is H-convex (K0= H 0)' it is said to be holomorphically convex.
We shall say that P -convex domains are polynomially convex. Obviously. every K -convex domain is holomorphically convex. (Ka = P)
We note that if G is a K-convex domain. the set F' A is compact and (40) Proof: Since FAg O. FA is bounded and (being the intersection of closed sets) closed. Therefore, FA is compact. Let us prove Eq. (40). Since AcFA• we have 8a (F A>'>- r (see Fig. 31). Consider a point ZO EFA. Since inequalities (33) hold for all functions IE Ko. we conclude from the theorem on simultaneous continuation of functions of class Ka that I (z) is holomorphic in the polydisk S(ZO, rJ) and satisfies inequality (34) for all p- r for arbitrary Zo EFA. This means that 80 (F A) >- r which completes the proof. It follows from this that every component of the interior of the intersec-
tion K-convex domains is a K-convex domain. To see this. let 0 be a component of int
no. and suppose that
Ac:O. We need to show that FAiG G. Since AiGoGa and the O. are Kconvex domains. we have
We define the class Ka as the smallest class of holomorphic functions in 0 that contains U Ko•• Since Oc 0 and Ka=:JKc-•• the 0
set FA (for the domain 0) is contained in all the sets FA,. and hence is bounded. Therefore. it remains to show that 80 (FA) > O. But this follows from (3) [see section 1.3] and (40):
138
DOMAINS AND ENVEL.OPES OF HOLOMORPHY
Suppose that K and KI are sets of functions that are ho1omorphic in a domain Q. We shall say that K, is dense in K if. for arbitrary fEK. O'eO. and £>0. there exists a functionflEKlsuchthat
For K I to be dense in K, it is necessary and sufficient that, for an arbitrary function f E K, there exists a sequence of functions fa E KI, for a = 1,2, ... , that converges uniformly to f in the domain G (in the sense of section 1.6).
The sufficiency is obvious. Let us prove the necessity. Suppose that O. c: O. + I that U O. = Q and that E. _ 0 as a - 00. The functions f.EKI
such that
.
If(z)-f.(z)1 <E ••
zEO••
11=1. 2, ....
form a sequence of the type required. LEMMA. If a domain G is K-convex and if some class K~ is dense in KG' the domain G is K~convex. Proof: Suppose that A ~ O. Then, there exists a set F A such that Ac.FAff!;O and, for every point zOEO'\F A, there exists a function IE: Ko satisfying Inequality (39). Therefore, there exists a 8 0,
>
such that o+supl/(z)1 zEA
< I/(zO)I·
(41)
Since K~ is dense in Ka. let us take a function II E K~ such that a
If(z)-fl(z)1 Ilk-I such that F k-I U{alA-ill cA,•• For the set A,., there exist a set F", a point alt Ea '\ F. and a function I. EHa such that A,,,, c F k C a, I a lk ) - Zlk) I < Ii' and sup
Ilk (z) I = I. If k (a(k» I > 1
(45)
zEA'k
and so on. Since, by construction, 1/.(a,k»1 > I, there exiSts a sequence of integers I", for k = 1, 2. "0, where (II = 1), that satisfies the inequalities (46)
1 I f k
(
Ik)}I'·
ak2
~
_
. 11, this sequence of functions I a (z) for k = O. 1 .•. converges uniformly in itself since. on the basis of (52), the inequalities P
P
1=1
1=1
I/HP- la 1- v + 1 - 01. Consequently, this sequence converges uniformly in a. and. thus (since a. is arbitrary) defines a holomorpbic function I(z) in the domain G (see section 4.4). Obviously, the function I (z) can be written as a series of the form
1=/0+(/1-/0)+(/2-/1)+ .... Therefore, by virtue of (51) and (52), we obtain 00
I/(z)- lo(z) 1 - o.
Consequently. from some a. on, we have r. > t. Therefore. since 8. -+ SUI we obtain the inclusion
.
.
q.e.d.
SOC:: U(8«). c:: U(8..),..
Remark: The basic difference in the theory of domains of holomorphy of functions of one and of several complex variables consists in the following: In the case of a single variable. bounded sets S and T for which the maximum principle holds with respect to the moduli of functions that are holomorphic in Q are subdomains ofSE6Q and their botmdary T=oS. The condition Oa(8):> 0a(08)imposes no restrictions at all on the domain O. On the other hand. in the case of several variables. it is possible to exhibit domains 8 G Q lying. for example, on two-dimensional analytic surfaces for which 00 (8) < 00 (a8) (see Fig. 34) if Q is not a domain of holomorphy. Therefore, the condition 0 (8):;9 &0 (oS) is valid only for a certain class of domains in en (for 11:> 2) (in every case for domains of holomorphy [see section
°
16.7]).
2. "Disk" theorems We note the following special cases of the ccntinuity theorem. They are referred to as "disk" theorems. These theorems are very convenient for applications. Fip;. 34
Suppose that (z. w) € en + m and z = z . where 0 5! .t.s 1. is a continuous curve in Suppose also that OW, for 0 t ~ I, is a family of domains in possessing the property that, for an arbitrary compact set K' c 0(0), there exists a number" = ,,(K'), in (0,1] such that K' c 0 (t) for all 0 5! t < 'I, there.
em
en.
.s
CONTINUITY PRINCIPLES
151
Suppose that the domain of holomorphy G contains the cdisks· [(z, w): z = z is holomorphic with respect
to A in the domain g for every t
€
[0,11, it is continuous, and az "lOin
aA
g(d x [0,11; (2) the domain of holomorphy G contains "disks" D (t) for 0 < t .s. 1 and the boundary of the limit "disk" aD(o) = [z: z = z(a,O>, A € ag(O»). Then, D(o)c=G.
Analogous "disk" theorems hold for a specific function fez). For example, corresponding to Theorem n is the THEOREM. Suppose that a function fez) is holomorphic at points of the set
U D (t> UOD(O). 0 x
for some r l faces
-< r
r) , it
> o.
follows that (58)
and x> O. Let us take a sequence of analytic sur-
p.=[Z:
P,I O.
If we apply Taylor'S formula to the function 'P [z (A) - ae) for small
values of a> 0 and use inequalities (57) and (58), we obtain, for all sufficiently small IAI PI < p,
-
(z)Ec(2) and grad tp (z) +- 0 in U (zo. r). Suppose that p < r 13. Then. in the neighborhood U (z'. 2p) of each point z' ES nU (zu. p) (see Fig. 39). the Taylor expansion cp(z) = 2 Re (z - z'. grad tp (z'» + =[z:
~
02'1' (z')
,
,
+Re~ oZloz" (zl-zl)(z,,-z,,)+
(64)
I.k
+(H(z'; cr)(z-z'). Z-z')+0(lz-z'1 2)
is valid. Let us consider the function cp on the (2" - 2)-dimensional analytic surface P z' (z)=2(z - z'. gradcp(z'»+ ~ + ....
0 2" (Z') ( ')( _ , Oz OZ ZI-ZI Z,,-Zk)=O
I."
I
k
which passes through the point z'. On the basis of (64). on this surface in the neighborhood U (z'. 2p), we have
= (H (z; cp) (z - z'). z-- z') + 0 ( Iz - z'1 2) = _ r(H(z;,,) (z-z'). z-z')
'? (z)
-l
Iz-z'1 2
+
(65)
+0(1)] jz-z'1 2. Furthermore. we conclude from condition (62) that (H (ZO; cp) a. a) >-" > 0 for some a and for all a such that Ia 1= 1 which satisfy condition (61). But on the surface Pz' (z) = 0 in the hypersphere U Fill' 39 (z'. 2p) condition (61) is approxima~ely satisfied for the vectors a = ,: - z, up to a quantity of the order ,z-z 1 p. SpeCl°fic ally. IC;=;:I'
grad~(zO»)I 0 means that the function -In R (Z2) is subharmonic in B; that is. the domain Gis pseudoconvex (see section 12.3). (3) Suppose that a portion of the boundary of the tubular domain T B = B + IR2 is 'I' = XI - ~ (x 2) = 0 (see section 7.5). Then. 1
0
"2
1
"2
0
-~V
0
L(cp)=-
-~~' 1 ~" . =-16
0 I
,II
-"4'i'
The condition L (r) > 0 means that tjI" -< 0; that is. the curve XI is convex (cf. section 13.5). (4) For the semitubular domain
G=[z:
V(Z2)
< XI < V (Z2)'
z2EB.
L(1'I)= -
1
2"
oV
oz,
2"
OV oz.
0
0
0
-~V
1
1
=-"4t:.V.
(Xl)
Iyd (z) < Q) side, then F' is an analytic surface (see Kneser (54). Proof: Since the hypersurface
8
H (0)
(G')
> 8 (G') == r 0
for all ZO EH (0'), which completes the proof. COROLLARY. 1(0'= 0, then H(G') = H(G), THEOREM. Suppose that a function f(z) is holomorphic in a bounded domain
o and possesses
a singularity at a point zO E aH (G). Then, this function Ims a singularity at some point of the boundary Proof: Suppose that this theorem is not true. Then, the function I(z) is holomorphic in some neighborhood Uofthe domain Consequently, 1 (z) is also holomorphic in the envelope of holomorphic H(U).. Since 0 is a bounded domain. we have Oc:::U. From this we conclude that H(O)ff:iH(U), which contradicts the fact that I(z) has a singularity at the point ZO EaH (0). This contradiction proves the
ao.
a.
theorem. It follows from this theorem that if, to every boundary point zO of a bounded domain of holomorphy 0 there corresponds a {zO}, then there is no subfunction I .. (z) that is holomorphic in domain O'c:O such that H(O') = 0 and ao ,\iJO' +- 0. For example, for the hypersphere Iz I < I, the function (I - ZZO)-I would be such a function 111' (z). Therefore, the hyper sphere is not
a"
180
DOMAINS AND ENVELOPES OF HOLOMORPHY
a domain of holomorphy of any of its subdomains (z : Iz I = II '\ ao'
Of
+ 0.
for which
5. The Cartan-ThulZen theorem As a preliminary we prove:
a
Suppose that a biholomorphic mapping f = f (z) maps a domain onto a domain a1. Suppose also that it maps H0
/.(z).".
Therefore, we shall not stop to write out the proof. It appears in the book by Bochner and Martin [3]. Chapter vn. We have the formula (84) H(Q X D)=H(O) X H(D).
CONSTRUCTION OF ENVELOPES OF HOLOMORPHV
181
Proof: By twice applying the theorem just given. we see that every fWlction I (z. w) that is holomorphic in a x D. is holomorphic in H(O) X H(D). Since the domains H(a) and H(D) are domains of holomorphy, it follows that H (a) X H (D) is a domain of holomorphy. The validity of Eq. (84) follows from this.
21. CONSTRUCTION OF ENVELOPES OF HOLOMORPHY In this section. we shall look at a number of examples of the construction of envelopes of holomorphy.
1. Envelopes of holomorphy of i:u.bular domains The envelope of holomorphy H(TB) of a tubular domain Ta coincides with the convex envelope O{TB) of the domain TB; that is, H(T B)= OCTB) = To (.8).
Proof: It follows from Bochner's theorem (see section 17.5) that
o(TB)c.H (T B). But 0 (TB) = To (B). which completes the proof.
If we assume in advance that H (T B) is single-sheeted, this theorem can be proven in a different manner without using Bochner's theorem (see. Bremermann [18]). Specifically. since 0 (T B) is a domain of holomorphy and 0 (T B)=:J T B' we have H (T B) c 0 (T B). Let T B, be the largest tubular domain contained in H(TB). (Since H(TB ) is assumed to be single-sheeted, the domain TBI exists.) From the construction, we have TBcTB,cll (TB). Let us show that T B, -4=H(TB).. Let us suppose, to the contrary, that T B, = H (T B). Then. on iJT B,. there exists a point ZO in the interior of H(TB ) such that AH(TB)(ZO»O. From the Cartan-Thullen theorem. H(T B ) is a tubular domain. Therefore. AH(TB)(xO+IY)=~H{TB)(ZO)
> 0,
lyl < 00.
From this it follows that the domain TBI can be enlarged somewhat in such a way that the enlarged domain will be tubular and will be contained in H (T B). But this contradicts the maximality of T B, • Therefore. HlTB)=TB, • This means that TBI is a domain of holomorphy and hence B1 is a convex domain (see section 19.3). From this and from the inclusions TBCT B,= H(TB)cO(TB)=To(B/. it follows that B = 0 (B). which completes the proof. j
EXAMPLE (Ruelle's lemma [50]). The envelope of holomorphy of a domain TC\f, where ~ = R 3 + iC is a tubular domain with ba,se 0, j = I, 2, 3] and f = [z: 0 $. argzl ~ argz2 ~ argz3 $. 7T], coincides with TC. Proof: Consider the tubular domain TIJ=RJ+IB in the space C3 of the variables CJ = ~j l"fjJ • for j = 1. 2. 3 where
+
182
DOMAINS AND ENVELOPES OF HOLOMORPHY
B=("r 0< "lj < 'It, j=l. 2. 3)'d"l: 0- IY211xl < 00) and D is some complex neighborhood of the set of real points [x: Ixd
(1)
where x = (Xl' -'=2' •••• xn) and the coefficients C1.J, ... Jk (x) are continuous functions in some domain of the space Rn. Two forms are considered equal if each can be transformed into the other by a transformation of the products of the differentials according to the formula (2)
With the use of rule (2), every kth-order form can be reduced to the canonical form (3)
The exterior product C1.~ of two differential forms a and {3 of orders and n~ respectively, is the form of order k+ m obtained by formal algebraic multiplication of the forms C1. and ~ by use of rule (2). Clearly, ry~=(_I)km~?:; ,,~=O if k+m >n. Under a continuously differentiable coordinate transformation x;=Xj(x;, .... x~}, for j = I, 2, ••. , n, the form (3) is mapped into the form k
(4)
where al'
I'"
J (x')= k
(5)
In particular, for an nth-order form, formulas (4)-(5) take the form a(x)dxI ... dx l1 = Xl' . . . .
x n\
XI' .•
xn
( =a[x(x')jD,
,)dx; '"
01
dx~.
(6)
FACTS FROM THE THEORY OF DIFFERENTIAL FORMS
193
The exterior differential da of a kth-order differential form IX (where 0 and holomorphic in U (zo, r)\Ii', where Ii' is a (2n - 2)-dimensionalanalytic surface passing through the point zoo Then, fez) is holomorphic and U (zo, r).
24. THE BERGMAN-WElL INTEGRAL REPRESENTATION The Bergman-Weil integral representation (see Bergman [68 Jand Weil [69]) has to do with functions that are holomorphic in a neighborhood of a Weil domain 0=
[z:
Ix. (z)1 < 1.
j = 1. 2•.. " N; z EV
('0)1.
This representation is derived from the Martinelli-Bochner formula. The derivation that we give is due to Sommer [56]. -The function F (z. w) is actually independent of .. (see section 23.1).
THE BERGMAN-WElL INTEGRAL REPRESENTATION
205
1. The hull oj a Weil domain For simplicity of exposition, we shall consider Weil domains possessing the property that grad X" (z)
rank
=k. grad v
""k
I OJ if ~ E C', then POe (e) Fig. 54),
0 for all eE pr 0 (- e). Thus, the continuous function I-'-c m is positive on the compact set pr 0 (- e) and therefore. I-'-c (e) = a
Inf
> O.
eEprO(-C)
Suppose that O(el=R·. In this case. pre.=le: lei = II.and it follows from (55) that Pc = 1/ a + which proves the lemma. Suppose now that 0 (e) =!= R". In this case, we may assume that o (e) and CO lie in the semispace YI ~ 0 (this is true because there exists a point y" EO (el such that y"y ~ 0 for ally E0 (C). and since y" E 0 (el, it follows that y"e ~ 0 for all eECO [see Fig. 57] ). y,
=.
Let us prove that P-c
en
m=
1-'-0 (e)
(e) for
el > O.
(56)
the basis of Eq. (54), it will
--~--~~~---+---~=O
be sufficient to prove this equation for I-'-c(e) > O. where el > O.
Suppose that
(Joc (e) = -eyO > o. yOEprC; (e) = - ey' > 0, y' Epr 0 (e)" (57)
1'0 eC)
We introduce the closed set (see Fig. 57) (58)
which contains the point yO. Since the cone 0 (el lies in the semispace YI ~ 0 , it follows that
By consideration of (57)-(59). we obtain Eq. (56): P-c(~)=
-eya =
sup (-ey)=
=
sup (-ey) = JlEC,
sup
(-ey) =
sup YEC~, IYI-I
YEprC
(-ey)=
JlEO(Ct)
sup
(-Ey)=
yEO (C E). IYlal
= - e,)"
=1'0 (C) (e).
Since the cone e is open, the cone CO is compact in the cone IE:el>OI (see Fig. 57). Therefore, the function I-'-cm whichis continuous on the compact set Ie: EI < o. lei = I I. is positive. so that inf
E. O.
222
INTEGRAL REPRESENTATIONS
From this and from Eq. (56), we conclude that Pc =..!. which completes the proof. a
< + 00.
EXAMPLES (1) C=R". ThenR"'=[O), fIoRn(E) = lEI. and PRn=I._ (2) C = r+ is a future light cone. Then. r H = r. Pr += 1. and
EE- r+, fIor+(E)=t ~ 0 and an open cone e" = en (C') (see Fig. 59> containing the cone e· such that
(60)
yEC',
Proof; Since the cone C' consists of interior points of the cone C.. = 0 (C), we have y~ > 0 for all ~ E pr C- and all y E pr C' (see section 25.1). Inequality (60) for some a> 0 and C" follows from this and from the homogeneity and continuity of the form y~. In this case, the cone C· is compact in C", which completes the proof. For tubular radial domain TC, we introduce the kernel K(z)=
f ,lzEd~.
(61)
e· The function K (z) is holomorphic in TO (e). For all y in the cone C' that is compact in 0 (C),
(62) Proof; If the cone 0 (C) contains an entire straight line, it follows from Lemma 1 that mes CO = 0 and the assertion is trivial. Therefore, let us assume that O(C) does not contain any entire line. Suppose that U (ZO, 8) Iii: TO (el. Then, it follows from inequality (60) (see Lemma 2) that ~y > I; I a, where a > 0 , for all ~ E CO and z EU (ZO, 8). From this we conclude that the integral (61) and all its derivatives D'K(z)=
f e-y O. J =
1,
2..... nJ. Then. ~ =
~
and co
K(z)=
IE 'Iel
J... Je J o 00
del'"
den =
(2) Suppose that
For n
c=r'.
~ Z,Z. '"
0
then
r+" =r+
. ZII
and (see section 30)
= 1. these kernels can be transformed into the Cauchy kernel.
3. Functions that are holomorphic in tubular radial domains Suppose that a function f (z) is holomorphic in a tubular radial domain rc and that. for all y in a cone C' that is compact in C. (64)
for any
£
> o.
BOCHNER'S INTEGRAL REPRESENTATION
225
Let us show that the function gy(~)=(2tt)-nety
f !(x+ly)e-I(xdx= = (2tt)-n f / (z) e-1tzdz
l ,',
(65)
dZ n
z:x+ly
is independent of y E C. It follows from condition (64) that. for an arbitrary subdomain BrsC.
f 1!(x+ly)1 dy_O 2
as
Ixl_oo,
B
From this and from the Cauchy-Poincare theorem (see section 22.6). we conclude that the integral (65) is independent of the plane of integration z = x + ly. Ix I < + 00 no matter how fast y varies in the domain B. This means that the function gym is independent of y E B. Since B is an arbitrary compact subdomain of the cone C • .g y (~) is independent of y for all y in C. We now denote the function g y (E) by g (E). This function is measurable and. on the basis of (65). it possesses the properties gme-tYE~
/ (z)
for all yEC;
= f g (~) eiz, d~
(66)
for all z ETe,
(67)
Here. Parseval's equation holds: yEC.
Let us show that g (~) E obtain from (68)
~
(68)
• On the basiS of inequality (64). we yEC',
(69)
From this and from Fatou's lemma (see section 2.8). we conclude that
f Itl-
>- f which is impossible for everywhere in C•• Thus. the function
f
Ig(OI2e-2tY"d~>-e2t.
1- 0 for ~ E C· and y E C. by using Lebesgue's theorem on taking the limit under the Lebesgue integral sign (see section 2.5). we obtain from (71)
f If(x+ly)-f(x+tO)1 dx-+O 2
for y-+O,
yEC.
Thus, we have obtained the following result (see Bochner [57]): Every
c. Fig. 60
function f(z) that is holomorphic in a tubular radial domain T C and satisfies conditions (64) has thf' limiting value
f(x+tO)=
lim
f(x+ly),
y .... O. yE c
°
u'hich is independent of the sequence y ~ for), E C. Convergence here is understood in the L2 sense. Tf the cone O(e) contains an entire straight line,f(z) '" O.
BOCHNER'S INTEGRAL REPRESENTATION
227
Remark: An investigation will be made in section26.3 of boundary values of functions that are holomorphic in TC and that satisfy a weaker condition than (64). In Bochner's work [57]. a study is made of the class of functions satisfying a more restrictive condition than (64): II/ (x + ly)11 -< M,. It is assumed in this connection that the cone C is convex and that C does not contain any complete straight line.
4. Bochner's formula Every function f(z) that is holomorphic in a tubular radial domain T C and that satisfies inequality (64) can be represented in the form of an integral / (z) = (2!)n
fK
(z - x') / (x'
+ to) dx'.
(72)
where K (z) is the kernel of T C and f(x + iOl is the boundary value of the function fez) as y ... 0 for y E C. Proof: Since g = 0 almost everywhere in R n '\ C', formula (67)
m
can be represented in the form (73) If we apply the convolution theorem to the integral (73) and make use of formulas (61) and (70), we obtain formula (72). Corollary. Every function f(z) that is holomorphic in a tubular radial domain T C and that satisfies equality (63) is holormophic in the convex envelope 0 (T c ) = TO (e). For all y in a cone C' that is compact in O(C), it satisfies an inequality of the form n
II/(x +ly)11
I
E
D* to be the spectral flDlCtion of a function
(z) = F [e-£Yg (m (x).
that is holomorphic in a domain TO (B), it is necessary and sufficient that e-ey gCe> E s* for y E 8.
The necessity is obvious. Let us prove the sufficiency. Consider a compact set K 0 (B). As a preliminary, let us show that, for some £ = a (K) > 0, the set of generalized functions
=
[T:T(E)=g(E)e-£y+d'l+l£I'.
yEK]
(5)
is bounded SO (see section 3.7). Since K cO (B). it follows that K' e 0 (B) for sufficiently small a> O. Therefore. the domain B contains a finite number of pointsy(l). y(2). • • •• y(1I the closed convex envelope of which (i.e •• the polyhedron 1t contains K' (see Fig. 61). Let us represent functions belonging to the set (5) in the form g (E) e-(Y+'Yl+ifji' =
~ l 1 and the cone C*. then all the derivatives D', (z) of its Fourier- Laplace transform / (z) belong to the class H p (p~a 8; o (C»).
+
COROLLARY 1. If f(z) € Hp(a + f; C), then Oaf(z) € Hp(pga + f; OW». COROLLARY 2·. For an entire function f(z) to have a spectral function g(~) of the form (19), where the continuous functions G/«() satisfy the inequality
lal(~)1 -< M;e-ca'-'IIEI P '
(p'
> 1.
a' > 0),
it is necessary and sufficient that, for some nonnegative 13, the function f(z) sat isfy the inequality (22)
THEOREM 2'*· Suppose that f(z) € Hl (a + l; C), where C is a connected cone and a ~ O. Then, its spectral function g(,f) € S· and
gm=o,
!1c(E)
> a.
Conversely, if g € S· vanishes in the domain Pe «() ~ a for some a > 0 and some cone C, then all the derivatives oaf(z) of its Fourier-Laplace transform f(z) belong fo the class Hl (Pea; OW». COROLLARY 1. If f(z), then Oaf(z) € Hl (Pea; OW». In particular, if C is a convex cone, then Hl (a + £; C) = Hl (a; C). COROLLARY 2. (Paley-Wiener-Schwartz [11]). For the spectrum of an entire function f(z) to be contained in a sphere 1(I ~ a, it is necessary and sufficient that f(z) satisfy inequality (22) for p = 1. Remark: It follows from Corollaries 1 to Theorems 1 and 2 that the sets Hp(a'; C) and HoeC) constitute classes in the sense of
U
a' ;;'a
section 16.1.
-See Eskin [86]. ·*Schwanz [45]. Uons [46]. Bogolyubov and Parasyuk [78]. and Vladimorov [42]. For functions in L 2 • the corresponding results were obtained by Bochner and Martin [3]. Cllapter VL See also Paley and Wiener [155] and Planchere1 and POlya [156].
240
SOME APPLICATIONS OF THE THEORY
5. Proof of Theorem 1 Suppose that fez) EHp(a+£. C). where p> I and a> 0, that is, suppose that fez) satisfies inequality (18) for some nonnegative values of a. and ~. Then, on the basis of section 26.3, there exists a boundary value f(x)=
lim y-+o. yE C
(23)
f(x+'y).
which is a weak limit in the space S(ml: where m=a+~+n+3. Let us choose an integer N such that 2N > m and let us' set
a=[~]+l.
c=L~r.
(24)
where (x] denotes the greatest integer not exceeding x. Let g = rr U ) be the spectral function of f (z). We have (see section 26.2) g(a),=eey rl (f(x+'y)] (a) = = F-1[ (2 + c2 ,e,20 +(x+iy)(x +iy»)N f(x+ly) eey] (~).
{2 + c2,e,2o+ (x + iy) (x + ly)I N
By using the properties of the Fourier transformation in S· (see section 3.9), we may represent this last expression in the form of a finite sum (19) with functions 0, of the form O,(a)=p,m
J
!(x+iy)e-Ie(X+/Yldx (2+c l ,e,2'+(X+ly)(X+ly)Jn , '
(25)
>-
where the P, are polynomials, the n, are integers N, and y is an arbitrary member of C. Let us show that the integral in (25) is independent of y for all yEC such that lyl2 < c2IH-+ 1. For such y, 12+c
lal2G +(x + ly)(x+ ly) I >-
>- 12+c2IEI2°+1 xl2 _I yl21 >- I + I
X
12.
(26)
Therefore, the integrand in (25) is holomorphic in the domain R" + I [y : y EC. y2 < e2 1 a 12• + 1]
and, by virtue of (18~, it does not exceed in absolute value the quality c" (y) (1 I x I )-n-. Consequently, on the basis of the CauchyPoincare" theorem, the integral in (25) does not change if the plane of integration z = x + ly (where y = const) is disflaced along y while remaining in the region y EC, where I y 12 < e21e I + I.
+
241
FUNCTIONS THAT ARE HOLDMORPHIC
Thus, if l.v I < 1 and y EC, the representation (25) is independent of .Y for all e and defines a continuous function e in R". Furthermore, on the basis of the choice of the number Nand the inequalities n, ~ N, we have y_O, yEC
,-tu
,-I, (x+ly)
- - - - ' - - - - - - - : : - -+ [2 e2 1 12• (x iy)2t' [2
+ e + +
+ c 1e 1 + I X 2)", 2
2•
0). Since P.c (e) is a continuous homogeneous function of first degree, there exists a constant 1 = 1 (cJ such that (28)
Let us now estabUsh the
= [~: j.lc (~) > 0]. For an arbitrary number "I in CO,D, there exists a cone C' = c' 1 > 0 for all eEpr C:. For the cone C' we take a cone for which (see Fig. 62) pr C' =
n
,'E'pr Be
[y: l.v - y/l > li
' JI Epr c].
242
SOME APPLICATIONS OF THE THEORY
where the number M ~ 1 is taken sufficiently large that prC'"p 0. Suppose that rCI(l
[1'~~e)Jp} X
+ IE!>· X
X exp{ -a'( 1- P''f/ - ~:)[II-e (E){}-< -< M; (CJ exp {- (a' - a) [II-e (E)t}. Obviously, on the basis of inequalities (28), inequality (20) is also valid fori EI < ape Conversely, suppose that g (~) is the finite sum (19) with continuous functions a, (e) of power increase that satisfy inequality (20) in the cone C * for some a' > 0 and p' > 1. Let y range over a cone C' that is compact in a (C). From Lemma 2 in section 25.2, there exists a positive number 8 and a cone C" such that ey~8IeIIYI.
(~.
(32)
y)EC"XC'.
Here, the cone C· is compact in C ** (see Fig. 59). Denoting the highest power of increase of the functions a, (E) by r, we derive from (32) the following inequality, valid for all (e. y) EC" X C': (33) where d is some positive number. On the other hand, if eEC" , it follows from the inclusion C"=sC* that E ranges over a cone C' = R",C: which is compact in C*. In this case, inequality (20) yieids
for arbitrary s > O. Let us recall (see section 25.1) that
- Ey -< 11-0 (e) (e) I y I.
11-0 (e) (e) -< pcp.e (e).
eEc•.
(35)
From inequalities (34) and (35), we conclude that
10, (E) ,-£11-< M;I(CJexp {-(a' -s)[lI-e (E»)p' +
+ Pell-e (E) Iy il, 'f/ > 0 for all (E.
y) EC:
X
C:
(36)
244
SOME APPLICATIONS OF THE THEORY
In particular, it follows from inequalities (33) and (36) and from the representation (19) that g (e) ,-EY E so for all yEO (C). Therefore, the function f (z), the Fourier- Laplace transform of gee), is holomorphic in TO ICl (see section 26.2) and can be represented in the form (see section 3.9)
~(-IZ>'
,
f(z)=
JO,(e)e'~de.
(37)
zETO(C).
Let us find a bound for each individual integral in the sum (37). By using inequalities (33). (36}. and (28) and Eqs. (21). we obtain the following chain of inequalities for z ETc':
1/ O,(e)elF.zde/< /lo,ml,-EYde+ / c·
la, (e)1
de
+ +
where N is an integer such that 2N ~ n 1. and a is the Laplacian operator. Since eo Ec., it follows from the lemma in section 26.5 that, for an arbitrary sufficiently small number 'Ij', there exists a vector yO Epr C (see Fig. 62) such that (39)
Let us set Y=AYo, where A is an arbitrary positive number, in Eqs. (38). Assuming that 'Ij and .,.' are sufficiently small so that
we obtain from (39), for all eE Sf' the inequality ye =>"[yo~o+ yO(~_~O») 0 and). > 0,
K:
where the constant ('f, eO) is independent of A (though dependent on the cone e', that is, on yU = yO (eO». If we take E < 't/ in this inequality and let A approach infinity. we obtain (g. 'f) = O. that is. g = Ofor EU (~o. 't/)as asserted. Conversely, suppose that g (e) ES· and g m= 0 for POe m> a • Therefore, the carrier Sg is contained in the closed set (see Fig.
m
e
59) F
= Ie:
POe (~) a).
But we
10 1Y 12 for 0 > 1. Suppose also that, for an arbitrary function rP E D [U I)
(xO, &",/)
since (by virtue of the
Continuing in this manner for arbitrary xU EQ and remembering that '1j = 40 (xO), we conclude on the basis of the holomorphic continuation theorem (see section 6.1) that I(z) is holomorphic in the domain Tk U G.• Here, we take & = ~I • This completes the proof. Thus, we only need to prove the special case of the "edge of the wedge" theorem. We shall do this in two stages.
2. The case of continuous boundary values Suppose that a function fez) is holornorphic in T r and continuous in Tr 2 '" 2 TJ TJ • where r = rl is the light cone Yl > I Y I . Then. fez) is holomorphic in the and hypersphere I z I ~
i
254
SOME APPLICATIONS OF THE THEOREM
max If(z)1 ~
-< max If(z)l· -r
(61)
zET~
Izl I. 'Ij
(62)
and we write the inclusion
This is true because fI = psi n 6 4= 0 and, hence, from (62) we have (1m W1)2 -llm.w 12=p2 sin26[(."o-2PXl cos 6)2_4p2/x/2 cos2 6]== = p2 sin26 (.,,0 - 2px) cos 6 - 2p I Icos (3) (Tjo - 2px) cos 0 (64) + 2p I Icos 0):> p2 sin 2 6 (Tjo - 2p I x 1 / - 2p I 1)2 ~ ~ p2 sin2 e(Tjtl_ 2 V2 pi x I f ~ p2sin 20 (rp _pTj,)2 > O.
x
x
Furthermore, for I z 1-
(i')dx'.
It follows from this definition that, for every cp ED [ U
(0, Vi"lj) ],
the function (P. 7) is continuous with respect to y and hence bounded on every compact set contained in the set[y: y Er, U [OJ. Iyl In particular,
I(T'. ",)1 I); 121 is a light cone. (It was shown in section 27.1 that the case of an arDitrary cone C, where en (- C) =1= 0 can be reduced to a light cone at the price of poorer approximations.) We recall that a cone r consists of two components, namely, a future cone r+ = [y: )II > l.vl ) and a past ccme r- = [y: )II < -I yl ). Thus, the function / (z) is holomorphic in the tubular cone rT: and the limit (59) exists for arbitrarn ED (0). According to the "edge of the wedge" theorem (see section 27.1).!(z)is holomorphic in the domain T r U6. where 6=
U lz:
.reQ
Iz -xol
< II~Q(xO)).
0
< II < 3~ .
260
SOME APPLICATIONS OF THE THEOREM
The question arises of constructing an envelope of holomorphy
H (1'1' U6) of the domain 1'1' U6. This problem has been solved only for a particular kind of domain O. Examples will be given
below. Let us suppose that H (1'1' U 0) is single-sheeted. Then. by using the "disk" theorem (see section 17.2). we can construct the holomorphic extension of the domain 1'1' U Consider a point xO EO. Then. U(xtl. "I)IEO for arbitrary "I < ~o(xO). We define b=("I. 0) and a =(0. a), where the complex vector a is such that laI2="i~. Let us show that the circle lei < I. which lies on the two-dimensional analytic surface*
a.
is, for sufficiently small A such that 0 - V32(m-lxm2-(XI-X~)2- Y~
(80)
in the circle (81)
for all 1x~1 x~ -
>
m2 - 2(xl -x~?+ 2&2(m -I x~ 1)2
3y~ -
(82)
in the circle (81). But this Circle lies in the strip IxIi m;
y)-lyl=o(l).
is unique and is given by the formula
Izl_oo
265
A THEOREM ON C-CONVEX ENVELOPES
u(x. y)=llm
V z2_m21.
(2) For n = 4 formula (77) enables us to prove the dispersion relations for a 1t-meson nucleonic dispersion (see section 33.4) in the interval 0 < - t < 8p.2 of variation of the square of the impulse transfer- t, where /.l is the mass of the 1t -meson (see Bogolyubov and Vladimirov (39) and Bremermann, Oehme, and Taylor (21).
28. A THEOREM ON C-CONVEX ENVELOPES
1. Definition oj a C-Convex Envelope Suppose that an open C consists of connected component cones Cl' .... Ct· For the cone C, let us construct new cones CO and C as follows: if the cones Cat for k = 1. 2..... t. are not all convex. let us consider the cone C(J) = U 0 (CII) and let us denote its components by C\l). .... C~:), for It~ If the cones C~) are not all convex, we introduce the components C~) • •••• C~:, where t2 < t l , of the cone C(2)= O(C~)) t etc. (see Fig. 68). After a finite number of
0). These constitute the cone r = [y ~ 0). Therefore Br(S) = S for n = 1. _ If C= R". then, obviously.Be(S)=R". If C = 0. then Be (S) = S • In all cases. either Be (S) = S or Be (S) is n-dimensional;Be [Be (0») = Be (0).
Fig. 73
Fig. 72
Suppose that c = r is a light cone. We shall give three examples of open r-regular sets Oc.R" for which the envelope Br(O) is essentially greater than 0: Oo=[x:
x~ < 6IxI2];
Br (Oo)=R".
6 > 1. Br (00) = Os. 0 < 6 -< 1 (see Fig. 73); Ot=[x: x~< IxI2+1]; Br (0 1)=00• 1- 0 (see Fig. 74);
in particular, Br(O_oo.oo)=R".
BdQ-oo.o)=[x: xl+lil<E. x1. forj=l. 2, .... n.where O 1. a> 0;
(90)
p= 1. a::;;;"O.
Proof: From Theorem 1. the function I(z) is an entire function. Therefore. the generalized functions 1k (x) coincide in RIt. From the theorem in section 26.7. f (z) satisfies inequalities (90), which completes the proof. COROLLARY (a generalization of Liouville's theorem). If p = 1 and a
=
Oin the conditions of Theorem 2, the function f(z) is a polynomial. Proof: It follows from Theorem 2 that f (z) is an entire polynom-
ially bounded function. From Liouville's classical theorem (see section 6.3), fez) is a polynomial of degree not exceeding ~.
SOME APPLICATIONS OF THE PRECEDING RESULTS
275
Remarks: (1) On the basis of the remark to the "edge of the wedge" theorem made in section 27.1. the condition that Cn (- C) ~ 0 in Theorems 1 and 2 can be weakened and replaced with the condition that O(C) = W. (2) The condition that Be (0) = R" is sufficient but not necessary for Theorems 1 and 2. The problem arises as to how we may characterize all open sets 0 that verify these theorems. In section 32.4. this will be done for a light cone r. for functions belonging to· the class Ho(r), and for r.-regular open sets C (see section 28.1).
2. The class Hl (a; C) For the definition of functions in the class H) (a; C). see section 26.4. Suppose that a cone 0 is connected. According to Theorem 2 (section 26.4). for a function /(z) to belong to H.(a; C). it is necessary and sufficient that its spectral function g mvanish for IlC<E) > Q. From Corollary 1 to this theorem. the class H) (a; C) coincides with the class H.(apc; O(C». From this and from Theorem 1 in section 29.1. we have the following THEOREM.* For a function f(z) to belong to 111 (a; C), where the cone C consists of the components C1. C2 • ...• C t and C (-C) "I CJ. to be holomorphic in the domain rC U G, it is necessary and sufficient that its boundary values fk (x). for k = 1. 2 •...• t. coincide in G and that its spectral functions gk «() = [fkl vanish when Jl.C k «() > a. Here. it is necessary that f(z) be holomorphic in TCo U Bc (G) and hence that the f k (x) coincide in Bc (G).
n
,..-1
In what follows. we shall asswne for simplicity that the cone C consists of two convex components in C± such that C- = - C+. We shall denote the corresponding boundary values and spectral functions by j± (x±IO)=
lim y-+O, yE c'"
j± (x
+ ly),
g'"
m=
(91)
Consider the case of a = O. In this case. the class HI (0; C) =Ho (C) constitutes a class in the sense of section 16.1 (see remark in section 26.4). Therefore. the set of class Ho(C) functions that are holomorphic in the domain rcUQalso constitutes aclass K=K cU(in the sense of section 16.1). T a We shall refer to the largest domain in which every class K function is holomorphic (and single-valued) as the envelope of TC U and shall denote it by K (TC U0). The following questions arise in connection with this definition:
a
-See Vladimirov (30).
276
SOME APPLICATIONS OF THE THEOREM
(1) When do the envelopes H (rc u 0) and K (rc u 0) exist, that is, when are they single-sheeted? We note that these envelopes always exist in the class of nonsingle-sheeted domains (see section 20.1) and that H (TC U O)c.K (TC U 0).
(92)
(2) When does equality hold in the inclusion (92t? _ (3) How can we construct the envelopes H (TC U0) and K (TC UO)? In the example that we considered in section 27.5 (where c=r and O=!1 x II < mD, the envelopes H and K were single-sheeted and they coincided. A second example of the construction of the envelopes Hand K is given by Theorem 2 of section 29.1: if BC 0). LEMMA. Every generalized function g ESO with carrier S6cF can be represented in the form of the difference g = g+ - g- of the generalized function gi ES" with carrier SgtcFt. This representation is not unique: the gt are defined up to an arbitrary generalized function 1j (the same for g+ andg-) with carrier contained in the compact set F+ n r. In particular, for a = 0, this 1j is an arbitrary finite combination of the P.-function and its derivatives at zero. Fi~. 81 Since F is a regular set, the generalized function g can, by the theorem in section 3.8, be represented in the form g=
~ D~
I_I
",,If
•
<e>.
where the I-'.m are measures of power increase with carrier in F. If we set
278
SOME APPLICATIONS OF THE THEOREM
11': (e) = 11'. (e n F+), 11'; (e) = 11': (e) - flo. (e), for an arbitrary bO\Ulded Borel set e. we obtain the desired generalized functions in the form
Let us suppose that there are two representations: g=gt -g) =g; -gi, S +cF+,
k= 1. 2.
S _cF-,
Kk
Kk
Then. the carrier of the generalized function '1}=gt -g; =g)-gi
is contained in F+
nr = [~ : I-'c+ (E) < a and I1'c (E) a.
This means that IE LfJ (e). Conversely, suppose that IE LfJ (C). From the preceding lemma, we can represent the generalized function rllll ESO in the form of the difference (99) of generalized functions g± ESO such thatSgi cp%. According to the theorem in section 29.2,
I± (z) =
F
rg± (~) e-EYI (x) EHI (a;
C±).
By considering formulas (91), we obtain from this equation the desired representation (96). Since the ftmctions g± are defined up to an arbitrary generalized function with compact carrier in p+ nP-, the functions I± (z) are defined up to the corresponding arbitrary entire function. In particular, for j± (z), the functions a = 0 are defined up to an arbitrary polynomial. Let us now prove the second part of the theorem. Suppose that !(x)=O. Then, on the basis of (96), we have
t+ (x + iO) =
r
(x - fO).
x EO.
By applying the theorem in section 29.2, we then obtain the remaining assertions of the theorem. It follows from this theorem that generalized functions belonging to the class La (e) possess the property of quasi-analyticity: vanishing of a function I (x) in this class in a domain a implies that it vanishes in the C-convex envelopes Be (0) of the domain O. This kind of quasi-analyticity for a causal commutator in quantum field theory was predicted by Jost and proven by Dyson in [41] (see also Bogolyuhov and Vladimirov [97], Vladimirov [92, 30, 40,42], Borchers [98]. and Araki [99]). The reason for this phenomenon of quasi-analyticity consists in the fact that although l(x)E La (e) is a generalized function. it can nonetheless be represented in the form of the difference (97) of the values I (x ± iO) of some function I (z) belonging to the class U HI (a'; e) (see section 26.4) that is holomorphic in the domain
fJ'~4
a
U and hence holomorphic in the corresponding K -convex envelope. which cannot be arbitrary (it must in any case be pseudoconvex [see section 17.1]. Therefore. at its real points (in which, by virtue of (97). I(x) must vanish). it cannotbe arbitrary.
]'C
280
SOME APPLICATIONS OF THE THEOREM
4. Quasi-analyticity oj the solutions oj a certain class oj linear equntions Let us use the results of the preceding section to establish the quasi-analytic nature of solutions u (x) in the space S· of the convolution equation (100)
with the hypotheses
IE La (C). 10E O~. for
F
lIolm +- 0
min [I-'c+ (;). I-'c- (~)]
> a.
(101)
We note that, since 10 is a convolutor in So. the function F 1101, is a multiplier in S', and all its derivatives are infinitely differentiable functions of power increase (see section 3.9). Convolution equations include differential equations with constant coefficients P(LD)u =/(x).
10 = P (LD) B(x).
(102)
difference equations
integral equations
f K(x-x')u(x')dx'=/(x).
10=K(x)
(104)
and various combinations of these: differential equations with displaced arguments, integro-differential equations, and others. The function F [/01 (e) takes the forms
pm.
~
eke IX• t •
f K(x)e1x a.: that is, u ELa (C). If we now apply the theorem in section 29.3, we get the following result (see [30, 93]).
r
SOME APPLICATIONS OF THE PRECEDING RESULTS
281
THEOREM 1. ";very solution u € g" of the equation fo • u = f that satisfies conditions (101) and vanishes in an open set G can be represented in the form of the difference (97) of the boundary values fez t iO) of a function fez) € HI (a; C) that is holomorphic in the domain T e U Be (G>. This solution therefore vanishes in the larger domain constituting the envelope Be (G). . COROLLARY 1. I( two solutIOns of Eq. 000> that belong to S coincide in the domain G, they also coincide in BeW>'
.
From Remark 1 in section 29.3. we obtain the COROLLARY 2. Suppose that F [fo]- (~) = 0 outside two closed convex cones L + and L - (where L - = -V) that do not contain any entire straight line. Then, in the representation (97) of each solution u € S* of the equation fo • u = O. the function fez) € HoW). where C = C+ U C- and C t = intU· For example. every solution u ESO of the iterated wave equation Omu = 0 (for In:> I that vanishes in an open set 0 also vanishes in Br(O). that is. in the convex envelope of 0 with respect to timelike curves. Proof: In this case. fo = Omil (x). FlfoJm=(-e~+~+
r±.
...
+E~t
r
and F± = that is, u ELo(l'). Noting that int u = r± and I' = r± u rand using Corollary 2. we obtain the assertion made. THEOREM 2. Suppose that the equation
(105) can be solved for the highest derivative with respect to Xl and that this equation satisfies condition (101):
(106) Let u € S* denote a solution of Eq. (l05). Suppose that this solution satisfies the conditions
u (0.
x) =
-=-Ou--,(,-,-,O.,-x~)
(107)
oX I
w~re
g is an open set lying on the hyperplane Xl = O. Then, u(x) vanishes in the envelope Be (g) and. in particular, Eqs. (107) are valid in a larger set than g, namely, in the intersection of Be (g) and the plane Xl = O. Proof: Since an arbitrary derivative D"u (x) belongs to S· and
satisfies Eq. (105) and since the differential operator t"Df-p(/D) is hypoelliptic with respect to Xl' we conclude that D"u (a. x) ES' for all a (see section 3.11). Therefore, conditions (107) are meaningful. Suppose that the function ,?(x)E D(U{O. a» approaches a(x) as 6: -> + O. Since u (x) E S· satisfies Eq. (105) and conditions (107), it follows that
282
SOME APPLICATIONS OF THE THEOREM
satisfies this equation in the ordinary sense and satisfies the conditions - -0'u . , (0, x)=O, dxj ,
(108)
a=O,l, .... k-l,
where g (e) is the set of points in the domain g that lie at a distance from oggreater than e (see Fig, 82). By using Holmgren's theorem (see Petrovskiy [105J. p. 49). we conclude from (108) thatu~ (x)=O in some n-dimensional neighborhood g' (e) of the (n - l)-dimen'sional set g (e). From this. we conclude on the basis of Theorem 1 that u" (x)=O in the envelope Be Ig' (e») and a fortiori in the envelope Be Ig (e») c: Be Ig' (e»). Now let e -. O. Then. in S-
+
u,,(x)_u(x),
_
~---_X,
d·" (0, x) ax;
a·"",• (0, x) axj
_
(109)
,a=O, I,., ..
Here. the sequence of open sets g (e) increases monotonically and g = U g (e). .>0
,Xl
According to the lemma in section 28.1. the sequence Be Ig (e») also increases monotonically and Be (g) =
U Be Ig (e»).
(110)
.>0
Since /I~. (x) = 0 in Be Ig (e»). it follows from (109) and (110) that u,.(x)=Ofor Be Ig(e»). Analogously. it follows from (109) and (110) that Eqs. (107) are valid in the intersection of Be (e) and the plane Xl = O. which completes the proof. The quasi-analytic nature of the solutions of the wave equation was pointed out in the book by Courant and Hilbert [88] (Chapter VI. § 8). For example. if a solution vanishes identically in a neighborhood of the interval 1 x I I -< I. = o. then it must also vanish in the double cone IXII + 1.;1 < 1(see Fig. 75). In connection with this. it is stated in that book that we encounter the remarkable phenomenon of the existence of nonanalytic functions whose values in some arbitrarily thin region tmiquely determine the behavior of the function in a much larger region. Similar phenomena of quasi-analyticity have been studied by John [89] Nirenberg[90] and Broda [91] in connection with questions on the Wliqueness of solutions of partial differential equations (in particular, those with constant coefficients). Nirenberg [90] offered the hypothesis that if a sufficiently smooth solution u (x) of a
x
SOME APPL.ICATIONS OF THE PRECEDING RESUL.TS
283
differential equation P(tD) /I = 0 vanishes in a domain O. it also vanishes in the domain BP (0). that is. the convex envelope with respect to timelike curves. A curve of class ell) is said to be timelike with respect to the operator P (tD) if. at an arbitrary point ~o of the curve. the equations et = O. Pu = 0 are simultaneously satisfied only by ; = o. where t is the tangent vector to that curve at ~o (see Fig. 83) and Po mis the principal part of the polynomial P <e>. We note that. for the wave operator D. timelike and r -like curves coincide. so that BO (0) = Br (0). These results prove that Nirenberg's hypothesis remains true for solutions in the space S· when the envelope BP(O) is replaced by the envelope Be (0). It would be interesting to extend these results to the space D' (0). The method of studying regions of uniqueness of solutions of differential equations with constant coefficients that Fig. 83 we have expounded here is based on the property of pseudoconvexity of domains of holomorphy. Therefore. it is quite different from previously used methods of studying such questions (see [88-91]).
m
5. A boundary-value problem jor class holomorphic junctions
Ho (C)
The results obtained in section 29.3 enable us to solve the following problem concerning the conjugacy of two holomorphic functions. Suppose that an open cone e c R" consists of two convex components C+ and C-. where c- = - C+. Let j (x) denote a function belonging to S·. Find two functions (z) belonging respectively to the classes Ho(C±) such that
r·
j+ (x +10) =
r
(x -
10)+ I(x).
(111)
This problem is a generalization of the well-known classical problem of determining a piecewise-holomorphic function from a given jump on the real line (see. for example. Muskhelishvili [107] to the case of many complex variables and to a broader class of admissible jumps in I. For n = I. problem (111) and its analogs have been studied in various spaces of generalized functions by KOthe [130]. Tillman-[131-133]. Parasyuk [108]. Sato [134. 135]. Bremermann and Durand [106]. Luszczki and Zielezny (136). and Rogozhin [137]. The similar problem (for /I >- 1) on the representation of an arbitrary generalized function in the form of the sum of 2" bmmdary values of a function that is holomorphic in the open set
284
SOME APPLICATIONS OF THE THEORY
(Cl " Rl) X (Cl " Rl) X ,,' X (Cl " Rl)
has been studied by Tillman (131-133). The theorem in section 29.3 provides the following condition for solvability of problem (111). For problem (111) to have a solution. it is necessary and sufficient that f (x) ELo (C)i this solution is not unique but is determined up to an arbitrary polynomial. Thus. the conditions under which the problem has a solution are quite different in the spaces Cl and en. where n ~ 2i in the space Cl. we know from Remark 2 of section 29.3 that the problem has a solution for all f ES' whereas this is not the case in cn for n >: 2. We shall now give an algorithm for solving this conjugacy problem. Suppose that f ELo (C) and g = F- l If). By hypothesis. SgcCHuC-o. where C-*=-C+o and C'= (OJ =c+onC-o. Therefore. there exists a hyperplane Ee = 0 (where e ECT) such that ± ~e > 0 for eEe±' and h t O. Let N denote the order of g ES·. Then. a±
m=
± (Ee)N+1a (± ee) g (E) ES',
Sa± c e±'
and (112) Since division by a polynomial is always possible in the space SO (see section 3.10), let us divide Eq. (112) by (ee)N+i. This gives us a representation of g in the form g(;)=g+(e)-g" (e),
(113)
where the generalized functions g± ES· are chosen in such a way that (114) Setting
we obtain. by virtue of (113). the solution to our problem. The spectral 'functions g± in the representation (113) subject to conditions (114) are not unique. Let us find their general form. Let g = gt - g) be another such representation. Then. '7j = g+ - gt = g- - g). Since the carrier g± is contained in c H • the carrier'7j is
gr
SOME APPLICATIONS OF THE PRECEDING RESULTS
285
From this. we conclude on the basis of formula (16) of Chapter 1 that the generalized function '1/ can be represented in the form '1/ =
~
cpoa (e).
I'j :1: (z) is a solution of the problem (111). then
r
j -( )_ /I.I-(z) z -
P(z)
is a solution of the problem (115). To generalize what has been said. it will be sufficient to show that it is always possible to divide by a polynomial P (z) such that P(z) f- 0 (for z ETc in the ring Ho (e-). that is. that the equation P(z)x.(z)=1I> (z) has a (Wlique) solution x.(z)in the ring Ho(e-) for any right member II> (z) in Ho (e-). This assertion follows from the LEMMA. If the polynomial P (z) t- 0 in the tubular radial domain T C = Rn + iC. I / P (z) E No (e). Proof: If P(z).;. 0 in TC then P(z) =1= 0 in its envelope of holomorphy TO (C) (see section 20.3). Therefore we may assume e to be a con-
then
vex cone. Since P (z) i= 0 in re. the function P ~z) is holomorphic in TC. Thus. it remains only to show that this function satisfies inequality (IS) (for a = 0). For this, we use the second algebraic lemma in the article by Hormander [81]: Let Q(x) be a polynomial with real coeffiCients. let N be the set of its real zeros, and denote the comment of N by eN = R n '\ N. Then. for certain values of a. b. and c (where c> 0), IQ(x)1
>- e(\ + Ixj)a ~h(x).
(For N = 0. instead OB~N(X), we may take 1.) Now let P(z) denote an arbitrary polynomial and let M denote the set of its zeros. Again. eM = en '\ M. If we apply H"ormander's
286
SOME APPL.ICATIONS OF THE THEORY
lemma to the polynomial Q(x. y)= IP(z)12 (with real coefficients) the real zeros of which coinc ide with M. we see that this lemma remains valid in this case also:
Suppose that the polynomial P(z):f= 0 in TC. If M = 0. then. on the basis of inequality (116). we have P~Z'1 EHo(C). Suppose thatM '*- 0. If ~ o. in (116). we conclude. bn the basis of the inequality
-
o. Since the cone C is convex. for all z ETC we have (see section 25.1) C"=C intC=C. and
Let C' be a cone that is compact in C. It follows from Lemma 2 of section 25.2 that yEC',
a=a(C') > o.
From this inequality and inequalities (117) and (116), we conclude that
that is,
P ~Z)
EHo (C). This completes the proof of the lemma.
6. The impossibility oj nonlocal theories oj a certain type Wightman posed the question [24}: Does vanishing of the commutator [A (x). B (y») for (x - )')2 < _[2 imply'" that it vanishes for (x - y)2 < o. This question was answered in the affirmative by Petrina [43} by means of the theorem of a r-convex envelope [92}. Another proof of this result appears in Wightman's article [32}. We shall present the solution of this question, following Petrina. *Here, we use the following notations, which are employed in quantum mechanics: (337m) (see section 30).
SOME APPLICATIONS OF THE PRECEDING RESULTS
287
We are dealing with quantum field theory. inwhich the following axioms are assumed satisfied: (a) In the Hilbert space ~ of amplitudes of state I). the linear operators of the field A (x). B (y) . ...• are given. These are operatorvalued generalized functions in S· with a common domain of definition D that is dense in .p. Here. AD c D. (b) Invariance under the group of translations. This means that in the space .p there is a unitary representation U(a)D c:: D
of the group of translations !a}.whereP = (Po. PI' P 2• Pa)is the energymomentum operator and E (p) is the corresponding spectral measure. (The operators Pi are self-adjoint in .p and they commute with each other.) This representation transforms the operators of the field A (x) according to the rule U (a) A (x)U(- a)= A(x - a).
(c) Completeness of the system of (generalized) * eigenamplitudes of the state Ip) of the energy-momentum operator p. Pip) = pip). P=(Po' PI' P2' pa) = (Po' p) and nonnegativeness of the spectrum of the energy operator Po- Thus. the carrier of the spectral measure E (p) is contained in the closure of the future light cone Ii> = [p : Po ~ O. p2 ~ 0[; that is. E (p) is a retarded function. (d) Nonlocal commutativity: For any two field operators A(x) and B (y). the commutator [A (x), B (y») = 0
if
(x - )/)2
0 respectively. Ther~fore, the boundary values of the function l/zk as y~ ±O exists in S' (see section 26.3). We denote them respectively by 1/ (x ± iO)k. From the definition (119) ofthe product of generalized functions, it follows that (
l)k x ± iO
=
1 (x ± iO)k .
293
SOME APPLICATIONS OF THE PRECEDING RESULTS
Suppose that k = 1. Using the definition of the Cauchy principal value (at the points x = 0 and x = 00), we obtain for all Cf' ES,
J Ixll and a past light coner-=[x: xl 2), by the formula
« n-I )/2, (where
J
0 (± x o) a(t, (x 2)'P (x) dx = = (-1)"/
(_iJ_)t ('I' (~o. x) ) I 2xooxo 2.xo.
dx. cp ES.
(124)
x.-±Ixi
For
k
E (see section 31.2). Since Dna = 0, the first of formulas (156) follows from the formula just obtained and from (154). The second of formulas (156) follows from the first and from the equation a (x) =
u(x) + a (x).
(158)
Since the carriers of the elementary solutions £(±) of the wave equation [see section 30, formulas (146)] are con~ed in r and
308
SOME APPLICATIONS OF THE THEORY
since the carrier of I\In is contained in E, their convolutions 8 n:l: I * I\In exist in D'. This is a consequence of the following property of a strictly spacelike hyper surface E: For any compact setK, the set of points
is bounded (see Schwartz [11], Vol. II, p. 26). Therefore, the representations U=E- o. Therefore. to prove the asserted properties of the convolution
it will be sufficient to show the existence of numbers N ~ 0 and K>O such that, for all If Es.
III" cp IIm.F-I"'I>' 1+lxl )p+m ~K2I1cpIiN X)X~L, ( 1+Y2(1-a)lxllx'l
• OU -,,(xo) oXo
~u. + -:(8 (x o) u.)_ uXo
(166)
uXo
+ oXoo.(,,(xo)u)='1'o(u: x).
e_+O.
Furthermore. from formulas (154) and (157). we obtain
"'E (u.: x) =
0 (E)
=
~~ + o~
[0 (1:)
(0 (E) u,) =
°o~n + o~
(167) (0 (E) DOl)]
* 'fo (u.:
x).
Since the carrier of the generalized function in the square brackets (in 167) is contained in f and since the carrier 0, they satisfy the inequality If±(z)l, I. (The case in which n = 1 is easily examined, but we do not need it.) As a preliminary, let us find an integral representation for all derivatives D1f± (z) of sufficiently high order 1 such that Ill>-s. where s is an integer to be chosen later. If we differentiate the representation (168) and use formula (143). we obtain
(172)
where 1
cn=2i(n_I)~
-!!.!..!. 2
n+l r(-2-)'
In what fOllows, we shall need the following LEMMA. 1. Suppose that a space like hypersurface 1:. is such that - 00 < a ~ xo ~ b < +00 for all x E 1:.. Then, for arbitrar:y m ~ 0, there exists a nonnegative
number s such that
312
SOME APPLICATIONS OF THE THEORY
(or all y in an arbitrary subcone I YO I ~ (1 + 0) I y I, where 5 > 0 and all 1 such that III ~ s. Here. a ~ 0 and 8 ~ 0 do not depend on O. Also. (or all
0, where y
E ~.
According to the lemma in section 30. the function (Z-X')2 400 for all z ET and x' ERn+l. Therefore, the function [_(z-x')2J- 1 is infinitely differentiable with respect to x'. Suppose that Yo > O. To prove inequality (173). we use (141). Assuming s = s (m) sufficiently large and remembering that Proof;
for
yo>-(I+o)lyl. we obtain a chain of inequalities for III>-s:
~DI[_(Z_X')2J-n;1 = KI
(l
sup
Ill<m,x'EI
-< K2
t.I =
+ Ix'l)m If 6 (~o) 0 (~2) ~T+l e
l (z-x'),
d~
1-
O. According to the lemma in section 30, the furiction [-(Z-X')2)-1 is holomorphic in T= for all x'. Suppose that s=s (m) > 0 is the integer defined in Lemma 1. Then, it follows from inequalities (173) that, for all 1 and) (where Ill>s, and 0 ~J 0). &
-!lo, Im~=O, that is, in a simply connected domain. Therefore, it can be approximated to any deSired degree of accuracy by means of polynomials in this cut plane (see section 24.B). But then, the function ,!,p. (Z2) can also be approximated by polynomials to -The idea of introducing such a class is due to Shinnbekov.
330
SOME APPLICATIONS OF THE THEORY
any desired degree of accuracy in the domain ". It follows from this that the class of polynomials is dense in the class K (in the sense of section 16.2). According to the lemma in section 16.2. the domain " is polynomially convex. which completes the proof.
2. Integral representation oj junctions of class
lio(r)
We shall say that a r -regular open set 0 satisfies condition (A') if N (0) # 0 and there exists a spacelike hyper surface E contained in the set Br (a) u N (a) and in some strip a -)Da-(z-x'Y+>-2r2 dx'dA+
(220)
+Pl(z, 0), III >s
are holomorphic in K' (T U 0) if s is sufficiently great. It follows from the definition of the envelope K' (T U0) (see section 33.1) that
Therefore, the functions n
Cf'/(z; x', >-)=D~ !-(Z-X')2+ A2)-1"
are holomorphic with respect to z in b < differentiable with respect to (x', A) EN (a). We now need the
+ =)
and are infinitely
LEMMA. F'or an arbitrary nonnegative integer m and subdomain C'
=
K'(T U
C\
(221) for all 1 such that III ~ s, where s = max [0, m - nl and the set F' is defined in (219). Proof: Let z range over A' <sK' (T U 0). Then, for (x', A) in the strip a x~ and K' (T U 0), we have the inequality
-
.(C2),
where the p. are polynomials and the flUlctions Il> (w) are holomorphic and polynomially bounded in the w-plane withcut along the line 1m w = 0, max (/)2, c2) -
