Exponential functionals

which by formulas (5.4) and (5.6) means that the definition of A(D)u(z) does not depend on the form of representing u(z). This is what was required. In conclusion we shall show that the Runge condition is essential. Counterexample. Let ~-~-CI~{0}, A (D)----I/D. Obviously, the symbol A(~) = I/~ is analytic in ~. We shall show that the operator A(D) is not single-valued in the space Expa(C~). Indeed, for any ~ ~ 0 we have by Cauchy's formula e} z--~ . 1

C

e nz

_,

and hence, breaking the contour of integration into N sufficiently small parts, we find that N

N

]=0 ~i

7=0

where 1

~j (z)= ~

J

~_~

d~.

r]

Obviously, rj can be taken so small that ]~](z)]:. 0 are constants. Thus, formula (5.7) gives a representation of the fucntion e xz in the space Expn(C1z) in the form (5.1). In correspondence with the definition of A(D)u(z) we have, on the one hand, [I/D] exp kz = X-lexpXz. On the other hand, proceeding from (5.7), we find that e rlZ d rl __~eZz_ 2zi 1 0 =~-~(exp~z--1). n (n--~) Ini=~ - -

Thus, in the space Expa(C~)if ~ is not a Runge domain the operator A(D), generally speaking, may be multivalued. In conclusion we note that an analogous representation holds for zn, n = 0, i,~

z=EExpa (C~), a = C : \ { 0 } : , 6.

Exponential

Functionals A continuous

The s p a c e o f a l l ponential functionals

e x p o n e n t i a l f u n c t i o n a l we d e n o t e by Exp~(C~) and c a l l a s s o c i a t e d w i t h t h e domain a.

.

1.

linear

functional

on t h e s p a c e E x p a (C z )n i s c a l l e d

D e f i n i t i o n 6 1. nential functional.

Ex_~le

i.e.,

and [I/Dlz.=z~+:/(n_l-1).

an e x p o -

the space of ex-

6(z) is obviously an exponential functional for any domain a.

Example 2._ Let A(~)Eg?(Q). Then h(z)=A(--D)8(z) (--D=(--0/aZl ..... --a/Ozn))is an expon nential functional acting on v(z)@ n xps(Cjaccording to the formula def

(h(z),v(z))--

(A(--D) 6(z),v(z))_--<6(z),A(O)v(z)).

We shall show that Example 2 exhausts all exponential functionals. THEOREM 6. i. h(z) = A(-D)6(z). Proof.

Let h(z)EExp'~(C~).

Then there exists a unique function ~4 (~)EO(~) such that

We set A(~)------(/z(Z),exp ~Z >, ~G~2. Obviously, A (~)66~(~2), and it is clear that aef

< A (--D)g(z), exp ;z > ----- < 8(z), A (O)exp ;z ) -----A(;),

i.e., the action of h(z) coincides with the action of the functional A(-D)6(z) on functions exp ;z, ~G~2. From the density lemma (Sec. 3) we conclude that the functionals h(z) and A(-D)6(z) are equal as exponential functionals. The uniqueness of this representation is obvious. The theorem is proved. Example 3. We consider the exponential functional ~(z)----exp(aD 2)8 (z), where a 6 Cn is a parameter, and aD==-g~D~-~ ...@gnD~. It can be shown (el. [ii, p. ii0]) that for any function 2757

v (z)6Exp (C~) and aj ~ 0, j = 1 , . . . ,n

where the integration goes over the lines (--VaToo. V~oo). passing through the points _+Va) of the complex plane sj. From this we immediately find that

-r

--r

i.e., ~(z) is the regular functional defined by the function contours indicated.

exp(--z2/4a) and

the family of

Remark. Another form can be given to the general form of exponential functionals. C n mxpa(C2)>we have Namely, as follows from Example 3 within the duality framework < Exp,(~), the formula 6 (z) = (2 V~) -~ exp ( -- D 2) exp ( -- Z214), ( 6.1 ) where exp ( - z 2 / 4 ) i s t h e r e g u l a r e x p o n e n t i a l f u n c t i o n a l d e f i n e d by t h i s f u n c t i o n and t h e r e a l c o n t o u r s of i n t e g r a t i o n - ~ < xj < ~, j = 1 , . . . , n . Hence, any e x p o n e n t i a l f u n c t i o n a l can be represented in the form h (z)----[A (-- D) exp (--D2)]o(2~)-~ exp (-- z~/4),

i.e., in the form of the action of a p/d operator with analytic symbol on the regular functional (2~/~)-nexp(--z2/4). To conclude the section we consider the algebra of p/d operators dual to the algebra ~(~)). For this we note that since the space ExPu(C2) is invariant relative to p/d operators A(D) with symbols A(~)60(e), the space Exp~ (C~) is invariant relative to p/d operators A(-D), i.e., the mapping A (-- m):Exp~ (C" ~ ) -~ Exp~ (C~) is continuous.

We denote b y ~t(fl-)the algebra of p/d operators A(D) with symbols A (~)6t~(~-), where ~- = {~:,~Ee},and with domain Exp~(C~). From what has been said we obtain THEOREM 6.2.

7.

The following algebraic isomorphism holds:

Operator Method i.

P/D Equations in the Entire Space C~_. We study the equation A (D) g (z) = h (z),

zEC~,

( 7. i )

where A(D) i s a p / d o p e r a t o r w i t h symbol A(~). THEOREM 7 . 1 . Suppose in some Runge domain f~cC~ t h e f u n c t i o n s A(r and A - I ( ~ ) a r e a n a lytic. Then f o r any f u n c t i o n h(z)~Exp~ (C~ ~) t h e r e e x i s t s a u n i q u e s o l u t i o n of e q u a t i o n ( 7 . 1 ) u(z)CExp.q( z] and t h i s s o l u t i o n i s g i v e n by t h e f o r m u l a

(z) A dual result

A@D) h (z).

(7.2)

can be f o r m u l a t e d s i m i l a r l y .

THEOREM 7 . 2 . Under t h e c o n d i t i o n s o f t h e p r e c e d i n g t h e o r e m f o r any f u n c t i o n a l h ( z ) (CA t h e r e e x i s t s a unique s o l u t i o n g(z)CExp'~-(C~) o f Eq. ( 7 . 1 ) , and t h i s s o l u t i o n ~s g i v e n E x p e l '* by f o r m u l a ( 7 . 2 ) . , C ~ the The proof of both theorems is obvious, since in the spaces Expa(C~)and Exp,-(~) operator I/A(D) is the inverse operator to the operator A(D).

Example i.

2758

We consider the complex Helmholtz equation Au (z) + ~o~u(z) = h (z), zCeC",

(7.3)