The ∗ -product of domains in several complex variables
aa r X i v : . [ m a t h . C V ] M a r THE ∗ -PRODUCT OF DOMAINS IN SEVERAL COMPLEXVARIABLES SYLWESTER ZAJĄC
Abstract.
In this article we continue the research, carried out in [28],on computing the ∗ -product of domains in C N . Assuming that ∈ G ⊂ C N is an arbitrary Runge domain and ∈ D ⊂ C N is a bounded, smoothand linearly convex domain (or a non-decreasing union of such ones), weestablish a geometric relation between D ∗ G and another domain in C N which is ’extremal’ (in an appropriate sense) with respect to a specialcoefficient multiplier dependent only on the dimension N . Next, for N =2 , we derive a characterization of the latter domain expressed in termsof planar geometry. These two results, when combined together, give aformula which allows to calculate D ∗ G for two-dimensional domains D and G satisfying the outlined assumptions. Introduction
Let O be the set of all germs of holomorphic functions at the origin of C N and let O ,D , for a domain ∈ D ⊂ C N , be the subset of O consistingof all germs of elements of O ( D ) . The latter symbol denotes, as usually, theFréchet space of all holomorphic functions on D equipped with the compact-open topology. The Hadamard product, called also the ∗ -product, can beregarded as a bilinear mapping from O × O to O given by the formula X α ∈ N N f α z α ∗ X α ∈ N N g α z α := X α ∈ N N f α g α z α . It was extensively studied in various aspects: as a bilinear form, as a linearoperator with one factor fixed (see the survey [22] and, for instance, thepapers [2], [3], [4], [5], [14], [17], [18], [19], [20], [21], [23], [24], [25], [28]),and also as a map acting on spaces of real analytic functions (see [7], [8], [9],[10], [11], [12]). The problem of similar nature as here, that is of extendingthe ∗ -products to as large domain as possible, was investigated, for instance,in [1], [13] and [16] for weighted Hadamard products with certain weights,where results were obtained for starlike domains and for so-called p -convexdomains (we refer the reader to [13]). Mathematics Subject Classification.
Primary: 32A05, 32D15.
Key words and phrases.
Hadamard product, analytic continuation, spaces of holomor-phic functions.The research was supported by NCN grant SONATA BIS no. 2017/26/E/ST1/00723of the National Science Centre, Poland.
In [28] it was shown (see Proposition 4.1 there) that if at least one ofdomains ∈ D, G ⊂ C N is Runge domain, then there exists the largestdomain ∈ Ω ⊂ C N having the property that the image of O ,D × O ,G under the ∗ -product lies in O , Ω (here, as well as in [28], we follow [15,Definition 2.7.1] and consider Runge domains to be pseudoconvex). Thislargest Ω , denoted as D ∗ G , was the subject of research carried out in [28],which concluded in finding a description of D ∗ G for domains of a specialclass. The methodology employed in [28], as well as the results achievedthere, were of symmetrical nature: both D and G were assumed to belong tothe same family of domains and the fundamental integral formula for f ∗ g relied on geometric properties of D to the same extent as on those of G .The approach presented in this paper is substantially different. Startingfrom a non-symmetric integral expressing f ∗ g , we investigate D ∗ G with only D being of the same particular class as in [28] and G being an arbitrary Rungedomain. These considerations conclude in Theorem 3.4, which establishes arelation, expressed in geometric terms, between D ∗ G and h N ∗ G . Here h N ( z ) = (1 − z − . . . − z N ) − N and the set h N ∗ G is the largest domaincontaining the origin on which every product h N ∗ g , for g ∈ O ,G , can beanalytically continued. When G is Runge domain, existence of h N ∗ G isguaranteed by the aforementioned [28, Proposition 4.1]. To calculate D ∗ G we must, however, face the problem of computing h N ∗ G . We deal with thistopic in Section 4, where, in Theorem 4.1, we derive a nice geometric formulafor h N ∗ G , but, unfortunately, only for N = 2 . It is worthy to emphasizethat although obtaining a candidate for this set was quite straightforwardand relied mainly of certain integral formula, main difficulties were met indemonstrating that this candidate is the largest one on which all h ∗ g ’sextend. Combining this result with Theorem 3.4 announced above leads toa complete description of D ∗ G for two-dimensional domains satisfying thelisted assumptions. Nevertheless, the question for higher dimensions remainsopen. 2. Preliminaries
We begin by introducing basic concepts and notation setting grounds forthis study. We use the standard symbols D , T , C ∗ and b C to denote, respec-tively, the unit disc in C , its boundary, the punctured plane C \ { } and theRiemann sphere C ∪ {∞} . We assume that the set N of all natural numberscontains . By P N ( z, r ) and P N ( z, r ) we mean the open and closed polydiscscentered at z and having the radius r .To shorten notation, we will use the word ’loop’ to denote a continuousmap defined on T and the word ’smooth’ to declare being of the C ∞ class.For points z = ( z , . . . , z N ) , w = ( w , . . . , w N ) ∈ C N , by z • w we denotethe product z w + . . . + z N w N and by z · w or zw we denote their coordinate-wise product, that is, the point ( z w , . . . , z N w N ) . The identity element ofthe latter multiplication, (1 , , . . . , ( N times), is called N . Given two sets HE ∗ -PRODUCT OF DOMAINS IN C A, B ⊂ C N , by AB or A · B we mean their algebraic product, i.e. the set { ab : a ∈ A, b ∈ B } . We use the classical notation of exponentiation, wherefor an α = ( α , . . . , α N ) ∈ Z N the symbol z α denotes z α . . . z α N N , holdingthe convention that z j = 1 . Moreover, α ! and | α | will, as usual, denote theproduct α ! . . . α N ! and the sum α + . . . + α N .For compact sets K ⊂ C N and L ⊂ Ω we use the standard notation b K and b L Ω for the polynomial hull of K and the holomorphic hull of L with respectto a domain Ω . The supremum of modulus of a complex-valued functionover a set A ⊂ C N is denoted by k f k A . Finally, if ∈ A , then by cc A weunderstand the connected component of A containing .2.1. Integral formula for the ∗ product. Take two power series f ( z ) = X α ∈ N N f α z α , g ( z ) = X α ∈ N N g α z α convergent in neighbourhoods of polydiscs P N (0 , r ) and P N (0 , ρ ) , respec-tively. A straightforward calculation allows us to derive the equality(1) ( f ∗ g )( z ) = (cid:18) πi (cid:19) N Z ρ − T N f ( zζ ) g (cid:0) ζ − , . . . , ζ − N (cid:1) dζ ζ . . . dζ N ζ N for z = ( z , . . . , z N ) ∈ P N (0 , rρ ) . Its one-dimensional version was extensivelyused in research of Hadamard product in one complex variable. Althoughin this paper we mostly rely on different tools, the above fact will come inuseful at some point in Section 4.2.2. Integral formula for the ∗ product with special weights. Takea bounded smooth domain ∈ D ⊂ C N , a neighbourhood V of ∂D and asmooth map ϕ : V → C N such that ζ • ϕ ( ζ ) = 1 for all ζ ∈ V . If a function f is holomorphic in a neighbourhood of D and P α ∈ N N f α z α is its Taylor seriesexpansion at the origin, then from the considerations made in [28, Section2.1] it follows that f α = c N ( N + | α | − α ! Z ∂D f ( ζ ) ϕ ( ζ ) α ω ϕ ( ζ ) , where c N is a constant dependent only on N and ω ϕ is certain smooth ( N, N − form on V dependent only on N and ϕ . Now, if g ∈ O has theTaylor series expansion P α ∈ N N g α z α , then for z close to one has that(2) X α ∈ N N α !( N − | α | + N − f α g α z α = c N ( N − Z ∂D f ( ζ ) g ( ϕ ( ζ ) z ) ω ϕ ( ζ ) . We will make use of this equality in Section 3.
SYLWESTER ZAJĄC Description of D ∗ G for D being of special class and G being an arbitrary Runge domain Our goal in this part of the paper is to demonstrate Theorem 3.4. For theReader’s convenience, we begin by recalling definition and elementary factsregarding the ∗ -product of domains. Assumption.
Throughout this section we assume that N ≥ and D , G aredomains in C N containing the origin. Definition.
Assume that at least one of D , G is Runge domain. Proposition4.1 from [28] establishes existence of the largest domain ∈ Ω ⊂ C N havingthe property that the image of O ,D × O ,G by ∗ lies in O , Ω . We define D ∗ G as this largest Ω . The referenced fact guarantees that D ∗ G itself isRunge domain.For every pair of functions f ∈ O ( D ) and g ∈ O ( G ) there exists the onlyelement of O ( D ∗ G ) equal to f ∗ g near the origin. Following [28], we denoteit by f ∗ D,G g . We then obtain the bilinear mapping ∗ D,G : O ( D ) × O ( G ) ∋ ( f, g ) f ∗ D,G g ∈ O ( D ∗ G ) . From the closed graph theorem it follows that ∗ D,G is separately continuous.This, together with [26, page 88, Corollary 1], guarantees that it is jointlycontinuous.In a similar fashion we introduce the ∗ -product of a germ from O and adomain containing the origin. Definition.
Assume that G is Runge domain. If τ ∈ O , then, in virtue of[28, Proposition 4.1], there exists the largest domain ∈ Ω ⊂ C N such thatthe image of O ,G by the map g τ ∗ g lies in O , Ω . We denote this largest Ω by τ ∗ G . As previously, τ ∗ G is Runge domain.For each g ∈ O ( G ) there exists the only function from O ( τ ∗ G ) equalto τ ∗ g near the origin. Denote this function by τ ∗ G g . The closed graphtheorem yields that the linear operator O ( G ) ∋ g τ ∗ G g ∈ O ( τ ∗ G ) is continuous. Definition.
Similarly as in [28] we introduce the compact set D ∗ := (cid:8) ξ ∈ C N : ξ • z = 1 for all z ∈ D (cid:9) . If ξ ∈ D ∗ , then the function(3) h ξ ( z ) := (1 − z • ξ ) − N . belongs to O ( D ) and(4) h ξ ( z ) = X α ∈ N N ( | α | + N − α !( N − ξ α z α is its Taylor series expansion at the origin. HE ∗ -PRODUCT OF DOMAINS IN C Lemma 3.1.
Assume that G is Runge domain and U ⊂ C N is an open set.If a domain Ω ⊂ C N contains the origin and h ξ ∗ g ∈ O , Ω for all ξ ∈ U and g ∈ O ,G , then U · Ω ⊂ h N ∗ G. Proof.
We need to show that ξ Ω ⊂ h N ∗ G for each ξ ∈ U . If ξ ∈ U ∩ ( C ∗ ) N ,then, in view of (4), for g ∈ O ,G and z close to we have ( h ξ ∗ g )( z ) = ( h N ∗ g )( ξz ) , so the germ of the function on the right hand side belongs to O , Ω . Hence, h N ∗ g ∈ O ,ξ Ω . This holds for every g ∈ O ,G , so the definition of h N ∗ G yields that it indeed contains ξ Ω . On the other hand, if ξ ∈ U \ ( C ∗ ) N , thenwe can take a number r > so that ξ + r T N ⊂ U ∩ ( C ∗ ) N . From the previousconsiderations it follows that ( ξ + r T N ) · Ω ⊂ h N ∗ G. Since the set on the right hand side is Runge domain, for each z ∈ Ω itcontains the polynomial hull of ( ξ + r T N ) · z and, in particular, the point ξ · z itself. (cid:3) Lemma 3.2. If D is a pseudoconvex domain and G is Runge domain, then D ∗ · ( D ∗ G ) ⊂ h N ∗ G. Consequently, D ∗ G ⊂ cc { z ∈ C N : zD ∗ ⊂ h N ∗ G } . It is worthy to note that, as D ∗ is compact, the set under cc above is open. Proof.
Fix an arbitrary domain ∈ Ω ⊂⊂ D ∗ G . From the continuity of ∗ D,G it follows that we can find a constant
C > and compact sets K ⊂ D and L ⊂ G such that K is holomorphically convex in D , ∈ int K and(5) k f ∗ D,G g k Ω ≤ C k f k K k g k L for all f ∈ O ( D ) and g ∈ O ( G ) .Take a neighbourhood U of D ∗ such that η • z = 1 for all z ∈ K and η ∈ U . If η ∈ U , then h η is holomorphic in a neighbourhood of K , so, by theOka-Weil theorem, it can be approximated uniformly on K by a sequence ( f n ) n ∈ N ⊂ O ( D ) . Hence, if g ∈ O ( G ) , then (5) gives that the functions f n ∗ D,G g form a Cauchy sequence with respect to the supremum norm on Ω . This means that they converge in O (Ω) and, thanks to the fact that ∈ int K , the limit has to be equal to h η ∗ g near the origin. Consequently, h η ∗ g ∈ O , Ω for every g ∈ O ,G and η ∈ U . Now, Lemma 3.1 allows us toconclude that D ∗ · Ω ⊂ U · Ω ⊂ h N ∗ G , what completes the proof. (cid:3) Lemma 3.3.
Let ∈ W and ∈ D ⊂ D ⊂ D ⊂ . . . be domains in C N such that S n ∈ N D n = D . Set Ω n = cc { z ∈ C N : zD ∗ n ⊂ W } , Ω = cc { z ∈ C N : zD ∗ ⊂ W } . Then Ω ⊂ Ω ⊂ Ω ⊂ . . . and S n ∈ N Ω n = Ω . SYLWESTER ZAJĄC
Proof.
Clearly, D ∗ ⊂ D ∗ n +1 ⊂ D ∗ n , so Ω n ⊂ Ω n +1 ⊂ Ω . To show that Ω iscontained in S n ∈ N Ω n , take an arbitrary connected compact set ∈ K ⊂ Ω .One has that K · D ∗ ⊂ W , so K · U ⊂ W for a neighbourhood U of D ∗ . Thesequence ( D ∗ n ) n ∈ N decreases and it is straightforward to check that D ∗ = T n ∈ N D ∗ n . Therefore, D ∗ n ⊂ U for some n , what gives that K · D ∗ n ⊂ W .Hence, K ⊂ Ω n , because K is connected and contains the origin. From thiswe conclude that Ω ⊂ S n ∈ N Ω n . (cid:3) Definition.
Similarly as in [28] we define D N as the family of all domainsin C N containing the origin which are countable unions of non-decreasingsequences of bounded smooth linearly convex domains.Recall that D is called linearly convex if through every point of C N \ D onecan pass an affine complex hyperplane disjoint from D . As it is described in[28, Remark 4.11], each element of D N , being a union of a non-decreasingsequence of Runge domains, is also Runge domain. Theorem 3.4. If D ∈ D N and G is Runge domain, then D ∗ G = cc { z ∈ C N : zD ∗ ⊂ h N ∗ G } . Proof.
The left-to-right inclusion was established in Lemma 3.2, so it remainsto prove the opposite one. In virtue of Lemma 3.3 and [28, Proposition 4.5]it suffices to restrict our considerations to the case when D is bounded,smooth and linearly convex. Then there exists a smooth map ν D from aneighbourhood of ∂D to C N such that at each point w ∈ ∂D its value ν D ( w ) is the unit outward normal vector for D at w . Hence, for w ∈ ∂D theequation ( z − w ) • ν D ( w ) = 0 describes the only complex hyperplane passingthrough w and disjoint from D . In particular, w • ν D ( w ) = 0 , as ∈ D .This means that the mapping ϕ : w ν D ( w ) · ( w • ν D ( w )) − is well-defined and smooth in a neighbourhood V of ∂D . Moreover, we havethat ϕ ( ∂D ) ⊂ D ∗ and w • ϕ ( w ) = 1 for w ∈ V .Denote by Ω the set on the right hand side of the conclusion and take adomain ∈ U ⊂⊂ Ω . From the definition of Ω it follows that U · ϕ ( ∂D ′ ) ⊂ h N ∗ G for a sufficiently large smooth domain ∈ D ′ ⊂⊂ D such that ∂D ′ ⊂ V . If f ∈ O ( D ) and g ∈ O ( G ) , then, by (2), for z lying near theorigin it holds that ( f ∗ g )( z ) = c N ( N − Z ∂D ′ f ( ζ )( h N ∗ G g )( ϕ ( ζ ) z ) ω ϕ ( ζ ) . The integral on the right hand side defines a function of the variable z whichis holomorphic on U . Consequently, f ∗ g ∈ O ,U . Since f , g and U weretaken arbitrarily, we conclude that Ω ⊂ D ∗ G . (cid:3) HE ∗ -PRODUCT OF DOMAINS IN C Description of h (1 , ∗ D for Runge domains This part is devoted to demonstration of Theorem 4.1, which completes,although only in the two-dimensional case, the description of the star productof domains established in Theorem 3.4. Recall that h (1 , is the function givenby the formula (3), that is, h (1 , ( z , z ) = (1 − z − z ) − = X α ,α ∈ N ( α + α + 1)! α ! α ! z α z α . Assumption.
In this section we assume that D is a domain in C containingthe origin. Definition.
To simplify certain statements in this section, let us say thatan open set U ⊂ C ∗ separates and ∞ if U contains a loop homotopic in C ∗ to the loop ζ ζ . This is equivalent to saying that and ∞ lie in differentconnected components of b C \ U .For a point z = ( z , z ) ∈ C introduce the mapping I z : C ∗ → C as I z ( ζ ) := ( z (1 + ζ ) , z (1 + ζ − )) . One has that I z ( −
1) = (0 , , so I − z ( D ) is non-empty. It is also importantthat I z is an injective proper map when z ∈ ( C ∗ ) . Theorem 4.1. If D is Runge domain, then h (1 , ∗ D = cc (cid:8) z ∈ C : the set I − z ( D ) separates and ∞ (cid:9) . Directly from the definition of separating is follows that the set under cc above is open. It also contains (0 , , because I − , ( D ) = C ∗ . Remark 4.2.
Assume that D is Runge domain and take z ∈ C . The openset I − z ( D ) is then O ( C ∗ ) -convex in the sense that b L C ∗ ⊂ I − z ( D ) for everycompact set L ⊂ I − z ( D ) . This means that every connected component of b C \ I − z ( D ) contains or ∞ (possibly both of them). Consequently, the latterset is connected if and only if I − z ( D ) does not separate and ∞ . Observation 4.3.
If an open set Ω ⊂ C and a point z ∈ ( C ∗ ) are suchthat I − z (Ω) does not separate and ∞ , then for every compact polynomiallyconvex set K ⊂ Ω the pre-image I − z ( K ) is either empty or polynomiallyconvex.Proof. First, note that the set L := I − z ( K ) , if non-empty, has to be com-pact and holomorphically convex in C ∗ , what means that every connectedcomponent of b C \ L contains or ∞ . But since I − z (Ω) does not separate and ∞ , these points lie in the same connected component of b C \ L . Hence,the set b C \ L is connected, so L is polynomially convex. (cid:3) Lemma 4.4.
Let K ⊂ C be a compact polynomially convex set, z ∈ ( C ∗ ) and ̺ ∈ (0 , ∞ ) . If I − z ( K ) is empty or polynomially convex and I − z ( K ) ∩ ̺ D = ∅ , SYLWESTER ZAJĄC then the union K ∪ I z ( ̺ T ) is polynomially convex.Proof. Write z = ( z , z ) and define µ ( w , w ) := ( w − z )( w − z ) − z z , ( w , w ) ∈ C . Clearly, M := µ − (0) = I z ( C ∗ ) is a complex submanifold of C and themap I z : C ∗ → M is a biholomorphism. By the assumptions, the union I − z ( K ) ∪ ̺ T is holomorphically convex in C ∗ . This implies that the set ( K ∩ M ) ∪ I z ( ̺ T ) , being its image by I z , is holomorphically convex in M and thus polynomially convex in C (use e.g. [15, Theorem 7.4.8]). Theconclusion now follows directly from the subsequent general lemma. (cid:3) Lemma 4.5.
Let V be an analytic subset of C N and let K ⊂ C N , L ⊂ V becompact sets. If both K and ( K ∩ V ) ∪ L are polynomially convex, then sois K ∪ L . Note that the sets K and L do not have to be disjoint. Proof.
It is known (see e.g. [15, Theorems 6.5.2 and 7.1.5]) that the sheafof germs of holomorphic functions vanishing on V is a coherent analyticsheaf on C N . Therefore, if X ⊂ C N is a compact, polynomially convexset (intersecting V or not), U is a neighbourhood of X , F ∈ O ( U ) and F | U ∩ V ≡ , then F is a section of this sheaf and it can be approximateduniformly on X by global sections, that is, by elements of O ( C N ) vanishingon V . This essential fact is a consequence of [15, Theorem 7.2.7].Fix a point z ∈ C N \ ( K ∪ L ) . We are going to show that z \ K ∪ L .If z V , then [15, Theorem 7.2.11] provides f ∈ O ( C N ) vanishing on V and such that f ( z ) = 1 . On the other hand, there exists g ∈ O ( C N ) having g ( z ) = 1 and k g k K < . Hence, for sufficiently large number n the function g n f maps z to and K ∪ L into D .It remains to consider the case when z ∈ V . Since ( K ∩ V ) ∪ L ispolynomially convex, one can find another compact polynomially convex set A ⊂ C N such that ( K ∩ V ) ∪ L ⊂ int A and z A. Take a sequence ( p n ) n ∈ N ⊂ O ( C N ) such that p n ( z ) = 1 and k p n k A → when n → ∞ . The set ( C N \ V ) ∪ int A is an open neighbourhood of K , so it has a pseu-doconvex open subset Ω containing K . This means that Ω ∩ V ⊂ int A and hence p n → on Ω ∩ V . In virtue of [6, Theorem 13.1], there existsa continuous linear extension operator from the Banach space of boundedholomorphic functions on Ω ∩ V into O (Ω) . Applying it to p n ’s we obtain asequence ( g n ) n ∈ N ⊂ O (Ω) convergent to in O (Ω) and such that g n − p n = 0 on Ω ∩ V . As it was described in the first paragraph of the proof, for every n one can find a function q n ∈ O ( C N ) so that q n | V ≡ and k q n + p n − g n k K < n . HE ∗ -PRODUCT OF DOMAINS IN C Finally, set f n := p n + q n . Every f n is an entire function and f n = p n on V and k f n − g n k K < n . This implies that f n ( z ) = 1 and f n → on K ∪ L uniformly when n → ∞ ,so for large n it holds that | f n ( z ) | > k f n k K ∪ L , as desired. (cid:3) For a holomorphic function f of two variables define Λ( f )( z , z ) := f ( z , z ) + z ∂f∂z ( z , z ) . For every domain Ω ⊂ C the mapping f Λ( f ) defines a continuous linearoperator on O (Ω) . Lemma 4.6. If γ is a loop in C ∗ homotopic to the loop ζ ζ and f is apolynomial in C , then ( h (1 , ∗ C f )( z ) = 12 πi Z γ (1 + ζ − ) Λ( f )( I z ( ζ )) dζ for all z ∈ C .Proof. Fix a number ρ ∈ (0 , ) and a polynomial f . From (1) it follows that ( h (1 , ∗ C f )( z ) = (cid:18) πi (cid:19) Z ρ − T f ( z ζ , z ζ ) ζ ζ ( ζ − ( ζ − ζ ( ζ − − ) dζ dζ , when z = ( z , z ) ∈ C . If ζ ∈ ρ − T , then the point ζ ( ζ − − lies in ρ − D , so, in view of the Cauchy formula, πi Z ρ − T f ( z ζ , z ζ ) ζ ( ζ − ζ ( ζ − − ) dζ = ddζ (cid:16) f ( z ζ , z ζ ) ζ (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) ζ = ζ ( ζ − − = Λ( f ) (cid:18) z ζ ζ − , z ζ (cid:19) . Therefore, ( h (1 , ∗ C f )( z ) = 12 πi Z ρ − T ζ ( ζ − Λ( f ) (cid:18) z ζ ζ − , z ζ (cid:19) dζ . A homotopy argument allows us to integrate over ρ − T + 1 instead of ρ − T .Then, after changing the variable via ζ := ( ζ − − , we obtain the equalityfrom the conclusion with the integral over ρ T . It is not affected by replacing ρ T by γ , because the integrated function of ζ is holomorphic on C ∗ . (cid:3) Proof of Theorem 4.1.
Denote by Ω the set on the right hand side of theconclusion, that is, Ω := cc (cid:8) z ∈ C : the set I − z ( D ) separates and ∞ (cid:9) . The proof of the equality h (1 , ∗ D = Ω is divided into a few steps. Step 1:
We show the inclusion Ω ⊂ h (1 , ∗ D . Fix a function f ∈ O ( D ) and take a sequence ( f n ) n ∈ N of polynomialsconvergent to f locally uniformly on D . Then the functions h (1 , ∗ C f n tend to h (1 , ∗ D f in the same manner on h (1 , ∗ D . We claim that theyform a sequence convergent on Ω as well. Fix a point a ∈ Ω and take a loop γ : T → I − a ( D ) homotopic in C ∗ to the identity loop. Choose a closed ball B ⊂ Ω centered at a so that I z ( γ ( T )) ⊂ D for all z ∈ B . From Lemma 4.6it follows that ( h (1 , ∗ C f n )( z ) = 12 πi Z γ (1 + ζ − ) Λ( f n )( I z ( ζ )) dζ for all z ∈ C , n ∈ N . Since Λ( f n ) → Λ( f ) in O ( D ) as n → ∞ , the integralson the right hand side of the above equality converge uniformly with respectto z ∈ B to the identical integral with f n repalced by f . Consequently, thesequence of polynomials h (1 , ∗ C f n is uniformly convergent on B . Hence, itdoes converge locally uniformly on Ω , because a ∈ Ω was chosen arbitrarily.If g ∈ O (Ω) is the limit, then clearly g = h (1 , ∗ f near the origin. Thismeans that h (1 , ∗ f ∈ O , Ω for every f ∈ O ( D ) , so Ω ⊂ h (1 , ∗ D . Step 2:
We prove the inclusion ( h (1 , ∗ D ) ∩ ( C ∗ ) ⊂ Ω .Suppose, to the contrary, that it is not valid, and choose a point z ∈ ( h (1 , ∗ D ) ∩ ( C ∗ ) ∩ ∂ Ω . There exist a constant C > and a polynomiallyconvex compact set K ⊂ D such that(6) | ( h (1 , ∗ D f )( z ) | ≤ C k f k K , f ∈ O ( D ) . Take a number ̺ > so that I − z ( K ) ∩ ̺ D = ∅ . Since z ∈ ∂ Ω , the set I − z ( D ) does not separate and ∞ , so from Observation 4.3 and Lemma 4.4it follows that the union K ∪ I z ( ̺ T ) is polynomially convex. Hence, thereexists a compact set L ⊂ C such that I z ( ̺ T ) ⊂ int L , K ∩ L = ∅ and K ∪ L is polynomially convex (one can justify it making use, for example,of the Kallin Lemma [27, Theorem 1.6.19]). This allows us to employ theOka-Weil theorem to get a sequence ( f n ) n ∈ N of polynomials in C uniformlyconvergent to on K and to on L . Lemma 4.6 implies that ( h (1 , ∗ D f n )( z ) = 12 πi Z ̺ T (1 + ζ − ) Λ( f n )( I z ( ζ )) dζ. If n → ∞ , then the left hand side goes to , in view of (6). On the other hand,the integrals converge to , because Λ( f n ) → on I z ( ̺ T ) . A contradiction. Step 3:
We show that if ( z , ∈ h (1 , ∗ D , then ( z , ∈ D . Thanks toevident symmetry, it will also mean that (0 , z ) ∈ D when (0 , z ) ∈ h (1 , ∗ D .Suppose, to the contrary, that ( z , D . One can find a constant C > and a compact polynomially convex set K ⊂ D satisfying(7) | ( h (1 , ∗ D f )( z , | ≤ C k f k K , f ∈ O ( D ) . Take a closed ball B centered at ( z , so small that K ∩ B = ∅ and K ∪ B is polynomially convex. In view of the Oka-Weil theorem, there exists a HE ∗ -PRODUCT OF DOMAINS IN C sequence ( f n ) n ∈ N of polynomials in C uniformly convergent to on K andto on B .From Lemma 4.6 is follows that(8) ( h (1 , ∗ C f n )( z ,
0) = Λ( f n )( z ,
0) = f n ( z ,
0) + z ∂f n ∂z ( z , . Now, if n → ∞ , then (7) implies that the left hand side of (8) goes to ,while the right hand side converges to . This contradiction completes theproof of this step. Step 4:
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Institute of Mathematics, Faculty of Mathematics and Computer Science,Jagiellonian University, Łojasiewicza 6, 30-348 Kraków, Poland
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