• Nie Znaleziono Wyników

Let Ω be a domain (open, bounded and connected) in Cn

N/A
N/A
Protected

Academic year: 2021

Share "Let Ω be a domain (open, bounded and connected) in Cn"

Copied!
7
0
0

Pełen tekst

(1)

POLONICI MATHEMATICI LVIII.2 (1993)

On the spectrum of A(Ω) and H(Ω) by Urban Cegrell (Ume˚a)

Abstract. We study some properties of the maximal ideal space of the bounded holomorphic functions in several variables. Two examples of bounded balanced domains are introduced, both having non-trivial maximal ideals.

1. Introduction. Let Ω be a domain (open, bounded and connected) in Cn. Denote by H(Ω) the analytic functions on Ω, H(Ω) = H(Ω)∩L(Ω) and by A(Ω) the functions in H(Ω) that are continuously extendable to Ω.

Let M be the spectrum of H(Ω) and MA the spectrum of A(Ω). If m ∈ MA, we have the projection πm = (m(z1), . . . , m(zn)), eΩ = {m ∈ M : πm ∈ Ω} and X = e\ eΩ where we take the closure in the Gelfand topology. We write Sh M (Sh MA) for the Shilov boundary of M (MA) and bf for the Gelfand transform of f ∈ H(Ω). Note that Sh MA ⊂ X, πX ⊂ ∂Ω and that π(M ) ⊂ π(MA) for if m ∈ M then m operates on A so that z ∈ π(M ) ⇒ z ∈ π(MA).

The purpose of this paper is to study the following statements:

1. πM = Ω.

10. πMA= Ω.

2. If πm ∈ Ω, then mf = f (πm), ∀f ∈ H(Ω).

20. If πm ∈ Ω, then mf = f (πm), ∀f ∈ A(Ω).

3. If πm ∈ ∂Ω, f ∈ H(Ω), then there is an m0∈ X so that π(m0) = π(m) and bf (m0) = bf (m).

The Gleason problem is to decide if the coordinate functions z1− z10, . . . . . . , zn− zn0 generate every maximal ideal {f ∈ A(Ω) : f (z0) = 0}, z0 ∈ Ω.

An obvious obstruction to the Gleason problem is the failure of 20, which is one of the motivations for us to study the statements above.

1991 Mathematics Subject Classification: 32E25, 32D10, 32A07.

Key words and phrases: bounded analytic function, spectrum, Gleason problem, bal- anced domain.

(2)

2. The case Ωb C. In the case when Ω is a domain in C, we will prove that all statements are true.

Proposition 1. If ψ ∈ L(Ω) then V (z) = R

ψ(ξ) dξ ∧ d ¯ξ ξ − z is continuous on C and ∂V /∂z = ψ on Ω.

P r o o f o f 1 a n d 10. If m ∈ MA then mz ∈ Ω for otherwise z−mz1 A(Ω) and so

1 = m1 = m(z − mz)m

 1

z − mz



= 0 , which is a contradiction.

P r o o f o f 2. If mz = ξ ∈ Ω, then

f (z) = f (ξ) + (z − ξ)f (z) − f (ξ) z − ξ where f (z)−f (ξ)

z−ξ ∈ H(Ω). Hence, mf = f (ξ) = f (πm), ∀f ∈ H(Ω).

In the same way 20 follows.

P r o o f o f 3. Assume that 3 is not true. Then there exist m ∈ M with πm = ξ ∈ ∂Ω and g ∈ H(Ω) with mg = 0 but |g(z)| ≥ d near ξ in Ω. Choose χ ∈ D(B(ξ, r)) so that χ ≡ 1 near ξ and |g(z)| ≥ δ > 0 on Ω ∩ B(ξ, r). Then

∂χ

∂z

(z − πm)g ∈ C(Ω) so by Proposition 1, there is a λ ∈ C(C) with

∂λ

∂z =

∂χ

∂z (z − πm)g. Define

g1= χ

g − (z − πm)λ , g2= 1 − χ

z − πm + λg .

Then ∂g1/∂z = ∂g2/∂z = 0 so g1, g2∈ H(Ω) and g1g + g2(z − πm) = 1, which is a contradiction since mg = 0.

Corollary. Suppose f1, . . . , fm−1 ∈ A(Ω) and fm∈ H(Ω) such that Pm

j=1|fj|2≥ δ > 0 on Ω. Then there exist g1, . . . , gm ∈ H(Ω) such that Pm

j=1fjgj = 1.

(3)

3. The case Ωb Cn

Theorem 1. Suppose Ω is a domain in Cn such that (i) πMA= Ω,

(ii) for every z0 ∈ ∂Ω, there is a ball B(z0, r) such that the analytic polynomials are dense in A(Ω ∩ B(z0, r)).

Then MA= Ω.

P r o o f. Given f ∈ A(Ω), consider the uniform algebra B on MA gener- ated by f ◦ π and A(Ω). It then follows from [3, Lemma 9.1, p. 93] that B and A(Ω) have the same Shilov boundary. But f = f ◦ π on Sh MA, which proves the theorem.

R e m a r k 1. If Ω is pseudoconvex with C1-boundary, then by [5, Lemma 3], (ii) holds true everywhere.

R e m a r k 2. If Ω is pseudoconvex with smooth boundary, then using the estimates for ∂ from [6] and the Koszul complex one can prove that (i) holds true (cf. [5]). Therefore, in this case MA= Ω.

R e m a r k 3. Let Ω be strictly pseudoconvex with C3-boundary. Then 2 and 20hold (cf. [8, pp. 292 and 350]). Note that, via Remark 2, 10 holds for all pseudoconvex sets of type Sδ.

Also, every boundary point is a peak point ([8, 1.14]) so Theorem 2 of [1] applies, which means that 3 is true in this case.

4. The case Ωb C2. In this section, we consider domains in C2. Proposition 2. Suppose that Ω has the property that for every ∂- closed (0, 1)-form t with coefficients in L(Ω) (C(Ω)), there is a function T ∈ L(Ω) (C(Ω)) with ∂T = t. Then 1 and 2 (10 and 20) hold true.

Furthermore, 3 holds at all peak points for A(Ω).

P r o o f. Let (z10, z20) 6∈ Ω be given. Consider g1= z1− z01

|z1− z10|2+ |z2− z20|2 − (z2− z20)λ , g2= z2− z02

|z1− z10|2+ |z2− z20|2 + (z1− z10 where λ ∈ L(Ω) and solves

∂λ = t = z2− z20

(|z1− z10|2+ |z2− z20|2)2dz1 z1− z10

(|z1− z10|2+ |z2− z20|2)2dz2. Since ∂t = 0 and since the coefficients of t are uniformly bounded on Ω, λ exists.

(4)

This means that g1and g2are analytic and bounded on Ω. Furthermore, 1 = g1(z1− z10) + g2(z2− z20) so (z01, z02) 6∈ πM .

To prove 2, we prove that Gleason’s problem can be solved. Assume that πm = (0, 0) ∈ Ω, f ∈ H(Ω) and f (0) = 0. We wish to prove that mf = 0.

If p is the Taylor expansion of f around zero of degree three, then f − p vanishes to order four near zero. Therefore, the ∂-closed (0, 1)-form

t = (f − p)z2

(|z1|2+ |z2|2)2dz1 (f − p)z1

(|z1|2+ |z2|2)2dz2

has uniformly bounded coefficients on Ω.

By assumption, there is a function λ ∈ L(Ω) so that ∂λ = t. Therefore, the functions

g1= (f − p)z1

|z1|2+ |z2|2 − z2λ , g2= (f − p)z2

|z1|2+ |z2|2 + z1λ

are in H(Ω). Furthermore, f − p = z1g1+ z2g2 so f = p + z1g1+ z2g2, which solves the Gleason problem, gives mf = 0 and completes the proof of 2.

10 and 20 are proved analogously. The proof of 3 is in [1].

R e m a r k 4. Fornæss and Øvrelid [2] proved the Gleason property for A(Ω) when Ω is real-analytic, and Noell [7] when Ω is of finite type. Thus 20 holds true in those cases.

5. Balanced H-domains. A domain in Cn is called an H-domain if it is its own H-envelope of holomorphy. A subset Ω in Cn is said to be balanced if λΩ ⊂ Ω, ∀λ ∈ C, |λ| ≤ 1.

Let Hm denote the homogeneous polynomials of degree m and Ho(Cn) = {ϕ ∈ PSH(Cn) : ϕ 6= 0, ϕ(λx) = |λ|ϕ(x), ∀λ ∈ C, ∀x ∈ Cn} where PSH(Cn) denote the plurisubharmonic functions on Cn. For a domain in Cn containing 0 we define the homogeneous extremal function

ψ = sup{|Q|1/ deg Q : Q ∈ Hdeg Q, |Q| ≤ 1 on Ω} . Then ψ ∈ Ho where ∗ denotes the usual regularization.

Proposition 3. Suppose Ω is a balanced domain. Then every f ∈ H(Ω) extends to Ω = {z ∈ Cn : ψ < 1} and Ω is an H-domain.

Every f ∈ A(Ω) can be uniformly approximated on Ω by polynomials.

P r o o f. Let f ∈ H(Ω) . We can assume that f is non-constant and

|f | ≤ 1 on Ω. Near zero, we have f (z) =

X

j=0

 X

|α|=j

aαzα

 .

(5)

Let z ∈ Ω; then

f (λz) =

X

j=0

X

|α|=j

(aαzαj, |λ| ≤ 1, λ ∈ C . By Cauchy’s inequality,

X

|α|=j

aαzα

≤ sup

|λ|=1

|f (λz)| < 1 so |P

|α|=jaαzα|1/j < 1 on Ω and therefore on Ω by the definition of ψ. Hence, the series P

j=0

P

|α|=jaαzα converges normally in Ω and extends f to H(Ω), which proves the first part of the proposition.

Assume now that f ∈ A(Ω). Let ε > 0 be given and choose 0 < r < 1 such that supz∈ ¯|f (z) − f (rz)| < ε. By the above, supz∈Ω|P aαzα| ≤ 1, therefore, Pp

j=0

P

|α|=j(aαzα)rj uniformly convergs on Ω to f (rz). The proposition is proved.

Theorem 2 (Siciak). Let h ∈ Ho(Cn) and consider D = {z ∈ Cn : h < 1} where we assume D to be bounded. Then the following conditions are equivalent.

(i) D is an H-domain, (ii) h = ψD,

(iii) D = ( bD),

(iv) the set N (h) = {a ∈ Cn : h is discontinuous at a} is pluripolar.

Lemma 1. Assume U is a subharmonic function and that the Lebesgue measure of N (U ) = {a ∈ Cn : U discontinuous at a} is zero. Then {U < 1}= {U < 1}.

The proof is left to the reader. It is based on the fact that if two sub- harmonic functions are equal almost everywhere, they are equal.

We now construct two examples.

Example 1. Let (aj) be a sequence of complex numbers contained and dense in {x ∈ C : |x| = 1}. Define

h(x, y) = eΣj=1αjlog |x−ajy|, (x, y) ∈ C2, where (αj)j=1 is a sequence of positive numbers with P

j=1αj = 1. Then h ∈ Ho(C2) and

N (h) =n

(x, y) ∈ C2: |x| = |y| and

X

j=1

αjlog |x − ajy| > −∞o so N (h) is a non-pluripolar set of vanishing Lebesgue measure. If we define D = {(x, y) ∈ C2: h(x, y) + max(|x|, |y|) < 1} then

(6)

(i) D is pseudoconvex and balanced, (ii) D = D by Lemma 1,

(iii) D is not an H-domain, by Theorem 2. The H-envelope is the bidisc by Proposition 3.

An example of this nature was given in Siciak [10, Ex. 5.3] (see also Sibony [9, Prop. 1] where a Hartogs domain Ω is given with properties (i) and (ii)) but our construction is much more elementary.

Example 2. Let (aj)j=1 and (αj)j=1 be as in Example 1. Define W (x, y, z) = eΣj=1αjmax(log |x−ajy|,log |z|)+ max(|x|, |y|, |z|) ,

(x, y, z) ∈ C3. Then W ∈ Ho(C3) so D = {W < 1} is a balanced, pseudo- convex set and N (W ) ⊂ {z = 0}, a pluripolar set. Thus

(i)D= D by Lemma 1,

(ii) D is an H-domain by Theorem 2,

(iii) bD 6= D since {(x, y, 0) : |x| ≤ 1, |y| ≤ 1} ⊂ bD by Example 1.

This gives a counterexample to a problem of Siciak [10, Problem 4.2], who proved [Th. 4.1] that such an example cannot be found in C2.

R e m a r k 5. Let Ω be a balanced domain in C2. Then 20holds true and MA= {z ∈ Cn: ψ < 1}.

R e m a r k 6. Let Ω be a balanced domain in Cn. Then 20holds true. By Example 2, πMA may be strictly larger than {z ∈ Cn: ψ < 1}.

References

[1] U. C e g r e l l, Representing measures in the spectrum of H(Ω), in: Complex Analy- sis, Proc. Internat. Workshop, Wuppertal 1990, K. Diederich (ed.), Aspects of Math.

E17, Vieweg, 1991, 77–80.

[2] J. E. F o r n æ s s and N. Ø v r e l i d, Finitely generated ideals in A(Ω), Ann. Inst.

Fourier (Grenoble) 33 (2) (1983), 77–85.

[3] T. W. G a m e l i n, Uniform Algebras, Prentice-Hall, Englewood Cliffs, N.J., 1969.

[4] —, Uniform Algebras and Jensen Measures, Cambridge Univ. Press, 1978.

[5] M. H a k i m et N. S i b o n y, Spectre de A(Ω) pour les domaines born´es faiblement pseudoconvexes r´eguliers, J. Funct. Anal. 37 (1980), 127–135.

[6] J. J. K o h n, Global regularity for ∂ on weakly pseudo-convex manifolds, Trans.

Amer. Math. Soc. 181 (1973), 273–292.

[7] A. N o e l l, The Gleason problem for domains of finite type, Complex Variables 4 (1985), 233–241.

[8] M. R a n g e, Holomorphic Functions and Integral Representations in Several Com- plex Variables, Springer, 1986.

[9] N. S i b o n y, Prolongement analytique des fonctions holomorphes born´ees, in: S´em.

Pierre Lelong 1972–73, Lecture Notes in Math. 410, Springer, 1974, 44–66.

(7)

[10] J. S i c i a k, Balanced domains of holomorphy of type H, Mat. Vesnik 37 (1985), 134–144.

[11] N. Ø v r e l i d, Generators of the maximal ideals of A(D), Pacific J. Math. 39 (1971), 219–223.

DEPARTMENT OF MATHEMATICS UME˚A UNIVERSITY

S-901 87 UME˚A, SWEDEN

Re¸cu par la R´edaction le 5.11.1992

Cytaty

Powiązane dokumenty

On Some Generalization of the Well-known Class of Bounded Univalent Functions 51 is an extremal function in many questions investigated in the class S(M).. It is evident that,

As every plurisubharmonic function can be approximated by a de- creasing sequence of smooth plurisubharmonic functions, algebraic properties of the Monge–Amp` ere operator acting

The goal of this paper is to show the existence of smooth (C ∞ ) convex exhaus- tions with additional properties in arbitrary convex and strictly convex domains.. Recall that a

[15] —, On nonstationary motion of a compressible barotropic viscous capillary fluid bounded by a free surface, to appear. [16] —, Local existence of solutions for free

This idea was used in [6] to obtain some properties of derivatives and integrals relative to such a function in addition to a suitable generation of signed

Moreover, it is shown that for modular classes of functions of bounded &lt;p-variation an equality of Lesniewicz type holds (cf.. Regulated functions are bounded.

1. Lambrinos [2] introduced the concept of bounded sets in a topological space. Quite recently, in [3] and [4], he has also defined almost-bounded sets and

This note contains the representation theorem of a bounded linear functional in a subspace of a symmetric space made of functions with absolutely continuous