- Email: [email protected]
Bounded holomorphic functions and maps with some boundary behavior
J. Math. Anal. Appl. 285 (2003) 691–707 www.elsevier.com/locate/jmaa
Bounded holomorphic functions and maps with some boundary behavior Toshio Matsushima Ishikawa National College of Technology, Kita-Chuhjoh, Tsubata, Ishikawa 929-0392, Japan Received 26 November 2002 Submitted by J. Noguchi
Abstract Let Ω be a domain of Cn . As a study of boundary behavior of functions and maps in Ω, we consider “the linear cluster set,” which is the cluster set along segment terminating at a boundary point of Ω. We prove that there exist bounded holomorphic functions and maps defined in Ω which have the linear cluster sets of positive measure at every point of a discrete subset of the boundary of Ω under some conditions. 2003 Elsevier Inc. All rights reserved. Keywords: Boundary behavior; Holomorphic mapping; Bounded holomorphic function
1. Introduction and main results It is well known that the bounded holomorphic functions in the unit disk of C have some kind of mild boundary behavior. Indeed, we have the following theorem due to Fatou [3]. Theorem 1.1 (Fatou). If f (z) is a bounded holomorphic function in the unit disk of C, then the limit lim f (rζ )
r→1
exists for almost every point ζ on the unit circle of C. We also have the similar theorem for bounded holomorphic functions in the unit ball of Cn for n 2 [6]. This kind of mildness of the boundary behavior of bounded holomorphic E-mail address: [email protected]. 0022-247X/$ – see front matter 2003 Elsevier Inc. All rights reserved. doi:10.1016/S0022-247X(03)00471-2
692
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
functions was one of the backgrounds of the inner function conjecture (for details, see [6]). The existence of an inner function in the unit ball of Cn for n 2 was, however, proved by Aleksandrov [1] and Löw [4]. This fact causes interest in the existence of the bounded holomorphic functions with wild boundary behavior. The author has shown the following theorem in [5]. Theorem 1.2. Let {ζk }m k=1 be an arbitrary discrete subset of the boundary of the unit ball of Cn , where 1 m +∞, n 1 and ζk = ζl if k = l. Then there exists a bounded holomorphic function f (z) in the unit ball of Cn whose radial cluster set at ζk f (tζk ): T < t < 1 T <1
includes a closed disk of positive radius for all k. The first aim of this article is to generalize this result. As a study of boundary behavior of functions f defined in the domain Ω of Cn , we consider the cluster set along a segment terminating at a boundary point of Ω. We start with giving some definitions. Throughout this paper, let Ω be a domain of Cn with n 1, and let ∂Ω be the boundary of Ω. Moreover, let f Ω = sup f (z), z∈Ω
where f is a function in Ω. Definition 1.3. Let ζ be a boundary point of Ω. (1) A point ζ is said to be linearly accessible in Ω if there exists a nonzero vector v such that {z ∈ Cn : z = ζ + sv, 0 < s 1} ⊂ Ω. We denote by V (ζ ) the set which consists of such vectors. (2) Let f (z) be a function defined in Ω, and assume that ζ is linearly accessible in Ω. Define the linear cluster set of f at ζ by f (1 − t)v + ζ : T < t < 1 , CL(f : ζ, v) = T <1
where v ∈ V (ζ ). (3) Let F be a map from Ω to Cg and ζ , and let v be as in (2). Define the linear cluster set of F at ζ by F (1 − t)v + ζ : T < t < 1 . CL(F : ζ, v) = T <1
We say that F is bounded if the image of Ω by F is bounded. Definition 1.4. A boundary point ζ of Ω is said to be a C-point if there exists a complex hyperplane of Cn that contains ζ and does not intersect Ω. We say that Ω is a C-domain if each point of ∂Ω is a C-point. When n = 1, we consider {ζ } as the hyperplane.
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
693
Definition 1.5. A domain D of C is said to be L-connected if D satisfies the following three conditions: (1) D is simply connected. (2) 0 ∈ / D. (3) If z ∈ D, then arg z Θ for some constant Θ, where arg z is defined as a single valued continuous function in D. Definition 1.6. Let ζ be a C-point on ∂Ω. Assume that H = z = (z1 , . . . , zn ) ∈ Cn : c1 z1 + · · · + cn zn + c0 = 0 is a hyperplane which goes through ζ and does not intersect Ω, and let Πζ (z) = c1 z1 + · · · + cn zn + c0 for z = (z1 , . . . , zn ) ∈ Cn . Then we say that ζ satisfies CL-condition by H if Πζ (Ω) = {w ∈ C: w = Πζ (z), z ∈ Ω} is L-connected. Note that 0 = Πζ (ζ ). We remark that CL-condition by H does not depend on the choice of the polynomial Πζ (z). Example 1.7. √ (1) Let z, w = nk=1 zk w¯ k and |z| = z, z for the points z = (z1 , . . . , zn ) and w = (w1 , . . . , wn ) in Cn . Consider the unit ball Bn = {z ∈ Cn : |z| < 1} of Cn . It is trivial that every boundary point ζ of Bn is linearly accessible. If we choose Πζ (z) =
z, ζ − 1, then it is also clear that ζ is a C-point and ζ satisfies CL-condition by {z: Πζ (z) = 0}. (2) Let Ω = {z = (z1 , . . . , zn ) ∈ Cn : Im z1 > 0}. Clearly every boundary point ζ = (ζ1 , . . . , ζn ) ∈ Cn with ζ1 ∈ R is linearly accessible. If we choose Πζ (z) = z1 − ζ1 , we can see that ζ is a C-point and ζ satisfies CL-condition by {z: Πζ (z) = 0}. (3) Let Ω = {z = (z1 , . . . , zn ) ∈ Cn : Im z1 > −(Re z1 )2 }. We can easily see that every boundary point ζ = (ζ1 , . . . , ζn ) ∈ Cn with Im ζ1 = −(Re ζ1 )2 is a C-point and ζ satisfies CL-condition by {z: Πζ (z) = 0}, where Πζ (z) = z1 − ζ1 . This example shows that a C-domain is not necessarily geometrically convex. Definition 1.8. Let A be an annulus {z ∈ C: a |z − c| b, 0 < a < b} in C. We define the thickness of A by b − a, and radius of A by (a + b)/2. Moreover, we define the center of A by c. Our main result in terms of the first aim is the following theorem. Theorem 1.9. Let {ζk }m k=1 be a discrete subset of the boundary of Ω, where ζk is linearly accessible for all k and 1 m +∞. Suppose that there exists a complex hyperplane Hk for each k that goes through ζk and does not intersect Ω such that every ζk satisfies CL-condition by Hk . Assume that Hk satisfies either Hk ∩ Hl ζk , ζl
(1.0)
694
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
or Hk = Hl
(1.1)
if k = l. Then there exists a bounded holomorphic function f (z) in Ω whose arbitrary linear cluster set at ζk contains a closed disk of positive radius, or a closed annulus of positive thickness. This result gives some type of counter part of Fatou’s theorem of arbitrary dimension, and comprehends Theorem 1.2 as a particular case. In fact, an arbitrary discrete subset of ∂Bn satisfies the assumption of Theorem 1.9. It is known that the Fatou-type theorem holds on the bounded domain with C 2 -boundary in Cn [7]. Though we also need some assumptions for the boundary of Ω, it is not necessary for us to assume the boundedness and the smoothness of the boundary of the domain. The second aim is to show the existence of a bounded holomorphic map with wild boundary behavior. The next theorem is our main result in terms of the bounded holomorphic map. Theorem 1.10. Let {ζk }m k=1 be a discrete subset of the boundary of Ω which satisfies the assumption of Theorem 1.9. Then, there exists a bounded holomorphic map F (z) = (f 1 (z), f 2 (z), . . . , f g (z)) : Ω → Cg for all g < +∞ whose arbitrary linear cluster set at ζk contains a direct product of g closed disks of positive radii, or of g closed annuli of positive thicknesses. We prove Theorem 1.9 in Section 3, and Theorem 1.10 in Section 4.
2. Lemmas Lemma 2.1. Let (2πT)n = Rn /(2πZ)n be a torus of dimension n, and let [x1 , x2 , . . . , xn ] ∈ (2πT)n denote the residue class of modulus (2πZ)n to which (x1 , x2 , . . . , xn ) belongs. For (ω1 , ω2 , . . . , ωn ) ∈ Rn , define the map ϕ : [0, +∞) t → [x1 + 2πω1 t, x2 + 2πω2 t, . . . , xn + 2πωn t] ∈ (2πT)n . for arbitrarily fixed (x1 , x2 , . . . , xn ) ∈ Rn . Then the image of ϕ is dense in (2πT)n if and only if ω1 , ω2 , . . . , ωn are linearly independent over Z. This map is classically known as Kronecker flow. See, for example, [2] for Lemma 2.1. Lemma 2.2. Let g1 (z) and g2 (z) be functions defined in Ω, and let ζ be a linearly accessible in Ω. Suppose that Λ = CL(g1 : ζ, v) for v ∈ V (ζ ), and g2 (z) has a limit α along the segment {z ∈ Ω: z = (1 − t)v + ζ, 0 t < 1} at ζ , i.e., lim g2 (1 − t)v + ζ = α. t →1−0
Then CL(g1 + g2 : ζ, v) = Λ + α.
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
695
Proof. Let ε be an arbitrary positive real number and b an arbitrary point in Λ. By assumptions of Lemma 2.2, there exist a sequence {tk }∞ k=1 with tk ↑ 1 as k → ∞ and a natural number N such that if k > N , then g1 (1 − tk )v + ζ − b < ε and g2 (1 − tk )v + ζ − α < ε . 2 2 Hence if k > N , then (b + α) − (g1 + g2 ) (1 − tk )v + ζ b − g1 (1 − tk )v + ζ + α − g2 (1 − tk )v + ζ < ε, which yields that b + α is a point of CL(g1 + g2 : ζ, v). Consequently, we see Λ + α ⊂ CL(g1 + g2 : ζ, v). Conversely, suppose that s is a point of CL(g1 + g2 , ζ, v), which is not in Λ + α. Then r := s − α ∈ / Λ, and so we can find some positive number d such that {w: |w − r| 2d} ∩ Λ = φ. Take an arbitrary monotone increasing sequence {qk }∞ k=1 of nonnegative numbers which converges to 1. Then, there exists a natural number Nd such that |g1 ((1 − qk )v + ζ ) − r| 2d if k > Nd . Thus if k > Nd , then (g1 + g2 ) (1 − qk )v + ζ − s = g1 (1 − qk )v + ζ + g2 (1 − qk )v + ζ − (r + α) = g1 (1 − qk )v + ζ − r + g2 (1 − qk )v + ζ − α g1 (1 − qk )v + ζ − r − g2 (1 − qk )v + ζ − α . It is clear that |g2 ((1 − qk )v + ζ ) − α| < ε for an arbitrary positive number ε if k is sufficiently large. If we choose ε less than d, (g1 + g2 ) (1 − qk )v + ζ − s 2d − ε > d, which contradicts the fact that s is a point of the linear cluster set of g1 + g2 at ζ . Thus we see CL(g1 + g2 : ζ, v) ⊂ Λ + α. Therefore we conclude that CL(g1 + g2 : ζ, v) = Λ + α.
✷
Lemma 2.3. Suppose that D is a L-connected domain of C and 0 is linearly accessible in D. Then there exists a bounded holomorphic function f˜(z) in D which maps a segment in D terminating at 0 to the dense subset of a closed annulus of positive thickness, or of a closed disk of positive radius in C. Proof. By the assumption of Lemma 2.3, we see that there exists a constant Θ such that arg z Θ for an arbitrary z in D. We can also set the segment terminating at 0 in D as follows: (1 − t)eiθ : d t < 1 ,
696
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
where θ < Θ and d is some constant. Let K and L be positive real numbers which are linearly independent over Z with K > L. Define the function f˜ by f˜(z) = exp(−2πiK log z) + exp(−2πiL log z)
(2.0)
for z ∈ D. Since D is L-connected, log z is a single valued holomorphic function in z. Thus we see that f˜ is holomorphic in D. Furthermore f˜(z) exp 2πKIm(log z) + exp 2πLIm(log z) = exp(2πK arg z) + exp(2πL arg z) exp(2πKΘ) + exp(2πLΘ)
(2.1)
for z ∈ D. This shows that f˜D exp(2πKΘ) + exp(2πLΘ). Therefore f˜(z) is a bounded holomorphic function in D. Next, we discuss the image of the segment s0 in D terminating at 0 by f˜. Put s0 = {(1 − t) exp(iθ ): d t < 1}, where θ Θ and d is some constant. It is clear that we can represent an arbitrary segment in D terminating at 0 as s0 . Since f˜ (1 − t)eiθ = exp −2πiK log (1 − t)eiθ + exp −2πiL log (1 − t)eiθ = exp −2πiK log(1 − t) + iθ + exp −2πiL log(1 − t) + iθ = exp(2π Kθ ) exp −2πiK log(1 − t) + exp(2π Lθ ) exp −2πiL log(1 − t) , by setting u = − log(1 − t), the image of s0 by f˜ is w ∈ C: w = exp(2π Kθ ) exp(i · 2πKu) + exp(2π Lθ ) exp(i · 2πLu), d t < 1 . It is trivial that u → +∞ as t → 1. Case 1. 0 < θ < Θ. Clearly exp(2π Kθ ) > exp(2π Lθ ) because K > L > 0. Let p be an arbitrary point of the annulus in C with center 0 whose radius is exp(2π Kθ ) and thickness is 2 exp(2π Lθ ). It is obvious that there exist z1 = exp(2π Kθ ) exp(i · 2πα) and z2 = exp(2π Lθ ) exp(i · 2πβ) for some real numbers α and β such that z1 + z2 = p. By Lemma 2.1, we can determine u with u − log(1 − d) such that [2πKu, 2πLu] is sufficiently close to [2πα, 2πβ] in (2πT)2 . Thus we have some u such that p − exp(2π Kθ ) exp(i · 2πKu) + exp(2π Lθ ) exp(i · 2πLu) < ε for an arbitrary positive real number ε. This shows that there exists a real number t with d t < 1 such that f˜((1 − t) exp(iθ )) is sufficiently close to p. Therefore the image of the segment s0 by f˜ is the dense subset of the closed annulus of positive thickness w ∈ C: exp(2π Kθ ) − exp(2π Lθ ) |w| exp(2π Kθ ) + exp(2π Lθ ) . Case 2. θ = 0. Note that exp(2π Kθ ) = exp(2π Lθ ) = 1. Then it is clear that there exist z1 = exp(i · 2πα) and z2 = exp(i · 2πβ) for some real numbers α and β such that z1 + z2 = p for an arbitrary point p of the closed disk in C with center 0 whose radius is 2. By Lemma 2.1, we can determine u with u − log(1 − d) such that [2πKu, 2πLu] is sufficiently close to [2πα, 2πβ] in (2πT)2 . Thus we have a real number u such that p − exp(i · 2πKu) + exp(i · 2πLu) < ε
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
697
for an arbitrary positive number ε. This shows that there exists some t with d t < 1 such that f˜((1 − t) exp(iθ )) is sufficiently close to p. Therefore the image of the segment s0 by f˜ is a dense subset of the closed disk {w ∈ C: |w| 2}. Case 3. θ < 0. Since exp(2π Kθ ) < exp(2π Lθ ) for K > L > 0, it is obvious that the image of the segment s0 by f˜ is a dense subset of the closed annulus w ∈ C: exp(2π Lθ ) − exp(2π Kθ ) |w| exp(2π Kθ ) + exp(2π Lθ ) by an argument similar to Case 1. This completes the proof. ✷ Remark 2.4. (1) For an arbitrary positive number M, let h(z) =
M f˜(z) exp(2πKΘ) + exp(2πLΘ)
for z ∈ D.
Suppose that K and L are positive real numbers with K > L which are linearly independent over Z. Then, by (2.1), hD M and the image of a segment in D terminating at 0 by h is a dense subset of one of the following three sets: w ∈ C: C exp(2π Kθ ) − exp(2π Lθ ) |w| C exp(2π Kθ ) + exp(2π Lθ ) for some θ with 0 < θ Θ, w ∈ C: |w| 2C for θ = 0, and
w ∈ C: C exp(2π Lθ ) − exp(2π Kθ ) |w| C exp(2π Kθ ) + exp(2π Lθ )
for some θ with θ < 0, where C = M/(exp(2πKΘ) + exp(2πLΘ)). (2) It is clear that h(z) has a limit along the segment in D terminating at every point of ¯ D/{0} and the absolute values of these limits are all less than or equal to M. (3) The linear cluster set of h at 0 is one of the three sets in (1).
3. Construction of the bounded holomorphic function Proof of Theorem 1.9. By the assumption of Theorem 1.9, there is a hyperplane Hk that goes through ζk and does not intersect Ω for all k with 1 k m. We set Hk = z ∈ Cn : Πζk (z) = 0 , where Πζk (z) = c1k z1 + c2k z2 + · · · + cnk zn + c0k for z = (z1 , z2 , . . . , zn ). Since ζk is linearly accessible, we can find a point pk in Ω for each v ∈ V (ζk ) such that {z: z = (1−t)pk +tζk , 0 t < 1} ⊂ Ω. Let s k = z: z = (1 − t)pk + tζk , pk ∈ Ω, 0 t < 1 , which is the segment in Ω terminating at ζk . We put ζk = (ζ1k , ζ2k , . . . , ζnk ) and pk = (p1k , p2k , . . . , pnk ). If w ∈ Πζk (s k ), then
698
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
w=
n
n n cjk (1 − t)pjk + tζjk + c0k = (1 − t) cjk pjk + t cjk ζjk + c0k
j =1
= (1 − t)
n
cjk pjk + c0k + t
j =1
j =1 n
j =1
cjk ζjk + c0k = (1 − t)Πζk (pk ) + tΠζk (ζk )
j =1
= (1 − t)Πζk (p ), k
since Πζk (ζk ) = 0. This yields that Πζk (s k ) is a segment in Πζk (Ω) terminating at 0. Since 0∈ / Πζk (Ω), 0 belongs to the boundary of Πζk (Ω). Thus 0 is linearly accessible in Πζk (Ω). Then, by Lemma 2.3, f˜ ◦ Πζk (z) = f˜(Πζk (z)) is a bounded holomorphic function in Ω, since Πζk (Ω) is L-connected. Moreover, the image of the segment in Ω terminating at ζk by f˜ ◦ Πζk is a dense subset of a closed disk of positive radius, or of a closed annulus of positive thickness. Let M be an arbitrary positive number, and define the function fk (z) by fk (z) =
M f˜ ◦ Πζk (z) exp(2πKΘ) + exp(2πLΘ)
(3.0)
for z ∈ Ω, where K and L are positive real numbers which are linearly independent over Z with K > L and Θ is a constant such that arg w Θ for w ∈ Πζk (Ω). By (1) in Remark 2.4, we see that fk Ω M and the image of the segment in Ω terminating at ζk by fk is a dense subset of a closed disk, or of a closed annulus. Thus CL(fk : ζk , v k ) is a closed disk or a closed annulus, where v k ∈ V (ζk ). When m is finite, let 1 fk (z). m m
f (z) =
k=1
Then f (z) is holomorphic in Ω and f Ω (1/m) m k=1 fk Ω M. Next, we discuss the linear cluster set of f at ζk . Without loss of generality, we may assume k = 1. Let A = {k: Πζ1 (ζk ) = 0}. Then it is trivial that A = φ because 1 ∈ A. We decompose f (z) into a sum 1 1 fk (z) + fk (z). f (z) = m m k∈A
k∈ /A
Suppose k ∈ A. Then we see ζk ∈ {z: Πζ1 (z) = 0}, thus ζk ∈ {z: Πζ1 (z) = 0} ∩ {z: Πζk (z) = 0}. By the assumption of Theorem 1.9, this indicates that the hyperplane {z: Πζk (z) = 0} is equal to {z: Πζ1 (z) = 0}. Then we deduce that we can choose Πζ1 (z) = 0 as the defining equation of the hyperplane Hk = {z: Πζk (z) = 0}. Thus we have fk (z) = C f˜(Πζ1 (z)) = f1 (z) for k ∈ A and z ∈ Ω, where C = M/(exp(2πKΘ) + exp(2πLΘ)). Let a be the number of the elements of A. Then a 1 fk (z) = f1 (z). m m k∈A This yields that CL((1/m) k∈A fk : ζ1 , v 1 ) is one of the following sets: w ∈ C: QC exp(2π Kθ ) − exp(2π Lθ ) |w| QC exp(2π Kθ ) + exp(2π Lθ )
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
699
for some θ with 0 < θ < Θ, w ∈ C: |w| 2QC , and
w ∈ C: QC exp(2π Lθ ) − exp(2π Kθ ) |w| QC exp(2π Kθ ) + exp(2π Lθ )
for some θ with θ < 0, where Q = a/m and C = M/(exp(2πKΘ) + exp(2πLΘ)). Note that Q is positive, since a 1. Suppose k ∈ / A. Note that fk (z) = C f˜(Πζk (z)) = h(Πζk (z)), where h is as in Remark 2.4. Since Πζk (ζ1 ) = 0 and Πζk (s 1 ) is either the segment in Πζk (Ω) terminating 1 at Πζk (ζ1 ) or one point Πζk (ζ1 ) in Π ζk (Ω), fk has a limit along s at ζ1 by (2) in Re mark 2.4. Then the finite sum (1/m) k ∈/ A fk (z) also has a limit along s 1 at ζ1 . Thus, by Lemma 2.2, we conclude that CL(f : ζ1 , v 1 ) is a closed disk of positive radius, or a closed annulus of positive thickness. When m is infinite, let f (z) =
∞
ak fk (z),
k=1
where {ak }∞ k=1 is a sequence of positive numbers satisfying f Ω
∞
ak fk Ω
k=1
∞ k=1
ak fk Ω M
∞
∞
k=1 ak
= 1. Then
ak = M < +∞,
k=1
which shows that f (z) is a bounded holomorphic function in Ω. Next, we see the linear cluster set of f (z) at ζk . Without loss of generality, we may assume k = 1. Decompose f (z) into ak fk (z) + ak fk (z), f (z) = k∈A
k∈ /A
where A = {k: Πζ1 (ζk ) = 0}. As in the case where m is finite,A is not empty, and if k ∈ A, then fk (z) = f1 (z) for z ∈ Ω. Thus, k∈A ak fk (z) = ( k∈A ak )f1 (z) = σf1 (z) letting the positive number σ = k∈A ak . This yields that CL( k∈A fk : ζ1 , v 1 ) is one of the following sets: w ∈ C: σ C exp(2π Kθ ) − exp(2π Lθ ) |w| σ C exp(2π Kθ ) + exp(2π Lθ ) for some θ with 0 < θ < Θ, w ∈ C: |w| 2σ C , and
w ∈ C: σ C exp(2π Lθ ) − exp(2π Kθ ) |w| σ C exp(2π Kθ ) + exp(2π Lθ )
for some θ with θ < 0, where C = M/(exp(2πKΘ) + exp(2πLΘ)). Let k ∈ / A. Then, every fk has a limit along s 1 at ζ1 as we have seen in the case where m is finite. Thus, if {k: k ∈ / A} is a finite set, then k ∈/ A ak fk (z) also has a limit along s 1
700
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
at ζ1 . Therefore, by Lemma 2.2, we see that CL(f : ζ1 , v 1 ) is a closed disk of positive radius, or a closed annulus of positive thickness. Suppose that {k: k ∈ / A} is an infinite set. Assume {k: k ∈ / A} = {k(1), k(2), . . .}, and let Fl (z) =
ak fk (z) +
l
ak(j ) fk(j ) (z).
j =1
k∈A
Since Fl (z) absolutely and uniformly converges to f (z) on Ω, there exists a natural number Nε for any positive real number ε such that f − Fτ Ω < ε
for τ > Nε
(3.1)
and Fµ − Fτ Ω < ε
for µ, τ > Nε .
(3.2)
Let C = M/(exp(2πKΘ) + exp(2πLΘ)). Case 1. CL( k∈A ak fk : ζ1 , v 1 ) is closed disk {w: |w| 2σ C}. Put δ = (1/10) · 2σ C. By (3.2), we have a natural number N such that Fl − FN+1 Ω < δ
for l > N + 1.
(3.3)
Decompose FN+1 (z) into FN+1 (z) =
ak fk (z) +
N+1
ak(j ) fk(j ) (z).
j =1
k∈A
Since the second term has a limit α along s 1 at ζ1 , we see that the linear cluster set of FN+1 (z) along s 1 at ζ1 is a closed disk D¯ N+1 = w ∈ C: |w − α| 2σ C by Lemma 2.2. Also, when l > N + 1, the linear cluster set of Fl (z) along s 1 at ζ1 is the closed disk D¯ l = w ∈ C: |w − βl | 2σ C , where βl is the limit of lj =1 ak(j )fk(j ) (z) along s 1 at ζ1 . Notice that each D¯ l and D¯ N+1 have the same radius 10δ. We also note that, if l > N + 1, the distance between the centers of D¯ l and D¯ N+1 is l ak(j ) fk(j ) (1 − t)p1 + tζ1 |α − βl | = lim t →1−0 j =1 N+1 1 − lim ak(j )fk(j ) (1 − t)p + tζ1 t →1−0 j =1 l = lim ak(j ) fk(j ) (1 − t)p1 + tζ1 . t →1−0 j =N+2
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
701
On the other hand, by (3.3), we have l 1 ak(j ) fk(j ) (1 − t)p + tζ1 = Fl (1 − t)p1 + tζ1 j =N+2 − FN+1 (1 − t)p1 + tζ1 < δ for all t with 0 t < 1. Since lj =N+2 ak(j ) fk(j ) ((1 − t)p1 + tζ1 ) has the limit as t → 1 − 0, we obtain that |α − βl | δ. Hence the distance between the centers of the disks D¯ l and D¯ N+1 is less than or equal to δ for l > N + 1. This indicates that every D¯ l contains the closed disk D¯ = w ∈ C: |w − α| δ , when l > N + 1. Therefore, we deduce that the closure of the image of s 1 by Fl (z) con¯ tains D. Take an arbitrary point ρ in D¯ and an arbitrary positive number ε. By (3.1), we can choose some natural number Nε such that f (1 − t)p1 + tζ1 − Fl (1 − t)p1 + tζ1 < ε (3.4) 2 for all t with 0 t < 1 and all l > Nε . Set l = max{N + 1, Nε } + 1. Then, since ρ belongs to D¯ l , there exist a sequence {tr }∞ r=1 with tr ↑ 1 and a natural number Aε such that Fl (1 − tr )p1 + tr ζ1 − ρ < ε for r > Aε . (3.5) 2 Thus, for r > Aε , by (3.4) and (3.5), we have f (1 − tr )p1 + tr ζ1 − ρ = f (1 − tr )p1 + tr ζ1 − Fl (1 − tr )p1 + tr ζ1 + Fl (1 − tr )p1 + tr ζ1 − ρ = f (1 − tr )p1 + tr ζ1 − Fl (1 − tr )p1 + tr ζ1 + Fl (1 − tr )p1 + tr ζ1 − ρ < ε. This shows that ρ belongs to the linear cluster set of f (z) at ζ1 . Hence ¯ CL(f : ζ1 , v 1 ) ⊃ D. Case 2. CL( k∈A ak fk : ζ1 , v 1 ) is closed annulus {w ∈ C: σ C(exp(2π Kθ ) − exp(2π Lθ )) |w| σ C(exp(2π Kθ ) + exp(2π Lθ ))} or {w ∈ C: σ C(exp(2π Lθ ) − exp(2π Kθ )) |w| σ C(exp(2π Kθ ) + exp(2π Lθ ))}. Let {w ∈ C: C1 |w| C2 } be an annulus as a linear cluster set as above. Put δ = (1/10)(C2 − C1 ). Notice that (3.3) also holds for this δ. By Lemma 2.2 and the similar argument in Case 1, we see that CL(FN+1 : ζ1 , v 1 ) is a closed annulus E¯ N+1 = w ∈ C: C1 |w − α| C2 ,
702
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
where N is as in (3.3) and α is the limit of that CL(Fl : ζ1 , v 1 ) is a closed annulus E¯ l = w ∈ C: C1 |w − βl | C2 ,
N+1 j =1
ak(j )fk(j ) (z) along s 1 at ζ1 . We also see
where l > N + 1 with N as in (3.3), and βl is the limit of lj =1 ak(j )fk(j ) (z) along s 1 at ζ1 . Note that E¯ N+1 and E¯ l have the same radius and thickness. We also note that, if l > N + 1, the distance between the centers of E¯ N+1 and E¯ l is l |α − βl | = lim ak(j ) fk(j ) (1 − t)p1 + tζ1 t →1−0 j =1 N+1 1 − lim ak(j )fk(j ) (1 − t)p + tζ1 t →1−0 j =1 l = lim ak(j ) fk(j ) (1 − t)p1 + tζ1 . t →1−0 j =N+2
Then, by the same argument in Case 1, we can see that the distance between the centers of E¯ N+1 and E¯ l is less than or equal to δ for l > N + 1. This indicates that every E¯ l contains the closed annulus E¯ = w ∈ C: C1 + 2δ |w − α| C2 − 2δ , when l > N + 1. It is clear that the thickness of E¯ is positive. Hence, the closure of the ¯ image of s 1 by Fl (z) contains E. Take an arbitrary point ρ ∈ E¯ and an arbitrary positive number ε. Then, by similar way of Case 1, we can find a sequence {tr }∞ r=1 with tr ↑ 1 and a natural number Aε such that f (1 − tr )p1 + tr ζ1 − ρ = f (1 − tr )p1 + tr ζ1 − Fl (1 − tr )p1 + tr ζ1 + Fl (1 − tr )p1 + tr ζ1 − ρ = f (1 − tr )p1 + tr ζ1 − Fl (1 − tr )p1 + tr ζ1 + Fl (1 − tr )p1 + tr ζ1 − ρ < ε for r > Aε . Thus, we see that ρ belongs to the linear cluster set of f (z) at ζ1 . Hence ¯ CL(f : ζ1 , v 1 ) ⊃ E. This completes the proof. ✷
4. Construction of the bounded holomorphic map Proof of Theorem 1.10. Let M be an arbitrary positive number, and K1 , L1 , K2 , L2 , . . . , Kg , Lg be positive real numbers which are linearly independent over Z with Kj > Lj
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
703
for 1 j g. We set f˜j (z) = exp(−2πiKj log z) + exp(−2πiLj log z) and define the j function fk (z) by j
fk (z) =
M f˜j ◦ Πζk (z) exp(2πKj Θ) + exp(2πLj Θ)
(4.0)
for z ∈ Ω and 1 k m, where Θ is a constant such that arg w Θ for w ∈ Πζk (Ω). j Notice that fk (z) is the function fk (z) of (3.0) substituting Kj and Lj for K and L. j j Then, we can see that fk (z) is holomorphic in Ω, and fk Ω M for all 1 k m and 1 j g by the similar argument in Section 3. When m is finite, let 1 j fk (z) m m
f j (z) =
for z ∈ Ω,
k=1
where 1 j g, and define the map F : Ω → Cg by F (z) = f 1 (z), f 2 (z), . . . , f g (z) . Clearly F is a bounded holomorphic map from Ω to Cg . Next, we discuss the linear cluster set of F along v k ∈ V (ζk ) at ζk . Without loss of generality, we may assume k = 1. Let A = {k: Πζ1 (ζk ) = 0} and a be the number of the elements of A. Note that 1 a m, since 1 ∈ A. Then, by the similar argument in the proof of Theorem 1.9, we may consider the hyperplane {z ∈ Cn : Πζk (z) = 0} as {z ∈ Cn : Πζ1 (z) = 0} if k ∈ A. Thus
m m 1 1 1 g F (z) = fk (z), . . . , fk (z) m m k=1 k=1 1 1 1 1 g 1 g 1 = fk (z) + fk (z), . . . , fk (z) + fk (z) m m m m k∈A k∈ /A k∈A k∈ /A a 1 1 1 a g 1 g f (z) + = fk (z), . . . , f1 (z) + fk (z) . m 1 m m m k∈ /A
k∈ /A
Since ζ1 in Ω such that {z ∈ Ω: z = (1 −t)p1 j 1 + tζ1 , p ∈ Ω, 0 t < 1} ⊂ Ω. If k ∈ / A, then fk (z) has a limit along the segment s 1 = {z ∈ Ω: z = (1 − t)p1 + tζ1 , p1 ∈ Ω, 0 t < 1} at ζ1 , which we have seen in j the proof of Theorem 1.9. Thus (1/m) k ∈/ A fk (z) also has a limit αj along s 1 at ζ1 for j 1 j g, since k ∈/ A fk (z) is a finite sum. j On the other hand, by (4.0), we see that the image of s 1 by f1 consists of all points j f1 (1 − t)p1 + tζ1 = Cj f˜ (1 − t)Πζ1 (p1 ) = Cj exp −2πiKj log (1 − t)Πζ1 (p1 ) + exp −2πiLj log (1 − t)Πζ1 (p1 ) = Cj exp 2πKj arg Πζ1 (p1 ) exp −2πiKj log (1 − t)Πζ1 (p1 ) + exp 2πLj arg Πζ1 (p1 ) exp −2πiLj log (1 − t)Πζ1 (p1 ) = Cj exp(2πKj θ ) exp(2πKj u) + exp(2πLj θ ) exp(2πLj u) is linearly accessible in Ω, there is a point p1
704
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
for 1 j g, where Cj = M/(exp(2πKj Θ) + exp(2πLj Θ)), θ = arg(Πζ1 (p1 )) with θ Θ and u = − log((1 − t)|Πζ1 (p1 )|). Notice that Πζ1 (p1 ) = 0 because p1 ∈ Ω. It j follows that the image of s 1 by f1 consists of all points M exp(2πKj θ ) exp(i · 2πKj u) exp(2πKj Θ) + exp(2πLj Θ)
+ exp(2πLj θ ) exp(i · 2πLj u) ,
such that u − log d for some positive number d, where 1 j g and θ Θ. Note that this image is a dense subset of Γj , where Γj is a closed disk of positive radius when θ = 0, or a closed annulus of positive thickness when θ = 0, as we have seen in the proof of Lemma 2.3. Take an arbitrary point pj ∈ Γj for 1 j g. Then, we can put pj =
M exp(2πKj θ ) exp(i · 2πvj ) exp(2πKj Θ) + exp(2πLj Θ)
+ exp(2πLj θ ) exp(i · 2πwj ) ,
by choosing real constants vj and wj . Let [2πv1 , 2πw1 , . . . , 2πvg , 2πwg ] ∈ (2πT)2g . Then, by Lemma 2.1, we can find a sequence {ur }∞ r=1 with ur > 0 and ur ↑ +∞ such that [2πK1 ur , 2πL1 ur , . . . , 2πKg ur , 2πLg ur ] → [2πv1 , 2πw1 , . . . , 2πvg , 2πwg ] (4.1) in (2πT)2g as r → +∞. Let tr satisfy − log((1 − tr )|Πζ1 (p1 )|) = ur , and let ε be an arbitrary positive real number. Then we have a sequence {tr }∞ r=1 with tr > 0, tr ↑ 1 and natural numbers A1 , A2 for all 1 j g such that j f (1 − tr )p1 + tr ζ1 − pj < ε (4.2) 1 2 for r > A1 by (4.1), and 1 ε j 1 (1 − t − α fk r )p + tr ζ1 j < m 2 k∈ /A
for r > A2 . Thus, letting Aε = max{A1 , A2 }, we deduce that there exist a sequence {tr }∞ r=1 with tr > 0 and tr ↑ 1 such that j f (1 − tr )p1 + tr ζ1 − a pj + αj m a j a 1 j = f1 (1 − tr )p1 + tr ζ1 − pj + fk (1 − tr )p1 + tr ζ1 − αj m m m k∈ /A a j 1 j f1 (1 − tr )p1 + tr ζ1 − pj + fk (1 − tr )p1 + tr ζ1 − αj < ε m m k∈ /A
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
705
if r > Aε for all 1 j g, since 0 < a/m 1. This yields that a a p1 + α1 , . . . , pg + αg ∈ CL(F : ζ1 , v 1 ) for v 1 ∈ V (ζ1 ). m m It is clear that (a/m)Γj is a closed disk of positive radius, or a closed annulus of positive thickness. Therefore we conclude that CL(F : ζ1 , v 1 ) contains a direct product of closed disks of positive radii, or of closed annuli of positive thicknesses. When m is infinite, let f j (z) =
∞
j
ak fk (z),
k=1
where 1 j g, {ak }∞ k=1 is a sequence of positive real numbers satisfying j and fk is as in (4.0). Define the map F : Ω → Cg by F (z) = f 1 (z), . . . , f g (z) .
∞
k=1 ak
= 1,
Then we see f j Ω
∞ ∞ ∞
j
ak f j
f a M ak = M < +∞ k k Ω k Ω k=1
k=1
k=1
for 1 j g, which shows that F (z) is a bounded holomorphic map from Ω to Cg . Next, we study the linear cluster set of F (z) at ζk . Without loss of generality, we may assume k = 1. Let A = {k: Πζ1 (ζk ) = 0}. Then, as in the case where m is finite, we decompose f j (z) into f j (z) =
j
ak fk (z) +
k∈A j = σf1 (z) +
j
ak fk (z) =
k∈ /A
j
ak f1 (z) +
j
ak fk (z)
k∈ /A
k∈A
j ak fk (z),
k∈ /A
j where σ = k∈A ak for 1 j g. Note that CL(σf1 : ζ1 , v 1 ) is a closed disk of positive radius, or a closed annulus of positive thickness. j j When k ∈/ A ak fk (z) is a finite sum, k ∈/ A ak fk (z) has a limit along s 1 at ζ1 , since j / A. Thus we see that CL(F : ζ1 , v 1 ) contains a direct product of ak fk (z) has a limit if k ∈ closed disks of positive radii, or of closed annuli of positive thicknesses by the similar way in the case where m is finite. j Suppose that / A} = {k(1), k∈ / A ak fk (z) is an infinite series. We assume {k: k ∈ j k(2), . . .}, and define the function Fl (z) by j
Fl (z) =
k∈A
j
ak f1 (z) +
l ν=1
j
j
ak(ν)fk(ν) (z) = σf1 (z) +
l ν=1
j
ak(ν)fk(ν) (z)
706
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707 j
for z ∈ Ω and 1 j g. Let ε be an arbitrary positive real number. Since Fl (z) absolutely and uniformly converges to f j (z) on Ω for all 1 j g, we can choose a natural number Nε such that
j
f − F j < ε for l > Nε (4.3) l Ω 2 and a natural number Nε such that
j
F − F j < ε for µ, τ > N µ τ Ω ε
(4.4)
for all 1 j g. In particular, for a positive constant δj , we can find a natural number N such that
j
F − F j < δj for l > N (4.5) l N+1 Ω l j 1 for 1 j g. It is clear that ν=1 ak(ν) fk(ν) (z) has a limit αl along s at ζ1 and j CL(σf1 : ζ1 , v 1 ) is σ Γj , where Γj is a closed disk or a closed annulus as in the case j j where m is finite. Then, by Lemma 2.2, we see that CL(Fl : ζ1 , v 1 ) is σ Γj + αl which is a closed disk of positive radius, or a closed annulus of positive thickness. By (4.5) and the similar argument which derives the existence of the closed disk D¯ and the closed annulus E¯ in the proof of Theorem 1.9, we can derive that there exist the natural number N and ¯ j for 1 j g such that G ¯ j is either a closed disk of positive radius or a closed the set G j annulus of positive thickness, which is the subset of CL(FN+1 : ζ1 , v 1 ), and is contained in j
CL(Fl : ζ1 , v 1 ) for all l > N . ¯ j for 1 j g and an arbitrary positive number ε. Set Take an arbitrary point bj ∈ G j j l = max{Nε , N + 1} + 1. Since bj ∈ CL(Fl : ζ1 , v 1 ), we can put bj = σpj + αl , where ∞ pj ∈ Γj . We can find a sequence {tr }r=1 with tr > 0, tr ↑ 1 and natural numbers A3 and A4 such that j σf (1 − tr )p1 + tr ζ1 − σpj < ε for r > A3 , (4.6) 1 4 by the similar way deducing (4.2), and l ε j j 1 for r > A4 , ak(ν)fk(ν) (1 − tr )p + tr ζ1 < (4.7) αl − 4 ν=1
j j since αl is the limit of lν=1 ak(ν) fk(ν) (z) along s 1 at ζ1 . Thus we have a natural number N˜ ε such that j F (1 − tr )p1 + tr ζ1 − bj < ε for r > N˜ ε . (4.8) l 2 In fact, by (4.6) and (4.7), j F (1 − tr )p1 + tr ζ1 − bj l
l j j j 1 = σf1 + ak(ν)fk(ν) (1 − tr )p + tr ζ1 − σpj + αl ν=1
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
707
l j j j 1 1 σf1 (1 − tr )p + tr ζ1 − σpj + ak(ν) fk(ν) (1 − tr )p + tr ζ1 − αl ν=1
ε ε ε < + = 4 4 2 for r > N˜ ε = max{A3 , A4 }. Hence, by (4.3) and (4.8), we deduce that j f (1 − tr )p1 + tr ζ1 − bj f j (1 − tr )p1 + tr ζ1 − F j (1 − tr )p1 + tr ζ1 l j + Fl (1 − tr )p1 + tr ζ1 − bj ε ε < + =ε 2 2 ¯ 1 × ···× G ¯ g. for r > N˜ ε and 1 j g. Thus we conclude that CL(F : ζ1 , v 1 ) ⊃ G Therefore we have our assertion. ✷
Acknowledgments The author thanks Professor Hirotaka Fujimoto for his warm encouragement for a long time and smart advice. The author also thanks Professors Yoshihiro Aihara, Kazuo Azukawa, Ei-ichi Hanzawa, Mitsuru Harita, Akio Kodama, Masatake Kuranishi, Junjiro Noguchi, Chikara Watanabe, and the late Professors Michitake Kita, Eiichi Sakai, and Yoshikazu Uno for their warm encouragement and useful suggestions.
References [1] A.B. Aleksandrov, Existence of inner functions, Mat. Sb. 117 (1982) 147–163; English transl. in: Math. USSR Sb. 46 (1983) 143–159. [2] V.I. Arnold, A. Avez, Problèm ergodiques de la mécanique classique, Gauthier–Villars, Paris, 1967. [3] P. Fatou, Séries trigonométriques et séries de Taylor, Acta Math. 30 (1906) 335–400. [4] E. Löw, A construction of inner functions on the unit ball in Cp , Invent. Math. 67 (1982) 223–229. [5] T. Matsushima, Bounded holomorphic function with some boundary behavior in the unit ball of Cn , Kodai Math. J. 24 (2001) 305–312. [6] W. Rudin, Function Theory in the Unit Ball of Cn , Springer, Berlin, 1980. [7] E.M. Stein, Boundary behavior of holomorphic functions of several complex variables, in: Mathematical Notes, Princeton Univ. Press, Princeton, NJ, 1972.
Bounded holomorphic functions and maps with some boundary behavior Toshio Matsushima Ishikawa National College of Technology, Kita-Chuhjoh, Tsubata, Ishikawa 929-0392, Japan Received 26 November 2002 Submitted by J. Noguchi
Abstract Let Ω be a domain of Cn . As a study of boundary behavior of functions and maps in Ω, we consider “the linear cluster set,” which is the cluster set along segment terminating at a boundary point of Ω. We prove that there exist bounded holomorphic functions and maps defined in Ω which have the linear cluster sets of positive measure at every point of a discrete subset of the boundary of Ω under some conditions. 2003 Elsevier Inc. All rights reserved. Keywords: Boundary behavior; Holomorphic mapping; Bounded holomorphic function
1. Introduction and main results It is well known that the bounded holomorphic functions in the unit disk of C have some kind of mild boundary behavior. Indeed, we have the following theorem due to Fatou [3]. Theorem 1.1 (Fatou). If f (z) is a bounded holomorphic function in the unit disk of C, then the limit lim f (rζ )
r→1
exists for almost every point ζ on the unit circle of C. We also have the similar theorem for bounded holomorphic functions in the unit ball of Cn for n 2 [6]. This kind of mildness of the boundary behavior of bounded holomorphic E-mail address: [email protected]. 0022-247X/$ – see front matter 2003 Elsevier Inc. All rights reserved. doi:10.1016/S0022-247X(03)00471-2
692
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
functions was one of the backgrounds of the inner function conjecture (for details, see [6]). The existence of an inner function in the unit ball of Cn for n 2 was, however, proved by Aleksandrov [1] and Löw [4]. This fact causes interest in the existence of the bounded holomorphic functions with wild boundary behavior. The author has shown the following theorem in [5]. Theorem 1.2. Let {ζk }m k=1 be an arbitrary discrete subset of the boundary of the unit ball of Cn , where 1 m +∞, n 1 and ζk = ζl if k = l. Then there exists a bounded holomorphic function f (z) in the unit ball of Cn whose radial cluster set at ζk f (tζk ): T < t < 1 T <1
includes a closed disk of positive radius for all k. The first aim of this article is to generalize this result. As a study of boundary behavior of functions f defined in the domain Ω of Cn , we consider the cluster set along a segment terminating at a boundary point of Ω. We start with giving some definitions. Throughout this paper, let Ω be a domain of Cn with n 1, and let ∂Ω be the boundary of Ω. Moreover, let f Ω = sup f (z), z∈Ω
where f is a function in Ω. Definition 1.3. Let ζ be a boundary point of Ω. (1) A point ζ is said to be linearly accessible in Ω if there exists a nonzero vector v such that {z ∈ Cn : z = ζ + sv, 0 < s 1} ⊂ Ω. We denote by V (ζ ) the set which consists of such vectors. (2) Let f (z) be a function defined in Ω, and assume that ζ is linearly accessible in Ω. Define the linear cluster set of f at ζ by f (1 − t)v + ζ : T < t < 1 , CL(f : ζ, v) = T <1
where v ∈ V (ζ ). (3) Let F be a map from Ω to Cg and ζ , and let v be as in (2). Define the linear cluster set of F at ζ by F (1 − t)v + ζ : T < t < 1 . CL(F : ζ, v) = T <1
We say that F is bounded if the image of Ω by F is bounded. Definition 1.4. A boundary point ζ of Ω is said to be a C-point if there exists a complex hyperplane of Cn that contains ζ and does not intersect Ω. We say that Ω is a C-domain if each point of ∂Ω is a C-point. When n = 1, we consider {ζ } as the hyperplane.
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
693
Definition 1.5. A domain D of C is said to be L-connected if D satisfies the following three conditions: (1) D is simply connected. (2) 0 ∈ / D. (3) If z ∈ D, then arg z Θ for some constant Θ, where arg z is defined as a single valued continuous function in D. Definition 1.6. Let ζ be a C-point on ∂Ω. Assume that H = z = (z1 , . . . , zn ) ∈ Cn : c1 z1 + · · · + cn zn + c0 = 0 is a hyperplane which goes through ζ and does not intersect Ω, and let Πζ (z) = c1 z1 + · · · + cn zn + c0 for z = (z1 , . . . , zn ) ∈ Cn . Then we say that ζ satisfies CL-condition by H if Πζ (Ω) = {w ∈ C: w = Πζ (z), z ∈ Ω} is L-connected. Note that 0 = Πζ (ζ ). We remark that CL-condition by H does not depend on the choice of the polynomial Πζ (z). Example 1.7. √ (1) Let z, w = nk=1 zk w¯ k and |z| = z, z for the points z = (z1 , . . . , zn ) and w = (w1 , . . . , wn ) in Cn . Consider the unit ball Bn = {z ∈ Cn : |z| < 1} of Cn . It is trivial that every boundary point ζ of Bn is linearly accessible. If we choose Πζ (z) =
z, ζ − 1, then it is also clear that ζ is a C-point and ζ satisfies CL-condition by {z: Πζ (z) = 0}. (2) Let Ω = {z = (z1 , . . . , zn ) ∈ Cn : Im z1 > 0}. Clearly every boundary point ζ = (ζ1 , . . . , ζn ) ∈ Cn with ζ1 ∈ R is linearly accessible. If we choose Πζ (z) = z1 − ζ1 , we can see that ζ is a C-point and ζ satisfies CL-condition by {z: Πζ (z) = 0}. (3) Let Ω = {z = (z1 , . . . , zn ) ∈ Cn : Im z1 > −(Re z1 )2 }. We can easily see that every boundary point ζ = (ζ1 , . . . , ζn ) ∈ Cn with Im ζ1 = −(Re ζ1 )2 is a C-point and ζ satisfies CL-condition by {z: Πζ (z) = 0}, where Πζ (z) = z1 − ζ1 . This example shows that a C-domain is not necessarily geometrically convex. Definition 1.8. Let A be an annulus {z ∈ C: a |z − c| b, 0 < a < b} in C. We define the thickness of A by b − a, and radius of A by (a + b)/2. Moreover, we define the center of A by c. Our main result in terms of the first aim is the following theorem. Theorem 1.9. Let {ζk }m k=1 be a discrete subset of the boundary of Ω, where ζk is linearly accessible for all k and 1 m +∞. Suppose that there exists a complex hyperplane Hk for each k that goes through ζk and does not intersect Ω such that every ζk satisfies CL-condition by Hk . Assume that Hk satisfies either Hk ∩ Hl ζk , ζl
(1.0)
694
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
or Hk = Hl
(1.1)
if k = l. Then there exists a bounded holomorphic function f (z) in Ω whose arbitrary linear cluster set at ζk contains a closed disk of positive radius, or a closed annulus of positive thickness. This result gives some type of counter part of Fatou’s theorem of arbitrary dimension, and comprehends Theorem 1.2 as a particular case. In fact, an arbitrary discrete subset of ∂Bn satisfies the assumption of Theorem 1.9. It is known that the Fatou-type theorem holds on the bounded domain with C 2 -boundary in Cn [7]. Though we also need some assumptions for the boundary of Ω, it is not necessary for us to assume the boundedness and the smoothness of the boundary of the domain. The second aim is to show the existence of a bounded holomorphic map with wild boundary behavior. The next theorem is our main result in terms of the bounded holomorphic map. Theorem 1.10. Let {ζk }m k=1 be a discrete subset of the boundary of Ω which satisfies the assumption of Theorem 1.9. Then, there exists a bounded holomorphic map F (z) = (f 1 (z), f 2 (z), . . . , f g (z)) : Ω → Cg for all g < +∞ whose arbitrary linear cluster set at ζk contains a direct product of g closed disks of positive radii, or of g closed annuli of positive thicknesses. We prove Theorem 1.9 in Section 3, and Theorem 1.10 in Section 4.
2. Lemmas Lemma 2.1. Let (2πT)n = Rn /(2πZ)n be a torus of dimension n, and let [x1 , x2 , . . . , xn ] ∈ (2πT)n denote the residue class of modulus (2πZ)n to which (x1 , x2 , . . . , xn ) belongs. For (ω1 , ω2 , . . . , ωn ) ∈ Rn , define the map ϕ : [0, +∞) t → [x1 + 2πω1 t, x2 + 2πω2 t, . . . , xn + 2πωn t] ∈ (2πT)n . for arbitrarily fixed (x1 , x2 , . . . , xn ) ∈ Rn . Then the image of ϕ is dense in (2πT)n if and only if ω1 , ω2 , . . . , ωn are linearly independent over Z. This map is classically known as Kronecker flow. See, for example, [2] for Lemma 2.1. Lemma 2.2. Let g1 (z) and g2 (z) be functions defined in Ω, and let ζ be a linearly accessible in Ω. Suppose that Λ = CL(g1 : ζ, v) for v ∈ V (ζ ), and g2 (z) has a limit α along the segment {z ∈ Ω: z = (1 − t)v + ζ, 0 t < 1} at ζ , i.e., lim g2 (1 − t)v + ζ = α. t →1−0
Then CL(g1 + g2 : ζ, v) = Λ + α.
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
695
Proof. Let ε be an arbitrary positive real number and b an arbitrary point in Λ. By assumptions of Lemma 2.2, there exist a sequence {tk }∞ k=1 with tk ↑ 1 as k → ∞ and a natural number N such that if k > N , then g1 (1 − tk )v + ζ − b < ε and g2 (1 − tk )v + ζ − α < ε . 2 2 Hence if k > N , then (b + α) − (g1 + g2 ) (1 − tk )v + ζ b − g1 (1 − tk )v + ζ + α − g2 (1 − tk )v + ζ < ε, which yields that b + α is a point of CL(g1 + g2 : ζ, v). Consequently, we see Λ + α ⊂ CL(g1 + g2 : ζ, v). Conversely, suppose that s is a point of CL(g1 + g2 , ζ, v), which is not in Λ + α. Then r := s − α ∈ / Λ, and so we can find some positive number d such that {w: |w − r| 2d} ∩ Λ = φ. Take an arbitrary monotone increasing sequence {qk }∞ k=1 of nonnegative numbers which converges to 1. Then, there exists a natural number Nd such that |g1 ((1 − qk )v + ζ ) − r| 2d if k > Nd . Thus if k > Nd , then (g1 + g2 ) (1 − qk )v + ζ − s = g1 (1 − qk )v + ζ + g2 (1 − qk )v + ζ − (r + α) = g1 (1 − qk )v + ζ − r + g2 (1 − qk )v + ζ − α g1 (1 − qk )v + ζ − r − g2 (1 − qk )v + ζ − α . It is clear that |g2 ((1 − qk )v + ζ ) − α| < ε for an arbitrary positive number ε if k is sufficiently large. If we choose ε less than d, (g1 + g2 ) (1 − qk )v + ζ − s 2d − ε > d, which contradicts the fact that s is a point of the linear cluster set of g1 + g2 at ζ . Thus we see CL(g1 + g2 : ζ, v) ⊂ Λ + α. Therefore we conclude that CL(g1 + g2 : ζ, v) = Λ + α.
✷
Lemma 2.3. Suppose that D is a L-connected domain of C and 0 is linearly accessible in D. Then there exists a bounded holomorphic function f˜(z) in D which maps a segment in D terminating at 0 to the dense subset of a closed annulus of positive thickness, or of a closed disk of positive radius in C. Proof. By the assumption of Lemma 2.3, we see that there exists a constant Θ such that arg z Θ for an arbitrary z in D. We can also set the segment terminating at 0 in D as follows: (1 − t)eiθ : d t < 1 ,
696
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
where θ < Θ and d is some constant. Let K and L be positive real numbers which are linearly independent over Z with K > L. Define the function f˜ by f˜(z) = exp(−2πiK log z) + exp(−2πiL log z)
(2.0)
for z ∈ D. Since D is L-connected, log z is a single valued holomorphic function in z. Thus we see that f˜ is holomorphic in D. Furthermore f˜(z) exp 2πKIm(log z) + exp 2πLIm(log z) = exp(2πK arg z) + exp(2πL arg z) exp(2πKΘ) + exp(2πLΘ)
(2.1)
for z ∈ D. This shows that f˜D exp(2πKΘ) + exp(2πLΘ). Therefore f˜(z) is a bounded holomorphic function in D. Next, we discuss the image of the segment s0 in D terminating at 0 by f˜. Put s0 = {(1 − t) exp(iθ ): d t < 1}, where θ Θ and d is some constant. It is clear that we can represent an arbitrary segment in D terminating at 0 as s0 . Since f˜ (1 − t)eiθ = exp −2πiK log (1 − t)eiθ + exp −2πiL log (1 − t)eiθ = exp −2πiK log(1 − t) + iθ + exp −2πiL log(1 − t) + iθ = exp(2π Kθ ) exp −2πiK log(1 − t) + exp(2π Lθ ) exp −2πiL log(1 − t) , by setting u = − log(1 − t), the image of s0 by f˜ is w ∈ C: w = exp(2π Kθ ) exp(i · 2πKu) + exp(2π Lθ ) exp(i · 2πLu), d t < 1 . It is trivial that u → +∞ as t → 1. Case 1. 0 < θ < Θ. Clearly exp(2π Kθ ) > exp(2π Lθ ) because K > L > 0. Let p be an arbitrary point of the annulus in C with center 0 whose radius is exp(2π Kθ ) and thickness is 2 exp(2π Lθ ). It is obvious that there exist z1 = exp(2π Kθ ) exp(i · 2πα) and z2 = exp(2π Lθ ) exp(i · 2πβ) for some real numbers α and β such that z1 + z2 = p. By Lemma 2.1, we can determine u with u − log(1 − d) such that [2πKu, 2πLu] is sufficiently close to [2πα, 2πβ] in (2πT)2 . Thus we have some u such that p − exp(2π Kθ ) exp(i · 2πKu) + exp(2π Lθ ) exp(i · 2πLu) < ε for an arbitrary positive real number ε. This shows that there exists a real number t with d t < 1 such that f˜((1 − t) exp(iθ )) is sufficiently close to p. Therefore the image of the segment s0 by f˜ is the dense subset of the closed annulus of positive thickness w ∈ C: exp(2π Kθ ) − exp(2π Lθ ) |w| exp(2π Kθ ) + exp(2π Lθ ) . Case 2. θ = 0. Note that exp(2π Kθ ) = exp(2π Lθ ) = 1. Then it is clear that there exist z1 = exp(i · 2πα) and z2 = exp(i · 2πβ) for some real numbers α and β such that z1 + z2 = p for an arbitrary point p of the closed disk in C with center 0 whose radius is 2. By Lemma 2.1, we can determine u with u − log(1 − d) such that [2πKu, 2πLu] is sufficiently close to [2πα, 2πβ] in (2πT)2 . Thus we have a real number u such that p − exp(i · 2πKu) + exp(i · 2πLu) < ε
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
697
for an arbitrary positive number ε. This shows that there exists some t with d t < 1 such that f˜((1 − t) exp(iθ )) is sufficiently close to p. Therefore the image of the segment s0 by f˜ is a dense subset of the closed disk {w ∈ C: |w| 2}. Case 3. θ < 0. Since exp(2π Kθ ) < exp(2π Lθ ) for K > L > 0, it is obvious that the image of the segment s0 by f˜ is a dense subset of the closed annulus w ∈ C: exp(2π Lθ ) − exp(2π Kθ ) |w| exp(2π Kθ ) + exp(2π Lθ ) by an argument similar to Case 1. This completes the proof. ✷ Remark 2.4. (1) For an arbitrary positive number M, let h(z) =
M f˜(z) exp(2πKΘ) + exp(2πLΘ)
for z ∈ D.
Suppose that K and L are positive real numbers with K > L which are linearly independent over Z. Then, by (2.1), hD M and the image of a segment in D terminating at 0 by h is a dense subset of one of the following three sets: w ∈ C: C exp(2π Kθ ) − exp(2π Lθ ) |w| C exp(2π Kθ ) + exp(2π Lθ ) for some θ with 0 < θ Θ, w ∈ C: |w| 2C for θ = 0, and
w ∈ C: C exp(2π Lθ ) − exp(2π Kθ ) |w| C exp(2π Kθ ) + exp(2π Lθ )
for some θ with θ < 0, where C = M/(exp(2πKΘ) + exp(2πLΘ)). (2) It is clear that h(z) has a limit along the segment in D terminating at every point of ¯ D/{0} and the absolute values of these limits are all less than or equal to M. (3) The linear cluster set of h at 0 is one of the three sets in (1).
3. Construction of the bounded holomorphic function Proof of Theorem 1.9. By the assumption of Theorem 1.9, there is a hyperplane Hk that goes through ζk and does not intersect Ω for all k with 1 k m. We set Hk = z ∈ Cn : Πζk (z) = 0 , where Πζk (z) = c1k z1 + c2k z2 + · · · + cnk zn + c0k for z = (z1 , z2 , . . . , zn ). Since ζk is linearly accessible, we can find a point pk in Ω for each v ∈ V (ζk ) such that {z: z = (1−t)pk +tζk , 0 t < 1} ⊂ Ω. Let s k = z: z = (1 − t)pk + tζk , pk ∈ Ω, 0 t < 1 , which is the segment in Ω terminating at ζk . We put ζk = (ζ1k , ζ2k , . . . , ζnk ) and pk = (p1k , p2k , . . . , pnk ). If w ∈ Πζk (s k ), then
698
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
w=
n
n n cjk (1 − t)pjk + tζjk + c0k = (1 − t) cjk pjk + t cjk ζjk + c0k
j =1
= (1 − t)
n
cjk pjk + c0k + t
j =1
j =1 n
j =1
cjk ζjk + c0k = (1 − t)Πζk (pk ) + tΠζk (ζk )
j =1
= (1 − t)Πζk (p ), k
since Πζk (ζk ) = 0. This yields that Πζk (s k ) is a segment in Πζk (Ω) terminating at 0. Since 0∈ / Πζk (Ω), 0 belongs to the boundary of Πζk (Ω). Thus 0 is linearly accessible in Πζk (Ω). Then, by Lemma 2.3, f˜ ◦ Πζk (z) = f˜(Πζk (z)) is a bounded holomorphic function in Ω, since Πζk (Ω) is L-connected. Moreover, the image of the segment in Ω terminating at ζk by f˜ ◦ Πζk is a dense subset of a closed disk of positive radius, or of a closed annulus of positive thickness. Let M be an arbitrary positive number, and define the function fk (z) by fk (z) =
M f˜ ◦ Πζk (z) exp(2πKΘ) + exp(2πLΘ)
(3.0)
for z ∈ Ω, where K and L are positive real numbers which are linearly independent over Z with K > L and Θ is a constant such that arg w Θ for w ∈ Πζk (Ω). By (1) in Remark 2.4, we see that fk Ω M and the image of the segment in Ω terminating at ζk by fk is a dense subset of a closed disk, or of a closed annulus. Thus CL(fk : ζk , v k ) is a closed disk or a closed annulus, where v k ∈ V (ζk ). When m is finite, let 1 fk (z). m m
f (z) =
k=1
Then f (z) is holomorphic in Ω and f Ω (1/m) m k=1 fk Ω M. Next, we discuss the linear cluster set of f at ζk . Without loss of generality, we may assume k = 1. Let A = {k: Πζ1 (ζk ) = 0}. Then it is trivial that A = φ because 1 ∈ A. We decompose f (z) into a sum 1 1 fk (z) + fk (z). f (z) = m m k∈A
k∈ /A
Suppose k ∈ A. Then we see ζk ∈ {z: Πζ1 (z) = 0}, thus ζk ∈ {z: Πζ1 (z) = 0} ∩ {z: Πζk (z) = 0}. By the assumption of Theorem 1.9, this indicates that the hyperplane {z: Πζk (z) = 0} is equal to {z: Πζ1 (z) = 0}. Then we deduce that we can choose Πζ1 (z) = 0 as the defining equation of the hyperplane Hk = {z: Πζk (z) = 0}. Thus we have fk (z) = C f˜(Πζ1 (z)) = f1 (z) for k ∈ A and z ∈ Ω, where C = M/(exp(2πKΘ) + exp(2πLΘ)). Let a be the number of the elements of A. Then a 1 fk (z) = f1 (z). m m k∈A This yields that CL((1/m) k∈A fk : ζ1 , v 1 ) is one of the following sets: w ∈ C: QC exp(2π Kθ ) − exp(2π Lθ ) |w| QC exp(2π Kθ ) + exp(2π Lθ )
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
699
for some θ with 0 < θ < Θ, w ∈ C: |w| 2QC , and
w ∈ C: QC exp(2π Lθ ) − exp(2π Kθ ) |w| QC exp(2π Kθ ) + exp(2π Lθ )
for some θ with θ < 0, where Q = a/m and C = M/(exp(2πKΘ) + exp(2πLΘ)). Note that Q is positive, since a 1. Suppose k ∈ / A. Note that fk (z) = C f˜(Πζk (z)) = h(Πζk (z)), where h is as in Remark 2.4. Since Πζk (ζ1 ) = 0 and Πζk (s 1 ) is either the segment in Πζk (Ω) terminating 1 at Πζk (ζ1 ) or one point Πζk (ζ1 ) in Π ζk (Ω), fk has a limit along s at ζ1 by (2) in Re mark 2.4. Then the finite sum (1/m) k ∈/ A fk (z) also has a limit along s 1 at ζ1 . Thus, by Lemma 2.2, we conclude that CL(f : ζ1 , v 1 ) is a closed disk of positive radius, or a closed annulus of positive thickness. When m is infinite, let f (z) =
∞
ak fk (z),
k=1
where {ak }∞ k=1 is a sequence of positive numbers satisfying f Ω
∞
ak fk Ω
k=1
∞ k=1
ak fk Ω M
∞
∞
k=1 ak
= 1. Then
ak = M < +∞,
k=1
which shows that f (z) is a bounded holomorphic function in Ω. Next, we see the linear cluster set of f (z) at ζk . Without loss of generality, we may assume k = 1. Decompose f (z) into ak fk (z) + ak fk (z), f (z) = k∈A
k∈ /A
where A = {k: Πζ1 (ζk ) = 0}. As in the case where m is finite,A is not empty, and if k ∈ A, then fk (z) = f1 (z) for z ∈ Ω. Thus, k∈A ak fk (z) = ( k∈A ak )f1 (z) = σf1 (z) letting the positive number σ = k∈A ak . This yields that CL( k∈A fk : ζ1 , v 1 ) is one of the following sets: w ∈ C: σ C exp(2π Kθ ) − exp(2π Lθ ) |w| σ C exp(2π Kθ ) + exp(2π Lθ ) for some θ with 0 < θ < Θ, w ∈ C: |w| 2σ C , and
w ∈ C: σ C exp(2π Lθ ) − exp(2π Kθ ) |w| σ C exp(2π Kθ ) + exp(2π Lθ )
for some θ with θ < 0, where C = M/(exp(2πKΘ) + exp(2πLΘ)). Let k ∈ / A. Then, every fk has a limit along s 1 at ζ1 as we have seen in the case where m is finite. Thus, if {k: k ∈ / A} is a finite set, then k ∈/ A ak fk (z) also has a limit along s 1
700
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
at ζ1 . Therefore, by Lemma 2.2, we see that CL(f : ζ1 , v 1 ) is a closed disk of positive radius, or a closed annulus of positive thickness. Suppose that {k: k ∈ / A} is an infinite set. Assume {k: k ∈ / A} = {k(1), k(2), . . .}, and let Fl (z) =
ak fk (z) +
l
ak(j ) fk(j ) (z).
j =1
k∈A
Since Fl (z) absolutely and uniformly converges to f (z) on Ω, there exists a natural number Nε for any positive real number ε such that f − Fτ Ω < ε
for τ > Nε
(3.1)
and Fµ − Fτ Ω < ε
for µ, τ > Nε .
(3.2)
Let C = M/(exp(2πKΘ) + exp(2πLΘ)). Case 1. CL( k∈A ak fk : ζ1 , v 1 ) is closed disk {w: |w| 2σ C}. Put δ = (1/10) · 2σ C. By (3.2), we have a natural number N such that Fl − FN+1 Ω < δ
for l > N + 1.
(3.3)
Decompose FN+1 (z) into FN+1 (z) =
ak fk (z) +
N+1
ak(j ) fk(j ) (z).
j =1
k∈A
Since the second term has a limit α along s 1 at ζ1 , we see that the linear cluster set of FN+1 (z) along s 1 at ζ1 is a closed disk D¯ N+1 = w ∈ C: |w − α| 2σ C by Lemma 2.2. Also, when l > N + 1, the linear cluster set of Fl (z) along s 1 at ζ1 is the closed disk D¯ l = w ∈ C: |w − βl | 2σ C , where βl is the limit of lj =1 ak(j )fk(j ) (z) along s 1 at ζ1 . Notice that each D¯ l and D¯ N+1 have the same radius 10δ. We also note that, if l > N + 1, the distance between the centers of D¯ l and D¯ N+1 is l ak(j ) fk(j ) (1 − t)p1 + tζ1 |α − βl | = lim t →1−0 j =1 N+1 1 − lim ak(j )fk(j ) (1 − t)p + tζ1 t →1−0 j =1 l = lim ak(j ) fk(j ) (1 − t)p1 + tζ1 . t →1−0 j =N+2
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
701
On the other hand, by (3.3), we have l 1 ak(j ) fk(j ) (1 − t)p + tζ1 = Fl (1 − t)p1 + tζ1 j =N+2 − FN+1 (1 − t)p1 + tζ1 < δ for all t with 0 t < 1. Since lj =N+2 ak(j ) fk(j ) ((1 − t)p1 + tζ1 ) has the limit as t → 1 − 0, we obtain that |α − βl | δ. Hence the distance between the centers of the disks D¯ l and D¯ N+1 is less than or equal to δ for l > N + 1. This indicates that every D¯ l contains the closed disk D¯ = w ∈ C: |w − α| δ , when l > N + 1. Therefore, we deduce that the closure of the image of s 1 by Fl (z) con¯ tains D. Take an arbitrary point ρ in D¯ and an arbitrary positive number ε. By (3.1), we can choose some natural number Nε such that f (1 − t)p1 + tζ1 − Fl (1 − t)p1 + tζ1 < ε (3.4) 2 for all t with 0 t < 1 and all l > Nε . Set l = max{N + 1, Nε } + 1. Then, since ρ belongs to D¯ l , there exist a sequence {tr }∞ r=1 with tr ↑ 1 and a natural number Aε such that Fl (1 − tr )p1 + tr ζ1 − ρ < ε for r > Aε . (3.5) 2 Thus, for r > Aε , by (3.4) and (3.5), we have f (1 − tr )p1 + tr ζ1 − ρ = f (1 − tr )p1 + tr ζ1 − Fl (1 − tr )p1 + tr ζ1 + Fl (1 − tr )p1 + tr ζ1 − ρ = f (1 − tr )p1 + tr ζ1 − Fl (1 − tr )p1 + tr ζ1 + Fl (1 − tr )p1 + tr ζ1 − ρ < ε. This shows that ρ belongs to the linear cluster set of f (z) at ζ1 . Hence ¯ CL(f : ζ1 , v 1 ) ⊃ D. Case 2. CL( k∈A ak fk : ζ1 , v 1 ) is closed annulus {w ∈ C: σ C(exp(2π Kθ ) − exp(2π Lθ )) |w| σ C(exp(2π Kθ ) + exp(2π Lθ ))} or {w ∈ C: σ C(exp(2π Lθ ) − exp(2π Kθ )) |w| σ C(exp(2π Kθ ) + exp(2π Lθ ))}. Let {w ∈ C: C1 |w| C2 } be an annulus as a linear cluster set as above. Put δ = (1/10)(C2 − C1 ). Notice that (3.3) also holds for this δ. By Lemma 2.2 and the similar argument in Case 1, we see that CL(FN+1 : ζ1 , v 1 ) is a closed annulus E¯ N+1 = w ∈ C: C1 |w − α| C2 ,
702
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
where N is as in (3.3) and α is the limit of that CL(Fl : ζ1 , v 1 ) is a closed annulus E¯ l = w ∈ C: C1 |w − βl | C2 ,
N+1 j =1
ak(j )fk(j ) (z) along s 1 at ζ1 . We also see
where l > N + 1 with N as in (3.3), and βl is the limit of lj =1 ak(j )fk(j ) (z) along s 1 at ζ1 . Note that E¯ N+1 and E¯ l have the same radius and thickness. We also note that, if l > N + 1, the distance between the centers of E¯ N+1 and E¯ l is l |α − βl | = lim ak(j ) fk(j ) (1 − t)p1 + tζ1 t →1−0 j =1 N+1 1 − lim ak(j )fk(j ) (1 − t)p + tζ1 t →1−0 j =1 l = lim ak(j ) fk(j ) (1 − t)p1 + tζ1 . t →1−0 j =N+2
Then, by the same argument in Case 1, we can see that the distance between the centers of E¯ N+1 and E¯ l is less than or equal to δ for l > N + 1. This indicates that every E¯ l contains the closed annulus E¯ = w ∈ C: C1 + 2δ |w − α| C2 − 2δ , when l > N + 1. It is clear that the thickness of E¯ is positive. Hence, the closure of the ¯ image of s 1 by Fl (z) contains E. Take an arbitrary point ρ ∈ E¯ and an arbitrary positive number ε. Then, by similar way of Case 1, we can find a sequence {tr }∞ r=1 with tr ↑ 1 and a natural number Aε such that f (1 − tr )p1 + tr ζ1 − ρ = f (1 − tr )p1 + tr ζ1 − Fl (1 − tr )p1 + tr ζ1 + Fl (1 − tr )p1 + tr ζ1 − ρ = f (1 − tr )p1 + tr ζ1 − Fl (1 − tr )p1 + tr ζ1 + Fl (1 − tr )p1 + tr ζ1 − ρ < ε for r > Aε . Thus, we see that ρ belongs to the linear cluster set of f (z) at ζ1 . Hence ¯ CL(f : ζ1 , v 1 ) ⊃ E. This completes the proof. ✷
4. Construction of the bounded holomorphic map Proof of Theorem 1.10. Let M be an arbitrary positive number, and K1 , L1 , K2 , L2 , . . . , Kg , Lg be positive real numbers which are linearly independent over Z with Kj > Lj
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
703
for 1 j g. We set f˜j (z) = exp(−2πiKj log z) + exp(−2πiLj log z) and define the j function fk (z) by j
fk (z) =
M f˜j ◦ Πζk (z) exp(2πKj Θ) + exp(2πLj Θ)
(4.0)
for z ∈ Ω and 1 k m, where Θ is a constant such that arg w Θ for w ∈ Πζk (Ω). j Notice that fk (z) is the function fk (z) of (3.0) substituting Kj and Lj for K and L. j j Then, we can see that fk (z) is holomorphic in Ω, and fk Ω M for all 1 k m and 1 j g by the similar argument in Section 3. When m is finite, let 1 j fk (z) m m
f j (z) =
for z ∈ Ω,
k=1
where 1 j g, and define the map F : Ω → Cg by F (z) = f 1 (z), f 2 (z), . . . , f g (z) . Clearly F is a bounded holomorphic map from Ω to Cg . Next, we discuss the linear cluster set of F along v k ∈ V (ζk ) at ζk . Without loss of generality, we may assume k = 1. Let A = {k: Πζ1 (ζk ) = 0} and a be the number of the elements of A. Note that 1 a m, since 1 ∈ A. Then, by the similar argument in the proof of Theorem 1.9, we may consider the hyperplane {z ∈ Cn : Πζk (z) = 0} as {z ∈ Cn : Πζ1 (z) = 0} if k ∈ A. Thus
m m 1 1 1 g F (z) = fk (z), . . . , fk (z) m m k=1 k=1 1 1 1 1 g 1 g 1 = fk (z) + fk (z), . . . , fk (z) + fk (z) m m m m k∈A k∈ /A k∈A k∈ /A a 1 1 1 a g 1 g f (z) + = fk (z), . . . , f1 (z) + fk (z) . m 1 m m m k∈ /A
k∈ /A
Since ζ1 in Ω such that {z ∈ Ω: z = (1 −t)p1 j 1 + tζ1 , p ∈ Ω, 0 t < 1} ⊂ Ω. If k ∈ / A, then fk (z) has a limit along the segment s 1 = {z ∈ Ω: z = (1 − t)p1 + tζ1 , p1 ∈ Ω, 0 t < 1} at ζ1 , which we have seen in j the proof of Theorem 1.9. Thus (1/m) k ∈/ A fk (z) also has a limit αj along s 1 at ζ1 for j 1 j g, since k ∈/ A fk (z) is a finite sum. j On the other hand, by (4.0), we see that the image of s 1 by f1 consists of all points j f1 (1 − t)p1 + tζ1 = Cj f˜ (1 − t)Πζ1 (p1 ) = Cj exp −2πiKj log (1 − t)Πζ1 (p1 ) + exp −2πiLj log (1 − t)Πζ1 (p1 ) = Cj exp 2πKj arg Πζ1 (p1 ) exp −2πiKj log (1 − t)Πζ1 (p1 ) + exp 2πLj arg Πζ1 (p1 ) exp −2πiLj log (1 − t)Πζ1 (p1 ) = Cj exp(2πKj θ ) exp(2πKj u) + exp(2πLj θ ) exp(2πLj u) is linearly accessible in Ω, there is a point p1
704
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
for 1 j g, where Cj = M/(exp(2πKj Θ) + exp(2πLj Θ)), θ = arg(Πζ1 (p1 )) with θ Θ and u = − log((1 − t)|Πζ1 (p1 )|). Notice that Πζ1 (p1 ) = 0 because p1 ∈ Ω. It j follows that the image of s 1 by f1 consists of all points M exp(2πKj θ ) exp(i · 2πKj u) exp(2πKj Θ) + exp(2πLj Θ)
+ exp(2πLj θ ) exp(i · 2πLj u) ,
such that u − log d for some positive number d, where 1 j g and θ Θ. Note that this image is a dense subset of Γj , where Γj is a closed disk of positive radius when θ = 0, or a closed annulus of positive thickness when θ = 0, as we have seen in the proof of Lemma 2.3. Take an arbitrary point pj ∈ Γj for 1 j g. Then, we can put pj =
M exp(2πKj θ ) exp(i · 2πvj ) exp(2πKj Θ) + exp(2πLj Θ)
+ exp(2πLj θ ) exp(i · 2πwj ) ,
by choosing real constants vj and wj . Let [2πv1 , 2πw1 , . . . , 2πvg , 2πwg ] ∈ (2πT)2g . Then, by Lemma 2.1, we can find a sequence {ur }∞ r=1 with ur > 0 and ur ↑ +∞ such that [2πK1 ur , 2πL1 ur , . . . , 2πKg ur , 2πLg ur ] → [2πv1 , 2πw1 , . . . , 2πvg , 2πwg ] (4.1) in (2πT)2g as r → +∞. Let tr satisfy − log((1 − tr )|Πζ1 (p1 )|) = ur , and let ε be an arbitrary positive real number. Then we have a sequence {tr }∞ r=1 with tr > 0, tr ↑ 1 and natural numbers A1 , A2 for all 1 j g such that j f (1 − tr )p1 + tr ζ1 − pj < ε (4.2) 1 2 for r > A1 by (4.1), and 1 ε j 1 (1 − t − α fk r )p + tr ζ1 j < m 2 k∈ /A
for r > A2 . Thus, letting Aε = max{A1 , A2 }, we deduce that there exist a sequence {tr }∞ r=1 with tr > 0 and tr ↑ 1 such that j f (1 − tr )p1 + tr ζ1 − a pj + αj m a j a 1 j = f1 (1 − tr )p1 + tr ζ1 − pj + fk (1 − tr )p1 + tr ζ1 − αj m m m k∈ /A a j 1 j f1 (1 − tr )p1 + tr ζ1 − pj + fk (1 − tr )p1 + tr ζ1 − αj < ε m m k∈ /A
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
705
if r > Aε for all 1 j g, since 0 < a/m 1. This yields that a a p1 + α1 , . . . , pg + αg ∈ CL(F : ζ1 , v 1 ) for v 1 ∈ V (ζ1 ). m m It is clear that (a/m)Γj is a closed disk of positive radius, or a closed annulus of positive thickness. Therefore we conclude that CL(F : ζ1 , v 1 ) contains a direct product of closed disks of positive radii, or of closed annuli of positive thicknesses. When m is infinite, let f j (z) =
∞
j
ak fk (z),
k=1
where 1 j g, {ak }∞ k=1 is a sequence of positive real numbers satisfying j and fk is as in (4.0). Define the map F : Ω → Cg by F (z) = f 1 (z), . . . , f g (z) .
∞
k=1 ak
= 1,
Then we see f j Ω
∞ ∞ ∞
j
ak f j
f a M ak = M < +∞ k k Ω k Ω k=1
k=1
k=1
for 1 j g, which shows that F (z) is a bounded holomorphic map from Ω to Cg . Next, we study the linear cluster set of F (z) at ζk . Without loss of generality, we may assume k = 1. Let A = {k: Πζ1 (ζk ) = 0}. Then, as in the case where m is finite, we decompose f j (z) into f j (z) =
j
ak fk (z) +
k∈A j = σf1 (z) +
j
ak fk (z) =
k∈ /A
j
ak f1 (z) +
j
ak fk (z)
k∈ /A
k∈A
j ak fk (z),
k∈ /A
j where σ = k∈A ak for 1 j g. Note that CL(σf1 : ζ1 , v 1 ) is a closed disk of positive radius, or a closed annulus of positive thickness. j j When k ∈/ A ak fk (z) is a finite sum, k ∈/ A ak fk (z) has a limit along s 1 at ζ1 , since j / A. Thus we see that CL(F : ζ1 , v 1 ) contains a direct product of ak fk (z) has a limit if k ∈ closed disks of positive radii, or of closed annuli of positive thicknesses by the similar way in the case where m is finite. j Suppose that / A} = {k(1), k∈ / A ak fk (z) is an infinite series. We assume {k: k ∈ j k(2), . . .}, and define the function Fl (z) by j
Fl (z) =
k∈A
j
ak f1 (z) +
l ν=1
j
j
ak(ν)fk(ν) (z) = σf1 (z) +
l ν=1
j
ak(ν)fk(ν) (z)
706
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707 j
for z ∈ Ω and 1 j g. Let ε be an arbitrary positive real number. Since Fl (z) absolutely and uniformly converges to f j (z) on Ω for all 1 j g, we can choose a natural number Nε such that
j
f − F j < ε for l > Nε (4.3) l Ω 2 and a natural number Nε such that
j
F − F j < ε for µ, τ > N µ τ Ω ε
(4.4)
for all 1 j g. In particular, for a positive constant δj , we can find a natural number N such that
j
F − F j < δj for l > N (4.5) l N+1 Ω l j 1 for 1 j g. It is clear that ν=1 ak(ν) fk(ν) (z) has a limit αl along s at ζ1 and j CL(σf1 : ζ1 , v 1 ) is σ Γj , where Γj is a closed disk or a closed annulus as in the case j j where m is finite. Then, by Lemma 2.2, we see that CL(Fl : ζ1 , v 1 ) is σ Γj + αl which is a closed disk of positive radius, or a closed annulus of positive thickness. By (4.5) and the similar argument which derives the existence of the closed disk D¯ and the closed annulus E¯ in the proof of Theorem 1.9, we can derive that there exist the natural number N and ¯ j for 1 j g such that G ¯ j is either a closed disk of positive radius or a closed the set G j annulus of positive thickness, which is the subset of CL(FN+1 : ζ1 , v 1 ), and is contained in j
CL(Fl : ζ1 , v 1 ) for all l > N . ¯ j for 1 j g and an arbitrary positive number ε. Set Take an arbitrary point bj ∈ G j j l = max{Nε , N + 1} + 1. Since bj ∈ CL(Fl : ζ1 , v 1 ), we can put bj = σpj + αl , where ∞ pj ∈ Γj . We can find a sequence {tr }r=1 with tr > 0, tr ↑ 1 and natural numbers A3 and A4 such that j σf (1 − tr )p1 + tr ζ1 − σpj < ε for r > A3 , (4.6) 1 4 by the similar way deducing (4.2), and l ε j j 1 for r > A4 , ak(ν)fk(ν) (1 − tr )p + tr ζ1 < (4.7) αl − 4 ν=1
j j since αl is the limit of lν=1 ak(ν) fk(ν) (z) along s 1 at ζ1 . Thus we have a natural number N˜ ε such that j F (1 − tr )p1 + tr ζ1 − bj < ε for r > N˜ ε . (4.8) l 2 In fact, by (4.6) and (4.7), j F (1 − tr )p1 + tr ζ1 − bj l
l j j j 1 = σf1 + ak(ν)fk(ν) (1 − tr )p + tr ζ1 − σpj + αl ν=1
T. Matsushima / J. Math. Anal. Appl. 285 (2003) 691–707
707
l j j j 1 1 σf1 (1 − tr )p + tr ζ1 − σpj + ak(ν) fk(ν) (1 − tr )p + tr ζ1 − αl ν=1
ε ε ε < + = 4 4 2 for r > N˜ ε = max{A3 , A4 }. Hence, by (4.3) and (4.8), we deduce that j f (1 − tr )p1 + tr ζ1 − bj f j (1 − tr )p1 + tr ζ1 − F j (1 − tr )p1 + tr ζ1 l j + Fl (1 − tr )p1 + tr ζ1 − bj ε ε < + =ε 2 2 ¯ 1 × ···× G ¯ g. for r > N˜ ε and 1 j g. Thus we conclude that CL(F : ζ1 , v 1 ) ⊃ G Therefore we have our assertion. ✷
Acknowledgments The author thanks Professor Hirotaka Fujimoto for his warm encouragement for a long time and smart advice. The author also thanks Professors Yoshihiro Aihara, Kazuo Azukawa, Ei-ichi Hanzawa, Mitsuru Harita, Akio Kodama, Masatake Kuranishi, Junjiro Noguchi, Chikara Watanabe, and the late Professors Michitake Kita, Eiichi Sakai, and Yoshikazu Uno for their warm encouragement and useful suggestions.
References [1] A.B. Aleksandrov, Existence of inner functions, Mat. Sb. 117 (1982) 147–163; English transl. in: Math. USSR Sb. 46 (1983) 143–159. [2] V.I. Arnold, A. Avez, Problèm ergodiques de la mécanique classique, Gauthier–Villars, Paris, 1967. [3] P. Fatou, Séries trigonométriques et séries de Taylor, Acta Math. 30 (1906) 335–400. [4] E. Löw, A construction of inner functions on the unit ball in Cp , Invent. Math. 67 (1982) 223–229. [5] T. Matsushima, Bounded holomorphic function with some boundary behavior in the unit ball of Cn , Kodai Math. J. 24 (2001) 305–312. [6] W. Rudin, Function Theory in the Unit Ball of Cn , Springer, Berlin, 1980. [7] E.M. Stein, Boundary behavior of holomorphic functions of several complex variables, in: Mathematical Notes, Princeton Univ. Press, Princeton, NJ, 1972.