BOUNDED HOLOMORPHIC FUNCTIONS WITH GIVEN MAXIMUM ...

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 137, Number 1, January 2009, Pages 179–187 S 0002-9939(08)09468-9 Article electronically published on July 31, 2008

BOUNDED HOLOMORPHIC FUNCTIONS WITH GIVEN MAXIMUM MODULUS ON ALL CIRCLES PIOTR KOT (Communicated by Mei-Chi Shaw) Abstract. We study Ω ⊂ Cd , a circular, bounded, strictly convex domain with C 2 boundary. Let g and h be continuous functions on ∂Ω with |g(z)| < h(z) = h(λz) for z ∈ ∂Ω and |λ| = 1. First we prove that h can be approximated by the maximum modulus values of K homogeneous polynomials, where K is independent from h. Next we construct f1 ∈ A(Ω) such that max |(g + f1 )(λz)| = h(z)

|λ|=1

for z ∈ ∂Ω. Moreover we can choose f2 ∈ O(Ω) with |f2∗ (z)| = h(z) for almost all z ∈ ∂Ω and max|λ| 1 a holomorphic non-constant function f ∈ A(Bd ) such that |f (z)| = 1 for all z ∈ ∂Bd does not exist. Therefore maximum modulus sets and peak sets are extensively considered by many authors. Topologically, peak sets and maximum modulus sets are small in strictly pseudoconvex domains, as was proved by Stout and Duchamp. The real topological dimension of a maximum modulus set is no more than d [7] and for a peak set is no more than d − 1 [6]. In particular, the maximum modulus set must have an empty interior. However, from the measure-theoretic point of view, peak sets and maximum modulus sets no longer have to be small. Stens¨ ones Henriksen has proved [5] that every strictly pseudoconvex domain with C ∞ boundary in C d has a peak set with a Hausdorff dimension 2d − 1. However, Eric Løv has proved [4] that a maximum modulus set can have positive (2d − 1)-dimensional Hausdorff measure. We construct a maximum modulus set which crosses any circle in ∂Ω with the center at zero. Therefore the Hausdorff dimension is at least 2d − 2. It is also Received by the editors September 11, 2007, and, in revised form, December 12, 2007. 2000 Mathematics Subject Classification. Primary 32A05, 32A35. Key words and phrases. Homogeneous polynomials, maximum modulus set, inner function. c 2008 American Mathematical Society Reverts to public domain 28 years from publication

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possible to combine our construction with the construction of an inner function. Other constructions of inner functions can be found in [1, 3, 4]. 2. Homogeneous polynomials In our constructions we use the homogeneous polynomials from the paper [2], and thus we consider the same (as in [2]) unitarily invariant pseudo-metric ρ(z, w) = min z − λw . |λ|=1

1 Now for ξ ∈ ∂Ω we define a vector νξ = ξ,η ηξ , where η is a defining function ξ ∂η ∂η of class C 2 for Ω and ηξ = ∂z (ξ), ..., ∂z (ξ) . Recall that: 1 d

Lemma ([2, Lemma 2.1]). There exist constants c1 , c2 > 0 such that (1)

c1 ρ2 (z, ξ) ≤ 1 − |z, νξ | ≤ c2 ρ2 (z, ξ)

for ξ, z ∈ ∂Ω. Let α > 0. A subset A ⊂ Cd is called α-separated if ρ(z1 , z2 ) > α for all distinct elements z1 and z2 of A. It is clear that for α > 0 each α-separated subset of ∂Ω is finite. We need the following lemmas: Lemma 2.1 ([2, Lemma 2.2]). For z ∈ ∂Ω let Ak (z) := {ξ ∈ A : αkt ≤ ρ(z, ξ) ≤ α(k + 1)t} . There exists a constant c > 0 such that if A = {ξ1 , ..., ξs } is a 2αt-separated subset of ∂Ω, then the set Ak (z) has at most c(k + 2)2d−1 elements. The set A0 has at most 1 element and s ≤ c(αt)1−2d . Lemma 2.2 ([2, Lemma 2.3]). If A ⊂ ∂Ω is αt-separated, then for each β > α > 0 there exists an integer K = K(α, β) such that A can be partitioned into K disjoint βt-separated sets. Lemma 2.3 ([2, Lemma 2.5]). Let 0 < c1 < c2 be constants from (1). For a given a ∈ (0, 0.5) there √ exist constants C > 2 and N0 ∈ N such that for all integers subset A of ∂Ω and each integer m with N ≥ N0 , for each C/ c1 N -separated m N ≤ m ≤ 2N , the polynomial pm (z) := ξ∈A z, νξ satisfies the following: (1) If z ∈ ∂Ω, Q(z) := ξ ∈ A : ρ(z, ξ) ≥ 2√Cc N , then ξ∈Q(z) |z, νξ |m < a. 1 (2) If z ∈ ∂Ω, then A \ Q(z) has at most one element. (3) If ξ0 ∈ A, z ∈ ∂Ω are such that ρ(z, ξ0 ) ≤ √ca N , then 2 (a) Q(z) = A \ {ξ0 }, (b) |z, νξ0 |m > 1 − 2a2 , 2 (c) |pm (z)| > 1 − 2a −ma. (4) |pm (z)| ≤ ξ∈A |z, νξ | < 1 + a for all z ∈ ∂Ω. Now we can prove the following result: Lemma 2.4. For a given θ ∈ (0, 1) there exists a constant C > 2 such that if h is a continuous strictly positive function on ∂Ω with the same values on the circles, then there exists N1 ∈ N such that for all integers N ≥ N1 , for each √cC N -separated 1 subset A of m with N ≤ m ≤ 2N the homogeneous polynomial ∂Ω and each integer pm (z) := ξ∈A h(ξ) z, νξ m satisfies the following:


BOUNDED HOLOMORPHIC FUNCTIONS

181

(1) |pm (z)| < (1 + θ)h(z) for all z ∈ ∂Ω, (2) |pm (z)| > (1 − θ)h(z) for each z ∈ ∂Ω such that ρ(z, ξ) ≤ ξ ∈ A.

√θ 4 c2 N

for some

Proof. We can choose C > 2 and N0 ∈ N from Lemma 2.3 used for a = θ4 . Since h has the same values on the circles, there exists ε ∈ (0, 1) such that we obtain the following property for z, ξ ∈ ∂Ω: 1 + 2a 1 h(z) ≤ h(ξ) ≤ h(z). (2) c1 ρ2 (ξ, z) < ε ⇒ 1+a 1+a Let Γ(z) := ξ ∈ A : c1 ρ2 (ξ, z) < ε . Now let us denote rm (z) := h(ξ) z, νξ m , ξ∈A\Γ(z)

qm (z)

:=

m

h(ξ) z, νξ .

ξ∈Γ(z)

In particular due to Lemma 2.1 we may estimate (1) m |rm (z)| ≤ h |z, νξ | ≤ h ξ∈A\Γ(z)

1−2d

(1 − ε)N

ξ∈A\Γ(z)

C √ (1 − ε)N . 2 c1 N Now we may conclude that there exists N1 > N0 such that for N1 ≤ N ≤ m ≤ 2N and z ∈ ∂Ω we have L2.1

≤

h c

|rm (z)| < a min h(ξ) ≤ ah(z).

(3)

ξ∈∂Ω

Due to Lemma 2.3(4) we may obtain the property (1): (3) |pm (z)| ≤ |qm (z)| + |rm (z)| < h(ξ) |z, νξ |m + ah(z) ξ∈Γ(z) L2.3(4) 1 + 2a h(z) |z, νξ |m + ah(z) ≤ (1 + 3a)h(z) ≤ (1 + θ)h(z). 1+a

(2)

|pm (z)| ≥ |qm (z)| − |rm (z)| > (1 − 3a)h(z) − ah(z) ≥ (1 − θ)h(z). We improve the above lemma to get a more useful tool in applications.


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Theorem 2.5. For a given constant a ∈ (0, 1) there exists a natural number K such that if h is a continuous strictly positive function on ∂Ω with the same values on the circles, then there exists N1 ∈ N such that for all the integers N and m1 , ..., mK with N1 ≤ N ≤ m1 ≤ ... ≤ mK ≤ 2N there exist homogeneous polynomials pm1 , ..., pmK of a degree m1 , ..., mK respectively such that ah(z) < maxi=1,...,K |pmi (z)| < h(z) for all z ∈ ∂Ω. Proof. Let 0 < c1 < c2 be from (1). For a constant θ = 1−a 1+a we can choose C, N1 from Lemma 2.4. Let us consider N ≥ N1 and denote AN - a maximal √θ -separated subset of ∂Ω. Due to Lemma 2.2 there exists a natural number 4 c2 N K = K 4√θc2 , √Cc1 such that AN can be partitioned into A1N , ..., AK N disjoint √ C -separated c1 N

sets. Now using Lemma 2.4 for a function h and set AiN we may conclude that there exists a homogeneous polynomial pmi of mi degree such that • (1 + θ)|pmi (z)| < (1 + θ)h(z) for all z ∈ ∂Ω. • (1 + θ)|pmi (z)| > (1 − θ)h(z) for each z ∈ ∂Ω such that ρ(z, ξ) ≤ some ξ ∈ AiN .

√θ 4 c2 N

for

Now let z ∈ ∂Ω. Since AN is a maximal 4√θc N -separated subset of ∂Ω, there 2 √ i exists ξ ∈ AN such that ρ(z, ξ) ≤ 1/(4 c2 N ). Due to AN = K i=1 AN there exists i0 1−θ i0 ∈ {1, ..., K} such that ξ ∈ AN . In particular |pmi (z)| > 1+θ h(z) = ah(z) and therefore maxi=1,...,K |pmi (z)| > ah(z), which finishes the proof. Now we can prove the following fact, which will enable us to construct the bounded holomorphic function with a desired property. Lemma 2.6. There exists θ ∈ (0, 1) such that if δ1 , δ2 ∈ (0, 1), T is a compact subset of Ω, g is a complex continuous function on ∂Ω and h is a continuous strictly positive function on ∂Ω such that |g(z)| < h(z) = h(λz) for |λ| = 1, z ∈ ∂Ω, then there exists a polynomial q with the following properties: (1) supz∈T |q(z)| < δ1 , (2) |q(z)| < H(z) for all z ∈ ∂Ω, (3) max−δ2 ≤ϕ≤δ2 g(eiϕ z) + q(eiϕ z) > |g(z)| + θH(z) for z ∈ ∂Ω, where H(z) = h(z) − max|λ|=1 |g(λz)|. Proof. Let a = 12 and K be a natural number from Theorem 2.5. There ex ists ε ∈ (0, 12 δ2 ) such that πε ∈ N and max−2ε≤ϕ≤2ε g(z) − g(eiϕ z) ≤ H(z) 8K for 1 z ∈ ∂Ω. Let n = πε . We show that we can choose θ = 4K . Due to Theorem 2.5 we conclude that there exists N0 ∈ N such that for all m ∈ N with N0 ≤ mn < (m + 1)n < ... < (m + K)n < 2mn there exist homogeneous polynomials p(m+1)n , ..., p(m+K)n of ..., (m + K)n degree respectively such (m + 1)n, that 12 H(z) < K maxj=1,...,K p(m+j)n (z) < H(z) for all z ∈ ∂Ω. Let us choose K qm := j=1 p(m+j)n . We show that it is enough to choose q = qm for m large enough. First we observe that qm fulfills the properties (1)-(2) for m large enough

π so we only prove property (3). Since −nπ exp (−itkn) dt = 0 for k ∈ N we may n



estimate √ 2ε max qm (eiϕ z) −ε≤ϕ≤ε

≥

ε

−ε

iϕ |qm (e z)| dϕ = 2

π n

183

K p(m+j)n (z) 2 dϕ

−π n j=1

2π H(z) √ H(z) = 2ε n 2K 2K for z ∈ ∂Ω. There exists ηz ∈ [−ε, ε] such that qm (eiηz z) = max−ε≤ϕ≤ε qm (eiϕ z) . In particular qm (eiηz z) ≥ H(z) 2K . Since K K ijnϕ ijnϕ e e qm (eiϕ z) − eimnϕ qm (z) ≤ − 1 p(m+j)n (z) ≤ − 1 H(z) ≥

j=1

j=1

there exists ε2 ∈ (0, ε) such that qm (eiϕ z) − eimnϕ qm (z) < H(z) 8K for z ∈ ∂Ω, m ∈ N and −ε2 ≤ ϕ ≤ ε2 . Moreover we may observe that if mnε2 > π, then there exists ϕz ∈ [−ε2 , ε2 ] such that g(z) + eimnϕz qm (eiηz z) = |g(z)| + qm (eiηz z) . In particular we may estimate g(z) + qm (eiϕ z) − H(z) −2ε≤ϕ≤2ε −δ2 ≤ϕ≤δ2 8K H(z) ≥ max g(z) + qm (ei(ηz +ϕ) z) − −ε2 ≤ϕ≤ε2 8K H(z) > max g(z) + eimnϕ qm (eiηz z) − −ε2 ≤ϕ≤ε2 4K H(z) ≥ |g(z)| + qm (eiηz z) − 4K H(z) ≥ |g(z)| + . 4K We have just proved that it is enough to define q = qm for m large enough. max

iϕ g(e z) + qm (eiϕ z) ≥

max

Now we construct an example of a maximum modulus set K which crosses any circle in ∂Ω with the center at zero. In particular Hausdorff dimension of K is at least 2d − 2. Theorem 2.7. Let ε > 0, T be a compact subset of Ω, g be a complex continuous function on ∂Ω and h be a positive continuous function on ∂Ω such that |g(z)| < h(z) = h(λz) for |λ| = 1, z ∈ ∂Ω. Then there exists f ∈ A(Ω) such that f T < ε and max|λ|=1 |(g + f )(λz)| = h(z) for z ∈ ∂Ω. Proof. Let θ ∈ (0, 1) be a constant from Lemma 2.6. We construct the sequence of holomorphic polynomials fn ∈ A(Ω) with the following properties: ε , (1) fn T < 2n+1 (2) |fn | < Hn on ∂Ω, (3) max|λ|=1 |(gn + fn )(λz)| > |gn (z)| + θHn (z) for z ∈ ∂Ω,


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PIOTR KOT

where g1 = g, gn+1 = gn + fn , Hn (z) = h(z) − max|λ|=1 |gn (λz)| for z ∈ ∂Ω. To construct f1 it is enough to use Lemma 2.6. If f1 , ..., fn are constructed, we can estimate |gn+1 (z)| = |(gn + fn )(z)| ≤ |gn (z)| + |fn (z)| < |gn (z)| + Hn (z) ≤ h(z). Therefore to construct fn+1 it suffices to use Lemma 2.6 once again. Since H(z) = H(λz) for |λ| = 1, due to (3) we can conclude max |(gn + fn )(λz)| > max |gn (λz)| + θHn (z) ≥ θh(z) + (1 − θ) max |gn (λz)|,

|λ|=1

|λ|=1

|λ|=1

which implies

Hn+1 (z) ≤ h(z) − max |(gn + fn )(λz)| < (1 − θ) h(z) − max |gn (λz)| |λ|=1

|λ|=1

≤ (1 − θ)Hn (z). The last inequality gives us limn→∞ Hn (z) = 0. Moreover we can estimate ∞ n=m

|fn |
1 − 2εm+1 ,



185

where H1 = h and Hm+1 := Hm − |ωm+1 | + |ωm |. First we construct (ε2 , ω2 , V1 ) so that (1)-(7) are fulfilled for m = 1. Let ε2 ∈ (0, 0.25). Due to Theorem 2.7 there exists g ∈ A(Ω) such that gT1 < 12 ε2 and max|λ|=1 |g(λz)| = 34 h(z). There exists δ ∈ (0, 1) so that |g(z) − g(δz)| < 1 8 minw∈∂Ω h(w) for z ∈ ∂Ω. Now due to [3, Lemma 3.7] there exists f ∈ A(Ω) and V1 , an open subset of ∂Ω such that • • • • •

f T1 < 12 ε2 , |f (δz)| < 18 minw∈∂Ω h(w) for z ∈ ∂Ω, |f + g| − |g| < h − |g| on ∂Ω, |f + g| − |g| > (1 − ε2 )(h − |g|) on V1 , σ(V 1 ) = σ(V1 ) > 1 − 2ε2 .

Let us choose ω2 = f + g. We show that properties (1)-(7) are fulfilled for m = 1. In fact it is enough to prove only (4) and (6). The remaining properties are trivial. We may estimate max |ω2 (λz)| ≥

|λ|=1

max |f (λδz) + g(λδz)| ≥ max (|g(λδz)| − |f (λδz)|)

|λ|=1

|λ|=1

1 1 1 > max |g(λz)| − min h(w) − min h(w) ≥ h(z) 8 w∈∂Ω 8 w∈∂Ω 2 |λ|=1 for z ∈ ∂Ω, which implies (4). Moreover we have |ω2 | > |g| + (1 − ε2 )(h − |g|) ≥ (1 − ε2 )h + ε2 |g| ≥ (1 − ε2 )h on V1 , which gives (6). Now suppose that we have constructed ε2 ...εm , ω2 ...ωm , V1 ...Vm−1 . Let us observe that Hm > 0 and h − Hm =

m−1

m−1

k=1

k=1

(Hk − Hk+1 ) =

(|ωk+1 | − |ωk |) = |ωm |.

In particular there exists εm+1 so that (1)-(2) are fulfilled. Since |ωm | < h, due to Theorem 2.7 there exists g ∈ A(Ω) such that g Tm < 12 εm+1 and (1−2−m−1 )h(z) < g (λz) + ωm (λz)| < h(z) for z ∈ ∂Ω. There exists δ ∈ (0, 1) so that max|λ|=1 | (1 − 2−m−1 )h(z) < (1 − εm+1 )Hm (z)

max |( g + ωm )(δλz)|,

|λ|=1

m (z) − (1 − δ)|ω m (z)| < δH

for z ∈ ∂Ω. Now due to [3, Lemma 3.7] there exists f ∈ A(Ω) and Vm an open subset of ∂Ω such that • f < 12 εm+1 , Tm

• • • •

max|λ|=1 |f(δλz)| < 2−m−1 h(z) for z ∈ ∂Ω, g + ωm | < h − | g + ωm | on ∂Ω, |f + g + ωm | − | − | |f + g + ωm | − | g + ωm | > δ(h g + ωm |) on Vm , σ(V m ) = σ(Vm ) > 1 − 2εm+1 .


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PIOTR KOT

We show that it is enough to choose ωm+1 = f+ g + ωm . In fact it suffices to prove the properties (4), (5), and (6). We may estimate > max |( g + ωm )(δλz)| − 2−m−1 h(z) h(z) > max |ωm+1 (λz)| ≥ max |ωm+1 (δλz)| |λ|=1

|λ|=1

−m−1

≥ (1 − 2

|λ|=1

−m−1

)h(z) − 2

h(z) > (1 − 2−m )h(z)

for z ∈ ∂Ω, which gives (4). We now prove (5): |ωm+1 | − |ωm | < h − |ωm | = Hm + |ωm | − |ωm | = Hm . To prove (6) we may estimate on Vm in the following way: − | |ωm+1 | − |ωm | ≥ |ωm+1 | − | g + ωm | + | g + ωm | − |ωm | > δ(h g + ωm |) +| g + ωm | − |ωm | ≥ δh − |ωm | m − (1 − δ)|ω m | > (1 − εm+1 )Hm . ≥ δH We can start to construct a bounded holomorphic function. Property (2) implies that {ωm }m∈N is a Cauchy sequence, so there exists a function F ∈ O(Ω) such that ωm → F uniformly on the compact subsets of Ω. We prove the maximum property of F . Since |ωm+1 | < Hm + |ωm | = h we can conclude |F | ≤ h on ∂Ω. Moreover using the properties (2), (3), and (4) we may conclude that ∞ max |F ((1 − εm+1 )λz)| ≥ max |ωm ((1 − εm+1 )λz)| − ωk − ωk+1 Tm |λ|