Modulus of continuity on parts of the boundary and solid modulus of continuity

Modulus of continuity on parts of the boundary and solid modulus of continuity

Salla Franz´ en

U.U.D.M. Project Report 2002:P7

Examensarbete i matematik, 20 po¨ang Handledare och examinator: Burglind Juhl-J¨oricke

Augusti 2002

Department of Mathematics

Our starting point is a theorem of Hardy and Littlewood, which asserts that an analytic function in the open unit disc D satisfies a H¨older condition of order α ∈ (0, 1), if its boundary values on the unit circle ∂D have this property. More precisely, denote by A(D) the algebra of analytic functions in the unit disc which are continuous in its closure D . The following theorem holds:

Theorem 1 Let f ∈ A(D). Suppose for a number α ∈ (0, 1] and a positive constant C

|f (w) − f (z)| ≤ C|w − z|^α for all w, z ∈ ∂D.

Then

|f (w) − f (z)| ≤ C|w − z|^α for all w, z ∈ D.

Here we stated the theorem also for α = 1. This case was proved later. For completeness we will give here a proof which follows [R-S-T] and works for α ∈ (0, 1]. We start with the following lemma which works for arbitrary domains G in Cⁿ. Denote by A(G) the algebra of analytic functions in G which are continuous in G.

Lemma 2 Let G be a domain in Cⁿ and let f ∈ A(G). Then for every δ > 0 the following supremum

sup{|f (w) − f (z)| : w, z ∈ G, |w − z| < δ}

is attained when one of the points w or z is contained in the boundary.

Proof of Lemma 2 Denote w = z + h. Both points z and z + h are contained in G iff z ∈ G^T(G − h). Fix the complex number h and apply the maximum principle to the analytic function f (z + h) − f (z) in G^T(G − h).

Since the boundary of the set G^T(G − h) is contained in the union of the boundaries ∂G and ∂(G − h), sup{|f (z + h) − f (z)| : z ∈ G^T(G − h)} is attained if either z ∈ ∂G or z ∈ ∂(G − h), i.e. w = z + h ∈ ∂G.

Proof of Theorem 1 By Lemma 2 we have to prove that for w ∈ ∂D and z ∈ D, |w − z| < δ, the inequality

|f (w) − f (z)| ≤ C|w − z|^α

holds. We may assume that z is in the open unit disc. Consider a holomor- phic branch of φ_w(z) = _(w−z)¹ α, z ∈ D. This function is in the Hardy space H^p for each p ∈ (0,_α¹]. Indeed, if w = e^iψ∈ ∂D and z = re^iϕ∈ D, then

1

|w − z|^pα = 1

|e^iψ− re^iϕ|^pα ≤ 2

|e^iψ− e^iϕ|^pα

and for a fixed ψ the latter function defines an integrable function on ∂D.

It follows that the function gw(z) = (f (w) − f (z))φw(z) is in H^p and is continuous on D\{w}. Since f is H¨older continuous on ∂D, |gw(z)| ≤ C for all z ∈ ∂D\{w}. Hence |g_w(z)| ≤ C on D\{w}. This is exactly the inequality we wanted to prove.

For our purposes we need the corresponding statement for arbitrary sim- ply connected (maybe unbounded) domains in C (see, e.g. [R-S-T]).

Theorem 3 Let G ⊂ C be a simply connected domain and α ∈ (0, 1].

Suppose f ∈ A(G) and for some number C

|f (w) − f (z)| ≤ C|w − z|^α for all w,z ∈ ∂G. (1) Then

|f (w) − f (z)| ≤ AC|w − z|^α for all w,z ∈ G. (2) Here A is a constant which does not depend on G or on f . For a proof we refer to [R-S-T].

Theorem 3 can be proven for a very general class of domains in Cⁿ, n > 1. However, one can expect, that for some domains one can replace the boundary by the Shilov boundary of the domain. The Shilov boundary of a domain G ⊂ Cⁿ is the smallest closed subset S(G) of G such that the maximum principle

|f (z)| ≤ max

S(G)

|f | for all z ∈ G and all f ∈ A(G) holds. Such a set always exists.

In [J¨o] a class of domains G is described for which (1) with ∂G replaced by the Shilov boundary S(G) of G implies (2). On the other hand examples are given of domains G for which this is not the case. It is not known if, e.g., for bounded pseudoconvex domains of holomorphy, the inequality (1) for ∂G replaced by an arbitrary open subset of ∂G which contains the Shilov boundary implies (2).

Here we will consider bounded complete Reinhardt domains in C² which are pseudoconvex and have C^∞boundary. Recall that a complete Reinhardt domain G ⊂ Cⁿ is a domain for which z = (z₁, ..., z_n) ∈ G implies that (z1ζ1, ..., znζn) ∈ G for arbitrary ζj ∈ D, j = 1, ..., n. A domain G has C^∞ boundary, if there is a C^∞ function ρ in Cⁿ (called defining function of G) such that ρ < 0 on G, ρ > 0 on Cⁿ\G, ρ = 0 on ∂G and dρ 6= 0 on ∂G.

Such a domain is pseudoconvex, if for each p ∈ ∂G the Levi form Lpρ of ρ at the point p is nonnegative in complex tangent directions:

Lpρ(w) =

n

X

j,k=1

( ∂²

∂z_j∂z_kρ)(p)wjw_k ≥ 0

for all w ∈ Cⁿ such that ^Pn_j=1∂z^∂ρ

j(p)w_j = 0.

Note that a domain G with C^∞ boundary satisfies this condition iff it is a domain of holomorphy, i.e. iff there exists an analytic function in G which does not extend analytically across the boundary. For a domain G ⊂ Cⁿ with C² boundary a point p ∈ ∂G is called a strictly pseudoconvex boundary point if the Levi form Lpρ(w) > 0 for all w ∈ Cⁿ\{0} such that Pn

j=1

∂ρ

∂zj(p)wj = 0. The conditions do not depend on the choice of the defining function ρ.

The following theorem of Basener and Pflug describes the Shilov boundary for smoothly bounded pseudoconvex domains.

Theorem 4 [Ba], [Pf ]. Let G be a bounded pseudoconvex domain in Cⁿ with C^∞ boundary. Then the Shilov boundary S(G) is equal to the closure of the set of strictly pseudoconvex boundary points of G.

To understand the properties of a complete Reinhardt domain D ⊂ C² it is useful to associate with D the domain D^∗ in R² defined in the following way: D^∗= {(ξ₁, ξ₂) ∈ R² : (e^ξ1, e^ξ2) ∈ D}. By the definition of a complete Reinhardt domain, D is the union of all bidiscs of the form

{(z₁, z₂) ∈ C² : |z₁| < e^ξ1, |z₂| < e^ξ2} for some (ξ₁, ξ₂) ∈ D^∗.

A complete Reinhardt domain D in C² with C^∞ boundary is pseudoconvex iff D^∗ is convex. A point z = (z₁, z₂) ∈ ∂D is a strictly pseudoconvex boundary point iff (ξ1, ξ2) = (log|z1|, log|z₂|) is a strictly convex boundary point of D^∗.

Let now D be a bounded complete pseudoconvex Reinhardt domain in C² with C^∞boundary. Denote by P the set of all strictly pseudoconvex bound- ary points of D^∗. Then

∂D^∗\P =^[

k

ω_k^∗

for an at most countable number of open line segments ω^∗_k in R². Moreover

∂D\S(D) =^[

k

ω_k,

where ω_k is the connected relatively open subset of ∂D which corresponds to ω_k^∗. If ω_k^∗ is bounded this correspondence has the following form:

ωk= {(z1, z2) ∈ C² : log|z1| = ξ₁, log|z2| = ξ₂, (ξ1, ξ2) ∈ ω^∗_k} (3) Note that in this case

∂ω_k= {(z₁, z₂) ∈ C² : (ξ₁, ξ₂) = (log|z₁|, log|z₂) ∈ ∂ω^∗_k}, (4)

and this set is contained in the Shilov boundary. We will consider here only the case when the ω_k^∗ are uniformly bounded. This is equivalent to the following condition:

For z = (z₁, z₂) ∈ ∂D\S(D), c_j ≤ |z_j| ≤ C_j (5) for j=1,2 and some constants c_j, C_j, 0 < c_j < C_j.

The main goal of this work is to prove the following theorem.

Theorem 5 Let D be a bounded complete pseudoconvex Reinhardt domain in C² with smooth boundary for which condition (5) holds. Let α ∈ (0, 1].

Then there is a constant A depending only on D such that for any function f ∈ A(D) the inequality

|f (w) − f (z)| ≤ C|w − z|^α for all w, z ∈ S(D) (6) implies

|f (w) − f (z)| ≤ CA|w − z|^α for all w, z ∈ D. (7) We think that condition (5) can be removed.

Before proving theorem 5 we state and prove the version of it for the bidisc, which will be used later. Let

Dⁿ= {z ∈ Cⁿ: |zj| < r_j, j = 1, ..., n}

be a polydisc. The Shilov boundary of Dⁿ is the direct product of the boundaries of the onedimensional discs {|z_j| = r_j}, i.e.

S(Dⁿ) = {z ∈ Cⁿ: |z_j| = r_j, j = 1, ..., n}.

Proposition 6 Let Dⁿ be an arbitrary polydisc in Cⁿ and f ∈ A(Dⁿ).

Suppose

|f (w) − f (z)| ≤ C|w − z|^α if w, z ∈ S(Dⁿ) Then

|f (w) − f (z)| ≤ Cγ|w − z|^α for all w, z ∈ Dⁿ. γ is a constant not depending on f nor on the radii r_j.

Proof of Proposition 6 Write z = (z₁, ..., z_n) ∈ Cⁿ, w = (w₁, ..., w_n) ∈ Cⁿ. Denote z⁰ = (z₂, ..., z_n), D⁰ = {z⁰ : |z₂| < r₂, ..., |z_n| < r_n}. The function

f^z0(ζ) = f (ζ, z⁰), |ζ| < r₁,

is analytic for any fixed z⁰ with |z2| ≤ r₂, ..., |zn| ≤ r_n. Indeed, the function is the uniform limit of the functions

f^rz0(ζ) = f (ζ, rz⁰), r ∈ (0, 1), r → 1.

Here rz⁰ = (rz₂, ..., rz_n). Analyticity of the function f^rz0 follows from the fact that the point (ζ, rz⁰) ∈ D for r ∈ (0, 1), |ζ| < r1 and, moreover, the function f^z0 ∈ A(|ζ| < r₁) for each z⁰ ∈ D⁰. By the same argument for each z₁, w₁, |z₁| ≤ r₁, |w₁| ≤ r₁, the function z⁰ → f (w₁, z⁰) − f (z₁, z⁰) is in A(D⁰). Hence, the maximum of this function is obtained for z⁰ ∈ S(D⁰).

Now theorem 1 gives

|f (w₁, z⁰) − f (z₁, z⁰)| ≤ C|w₁− z₁|^α. Applying the same argument to the function

ζ → f (w₁, ..., w_j−1, ζ, z_j+1, ..., z_n), j = 2, ..., n, we get the assertion.

Proof of Theorem 5 The first step of the proof is the following Lemma.

Lemma 7 (6) implies

|f (w) − f (z)| ≤ Cb|w − z|^α for w, z ∈ ∂D (8) for a constant b which depends only on D.

The second step of the proof consists of the following Lemma.

Lemma 8 (8) implies (7).

Lemma 8 is known for a large class of domains. For completeness in the end of the paper we give a short argument which works in our situation.

Lemma 9 For each k,

|f (w) − f (z)| ≤ CB|w − z|^α for z, w ∈ ωk for a constant B depending only on D.

Lemma 9 implies Lemma 7. This is a consequence of the fact that D has smooth boundary. We will not give the arguments here.

Proof of Lemma 9 The set ω_k^∗ is an open line segment in R², say

ω_k^∗= {(ξ₁, ξ₂) ∈ R² : ξ₂ = a_kξ₁+ b_k, ξ₁ ∈ (σ₁, σ₂)} (9) (or with the roles of ξ₁ and ξ₂ interchanged). Here a_k, b_k, σ₁ and σ₂ are real numbers. We may assume that |a_k| ≤ 1.

Let w, z ∈ ω_k, |w − z| < δ for a small positive number δ. (10)

Using condition (5) we get

|1 − w_j zj

| < a⁰δ, j = 1, 2, (11)

where a⁰= max(_c¹

1,_c¹

2) depends only on D.

Write z in the form z = (z₁, z₂) = (e^log|z1|+iϕ1, e^log|z2|+iϕ2) = (e^ζ1, e^ζ2). Since z ∈ ω_k, we have Reζ₂ = a_kReζ₁ + b_k, see (3). Hence (after the branches of the arguments ϕ1 and ϕ2 are chosen) there is a uniquely defined β ∈ R, β = β(ζ), such that ζ₂ = a_kζ₁+ b_k+ iβ, namely β = ϕ₂− a_kϕ₁.

By (11) we can find (the branches of) arguments ϕ⁰₁ and ϕ⁰₂ of w₁, w₂ such that

w = (w₁, w₂) = (|w₁|e^iϕ01), |w₂|e^iϕ02)) = (e^ζ10, e^ζ20) and (12)

|ϕ⁰₁− ϕ₁| ≤ λ₁δ, |ϕ⁰₂− ϕ₂| ≤ λ₂δ (13) for constants λ1, λ2 depending only on D. As above ζ₂⁰ = akζ₁⁰ + bk+ iβ⁰, where β⁰ = ϕ⁰₂− a_kϕ⁰₁. Inequalities (10) and (13) give

|ζ − ζ⁰| ≤ c⁰δ (14)

|β − β⁰| ≤ c⁰⁰δ (15)

For β ∈ R consider the analytic function

φβ(ζ) = akζ + bk+ iβ, Reζ ∈ (σ1, σ2).

So, z = (e^ζ, e^φβ(ζ)) for some ζ, Reζ ∈ (σ1, σ2), (16) w = (e^ζ0, e^φβ0(ζ0)) for some ζ⁰, Reζ⁰ ∈ (σ₁, σ2),

with |ζ − ζ⁰| ≤ c⁰δ, |β − β⁰| ≤ c⁰⁰δ.

The following Lemmas hold.

Lemma 10 There is a constant M depending only on D such that for the function g_β(ζ) = f (e^ζ, e^φβ(ζ)), Reζ ∈ (σ₁, σ₂) the following inequality holds

|g_β(ζ) − g_β( ˜ζ)| ≤ M |ζ − ˜ζ| for all ζ, ˜ζ ∈ C, Reζ, Re ˜ζ ∈ (σ1, σ2).

Lemma 11 For all ζ ∈ C, Reζ ∈ (σ1, σ2),

|g_β(ζ) − gβ⁰(ζ)| ≤ mδ^α for a constant m depending only on D.

We postpone the proof of the Lemmas 10 and 11 and finish the proof of Lemma 9.

Proof of Lemma 9 continued. Write the points z and w as in (16), with

|ζ − ζ⁰| ≤ c⁰δ and |β − β⁰| ≤ c⁰⁰δ. By Lemma 10 we have

|f (e^ζ, e^φβ(ζ)) − f (e^ζ0, e^φβ(ζ0))| = |g_β(ζ) − g_β(ζ⁰)| ≤ M (c⁰)^αδ^α. By Lemma 11 we have

|f (e^ζ0, e^φβ(ζ0)) − f (e^ζ0, e^φβ0(ζ0))| ≤ mδ^α. The two inequalities prove Lemma 9.

Proof of Lemma 10 The function g_β(ζ), Reζ ∈ (σ1, σ2), is a bounded analytic function as the uniform limit of the analytic functions

g^r_β(ζ) = f (re^ζ, re^φβ(ζ)), Reζ ∈ (σ₁, σ₂). Here r ∈ (0, 1), r → 1.

By the same reason it is continuous in Reζ ∈ [σ1, σ2]. By Theorem 3 it is enough to prove that

|g_β(ζ) − g_β( ˜ζ)| ≤ M⁰|ζ − ˜ζ|^α,

if ζ, ˜ζ ∈ C and Reζ equals σ1 or σ2, Re ˜ζ equals σ1 or σ2. Here M⁰ depends only on D. This follows from (6) and the following two facts:

|(e^ζ, e^φβ(ζ)) − (e^ζ˜, e^{φβ( ˜ζ)})| ≤ ˜a|ζ − ˜ζ|

for Reζ, Re ˜ζ ∈ [σ1, σ2] and a constant ˜a not depending on β (this follows from the fact, that φ_β has uniformly bounded derivative for Reζ ∈ [σ₁, σ₂]).

Moreover, if Reζ equals σ1 or σ2, then (e^ζ, e^φβ(ζ)) ∈ S(D).

Proof of Lemma 11 If Reζ ∈ [σ₁, σ₂], then

|(e^ζ, e^φβ(ζ)) − (e^ζ, e^φβ0(ζ))| < b⁰|β − β⁰| ≤ b⁰c⁰⁰δ.

Since for Reζ equal to σ₁ or σ₂ both points (e^ζ, e^φβ(ζ)) and (e^ζ, e^φβ0(ζ)) are in S(D) the maximum principle applied to the analytic function g_β − g_β⁰, gives

|g_β(ζ) − g_β⁰(ζ)| = |f (e^ζ, e^φβ(ζ)) − f (e^ζ, e^φβ0(ζ))| ≤ C(b⁰c⁰⁰δ)^α = mδ^α. Proof of Lemma 8 Let δ > 0 and z = (z1, z2), w = (w1, w2) ∈ D such that |w − z| < δ. Say |z₂| ≤ |w₂|. Let r_z, r_w > 0 be such that (z₁, r_z), (w1, rw) ∈ ∂D. Then by Proposition 6, f is H¨older continuous of order α on the closure of the bidiscs Dz and Dw, where:

D_z = {ζ ∈ C : |ζ₁| < |z₁|, |ζ₂| < r_z}, D_w = {ζ ∈ C : |ζ₁| < |w₁|, |ζ₂| < r_w}.

We have two cases. If |z1| ≤ |w₁|, then (z₁, z2) ∈ Dw, so |f (w) − f (z)| ≤ C|w − z|^α. If |z1| ≥ |w₁| then by (18), (w₁, z2) ∈ DzT

Dw and we have

|f (w) − f (z)| ≤ |f (w) − f (w₁, z₂)| + |f (w₁, z₂) − f (z)|, from which (7) now follows.

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