SOME CHARACTERIZATIONS OF C(M)

CHRISTOPHER J. BISHOP

May 20, 1996

Abstract. We show that a function f on the unit disk extends continuously to M, the maximal ideal space of H^{∞}(𝔻) iff it is uniformly continuous (in the hyperbolic metric) and close to constant on the complementary components of some Carleson contour.

1991 Mathematics Subject Classification. 46J10. Key words and phrases. Harmonic functions, holomorphic functions, function algebras, maximal

ideal space, Carleson measures, uniform approximation.

The author is partially supported by NSF Grant DMS 92-04092 and an Alfred P. Sloan research fellowship.

CHARACTERIZATIONS OF C(M) 1

1. Introduction

Let H^{∞}(𝔻) denote the algebra of bounded holomorphic functions on the unit disk, 𝔻. Let M denote the maximal ideal space of H^{∞}(𝔻) (see [3] for definitions). Hoffman proved that every bounded harmonic function on the disk extends continuously to M and that such functions generate C(M) as a closed algebra (closed with respect to the sup norm). The purpose of this note is to give a more geometric characterization of the functions on 𝔻 which extend continuously to M. Recall that a positive measure μ on the disk is a Carleson measure if there is a C > 0 such that μ(D(a, r)) ≤ Cr for every a ∈ 𝕋 = ∂𝔻 and r > 0. A union of curves Γ ⊂ 𝔻 is called a Carleson contour if arclength measure on Γ is Carleson.

Theorem 1.1. For a bounded, continuous function g on 𝔻 the following are equivalent.

(1) g extends continuously to M.

(2) For every ϵ > 0 there is a smooth function φ on 𝔻 so that ∥g − φ∥_{∞} ≤ ϵ, sup_{z} |∇φ(z)|(1 − |z|^{2}) < ∞ and |∇φ|dxdy is a Carleson measure.

(3) g is uniformly continuous with respect to the hyperbolic metric on 𝔻 and for every ϵ > 0 there is a Carleson contour Γ so that g is within ϵ of a constant on each connected component of 𝔻\Γ.

Recall that a Douglas algebra is any closed algebra H^{∞}(𝕋) ⊂ A ⊂ L^{∞}(𝕋). Our first application of Theorem 1.1 is

Corollary 1.2. Suppose f is bounded and continuous on the disk and that H^{∞}(𝕋) ⊆ A ⊆ L^{∞}(𝕋) is a Douglas algebra. Let M_{A} ⊂ M denote the maximal ideal space of A. Then the following are equivalent.

2 CHRISTOPHER J. BISHOP

(1) f extends continuously to M_{A}.

(2) For every ϵ > 0 there is an inner function I, such that Ī ∈ A, and u ∈ C(M) on 𝔻 so that and |(f(z) − u(z))I(z)| < ϵ on 𝔻.

(3) For every ϵ > 0 there is an inner function I, Ī ∈ A and a Carleson contour Γ so that f differs from a constant by at most ϵ on every component 𝔻\Γ which hits {|I| > 1/2}.

At one extreme A = H^{∞} and the only allowable I’s are constants. In this case, Corollary 1.2 is just Theorem 1.1. The other extreme occurs for A = L^{∞}. Axler and Shields [1] showed that any function extending to the maximal ideal space X of L^{∞} (X is also the Shilov boundary of H^{∞}) has nontangential limits almost everywhere and asked if the converse is true. This was proven independently by O.V. Ivanov [5]. Corollary 1.3. A bounded, continuous function f on 𝔻 has a continuous extension to X (the Shilov boundary of H^{∞}) iff f has nontangential limits a.e. on 𝕋.

I am grateful to Allen Shields for mentioning the problem on nontangential limits to me. This problem led me to think about the others. I also thank Sheldon Axler for bringing Ivanov’s result to my attention and the referee for many helpful comments.

2. Proof of Theorem 1.1

We shall show (1) ⇒ (2) ⇒ (3) ⇒ (1). The first implication is known, the second is a minor variation of a known technique, but the third seems to be new and depends on a recent result from [2].

Proof of (1) ⇒ (2): Let A denote the set of smooth, bounded functions on the disk described by condition (2). A is an algebra and by [3], Theorem VIII.6.1 every bounded harmonic function can be uniformly approximated by functions in A. By

CHARACTERIZATIONS OF C(M) 3

Hoffman’s result ([4], Lemma 4.4) bounded harmonic functions generate C(M), so the sup norm closure of A contains C(M).

Proof of (2) ⇒ (3): This is a standard “stopping time argument”, but since I do not know a reference for this precise application, I will sketch the proof.

Suppose g satisfies condition (2) of Theorem 1.1. First we show g is uniformly continuous with respect to the hyperbolic metric, ρ. Given ϵ > 0 suppose ∥g −φ∥_{∞} < ϵ/4 and φ satisfies the conditions in (2). The estimate |∇φ(z)| ≤ C(1 − |z|) implies φ is uniformly continuous (in fact, Lipschitz) with respect to the ρ, so there is a δ = δ(ϵ) so that ρ(z, w) ≤ δ implies |φ(z) − φ(w)| ≤ ϵ/2. Thus

|g(z) − g(w)| ≤ |g(z) − φ(z)| + |φ(z) − φ(w)| + |φ(w) − g(w)| ≤ ϵ,

so g is uniformly continuous with respect to ρ.

Next we will construct a Carleson contour Γ so that φ is within ϵ of a constant on each component of 𝔻 \ Γ (and therefore g is within 2ϵ of a constant on each component). Given ϵ > 0, choose δ > 0 so that φ varies by less than ϵ/8 over any hyperbolic ball of diameter 10δ. It will also be convenient to take δ = 2^{−N}, a power of 2. To simplify the construction we move to the upper half plane. Clearly it is enough to consider what happens in one square, say Q_{0} = [0, 1] × [0, 1]. Now divide Q_{0} into rectangles of the form

Q = {z = x + iy : j2^{−k} < x < (j + 1)2^{−k}, 2^{−k−1} < y < 2^{−k}}.

Subdivide each such Q into 2δ^{−2} equal subsquares of side length δ2^{−k−1}. Each of these will then have hyperbolic diameter less than 10δ, so φ varies by at most ϵ/8 on each such square.

For each square S let z_{S} denote the center of the square. Each square S in the

Figure 2.1. Stopping time regions in the proof of (2) ⇒ (3)

partition has either 1 or 2 squares adjacent and underneath it; let G_{1}(S) denote these squares. In general, let

G_{n}(S) = ⋃{S′′ ∈ G_{1}(S′) : S′ ∈ G_{n−1}(Q) and |φ(z_{S}) − φ(z_{S′′})| ≤ ϵ/2}.

Let D_{S} = S ∪ ∪_{n=1}^{∞} ∪_{S′∈G_{n}(S)} S′.

Choose a square S on the “top row” of Q_{0} and let D_{1} = D_{S}. In general, let S be a square of maximal size not contained in ∪_{j=1}^{n}D_{j} and let D_{n+1} = D(S). This gives a countable number of disjoint (except for boundaries) regions which exhaust Q_{0}. For each j let I_{j} be the vertical projection of D_{j} onto ℝ, F_{j} = ∂D_{j} ∩ ℝ ⊂ I_{j} and let z_{j} denote the center of the top edge of D_{j}. Let Γ = ∪_{j}∂D_{j}.

Note that for each such region D = D_{j} with “base” I = I_{j}, ℓ(∂D) ≤ 4(1 + δ^{−1})|I|, where ℓ denotes arclength. To see this, note that we can write the region D as D = R \ ∪_{j}R_{j} where R is a |I| × 2δ^{−1}|I| vertical rectangle and the R_{j} are similar

CHARACTERIZATIONS OF C(M) 5

rectangles with disjoint bases in I. Thus

ℓ(∂D) ≤ ℓ(∂R_{0}) + ∑_{j} ℓ(∂R_{j}) ≤ 4(1 + δ^{−1})|I|, (2.1)

as desired.

By construction, |φ − φ(z_{j})| ≤ ϵ on D_{j}, so we need only check that arclength on Γ is a Carleson measure. We start by showing

length(∂D_{j}) ≤ C_{1}|F_{j}| + C_{2} ∫∫_{D_{j}} |∇φ|dxdy, (2.2)

This is proved by considering two cases. First suppose |F_{j}| ≥ |I_{j}|/2. Then by (2.1) equation (2.2) holds with C_{1} = 8(1 + δ^{−1}). On the other hand, if |F_{j}| < |I_{j}|/2, then we can find a set E of length ≥ ℓ(I_{j})/2 along the top edge of D_{j} so that if we follow any vertical line down from E it eventually enters a different D_{k}. Consider the segment formed by intersecting the line with D_{j}. By the definition of D_{j} (and the fact that φ varies by less than ϵ/8 over each square S) the values of φ at the endpoints of this segment differ by at least ϵ/4. Thus the integral of |∇φ| along this segment is at least ϵ/4. Hence,

∫∫_{D_{j}} |∇φ|dxdy ≥ \frac{ϵ}{4}|E| ≥ \frac{ϵ}{8}|I_{j}|.

Thus (2.2) holds with C_{2} = 8/ϵ.

Finally we want to deduce arclength on Γ is a Carleson measure. Consider a dyadic Carleson square

Q = {z = x + iy : j2^{−k} < x < (j + 1)2^{−k}, 0 < y < 2^{−k}},

with base I = [j2^{−k}, (j + 1)2^{−k}]. It is sufficient to show that ℓ(Γ ∩ Q) ≤ C|I|, with some absolute C independent of Q.

6 CHRISTOPHER J. BISHOP

First note that there are at most 2δ^{−1} regions D which hit Q but are not contained in Q (because any such must contain a square S from the “top row” of Q and there are only 2δ^{−1} such squares). For each such region, the proof of (2.1) shows ℓ(∂D ∩ Q) ≤ 4(1 + δ^{−1})|I|. On the other hand, if D_{j} ⊂ Q then F_{j} ⊂ I, so

∑_{j:D_{j}⊂Q} |F_{j}| ≤ |I|.

Furthermore,

∑_{j:D_{j}⊂Q} ∫∫_{D_{j}} |∇φ(z)|dxdy ≤ ∫∫_{Q} |∇φ(z)|dxdy ≤ C|I|,

by assumption. Thus

ℓ(Γ ∩ Q) ≤ ∑_{j:D_{j}∩Q≠∅,D_{j}⊄Q} ℓ(∂D_{j} ∩ Q) + ∑_{j:D_{j}⊂Q} ℓ(∂D_{j})
≤ 8δ^{−1}(1 + δ^{−1})|I| + ∑_{j:D_{j}⊂Q}(C_{1}|F_{j}| + C_{2} ∫∫_{D_{j}} |∇φ|dxdy)
≤ C(δ, ϵ)|I|,

as desired.

Proof of (3) ⇒ (1): We start with some definitions. For f on the disk, a ∈ ℂ and δ > 0, let Ω_{f}(a, δ) = f^{−1}(D(a, δ)) = {z : |f(z) − a| < δ}. For convenience, given an open set Ω ⊂ 𝔻 we will write

dist(g, H^{∞}(Ω)) = inf_{G∈H^{∞}(Ω)} ∥g|_{Ω} − G∥_{L^{∞}(Ω,dxdy)}.

If Ω is empty, we interpret the distance as 0. Let H^{∞}(𝔻)[f] denote the smallest algebra containing H^{∞}(𝔻) ∪ f which is closed with respect to the sup norm on 𝔻. If f is bounded and harmonic, then it is clear that H^{∞}(𝔻)[f] ⊂ C(M).

The following is Theorem 1.1 of [2].

CHARACTERIZATIONS OF C(M) 7

Proposition 2.1. Suppose f̄ ∈ H^{∞}(𝔻) and g ∈ L^{∞}(𝔻). Then

dist(g, H^{∞}(𝔻)[f]) = inf_{δ>0} sup_{a∈ℂ}dist(g, H^{∞}(Ω_{f}(a, δ))).

(Distance is in the L^{∞} norm on 𝔻.)

We will now use this to prove (3) ⇒ (1). Suppose g satisfies condition (3) in Theorem 1.1. Fix ϵ > 0 and let Γ be the Carleson contour given by part (3) of the theorem. Choose a sequence {z_{n}} ⊂ Γ so that ρ(z_{n}, z_{m}) ≥ 1/4 for all n ≠ m and so that every z ∈ Γ is within hyperbolic distance 1/2 of one of the z_{n}’s. Then {z_{n}} is an interpolating sequence because it is hyperbolically separated and satisfies

∑_{z_{n}∈D(x,r)}(1 − |z_{n}|) ≤ C ∑_{z_{n}∈D(x,r)} ℓ(Γ ∩ D(z_{n}, \frac{1}{100}(1 − |z_{n}|)))
≤ Cℓ(Γ ∩ D(x, 2r))
≤ Cr.

for every disk D(x, r) centered on ∂𝔻. (See [3], Chapter VII). Let

B(z) = ∏_{n=1}^{∞} \frac{z_{n} − z}{1 − z̄_{n}z} \frac{|z_{n}|}{z_{n}},

be the Blaschke product with zeros {z_{n}}. Note that Γ ⊂ {z : |B(z)| < 1/2}. By Exercise VII.1 of [3], we also have |B′(z_{n})|(1 − |z_{n}|^{2}) > C_{0} for all n and some C_{0} > 0 (depending only on the Carleson norm of arclength on Γ).

We claim that for any δ there is a ν(δ) > 0 such that any component of Ω_{B}(a, δ) which hits Γ has hyperbolic diameter at most ν(δ). Suppose not. Then there is a ν_{0} > 0 so that for any δ > 0 we can choose a ∈ ℂ and a component Ω of Ω_{B}(a, δ) such that

(1) Ω ∩ Γ ≠ ∅.

(2) The hyperbolic diameter of Ω is ≥ ν_{0}.

8 CHRISTOPHER J. BISHOP

Choose a sequence δ_{j} → 0 and let {Ω_{j}} be the corresponding components. Each such Ω_{j} is within hyperbolic distance 1/2 of some point z_{j} in our sequence. Renormalize by Möbius transformations so that z_{j} = 0 for all j. We now have a sequence of uniformly bounded functions {B_{j}} on 𝔻 such that |B′_{j}(0)| > C_{0} and B_{j} is within δ_{j} of a constant on a connected set of hyperbolic diameter ≥ ν_{0} which intersects the disk D(0, 1/2). By normal families there is subsequence which converges to a constant function uniformly on compacta, but this contradicts the condition |B′_{j}(0)| > C_{0}. This proves the claim.

We can now apply the proposition. Since g is uniformly continuous with respect to the hyperbolic metric, there is a η = η(g, ϵ) so that g is within ϵ of a constant on any set of hyperbolic diameter ≤ η. Choose δ so small that ν(δ) ≤ η(g, ϵ). Suppose a ∈ ℂ and Ω is a connected component of Ω_{B}(a, δ). If Ω has hyperbolic diameter ≤ η, then g is within ϵ of a constant on Ω by our choice of η. If Ω has hyperbolic diameter ≥ η, then Ω cannot meet Γ, so it lies completely within a component of 𝔻 \ Γ. Thus g differs by less than ϵ from a constant by hypothesis. Since a piecewise constant function is holomorphic on Ω_{f}(a, δ) we have

dist(g, H^{∞}(Ω_{f}(a, δ))) ≤ ϵ,

for all a. Hence by the proposition,

dist(g, C(M)) ≤ dist(g, H^{∞}(𝔻)[f]) ≤ ϵ.

Since g can be uniformly approximated by elements of C(M), we have g ∈ C(M). This completes the proof of Theorem 1.1.

Remark: The proof of the final implication shows that any function g ∈ C(M) is contained in an closed subalgebra generated by H^{∞}(𝔻) and a countable collection

CHARACTERIZATIONS OF C(M) 9

of anti-holomorphic functions. However, I don’t know if g must be in a subalgebra generated by H^{∞}(𝔻) and a single anti-holomorphic function (even if g itself is harmonic).

3. Proof of the corollaries

Proof of Corollary 1.2: Given Theorem 1.1, Corollary 1.2 is very easy. To prove (1) ⇒ (2) apply the Tietze extension theorem (which we can do since M is a compact Hausdorff space, hence normal) to f restricted to M_{A} to get u ∈ C(M) with u = f on M_{A}. By [3], Theorem IX.1.3 and Marshall’s theorem ([3], Theorem IX.3.2)

M_{A} = ∩_{I:Ī∈A}{m ∈ M : |I(m)| = 1},

which implies sets of the form {|I(m)| > 1 − ϵ} with ϵ > 0, Ī ∈ A form a base of neighborhoods of M_{A}. Since f − u vanishes on M_{A} we can find an ϵ and I so that |f −u| ≤ ϵ on {|I| > 1−ϵ}. Replacing I by a high power of itself gives (2). Conversely,

(2) ⇒ (1) because by taking ϵ → 0 we can obtain a sequence of continuous functions on M (and hence continuous on M_{A}) which converge uniformly to f on M_{A}.

Next we prove (2) ⇒ (3). By (2) (applied with ϵ^{2}) we can find u and I so that |f − u| < ϵ on {|I| > ϵ}. Take the contour corresponding to g = u and ϵ in part (3) of Theorem 1.1. We also wish to insure that any component that intersects {|I| > 1/2} is contained in {|I| > ϵ} as long as ϵ is less than some absolute constant ϵ_{0}. Carleson’s construction (e.g., proof of [3], Theorem VIII.4.1) says there is an ϵ_{0} > 0 and a Carleson contour Γ which separates {|I| = 1/2} from {|I| = ϵ_{0}}. By adding this contour to the one for u we get the desired property. Thus we have |f − u| ≤ ϵ on any component which hits {|I| > 1/2}, as required. Conversely, (3) ⇒ (2) holds because given the piecewise constant approximation we extend it to the

10 CHRISTOPHER J. BISHOP

whole disk by setting it equal to zero outside the contours which hit {|I| > 1/2} and smoothing it (see e.g., [3], page 357). The resulting function u satisfies (2) if we replace I by a high power of itself. This proves Corollary 1.2.

Proof of Corollary 1.3: If f extends continuously to X = M_{L^{∞}(𝕋)} then by part (2) of Corollary 1.2 for every ϵ > 0 we can find some u ∈ C(M) and some inner function I so that |f − u| < 2ϵ on {|I| > 1/2}. Every function in C(M) has nontangential limits almost everywhere (they can all be uniformly approximated by products and sums of bounded harmonic functions) and for any fixed θ < π the set {|I| > 1/2} contains cones with vertex angle θ at almost every point of the circle. Thus the nontangential oscillation of f in cones of angle θ is less than ϵ almost everywhere. Taking ϵ → 0 and θ → π shows f has nontangential limits almost everywhere.

Conversely, suppose f has nontangential limits almost everywhere. By subtracting the harmonic function v with the same limits we may assume f has limit 0 almost everywhere. Since v extends continuously to M (hence to X), this does not alter the problem. Suppose ϵ > 0 and fix a cone angle, say θ = π/2 and for each n = 1, 2, … let K_{n} be the set of points x on the circle such that |f| < ϵ in the cone W (x, \frac{π}{2}, n) with vertex x, angle \frac{π}{2} and diameter 2^{−n}. Note that |K_{n}| → 2π as n → ∞. Choose compact sets E_{n} ⊂ K_{n} so that |E_{n}| ≥ |K_{n}| − \frac{1}{n}.

Next, choose n_{0} so that |E_{n_{0}}| > 1/2 and let W_{0} = ∪_{x∈E_{n_{0}}}W (x, \frac{π}{2}, n_{0}). For each complementary interval J of E_{n_{0}} we can choose a n = n_{J} such that |F_{J}| = |E_{n} ∩ J| > \frac{1}{2}|J| and 2^{−n} ≤ |J|. Define W_{J} = ∪_{x∈F_{J}} W (x, \frac{π}{2}, n_{J}). Then W is a sawtooth domain with boundary length at most 6|J|.

Continuing in this way we obtain by induction a collection of intervals {J_{k}}, sets

CHARACTERIZATIONS OF C(M) 11

F_{k} ⊂ J_{k} and sawtooth regions W_{k} such that (1) |F_{k}| ≥ \frac{1}{2}|J_{k}|.

(2) The {F_{k}} are pairwise disjoint.

(3) ∂W_{k} ∩ ∂𝔻 = F_{k}.

(4) | ∪_{k} F_{k}| = |∂𝔻|.

(5) ℓ(∂W_{k}) ≤ C|J_{k}| ≤ 2C|F_{k}|. In particular, ∑_{k} ℓ(∂W_{k}) < ∞.

Define a sequence {z_{n}} in the disk by placing points along each arc in each ∂W_{k} at hyperbolic distance ϵ from each other. The sequence is Blaschke, i.e., satisfies ∑_{n}(1 − |z_{n}|) < ∞ because the sum along any arc of ∂W_{k} is easily dominated by a multiple of the length of that arc, and we observed above that ∑ ℓ(∂W_{k}) < ∞. Now let I(z) be the Blaschke product corresponding to the sequence {z_{n}}. Then |I| ≤ ϵ on each ∂W_{k}’s (because we are never more than distance ϵ from a zero) and is therefore less than ϵ outside the union of the W_{k} (because any component of 𝔻\ ∪_{k} \overline{W_{k}} hits the unit circle in zero measure). Therefore f differs from zero by less than ϵ on every component of {|I| > ϵ}. Taking u to be the zero function and applying part (2) of Corollary 1.2 completes the proof (since A = L^{∞} we automatically get Ī ∈ A).

References

1. S. Axler and A. Shields Extensions of harmonic and analytic functions, Pacific J. Math. 145:1–15, 1990.

2. C.J. Bishop A distance formula for algebras on the disk, preprint.

3. J.B. Garnett Bounded analytic functions, Academic Press 1981.

4. K. Hoffman Bounded analytic functions and Gleason parts, Ann. Math. 86:74–111, 1967.

5. O.V. Ivanov Nontangential limits and Shilov boundary of the algebra H^{∞}, Dokl. Akad. Nauk. Ukraine SSR Ser. A 5–8, 1991. C.J. Bishop, Math. Dept., SUNY at Stony Brook, Stony Brook, NY 11794-3651 E-mail address: bishop@math.sunysb.edu