ORTHOGONAL FUNCTIONS IN H^{∞}

CHRISTOPHER J. BISHOP

Abstract. We construct examples of H^{∞} functions f on the unit disk, so that the push forward of Lebesgue measure on the circle is a radially symmetric measure μ_{f} in the plane and characterize which symmetric measures can occur in this way. Such functions have the property that {f^{n}} is orthogonal in H^{2}, and provide counterexamples to a conjecture of W. Rudin, originally disproved by Carl Sundberg. Among the consequences is that there is f in the unit ball of H^{∞} so that the corresponding composition operator maps the Bergman space isometrically into a closed subspace of the Hardy space.

1This is a preliminary version. Please do not distribute copies without author’s consent Date: August 20, 1998. 1991 Mathematics Subject Classification. Primary: 30H05 Secondary: 30D35, 30D55, 47B38 . The author is partially supported by NSF Grant DMS 98-00924.

ORTHOGONAL FUNCTIONS IN H^{∞} 1

1. Introduction

Let H^{∞} denote the algebra of bounded holomorphic functions on the unit disk, 𝔻, let U be the closed unit ball of H^{∞} and let U_{0} = {f ∈ U : f(0) = 0}. If f ∈ H^{∞} then it has radial boundary values (which we also call f) almost everywhere on the unit circle, 𝕋. We say that f is orthogonal if the sequence of powers, {f^{n}}, n = 0, 1, … is orthogonal, i.e., if

∫_{𝕋} f^{n}f̄^{m}dθ = 0,

whenever n ≠ m. In this paper we will characterize orthogonal functions in H^{∞} in terms of the Borel probability measure μ_{f}(E) = |f^{−1}(E)|, where | · | denotes Lebesgue measure on 𝕋, normalized to have mass 1. Moreover, we will determine exactly which measures arise in this way. More precisely,

Theorem 1.1. If f ∈ U_{0} then {f^{n}}, n = 0, 1, 2, … is an orthogonal sequence iff μ_{f} is a radially symmetric probability measure supported in the closed unit disk and satisfies

∫_{|z|≤1} log \frac{1}{|z|}dμ_{f}(z) < ∞. (1.1)

Moreover, given any measure μ satisfying these conditions there is a f ∈ U_{0} such that μ = μ_{f}.

The result is motivated by the observation that if f is an inner function (i.e., f ∈ H^{∞} and |f| = 1 a.e. on 𝕋) with f(0) = 0 then μ_{f} is normalized Lebesgue measure on 𝕋 (Lemma 2.3) and f is orthogonal since (if m > n),

∫_{𝕋} f^{n}f̄^{m}dθ = ∫_{𝕋} f^{n−m}dθ = 2πf^{n−m}(0) = 0.

At a 1988 MSRI conference Walter Rudin asked if the converse is true, i.e., are multiplies of inner functions the only orthogonal bounded holomorphic functions

2 CHRISTOPHER J. BISHOP

on the disk? In other words, is normalized Lebesgue measure on the circle the only radially symmetric measure which can occur as a μ_{f}? Our characterization shows that many other symmetric measures can occur and hence provide counterexamples to Rudin’s “orthogonality conjecture”. The conjecture was originally and independently disproved by Carl Sundberg.

The simplest example of a measure satisfying Theorem 1.1 (other than Lebesgue measure on a circle) is to take μ to be Lebesgue measure on the union of two circles \{z : |z| = \frac{1}{2}\} ∪ \{z : |z| = 1\}, normalized to give each mass \frac{1}{2}. The corresponding function f is orthogonal by the theorem, but is clearly not inner since |f| = \frac{1}{2} on a subset of 𝕋 of positive measure.

A more interesting example of a radially symmetric measure satisfying (1.1) is normalized area measure on the disk. Thus there is an f ∈ U_{0} so that μ_{f} is normalized area measure. We will show (Lemma 6.1) that for any holomorphic g on the disk, and f ∈ U_{0} orthogonal,

∥g ∘ f∥_{H^{p}}^{p} = ∫_{𝔻} |g|^{p}dμ_{f} + μ_{f}(𝕋)∥g∥_{H^{p}}^{p}, (1.2)

and hence

Corollary 1.2. There is an f ∈ U_{0} so that for any analytic g on 𝔻, g is in the Bergman space, A^{p}, iff g ∘ f is in the Hardy space, H^{p} (and the norms are equal).

Thus the subspace M_{f} spanned by the powers of f in H^{2} is isomorphic to the Bergman space and multiplication by f on M_{f} is isomorphic to multiplication by z on the Bergman space. Since both spaces are Hilbert spaces, of course one is isomorphic to a subspace of the other, but it is perhaps a little surprising that this isomorphism can be accomplished with a composition operator. Similar statements

ORTHOGONAL FUNCTIONS IN H^{∞} 3

can be made for Bergman spaces with respect to radial weights wdxdy = dμ of finite mass which satisfy (1.1).

More generally, it would be interesting to know for which pair of spaces X, Y of analytic functions on 𝔻 there is an f ∈ U_{0} so that g ∈ X iff g ∘ f ∈ Y, and to characterize such f’s when they exist. The latter problem is interesting even when X = Y, e.g., see [7]. In Corollary 6.3 we characterize orthogonal functions with this property when X = Y = H^{p} (it is true iff μ_{f}(𝕋) > 0). In particular, all inner functions have this property (as claimed in [7]).

Paul Bourdon has pointed out that (1.2) implies that orthogonal functions f such that μ_{f}(𝕋) > 0 give examples of composition operators with closed range. See [9] and [25] for characterizations of such functions.

The radial symmetry of a “Rudin counterexample” has also been noted by Paul Bourdon [4]. He showed that f is orthogonal iff the Nevanlinna counting function,

N_{f}(w) = ∑_{f(z)=w} log(\frac{1}{|z|})

is almost everywhere constant on each circle centered on the origin. He also showed the answer to Rudin’s question is “yes” if f is univalent and that if f is orthogonal, the closure of the range of f is a disk (since the range of f equals the set where N_{f} is positive). The Nevanlinna function N_{f} is related to μ_{f} by the formula

N_{f}(w) = log \frac{1}{|w|} − ∫ log \frac{1}{|z − w|}dμ_{f}(z),

(except possibly on a set of logarithmic capacity zero). This is due to W. Rudin in [14] but we shall give a proof for completeness (Lemma 3.1).

Corollary 1.3. If f ∈ U_{0} is non-constant and orthogonal then N_{f}(w) = N(|w|) for all w outside an exceptional set of zero logarithmic capacity, where

N(r) = ∫_{r}^{1} \frac{1 − μ(t)}{t}dt,

for some increasing function μ on [0, 1] with μ(0) = 0, μ(1) = 1 and ∫_{0}^{1} μ(t)\frac{dt}{t} < ∞ (in fact, μ(r) = μ_{f}(D(0, r))). Moreover, for every such N there is an f ∈ U_{0} so that N_{f}(w) = N(|w|) except possibly on a set of logarithmic capacity zero.

The “⇒” direction of this is due to Paul Bourdon [5]. The condition on N in the previous result has many equivalent formulations; for example, it holds iff M(r) = N(e^{r}) on (−∞, 0] is concave up, has M(0) = 0 and sup_{r<0} M(r) + r < ∞, or if N(|z|) is subharmonic on 𝔻 \ {0} and N(|z|) + log |z| is bounded above. The behavior of the composition operator C_{f} : g → g ∘ f can often be expressed in terms of N_{f}, e.g., [18], [20], [21]. The result above provides radial examples with any desired rate of decay faster than 1 − r as r → 1.

If f is orthogonal, then f(0) = 2π ∫ fdθ = 0, so f cannot be an outer function. However, our construction can be modified to give

Corollary 1.4. There is an orthogonal f so that f(z)/z is a non-constant outer function.

Thus, not only are there orthogonal functions which are not inner, there are examples with only the most trivial possible inner factor. I do not know whether there is an example so that f(z)/z is bounded away from zero on 𝔻 or which symmetric measures μ are of the form μ_{f} with f(z)/z outer.

ORTHOGONAL FUNCTIONS IN H^{∞} 5

One can also construct examples with other properties. For example, f ∈ U_{0} is said to be in the hyperbolic little Bloch class B_{0}^{h} if

lim_{|z|→1} \frac{(1 − |z|^{2})|f′(z)|}{1 − |f(z)|^{2}} = 0.

(This is contained in the usual little Bloch space, where only the numerator is required to go to zero.) We will show (Lemma 5.2) that if g is inner and f ∈ H^{∞} then μ_{f∘g} = μ_{f}. Thus taking g to be an inner function in the hyperbolic little Bloch class (which exists by a result of Wayne Smith [19] and independently of Aleksandrov, Anderson and Nicolau, [1]; also see Cantón’s paper [6]), we can deduce that

Corollary 1.5. Any of the measures in Theorem 1.1 is μ_{f} for some f ∈ B_{0}^{h}.

Cima, Korenblum and Stessin [8] also identified symmetric properties of orthogonal functions and showed the answer to Rudin’s question is “yes” if f is Hölder of order α > \frac{1}{2} on 𝕋. I do not know if there is an (non-inner) orthogonal function which is continuous up to the boundary, but expect that it might be possible to build one by modifying the construction in this paper. If there is a continuous orthogonal function, it would be very interesting to know if the result of Cima, Korenblum and Stessin is sharp, and if not, what the best modulus of continuity for such a function could be. What other natural conditions on an orthogonal function imply that it is actually inner?

I would like to thank Joseph Cima for helpful conversations about Rudin’s problem and composition operators. I am extremely grateful to Paul Bourdon for his carefully reading of an earlier version and the many helpful comments, suggestions and references he provided. Also thanks to Carl Sundberg for discussing his results with me.

6 CHRISTOPHER J. BISHOP

The remaining sections are organized as follows: Section 2: We describe some elementary properties of μ_{f} and prove it is radial

iff f is orthogonal. Section 3: We prove Corollary 1.3, (given Theorem 1.1). Section 4: We prove some results concerning the convergence of μ_{f}. Section 5: We prove Corollary 1.5 (given Theorem 1.1). Section 6: We prove Corollary 1.2 (given Theorem 1.1). Section 7: We construct a symmetric μ_{f} which is supported on two circles. Section 8: We construct all examples supported in \{\frac{1}{2} ≤ |z| ≤ 1\}. Section 9: We complete the proof of Theorem 1.1. Section 10: We prove Corollary 1.4.

2. Elementary properties of μ_{f}

We begin by recalling a few simple facts about analytic functions f and the corresponding measures μ_{f}. Many of these are well known but we include them for the convenience of the reader.

Lemma 2.1. If f ∈ H^{∞} then μ_{f} satisfies

∫ log \frac{1}{|z|}dμ_{f}(z) < ∞.

Proof. If f has a zero of order n at the origin, then g(z) = f(z)/z^{n} is holomorphic on the disk and |g| = |f| on 𝕋, so μ_{g}(A) = μ_{f}(A) for any annulus A = {z : r_{1} ≤ |z| ≤ r_{2}}. Thus

∫ φ(z)dμ_{f}(z) = ∫ φ(z)dμ_{g}(z),

ORTHOGONAL FUNCTIONS IN H^{∞} 7

for any radial function φ. Using Fatou’s lemma and the fact that log |g(z)|^{−1} is superharmonic on the disk ([11], page 35), we deduce

∫ log \frac{1}{|z|}dμ_{f}(z) = ∫ log \frac{1}{|z|}dμ_{g}(z)
= \frac{1}{2π} ∫ log |g(e^{iθ})|^{−1}dθ
= \frac{1}{2π} ∫ lim_{r→1} log |g(re^{iθ})|^{−1}dθ
≤ \frac{1}{2π} lim_{r→1} ∫ log |g(re^{iθ})|^{−1}dθ
≤ log |g(0)|^{−1}
< ∞,

as desired.

A similar estimate is true for other points, e.g.,

∫ log \frac{1}{|z − a|}dμ_{f}(z) < ∞.

In particular, this implies that the set where f has radial limit a must have measure zero (a well known fact).

Given an arc I ⊂ 𝕋 we define the Carleson box with base I as

Q = Q_{I} = {z ∈ 𝔻 : z/|z| ∈ I, 1 − |z| ≤ |I|}.

A positive measure μ is a Carleson measure if there is a C < ∞ so that μ(Q_{I}) ≤ C|I|, for every arc I ∈ 𝔻.

Lemma 2.2. If f ∈ U_{0} then μ_{f} is a Carleson measure with constant independent of f.

Proof. Define φ(z) = ω(z, Q, 𝔻 \ Q) for z ∈ 𝔻 \ Q and φ(z) = 1 for z ∈ Q. It is easy to see that ω(z, I, 𝔻) ≥ M^{−1} > 0 for every z ∈ ∂Q ∩ 𝔻 and some M < ∞

8 CHRISTOPHER J. BISHOP

(independent of I and z ∈ ∂Q), so the maximal principle implies

φ(0) ≤ Mω(0, I, 𝔻) ≤ M|I|.

Let f_{r}(z) = f(rz). Note that lim_{r→1} φ(f(rx)) = φ(f(x)) for almost every x ∈ 𝕋, because φ is continuous on the closed disk, except at two points and the set where f has a radial limit equal to one of these is measure zero (by the remark following Lemma 2.1). So by the Lebesgue dominated convergence theorem,

μ_{f}(Q) ≤ ∫ φdμ_{f} = \frac{1}{2π} ∫ φ ∘ fdθ = lim_{r→1} \frac{1}{2π} ∫ φ ∘ f_{r}dθ. (2.1)

Since φ is superharmonic on 𝔻, φ ∘ f is superharmonic on 𝔻, so the right hand side of (2.1) is ≤ φ(f(0)) = φ(0) ≤ M|I|.

If f(0) ≠ 0 then μ_{f} is still a Carleson measure, but with norm depending on |f(0)|.

One can think of the previous lemma as a weak version of the Littlewood subordination principle: that if f is an analytic self map of the disk then g ∈ H^{p} implies g ∘ f ∈ H^{p} (with smaller or equal norm). Formally, this implies,

∫ |g|dμ_{f} ≤ \frac{1}{2π} ∫_{𝕋} |g ∘ f|dθ = ∥g ∘ f∥_{H^{p}} ≤ ∥g∥_{H^{p}}

for every g ∈ H^{p}. This implies dμ_{f} is a Carleson measure with norm independent of f (e.g., Theorem I.5.6 of [11]).

The following result appears in many places (e.g. [12], Lemma 1 of [13], page 405 [16], Theorem VIII.30 [23]) and is sometimes called “Löwner’s lemma”. See [10] and its references for various generalizations.

Lemma 2.3. If f is an inner function such that f(0) = 0, then μ_{f} is normalized Lebesgue measure on the unit circle.

ORTHOGONAL FUNCTIONS IN H^{∞} 9

Proof. It is enough to check that μ_{f}(I) = |I| for arcs. Let I be an arc on the unit circle and let φ(z) = ω(z, I, 𝔻). Then φ ∘ f is bounded, harmonic, takes only radial boundary values 1 and 0 almost everywhere f has radial boundary values in I or its complement respectively (since f is inner). Thus

|I| = φ(0) = φ(f(0)) = \frac{1}{2π} ∫_{f^{−1}(I)} dθ = μ_{f}(I),

as desired.

As noted before, the following is similar to results in [4] and [8].

Lemma 2.4. Suppose f ∈ H^{∞}. Then the measure μ_{f} is radially symmetric iff {f^{n}} is orthogonal.

Proof. If μ_{f} is radially symmetric, it can be written so that

∫ g(z)dμ_{f}(z) = ∫_{0}^{2π} ∫_{0}^{∞} g(re^{iθ})dθdν(r),

for every g ∈ C_{c}(ℝ^{2}) (continuous functions of compact support) and for some measure ν on (0, ∞). Thus

∫_{𝕋} f^{n}f̄^{m}dθ = ∫_{ℂ} z^{n}z̄^{m}dμ_{f}(z) = ∫_{0}^{∞} ∫_{0}^{2π} r^{n+m}e^{i(n−m)θ}dθdν(r) = 0, if n ≠ m, so f is orthogonal. Conversely, if f is orthogonal, then μ_{f} satisfies

∫_{ℂ} z^{n}z̄^{m}dμ_{f}(z) = ∫_{0}^{2π} ∫_{0}^{∞} r^{n+m}e^{i(n−m)θ}dμ_{f}(re^{iθ}) = 0,

for n ≠ m. Thus for any polynomial P (z, z̄) = ∑_{n,m} a_{n,m}z^{n}z̄^{m} in z and z̄,

∫_{ℂ} P (z, z̄)dμ_{f}(z) = ∫_{𝔻} ∑_{n} a_{n,n}r^{2n}dμ_{f}(z),

and hence for any |λ| = 1, ∫_{ℂ} P (λz, λ̄z̄)dμ_{f}(z) = ∫_{ℂ} P (z, z̄)dμ_{f}(z).

10 CHRISTOPHER J. BISHOP

Since polynomials in z and z̄ are dense in the continuous functions on the closed unit disk, we deduce that for any g ∈ C_{c}(ℝ^{2}),

∫_{𝔻} g(z)dμ_{f}(z) = ∫_{𝔻} g(λz)dμ_{f}(z),

for any |λ| = 1. This implies μ_{f} is radially symmetric.

The following lemma greatly simplifies the construction of the basic example (where μ_{f} is supported on two circles). It says that if we can construct an example so that μ_{f} is radially symmetric on the smaller circle, then it automatically looks like Lebesgue measure on the larger one.

Lemma 2.5. Suppose f ∈ U_{0} and μ_{f} is supported on the circles C_{1/2} ∪C_{1} = {|z| = \frac{1}{2}\} ∪ \{|z| = 1\}. If μ_{f} restricted to C_{1/2} is a multiple of Lebesgue 1-dimensional measure, then so is μ_{f} restricted to C_{1}.

Proof. Suppose u is any bounded harmonic function on 𝔻. Then v(z) = u(f(z)) is also bounded and harmonic on 𝔻 and u(0) = v(0). Thus

u(0) = v(0) = \frac{1}{2π} ∫_{0}^{2π} u(f(e^{iθ}))dθ
= ∫ u(z)dμ_{f}(z)
= ∫_{C_{1/2}} u(z)dμ_{f}(z) + ∫_{C_{1}} u(z)dμ_{f}(z)
= μ_{f}(C_{1/2})u(0) + ∫_{C_{1}} u(z)dμ_{f}(z).

Hence ∫_{C_{1}} udμ_{f} = μ_{f}(C_{1})u(0), for any bounded harmonic function u on 𝔻. This easily implies μ_{f} restricted to C_{1} is a multiple of Lebesgue measure on C_{1}.

The same proof gives the following generalization of Lemma 2.5.

ORTHOGONAL FUNCTIONS IN H^{∞} 11

Lemma 2.6. Suppose f ∈ U_{0}. Then μ_{f} restricted to the unit circle is of the form \frac{1}{2π} (1 − g(θ))dθ, where g is the balayage of μ_{f} onto the circle, i.e.,

g(θ) = ∫_{𝔻} P_{z}(θ)dμ_{f}(z),

where P_{z}(θ) is the Poison kernel for 𝔻 with respect to the point z.

3. The Nevanlinna counting function

If f ∈ H^{∞}, the Nevanlinna counting function is defined as

N_{f}(w) = ∑_{f(z)=w} log \frac{1}{|z|}.

If f ∈ U_{0} then N_{f}(w) ≤ log |w|^{−1}. Clearly this is just the Greens function for the Riemann surface associated to f (projected to the plane by summing over sheets). Since μ_{f} is the projection of harmonic measure for the Riemann surface, the following is analogous to the standard result for Greens functions of planar domains. As noted in the introduction the following is due to W. Rudin in [14].

Lemma 3.1. If f ∈ U_{0} then ΔN_{f} = −δ_{0} + μ_{f} in the sense of distributions (Δ = ∂_{x}^{2} + ∂_{y}^{2} denotes the Laplacian and δ_{0} is the Dirac mass at the origin). Moreover,

N_{f}(w) = log \frac{1}{|w|} − ∫ log \frac{1}{|z − w|}dμ_{f}(z), (3.1)

for all w, except possibly for an exceptional set E of logarithmic capacity zero where “<” holds.

The exceptional set is required. For example, if f is the universal covering map of 𝔻 minus a compact set E of zero logarithmic capacity, then f is an inner function, μ_{f} is normalized Lebesgue measure on the circle and N_{f}(z) = χ_{𝔻\E}(z) log \frac{1}{|z|}. Proof. For 0 < r < 1, let f_{r}(z) = f(rz) and let γ_{r} = f_{r}(𝕋). If we choose r so that f′ never vanishes on the circle of radius r, then γ_{r} is a smooth curve and it

12 CHRISTOPHER J. BISHOP

is easy to check using Green’s theorem that ΔN_{f_{r}} = −δ_{0} + μ_{f_{r}}. To see that (3.1) holds for μ_{f_{r}}, note that both sides of the equation have the same distributional Laplacian, so they differ by a harmonic function. N_{f_{r}} vanishes outside the unit disk by definition, and the right side of (3.1) vanishes there because μ_{f_{r}} evaluates harmonic functions at 0. Hence the difference between the left and right sides is the constant zero function.

For any smooth φ with compact support,

∫ N_{f_{r}}Δφdxdy = −φ(0) + ∫ φdμ_{f_{r}}.

We shall see later that μ_{f_{r}} converges to μ_{f} weakly (Corollary 4.4) and clearly N_{f_{r}} ↗ N_{f} as r ↗ 1. Thus taking r → 1 and applying the monotone convergence theorem we get

∫ N_{f}Δφdxdy = −φ(0) + ∫ φdμ_{f}.

This proves the first claim of the lemma. Next we verify (3.1).

We already know that if we replace f by f_{r} then we have equality in (3.1) for all z and as r → 1, and we know N_{f_{r}}(z) ↗ N_{f}(z) for all z. Thus the question reduces to whether U_{r}(w) → U_{1}(w), r → 1 (3.2) for all w except a set E of logarithmic capacity zero, where

U_{r}(w) = ∫ log \frac{1}{|z − w|}dμ_{f_{r}}(z).

Note that by superharmonicity of log |f|^{−1}, U_{r} is decreasing in r and that U_{1} is bounded below by − log 2 (since |z − w| < 2 for points in the unit disk).

To prove (3.2) we follow the proof of Frostman’s theorem (e.g., Theorem II.6.4 in [11]). Suppose σ is a measure so that V (z) = ∫ log |z − w|^{−1}dσ(z) is bounded.

ORTHOGONAL FUNCTIONS IN H^{∞} 13

It suffices to show σ(E) = 0. By Fatou’s lemma

lim_{r→1} ∫ log \frac{1}{|z − w|}dμ_{f_{r}}(z) ≥ ∫ log \frac{1}{|z − w|}dμ_{f}(z),

so

lim_{r→1} U_{r}(w) ≥ U_{1}(w),

for all w. On the other hand, by Fatou’s lemma, Fubini’s theorem and the Lebesgue dominated convergence theorem

∫_{E} lim_{r→1} U_{r}(w)dσ(w) ≤ lim_{r→1} ∫_{E} U_{r}(w)dσ(w)
= lim_{r→1} \frac{1}{2π} ∫_{0}^{2π} V (f(re^{iθ}))dθ
= \frac{1}{2π} ∫_{0}^{2π} V (f(e^{iθ}))dθ
= \frac{1}{2π} ∫_{0}^{2π} ∫_{E} log \frac{1}{|f(e^{iθ}) − w|}dσ(w)dθ
= ∫_{E} U_{1}(z)dσ(w)

Thus we must have lim_{r→1} U_{r}(w) = U_{1}(w) except on a set of zero σ measure.

Thus we we see that {f^{n}} is an orthogonal sequence iff μ_{f} is radial iff N_{f} is radial except on a set of logarithmic capacity zero. This gives an alternate approach to the results of Bourdon in [4].

We can also compute exactly which radial functions can occur as N_{f} for some f ∈ U_{0}. Note that

\frac{1}{2π} ∫_{0}^{2π} log \frac{1}{|re^{iθ} − w|}dθ = \{ \begin{matrix}log \frac{1}{|w|}, & r ≤ |w|, \\ log \frac{1}{r}, & r ≥ |w|. \end{matrix}

Thus if μ_{f} is radially symmetric and we set μ(r) = μ_{f}(D(0, r)), then

N_{f}(w) = log \frac{1}{|w|} − ∫ log \frac{1}{|z − w|}dμ_{f}(z) = ∫_{|w|}^{1} \frac{1 − μ(r)}{r}dr.

Moreover, the integral condition

∫_{𝔻} log \frac{1}{|z|}dμ < ∞,

14 CHRISTOPHER J. BISHOP

becomes

∫_{0}^{1} μ(r)\frac{dr}{r} < ∞.

Thus Theorem 1.1 implies,

Corollary 3.2. Suppose N(r) = ∫_{r}^{1} \frac{(1−μ(t))}{t}dt for some increasing function μ so that μ(0) = 0, μ(1) = 1 and ∫_{0}^{1} μ(r)dr/r < ∞. Then there is a f ∈ U_{0} so that N_{f}(z) = N(|z|) except on a set of zero logarithmic capacity.

For example, if μ_{f} is normalized area measure on the unit disk then μ(r) = r^{2} and N_{f}(z) = log \frac{1}{r} − (1 − r) ≈ (1 − r)^{2} as r → 1.

4. Weak convergence of μ_{f}

We will obtain the functions f in Theorem 1.1 by a “cut and paste” construction of the corresponding Riemann surface. What this means is that we shall build a sequence of nested Riemann surfaces R_{0} ⊂ R_{1} ⊂ R_{2} ⊂ ⋯ ⊂ ∪R_{n} = R by identifying subdomains of the unit disk along common boundary arcs. The projection of R into the unit disk is a bounded holomorphic function on R, and hence R must be hyperbolic, i.e., its universal covering space is the unit disk, 𝔻. The desired map will be the covering map f : 𝔻 → R followed by the projection into the disk and the corresponding measure μ_{f} is simply the harmonic measure for the surface R, projected into the plane. In fact, we shall abuse notation and consider the covering map f : 𝔻 → R as actually mapping into the complex numbers (i.e., we identify the covering map and this map followed by the projection into the plane). By a similar abuse we shall think of harmonic measure on R and the corresponding projected measure μ_{f} as the same. Similarly, we will fix a point in R_{0} which projects to 0 and call it 0 as well. All our covering maps will be chosen to map 0 in the disk to

ORTHOGONAL FUNCTIONS IN H^{∞} 15

0 on the surface. See [2], [22], where a similar procedure has been used in different problems.

The main point we must be careful about is to show that the harmonic measure for R is the limit of the measures for R_{n}. To see that there might be a problem in general, consider what can happen when the surfaces are not nested, e.g., R_{n} is the unit disk minus the points \{z_{k}\} = \frac{1}{2} exp(i2πk2^{−n}), k = 1, …, 2^{n}. Then the universal covering map f_{n} : 𝔻 → R_{n} is an inner function (the isolated boundary points do not have any harmonic measure, so all the measure lives on the part of the boundary above the unit circle) and hence μ_{f_{n}} is Lebesgue measure on the unit circle. However, one can show (with some work) that f_{n}(z) → \frac{1}{2}z uniformly on compact sets of 𝔻 so that μ_{f} is Lebesgue measure on the circle of radius 1/2. However, if the Riemann surfaces are nested by (increasing) inclusion, then we will show the corresponding measures converge weakly, i.e.,

lim_{n→∞} ∫ gdμ_{n} = ∫ φdμ,

for any g ∈ C_{c}(ℝ^{2}).

Lemma 4.1. Suppose ϵ > 0 and D(0, ϵ) = R_{0} ⊂ R_{1} ⊂ … are obtained by identifying subdomains of the unit disk along boundary arcs. Let R = ∪_{n=1}^{∞}R_{n}. Choose covering maps f_{n} : 𝔻 → R_{n} and f : 𝔻 → R so that f_{n}(0) = f(0) = 0. Then μ_{f_{n}} converges weakly to μ_{f} on the closed unit disk.

The easiest way to see this is using Brownian motion; we shall first sketch such a proof and then give a more classical proof without using Brownian motion.

Let W be the Weiner space of continuous paths in ℂ starting at the origin. If R is a Riemann surface constructed as above then we can think of the paths as taking values in R and for each path w ∈ W, we define the stopping time t_{w} as the first

16 CHRISTOPHER J. BISHOP

time t such that w(t) ∉ R. Then w → t_{w} is measurable and the harmonic measure for R is simply the push forward of Wiener measure on W under the map given by w → w(t_{w}). Given a sequence of nested surfaces R_{0} ⊂ R_{1} ⊂ … as in the lemma, we get a corresponding sequence of maps g_{n} : W → ℂ. Moreover, if R = ∪_{n}R_{n} and g : W → ℂ is the corresponding map, then g(w) = lim_{n} g_{n}(w); this is because the inclusions imply that for any continuous path in the plane, the first time it leaves R is the limit of the first times it left R_{n}. Thus for any bounded, continuous function φ on the plane, φ(g_{n}(w)) → φ(g(w)) for all w, so the Lebesgue dominated convergence theorem implies

∫_{W} φ(g(w))dw = lim_{n} ∫_{W} φ(g_{n}(w))dw,

which is the desired weak convergence.

The sketch above is simple and explains why the result is true, but uses the existence of Wiener measure and deep connections between it and harmonic measure. It therefore seems desirable to provide a second proof which uses only function theory. Moreover, we will need some corollaries of the following classical proof for our applications to composition operators.

Let {R_{n}}, R, {f_{n}} and f be as in the lemma and let Ω_{n} = f^{−1}(R_{n}) ⊂ 𝔻. Then Ω_{0} ⊂ Ω_{1} ⊂ … and ∪_{n}Ω_{n} = 𝔻. Let ω_{n} be the harmonic measure for Ω_{n} with respect to the origin and let φ be any continuous function on the plane. We want to show that

lim_{n} ∫ φ(f(z))dω_{n}(z) = ∫_{𝕋} φ(f(e^{iθ}))dθ/2π.

ORTHOGONAL FUNCTIONS IN H^{∞} 17

We start by proving the much easier fact that ω_{n} converges weakly to normalized Lebesgue measure on the circle. (Since f need not be continuous up to the boundary, φ ∘ f need not be continuous either, so weak convergence of ω_{n} is not, by itself, enough to prove weak convergence of μ_{f_{n}}.)

Lemma 4.2. If {0} ∈ Ω_{0} ⊂ Ω_{1} ⊂ is a sequence of subdomains so that ∪_{n}Ω_{n} = 𝔻, and ω_{n} = ω(0, ·, Ω_{n}) is the corresponding harmonic measure with respect to the origin, then {ω_{n}} converges weakly to (normalized) Lebesgue measure on 𝕋. Moreover, the measures ω_{n} are all Carleson with a uniform constant.

Proof. The Carleson condition follows from Lemma 2.2 applied to the covering map onto Ω_{n}, so we need only prove weak convergence. Since ∩_{n}(\overline{𝔻} \ Ω_{n}) = 𝕋, there is a sequence {r_{n}} ↗ 1 such that D_{n} = {z : |z| < r_{n}} ⊂ Ω_{n} ⊂ 𝔻. Suppose I ⊂ 𝕋 is an open arc and let Q be the corresponding Carleson square, Q = {z ∈ 𝔻 : z/|z| ∈ I, 1 − |z| ≤ |I|}. To show ω_{n} converges weakly to normalized Lebesgue measure, it is clearly enough to show that ω_{n}(Q) → |I|.

Let U_{n} = D_{n} ∪ Q and V_{n} = 𝔻 \ (Q \ D_{n}). First observe that ω(0, I, U_{n}) → |I|. To see this simply note that

ω(0, I, U_{n}) = |I| − ∫_{∂U_{n}\I}[∫_{I} P_{z}(θ)dθ]dω(0, ·, U_{n}),

and that ∫_{I} P_{z}(θ)dθ → 1/2, as n → ∞ when z is in one of the radial part of ∂U_{n} and → 0 when z is in the interior of ∂U_{n} ∩ D_{n}. Since the radial parts of ∂U_{n} have length tending to zero as n → ∞ their harmonic measure also tends to 0 as n → ∞ (e.g., by Lavrentiev’s estimate of harmonic measure on rectifiable domains (e.g. page 80 of [3]; one could also use Beurling’s estimate that a set of diameter δ in a simply connected domain has harmonic measure ≲ \sqrt{δ}). Thus the Lebesgue

18 CHRISTOPHER J. BISHOP

dominated convergence theorem implies ω(0, I, U_{n}) → |I|. A similar argument shows ω(0, Q ∩ V_{n}, V_{n}) → |I| as n → ∞.

Thus by the monotonicity of harmonic measure,

ω_{n}(Q) ≥ ω(0, ∂Ω_{n} ∩ Q, U_{n}) ≥ ω(0, I, U_{n}) → |I|,

so lim inf_{n} ω_{n}(Q) ≥ |I|. On the other hand,

ω_{n}(Q) ≤ ω(0, Q ∩ ∂V_{n}, V_{n}) → |I|,

which proves ω_{n}(Q) → |I|. This proves the lemma.

Lemma 4.3. Suppose g is a bounded, continuous function on 𝔻 which has nontangential limit g(x) almost everywhere on 𝕋 and that ν_{n} is a sequence of probability measures on \overline{𝔻} which converge weakly to (normalized) Lebesgue measure on the circle and which are all Carleson measures with a uniform constant. Then

lim_{n} ∫\overline{𝔻} g(z)dν_{n}(z) = \frac{1}{2π} ∫_{𝕋} g(z)dθ.

Proof. We may assume that ∥g∥_{∞} = 1. Fix some ϵ > 0. Since g has non-tangential limits almost everywhere, given almost any x ∈ 𝕋 there is a δ(x) > 0 so that if I is any interval containing x with length less than δ(x), then g is within ϵ/2 of g(x) on the top half of the corresponding Carleson box Q. Fix a complex number a and a δ > 0 and assume E_{a} = {x ∈ 𝕋 : |g(x) − a| ≤ ϵ/2, δ(x) > δ} has positive Lebesgue measure. Using the Lebesgue density theorem choose a dyadic interval I of length less than δ so that |I ∩ E_{a}| ≥ (1 − ϵ)|I| and let Q_{k} be the collection of maximal dyadic subsquares with bases {I_{k}} ⊂ I, so that |g(z) − g(x)|| > ϵ for some z in the top half of Q_{k}. Let Q be the Carleson square with base I and let W = Q \ ∪Q_{k}. Then g is within ϵ of a constant on W, and |∂W ∩ I| ≥ (1 − ϵ)|I|.

ORTHOGONAL FUNCTIONS IN H^{∞} 19

We claim that for these domains lim_{n} ν_{n}(W) = |∂W ∩ 𝕋|. To prove this, we use the weak convergence of {ν_{n}} to deduce

lim_{n} ∫_{W} dν_{n} = lim_{n}[ν_{n}(Q) − ∑_{k} ν_{n}(Q_{k})]
= |I| − lim_{n} ∑_{k} ν_{n}(Q_{k})
= |I| − ∑_{k} lim_{n} ν_{n}(Q_{k})
= |I| − ∑ |I_{k}|
= |∂W ∩ 𝕋|,

where we used the Lebesgue dominated convergence theorem on the sequence space ℓ^{1} to interchange the limit and the infinite sum (our assumption that the measures are uniformly Carleson gives that ν_{n}(Q_{k}) ≤ C|I_{k}|, independent of n; this gives the ℓ^{1} upper bound).

Moreover, the intervals I with these properties form a Vitali cover of 𝕋 (see e.g., Section 7.3 [24]), so we can form disjoint cover of almost every point of 𝕋 using such intervals. Thus we can construct a finite number of disjoint domains W_{j} = Q_{j} \ ∪_{k}Q_{k}^{j}, where

1. Q_{j} is a Carleson square with base I_{j} and |∂W_{j} ∩ I_{j}| ≥ (1 − ϵ)|I_{j}|,

2. g is within ϵ of a constant c_{j} on each W_{j},

3. ∑_{j} |∂W_{j} ∩ 𝕋| ≥ 1 − ϵ.

Set W = ∪_{j}W_{j} be this finite union. Note that the weak convergence of {ν_{n}} implies that

lim sup_{n→∞} ν_{n}(𝔻 \ W) ≤ ϵ,

20 CHRISTOPHER J. BISHOP

and so

| lim_{n} ∫ gdν_{n} − ∫_{𝕋} gdθ/2π| ≤ lim_{n→∞} | ∫_{W} gdν_{n} − \frac{1}{2π} ∫_{∂W ∩𝕋} gdθ| + ∫_{𝔻\W} |g|dν_{n}

+\frac{1}{2π} ∫_{𝕋\∂W} |g|dθ

≤ ϵ ∑_{j} |∂W_{j} ∩ 𝕋| + 2|𝕋 \ ∪∂W_{j}|
≤ 4ϵ.

Taking ϵ → 0 proves Lemma 4.3 and thus our function theoretic proof of Lemma 4.1.

A very special (and easier) case of Lemma 4.3 is

Corollary 4.4. Suppose f ∈ U_{0} and let f_{r}(z) = f(rz), for r < 1. Then μ_{f_{r}} converges weakly to μ_{f} as r → 1.

Corollary 4.5. If f is inner and f(0) = 0 then μ_{f_{r}} converges weakly to normalized Lebesgue measure on 𝕋.

5. A change of variables

The following result was suggested by Paul Bourdon and simplifies certain arguments from an earlier version of the paper.

Lemma 5.1. Suppose g is a positive, continuous function on 𝔻 and has nontangential boundary values almost everywhere on 𝕋. Then for any f ∈ U,

∫ g(z)dμ_{f}(z) = \frac{1}{2π} ∫_{0}^{2π} g(f(e^{iθ}))dθ.

The integral on the left requires some interpretation since g is not necessarily continuous on the support of μ_{f}. On the interior of the disk, g is continuous and positive so the integral is well defined (possible infinite). On the circle, μ_{f} is

ORTHOGONAL FUNCTIONS IN H^{∞} 21

absolutely continuous with respect to Lebesgue measure and the boundary values of g are Borel, so the integral on the circle is also well defined. Proof. Using the monotone convergence theorem we can reduce to the case when g is bounded (just truncate and let the truncation tend to ∞). So assume g is bounded by M. For any ϵ > 0 we can easily construct a sawtooth region W so that |𝕋 ∩ ∂W | > 1 − ϵ and g extends continuously to the closure of W. Thus we can write g = (g − h) + h where h is continuous, bounded by M and g − h is zero on W. The lemma is true for continuous functions by the definition of μ_{f}, and

∫ (g − h)dμ_{f} ≤ 2Mμ_{f}(\overline{𝔻} \ W) ≤ 2MCϵ,

since μ_{f} is Carleson with a uniform constant. Similarly

∫ (g − h) ∘ f(e^{iθ})dθ ≤ 2MCϵ,

so taking ϵ → 0 proves the lemma.

The following is now immediate.

Lemma 5.2. If g ∈ H^{∞} and f is inner with f(0) = 0 then μ_{g} = μ_{g∘f}.

The hyperbolic little Bloch space, B_{0}^{h}, is defined as those holomorphic maps f ∈ U so that

lim_{|z|→1} \frac{(1 − |z|^{2})|f′(z)|}{1 − |f(z)|^{2}} = 0,

and is contained in the usual little Bloch space, B_{0}. Schwarz’s inequality implies the left side is bounded by 1 for any analytic self map of the disk, and from this it is easy to verify that g, f are both holomorphic selfmaps of the disk and f is hyperbolic little Bloch then so is g ∘ f. It is far from obvious that there is an inner function in the hyperbolic little Bloch space, but they do exist (see [1], [6], [19]). This and Lemma 5.2 thus imply,

22 CHRISTOPHER J. BISHOP

Corollary 5.3. If g ∈ U, then there is a f ∈ B_{0}^{h} so that μ_{f} = μ_{g}.

Recall that the Hardy space, H^{p} , is the set of holomorphic functions g such that

∥g∥_{H^{p}} = lim_{r→1}(\frac{1}{2π} ∫_{0}^{2π} |g(re^{iθ})|^{p}dθ)^{1/p} < ∞.

Such a function has radial boundary values almost everywhere on 𝕋 which we also denote by g. If we know g ∈ H^{p} for p > 1, then the radial maximal function of g is in L^{p} and so on can use the dominated convergence theorem to deduce that

∥g∥_{H^{p}} = \frac{1}{2π} ∫_{0}^{2π} |g(e^{iθ})|^{p}dθ.

In general, however, the right hand side might be finite but g ∉ H^{p} (there exist non-zero holomorphic functions on the disk that have radial value zero almost everywhere, hence are not in H^{p}). If f ∈ U then μ_{f} restricted to 𝕋 is absolutely continuous with respect to Lebesgue measure, so ∫\overline{𝔻} |g|^{p}dμ_{f} makes sense.

As another application of Lemma 5.1 we can show

Lemma 5.4. Suppose g ∈ H^{p} on the unit disk and f ∈ U_{0}. Then for any 0 < p < ∞,

∥g ∘ f∥_{H^{p}}^{p} = lim_{r→1} ∫_{𝔻} |g|^{p}dμ_{f_{r}} = ∫\overline{𝔻} |g|^{p}dμ_{f}. (5.1)

Proof. The first equality is the definition of the H^{p} norm, so we only have to prove the second. If g ∈ H^{p} and f ∈ U_{0} then by a result of Ryff [17], g ∘ f ∈ H^{p} with smaller or equal norm. Thus |g|^{p} is positive, continuous function on the disk which has non-tangential boundary values almost everywhere and so Lemma 5.1 applies to show

∫ |g(z)|^{p}dμ_{f} = \frac{1}{2π} ∫_{0}^{2π} |g(f(e^{iθ}))|^{p}dθ,

and since we already know g ∘ f ∈ H^{p}, we can deduce that the right hand side equals ∥g ∘ f∥_{H^{p}}.

ORTHOGONAL FUNCTIONS IN H^{∞} 23

6. Mapping the Bergman space into the Hardy space For our applications to composition operators, we need a version of Lemma 5.4

that works without the assumption that g ∈ H^{p}. The proof given above doesn’t work in general because if g ∉ H^{p} we can’t say that ∥g∥_{H^{p}} = ∫_{0}^{2π} |g|^{p}dθ/2π. In fact, we will not even assume g has boundary values on the circle so that this integral is not necessarily defined.

Lemma 6.1. Suppose g is holomorphic on the open unit disk, f ∈ U_{0} and μ_{f} is radially symmetric. Then for any 0 < p < ∞,

∥g ∘ f∥_{H^{p}}^{p} = lim_{r→1} ∫_{𝔻} |g|^{p}dμ_{f_{r}} = ∫_{𝔻} |g|^{p}dμ_{f} + μ_{f}(𝕋)∥g∥_{H^{p}}^{p}. (6.1)

Proof. Let g_{s}(z) = g(sz), for 0 < s < 1. First we want to show that for any 0 < p < ∞,

lim_{s→1} ∫ |g(sz)|^{p}dμ_{f} = ∫_{𝔻} |g(z)|^{p}dμ_{f} + μ_{f}(𝕋)∥g∥_{H^{p}}^{p}, (6.2)

Since g is holomorphic, |g|^{p}, 0 < p < ∞ is subharmonic (e.g., [11], page 35) and hence m(r) = \frac{1}{2π} ∫ |g(re^{iθ})|^{p}dθ, is defined on [0, 1) and is an increasing function of r (Corollary I.6.6, [11]). Therefore we can extend it to be defined at r = 1 by ∥g∥_{H^{p}}^{p} = m(1) = lim_{r→1} m(r). Thus m_{s}(r) ≡ m(sr) increases to m(r) as s → 1 for all r ∈ [0, 1]. Let ν be the measure on [0, 1] defined by ν(E) = μ_{f}({z : |z| ∈ E}). Since μ_{f} is radial we have

∫ φdμ_{f} = \frac{1}{2π} ∫_{0}^{1} ∫_{0}^{2π} φ(re^{iθ})dθdν(r).

Thus by the monotone convergence theorem,

lim_{s→1} ∫ |g_{s}|^{p}dμ_{f} = lim_{s→1} ∫ m_{s}(r)dν = ∫_{[0,1]} m(r)dν = ∫_{𝔻} |g|^{p}dμ_{f} + μ_{f}(𝕋)m(1).

This is (6.2).

24 CHRISTOPHER J. BISHOP

We will break the proof of (6.1) into three cases: (1) ∫_{𝔻} |g|^{p}dμ_{f} = ∞, (2) ∫_{𝔻} |g|^{p}dμ_{f} < ∞ and μ_{f}(𝕋) = 0 and finally (3) ∫_{𝔻} |g|^{p}dμ_{f} < ∞ and μ_{f}(𝕋) > 0.

Case 1: First suppose ∫ |g|^{p}dμ_{f} = ∞. For any M > 0 choose 0 < t < 1 so that ∫_{|z|<t} |g|^{p}dμ_{f} > 2M and write |g|^{p} = g_{1} + g_{2} where g_{1}, g_{2} are non-negative, g_{1} = |g|^{p} on |z| < t and g_{1} is continuous and compactly supported in 𝔻. Then

∫ |g|^{p}dμ_{f_{r}} ≥ ∫ g_{1}dμ_{f_{r}} > \frac{1}{2} ∫ g_{1}dμ_{f} ≥ M,

if r is close enough to 1. Thus ∫ |g|^{p}dμ_{f_{r}} → ∞ = ∫ |g|^{p}dμ_{f}.

Case 2: Now suppose μ_{f}(𝕋) = 0 and ∫ |g|^{p}dμ_{f} < ∞. Since μ_{f_{r}} converges weakly to μ_{f},

lim_{r→1} ∫ |g_{s}|^{p}dμ_{f_{r}} = ∫ |g_{s}|^{p}dμ_{f},

for any fixed s < 1. Since g_{s}(f(z)) is holomorphic on the open disk, |g_{s}(f(z))|^{p} is subharmonic. Thus ∫ |g_{s}|^{p}dμ_{f_{r}} is increasing in r. Thus

∫ |g_{s}|^{p}dμ_{f_{r}} ≤ ∫ |g_{s}|^{p}dμ_{f}.

Now take s → 1. For a fixed r, μ_{f_{r}} is compactly supported in the disk , so |g_{s}|^{p} is uniformly bounded on its support and so the left hand side converges to ∫ |g|^{p}dμ_{f_{r}}. Condition (6.2) implies that the right hand side converges to ∫ |g|^{p}dμ_{f}. Thus

∫ |g|^{p}dμ_{f_{r}} ≤ ∫ |g|^{p}dμ_{f},

for all r < 1.

Fix ϵ > 0 and choose 0 < t < 1 so that ∫_{t<|z|<1} |g|^{p}dμ_{f} < ϵ. Write |g|^{p} = g_{1} + g_{2} as in Case 1. Thus ∫ g_{2}μ_{f} < ϵ. Also, if r is close enough to 1 then by weak convergence,

| ∫ g_{1}dμ_{f} − ∫ g_{1}dμ_{f_{r}}| < ϵ.

ORTHOGONAL FUNCTIONS IN H^{∞} 25

Thus

∫ g_{2}dμ_{f_{r}} ≤ | ∫ g_{1}dμ_{f} − ∫ g_{1}dμ_{f_{r}}| + ∫ g_{2}dμ_{f} ≤ 2ϵ.

Hence

| ∫ |g|^{p}dμ_{f} − ∫ |g|^{p}dμ_{f_{r}}| ≤ ∫ g_{2}dμ_{f_{r}} + | ∫ g_{1}dμ_{f} − ∫ g_{1}dμ_{f_{r}}| + ∫ g_{2}dμ_{f}
≤ 4ϵ,

if r is close enough to 1.

Case 3: Now assume that ∫_{𝔻} |g|^{p}dμ_{f} < ∞, μ_{f}(𝕋) > 0. If lim_{r→1} ∫ |g|^{p}dμ_{f_{r}} = ∞ then by subharmonicity of |g ∘ f|^{p} we get ∫ |g|^{p}μ_{f} = ∞ so (6.1) holds. Thus we may assume lim_{r→1} ∫ |g|^{p}dμ_{f_{r}} < ∞, i.e., we may assume g ∘ f ∈ H^{p}, and hence that |g(f(z))|^{p} has a harmonic majorant u on 𝔻 (Lemma II.1.1, [11]).

First we show that g ∈ H^{p}. For 0 < r < 1 let D_{r} = D(0, r). Let Ω_{r} be the component of f^{−1}(D_{r}) which contains the origin and let ω_{r} be the harmonic measure on Ω_{r} with respect to the origin. Let ν_{r} be the push forward of ω_{r} under the map f. Then clearly ν_{r} is supported on \overline{D_{r}} and for any E ⊂ D_{r}, ν_{r}(E) ≤ μ_{f}(E). By Lemma 2.6 ν_{r} on C_{r} = ∂D_{r}, must be \frac{1}{2π} dθ minus the balayage of ν_{r} restricted to D_{r}. Since ν_{r} ≤ μ_{f}, this means that ν_{r} on C_{r} is greater or equal \frac{1}{2π} dθ minus the balayage of μ_{f} restricted to D_{r}. Since μ_{f} is radial, its balayage onto C_{r} is also radial, i.e., is \frac{1}{2π} μ_{f}(D_{r})dθ ≤ \frac{1}{2π} (1−μ_{f}(𝕋))dθ. Thus on C_{r}, ν_{r} ≥ \frac{1}{2π} μ_{f}(𝕋)dθ. Hence for any g holomorphic on 𝔻

\frac{1}{2π} ∫_{0}^{2π} |g(re^{iθ})|^{p}dθ ≤ \frac{1}{μ_{f}(𝕋)} ∫ |g|^{p}dν_{r} = \frac{1}{μ_{f}(𝕋)} ∫ |g ∘ f|^{p}dω_{r}.

Thus if u is a harmonic majorant of |g ∘ f|^{p} on 𝔻,

\frac{1}{2π} ∫_{0}^{2π} |g(re^{iθ})|^{p}dθ ≤ \frac{1}{μ_{f}(𝕋)} ∫ udω_{r} = \frac{u(0)}{μ_{f}(𝕋)} < ∞.

In other words, g ∈ H^{p} and thus (6.1) follows from Lemma 5.4

26 CHRISTOPHER J. BISHOP

Recall that the Bergman space, A^{p}, is defined as the set of holomorphic functions g on the disk 𝔻 such that

∥g∥_{A^{p}} = (\frac{1}{π} ∫_{𝔻} |g|^{p}dxdy)^{1/p} < ∞.

Corollary 6.2. If f ∈ H^{∞} is such that dμ_{f} = \frac{1}{π} χ_{𝔻}dxdy, then for any analytic function g on the disk, g is in the Bergman space iff g ∘f is in the Hardy space, and ∥g∥_{A^{p}} = ∥g ∘ f∥_{H^{p}}, i.e., the corresponding composition operator, C_{f} : A^{p} → H^{p}, is an isometry onto its range.

Proof. Using the Lemma 6.1 we see

∥g ∘ f∥_{H^{p}} = lim_{r→1}(\frac{1}{2π} ∫_{0}^{2π} |g(f(re^{iθ}))|^{p}dθ)^{1/p}
= lim_{r→1}(∫ |g|^{p}dμ_{f_{r}})^{1/p}
= (∫ |g|^{p}dμ_{f})^{1/p}
= ∥g∥_{A^{p}}.

The corollary may seem a little surprising since functions in H^{p} have nontangential limits almost everywhere, whereas those in A^{p} need not, but since f has almost all of its boundary values in the interior of the disk, this is not a contradiction. Of course, it still remains to show (Section 9) that there is an f ∈ H^{∞} so that μ_{f} is area measure.

The previous corollary obviously holds for any weighted Bergman space where the weight is a radial measure of finite mass satisfying the integral condition (1.1) in Theorem 1.1. If instead of an isometry, we merely want ∥g∥_{A_{p}} ≃ ∥g ∘ f∥_{H^{2}} we could take a much bigger class of f’s, e.g. μ_{f} = wdxdy for some weight w which is bounded above and below on an annulus {r < |z| < 1}. Constructing such

ORTHOGONAL FUNCTIONS IN H^{∞} 27

examples only requires the techniques of Section 8, not the full proof of Theorem 1.1.

Similarly, by appropriate choices of μ_{f} one can construct composition operators on H^{p} which satisfy conditions like

∥C_{f}(g)∥_{H^{p}}^{p} = \frac{1}{2}∥g∥_{H^{p}}^{p} + \frac{1}{2}∥g∥_{A^{p}}^{p}

or

∥C_{f}(g)∥_{H^{p}}^{p} = \frac{1}{2}∥g∥_{H^{p}}^{p} + \frac{1}{2}∥g_{1/2}∥_{H^{p}}^{p}.

In [7], a function f is said to have property (∗) relative to H^{p} if for any holomorphic g on 𝔻, g ∘ f ∈ H^{p} implies g ∈ H^{p}. Paul Bourdon has pointed out that for general f ∈ U, the condition μ_{f}(𝕋) = 0 implies this condition implies N_{f}(z) = o(1 − |z|), which by J. Shapiro’s theorem from [18], implies C_{f} is compact and hence does not have a bounded right inverse. Since f is non-constant, C_{f} is 1-to-1 and so C_{f} does not have closed range (this is a consequence of the open mapping theorem, e.g. Corollary 2.12c of [15]). Thus C_{f} does not have property (∗), since any function in \overline{C_{f}(H^{p})} \ C_{f}(H^{p}) is an H^{p} function without an H^{p} preimage. Lemma 6.1 clearly implies

Corollary 6.3. If f ∈ U_{0} is orthogonal, then f has property (∗) relative to H^{p} iff μ_{f}(𝕋) > 0.

Proof. If μ_{f}(𝕋) > 0 then the proof in Case 3 of the proof of Lemma 6.1 shows that g ∘ f ∈ H^{p} implies g ∈ H^{p}. Thus f has property (∗) with respect to H^{p}.

A special case of Corollary 6.3 is when μ_{f}(𝕋) = 1, i.e., all inner functions have property (∗). It would be very interesting to have a similar characterization of property (∗) for general functions in U_{0}.

28 CHRISTOPHER J. BISHOP

7. An example of μ_{f} supported on two circles

In this section we will construct an f ∈ H^{∞} so that μ_{f} is supported on the union of two circles C_{1/2} and C_{1} (where C_{r} = {z : |z| = r}) and is a multiple of Lebesgue measure on each. This example suffices to disprove Rudin’s orthogonality conjecture, and introduces the estimates and techniques needed for the general case of Theorem 1.1. In the next section we will show that any radially symmetric probability measure supported in \{\frac{1}{2} ≤ |z| ≤ 1\} can occur as a μ_{f}, and in Section 9 we will do the general case of measures supported on 𝔻.

Based on Lemmas 4.1 and 2.5 it suffices to build an increasing sequence of Riemann surfaces {R_{n}} so that the corresponding maps {f_{n}} satisfy f_{n}(0) = 0, μ_{f_{n}} is supported on the two circles C_{1/2} ∪ C_{1} and μ_{f_{n}} restricted to C_{1/2} is of the form \frac{1}{2π} g_{n}(θ)dθ where g_{n} converges to uniformly to a positive constant.

We start by taking f_{0}(z) = \frac{1}{2}z, i.e, f_{0} is the (trivial) Riemann mapping from 𝔻 to the disk R_{0} = {|z| < 1/2}. The corresponding measure μ_{0} = μ_{f_{0}} is normalized Lebesgue measure on the circle C_{1/2}, i.e., it is \frac{1}{2π} g_{0}(θ)dθ where g_{0}(θ) = 1.

Now we describe the idea of the construction of R_{1} (we will give the details later). First we replace R_{0} by a slightly smaller disk, S_{1}. We divide the boundary of S_{1} into a large number of alternating intervals which we call type I and J. Along each type I interval we attach a copy certain Riemann surface with boundary over C_{1/2} (different copies to different intervals) and along each type J interval we attach copies of certain surfaces with boundary over C_{1}. This gives the surface R_{1}. With appropriate choices of the parameters involved we can show that with high probability, the Brownian paths which first hit ∂S_{1} at a type I interval go on to hit the part of ∂R_{1} over C_{1/2} and the paths which hit the J intervals go on to hit

ORTHOGONAL FUNCTIONS IN H^{∞} 29

∂R_{1} over C_{1}. Thus we have “rerouted” a certain fraction of the harmonic measure on C_{1/2} out to C_{1}. By choosing various parameters correctly, we can make the harmonic measure over C_{1/2} in R_{1} be close to any multiple of Lebesgue measure we want (as long as the total mass is < 1). The resulting measure may not be radially symmetric, but by iterating the construction with variable size barriers, we can make harmonic measure as close to a multiple of Lebesgue measure as we wish, obtaining a radial measure in the limit.

Now we give the construction of R_{1} in more detail. Choose δ_{1} very small and let S_{1} = D(0, r_{1}), where r_{1} = \frac{1}{2} − δ_{1}. Obviously harmonic measure on S_{1} is just normalized Lebesgue measure on its boundary. Choose a large integer m_{1} and points {z_{j}} j = 1, …, m_{1} equally spaced on the circle C_{r_{1}}. Choose a continuous function 0 < η(x) < 1 on C_{r_{1}} and let I_{j} be an arc of ∂S_{1} of angle measure η(z_{j})2π/m_{1} and let {J_{j}} be the complementary arcs. For the first step of the construction we can take η(x) = η_{1} to be a constant for simiplcity, but in later steps we will have to use non-constant η’s.

Fix some 0 < τ_{1} < 1 and for each arc of the form I_{j} with endpoints {p, q}, choose a countable collection of points E = {w_{k}^{j}} ⊂ I_{j}, accumulating only at the endpoints of I_{j}, so that for any z ∈ I_{j},

(7.1) dist(z, E) ≤ τ_{1}dist(z, {p, q}).

Let the components of I_{j} \ E be denoted {I_{k}^{j}}. For each I_{k}^{j}, consider the (infinitely connected) planar domain 𝔻 \ E and the universal cover of the domain. Take a copy of the arc I_{k}^{j} in the universal cover; it is on the boundary of a disk D in the universal cover which covers D(0, \frac{1}{2}). The arc cuts the universal cover into two

30 CHRISTOPHER J. BISHOP

components and we let R_{k}^{j} denote the component which does not contain D. For each interval I_{k}^{j}, we attach a copy of R_{k}^{j} to S_{1} along the arc I_{k}^{j}.

For the intervals {J_{j}} we follow the same procedure, defining a set E ⊂ J_{j} and sub intervals \{J_{k}^{j}\}. The only difference is that except that D(0, \frac{1}{2}) is replaced by D(0, 1), i.e., we attach a component of the universal cover of D(0, 1) \ E, cut along J_{k}^{j}. Doing this for all j and k gives the surface R_{1}. The harmonic measure for R_{1} is now supported on C_{1/2} ∪ C_{1}, (the rest of the ideal boundary covers a countable set, so has zero measure) so we only need to check that it is still close to radial on C_{1/2}.

Now we want to discuss the two main estimates for describing the harmonic measure of R_{1}. The first says that a continuous convolution of the Poisson kernel is well approximated by a discrete version, if the sample points are sufficiently close together. The second says that the harmonic measure of I intervals is small when viewed from a J interval and vice versa.

Suppose D(0, r) is a disk and g is a continuous function on a smaller circle C_{s}, s < r. The balayage of g onto the circle C_{r} is

Bg(θ) = ∫_{0}^{2π} g(se^{it})P_{se^{it}}(θ)dt,

where P_{z}(θ) is the Poisson kernel for D(0, r) with respect to the point z.

Lemma 7.1. With the intervals {I_{j}} defined as above, and F = ∪_{j}I_{j}, for any continuous 0 < g < 1 on the circle C_{s}

B(gχ_{F})(θ) = ∫_{F} g(se^{it})P_{se^{it}}(θ)dt → B(gη)(θ),

uniformly as m_{1} → ∞.

ORTHOGONAL FUNCTIONS IN H^{∞} 31

Proof. Let K_{j} be the interval on C_{s} centered at z_{j} of angle measure 2π/m_{1} (choose them to be half-open, so that they form a disjoint cover of the circle). Define the piecewise constant functions a(x), b(x) on C_{r_{1}} by

a(x) = ∑_{j} χ_{K_{j}}(x)η(z_{j}),

b(x, θ) = ∑_{j} χ_{K_{j}}(x)g(z_{j})P_{z_{j}}(θ),

and define

A(m_{1}) = ∥η(z) − a(z)∥_{∞},
B(m_{1}) = ∥g(x)P_{x}(θ) − b(x, θ)∥_{∞}.

It is clear that by uniform continuity, both quantities tend to zero as m_{1} → ∞. Thus by using the fact that χ_{F} (x) − a(x) has mean value zero on each interval K_{j} where b(x, θ) is constant in x we get

|B(gχ_{F})(θ) − B(gη)(θ)|
= | ∫_{0}^{2π}(g(se^{it})P_{se^{it}}(θ) − b(se^{it}, θ) + b(se^{it}, θ))(χ_{F} − η(se^{it}))dt|
≤ B(m_{1}) ∫_{0}^{2π} |(χ_{F} − η(se^{it})|dt + ∫_{0}^{2π} b(se^{it}, θ)|a(se^{it}) − η(se^{it})|dt
≤ 2πB(m_{1}) + A(m_{1}) max |b|.

This clearly tends to zero as m_{1} → ∞, as desired.

Now for the second estimate. Next we want to show that the harmonic measure of C_{1/2} is much larger than that of C_{1} with respect to a point z ∈ I_{k}^{j}. More precisely,

Lemma 7.2. Suppose z ∈ I_{k}^{j} and suppose γ is a circular arc in S_{1} with endpoints in the corresponding set E such that dist(γ, z) ≃ dist(z, {p, q}) and which separates z from all the J-intervals. Let Ω be the component of R_{1} \γ which contains z. Then ω(z, γ, Ω) tends to 0 as τ_{1} does.

32 CHRISTOPHER J. BISHOP

Proof. Standard estimates of hyperbolic metric imply that γ is within a bounded hyperbolic distance of a geodesic in R_{1} and that the hyperbolic distance from γ to z is at least C log τ_{1}^{−1}. Lifted to the disk, this implies the harmonic measure of γ with respect to z is ≤ exp(C log τ_{1}) ≤ τ_{1}^{α}, for some α > 0, as desired. Obviously, the same estimate holds if we reverse the rolls of the I and J intervals.

The previous result has a simple explaination in terms of Brownian motion. Consider a Brownian motion on the Riemann surface started at z and run until it either hits γ or leaves R_{1}. The path will only hit γ if it stays on the correct sheet of R_{1}, but this is extremely unlikely because it will cross the arc I_{j} many times and each time it has a certain chance (which is large if τ is small) of becoming “tangled” and ending up on the wrong sheet.

We can now show that the harmonic measure of R_{1} on the circle C_{1/2} can be taken as close to a multiple of Lebesgue measure as we wish (depending on our choices of m_{1}, τ_{1} and η). The harmonic measure of R_{1} on the circle C_{1/2} will be the balayage of the harmonic measure of S_{1} restricted to the I intervals, with an error bounded by Cτ_{1}^{α}. The harmonic measure is (normalized) angle measure restricted to the I-intervals. Thus if m_{1} is large enough the harmonic measure on C_{1/2} will be of the form \frac{1}{2π} g_{1}(x)dθ, with g_{1} as close to a constant as we wish. Take \frac{1}{2} + \frac{1}{10} ≤ g_{1}(x) ≤ \frac{1}{2} + \frac{3}{10}, to be concrete.

Now suppose we have constructed R_{n−1}. To construct R_{n}, we follow the method above. We start passing to a subsurface S_{n} ⊂ R_{n−1} where the boundary circles over C_{1/2} are replaced by boundaries over C_{1/2−δ_{n}}. The parameter δ_{n} is chosen so small that every component of R_{n−1}\S_{n} is a regular cover of the annulus \frac{1}{2} −δ_{n} < |z| < \frac{1}{2}\} (which will be possible by the construction of R_{n−1}) and so that harmonic measure

ORTHOGONAL FUNCTIONS IN H^{∞} 33

μ_{S_{n}} on S_{n} is very close to harmonic measure on R_{n−1}, say

| ∫ φd(μ_{S_{n}} − μ_{R_{n−1}})| ≤ 2^{−n}, (7.2)

for every smooth φ with gradient bounded by n.

As before we choose m_{n} equally spaced points \{z_{j}^{n}\} points on C_{n} = C_{\frac{1}{2} −δ_{n}} and define intervals {I_{j}^{n}} of C_{n} centered at these points of angle measure 2πη_{n}(z_{j}^{n})/m_{n}, where

η_{n}(x) = (\frac{1}{2} + \frac{2}{10^{n}})/g_{n−1}(x).

The complementary intervals are denoted {J_{j}^{n}}. We choose a very small τ_{n} and sets E in each interval which satisfies (7.1) with τ_{n}. We then attach copies D(0, \frac{1}{2}) \ E to the copies of the I intervals in ∂S_{n} and copies of D(0, 1) \ E to the J intervals. Then if we choose δ_{n} and τ_{n} small enough and m_{n} large enough, we can get the harmonic measure of R_{n} over C_{1/2} to be g_{n}(x)\frac{dθ}{2π} with g_{n} as close to g_{n−1}η_{n} as we wish, say

\frac{1}{2} + \frac{1}{10^{n}} ≤ g_{n} ≤ \frac{1}{2} + \frac{3}{10^{n}} .

Continuing in this way we can clearly construct a sequence {R_{n}} of Riemann surfaces so that the harmonic measures over C_{1/2} converge to a multiple of Lebesgue measure. This almost finishes the proof, except that the surfaces {R_{n}} are not nested by inclusion. However, the subsurfaces {S_{n}} constructed as part of the induction are nested and their union is also R. Hence their harmonic measures converge to that of R. By (7.2), the weak limit for the measures on {S_{n}} and {R_{n}} must be the same, so we are done.

The same proof shows that we can build an f ∈ H^{∞} so that μ_{f}|_{C_{1/2}} = \frac{1}{2π} gdθ for any continuous g with 0 ≤ g < 1 (or any g which is the decreasing limit of such

34 CHRISTOPHER J. BISHOP

functions). Similarly, the circle can be replaced by any smooth curve γ, and g by a continuous function such that gds ≤ dω(0, ·, 𝔻 \ γ).

The construction in this section clearly generalizes as follows.

Lemma 7.3. Suppose R is a Riemann surface build by attaching subdomains of 𝔻 along boundary arcs. Let Π denote the corresponding projection of R into the plane. Suppose Π(∂R) hits C_{r} and there is a δ > 0 so that every component of Π^{−1}(C_{r}) in ∂R is the boundary of a domain in R which is a regular cover of the annulus {r − δ < |z| < r} (or {r < |z| < r + δ}). Suppose the harmonic measure of R over C_{r}, projects to a measure of the form \frac{1}{2π} gdθ on C_{r} where 0 < g < 1. Choose s < r very close to r (or s > r). Suppose we are given N functions {η_{k}} so that 0 < η_{k} < 1. Choose a large integer m and choose mN equally spaced points {z_{i}} on C_{s}. Let I_{j}^{k} be the interval of length 2πη_{k}(z_{k+jN})/mN centered at z_{k+jN}. Let J_{i} denote the components of C_{s} \ ∪_{j,k}I_{j}^{k}. Choose a small τ and choose sets E satisfying (7.1) in every interval. For k = 0, …, N, choose s_{k} < s < r_{k}. For each each arc in ∂R projecting to I_{j}^{k} attach a copy of A_{k} \ E = {s_{k} < |z| < r_{k}} \ E . To each arc projecting to a J_{i} attach a copy of A_{0} \ E = {s_{0} < |z| < r_{0}} \ E. Then if s is close enough to r, m is large enough and τ is small enough, then the projected harmonic measure of the new surface S on ∂S \ R is as close to ∑_{k} B_{k}(η_{k}g) as we wish, where B_{k} denotes balayage from C_{s} onto ∂A_{k}.

For the proof of Theorem 1.1, we can always take s_{k} = 0, i.e., we can attach disks instead of annuli. Only for the proof of Corollary 1.4, will we have to attach proper annuli.

ORTHOGONAL FUNCTIONS IN H^{∞} 35

8. Theorem 1.1 on an annulus

In this section we will show that any radially symmetric probability measure μ supported in the annulus \{z : \frac{1}{2} ≤ |z| ≤ 1\} is of the form μ_{f} for some f ∈ U_{0}, and in the next section we will extend this to the general case.

First some notation. For 0 < r < s < 1 let A(r, s) = {z : r ≤ |z| < s}. When s = 1, we let A(r, 1) = {z : r ≤ |z| ≤ 1}. For 0 < r < 1, let μ(r) = μ(A(0, r)). Let r^{0} = \frac{1}{2}, r_{0}^{1} = \frac{1}{2}, r_{1}^{1} = \frac{3}{4}, and more generally, r_{k}^{n} = \frac{1}{2} + k2^{−n−1}. Let μ_{k}^{n} = μ(A(r_{k}^{n}, r_{k+1}^{n})). Also let C_{k}^{n} = C_{r_{k}^{n}}.

By rescaling we may assume 𝕋 ⊂ supp(μ) ⊂ \overline{𝔻} and hence that μ^{n}2^{n} − 1 > 0 for all n.

We will construct a sequence of Riemann surfaces R_{0} ⊂ R_{1} ⊂ … such that the corresponding measure μ_{n} is supported on the union of 2^{n} circles, ∪_{k=0}^{2^{n}−1}C_{r_{k}^{n}}. On C_{r_{k}^{n}} the measure μ_{n} will have the form \frac{1}{2π} g_{k}^{n}dθ where μ_{k}^{n} < g_{k}^{n} ≤ μ_{k}^{n} + ϵ_{n} (8.1) for k = 0, …, 2^{n} − 2 and any ϵ_{n} > 0 we choose, and for k = 2^{n} − 1 we have μ_{2^{n+1}−2}^{n+1} < g_{k}^{n} ≤ μ_{2^{n}−1}^{n}. (8.2) Recall that since μ_{n} is a probability measure, if it gives too much mass to the first 2^{n} − 1 annuli, then it must give too little to the last one. It is obvious that such measures {μ_{n}} converge weakly to μ, so by the argument at the end of the previous section, the μ_{f} corresponding to the limiting surface R = ∪_{n}R_{n} must equal μ.

Thus it only remains to construct the surfaces. As in the previous section we start with R_{0} = D(0, \frac{1}{2}). To construct R_{1}, we will proceed exactly as in the previous section, except that instead of redirecting harmonic measure to the unit circle, we send it to the circle C_{3/4}. The estimates are all the same so we can obtain a surface

36 CHRISTOPHER J. BISHOP

R_{1} so that the corresponding μ_{1} is supported on C_{1/2} ∪ C_{3/4} and is of the form \frac{1}{2π} g_{0}^{1}dθ on C_{0}^{1} and \frac{1}{2π} g_{1}^{1}dθ on C_{1}^{1} where

μ_{0}^{1} < g_{0}^{1} < μ_{0}^{1} + ϵ_{1},
μ_{2}^{2} < g_{1}^{1} < μ_{1}^{1},

for any ϵ_{1} > 0 we choose.

To construct R_{n+1}, n ≥ 1, we just make one small change. The mass on the outermost circle C_{2^{n}−1}^{n} is redistributed to itself C_{2^{n}−1}^{n} = C_{2^{n+1}−2}^{n+1} and the outermost circle of the next stage, C_{2^{n+1}−1}^{n+1}. The mass of any other circle C_{j}^{n} is redistributed to three circles; itself, C_{j}^{n} = C_{2j}^{n+1}, the next circle out in the next generation, C_{2j+1}^{n+1} and the outermost circle of the next generation, C_{2^{n+1}−1}^{n+1}.

To do this we let C̃_{j}^{n} be the circle of radius r_{j}^{n} − δ_{n}, where δ_{n} < 2^{−n−10} is chosen so small that the harmonic measure on S_{n} (the subsurface of R_{n} bounded by the lifts of C̃_{j}^{n} which contains 0 and hence contains S_{n−1}) is as close as we wish to harmonic measure on R_{n−1}, i.e., it satisfies (7.2). We now just apply the construction of Lemma 7.3, with N = 2, s_{0} = s_{1} = s_{2} = 0, r_{0} = r_{2j}^{n+1}, r_{1} = r_{2j+1}^{n+1} and r_{2} = r_{2^{n+1}−1}^{n+1}. More precisely, suppose we have two continuous functions η_{1}, η_{2} on C̃_{j}^{n} such that η_{1} + η_{2} < 2 and m_{n} equidistributed points {z_{j}} on ∂S_{n}, choose intervals centered at these points. However, instead of having two types of intervals, we will have three; {I_{j}} of angle measure 2πη_{1}(θ)/m_{n} centered at z_{j} for j even, {K_{j}} of angle measure 2πη_{2}(θ)/m_{n} centered at z − j for j odd, and the remaining intervals {J_{j}}. We choose a very small τ_{n} and a countable set E in each interval which satisfies (7.1). Then along type I intervals we attach a copy of the universal cover of D(0, r_{j}^{n}) \ E; along the type K intervals we attach the universal cover of D(0, r_{2j+1}^{n+1}) \ E; and along the type J intervals we attach the universal cover of

ORTHOGONAL FUNCTIONS IN H^{∞} 37

D(0, r_{2^{n+1}−1}^{n+1}) \ E. Then if we take m_{n} large enough and δ_{n}, τ_{n} small enough, the harmonic measure of the surface R_{n+1} over C_{j}^{n} will be as close to the balayage of η_{1}g_{j}^{n} onto C_{j}^{n} as we wish and the harmonic measure over C_{2j+1}^{n+1} will be as close to the balayage of η_{2}g_{j}^{n} onto that circle as we wish, independent of what changes we make at circles other than C_{j}^{n}.

Now do a similar construction around each of the circles C_{j}^{n}, j = 0, … 2^{n} −2. At the outer most circle C_{2^{n}−1}^{n} we redirect the measure to only two circles; itself and the outermost circle of the next generation, C_{2^{n+1}−1}^{n+1}. By construction, condition (8.1) holds with any constant ϵ_{n} we want. Then by Lemma 2.6, μ_{n} on the outermost circle must be normalized Lebesgue measure minus the balayage of the measures on the inner circles. Since these measures have total mass is as close to (but larger than) μ(A(\frac{1}{2}, r_{2^{n}−1}^{n})) = ∑_{j=0}^{2^{n}−2} μ_{j}^{n} as we wish, the mass of the outermost circle is as close to (but less than) μ(A(r_{2^{n}−1}^{n}, 1) = μ_{2^{n}−1}^{n}. Moreover, since the measures on the inner circles are as close to radial as we wish, so is their balayage onto the outermost circle and hence so is μ_{f} restricted to the outermost circle (this condition defines our choice of ϵ_{n}). This gives condition (8.2). The proof is completed by taking limits just as in the two previous sections.

9. Theorem 1.1 on the whole disk

To complete the proof of Theorem 1.1 we need to show how to obtain any measure satisfying (1.1). As in the last section we can assume 𝕋 ⊂ supp(μ) ⊂ \overline{𝔻}. We can also simplify the situation slightly by observing that it is enough to assume that most of the mass of μ lives away from the origin, i.e.,

∫ log \frac{1}{|z|}dμ ≤ δ. (9.1)

38 CHRISTOPHER J. BISHOP

This is because for f ∈ H^{∞} the measure μ_{f^{d}} is the push forward under z → z^{d} of the measure μ_{f} and so

∫ log \frac{1}{|z|}dμ_{f} = \frac{1}{d} ∫ log \frac{1}{|z|}dμ_{f^{d}}.

By taking d large we can make the right hand side as small as we wish. Thus for any μ on the disk satisfying (1.1), it suffices to construct an f corresponding to the pull back of μ under z^{d}, i.e., it suffices to consider only measures satisfying (9.1) for any δ > 0 we choose.

Start by taking R_{0} = D(0, \frac{1}{4}). Let r_{n} = 2^{−n} for n = 0, 1, 2, … and μ_{n} = μ(A(r_{n}, r_{n−1})). Then

∑_{n>2}(n − 1)(log 2)μ_{n} ≤ ∫ log \frac{1}{|z|}dμ ≤ δ, (9.2)

so

μ_{n} ≤ \frac{δ}{(log 2)(n − 1)} ≤ \frac{δ′}{n} , (9.3)

where δ′ is as small as we wish.

We need two simple facts about harmonic measure on an annulus.

Lemma 9.1. Suppose A = {z : s < |z| < r} and s < t < r. Then ω(z, C_{s}, A) = u_{s,r}(z) = (log |z| − log r)/(log s − log r) for any z with |z| = t.

Proof. This is immediate since the given function is harmonic in A, equals 1 on C_{s} and equals 0 on C_{r}.

Lemma 9.2. Suppose s, t, r and A are as in previous lemma. Then if t ≥ 2s, there is an M < ∞ (independent of s, t, r) such that for |z| = t, ω(z, ·, A) restricted to C_{s} has the form \frac{1}{2π} gdθ and g satisfies max_{C_{s}} g ≤ M min_{C_{s}} g.

ORTHOGONAL FUNCTIONS IN H^{∞} 39

Proof. Recall that harmonic measure on ∂A is the normal derivative of Green’s function G with pole at z. Let t′ = \frac{2}{3}t > s. By Harnack’s inequality there is an M so that max_{C_{t′}} G ≤ M min_{C_{t′}} G, and hence there is a constant CE so that

C(1 − u_{s,t′}) ≤ G ≤ MC(1 − u_{s,t′}),

on {s < |z| < t′}. Since the normal derivative of u_{s,r′} is constant on C_{s} (since u is radial), this implies the normal derivative of G on C_{s} is trapped between two constants A and MA, as desired.

Consider the annulus A_{n} = {2^{−n} < |z| < 2^{−1}}, n = 3, 4, … and a point z with |z| = \frac{1}{3}. The two previous results imply there is an constant B so that harmonic measure for A on the circle C_{2^{−n}} is of the form \frac{1}{2π} gdθ where g ≥ \frac{B}{n}, for n ≥ 3. By (9.2) we can assume μ is chosen so that ∑_{n} nμ_{n} ≤ (2B)^{−1}. Thus ∑_{n} Bnμ_{n} ≤ \frac{1}{2}. Thus it is possible to choose disjoint, adjacent intervals {I_{n}}, n = 2, 3, 4 …, on C_{1/4} of angle measure 4πnμ_{n}/B. In each interval I_{n} choose a countable set E_{n} satisfying the “thickness” condition (7.1) with some τ_{n}, and attach to I_{n} a copy of the universal cover of A_{n+1} \ E_{n}. The resulting Riemann surface has harmonic measures supported over the union of circles ∪_{n}C_{2^{−n}}, n = 1, 3, 4, 5, . . .. Moreover, if we choose τ_{n} → 0 quickly enough, the harmonic measure of the circles corresponding to n = 3, 4, 5 …, is of the form \frac{1}{2π} g_{n}dθ with g_{n} > μ_{n−1} (but might not be close to radial).

For each such circle C_{2^{−n}}, choose I and J intervals in the usual way and attach copies of D(0, \frac{1}{2}) \ E and D(0, 2^{−n}) \ E respectively. As we have seen before, we can choose η, m and τ so that the harmonic measure \frac{1}{2π} g_{n}dθ on C_{2^{−n}} is as close to (but larger than) μ_{n} as we wish. Using Lemma 2.6 the harmonic measure of C_{1/2} will be as close to (but less than) μ_{1} as we wish, in particular, it is larger than

40 CHRISTOPHER J. BISHOP

μ(\{\frac{1}{2} ≤ |z| < \frac{3}{4}\}) (this is where we use the assumption that 𝕋 is in the support of μ).

The rest of the proof is now the same as the previous section. On each annulus we redistribute the harmonic measure from the circle into the annulus, sending any “extra” measure to the outermost circle, C_{1−2^{−n}}. In the limit, we obtain the desired measure μ.

10. An example which is almost an outer function In this section we will construct an orthogonal function f whose only inner factor

is the required zero at 0, i.e., f(z)/z is outer. We will construct f so that 0 is the only zero of f; thus f(z)/z has no Blaschke factor. In order to prove it has no singular inner factor, recall that if f(z)/z = gh with g outer and h a non trivial singular inner function then

log |f|^{−1} = log |g|^{−1} + log |h|^{−1},

and that the first term on the right is the Possion integral of its boundary values on 𝕋, but the second term is the Poisson integral of a singular measure on 𝕋 and has boundary value zero almost everywhere on 𝕋. Let H_{ϵ} = {z ∈ 𝔻 : |h(z)| < ϵ} and F_{ϵ} = \{z ∈ 𝔻 : |f(z)| < ϵ\}. Since log |h(0)|^{−1} = (log \frac{1}{ϵ})ω(0, H_{ϵ}, 𝔻 \ H_{ϵ}), we deduce

ω(0, H_{ϵ}, 𝔻 \ H_{ϵ}) ≥ C/ log \frac{1}{ϵ},

where C = log |h(0)|^{−1} and consequently, since H_{ϵ} ⊂ F_{ϵ},

ω(0, F_{ϵ}, 𝔻 \ F_{ϵ}) ≥ C/ log \frac{1}{ϵ}. (10.1)

We will construct R so that the harmonic measure of \{z ∈ R\D(0, \frac{1}{2}) : |z| ≤ 2^{−n}\} has harmonic measure (in R, with respect to 0) less than λ^{n} for some λ < 1. This contradicts (10.1), so the covering map has no singular inner factor.

ORTHOGONAL FUNCTIONS IN H^{∞} 41

Since we have already seen several constructions of this type in great detail, I will only sketch the construction. Start with R_{0} = D(0, \frac{1}{2}). Divide C_{1/2} into a finite collection of intervals {I_{n}} and in each choose a set E satisfying (7.1). Along each interval attach a copy of \{\frac{1}{4} < |z| < 1\} \ E. This gives R_{2}.

Lemmas 9.1 and 9.2 of the previous section imply that harmonic measure of R_{2} over C_{1/4} is of the form \frac{1}{2π} gdθ where the max of g is bounded by a universal constant times the minimum. Thus there is a constant c < min(g) and a λ < 1 so that

∫ (g − c)dθ ≤ λ ∫ gdθ.

In other words, we can truncate g to be a constant and still retain a fixed fraction of the harmonic measure.

Now do the (by now) standard construction of I and J intervals on C_{\frac{1}{4} +δ} attaching copies of \{\frac{1}{8} < |z| < 1\} and \{\frac{1}{4} < |z| < 1\} respectively, so that the new harmonic measure on C_{1/4} is very close to radially symmetric (say within ϵ_{1} of constant) and has mass at least (1 − λ) times the previous mass.

At the next stage we do the construction near both circles C_{1/4} and C_{1/8}. At C_{1/8} we repeat the process of the previous paragraph, making the harmonic measure above C_{1/8} as close to radial as we wish, while retaining at least (1 − λ) of the total mass, transferring the excess to C_{1} and C_{1/16}. On C_{1/4} we only make the measure within ϵ_{2} of constant (while losing at most ϵ_{1} of the mass), the excess being transferred to C_{1/8} and C_{1}.

We now iterate the process in the obvious way. At the nth stage we have a surface R_{n} which only covers the origin once, and so that the harmonic measure is supported on the circles {C_{2^{−k}}} with the kth circle getting mass at most λ^{k}. Thus the same is true for the limiting measure μ. Hence the harmonic measure of the set

{z ∈ R \ R_{0} : |z| < 2^{−n}} has harmonic measure less than Cλ^{n} in R. This proves

that f(z)/z is outer.

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C.J. Bishop, Mathematics Department, SUNY at Stony Brook, Stony Brook, NY 11794-3651

   E-mail address: bishop@math.sunysb.edu