Contemporary Mathematics

The law of the iterated logarithm for Kleinian groups

Christopher J. Bishop and Peter W. Jones

Abstract. Suppose G is an analytically finite, geometrically infinite Kleinian group and that there is a lower bound on the injectivity radius for M = 𝔹/G. Then the limit set Λ has positive Hausdorff measure with respect to the gauge function φ(t) = t^{2}\sqrt{log \frac{1}{t} log log log \frac{1}{t}} . If, in addition, the group is topological tame and Λ ≠ S^{2}, then the limit set has finite measure with respect to this gauge. We also obtain corollaries on the size of the conical limit set, the existence and ergodicity of conformal densities, and the differentiability of quasiconformal conjugacies.

1. Statement of results

Suppose G is a Kleinian group acting on the hyperbolic three ball 𝔹 and let Λ and Λ_{c} denote the limit set and conical limit set respectively (all definitions will be reviewed in the next section). The purpose of this paper is to prove the following result.

Theorem 1.1. Suppose G is an analytically finite but geometrically infinite Kleinian group such that the injectivity radius of M = 𝔹/G is bounded away from zero. Then Λ \ Λ_{c} has positive Hausdorff measure with respect to the function

φ(t) = t^{2}\sqrt{log \frac{1}{t} log log log \frac{1}{t}}. If, in addition, G is topologically tame and Λ ≠ S^{2}, then Λ has finite Hausdorff φ-measure and Λ_{c} has zero φ-measure.

This had been conjectured for certain examples by Sullivan in [26] and is extremely close to results of Makarov for conformal mappings [16]. The topological tameness is needed for the second conclusion since there are analytically finite groups with injectivity radius bounded away from zero such that the limit sets are nowhere dense and have positive area [20]. The first statement of the theorem

\begin{matrix}c \\ ◯ \end{matrix}0000 (copyright holder)

1991 Mathematics Subject Classification. 30F40 ; Secondary 28A78, 30C35, 31A15 .

Key words and phrases. Hausdorff dimension, Kleinian groups, limit set, degenerate group, geometrically infinite, conical limit set, Poincaré exponent, critical exponent, Martin boundary, conformal density.

The first author is partially supported by NSF Grant DMS 95-00577 and an Alfred P. Sloan research fellowship. The second author is partially supported by NSF grant DMS-94-23746.

strengthens the result from [5] that the limit set has dimension 2. The final statement strengthens the results of Bonahon, Canary and Thurston that limit sets of such groups have zero area (see [6]).

The lower bound in Theorem 1.1 can be easily deduced from known results of Makarov on laws of the iterated logarithm for Bloch functions and martingales [17]. The upper bound is in some sense new, but we point out that Makarov (Proposition 4.7 of [17]) proved a similar result in an analogous setting (he considered the support of Kahane’s measure). His results use classical estimates for random walks which do not seem to apply directly to our case.

If the injectivity radius of M is bounded away from zero, then G has no parabolic elements and in this case G is geometrically finite iff Λ = Λ_{c} (see [2], [4]). Thus

Corollary 1.2. Suppose G is an analytically finite Kleinian group such that the injectivity radius of M = 𝔹/G is bounded away from zero. Then Λ \ Λ_{c} is either empty or has dimension 2, depending on whether G is geometrically finite or infinite.

A conformal density of dimensional α is a positive measure μ on Λ with the property that

μ(g(E)) = ∫_{E} |g′|^{α}dμ, for all g ∈ G.

From this it follows that the Poisson extension of μ is a G-invariant, positive harmonic function on 𝔹. There is always a conformal density of dimension δ, where δ is the critical exponent of G (e.g., [23], Chapter 4). If G is topologically tame and Λ = S^{2} then δ = 2 and Lebesgue measure is the unique conformal density.

Corollary 1.3. Suppose G is topologically tame, geometrically infinite, has a positive lower bound for the injectivity radius, and Λ ≠ S^{2}. Then μ = H^{φ} is a conformal density of dimension 2 where H^{φ} is the Hausdorff measure associated to the guage function φ(t) = t^{2}\sqrt{log \frac{1}{t} log log log \frac{1}{t}}.

If M has only one geometrically infinite end and Λ ≠ S^{2}, then, up to constant multiples, there is only one conformal density and G will be ergodic with respect to this measure. If M has more than one geometrically infinite end then there will be a different conformal density associated to each geometrically infinite end of M and G will be ergodic with respect to each of them (Corollary 9.6). Examples of groups with one geometrically infinite end are well known (e.g., [22]) and examples with more than one such end can be obtained by combination theorems (e.g., [20]). In the latter case, H^{φ} on Λ is the sum of distinct conformal densities for each geometrically infinite end and thus is a conformal density for G which is not ergodic. Moreover, the measures corresponding to different ends are mutually singular, so that Theorem 4.2.1 of [23] is false (it says that a conformal density is ergodic iff it is unique up to positive multiples).

Since conformal densities for G correspond to positive harmonic functions on M, these remarks can also be interpreted in terms of the Martin boundary of M (e.g., [19]) Suppose G is topologically tame, has positive lower bound for the injectivity radius and Λ ≠ S^{2}. Then the Martin boundary for M is S ∪ E, where S is the compact surface (S^{2} \ Λ)/G and E is a finite set, one point corresponding to each geometrically infinite end of M.

It is a classical result of Hopf [12] that the conical limit set of a Fuchsian group has (normalized) Lebesgue measure either zero or one, and Sullivan proved the same in higher dimensions [26]. Λ_{c} always has either zero or full measure for a conformal density of dimension α (e.g., [23], Theorem 4.4.4), and the measure must be zero if G is convergence type for exponent α (e.g., [23], Theorem 4.4.1). If Λ ≠ S^{2} then G must be convergence type for the exponent 2.

Corollary 1.4. Suppose G is a topologically tame Kleinian group such that the injectivity radius of M = 𝔹/G is bounded away from zero. Then Λ_{c} has either full 2-dimensional Lebesgue measure or zero Hausdorff measure with respect to φ(t) = t^{2}\sqrt{log \frac{1}{t} log log log \frac{1}{t}}, depending on whether Λ = S^{2} or not. In the latter case, Λ_{c} even has zero Hausdorff measure with respect to the functions ψ(t) = t^{2} log \frac{1}{t}(log log \frac{1}{t})^{−1−ϵ}, for any ϵ > 0.

The geodesic flow is not ergodic for the groups we are considering since the conical limit set has zero μ measure (Theorem 8.3.5, [23]). One can also see this directly as follows. Given r > 0 one can consider the set of doubly infinite geodesics in C(M) which are never closer than distance r to ∂C(M). It is easy to check that such a set is G invariant and that we can choose r so that it has μ × μ measure strictly between 0 and 1.

Suppose G_{1} and G_{2} are two groups as above which are quasiconformally conjugate via a map Φ. McMullen asked if the image of a conformal density for G_{1} is singular or absolutely continuous with respect to a density for G_{2}. We shall show the latter is true.

Corollary 1.5. Suppose G_{1} and G_{2} are two groups which are topologically tame, geometrically infinite and have a positive lower bound for the injectivity radius. Let μ_{1} and μ_{2} be H^{φ} restricted to Λ_{1} and Λ_{2} respectively. Suppose also that the groups are quasiconformally conjugate, i.e., there is a quasiconformal map Φ on S^{2} so that Φ^{−1} ∘ G_{1} ∘ Φ = G_{2}. Then Φ^{∗}μ_{1} is absolutely continuous with respect to μ_{2}. In fact, Φ is differentiable and conformal at μ_{1} almost every point.

Here Φ^{∗}μ(E) = μ(Φ^{−1}(E)). The proof of the corollary uses McMullen’s result that the quasiconformal conjugacy on S^{2} extends to be a quasi-isometry on M with constant approaching 1 exponentially fast in the geometrically infinite ends, and the fact that μ almost every geodesic escapes into a geometrically infinite end sufficiently quickly. McMullen calls x ∈ Λ a deep point if the geodesic γ ending at x satisfies

(1.1)  lim inf_{t→∞} \frac{dist(γ(t), M \ C(M))}{t} > 0,

and proves the quasiconformal conjugacy Φ is differentiable at such points. In fact, one only needs

(1.2)  lim inf_{t→∞} \frac{dist(γ(t), M \ C(M))}{log t} = ∞,

to deduce Φ is differentiable at x. We shall show that for μ almost every geodesic,

(1.3)  lim inf_{t→∞} \frac{dist(γ(t), M \ C(M))}{\sqrt{t} log^{−2} t} = ∞,

which shows Φ is differentiable μ almost everywhere. The deep points in Λ are a set of dimension zero, and Corollary 1.5 improves this by saying Φ is differentiable on a subset of Λ having dimension 2.

We will now outline the proof of Theorem 1.1. The basic idea is that the lower bound for the dimension is deduced from the existence of a positive harmonic function U on M with bounded gradient, i.e., a harmonic function which can grow at most linearly in the geometrically infinite end of M. The upper bound for dimension can be deduced if we know that |∇U| is bounded away from zero (at least on average) in the geometrically infinite end, i.e., the growth is at least linear.

The function U is constructed by taking the Greens function on M and moving the pole to infinity through a geometrically infinite end, while normalizing the function to have value 1 at some fixed base point. To show the limit U has bounded gradient we need to show the Greens function does (at least away from the pole). A version of the following result is proved in Sullivan’s paper [26]; we shall give a different proof in this paper which applies in greater generality.

Lemma 1.6. Suppose M = 𝔹/G and that Greens function G(w, z) exists on M. Then

sup_{z∈M} sup_{ρ(w,z)≥1} |∇_{w}G(w, z)| ≤ \frac{A}{vol(B(w, 1))}

i.e., the hyperbolic gradient of Green’s function is bounded outside a neighborhood of the pole by the reciprocal of the local volume. In particular, if the injectivity radius is bounded away from zero, then the gradient is uniformly bounded away from the pole.

By moving the pole of Greens function off to infinity and passing to a limit we can deduce the following.

Lemma 1.7. Suppose G is analytically finite, M = 𝔹/G has injectivity radius bounded below by ϵ > 0 and that Greens function G(w, z) exists on M. Then there exists a positive harmonic function U on M such that

1. sup_{z∈M} |∇U(z)| ≤ 1,

2. U tends to zero in the geometrically finite ends of M. If, in addition, G is topologically tame then for any a_{0} > 0 there are constants a_{1} and a_{2} so that

∫_{B(z,a_{1})} |∇U|^{2}dV ≥ a_{2},

for every z such that dist(z, C(M)) ≤ a_{0}. Moreover, U(z) tends to +∞ in the geometrically infinite ends as dist(z, ∂C(M)) → ∞.

U lifts to a hyperbolically harmonic function (which we shall also call U) on 𝔹, and this function must be the Poisson integral of some positive measure μ which is supported on the limit set. Since U has bounded hyperbolic gradient, it satisfies the estimate

|U(z)| ≤ U(0) + A log \frac{1}{1 − |z|}.

The explicit formula for the Poisson integral implies

μ(D(x, r))(1 − r)^{−2} ≤ CU(z),

where x is the radial projection of z onto S^{2} and r = |z|. Thus

μ((D(x, r)) ≤ C(1 − r)^{2} log \frac{1}{1 − r} , ∀x ∈ S^{2}, 0 < r < 1. (1.4)

The mass distribution principle then implies that Λ has positive Hausdorff measure with respect to the function φ_{1}(t) = t^{2} log \frac{1}{t}. This is due to Sullivan [26].

This estimate is easily improved using a law of the iterated logarithm for harmonic functions as found, e.g., in the paper of Chang, Wilson and Wolff [8]. They show that a harmonic function with bounded square function satisfies an exponential integrability estimate. In our case, their estimate says there are absolute constants C_{1}, C_{2} depending only on the supremum of |∇_{H}U|, so that

∫_{S^{2}} exp(\frac{C_{1}}{n} U_{n}^{2}(x))dx ≤ C_{2},

where U_{n}(x) = U((1 − e^{−n})x). Tchebyshev’s inequality then implies

|E_{n}| = |\{x ∈ S^{2} : |U_{n}(x) − U(0)| > λ\sqrt{n}\}| ≤ C_{2} exp(−λ^{2}/C_{1}).

Taking λ = C_{1}C_{2}M log n and doing some arithmetic gives |E_{n}| ≤ n^{−M}. Since |∇U_{n}| ≤ Ce^{n} on S^{2}, the set E_{n} can be covered by Ce^{2n}n^{−M} disks of diameter e^{−n}. Thus the μ measure of E_{n} is bounded by

μ(E_{n}) ≤ Ce^{2n}n^{−M}e^{−2n}n = Cn^{M−1}.

Hence the μ measure of F_{n} = ∪_{k>n}E_{n} is bounded by Cn^{M−2} and hence F = ∩_{n}F_{n} has μ measure zero if M > 2. Thus

lim sup_{n→∞} \frac{|U_{n}(x) − U(0)|}{\sqrt{n log n}} < ∞

on a set of full μ measure. Since U has bounded hyperbolic gradient, this implies

(1.5)  lim sup_{r→1} \frac{|U(rx) − U(0)|}{\sqrt{log \frac{1}{1−r} log log \frac{1}{1−r}}} < ∞,

for μ-almost every x ∈ S^{2}. Thus

μ(D(x, t)) ≤ C_{x}t^{2}\sqrt{log \frac{1}{t} log log \frac{1}{t}},

for μ-almost every x. By the Frostman-Rogers-Taylor theorem, this means that Λ has positive measure for the gauge φ(t) = t^{2}\sqrt{log \frac{1}{t} log log \frac{1}{t}}.

To get Theorem 1.1 we need to improve (1.5) by proving the following

Theorem 1.8. Suppose U is the hyperbolic Poisson integral of a positive measure μ and has bounded hyperbolic gradient in 𝔹. Then for μ-almost every x ∈ S^{2},

(1.6)  lim sup_{r→1} \frac{|U(rx) − U(0)|}{\sqrt{log \frac{1}{1−r} log log log \frac{1}{1−r}}} < ∞.

It follows from the Chang-Wilson-Wolff theorem that this is true for Lebesgue almost every x. However, in our case we already know U has radial limit 0 Lebesgue almost everywhere and we really need the sharper version. This completes the sketch of how we prove the lower bound in Theorem 1.1. Next we describe how to prove the upper bound.

To show that Λ has finite H^{φ} measure, we need to find a cover {Q_{j}} of Λ by arbitrarily small squares so that ∑_{j} φ(diam(Q_{j})) is uniformly bounded. We do this by running a stopping time argument. Move to the upper half-space model ℝ_{+}^{3} and assume Λ is contained in the unit square Q_{0}. First we show there is an c_{1} > 0 so that if U(z) < c_{1} somewhere in the top half of a Carleson cube then the base of the cube does not intersect Λ. Next, we define a cover of Λ inductively as follows. For the first step, choose some integer n_{0} and take all the dyadic squares of side length 2^{−2^{n_{0}}} which intersect Λ. Call this collection G_{n_{0}}.

Fix an ϵ > 0 and for a general n > n_{0} and a collection of squares G_{n}, subdivide each square in the collection into dyadic subsquares of side length 2^{−2^{n+1}}. If U(z) < c_{1} at any point of the top half of the corresponding Carleson cube throw the square away. If U(z) > ϵ\sqrt{2^{n} log n} in the top half of the Carleson cube, put the square into a collection B_{n} for Λ. If neither condition holds, put the square into the new collection G_{n+1}. We stop the construction at some large value n = N − 1 and cover Λ by B ∪ G_{N}, where B = ∪_{n=n_{0}}^{N−1}B_{n}.

For squares in B we have μ(Q) ≥ \frac{C}{ϵ} φ(diam(Q)), so the φ-sum of such squares is uniformly bounded since μ is a finite measure. Thus we need to show that if ϵ is small enough, then the φ sum of squares in G_{N} is uniformly bounded (in fact, we will show it tends to zero). This argument is based on analogous estimates for random walks and Brownian motion (it roughly corresponds to computing the probability that a random walk started at 1 stays positive for n steps and is never larger than max(1, ϵ\sqrt{k log log k}) for n_{0} ≤ k ≤ n). This is the most technical part of the paper.

The remainder of the paper is organized as follows. Section 2: We review definitions and related results. Section 3: We prove some technical estimates used in the following sections. Section 4: We prove the lower bound in Theorem 1.1 Section 5: We prove the upper bound in Theorem 1.1. Section 6: We give an upper bound for the size of Λ_{c}. Section 7: We prove Corollary 1.5. Section 8: We prove Lemma 1.6. Section 9: We prove Lemma 1.7.

2. Definitions

If A, B are quantities that depend on some parameter we write A ≃ B if the ratios B/A and A/B are bounded uniformly independent of the parameter. If E ⊂ ℝ^{d} we let |E|_{d} denote the d-dimensional Lebesgue measure of E and use |E| if the dimension is clear from context. Suppose φ is a increasing continuous function from [0, ∞) to itself such that φ(0) = 0. We define the Hausdorff content of a set E ⊂ ℝ^{2} as

H_{∞}^{φ} = inf{∑ φ(r_{j}) : E ⊂ ∪_{j}D(x_{j}, r_{j})}.

If φ(t) = t^{α} we denote H^{φ} by H^{α}. The Hausdorff dimension of E is

dim(E) = inf{α : H_{∞}^{α}(E) = 0},

and the Hausdorff measure of E is

H^{φ}(E) = lim_{δ→0}[inf{∑ φ(r_{j}) : E ⊂ ∪_{j}D(x_{j}, r_{j}), r_{j} ≤ δ}].

A nth generation dyadic cube in ℝ^{d} is a product of intervals of the form (j2^{−n}, (j + 1)2^{−n}]. The collection of such cubes is denoted D_{n} (or D_{n}^{d} is the dimension is not clear from context). Given a point x ∈ ℝ^{d}, we let Q_{n}(x) denote the unique cube in D_{n} which contains x. If Q ∈ D_{n}, we let ℓ(Q) = 2^{−n} denote the side length of Q. The parent of a dyadic cube Q ∈ D_{n} is the unique dyadic cube in D_{n−1} which contains it; its children are the 2^{d} cubes of D_{n+1} which it contains. It is sometimes convenient to consider the Hausdorff measure defined using only coverings of a set only by dyadic cubes. The resulting content and measure are easily seen to be comparable to the usual definitions, with constants depending only on the dimension d.

If E supports a measure μ such that μ(D(x, r)) ≤ Cr^{α} for every disk D(x, r), then it is trivial to see that H^{α}(E) ≥ C^{−1}μ(E) > 0 and hence dim(E) ≥ α. This is called the mass distribution principle. A more sophisticated version of this is the Frostman-Rogers-Taylor density theorem, [24],

Theorem 2.1. Let μ be a finite measure on [0, 1]^{2} and let A ⊂ [0, 1]^{2} be a closed set. Then

1. If for all x ∈ A,

lim sup_{n→∞} \frac{μ(Q_{n}(x))}{φ(2^{−n})} ≥ C,

then H^{φ}(A) ≤ MC^{−1}μ(A).

2. If for all x ∈ A,

lim sup_{n→∞} \frac{μ(Q_{n}(x))}{φ(2^{−n})} ≤ C,

then H^{φ}(A) ≥ M^{−1}C^{−1}μ(A).

We let 𝔹^{n} denote the unit ball of ℝ^{n} equipped with the hyperbolic metric

dρ = \frac{2|dz|}{1 − |z|^{2}} .

In most of this paper we take n = 3 and simply write 𝔹 instead of 𝔹^{3}. We will also use the upper half space model ℝ_{+}^{n+1} = {(x_{1}, x_{2}, …, x_{n}, y) : y > 0} for hyperbolic space with the metric

dρ = \frac{|dz|}{y}.

The hyperbolic volume in ℝ_{+}^{n+1} will be denoted

dV = \frac{dx_{1}dx_{2} … dx_{n}dy}{y^{n}} .

The hyperbolic Laplacian on 𝔹^{n} is given by

Δ_{H} = \frac{(1 − r^{2})^{2}}{4} [Δ + \frac{2(n − 2)r}{1 − r^{2}} \frac{∂}{∂r}],

and on the upper half space ℝ_{+}^{n}it is

Δ_{H} = y^{2}[Δ − \frac{n − 2}{y} \frac{∂}{∂y}].

The hyperbolic Poisson kernel on 𝔹^{n} is

P (z, ζ) = w_{n}[\frac{1 − |z|^{2}}{|z − ζ|^{2}} ]^{n−1},

where w_{n} is chosen to make the integral 1. On the upper half space ℝ_{+}^{3} the hyperbolic Poisson kernel is

P (z, ζ) = v_{n}[\frac{y}{|z − ζ|^{2}} ]^{n−1},

again with v_{n} chosen to make the integral 1. See Chapter 5 of [23].

A discrete group G of isometries of the hyperbolic metric on 𝔹 is called Kleinian group. Such a group can also be considered as a group of linear fractional transformations on S^{2}. G is called elementary if it contains a finite index Abelian subgroup. In this paper we are only concerned with non-elementary groups. For a non-elementary G, the accumulation of any orbit in 𝔹 ∪ S^{2} is a closed set Λ ⊂ S^{2} which is independent of the particular orbit. This is the limit set. The conical limit set Λ_{c} is the set of points x ∈ Λ for which there is a sequence of orbit points converging to x within a non-tangential cone in 𝔹. It is easy to see that x is a conical limit point iff a geodesic ray in 𝔹 ending at x projects to a geodesic in M = 𝔹/G which returns to some compact set of M infinitely often.

We call Ω = S^{2} \ Λ the ordinary set of G. G acts discontinuously on Ω, and Ω/G is a (branched) Riemann surface. If it is a Riemann surface of finite type (i.e., is a finite union of compact surfaces each with at most a finite number of punctures) we call G analytically finite. The famous Ahlfors finiteness theorem [1], [3] says that any finitely generated Kleinian group is analytically finite.

G is called geometrically finite if there is a finite sided fundamental polygon for G’s action on 𝔹. One can show ([2], [4]) that G is geometrically finite iff every limit point of G is either a conical limit point or a fixed point of some parabolic element of G.

If G is a Kleinian group with limit set Λ we let C(Λ) ⊂ 𝔹 denote the convex hull of Λ and let C(M) = C(Λ)/G ⊂ M = 𝔹/G be the convex core of M. C(M) can be defined more intrinsically as the closed convex hull of the closed geodesics in M. If G is analytically finite then G is geometrically finite iff C(M) has finite hyperbolic volume. In general, G is geometrically finite iff the unit neighborhood of C(M) has finite volume.

If x ∈ M then the injectivity radius, inj(x), is half the length of the shortest loop through x; alternatively it is half the distance between the two closest distinct lifts of x to 𝔹. In this paper we will assume that the injectivity radius is bounded away from zero uniformly on M. In particular, this implies that G has no parabolic elements. With this extra assumption, G is geometrically finite iff C(M) is compact iff Λ = Λ_{c}. Theorem 1.1 gives a very strong converse of this; if G is not geometrically finite then “most” limit points are non-conical.

G is called topologically tame if M is homeomorphic to the interior of a compact 3-manifold with boundary. By a result of Canary [6], this implies that G is analytically tame, i.e., there are finite area surfaces which sweep out the geometrically infinite ends. In this case C(M) consists of a compact piece and a finite number of ends E_{j} each of which is topologically equivalent to ℝ_{+} × S for some compact surface S. The only property we will use is that for each x ∈ E_{j} there is a finite diameter smooth surface S_{j}(x) which is a bounded hyperbolic distance from x and which separates E_{j}.

Next we recall some facts about dyadic martingales (these can be found in [10]). Suppose Q_{0} is a unit cube in ℝ^{d}. We say that a sequence of functions {f_{n}}, n = 0, 1, … on Q_{0} is a dyadic martingale if each function is constant on each nth generation dyadic cube and satisfies

∫_{Q} f_{n+1} = ∫_{Q} f_{n},

if Q ∈ D_{n}. A martingale is said to be in the space H^{p} if

∥{f_{n}}∥ = (sup_{n} ∫_{Q_{0}} |f_{n}|^{p})^{1/p} < ∞.

The martingale maximal function is defined for x ∈ Q_{0} as

Mf(x) = sup_{n} |f_{n}(x)|,

and is a bounded operator from H^{p} to L^{p}(Q_{0}), 1 < p ≤ ∞, and from H^{1} to weak L^{1}. Moreover, if {f_{n}} ∈ H^{p}, 1 < p ≤ ∞, then there is an f ∈ L^{p} such that

f_{n}(x) = \frac{1}{|Q_{n}(x)|} ∫_{Q_{n}(x)} f(y)dy.

If {f_{n}} is a positive martingale then there is a positive, finite measure μ on Q_{0} such that

f_{n}(x) = \frac{1}{|Q_{n}(x)|} ∫_{Q_{n}(x)} dμ(y).

We let Δf_{n} = f_{n} − f_{n−1}. A martingale is called Bloch if sup_{n} ∥Δf_{n}∥_{∞} < ∞, i.e., the steps are uniformly bounded. We define the martingale square function as

S(f)(x) = (∑_{n=1}^{∞}[Δf_{n}(x)]^{2})^{1/2}.

This is a bounded operator from H^{p}, 1 < p < ∞ to L^{p}. Actually, the square function considered in the paper of Chang, Wilson and Wolff [8] which we will use is slightly different and defined by

S(f)(x) = (∑_{n=1}^{∞} ∥χ_{Q_{n}(x)}Δf_{n}∥_{∞}^{2})^{1/2}.

The two definitions agree when d = 1 and in general they are comparable up to a dimension dependent constant. The following can be found, e.g. in [8],

Theorem 2.2. Let {f_{n}} be a dyadic martingale on Q_{0} ⊂ ℝ^{d} with limit function f. Suppose ∥Sf∥_{∞} < ∞. Then for λ ≥ 0,

|\{x ∈ Q_{0} : f(x) − f_{0}(x) ≥ λ\}| ≤ exp(−\frac{1}{2}λ^{2}/∥Sf∥_{∞}^{2}).

In particular, there are constants C_{1}, C_{2} such that

∫_{Q} exp(\frac{C_{1}|f − f_{0}|^{2}}{∥Sf∥_{∞}^{2}} )dx ≤ C_{2}.

This theorem implies the following estimate for the maximal function.

Corollary 2.3. Suppose {f_{n}} is a martingale with limit function f and suppose Sf ∈ L^{∞}(ℝ^{2}). Then there are constants C_{1} > 0 and C_{2} < ∞ (depending only on α) so that

|\{x ∈ Q : |Mf(x) − \frac{1}{|Q|} ∫_{Q} f| > λ\}| ≤ C_{2}|Q| exp(−\frac{C_{2}}{4} \frac{λ^{2}}{∥Sf∥_{∞}^{2}} ).

Proof: Without loss of generality assume \frac{1}{|Q|} ∫_{Q} f = 0. The set \{Mf(z) > λ\} can be covered by dyadic squares {Q_{j}} such that

\frac{1}{|Q_{j}|} ∫_{Q_{j}} fdx > λ/2.

For such a square Q_{j} we get

exp(\frac{C_{1}λ^{2}}{4∥Sf∥_{∞}^{2}} ) ≤ exp(\frac{C_{1}}{∥Sf∥_{∞}^{2}} (\frac{1}{|Q_{j}|} ∫_{Q_{j}} f(y)dy)^{2}).

Now apply Cauchy-Schwarz to get

≤ exp(\frac{C_{1}}{∥Sf∥_{∞}^{2}} \frac{1}{|Q_{j}|} ∫_{Q_{j}} f^{2}(y)dy),

and Jensen’s inequality (exp(\frac{1}{|Q|} ∫_{Q} |f|^{2}) ≤ \frac{1}{|Q|} ∫_{Q} exp(|f|^{2})),

≤ \frac{1}{|Q_{j}|} ∫_{Q_{j}} exp(\frac{C_{1}}{∥Sf∥_{∞}^{2}} f^{2}(y))dy.

Thus by Theorem 2.2,

∑_{j} |Q_{j}| exp(\frac{C_{1}λ^{2}}{∥Sf∥_{∞}^{2}} ) ≤ ∑_{j} |Q_{j}|\frac{1}{|Q_{j}|} ∫_{Q_{j}} exp(\frac{C_{1}}{∥Sf∥_{∞}^{2}} f^{2}(y))dy

≤ ∫_{Q_{0}} exp(\frac{C_{1}}{∥Sf∥_{∞}^{2}} f^{2}(y))dy
≤ C_{2}.

Hence

|\{Mf > λ\}| ≤ ∑_{j} |Q_{j}| ≤ C_{2} exp(\frac{−C_{1}λ^{2}}{∥Sf∥_{∞}^{2}} ).

For x ∈ ℝ^{2} let Γ_{α}(x) be the non-tangential cone in ℝ_{+}^{3} with vertex x and angle α,

Γ_{α}(x) = {(w, y) ∈ ℝ^{2} × ℝ_{+}^{1} : y > α|x − w|}.

Define the square function

S_{α}f(x) = (∫_{Γ_{α}(x)} |∇_{H}f(y, t)|^{2}ydydt)^{1/2}.

An important property of the square function is that for 1 < p < ∞, ∥S_{α}(f)∥_{p} ≃ ∥f∥_{p} if f has mean value zero, (and the constants depend only on p and α). See Chapter IV of [25].

Next we recall some facts about the space BMO (“bounded mean oscillation”). The space BMO is defined to be the class of locally integrable functions f on ℝ^{d} so that

∥f∥_{BMO} = sup_{Q} \frac{1}{|Q|_{d}} ∫_{Q} |f(x) − \frac{1}{|Q|_{d}} ∫_{Q} f(y)dy|dx < ∞.

This space contains L^{∞} and is contained in every L^{p}, 1 < p < ∞. One of the most important estimates concerning BMO functions is the John-Nirenberg theorem ([14], or Theorem VI.2.1 of [9]). It says that there are constants C_{3}, C_{4} (depending only on d) so that if f ∈ BMO and Q ⊂ ℝ^{d} is a cube then

\{x ∈ Q : |f(x) − \frac{1}{|Q|} ∫_{Q} f| > λ\}| ≤ C_{3}|Q| exp(−C_{4} \frac{λ}{∥f∥_{BMO}} ). (2.1)

If Q is a cube in ℝ^{d}, then Q̂ = Q × [0, ℓ(Q)] is called the Carleson cube in ℝ_{+}^{d+1} with base Q. We let T (Q) = Q × [\frac{1}{2}ℓ(Q), ℓ(Q)] denote the top half of Q and let z_{Q} be the center of Q̂. A measure μ on ℝ_{+}^{d+1} is called Carleson measure if is a M < ∞ so that

∥μ∥_{C} = sup_{Q} \frac{μ(Q̂)}{|Q|_{d}} ≤ M,

where the sup is over all cubes Q ∈ ℝ^{d}. Suppose u is the hyperbolic harmonic extension of f from ℝ^{d} to the upper half-space ℝ_{+}^{d+1}. Then f ∈ BMO iff

μ = |∇_{E}u|^{2}ydx_{1} … dx_{n}dy = |∇_{H}u|^{2}y^{n−1}dV

is a Carleson measure. Moreover, ∥f∥_{BMO} ≃ \sqrt{∥μ∥_{C}}.

3. Some estimates for Bloch and BMO functions

In this section we record a few technical estimates that we will use in the following sections. The first is an estimate which says that various ways of averaging a harmonic function with bounded gradient all give essentially the same answer. The second is the so called “L^{2} − L^{4}” trick and the last allows us to control the L^{2} norm of a function in terms of its L^{1} and BMO norms.

The following was proven by Carleson (Lemma 3, [7]) in dimension two.

Lemma 3.1. Suppose U on ℝ_{+}^{3} is the (hyperbolic) Poisson integral of the positive measure μ and satisfies |∇_{H}U(z)| ≤ 1. For a square Q ∈ ℝ^{2}, let Q_{t} = {(x_{1}, x_{2}, t) : (x_{1}, x_{2}) ∈ Q}. Then there is an A_{1} < ∞ so that

|U(z_{Q}) − \frac{1}{|Q|_{2}} ∫_{Q} U(x_{1}, x_{2}, t)dx_{1}dx_{2}| ≤ A_{1},

for any 0 < t ≤ ℓ(Q) and

|U(z_{Q}) − \frac{1}{|Q|_{2}} ∫_{Q} dμ| ≤ A_{1}.

Proof: By rescaling we may assume z_{Q} = (0, 0, 1) and by replacing U by U − U(z_{Q}) we may assume U(z_{Q}) = 0. Let ψ ∈ C^{∞}(ℝ_{+}^{3}) be chosen with supp(ψ) ⊂ 3Q. Our first goal is to prove

∫∫ _{Q} ψ(x_{1}, x_{2}, t)U(x_{1}, x_{2}, t)dx_{1}dx_{2} = O(1),

with bounds depending only on the C^{2} norm of ψ (in particular, independent of t).

We begin by assuming U is smooth up to the boundary (later we will apply these estimates to functions of the form U(x_{1}, x_{2}, y+t)). Let W (z) be the hyperbolic Poisson extension of the characteristic function of Q to ℝ_{+}^{3} and let F (z) = y^{2}ψ(z)W (z). We now apply Greens theorem (in the hyperbolic setting) to U(z) and F (z) to get

∫∫ _{ℝ_{+}^{3}} UΔ_{H}F − F Δ_{H}UdV = ∫_{ℝ^{2}} U \frac{∂F}{∂n} − F \frac{∂U}{∂n} dA.

Since Δ_{H}U = 0 this becomes

∫∫ _{ℝ_{+}^{3}} UΔ_{H}F dV = ∫_{ℝ^{2}} U \frac{∂F}{∂n} − F \frac{∂U}{∂n} dA. (3.1)

First we estimate the right hand side. Since |∇_{H}U| = O(1) and F = O(y^{2}) we get,

∫_{ℝ^{2}} F \frac{∂U}{∂n} dA = O(1).

To estimate the first term on the right hand side of (3.1) note that

(3.2) \frac{∂F}{∂n} (z) = \frac{∂ψ(z)}{∂n} · y^{2} · W (z) + ψ(z) · \frac{∂y^{2}}{∂n} · W (z) + ψ(z) · y^{2} · \frac{∂W (z)}{∂n}
= O(y^{3}) + ψ(z)y^{2}W (z) + O(y^{2} min(W (z), 1 − W (z)).

The last term uses the fact that the hyperbolic gradient of a bounded harmonic function is bounded by a multiple of the function itself, applied to both W and 1 − W.

Now break the integral ∫ U \frac{∂F}{∂n} into three terms according to (3.2). The integral of the first term over the boundary tends to zero as y → 0. The integral of the second term gives

∫_{Q_{t}} U(x)ψ(x)dx.

Since min(W (z), 1 − W (z)) → 0 almost everywhere as y → 0, the Lebesgue dominated convergence theorem implies the integral of the third term also tends to zero. Thus we deduce

∫_{ℝ^{2}} U \frac{∂F}{∂n} − F \frac{∂U}{∂n} dx_{1}dx_{2} = ∫_{Q} U(x)ψ(x)dx_{1}dx_{2} + O(1). (3.3)

Now we estimate the left hand side of (3.1). Since ψ is C^{2} up to the boundary, we have ∇_{H}ψ = O(y) and Δ_{H}ψ = O(y^{2}). Thus

Δ_{H}F = Δ_{H}(ψ · y^{2} · W)
= Δ_{H}ψ · y^{2} · W + ψ · Δ_{H}y^{2} · W + ψ · y^{2} · Δ_{H}W
+∇_{H}ψ · ∇_{H}y^{2} · W + ∇_{H}ψ · y^{2} · ∇_{H}W + ψ · ∇_{H}y^{2} · ∇_{H}W
= O(y^{4}) + 0 + 0 + O(y^{3}) + O(y^{3}) + O(y^{2}∇_{H}W).

The integral ∫∫_{ℝ_{+}^{3}} UΔ_{H}F dV breaks into six terms, the first five of which are either

zero or are obviously bounded (since U(z) = O(log \frac{1}{y})). To bound the last term, we use the fact that |∇_{H}W | = O(y|∇W |) and the simple estimate

∫_{3Q} |∇W (x_{1}, x_{2}, y)|dx_{1}dx_{2} = O(1),

independent of y. This gives of order

∫∫ _{|z|<3} y^{2}|∇_{H}W | log \frac{1}{y} dV = ∫_{0<y<3} ∫_{|x_{1}|,|x_{2}|<3} y|∇W | log \frac{1}{y} dx_{1}dx_{2}dy
= ∫_{0<y<3} y log \frac{1}{y} y
= O(1).

Thus by (3.3), we have now proven that if U is hyperbolically harmonic, has bounded hyperbolic gradient and U(z_{Q}) = 0 then

∫_{Q} Uψdx_{1}dx_{2} = O(1), (3.4)

with bounds depending only on the C^{2} norm of ψ. To deduce the first part of the lemma, set F (z) = U(z + (0, 0, t)) − U(0, 0, 1 − t) and apply the preceding case to F. Choose ψ so that ψ ≡ 1 on Q. Then (3.4) implies

∫_{Q} U(x_{1}, x_{2}, t)dx_{1}dx_{2} = ∫_{Q} F dx_{1}dx_{2} + U(0, 0, 1 − ϵ) = U(z_{Q}) + O(1),

as desired. To prove the second part of the lemma we simply wish to note that

lim_{t→0} \frac{1}{|Q|_{2}} ∫_{Q} U(x_{1}, x_{2}, t)dx_{1}dx_{2} = \frac{1}{|Q|_{2}} ∫_{Q} dμ.

To prove this, we first observe that μ(∂Q) = 0. This is because we can cover ∂Q by 4 · 2^{n} squares of size 2^{−n}, and U is bounded by Cn in the tops of the corresponding Carleson boxes. Hence the μ measure of each square is at most C2^{−2n}n and the total μ measure of ∂Q is at most C2^{−n}n → 0. Next note that

∫_{Q} P_{H}(x_{1}, x_{2}, t)dx_{1}dx_{2},

converges to 1 inside Q and to 0 outside Q, and hence to χ_{Q} μ-almost everywhere. Now use the fact that U is the Poisson integral of μ, Fubini’s theorem and the Lebesgue dominated convergence theorem to get

\frac{1}{|Q|_{2}} ∫_{Q} U(x_{1}, x_{2}, t)dx_{1}dx_{2} = \frac{1}{|Q|_{2}} ∫_{Q}[∫_{ℝ^{2}} P_{H}(x_{1}, x_{2}, t)dμ]dx_{1}dx_{2}
= ∫_{ℝ^{2}}[\frac{1}{|Q|_{2}} ∫_{Q} P_{H}(x_{1}, x_{2}, t)dx_{1}dx_{2}]dμ
→ \frac{1}{|Q|_{2}} ∫_{ℝ^{2}} χ_{Q}dμ.

This completes the proof of the lemma.

The previous lemma says that if we form the dyadic martingale corresponding to the measure μ then the value of the martingale on a dyadic square Q is (up to an additive constant) the same as the value of the harmonic function U in the top half of the corresponding Carleson cube. Thus the boundary behavior of U exactly mimics the behavior of the martingale.

The second estimate which we wish to record is the standard “L^{2} − L^{4} trick”. This says that a function which has mean value zero, big L^{2} norm and small L^{4} norm necessarily satisfies certain pointwise estimates.

Lemma 3.2. Suppose μ is a probability measure on X. Suppose F is a real valued function on X so that ∫_{X} F dμ = 0 and ∥F ∥_{4} ≤ B∥F ∥_{2}. Then

μ(\{x : F (x) ≤ −\frac{1}{\sqrt{8}B^{2}} ∥F ∥_{2}\}) ≥ \frac{1}{64B^{12}} .

Proof: Without loss of generality we may assume ∥F ∥_{2} = 1. Hölder’s inequality implies that

1 = ∥F ∥_{2} ≤ ∥F ∥_{1}^{1/3}∥F ∥_{4}^{2/3} ≤ ∥F ∥_{1}^{1/3}B^{2/3},

which implies ∥F ∥_{1} ≥ B^{−2}. Write F = F_{+} − F_{−} as the difference of its positive and negative parts. Since F has mean value zero we must have ∥F_{−}∥_{1} ≥ \frac{1}{2}B^{−2}, and hence F (x) ≤ −\frac{1}{2}B^{−2} at some point.

To show F is negative on a set of large measure, let

E_{1} = \{x ∈ Q : F_{−} ≤ \frac{1}{\sqrt{8}B^{2}} \}, E_{2} = \{x ∈ Q : F_{−} ≥ \frac{1}{\sqrt{8}B^{2}} \}.

Since ∥F_{−}∥_{2}^{2} ≥ ∥F_{−}∥_{1}^{2} ≥ B^{−4}/4 and ∫_{E_{1}}(F_{−})^{2}dμ ≤ \frac{1}{8}B^{−4}, we deduce ∫_{E_{2}}(F_{−})^{2}dμ ≥ \frac{1}{8}B^{−4}. Thus if μ(E_{2}) < (64B^{12})^{−1}, we would get

\frac{1}{8}B^{−4} ≤ ∫_{E_{2}} F_{−}^{2}dμ ≤ (∫ F_{−}^{4}dμ)^{1/2}μ(E_{2})^{1/2} < B^{2} \frac{1}{8}B^{−6} = \frac{1}{8}B^{−4},

which is a contradiction. This proves the lemma.

Lemma 3.3. There is a C < ∞ so that if f ∈ L^{1} ∩ BMO and f has mean value zero then

∥f∥_{2} ≤ C(∥f∥_{1}∥f∥_{BMO})^{1/2}.

Proof. There are several ways to do this, e.g., use the fact that interpolating BMO and L^{1} gives L^{2} ([13]). A more direct proof uses the John-Nirenberg theorem as follows. Without loss of generality we may assume ∥f∥_{1} = 1 and set λ = ∥f∥_{BMO}. Write (using the formula for L^{p} norms on page 7 of [25])

∥f∥_{2}^{2} = 2 ∫_{0}^{∞} t|{x : |f(x)| > t}|dt
≤ 2 ∫_{0}^{λ} λ|{x : |f(x)| > t}|dt + 2 ∫_{λ}^{∞} t|{x : |f(x)| > t}|dt.

The first term is clearly bounded by 2λ∥f∥_{1} = 2λ. The second term can be estimated by the John-Nirenberg theorem as follows.

∫_{λ}^{∞} t|{x : |f(x)| > t}|dt ≤ ∑_{n=1}^{∞} λn|{x : |f(x)| > λn}|

≤ ∑_{n=1}^{∞} λn exp(−Cλn/∥f∥_{BMO})

≤ ∑_{n=1}^{∞} λn exp(−Cn)
≤ Cλ.

Thus ∥f∥_{2}^{2} ≤ C∥f∥_{1}∥f∥_{BMO}, as desired.

To see why we need mean value zero, consider the constant function 1; it has L^{1} and L^{2} norm 1 but BMO norm 0.

4. The lower bound in the law of the iterated logarithm

Suppose U is the hyperbolic Poisson integral of a positive measure μ on ℝ^{2} and has bounded hyperbolic gradient. We want to show the support of μ has positive Hausdorff measure for the function φ(t) = t^{2}\sqrt{log \frac{1}{t} log log log \frac{1}{t}}. This could be deduced from the methods of Makarov [17], but we include a proof for completeness. Define a dyadic martingale by

S_{n}(x) = \frac{μ(Q_{n}(x))}{|Q_{n}(x)|_{2}}.

Then by Lemma 3.1, S_{n} is a Bloch martingale. Therefore it is enough to prove the following (Lemma 4.4 of of [17]; Theorem 1 of [18]).

Theorem 4.1. Suppose Y_{n} is a positive Bloch martingale. Let E = {x ∈ [0, 1] :

Y_{n}(x) > 0 for all n\}. Then H^{φ}(E) > 0 for φ(t) = t\sqrt{log \frac{1}{t} log log log \frac{1}{t}}.

Proof: {Y_{n}} is a positive martingale and hence there is a positive measure μ on [0, 1] so that Y_{n}(x) = μ(Q_{n}(x))/|Q_{n}(x)|_{2} where Q_{n}(x) is the nth generation dyadic interval containing x. Without loss of generality, we may assume μ has mass 1. For μ-almost every x, Y_{n} > 0 for all n, so E has full μ measure. We claim that there is a A < ∞ so that for μ-almost every x, the estimate

μ(Q_{k}(x)) ≤ 2A2^{−2k}\sqrt{k log log k},

holds for all large enough k. If we can show this, then by the Frostman-Rogers-Taylor theorem, E has positive Hausdorff measure with respect to φ.

If 2^{n−1} ≤ k ≤ 2^{n}, and Q is a kth generation square such that

μ(Q) ≥ 2A2^{−2k}\sqrt{k log log k},

then Y_{k}(x) ≥ 2A\sqrt{k log log k}. Therefore the maximal function

MY_{2^{n}}(x) ≥ 2A\sqrt{k log log k}
≥ 2A\sqrt{2^{n−1} log log 2^{n−1}}
≥ A\sqrt{2^{n} log log 2^{n}}.

Thus, by the Borel-Cantelli lemma it will be enough to show

μ(F_{n}^{A}) ≡ μ(\{x : MY_{2^{n}}(x) ≥ A\sqrt{2^{n} log log 2^{n}}) ≤ n^{−2},

for all large enough n.

Fix an integer m ≥ n and let {Q_{j}} be the collection of dyadic squares Q_{k}(x) with k ≤ 2^{n}, which are maximal with respect to the property

μ(Q_{j}) ≥ \frac{A}{8}2^{m/2}|Q_{j}|_{2}.

Thus the total Lebesgue measure of such squares is less than

∑_{j} |Q_{j}|_{2} ≤ 2^{−m/2} \frac{8}{A} ∑_{j} μ(Q_{j}) ≤ 2^{−m/2} \frac{8}{A}.

Moreover, by maximality, μ(Q_{j}) ≤ \frac{A}{2}2^{m/2}|Q_{j}|_{2}, for such a square. Now fix such a square and define

F_{n}^{A}(Q_{j}) = \{x ∈ Q_{j} : MY_{2^{n}}(x) > A\sqrt{2^{m} log log 2^{m}}\}.

Since Y has average ≤ 2^{m/2}A/2 on Q_{j}, this is contained in the set

\{x ∈ Q_{j} : MY_{2^{n}}(x) − \frac{1}{|Q_{j}|_{2}} ∫_{Q_{j}} Y_{2^{n}}dx > \frac{A}{2}\sqrt{2^{m} log log 2^{m}}\}.

Since the square function of Y is bounded above by C2^{n/2}, Corollary 2.3 gives

|F_{n}^{A}(Q_{j})|_{2} ≤ C_{1}|Q_{j}|_{2} exp(−C_{2} \frac{1}{2}A2^{m−n} log log 2^{m}).

Since m ≥ n, then summing over j shows (if A is large enough),

|\{x : MY_{2^{n}}(x) ≥ \sqrt{m log log m}\}|_{2} ≤ C_{2} \frac{2}{A}2^{−m/2} exp(−C_{2} \frac{1}{2}A log log m)
≤ C_{2} \frac{2}{A}2^{−m/2}m^{−4}.

Now we want to convert this estimate of Lebesgue measure into an estimate of μ measure.

μ(\{x : MY_{2^{n}}(x) ≥ A\sqrt{2^{n} log log 2^{n}}\})

≤ ∑_{m=n}^{∞} μ(\{x : A\sqrt{2^{m} log log 2^{m}} ≤ MY_{2^{n}}(x) ≤ A\sqrt{2^{m+1} log log 2^{m+1}}\})

≤ ∑_{m=n}^{∞} |\{x : A\sqrt{2^{m} log log 2^{m}} ≤ MY_{2^{n}}(x)\}|_{2}(A\sqrt{2^{m+1} log log 2^{m+1}})

≤ ∑_{m=n}^{∞}(C2^{−m/2}m^{−4} \frac{1}{A})(\sqrt{2^{m+1} log log 2^{m+1}})

≤ ∑_{m=n}^{∞} C\sqrt{2} log(m + 1)m^{−4}

≤ Cn^{−2}.

This is what we wanted to prove, so we are done.

5. The upper bound in the law of the iterated logarithm

The proof in this section is stated in terms of harmonic functions in the upper half space, but would work equally well for positive Bloch martingales with an appropriate lower bound on the square function.

Suppose U is the positive harmonic function given by Lemma 1.7. By moving to the upper half space and conjugating the group, we may assume U is a hyperbolically harmonic positive function on ℝ_{+}^{3}, and that it is the Poisson integral of a positive measure μ supported on the limit set Λ ⊂ [0, 1]^{2}. We may also assume diam(Λ) ≥ 1/2. By multiplying U by a constant we may also assume |∇_{H}U(z)| ≤ 1. For a square Q ∈ ℝ^{2}, recall that T (Q) is the top half of the corresponding Carleson square in ℝ_{+}^{3}.

Lemma 5.1. With U as above, there is a 0 < c_{1} < ∞ such that U(z) ≤ c_{1} for some z ∈ T (Q) implies Q ∩ Λ = ∅.

Proof: Since ∂C(M) is compact and U and tends to ∞ in the geometrically infinite ends, U is bounded below on C(M), say by m_{0}. Thus if U(z) ≤ c_{0} ≪ m_{0} for some z ∈ T (Q), then this point must be a large distance from ∂C(Λ) and outside C(Λ) (since U ≥ m_{0} on all of C(Λ)). This easily implies the base of Q misses Λ.

The next lemma quantifies the fact that U oscillates everywhere in C(M) and hence has large square function.

Lemma 5.2. With U as above, let F_{n}(x_{1}, x_{2}, y) = U(x_{1}, x_{2}, 2^{−n} + y). There are constants α, b_{1}, b_{2} > 0 so that for any x = (x_{1}, x_{2}, 0) such that F_{n}(x) > c_{1}, we have

b_{1}\sqrt{n} ≤ S_{α}(F_{n})(x) ≤ b_{2}\sqrt{n}. (5.1)

The right hand side actually holds for every x ∈ ℝ^{2}.

Proof: Since |∇_{H}U| ≤ 1, we have |∇_{E}U(x_{1}, x_{2}, y)| ≤ Cy^{−1} and hence the right hand side of (5.1) holds for every x for some b_{2} depending only on α.

To prove the left hand side, note that if U(z) ≥ c_{1}, then dist(z, C(Λ)) is uniformly bounded and hence by the convexity of Λ, the axis of the Stolz cone with vertex z remains within a bounded distance a_{0} of C(Λ). Let a_{1}, a_{2} be the constants from Lemma 1.7 and choose α so small that the Stolz cone Γ_{α} contains an a_{1}- neighborhood of its axis. Then the Stolz cone will contain approximately n/(2a_{1}) disjoint balls of radius a_{1} and on each of them

∫ |∇_{H}U|^{2}dx_{1}dx_{2} \frac{dy}{y^{3}} ≥ a_{2},

by Lemma 1.7. Thus there is a b_{1}, such that S_{α}(F_{n})(x) ≥ b_{1}\sqrt{n}, as desired.

Fix a large integer n_{0}. Let ϵ > 0. A particular value for ϵ will be chosen following Lemma 5.8. We use induction to define two classes of dyadic cubes. First, we say Q ∈ B_{n_{0}} if Q ∈ D_{2^{n_{0}}} and U(z_{Q}) ≥ ϵ\sqrt{2^{n_{0}} log n_{0}}. Secondly, we say Q ∈ G_{n_{0}} if Q ∈ D_{2^{n_{0}}} , u(z_{Q}) ≥ c_{0} and Q ∉ B_{n_{0}}. In general, for n ≥ n_{0}, we let C_{n} ⊂ D_{2^{n}} be the dyadic squares of side length 2^{−2^{n}} which are contained in some square of G_{n−1}. Then set

B_{n} = \{Q ∈ C_{n} : U(z_{Q}) ≥ ϵ\sqrt{2^{n} log n}\},
G_{n} = \{Q ∈ C_{n} : c_{1} ≤ U(z_{Q}) ≤ ϵ\sqrt{2^{n} log n}\}.

If we fix a large integer N > n_{0}, then the collection F_{N} = B_{n_{0}} ∪ ⋯ ∪ B_{N} ∪ G_{N} is a cover of Λ. To see this, note that any square in D_{2^{N}} which is not covered by this collection has U(z_{Q}) < c_{1}, which implies it does not hit Λ.

We claim the φ-sum for this covering becomes uniformly bounded for large enough N (recall that φ(t) = t^{2}\sqrt{log \frac{1}{t} log log log \frac{1}{t}}). For the squares in B_{n} this is trivial, because for each such square Q,

(5.2)  φ(ℓ(Q)) ≤ \frac{C}{ϵ} U(z_{Q})|Q|_{2} ≤ \frac{C}{ϵ} μ(Q).

(Here we have used the fact U(z_{Q}) ≃ \frac{1}{|Q|_{2}} ∫_{Q} dμ when U is large, i.e., that the two terms are bounded by multiples of each other. In fact, more is true; they are within an additive constant of each other by Lemma 3.1.) Since μ is a finite measure, the sum over all such squares is bounded by C∥μ∥/ϵ.

Thus to finish the proof we need only prove the φ-sum over G_{N} is bounded for large N. In fact, we shall show this sum tends to zero as N → ∞ with the explicit estimate

∑_{Q∈G_{N}} φ(Q) ≤ A_{n_{0}}e^{−N^{1/4}}, (5.3)

if n_{0} is large enough (depending on ϵ). To prove (5.3) we start by splitting G_{n} into three smaller classes. Let c_{1} and A_{1} be the constants from Lemmas 5.1 and 3.1 and suppose δ > 0 is a small number (which will be fixed following Lemma 5.5). Suppose Q ∈ G_{n}. Then we let

Q ∈ \begin{cases}G_{n}^{0}, & if c_{1} ≤ U(z_{Q}) ≤ 4(A_{1} + c_{1}) \\ G_{n}^{1}, & if 4(A_{1} + c_{1}) < U(z_{Q}) ≤ δ2^{−n/2} \\ G_{n}^{2}, & if U(z_{Q}) ≥ δ2^{−n/2} \end{cases} .

It will be easy to bound the sum over G_{n}^{0} by the corresponding sum over G_{n}^{1}, so we really only have to control the size of the latter two collections.

Lemma 5.3. Let η > 0. If δ is small enough and Q ∈ G_{n}^{0} ∪ G_{n}^{1}, then

| ∪_{Q′∈G_{n+1}∪B_{n+1}} Q′ ∩ Q|_{2} ≤ η|Q|_{2}.

Proof: Suppose δ is very small, but that the conclusion fails for Q. We will apply the “L^{2} −L^{4} trick” (i.e., Lemma 3.2) to F = U_{n} −U_{n}(z_{Q}), where U_{n}((x, y) = U(x, 2^{−2^{n+1}} + y). By Lemma 5.2, the square function of U_{n} is bounded by b_{2}2^{n/2} on Q and hence \frac{1}{|Q|_{2}} ∫_{Q} F^{4} ≤ B_{2}2^{2n}. Similarly

∥F ∥_{L^{2}(Q)}^{2} ≃ ∥S_{α}(F)∥_{L^{2}(Q)}^{2}
≥ C ∫_{Q} S_{α}(U_{n})^{2}dx
≥ C ∫_{∪_{Q′∈G_{n+1}∪B_{n+1}} Q′∩Q} S_{α}(U_{n})^{2}dx

≥ C ∫_{∪_{Q′∈G_{n+1}∪B_{n+1}} Q′∩Q} 2^{n}dx
≥ Cη|Q|_{2}2^{n}.

The penultimate inequality above follows from the lower bound in Lemma 5.2. Thus ∥F ∥_{4} ≤ Cη^{−1/2}∥F ∥_{2}. Hence by Lemma 3.2, F = U − U(z_{Q}) ≤ −Cηη^{1/2}2^{n/2} at some point of Q. But this means

U(z) ≤ U(z_{Q}) − Cη^{3/2}\sqrt{2^{n}} ≤ δ\sqrt{2^{n}} − Cη^{3/2}\sqrt{2^{n}}.

Since U > 0, taking δ < Cη^{3/2} gives a contradiction and proves the lemma.

Corollary 5.4. If Q^{∗} ∈ G_{n}^{0} ∪ G_{n}^{1} then

∑_{Q∈G_{n+1}∪B_{n+1},Q⊂Q^{∗}} |Q|_{2} ≤ η|Q^{∗}|_{2},

and if η < 1/8 then

∑_{Q∈G_{n+1}Q^{∗},Q⊂Q^{∗}} φ(Q) ≤ \frac{1}{2}φ(Q^{∗}).

Proof: The first inequality is an obvious consequence of the previous lemma. To prove the second, note that Q^{∗} has side length 2^{−2^{n}} and Q has side length 2^{−2^{n+1}}, so

φ(Q^{∗}) = 2^{−2^{n+1}} \sqrt{2^{n} log log 2^{n}} = |Q^{∗}|_{2}\sqrt{2^{n} log log 2^{n}},
φ(Q) = 2^{−2^{n+2}} \sqrt{2^{n+1} log log 2^{n+1}} = |Q|_{2}\sqrt{2^{n+1} log log 2^{n+1}},

and hence

∑_{Q∈G_{n+1}∩Q^{∗}} φ(Q) ≤ η|Q^{∗}|_{2}\sqrt{2^{n+1} log log 2^{n+1}}

≤ ηφ(Q^{∗})\sqrt{2}\sqrt{\frac{log n + 1}{log n}}
≤ \frac{1}{2}φ(Q^{∗}),

if η < 1/8 and n ≥ 2.

Lemma 5.5. Suppose Q^{∗} ∈ G_{n}^{1}. If δ is small enough, then

∑_{Q∈G_{n+1}^{2}∪B_{n+1},Q⊂Q^{∗}} U(z_{Q})|Q|_{2} ≥ CU(z_{Q^{∗}})|Q^{∗}|_{2}.

Proof: Let a = U(z_{Q^{∗}}) and, as above, let U_{n}(x_{1}, x_{2}, y) = U(x_{1}, x_{2}, 2^{−2^{n+1}} + y). Then since U > 0, ∥U_{n}∥_{L^{1}(Q^{∗})} ≃ a. Moreover, using the Carleson measure characterization on BMO, and the fact that U has bounded hyperbolic gradient we get ∥U_{n}∥_{BMO(Q^{∗})} ≤ C\sqrt{2^{n}}. Now by Lemma 3.3 we get

∥U_{n}∥_{L^{2}(Q^{∗})} ≤ C(∥U∥_{1}∥U∥_{BMO})^{1/2} ≤ Ca^{1/2}2^{n/4}. (5.4)

(Actually we interpolate with the function U(z)−U(z_{Q^{∗}}) which has L^{1} norm ≤ 2a, because the lemma requires functions which vanish at the z_{Q^{∗}}.)

Let E = \{x ∈ Q^{∗} : U_{n}(x) > c_{1}\}. We claim that ∫_{E} U_{n}dx ≥ \frac{3}{4}a|Q^{∗}|_{2}. To prove this, first use Lemma 3.1 to deduce that since Q^{∗} ∈ G_{n}^{1},

∫_{Q^{∗}} U_{n}dx ≥ (U_{n}(z_{Q^{∗}}) − A_{1})|Q^{∗}|_{2}.

Furthermore,

∫_{Q\E} U_{n}dx ≤ c_{1}|Q^{∗}|_{2},

since U is uniformly bounded by c_{1} off E. Thus

∫_{E} U_{n}dx ≥ (U(z_{Q}) − A_{1} − c_{1})|Q^{∗}|_{2} ≥ \frac{3}{4}a|Q^{∗}|_{2},

which is the claim.

The square function S(U_{n}) > C\sqrt{2^{n}} on E by Lemma 5.2. Thus

|E|_{2}2^{n} ≤ C ∫_{E} S(U_{n})^{2}dx ≤ C ∫_{Q^{∗}} S(U_{n})^{2}dx ≃ ∫_{Q^{∗}} U_{n}^{2}dx ≤ Ca2^{n/2}|Q^{∗}|_{2},

where the last inequality follows by (5.4). Therefore |E|_{2} ≤ Ca2^{−n/2}|Q^{∗}|_{2}. Now define

E_{δ} = \{x ∈ E : U_{n}(x) ≤ δ\sqrt{2^{n}}\}.

Then

∫_{E_{δ}} U_{n} ≤ δ\sqrt{2^{n}}|E|_{2} ≤ Cδa|Q^{∗}|_{2} ≤ \frac{a}{2}|Q^{∗}|_{2},

if δ is small. Since ∫_{E} U_{n} ≥ \frac{3}{4}a|Q^{∗}|_{2}, we have ∫_{E\E_{δ}} U_{n} ≥ \frac{1}{4}a|Q^{∗}|_{2} = \frac{1}{4}U_{n}(z_{Q^{∗}})|Q^{∗}|_{2}, as desired.

We now fix a value of δ in the definition of G_{n}^{1} and G_{n}^{2} so that the previous lemmas hold with η = 1/8.

Lemma 5.6. Suppose Q ∈ G_{n}^{2}. Let y_{Q} = 2^{−2^{n}}. If γ_{0} is small enough (depending only on δ), then for any y_{0} such that 2^{−γ_{0}2^{n}}y_{Q} ≤ y_{0} ≤ y_{Q} we have

|\{z = (x_{1}, x_{2}, y_{0}) ∈ Q : U(z) > c_{1}\}|_{2} ≥ \frac{1}{2}|Q|_{2}.

Proof: Let A = − log_{2} y and define L = {z = (x_{1}, x_{2}, y_{0}) : (x_{1}, x_{2}) ∈ Q}. Then by the usual square function estimates,

∫_{L} |U − U(z_{Q})|^{2} ≤ CA|Q|_{2} ≤ Cγ_{0}2^{n}|Q|. (5.5)

Tchebyshev’s inequality implies

|{x : U > c_{1}}|_{2} ≤ |{x ∈ Q : |U − U(z_{Q})| > δ2^{n/2} − c_{1}}|_{2}
≤ Cγ_{0}2^{n}|Q|_{2}/(δ2^{n/2} − c_{1})^{2}
≤ \frac{1}{2}|Q|_{2},

if γ_{0} is small enough and n large enough.

Lemma 5.7. There are constants α, β > 0 such that if γ < γ_{0}, Q ∈ D_{k} with 2^{n} ≤ k ≤ 2^{n+1} and U(z_{Q}) ≥ δ2^{n/2}, then

|\{z = (x_{1}, x_{2}, 2^{−γ2^{n}}2^{−k}) : (x_{1}, x_{2}) ∈ Q, U(z) − U(z_{Q}) ≥ α\sqrt{γ2^{n}}\}|_{2} ≥ β|Q|_{2}.

Proof: By Lemma 5.6,

|\{z = (x_{1}, x_{2}, 2^{−γ2^{n}}y_{Q}) : U(z) > c_{1}\}|_{2} ≥ \frac{1}{2}|Q|_{2}.

The cones above these points lie near C(Λ), so the standard square function estimate for the L^{2} norm gives

C_{0}\sqrt{γ2^{n}} ≤ (\frac{1}{|Q|} ∫_{Q} |U(z) − U(z_{Q})|^{2})^{1/2} ≃ (\frac{1}{|L|} ∫_{L} |U(z) − U(z_{Q})|^{2})^{1/2},

and gives an L^{4} norm bounded above by C\sqrt{γ}2^{n/2}. Thus the L^{2} and L^{4} norms are comparable with constants independent of γ, so by the “L^{2} − L^{4} trick” (Lemma 3.2), the lemma follows.

We now iterate Lemma 5.7.

Lemma 5.8. Suppose Q^{∗} ∈ D_{2^{n}} and U(z_{Q^{∗}}) ≥ δ2^{n/2}. Then if ϵ is small enough,

∑ |Q|_{2} ≥ n^{−1/3}|Q^{∗}|_{2}, (5.6)

where the sum is over \{Q ∈ D_{2^{n+1}} : Q ⊂ Q^{∗} and U(z_{Q}) ≥ 2ϵ\sqrt{2^{n} log n}\}.

Proof: Let y_{0} = y_{Q^{∗}} (the y coordinate of z_{Q^{∗}}), y_{1} = 2^{−γ2^{n}}y_{0}, and in general y_{k} = 2^{−γ2^{n}}y_{k−1} for k = 1, …, M ≥ γ^{−1}. If we choose γ = 1/(c_{3} log n), then iterating Lemma 5.7 M times gives

|Q^{∗}|_{2}β^{M} ≤ |\{z = (x_{1}, x_{2}, y_{M}) : (x_{1}, x_{2}) ∈ Q^{∗}, U(z) − U(z_{Q^{∗}}) > Mα\sqrt{γ2^{n}}\}|_{2}
≤ |\{z = (x_{1}, x_{2}, y_{M}) : (x_{1}, x_{2}) ∈ Q^{∗}, U(z) > Mα\sqrt{γ2^{n}}\}|_{2}
= |\{z = (x_{1}, x_{2}, y_{M}) : (x_{1}, x_{2}) ∈ Q^{∗}, U(z) > c_{4}α\sqrt{2^{n} log n}\}|_{2},

where β^{M} ≥ β^{c_{4}c_{3} log n} = n^{c_{4}c_{3} log β} ≥ n^{−1/3}. By Harnack’s inequality, if U > a at some point of T (Q) then it is bigger than Ca on all of T (Q) (and in particular at z_{Q}). This gives the lemma if ϵ ≤ Cc_{4}α.

Fix ϵ in the definition of B and G so that the previous lemma holds. Let G_{n}^{i} = ∪_{Q∈G_{n}^{i}}Q for i = 0, 1, 2 and let G_{n} = G_{n}^{0} ∪ G_{n}^{1} ∪ G_{n}^{2}. Similarly, let B_{n} = ∪_{Q∈B_{n}}Q.

Lemma 5.9. If Q^{∗} ∈ G_{n}^{2} then μ(G_{n+1} ∩ Q^{∗}) ≤ (1 − n^{−1/2})μ(Q^{∗}).
Proof: By Lemma 5.8 |Q^{∗} ∩ B_{n+1}|_{2} ≥ n^{−1/3}|Q^{∗}|_{2}. Since Q^{∗} ∈ G_{n}^{2} we have
μ(Q^{∗}) ≤ ϵ2^{n/2}\sqrt{log n}|Q^{∗}|_{2}.

Therefore

μ(Q^{∗} \ G_{n+1}) ≥ ∑_{Q⊂Q^{∗}} μ(Q ∩ B_{n+1})
≥ Cn^{−1/3}|Q^{∗}|_{2}ϵ2^{(n+1)/2}\sqrt{log(n + 1)}
≥ Cn^{−1/3}μ(Q^{∗}).

The last term is ≥ n^{−1/2}μ(Q^{∗}) if n ≥ n_{0}(ϵ), so we are done.

Lemma 5.10. If Q^{∗} ∈ G_{n}^{1} then

μ(G_{n+1} ∩ Q^{∗}) ≤ (1 − n^{−1/2})μ(Q^{∗}).

Proof: Use the proof of Lemma 5.5 to deduce μ(∪Q) ≥ Cμ(Q^{∗}), where the union is over all dyadic Q ∈ Q^{∗} of side length = 2^{− \frac{9}{8} 2^{n}} such that U(z_{Q}) ≥ c_{5}\sqrt{2^{n}}. Now apply the argument of Lemma 5.9 to each Q to obtain

μ(G_{n+1} ∩ Q) ≤ (1 − n^{−1/3})μ(Q),
∫_{B_{n+1}∩Q} dμ ≥ n^{−1/3} ∫_{Q} dμ.

Thus

μ(B_{n+1} ∩ Q^{∗}) ≥ ∑_{Q} μ(B_{n+1} ∩ Q) ≥ ∑_{Q} n^{−1/3}μ(Q) ≥ Cn^{−1/3}μ(Q^{∗}).

Since Cn^{−1/3} ≥ n^{−1/2} for large n, this proves the lemma.

Lemma 5.11. μ(G_{n}^{0}) ≤ B2^{−n/2}.

Proof: Recall that the L^{1} norm of U_{n} = U(x_{1}, x_{2}, 2^{−2^{n}} + y) is ≃ 1 and the BMO norm is ≤ C2^{n/2}. Thus by Lemma 3.3 the L^{2} norm is at most C2^{n/4}. On the set G_{n}^{0}, the square function of U_{n} is bounded below by b_{1}2^{n/2} which means the L^{2} norm of U_{n} is bounded below by b_{1}\sqrt{2^{n}|G_{n}^{0}|_{2}}. Thus |G_{n}^{0}|_{2} ≤ C2^{−n/2}. For Q ∈ G_{n}^{0}, U(z_{Q}) ≃ 1, so μ(Q) ≤ C|Q|_{2}. This proves the lemma.

Lemma 5.12. μ(G_{n}) ≤ A_{n_{0}}e^{− \frac{1}{4} n^{1/2}}.

Proof: We prove this by induction. For k = n_{0} it is true with A_{n_{0}} = ∥μ∥e^{\sqrt[\frac{1}{4}]{n_{0}}}. Assume it is true for n ≥ n_{0}. From the previous three lemmas we can deduce

μ(G_{n+1}) ≤ μ(G_{n+1} ∩ G_{n}^{0}) + μ(G_{n+1} ∩ (G_{n}^{1} ∪ G_{n}^{2}))
≤ B2^{−n/2} + (1 − n^{−1/2})μ(G_{n}^{1} ∪ G_{n}^{2})
≤ B2^{−n/2} + (1 − n^{−1/2})A_{n_{0}} exp(−\frac{1}{4}n^{1/2}).

22 CHRISTOPHER J. BISHOP AND PETER W. JONES

Now if n_{0} is so large that n ≥ n_{0} implies B2^{−n/2} ≤ \frac{1}{2}n^{−1/2} exp(−\frac{1}{4}n^{1/2}) we get

≤ (1 − \frac{1}{2}n^{−1/2})A_{n_{0}} exp(−\frac{1}{4}n^{1/2})
≤ A_{n_{0}} exp(log(1 − \frac{1}{2}n^{−1/2}) − \frac{1}{4}n^{1/2})
≤ A_{n_{0}} exp(−\frac{1}{2}n^{−1/2} − \frac{1}{4}n^{1/2})
≤ A_{n_{0}} exp(−\frac{1}{4}(n + 1)^{1/2}),

where we have used the facts log(1 − x) ≤ −x and \sqrt{n + 1} − \sqrt{n} ≤ \frac{1}{2}n^{−1/2}. Lemma 5.13. ∑_{Q∈G_{n+1}} φ(Q) ≤ \frac{1}{2} ∑_{Q^{∗}∈G_{n}} φ(Q^{∗}) + A′_{n_{0}}e^{−n^{1/4}}. Proof: Note that by Corollary 5.4

∑_{Q^{∗}∈G_{n}^{0}∪G_{n}^{1}}[ ∑_{Q⊂Q^{∗},Q∈G_{n+1}} φ(Q)] ≤ \frac{1}{2} ∑_{Q^{∗}∈G_{n}^{0}∪G_{n}^{1}} φ(Q^{∗}).

If Q ∈ G_{n+1} and Q^{∗} ∈ G_{n}, then

φ(diam(Q)) ≤ \sqrt{2}(\frac{log n + 1}{log n})^{1/2}φ(diam(Q^{∗})).

Furthermore, since U ≥ δ\sqrt{2^{n}} on the top half of Q^{∗} ∈ G_{n}^{2},

φ(diam(Q^{∗})) ≤ \frac{C}{δ} \sqrt{log n}μ(Q^{∗}).

Thus

∑_{Q^{∗}∈G_{n}^{2}}[  ∑_{Q⊂Q^{∗},Q∈G_{n+1}} φ(diam(Q))] ≤ \sqrt{2}(\frac{log n + 1}{log n})^{1/2} ∑_{Q^{∗}∈G_{n}^{2}} φ(diam(Q^{∗}))
≤ \frac{C}{δ} \sqrt{log n} ∑_{Q^{∗}∈G_{n}^{2}} μ(Q^{∗})
≤ \frac{C}{δ} \sqrt{log n}A_{n_{0}}e^{− \frac{1}{4} n^{1/2}}
≤ A_{n_{0}}e^{−n^{1/3}},

if n_{0} is large enough.

Corollary 5.14. ∑_{Q∈G_{n}} φ(Q) ≤ A′′_{n_{0}}e^{−n^{1/4}}.

Proof: Let a_{n} = ∑_{Q∈G_{n}} φ(Q). Then the previous lemma says

a_{n+1} ≤ \frac{1}{2}a_{n} + A′_{n_{0}}e^{−n^{1/3}}.

To prove the corollary we use induction. It is true for n = n_{0} by an appropriate choice of A′′_{n_{0}}. In general we need only prove

\frac{1}{2}(A′′_{n_{0}} exp(−n^{1/4})) + A′_{n_{0}} exp(−n^{1/3}) ≤ A′′_{n_{0}} exp(−(n + 1)^{1/4}).

Assuming we have chosen A′′_{n_{0}} ≥ A′_{n_{0}} this reduces to showing

exp(−n^{1/3}) ≤ exp(−(n + 1)^{1/4}) − \frac{1}{2} exp(−n^{1/4}).

This in turn follows from the two following estimates

\frac{1}{2} exp(−n^{1/4}) ≤ \frac{2}{3} exp(−(n + 1)^{1/4}),

exp(−n^{1/3}) ≤ \frac{1}{3} exp(−(n + 1)^{1/4}),

each of which is easy and left to the reader.

6. The measure of the conical limit set

As noted in the introduction, the theory of conformal densities implies the conical limit set must have zero measure for the gauge φ(t) = t^{2}\sqrt{log \frac{1}{t} log log log \frac{1}{t}}. We can see this more directly as follows. Note that the proof in the previous section actually shows that for any square Q, H^{φ}(Λ∩Q) ≤ Cμ(Q). Since U has radial limit ∞ at μ almost every point of Λ (by standard differentiation theorems, e.g. Section 2.9 of [21]) but at no point of Λ_{c}, we see Λ_{c} must have zero μ measure and hence zero H^{φ} measure. In fact, it is not too hard to prove a better estimate.

Lemma 6.1. If G is topologically tame, Λ ≠ S^{2}, and the injectivity radius is bounded away from zero, then H^{ψ}(Λ_{c}) = 0 where ψ(t) = t^{2} log \frac{1}{t}(log log \frac{1}{t})^{−1−ϵ} for any ϵ > 0.

Proof. Fix a M < ∞ and let Λ_{c}^{M} be the limit points which are covered by infinitely many dyadic squares Q with U(z_{Q}) ≤ M. Then Λ_{c} is the countable union of the sets Λ_{c}^{1}, Λ_{c}^{2}, …, so it suffices to show each one has zero measure.

Let A_{n} be the collection of dyadic squares in ∪_{k=2^{n}}^{2^{n+1}}D_{k} which are maximal with respect to the property c_{1} ≤ U(z_{Q}) ≤ M. Let E_{n} be the squares in D_{2^{n}} where U(z_{Q}) ≥ c_{1} and let E_{n}(Q) be the squares Q′ ⊂ Q with diam(Q′) = 2^{−2^{n}}diam(Q) and U(z_{Q′}) ≥ c_{1}. We claim that |E_{n}|_{2} ≤ C2^{−n/2}. To prove this, let U_{n}(x, y) = U(x, 2^{−2^{n+1}} + y) and recall that the L^{1} norm of U ≃ 1 and the BMO norm is ≃ 2^{n/2}. Thus by Lemma 3.3 the L^{2} norm is bounded by C2^{n/4}. On the other hand, at points of E, the square function of U_{n} is bounded below by b_{1}2^{n/2} which means that the L^{2} norm of the square function is at least (b_{1}^{2}2^{n}|E|_{2})^{1/2}. Since the L^{2} norms of U and S_{α}U are equivalent, we deduce

|E_{n}|_{2} ≤ C2^{−n/2}, (6.1)

as desired. A similar argument shows |E_{n}(Q)|_{2} ≤ C2^{−n/2}|Q|_{2}, if U(z_{Q}) ≃ 1.

Cover Λ_{c}^{M} with the collection ∪_{n=n_{0}}^{∞} ∪_{Q∈A_{n}} E_{n}(Q). Note that ∪_{Q∈A_{n}}E_{n}(Q) ⊂ E_{n}. Thus

∑_{Q∈A_{n}} |E_{n}(Q)|_{2} ≤ C ∑_{Q∈A_{n}} 2^{−n/2}|Q|_{2} ≤ C2^{−n/2}|E_{n}|_{2} ≤ C2^{−n/2} · 2^{−n/2},

and therefore,

∑_{Q∈A_{n}} |E_{n}(Q)|_{2} ≤ C2^{−n},

which implies

∑_{Q′∈E_{n}(A_{n})} |Q′|_{2} log \frac{1}{diam(Q)} ≤ C2^{−n}2^{n} ≤ C.

Thus

∑_{Q′∈E_{n}(A_{n})} \frac{|Q′|_{2} log \frac{1}{diam(Q)}}{(log log \frac{1}{diam(Q)})^{1+ϵ}} ≤ Cn^{−1−ϵ}.

Since this is a convergent series, we have proved the lemma.

A similar proof shows Λ_{c} has zero measure for any gauge of the form ψ(t) = t^{2} log \frac{1}{t}η(t) where ∑_{n} η(2^{−2^{n}}) < ∞.

7. Absolute continuity of quasiconformal conjugacies

In this section we prove Corollary 1.5. There are two facts needed for the proof. The first is the following result of McMullen, Theorem 2.11 of [22]

Theorem 7.1. Suppose Φ is a K-quasiconformal conjugacy on S^{2} between two Kleinian groups G_{1} and G_{2}, with quotient manifolds M_{1} and M_{2}. Suppose the injectivity radius for M_{1} in the convex core lies in [R_{0}, R_{1}] with 0 < R_{1} ≤ R_{1} < ∞. Then Φ extends to a volume preserving, quasi-isometry Φ between M_{1} and M_{2} and the pointwise quasi-isometry constant L(Φ, p) satisfies

log L(Φ, p) ≤ C exp(−αdist(p, M \ C(M)).

the constants C and α > 0 depend only on K, R_{0}, R_{1}.

The second fact we need is that μ almost every geodesic tends to infinity in the convex core with a certain estimate.

Lemma 7.2. Suppose G is a geometrically infinite, topologically tame group and let μ be a conformal density for G. Then for μ almost every geodesic γ,

lim inf_{t→∞} \frac{dist(γ(t), M \ C(M))}{\sqrt{t}(log t)^{−2}} = +∞.

Proof. We use the notation of the two previous sections. Let E_{n} be the union of dyadic squares of generation 2^{n} which hit the Λ. Then Equation (6.1) says that |E_{n}|_{2} ≤ C2^{−n/2}. Let A > 0 and set

B_{n} = {Q : 2^{−n−1} ≤ log ℓ(Q)^{−1} ≤ 2^{−n} and μ(Q) ≤ An^{−2}2^{n/2}|Q|_{2}}.

Then if B_{n} = ∪_{QB_{n}}Q, we see

∑_{n} μ(B_{n}) ≤ A ∑_{n} n^{−2}2^{n/2}|E_{n}|_{2} ≤ CA ∑_{n} n^{−2} < ∞.

Thus by the Borel-Cantelli lemma, μ almost every point of Λ is in only finitely many of the sets B_{n}, which proves the lemma.

We can now prove Corollary 1.5. Let Φ be the conjugacy given by Theorem 7.1 lifted to 𝔹 and assume it maps the origin to itself. Let μ_{1} and μ_{2} be the restrictions of H^{φ} to Λ_{1} and Λ_{2} respectively. Let x ∈ Λ_{1} be a point where the conclusion of Lemma 7.2 holds and let B = B(x, t) denote the (Euclidean) ball of radius t around x in ℝ^{3}. Let y = x(1 − t). Since Φ is quasiconformal in ℝ^{3}, Φ(B) contains a ball of size at least C(1 − |Φ(y)|) around Φ(x). We claim this is comparable to 1−|y|. To see this, note that ρ(0, Φ(y)) can be estimated from above by integrating the pointwise quasi-isometric constant of Φ along the geodesic from 0 to y. By combining McMullen’s estimate and the previous lemma this is bounded by

∫_{0}^{ρ(0,y)} 1 + C exp(−αA_{x}\sqrt{t}/ log^{2} t)dt ≤ ρ(0, y) + C_{x}.

Thus for μ almost every point x ∈ Λ_{1} there is a constant C_{x} so that Φ(B(x, t)) contains the ball B(Φ(x), C_{x}t).

Finally we have to check that this implies Φ^{∗}μ_{1} is absolutely continuous with respect to μ_{2}. If not, then there is a compact set E of positive μ_{1} measure so that Φ(E) has zero μ_{2} measure, i.e., zero H^{φ} measure. By passing to a subset we may assume the same constant C = C_{x} works for all x ∈ E. Let {B_{j}} be a covering of Φ(E) with small φ-sum. Then {Φ^{−1}(B_{j})} is a covering of E by sets of comparable diameter, hence also has small φ-sum. Thus μ_{2}(Φ(E)) = 0 ⇒ H^{φ}(Φ(E)) = 0 ⇒ H^{φ}(E) = 0 ⇒ μ_{1}(E) = 0. This is a contradiction, so the first part of the corollary is proved.

Finally, we would like to show that Φ is differentiable μ_{1} almost everywhere. Let x ∈ Λ be a point where the corresponding geodesic γ satisfies

dist(γ(t), M \ C(M)) ≥ η(t),

for some function η which satisfies

∫_{a}^{∞} exp(−αη(t))dt < ∞. (7.1)

Then for points t_{1} > t_{2} on the geodesic, we have (using estimates for Φ and Φ^{−1})

ρ(Φ(γ(t_{1})), Φ(γ(t_{2}))) = |t_{2} − t_{1}| + O(∫_{t_{1}}^{t_{2}} exp(−αη(t))dt) = |t_{2} − t_{1}| + o(1),

uniformly as t_{1} → ∞. This implies the image of γ is asymptotically close to a geodesic ending at Φ(x).

Assume x = 0 ∈ Λ and consider the rays R_{0} = {(0, 0, t) : t > 0}, R_{1} = {(t, 0, t) : t > 0} and R_{2} = {(0, t, t) : t > 0} in the upper half space which end at x. Let z_{n}^{i} = R_{i}(e^{−n}), i = 0, 1, 2 and let w_{n}^{i} be the orthogonal projection of z_{n}^{i} onto ℝ^{2}. The position of z_{n+1}^{i} is determined, to within O(a_{n}) = O(exp(−αη(n))), by its distances to the points z_{n}^{i}, i = 0, 1, 2 (e.g., Lemma B.23 of [22]). In particular, the argument of w_{n+1}^{i}, i = 1, 2, is the same as the argument of w_{n}^{i}, to within O(a_{n}). Since this is a summable sequence we see Φ(R_{1}) and Φ(R_{2}) are asymptotic to rays at Φ(x) and the angles between Φ(R_{0}), Φ(R_{1}) and Φ(R_{2}) are the same as the corresponding angles between R_{0}, R_{1} and R_{2}. If we renormalize the maps so that z_{n}^{0} = (0, 0, 1), we see (by symmetry and equicontinuity) that the maps converge uniformly on compact sets of ℝ_{+}^{3} to a conformal map and hence converge uniformly on compact subsets of ℝ^{3} to a conformal map. This proves Φ is differentiable at x, as desired. Since all we need to prove Φ is differentiable was (7.1), we see that Equation (1.2) suffices. Thus Lemma 7.2 implies Φ is differentiable μ_{1} almost everywhere, finishing the proof.

It is conjectured (e.g., Equation (3.17) of [11]) that if f is a quasiconformal map in ℝ^{n} with real dilatation K_{f}, then

∫_{|x−x_{0}|<t} \frac{K_{f}(x) − 1}{|x − x_{0}|^{n}}dx < ∞,

implies f is differentiable at x_{0} and is conformal there (this is known for n = 2, [15]). Our mapping Φ satisfies such an estimate at μ_{1} almost every x, so this conjecture would give an alternate proof that Φ is differentiable μ_{1} almost everywhere.

The set where Φ is not differentiable is only slightly larger than the conical limit set. Indeed, if we repeat the proof of Lemma 6.1, but use the fact that U has mean value norm ≤ C log n on a square Q where U(z_{Q}) ≤ log n (ℓ(Q) = 2^{−2^{n}}), then we can show Φ is differentiable everywhere on Λ except on a set of zero Hausdorff measure for ψ(t) = t^{2} log \frac{1}{t}(log log \frac{1}{t})^{−1−ϵ} for any ϵ > 0. (The extra growth gives a log log log term, but this can be absorbed into the log log^{−1−ϵ} term.)

8. An estimate for the gradient of Green’s function

Lemma 8.1. There is a constant A < ∞ so that the following holds. Suppose x ∈ M = 𝔹/G and y_{1}, y_{2} ∈ M with ρ(x, y_{1}), ρ(x, y_{2}) ≥ 1 and ρ(y_{1}, y_{2}) ≤ 1. Then |G(y_{1}, x) − G(y_{2}, x)| ≤ Avol(B(1, y_{1}))^{−1}.

Proof. We will first consider the case when ρ(x, y_{1}) = ρ(x, y_{2}) = 1, and will deduce the general case using the maximum principle at the end of the proof.

By the symmetry of Greens function it is enough to prove

|G(x, y_{1}) − G(x, y_{2})| ≤ A/vol(B(1, y_{1})).

Let B_{r} = {z ∈ M : ρ(z, x) < r} and let S_{r} be the boundary of B_{r}. Let W_{y} be the space of Brownian paths on M started at y at time t = 0. For a path b ∈ W_{y}, define stopping times by

τ_{1}^{y} = inf t > 0 such that b(t) ∈ S_{1}.
σ_{1}^{y} = inf t > τ_{1}^{y} such that b(t) ∈ S_{2}.

In general, we let

τ_{n}^{y} = inf t > σ_{n−1}^{y} such that b(t) ∈ S_{1}.
σ_{n}^{y} = inf t > τ_{n}^{y} such that b(t) ∈ S_{2}.

Let

E_{n}^{y} = {b ∈ W : τ_{n} < finite },

and note that for b ∈ E_{n}^{y}, σ_{n}^{y} is finite almost surely. Let μ_{n}^{y} be the measure defined on S_{2} by the stopping time σ_{n}^{y}, i.e., for E ⊂ S_{2},

μ_{n}^{y}(E) = ℙ(σ_{n}^{y} < ∞ and b(σ_{n}^{y}) ∈ E).

Then clearly ∥μ_{n}^{y}∥ is decreasing in n and

∥μ_{n}^{y}∥ = ∫_{S_{2}} ω(w, S_{1}, M \ B_{1})dμ_{n−1}^{y}(w) = ∥μ_{n−1}^{y}∥ ∫_{S_{2}} ω(w, S_{1}, M \ B_{1})dμ̃_{n−1}^{y}(w).

where μ̃_{n}^{y} = μ_{n}^{y}/∥μ_{n}^{y}∥.

For y ∈ M let T (y) be the expected time a path started at y spends in B_{1} before σ_{1}^{y} (i.e., before the first time it hits S_{2} following its first intersection with B_{1}). This is harmonic on M \ B_{1} and is uniformly bounded. Then for y ∈ M \ B_{1}, let μ_{1}^{y} = ω(y, ·, M \ B_{1}) and

μ_{n}^{y} = L(μ̃_{n−1}^{y}) ∫_{S_{2}} ω(w, S_{1}, M \ B_{1})dμ_{n−1}^{y}(w).

Then if y ∈ B_{2} we get (by the mean value property)

(8.1) G(x, y)vol(B_{1}) = ∫_{B_{1}} G(x, y)dx
= expected time a path started at y spends in B_{1}
= T (y) + ∑_{n=1}^{∞} ∫_{S_{2}} T (w)dμ_{n}^{y}(w).

Thus to estimate G it will be enough to estimate ∥μ_{n}^{y}∥.

We claim that there is a probability measure μ on S_{2}, so that μ̃_{n}^{y} → μ exponentially fast, i.e., ∥μ − μ̃_{n}^{y}∥ ≤ λ^{n}, for some λ < 1. To prove this we will show there is a strict contraction L on the convex set of probability measures on S_{2} such that μ̃_{n}^{y} = L(μ̃_{n−1}^{y}). Then μ will simply be the fixed point of the contraction.

For u ∈ B_{2} and v ∈ S_{2}, let P_{B_{2}}(u, v) be the Poisson kernel for B_{2} with respect to the point u, i.e., for E ⊂ S_{2},

ω(u, E, B_{2}) = ∫_{S_{2}} χ_{E}(v)P_{B_{2}}(u, v)dv,

(where dv denotes area measure on S_{2}). Similarly, define a kernel P_{M\B_{1}}(w, u) on ∂B_{1} by

ω(u, E, M \ B_{1}) = ∫_{S_{1}} χ_{E}(v)P_{M\B_{1}}(w, u)du,

Define a kernel K(v, w) on S_{2} × S_{2} by

K(w, v) = P_{M\B_{1}}(w, u)P_{B_{2}}(u, v).

This corresponds to the first hitting probability on S_{2} of a Brownian motion started at v ∈ S_{2}, conditioned to first hit B_{1}. Let

K̃(w, v) = \frac{K(w, v)}{∫_{S_{2}} K(w, v)dw} .

Thus

∫_{S_{2}} K̃(w, v)dw = 1, (8.2)

We claim that area(S_{2})K(w, v) is bounded away from zero by some absolute constant (independent of x). This is easiest to see by lifting to 𝔹. Let Ω ⊂ 𝔹 be the lift of B_{2} and assume x lifts to the origin. Since B(0, 2) ⊂ Ω the Poisson kernel for Ω with respect to 0 ∈ Ω is uniformly bounded below on ∂B(0, 2) ∩ ∂Ω. If the volume of B_{2} is large we are done. If it is small then there are ≃ vol(B_{1})^{−1} lifts of x in B(0, 1) and by Harnack’s inequality, the Poisson kernel for each of these is comparable on ∂Ω to the kernel for 0. Thus summing over the orbit of 0 gives a sum which bounded below by area(S_{2})K(w, v) since area(S_{2}) ≃ vol(B_{1}).

Thus there is a C_{1} > 0 so that

∫_{S_{2}} |K̃(w, v) − C_{1}|dv ≤ λ, (8.3)

for some fixed λ < 1. Note that λ only depends of the behavior of the Dirichlet problem in the ball B_{2} and hence may be chosen independent of the point x (and even the group G).

For a probability measure ν on S_{2} define a measure L(ν) by

L(ν)dv = [∫_{S_{2}} K̃(w, v)dν(w)]dv.

Then by Equation 8.2, L maps probability measures to probability measures. To see L is a strict contraction, let ν_{1}, ν_{2} be two probability measures on S_{2}, Then using Equation 8.3 and the fact that ν_{1} − ν_{2} has mean value zero we get

∥L(ν_{1}) − L(ν_{2})∥ = | ∫_{S_{2}}[∫_{S_{2}} K̃(w, v)d(ν_{1} − ν_{2})(w)]dv|
≤ | ∫_{S_{2}}[∫_{S_{2}}(K̃(w, v) − C_{0})d(ν_{1} − ν_{2})(w)]dv
+ ∫_{S_{2}}[∫_{S_{2}} C_{0}d(ν_{1} − ν_{2})(w)]dv|
≤ | ∫_{S_{2}}[∫_{S_{2}}(K̃(w, v) − C_{0})d(ν_{1} − ν_{2})(w)]dv + 0|
≤ ∫_{S_{2}}[∫_{S_{2}} |K̃(w, v) − C_{0}|dvd|ν_{1} − ν_{2}|(w)
≤ λ∥ν_{1} − ν_{2}∥.

Thus L is a strict contraction. It is easy to check that μ̃_{n} = L(μ̃_{n−1}), so there is a probability measure μ on S_{2} which is a fixed point for L and so that μ̃_{n} converges to μ exponentially fast.

Define

η ≡ 1 − ∫_{S_{2}} ω(w, S_{1}, M \ B_{1})dμ(w).

Observe that η is bounded away from 1 uniformly and can be bounded away from zero depending only on dist(x, ∂C(M)). Fix a probability measure μ_{0} on S_{2} and let

μ_{n} = L(μ̃_{n−1})∥μ_{n−1}∥ ∫_{S_{2}} ω(w, S_{1}, M \ B_{1})dμ̃_{n−1}(w).

We want to estimate ∥μ_{n}∥. Note,

∥μ_{n}∥ = ∥μ_{n−1}∥ ∫_{S_{2}} ω(w, S_{1}, M \ B_{1})dμ̃_{n−1}(w)
= ∥μ_{n−1}∥[∫_{S_{2}} ω(w, S_{1}, M \ B_{1})dμ(w)
+ ∫_{S_{2}} ω(w, S_{1}, M \ B_{1})d(μ̃_{n−1} − μ)(w)]
= ∥μ_{n−1}∥[(1 − η) + ∫_{S_{2}} ω(w, S_{1}, M \ B_{1})d(μ̃_{n−1} − μ)(w)]
= ∥μ_{n−1}∥[(1 − η) + ∫_{S_{2}}((1 − 1) + ω(w, S_{1}, M \ B_{1}))d(μ̃_{n−1} − μ)(w)]
= ∥μ_{n−1}∥[(1 − η) − ∫_{S_{2}}(1 − ω(w, S_{1}, M \ B_{1}))d(μ̃_{n−1} − μ)(w)].

THE LIL FOR KLEINIAN GROUPS 29

By Harnack’s inequality we have Cη ≥ 1 − ω(w, S_{1}, M \ B_{1}), for all w ∈ S_{2}, so

[(1 − η) − Cη∥μ̃_{n−1} − μ∥] ≤ \frac{∥μ_{n}∥}{∥μ_{n−1}∥} ≤ [(1 − η) + Cη∥μ̃_{n−1} − μ∥],

which gives

[(1 − η) − Cηλ^{n}] ≤ \frac{∥μ_{n}∥}{∥μ_{n−1}∥} ≤ [(1 − η) + Cηλ^{n}].

Thus if we assume η < 1/2,

∏_{k=1}^{n}[1 − 2Cηλ^{k}] ≤ \frac{∥μ_{n}∥}{(1 − η)^{n}} ≤ ∏_{k=1}^{n}[1 + 2Cηλ^{k}].

1 − 2Cη ≤ \frac{∥μ_{n}∥}{(1 − η)^{n}} ≤ 1 + 2Cη.

Thus

|∥μ_{n}∥ − (1 − η)^{n}| ≤ Cη(1 − η)^{n}.

Next we want to show G does not oscillate much on S_{2}. Let

A = ∑_{n=1}^{∞} ∫_{S_{2}} T (w)(1 − η)^{n}dμ(w) ≃ \frac{1}{η} .

For y ∈ S_{2}, (8.1) implies

|G(x, y)vol(B_{1}) − A| ≤ T (y) + | ∑_{n=1}^{∞}[∫_{S_{2}} T (w)(1 − η)^{n}dμ(w) − ∫_{S_{2}} T (w)dμ_{n}^{y}(w)]|
≤ C + C ∑_{n=1}^{∞} ∥(1 − η)^{n}μ(w) − μ_{n}^{y}(w)∥
≤ C + C ∑_{n=1}^{∞}[∥(1 − η)^{n}μ(w) − (1 − η)^{n}μ̃_{n}^{y}(w)∥
+∥(1 − η)^{n}μ̃_{n}^{y}(w) − μ_{n}^{y}∥]
≤ C + C ∑_{n=1}^{∞}[(1 − η)^{n}λ^{n} + η(1 − η)^{n}]
≤ C + C[\frac{λ}{1 − λ} + \frac{η(1 − η)}{η}]
≤ C.

From this it follows that G(x, y)vol(B_{1}) oscillates by at most a uniform constant for y ∈ S_{2}. This is what we wanted to prove. Since A ≃ 1/η, this also shows

G(x, y) ≤ \frac{Cη^{−1}}{vol(B_{1}(x))}. (8.4)

If y_{1}, y_{2} are both distance ≥ 1 from x then the maximum principle,

max_{M\B(x,1)} G(y, x) = max_{∂B(x,1)} G(y, x),

implies

|G(y_{1}, x) − G(y_{2}, x)| = |G(x, y_{1}) − G(x, y_{2})|
≤ sup_{w∈S(2,y_{1})} |G(w, y_{1}) − G(w, y_{2})|
≤ sup_{w∈S(2,y_{1})} |G(y_{1}, w) − G(y_{2}, w)|
≤ Avol(B(1, y_{1}))^{−1}.

A standard estimate for harmonic functions says there is an absolute C < ∞ so that

|∇U(x)| ≤ C( sup_{B(x,1)} U(y) − inf_{B(x,1)} U(y)),

so we can now deduce,

Corollary 8.2. If dist(y, x) ≥ 1 then |∇_{y}G(y, x)| ≤ Cvol(B(1, y))^{−1} .

9. A positive harmonic function with linear growth on M

Using the bound for the gradient of Green’s function from the previous section, we now want to construct a positive harmonic function U on M by taking the limit as we push the pole off to infinity. We start with the following observations.

Lemma 9.1. Suppose ∂C(M) is compact and x ∈ C(M) with dist(x, ∂C(M)) ≥ 1. Then G(y, x) is bounded on ∂C(M) (independent of x).

Proof. By symmetry we only have to show that G(x, y) is bounded outside the ball of radius one around y. By the maximum principle it is enough to show it is bounded on the boundary of this ball. However, this follows from the (8.4).

Lemma 9.2. Suppose ∂C(M) is compact and area(Λ) = 0. Then there is an ϵ > 0 so that for any x ∈ C(M) with dist(x, ∂C(M)) > 10 there is a point y ∈ ∂C(M) where G(y, x) ≥ ϵ

Proof. This is obvious by Harnack’s inequality and the fact that

∫_{∂C(M)} |∇G|ds ≥ ∫_{∂C(M)} |\frac{∂G}{∂n}|ds ≥ 1.

Lemma 9.3. Suppose G is geometrically infinite, ∂C(M) is compact and there is a lower bound for the injectivity radius in C(M). Fix a point x_{0} ∈ ∂C(M). Then there is a positive harmonic function U on M = 𝔹/G so that U(x_{0}) = 1, U(z) → 0 in the geometrically finite ends and |∇U| ≤ A on all of M. In particular, U(z) ≤ C + Cdist(z, x_{0}) for z ∈ C(M).

Proof. If Λ has positive area, this is easy (take the Poisson integral of its characteristic function). Thus we may assume Λ has zero area. Since G is geometrically infinite we can choose a sequence of points x_{n} ∈ C(M) so that dist(x_{0}, x_{n}) → ∞. Then G(z, x_{n}) is uniformly bounded on any compact subset of C(M) and hence there is a subsequence which converges uniformly on compact subsets to a harmonic function U. Clearly U satisfies the same upper bounds as G(z, x_{n}) which gives the last two conditions. Finally, Lemma 9.2 and Harnack’s inequality give a lower bound for G(y, x_{n}) on ∂C(M) which is independent of x_{n}. Hence U also satisfies this lower bound, so by a bounded rescaling we can get U(x_{0}) = 1.

Lemma 9.4. Suppose G is topologically tame but geometrically infinite group and there is a positive lower bound for the injectivity radius. Then there is a positive harmonic function U satisfying the conclusions of Lemma 9.3 and in addition satisfies the following condition. For every a_{0} < ∞ there are positive constants a_{1} and a_{2} so that

∫_{B(z,a_{1})} |∇U|^{2}dV ≥ a_{2},

for every z ∈ M with dist(z, C(M)) ≤ a_{0}. Moreover, U tends to ∞ in the geometrically infinite ends.

Proof. By standard results (see [6]) if G is topologically tame and has a positive lower bound for the injectivity radius then C(M) can be written as a compact part K and a finite number of ends E_{i} each of which is topologically of the form E_{j} = S_{j} × ℝ^{+}, j = 1, …, N. For z ∈ K the estimate is true at each point (since U is not constant) and hence is true uniformly on K by compactness.

Next consider an end E_{j}. Choose a sequence of points converging to ∞ in this end and let U_{j} denote the limiting harmonic function, normalized to be 1 at a fixed point x_{0} ∈ ∂C(M). For any x ∈ E_{j} there is a surface S_{j}(x) containing x which divides E_{j} and which has uniformly bounded area and diameter. If z is a point in the unbounded component of E_{j} \S_{j}(x) then the integral over S_{j}(x) of the gradient of Greens function with pole at z is at least 1 (since in this case the limit set has zero area, hence zero harmonic measure with respect to z). Moreover, the same integral for a Greens function with pole in a different end E_{k}, k ≠ j, is zero over S_{j}(x). Letting z → ∞ we see the same is true for the limiting functions U_{j} and U_{k}.

Now let U = ∑_{j} U_{j}. Then for each end E_{j} and each x ∈ E_{j}, the integral of |∇U| over S_{j}(x) is at least 1. Thus there is a point on S_{j}(x) where the |∇U| ≥ (area(S_{j}(x))^{−1} is bounded uniformly away from zero. Because |∇U| is also bounded above uniformly, so are all second derivatives of U and hence |∇U| is actually bounded uniformly away from zero on some ball of fixed size centered at some point of S_{j}(x).

Finally we note that the gradient lower bound implies U cannot be bounded in any end E_{j}. Together with the maximum principle, Harnack’s inequality and the fact that the ends have bounded diameter cross sections, we easily see that U must tend to infinity in each end. This proves the lemma.

Lemma 9.5. Suppose M = 𝔹/G is topologically tame with N ≥ 1 geometrically infinite ends E_{1}, …, E_{N}, that there is a lower bound for the injectivity radius on M and that Λ ≠ S^{2}. Then for each j = 1, …, N, there is a unique (up to constant multiples) positive harmonic function U_{j} which tends to infinity in E_{j}, is bounded in the other geometrically infinite ends and tends to zero in the geometrically finite ends. Any positive harmonic function on M which tends to zero in the geometrically finite ends is of the form ∑ a_{j}U_{j}, where a_{j} ≥ 0, j = 1, …, N.

Proof. Suppose U is a positive harmonic function on M which tends to zero in the geometrically finite ends. Choose a base point z_{0} and with loss of generality assume U(z_{0}) = 1. First we will show that the gradient of U is uniformly bounded.

Suppose R is a large positive number and define F (z) = min(U(z), R). Then F is clearly superharmonic, so ΔF is a negative measure σ_{R}. Let H(z) be the convolution of σ_{R} with the Green’s function for M. Then H is bounded (since σ_{R} locally looks like surface measure on {U = R} and ΔH = σ_{R}. Thus F − H is a bounded harmonic function on M which tends to zero in the geometrically finite ends. Thus the lift of F − H to the ball is a bounded harmonic function with radial limits zero almost everywhere. Hence H = F. Since the Greens function with pole in the convex core of M is bounded below on ∂C(M), it will be uniformly bounded below at z_{0} by Harnack’s inequality. Thus we see that ∥σ_{R}∥ ≃ H(z_{0}) = 1 is uniformly bounded, independent of R. Thus ∇_{H}U is uniformly bounded since the gradient of Green’s function is uniformly bounded.

Now suppose U_{1} and U_{2} are two positive harmonic functions on M, tending to zero in the geometrically finite ends, and both are bounded on M \ E_{1}. Without loss of generality assume U_{1}(z_{0}) = U_{2}(z_{0}) = 1. Then F = U_{1} − U_{2} has uniformly bounded gradient in E_{1} and must be zero on every connected surface S which divides E_{1} (otherwise solving the Dirichlet problem on M \ S with boundary values F |_{S} would show F (z_{0}) ≠ 0). Since E has a family of dividing surfaces with bounded diameter we deduce that F is uniformly bounded in E_{1}. As above the lift of F to the ball is a bounded harmonic function with radial limits zero almost everywhere and hence U_{1} = U_{2}.

Given any positive harmonic function U on M which tends to zero in the geometrically finite ends, we can can write it as a sum of harmonic functions, each of which is unbounded in at most one end. To do this, suppose U is unbounded in E_{1} and choose a family of dividing surfaces S_{t} which tend to infinity in E_{1}. Solve the Dirichlet problem with boundary values U|_{S_{t}} (and zero elsewhere) and take the limit as t → ∞. This gives a positive harmonic function U_{1} on M. Moreover if C(M) = K ∪_{j} E_{j} where K is compact, let C be an upper bound for U on K. Then the maximum principle implies U − C ≤ U_{1} ≤ U in E_{1} and hence U − U_{1} is bounded in E_{1}. Now repeat this procedure in each of the other ends to obtain the desired decomposition. It is unique since two decompositions would differ by a bounded harmonic function in M which if U = ∑_{j} U_{j} = ∑_{j} U′_{j} then U_{j} −U′_{j} would be bounded and hence zero on M.

The previous result can be summarized using the language of Martin boundaries. It says that there is exactly one Martin boundary point corresponding to each geometrically infinite end. The result can also be interpreted in terms of conformal densities on the ball.

Corollary 9.6. Suppose G is a Kleinian group and M = 𝔹/G is as in the previous lemma. Then associated to each geometrically infinite end of M there is a unique (up to constant multiples) conformal density μ_{j} of dimension 2. G is ergodic with respect to each of these densities. The densities {μ_{j}} are mutually singular and any conformal density for G is of the form ∑_{j} a_{j}μ_{j}, a_{j} ≥ 0.

Proof. Conformal densities of dimension 2 correspond to positive harmonic functions on M via the Poisson extension, so existence and uniqueness follow from the previous lemma. Similarly for the decomposition of densities. The action is ergodic on each μ_{j} since an invariant subset E with measure 0 < μ_{j}(E) < μ_{j}(Λ) would imply the corresponding harmonic function U_{j} could be decomposed as the sum of two other positive harmonic functions (the Poisson extensions of μ_{j}|_{E} and μ_{j}|_{E^{c}}) which were not constant multiplies of each other (since they are Poisson integrals of mutually singular measures). To see that the densities corresponding to different ends are mutually singular recall that given harmonic functions U_{1} and U_{2} corresponding to different ends, U_{2} remains bounded along geodesics where U_{1} tends to infinity. Thus if we consider the corresponding conformal densities, the Poisson integral of μ_{1} tends to infinity μ_{1} almost everywhere and the Poisson integral of μ_{2} remains bounded μ_{1} almost everywhere. Hence μ_{1} and μ_{2} must be singular.

References

[1] L. Ahlfors. Finitely generated Kleinian groups. Amer. J. of Math., 86:413–429, 1964.

[2] A.F. Beardon and B. Maskit. Limit points of Kleinian groups and finite sided fundamental polyhedra. Acta Math., 132:1–12, 1974.

[3] L. Bers. Inequalities for finitely generated Kleinian groups. J. d’Analyse Math., 18:23–41, 1967.

[4] C.J. Bishop. Geometric exponents and Kleinian groups. Invent. Math., 127:33–50, 1997.

[5] C.J. Bishop and P.W. Jones. Hausdorff dimension and Kleinian group. To appear in Acta Math., also appeared as Stony Brook IMS preprint 1994/5.

[6] R.D. Canary. Ends of hyperbolic 3-manifolds. J. Amer. Math. Soc., 6:1–35, 1993.

[7] L. Carleson. On the support of harmonic measure for sets of Cantor type. Ann. Acad. Sci. Fenn., 10:113–123, 1985.

[8] S.-Y. A. Chang, J.M. Wilson, and T. Wolff. Some weighted norm inequalities concerning the Schrödinger operators. Comment. Math. Helv., 60:217–246, 1985.

[9] J.B. Garnett. Bounded analytic functions. Academic Press, 1981.

[10] A.M. Garsia. Martingale inequalities: seminar notes on recent progress. W.A. Benjamin, Inc., Reading, Massachusetts, 1973.

[11] V.Ya. Gutlyanskii and M. Vuorinen. On maps almost conformal at the boundary. 1995. Helsinki preprint 98, to appear in Complex Variables.

[12] E. Hopf. Fuchsian groups and ergodic theory. Trans. Amer. Math. Soc., 39:299–314, 1936.

[13] S. Janson and P.W. Jones. Interpolation between H^{p} spaces: the complex method. J. Funct. Analysis, 48:58–80, 1982.

[14] F. John and L. Nirenberg. On functions of bounded mean oscillation. Comm. Pure Appl. Math., 14:415–416, 1961.

[15] O. Lehto. On the differentiability of quasiconformal mappings with prescribed complex dilatation. Ann. Acad. Sci. Fenn, Ser. A I, 275, 1960.

[16] N.G. Makarov. On the distortion of boundary sets under conformal mappings. Proc. London Math. Soc., 51:369–384, 1985.

[17] N.G. Makarov. Probability methods in the theory of conformal mappings. Albegra i Analiz, 1:3–59, 1989. English translation in Leningrad J. Math. 1:1-56, 1990.

[18] N.G. Makarov. Smooth measures and the law of the iterated logarithm. Izv. Akad. Nauk SSSR Ser. Mat., 53:439–446, 1989. English translation in Math. USSR-Izv. 34:455–463, 1990.

[19] R.S. Martin. Minimal positive harmonic functions. Trans. Amer. Math. Soc., 49:137–172, 1941.

[20] B. Maskit. Kleinian groups. Springer-Verlag, 1988.

[21] P. Mattila. Geometry of sets and measures in Euclidean spaces: fractals and rectifiability. Cambridge University Press, 1995.

[22] C. McMullen. Renormalization and 3-manifolds which fiber over the circle. Annals of Math. Studies, 142. Princeton University Press, 1996.

[23] P.J. Nicholls. The ergodic theory of discrete groups. LMS Lecture Notes 143. Cambridge University Press, 1989.

[24] C.A. Rogers and S.J. Taylor. Functions continuous and singular with respect to a Hausdorff measure. Mathematika, 8:1–31, 1961.

[25] E. Stein. Singular integrals and differentiability properties of functions. Princeton University Press, 1971.

[26] D. Sullivan. Growth of positive harmonic functions and Kleinian group limit sets of planar measure 0 and Hausdorff dimension 2. In Geometry Symposium (Utrecht 1980), Lecture Notes in Math. 894, pages 127–144. Springer-Verlag, 1981.

C.J. Bishop, Mathematics Department, State University of New York at Stony Brook, Stony Brook, NY 11794-3651

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

P.W. Jones, Mathematics Department, Yale University, New Haven, CT 06520