Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 30, 2005, 143–158

A SHARP CONDITION FOR THE LOEWNER EQUATION TO GENERATE SLITS

Joan R. Lind

University of Washington, Department of Mathematics Box 354350, Seattle WA 98195-4350, U.S.A.; jlind@math.washington.edu

Abstract. D. Marshall and S. Rohde have recently shown that there exists C_{0} > 0 so that the Loewner equation generates slits whenever the driving term is Hölder continuous with exponent \frac{1}{2} and norm less than C_{0} [11]. In this paper, we show that the maximal value for C_{0} is 4.

1. Introduction

When Loewner introduced his namesake differential equation in 1923, it greatly impacted the theory of univalent functions. A univalent function f is a conformal map of the unit disk, normalized by f(0) = 0 and f′(0) = 1. In other words, it has the following power series representation in the unit disk:

f(z) = z + a_{2}z^{2} + a_{3}z^{3} + · · · .

In 1916 Bieberbach [2] had shown that |a_{2}| ≤ 2 and had conjectured that |a_{n}| ≤ n for all n. It was Loewner’s differential equation that led to a proof of the case n = 3 in 1923. See [1] or [5] for a proof of this and for more classical applications of the Loewner equation. When the Bieberbach conjecture finally was proved for general n in 1985 by de Branges [4], the Loewner equation again played a key role.

In addition to its importance in the theory of univalent functions, the Loewner differential equation has gained recent prominence with the introduction of a stochastic process called “Stochastic Loewner Evolution”, or SLE, by O. Schramm [13]. Many results in this fast-growing field can be found in the recent work of mathematicians such as Lawler, Rohde, Schramm, Smirnov, and Werner. See [7] for a survey paper with an extensive bibliography.

In the next two sections, we will introduce two formulations of the deterministic Loewner differential equation, the halfplane version and the disk version. This is followed by a discussion of some problems associated with the geometry of the solutions to the Loewner equation. The rest of the paper is concerned with proving Theorem 2 below, which builds upon D. Marshall and S. Rohde’s recent work [11] concerning when the Loewner equation can generate slits. The fifth section contains examples and lemmas related to a natural obstacle to generating slits, the sixth section includes lemmas about conformal welding and the Loewner equation, and the final section is the proof of Theorem 3, which is equivalent to Theorem 2. Some simplifications of the arguments in the fifth section were communicated to us by O. Schramm and are discussed in the Appendix.

2000 Mathematics Subject Classification: Primary 30C35.

2. The Loewner equation in the halfplane

Let γ(t) be a simple continuous curve in H∪{0} with γ(0) = 0 and t ∈ [0, T]. Then there is a unique conformal map g_{t}: H \ γ[0, t] → H with the following normalization, called the hydrodynamic normalization, near infinity:

g_{t}(z) = z + \frac{c(t)}{z} + O(\frac{1}{z^{2}} ).

It is an easy exercise to check that c(t) is continuously increasing in t and that c(0) = 0. Therefore γ can be reparametrized so that c(t) = 2t. Assuming this normalization, one can show that g_{t} satisfies the following form of Loewner’s differential equation: for all t ∈ [0, T] and all z ∈ H \ γ[0, t],

\frac{∂}{∂t}g_{t}(z) = \frac{2}{g_{t}(z) − λ(t)}, g_{0}(z) = z,

where λ is a continuous, real-valued function. Further, it can be shown that g_{t} extends continuously to γ(t) and g_{t}(γ(t)) equals λ(t).

On the other hand, if we start with a continuous λ: [0, T] → R, we can consider the following initial value problem for each z ∈ H:

(1)  \frac{∂}{∂t}g(t, z) = \frac{2}{g(t, z) − λ(t)},  g(0, z) = z.

For each z ∈ H there is some time interval [0, s) for which a solution g(t, z) exists. Let T_{z} = sup{s ∈ [0, T] : g(t, z) exists on [0, s)}. Set G_{t} = {z ∈ H : T_{z} > t} and g_{t}(z) = g(t, z). Then one can prove that the set G_{t} is a simply connected subdomain of H and g_{t} is the unique conformal map from G_{t} onto H with the following normalization near infinity:

g_{t}(z) = z + \frac{2t}{z} + O(\frac{1}{z^{2}} ).

The function λ(t) is called the driving term, and the domains G_{t} as well as the functions g_{t} are said to be generated by λ.

The domains G_{t} generated by a continuous driving term λ are not necessarily slit-halfplanes, i.e. domains of the form H \ γ[0, t], for some simple continuous curve γ in H ∪ {γ(0)} with γ(0) ∈ R. We will give an example later in the paper where a non-slit-halfplane is generated by a driving term which is not only continuous but also is in Lip(\frac{1}{2}). Recall that Lip(\frac{1}{2}) is the space of Hölder continuous functions with exponent \frac{1}{2} , that is the space of functions λ(t) satisfying |λ(s)−λ(t)| ≤ c|s−t|^{1/2} , with ∥λ∥_{1/2} denoting the smallest such c. The necessary and sufficient condition for a decreasing family of domains {G_{t}} to be generated by a continuous driving term can be found in Section 2.3 of [10].

3. The Loewner equation in the disk

The setup for the disk version of the Loewner equation is similar to that of the halfplane version, but the normalization will be at an interior point rather than at a boundary point. For the unit disk D slit by a simple curve γ(t) in D ∪ {1} with γ(0) = 1 and γ(t) ≠ 0 for any t, there is a unique family of conformal maps {g_{t}} so that g_{t}: D \ γ[0, t] → D with the normalizations g_{t}(0) = 0 and g′_{t}(0) > 0. Further, by reparametrizing γ if necessary, we can assume that g′_{t}(0) = e^{t}. If we again set λ(t) = g_{t}(γ(t)), then

(2)  \frac{∂}{∂t}g_{t}(z) = g_{t}(z)\frac{λ(t) + g_{t}(z)}{λ(t) − g_{t}(z)},  g_{0}(z) = z.

Given any continuous function λ: [0, T] → ∂D, we can solve the initial value problem (2) for z ∈ D. As in the halfplane version, this will generate a family of conformal maps {g_{t}} which map from a simply connected subdomain of the unit disk onto the unit disk and which are normalized by g_{t}(0) = 0 and g′_{t}(0) = e^{t}.

4. Some results

We return to the halfplane version of the Loewner equation, which will be the setting for the rest of this paper. For κ ≥ 0, set λ(t) = \sqrt{κ} B_{t}, where B_{t} is standard Brownian motion. Then chordal SLE_{κ} is the random family of conformal maps generated by λ, that is, the family of maps solving the following stochastic differential equation:

\frac{∂}{∂t}g_{t}(z) = \frac{2}{g_{t}(z) − \sqrt{κ} B_{t}} , g_{0}(z) = z.

For SLE, it is possible to define an almost surely continuous path γ: [0, ∞) → \overline{H} such that the domains G_{t} generated by λ(t) = \sqrt{κ} B_{t} are the unbounded components of H \ γ[0, t] for every t ≥ 0. See [12] and, for the case κ = 8, [10]. Further, S. Rohde and O. Schramm [12] have shown the following classification:

(1) For κ ∈ [0, 4], γ(t) is almost surely a simple path contained in H ∪ {0}.

(2) For κ ∈ (4, 8), γ(t) is almost surely a non-simple path.

(3) For κ ∈ [8, ∞), γ(t) is almost surely a space-filling curve.

This result motivates a question in the deterministic setting. Can we classify the kinds of domains generated by a driving term λ in terms of some characteristic of λ? There is only a partial understanding of this question. In the case of a domain slit by an analytic slit, the driving term is real analytic, and if the slit is C^{n} then the driving term is at least C^{n−1} . See [6] and [3].

D. Marshall and S. Rohde address the question of when the generated domains G_{t} are quasislit-halfplanes in [11], where a quasislit-halfplane is the image of H \ [0, i] under a quasiconformal mapping fixing H and ∞. They prove the following:

Theorem 1. If G_{t} is a quasislit-halfplane for all t, then λ ∈ Lip(\frac{1}{2}). Conversely, there exists C_{0} such that if the driving term λ ∈ Lip(\frac{1}{2}) with ∥λ∥_{1/2} < C_{0} , then G_{t} is a quasislit-halfplane for all t.

Although they work with the technically more challenging disk version of the Loewner equation, their techniques carry over to prove the result in the halfplane version as well. In the remainder of this paper, working with the halfplane version of Loewner’s equation, we will show that the maximal value for C_{0} is 4.

Theorem 2. If λ ∈ Lip(\frac{1}{2}) with ∥λ∥_{1/2} < 4, then the domains G_{t} generated by λ are quasislit-halfplanes.

Further, for each c ≥ 4, there exists a driving term λ ∈ Lip(\frac{1}{2}) with ∥λ∥_{1/2} = c so that λ does not generate slit-halfplanes. We will see examples of this in the next section. Similar examples were discovered independently by L. Kadanoff, W. Kager, and B. Nienhuis [8]. Their work also includes descriptions and pictures of the generated domains.

There is another version of the Loewner equation in the halfplane. Let ξ: [0, T] → R be continuous and consider the following initial value problem, in which a negative sign has been introduced on the right-hand side of (1):

(3)  \frac{∂}{∂t}f(t, z) = \frac{−2}{f(t, z) − ξ(t)},  f(0, z) = z

for z ∈ H. In this case, for each z ∈ H, the solution f(t, z) exists for all t ∈ [0, T]. Setting f_{t}(z) = f(t, z), we have that f_{t} is defined on all of H. As in the previous case, it can be shown that f_{t} is a conformal map from H into H, and near infinity it has the form

f_{t}(z) = z + \frac{−2t}{z} + O(\frac{1}{z^{2}} ).

We think of the functions f_{t} as being generated by “running time backward”.

These two forms of Loewner’s differential equation are related. Given a continuous function λ on [0, T], set ξ(t) = λ(T −t). Let g_{t} be the functions generated by λ from (1), and let f_{t} be the functions generated by ξ from (3). It is not true that f_{t}(z) = g_{t}^{−1}(z) for all t ∈ [0, T], but it is true that f_{T} (z) = g_{T}^{−1}(z). Therefore Theorem 2 is equivalent to the following:

Theorem 3. If ξ ∈ Lip(\frac{1}{2}) with ∥ξ∥_{1/2} < 4, then f_{t}(H) is a quasislithalfplane for all t, where f_{t} are the maps generated by ξ.

5. When the singularity catches solutions

Let λ ∈ Lip(\frac{1}{2}) and suppose that the domains G_{t} generated by λ are slithalfplanes. Then the maps g_{t} extend continuously to R \ {λ(0)}. Thus for each x_{0} ∈ R \ {λ(0)}, x(t) := g_{t}(x_{0}) is a solution to the following real-valued initial value problem:

(4)  \frac{∂}{∂t}x(t) = \frac{2}{x(t) − λ(t)},  x(0) = x_{0}.

Further, if λ is defined on [0, T], then x(t) ≠ λ(t) for any t ∈ [0, T], since otherwise (4) would fail to have a solution for all t ∈ [0, T].

Note that if x_{0} > λ(0), then (∂/∂t)x(t) > 0 as long as x(t) ≠ λ(t). So two things can happen: either x(t) continues to move to the right, staying strictly larger than the driving term, or the driving term moves fast enough to “catch” x(t) and there is some time t_{0} where x(t_{0}) = λ(t_{0}). The case when x_{0} < λ(0) is similar but with x(t) moving to the left. Thus, when the domains generated are slit-halfplanes, we see that λ(t) cannot “catch” any solution x(t) to (4).

To build our intuition, let us briefly consider a particular example. Let G_{t} = H \ γ[0, t], where γ parametrizes the upper half-circle of radius \frac{1}{2} centered at \frac{1}{2} , as pictured in Figure 1. In this case it is possible, although unpleasant, to compute the maps g_{t} and to ascertain that the driving term generating this scenario is the function λ(t) = \frac{3}{2} − \frac{3}{2} \sqrt{1 − 8t} , for t ∈ [0, \frac{1}{8}]. The time t = \frac{1}{8} corresponds to the moment that the circular arc touches back on the real line, and G_{1/8} = H\D(\frac{1}{2}, \frac{1}{2}).

Figure 1. One of the domains generated by λ(t) = \frac{3}{2} − \frac{3}{2} \sqrt{1 − 8t} .

For t ∈ [0, \frac{1}{8} − ε], the domains G_{t} are slit-halfplanes, and therefore for any x_{0} ≠ 0, the solutions x(t) to (4) exist on this time interval. What happens to these solutions when t = \frac{1}{8} ? Clearly, g_{1/8} extends only to R \ [0, 1]. That is, on [0, \frac{1}{8}], solutions to (4) exist only for x_{0} > 1 or x_{0} < 0. So if x_{0} ∈ (0, 1], the function x(t) resulting from (4) must be caught by λ at time t = \frac{1}{8} . For example, it is easy to check that the solution to (4) when x_{0} = 1 is x(t) = \frac{3}{2} − \frac{1}{2} \sqrt{1 − 8t} . Here we see that x(\frac{1}{8}) = \frac{3}{2} = λ(\frac{1}{8}).

To determine an upper bound on the constant C_{0} in Theorem 1, we can analyze the situations where this catching could occur, since this implies that the family of domains G_{t} is not a family of slit-halfplanes. In the example above, ∥λ∥_{1/2} = 3\sqrt{2} , which indicates that C_{0} ≤ 3\sqrt{2} . Moreover, for any c ≥ 4, it is easy to give an example of a driving term λ with ∥λ∥_{1/2} = c so that λ can catch a function x(t) generated by (4) for some x_{0} . Let λ(t) = c − c\sqrt{1 − t} and x(t) = c − a\sqrt{1 − t} where a = \frac{1}{2}(c + \sqrt{c^{2} − 16} ). In particular, when c = 4, then λ(t) = 4 − 4\sqrt{1 − t} and x(t) = 4 − 2\sqrt{1 − t} . One can check that x(t) is a solution to (4) with x_{0} = c − a > 0. However x(1) = c = λ(1). Therefore, since λ(t) has caught x(t), λ cannot generate slit-halfplanes. This implies that the constant C_{0} in Theorem 1 cannot be greater than 4.

In contrast to the examples above, the following lemma shows that if λ can catch some x(t), then ∥λ∥_{1/2} ≥ 4. To make things slightly simpler, we take advantage of the fact that the halfplane version of the Loewner equation satisfies a useful scaling property: If λ(t) and x(t) satisfy equation (4), then λ̂(t) := λ(r^{2}t)/r and x̂(t) := x(r^{2}t)/r also satisfy equation (4). Verifying this is an easy exercise. Using this scaling property, we can assume that if a catching occurs, then it happens at time 1. More precisely, if x(t_{0}) = λ(t_{0}) and x(t) ≠ λ(t) for t < t_{0} , then without loss of generality t_{0} = 1. Also, nothing is lost by assuming that λ(0) = 0 and x_{0} > 0.

Lemma 1. Let λ ∈ Lip(\frac{1}{2}) with λ(0) = 0 and let x_{0} > 0. Suppose that x(t) is a solution to (4) and that x(1) = λ(1). Then ∥λ∥_{1/2} ≥ 4.

Proof. Let c=∥λ∥_{1/2} . From (4), we have that x(t) is increasing in t. So then since λ ∈ Lip(\frac{1}{2}),

x(t) − λ(t) ≤ x(1) − λ(1) + c\sqrt{1 − t} ≤ c\sqrt{1 − t} .

From (4) we have

ẋ(t) ≥ \frac{2}{c\sqrt{1 − t}} .

Integrating gives that

x(1) − x(t) ≥ \frac{4}{c}\sqrt{1 − t} .

Letting t = 0 and using that x(1) − x_{0} < c, we see that c − 4/c > 0 and so c > 2. But we also have a better estimate for x(t):

x(t) ≤ x(1) − \frac{4}{c}\sqrt{1 − t} .

Now using this estimate, we can repeat the above argument. So

x(t) − λ(t) ≤ (c − \frac{4}{c})\sqrt{1 − t} ,

which leads to a new estimate for ẋ(t). Then by integration,

x(1) − x(t) ≥ \frac{4}{c − \frac{4}{c}} \sqrt{1 − t} .

This implies that c − (4/(c − 4/c)) > 0 and so c > 2\sqrt{2} . Again we also get an improved estimate for x(t):

x(t) ≤ x(1) − \frac{4}{c − \frac{4}{c}} \sqrt{1 − t} .

Repeating this procedure n times gives that h_{n}(c) > 0 where h_{n} is recursively defined as follows:

h_{1}(x) = x − \frac{4}{x}, h_{n}(x) = x − \frac{4}{h_{n−1}(x)}.

Note that h_{1}(x) is an increasing function from (0, ∞) onto R. It is easy to show inductively that we can define an increasing sequence {x_{n}} so that h_{n}(x_{n}) = 0, and h_{n+1}(x) is an increasing function from (x_{n}, ∞) onto R. Note that we have shown that x_{1} = 2 and x_{2} = 2\sqrt{2} . Since h_{n}(c) > 0 for all n, c > x_{n} for all n. It simply remains to show that x_{n} ↗ 4.

An easy inductive argument gives that h_{n}(4) ≥ 2 for all n. If 4 ∈ (x_{k−1}, x_{k}] for some k, then h_{k}(4) ≤ 0. Therefore, the increasing sequence {x_{n}} is bounded above by 4, and hence there exists some a ≤ 4 such that x_{n} ↗ a. Now, h_{n}(a) > h_{n}(x_{n}) = 0 for all n. If h_{k}(a) ≤ 1 for some k, then h_{k+1}(a) = a−(4/h_{n}(a)) ≤ 0. So we must have h_{n}(a) > 1 for all n. Since h_{n}(a) is decreasing in n and bounded below by 1, h_{n}(a) ↘ L for some L ≥ 1. So then,

L = lim_{n→∞} h_{n}(a) = lim_{n→∞} a − \frac{4}{h_{n−1}(a)} = a − \frac{4}{L}.

Solving the above for L gives that

L = \frac{a ± \sqrt{a^{2} − 16}}{2} .

Since we know the real-valued limit L exists, we must have a ≥ 4. Hence, a = 4, completing the proof.

Note that in the proof above, we have also shown the following: if h_{n}(c) > 0 for all n, then c ≥ 4. This follows since h_{n}(c) > 0 for all n implies that c > x_{n} for all n and since x_{n} ↗ 4. We mention this here, since we will use this fact in the proof of the next lemma.

Although Lemma 1 certainly suggests that the maximal value for C_{0} is 4, it is not a proof of Theorem 2. In theory, there may be more obstacles to generating quasislit-halfplanes than that of the driving term catching up to some solution to (4). However, we will see that this is basically the only obstacle. Refining the above argument gives Lemma 2, which combined with techniques in [11] will lead to the proof of Theorem 2. The idea of Lemma 2 is that if λ can get close to catching some x(t), then ∥λ∥_{1/2} must be close to being greater than or equal to 4.

Lemma 2. Let λ ∈ Lip(\frac{1}{2}) with λ(0) = 0 and ∥λ∥_{1/2} < 4. Then there exists ε = ε(∥λ∥_{1/2}) > 0 so that x(1) − λ(1) > ε, where x(t) is the solution to (4) with x_{0} > 0.

Proof. Suppose x(t) is a solution to (4) for some x_{0} > 0 so that x(1)−λ(1) ≤ ε. We will show that there exists some ε > 0 so that this leads to a contradiction. Again, let c = ∥λ∥_{1/2} . As in the previous proof, define h_{n} recursively by

h_{1}(c) = c − \frac{4}{c}, h_{n}(c) = c − \frac{4}{h_{n−1}(c)}.

Since c < 4, there is some minimal n so that h_{n}(c) ≤ 0 (see the comment following the proof of Lemma 1). If h_{n}(c) = 0, replace c with a slightly larger value, that is, recalling our notation from the previous proof, replace c with some number in the interval (x_{n}, x_{n+1}). Then h_{n+1}(c) < 0. We stop once we are in the case that h_{k}(c) < 0.

Also recursively define e_{n} by

e_{1}(c, ε) = ε + \frac{4ε}{c^{2}} ln(1 + \frac{c}{ε}), e_{n}(c, ε) = ε + \frac{4e_{n−1}(c, ε)}{(h_{n−1}(c))^{2}} ln(1 + \frac{h_{n−1}(c)}{e_{n−1}(c, ε)}).

The recursive definition for e_{n} is unpleasant, but all that we shall need is that for c and n fixed, e_{n}(c, ε) → 0 as ε → 0. This is easy to verify by induction.

To begin, we will prove by induction that

(5)  x(1) − x(t) ≥ ε − e_{n}(c, ε) + (c − h_{n}(c))\sqrt{1 − t} .

First we show equation (5) when n = 1. We have

x(t) − λ(t) ≤ x(1) − λ(1) + c\sqrt{1 − t} ≤ ε + c\sqrt{1 − t}

which implies that

ẋ(t) ≥ \frac{2}{ε + c\sqrt{1 − t}} .

Since

∫_{t}^{1} \frac{2}{a + b\sqrt{1 − s}} ds = \frac{4}{b}\sqrt{1 − t} − \frac{4a}{b^{2}} ln(1 + \frac{b}{a}\sqrt{1 − t} ),

integrating gives

x(1) − x(t) ≥ \frac{4}{c}\sqrt{1 − t} − \frac{4ε}{c^{2}} ln(1 + \frac{c}{ε}\sqrt{1 − t} ),

and so, as desired (5) holds for n = 1.

Next assume equation (5) holds for n = k. Then

x(t) ≤ x(1) − ε + e_{k}(c, ε) + (h_{k}(c) − c)\sqrt{1 − t} ,

and so

x(t) − λ(t) ≤ e_{k}(c, ε) + h_{k}(c)\sqrt{1 − t} .

This again gives us an estimate for ẋ(t) and integrating yields

x(1) − x(t) ≥ \frac{4}{h_{k}(c)}\sqrt{1 − t} − \frac{4e_{k}(c, ε)}{h_{k}(c)^{2}} ln(1 + \frac{h_{k}(c)}{e_{k}(c, ε)}\sqrt{1 − t} ).

Thus equation (5) holds for n = k + 1, completing our verification of (5) by induction.

Recall that x(1) ≤ c + ε. Thus letting t = 0 in equation (5) gives

h_{n}(c) + e_{n}(c, ε) > 0.

As mentioned before, by adjusting c slightly if necessary, there is some n such that h_{n}(c) < 0. Then since e_{n}(c, ε) → 0 as ε → 0, there exists some ε > 0 so that e_{n}(c, ε) < −h_{n}(c). But this contradicts the fact that h_{n}(c) + e_{n}(c, ε) > 0. Therefore, there exists ε > 0 so that x(1) − λ(1) > ε, for x(t) the solution to (4) with x_{0} > 0.

Now we wish to run time backward, and so we must consider the second form of the Loewner equation in the upper halfplane. Recall that from (3), the driving term ξ(t) generates conformal functions f_{t}, which map from H into H. If the image of f_{t} is a quasislit-halfplane, then we can extend f_{t} continuously to R, and for each x_{0} ∈ R \ {ξ(0)}, x(t) := f_{t}(x) is a solution to

(6)  \frac{∂}{∂t}x(t) = \frac{−2}{x(t) − ξ(t)},  x(0) = x_{0}.

Note that the solution x(t) might not exist for all time. Indeed, in the case that ∥ξ∥_{1/2} < 4, the following corollary shows that x(t) will hit the singularity ξ(t) in finite time. We define the hitting time T (x_{0}) to be the first time that x(t) equals ξ(t), that is, x(T (x_{0})) = ξ(T (x_{0})) and x(t) ≠ ξ(t) for t < T (x_{0}). If x(t) never equals ξ(t), then T (x_{0}) := ∞.

Corollary 1. Let ξ ∈ Lip(\frac{1}{2}) with ∥ξ∥_{1/2} < 4 and ξ(0) = 0. Suppose that x(t) is a solution to (6), with x_{0} ≠ 0. Then K_{1}x_{0}^{2} ≤ T (x_{0}) ≤ K_{2}x_{0}^{2} , where 0 < K_{i} = K_{i}(∥ξ∥_{1/2}) < ∞.

Proof. For c = ∥ξ∥_{1/2} , let ε = ε_{c} > 0 be given as in Lemma 2, and let x(t) be the solution to (6) with x(0) = ε. If T (ε) > 1, then λ(t) = ξ(1 − t) − ξ(1) and y(t) = x(1−t)−ξ(1) satisfy the differential equation (4), with y(0) = x(1)−ξ(1) > 0. Thus Lemma 2 implies that ε = y(1) − λ(1) > ε. This is a contradiction, and so T (ε) ≤ 1.

Now suppose x_{0} > 0, with x(t) again the corresponding solution to (6). Then by the scaling property, ξ̂(t) and x̂(t) satisfy equation (6), where

ξ̂(t) := \frac{ε}{x_{0}} ξ(\frac{x_{0}^{2}}{ε^{2}} t),

and

x̂(t) := \frac{ε}{x_{0}} x(\frac{x_{0}^{2}}{ε^{2}} t).

Note that x̂(0) = ε. Therefore

T (x_{0}) = \frac{x_{0}^{2}}{ε^{2}} T (ε) ≤ K_{2}x_{0}^{2}

where K_{2} = K_{2}(c) < ∞.

For the lower bound, assume first that x_{0} = 1, and assume that T (1) = δ is small. Then since ξ(t) ≤ c\sqrt{t} , we have x(δ) ≤ c\sqrt{δ} . Taking δ small enough so that c\sqrt{δ} < \frac{1}{2} , let t_{0} be the time when x(t) = \frac{1}{2} . Then,

−\frac{1}{2} = ∫_{0}^{t_{0}} \frac{−2}{x(s) − ξ(s)} ds ≥ \frac{−2t_{0}}{\frac{1}{2} − c\sqrt{δ}} ,

and so,

\frac{1}{2}(\frac{1}{2} − c\sqrt{δ} ) ≤ 2δ.

This leads to a contradiction if δ is sufficiently small. Therefore T (1) ≥ K_{1} for some K_{1} = K_{1}(c) > 0. Then by the scaling property, T (x_{0}) ≥ K_{1}x_{0}^{2} .

In the previous corollary, we saw that if ∥ξ∥_{1/2} < 4 then solutions x(t) to (6) will hit the singularity in finite time. Lemma 3 shows that there is more that is true. For each finite time, there are exactly two initial points, one on each side of the singularity, so that the solutions to (6) will hit the singularity at that time.

Lemma 3. Let ξ ∈ Lip(\frac{1}{2}) with ∥ξ∥_{1/2} < 4. For each T > 0, there exist exactly two real numbers x_{0}, x̂_{0} so that x(T) = x̂(T) = ξ(T).

Proof. First notice that no two points on the same side of the singularity can give rise to solutions to (6) that will hit at the same time. This follows from the fact that δ(t) := y(t) − x(t) is increasing in t for ξ(0) < x_{0} < y_{0} , since

δ̇(t) = 2\frac{y(t) − x(t)}{(y(t) − ξ(t))(x(t) − ξ(t))}.

Thus there are at most two points that can hit at time T.

Next we will show that there is one point x_{0} to the right of the singularity with x(T) = ξ(T). For each n ∈ N, set w_{n} = ξ(T) + 1/n. Now, starting at w_{n}, run time from T back to 0. This corresponds to solving (4) with initial value w_{n}. Since ∥ξ∥_{1/2} < 4, the driving term cannot catch up with this solution, g_{t}(w_{n}), by Lemma 1, and so it is well-defined up through time T. Thus, x_{n} := g_{T} (w_{n}) = f_{T}^{−1}(w_{n}) is well-defined. Further, by Corollary 1, x_{n} − ξ(0) ≥ K\sqrt{T} , for some K > 0. Therefore, \{x_{n}\} is a decreasing sequence bounded below by ξ(0)+K\sqrt{T} , and so it has a limit x_{0} . Then x_{0} > ξ(0) and clearly we have x(T) = ξ(T). This completes the proof.

6. Conformal welding with the Loewner equation

The previous lemma allows us to define the welding homeomorphism ϕ: R → R as the orientation-reversing map that satisfies ϕ(x) = y if and only if T (x) = T (y). Thus the welding homeomorphism interchanges the two points which hit the singularity at the same time. Note that if ξ is not defined for all time, but for a finite interval [0, T], the welding homeomorphism will not be defined on all R. However, we can overcome this technicality by setting ξ(t) := ξ(T) for t > T.

This next lemma is an analogue of Lemma 3.2 found in [11].

Lemma 4. Let ξ ∈ Lip(\frac{1}{2}) with ∥ξ∥_{1/2} < 4 and ξ(0) = 0. There exists some constant A_{0} > 0, depending only on ∥ξ∥_{1/2} , so that if 0 ≤ x < y < z with y − x = z − y, then

(7)  \frac{1}{A_{0}} ≤ \frac{ϕ(x) − ϕ(y)}{ϕ(y) − ϕ(z)} ≤ A_{0}.

To prove this lemma, we will need the following.

Lemma 5. Let c < 4 and 0 < ε < 1. Then there exists δ > 0 so that

\frac{ϕ(β)}{ϕ(α)} ≥ 1 + δ

for non-zero α and β satisfying β/α ≥ 1 + ε and for any Lip(\frac{1}{2}) driving term ξ with ∥ξ∥_{1/2} ≤ c.

Proof. Notice first that without loss of generality we can take α = −1 and β ≤ −(1 + ε) by the scaling property.

Suppose there is no such δ as in the statement of the lemma. Then for each n ∈ N there exists a driving term ξ_{n} and β_{n} ≤ −(1+ε) so that b_{n} < (1+1/n)a_{n}, where 0 < a_{n} := ϕ(−1) < b_{n} := ϕ(β_{n}). Set T_{n} = T (a_{n}) and S_{n} = T (b_{n}).

By Ascoli–Arzela, there exists a subsequence of {ξ_{n}} which converges locally uniformly to ξ. Note that ξ ∈ Lip(\frac{1}{2}) with ∥ξ∥_{1/2} ≤ c. Since T (x) ≍ x^{2} by Corollary 1, a_{n}, b_{n}, β_{n}, T_{n} and S_{n} are all bounded. Hence by taking subsequences and renaming to avoid notational hazards, we have a_{n} → a, b_{n} → b, β_{n} → β, T_{n} → T and S_{n} → S. Note that a = b since a_{n} < b_{n} < (1 + 1/n)a_{n}. If we had that T (a) = T = T (−1) and T (b) = S = T (β), this would give us the desired contradiction, since T (−1) < T (β). The same argument can be used to prove each of these four equalities, and so we will simply show that T (a) = T. Since ξ_{n} → ξ locally uniformly, ξ_{n}(T_{n}) → ξ(T). Hence lim_{n→∞} a_{n}(T_{n}) = lim_{n→∞} ξ_{n}(T_{n}) = ξ(T), where a_{n}(t) is the solution to (6) with a_{n}(0) = a_{n}. Thus it remains to show that a_{n}(T_{n}) → a(T).

Claim: Let ε > 0. Then a_{n}(T − ε) → a(T − ε).

Proof of Claim: We will assume without loss of generality that T_{n} ≥ T − \frac{1}{2}ε. Then, a_{n}(T − ε) is well-defined and is bounded away from ξ_{n}(T − ε) by a factor of \sqrt{ε} by Corollary 1.

Fix n for a moment. Then looking to solve the initial value problem (6) with the method of successive approximations, let ψ_{0}^{n} ≡ a_{n} and recursively define

ψ_{k+1}^{n}(t) = a_{n} + ∫_{0}^{t} \frac{−2}{ψ_{k}^{n}(s) − ξ_{n}(s)} ds.

Similarly, let ψ_{k} be the approximation for ξ with initial value a. Then for t ∈ [0, T − ε], ψ_{k}^{n}(t) ≥ a_{n}(t) and ψ_{k}(t) ≥ a(t). By an easy induction argument, we have that for t ∈ [0, T − ε],

|ψ_{k}^{n}(t) − ψ_{k}(t)| ≤ |a_{n} − a| + (|a_{n} − a| + ∥ξ_{n} − ξ∥_{∞}) ∑_{j=1}^{k} \frac{(Bt)^{j}}{j!}

where B depends only on ε. So, for t ∈ [0, T − ε],

|a_{n}(t) − a(t)| = lim_{k→∞} |ψ_{k}^{n}(t) − ψ_{k}(t)| ≤ |a_{n} − a| + (|a_{n} − a| + ∥ξ_{n} − ξ∥_{∞})(e^{Bt} − 1).

Therefore, a_{n}(T − ε) → a(T − ε), proving the claim.

Assuming T_{n} ∈ [T − \frac{1}{2}ε, T + \frac{1}{2}ε] and using Corollary 1, we have

\begin{aligned}0 & ≤ a_{n}(T − ε) − a_{n}(T_{n}) = (a_{n}(T − ε) − ξ_{n}(T − ε)) + (ξ_{n}(T − ε) − ξ_{n}(T_{n})) \\ & ≤ A\sqrt{ε} + c\sqrt{T_{n} − (T − ε)} ≤ A\sqrt{ε} \end{aligned}

where A is a constant depending only on c. So by the claim above,

0 ≤ a(T − ε) − lim_{n→∞} a_{n}(T_{n}) ≤ A\sqrt{ε}

implying that a_{n}(T_{n}) → a(T).

Now we are ready for the proof of Lemma 4.

Proof. In this proof, A ≥ 1 will stand for any constant which depends only on ∥ξ∥_{1/2} . Let z(t) be the solution to (6) with z(0) = z, and ẑ(t) the solution to (6) with ẑ(0) = ϕ(z). Define x(t), y(t), x̂(t) and ŷ(t) similarly.

First we consider the case x = 0. Instead of only taking z = 2y, we simply assume that z/y ∈ [1 + ε, 2], since we will reduce the next case to this setting. By the scaling invariance, we can assume that y = 1. Set T = T (1), and recall that K_{1} ≤ T ≤ K_{2} from Corollary 1. Then z(T) − ξ(T) ≤ 2 + c\sqrt{K}_{2} . Abusing notation a little, we have T (z) = T + T (z(T) − ξ(T)), where by T (z(T) − ξ(T)) we mean the hitting time for the solution to (6) with initial value z(T) and driving term ξ(T + t). By Corollary 1,

ϕ(z)^{2} ≤ \frac{1}{K_{1}} T (ϕ(z)) = \frac{1}{K_{1}} T (z) ≤ \frac{K_{2}}{K_{1}} (1 + (2 + c\sqrt{K}_{2} )^{2})

and similarly,

ϕ(1)^{2} ≥ \frac{1}{K_{2}} T (ϕ(1)) = \frac{1}{K_{2}} T (1) ≥ \frac{K_{1}}{K_{2}} .

Therefore,

\frac{ϕ(z)}{ϕ(1)} ≤ A.

By Lemma 5, we have

\frac{ϕ(z)}{ϕ(1)} ≥ 1 + δ

where δ depends only on c and ε. This gives (7) in the case x = 0.

Next we consider the case where x > 0 and z ≥ 2x. We will reduce this to case 1 by letting time run for T = T (x) at which point x(T) = ξ(T). Since

\frac{∂}{∂t} log(\frac{y(t) − x(t)}{z(t) − y(t)}) = 2\frac{z(t) − x(t)}{(x(t) − ξ(t))(y(t) − ξ(t))(z(t) − ξ(t))},

the quotient

q(t) := \frac{y(t) − x(t)}{z(t) − y(t)}

is increasing in t. Therefore q(T) > 1. Also,

q(T) = \frac{y(T) − x(T)}{z(T) − y(T)} ≤ \frac{y + c\sqrt{T}}{\frac{1}{2}(z − x)} ≤ \frac{(1 + c\sqrt{K}_{2} )z}{\frac{1}{4}z} ≤ A.

Now we are back to case 1, since we have

(1 + \frac{1}{A})(y(T) − ξ(T)) ≤ z(T) − ξ(T) ≤ 2(y(T) − ξ(T)).

Hence by case 1, there exists A depending only on c, so that

\frac{1}{A} ≤ \frac{x̂(T) − ŷ(T)}{ŷ(T) − ẑ(T)} ≤ A.

Now we would like to run time from T back to 0 to give (7) for case 2. Since the quotient will be decreasing in t as time run backward, we immediately get the upper bound. For the lower bound,

\frac{ϕ(x) − ϕ(y)}{ϕ(y) − ϕ(z)} ≥ \frac{ϕ(x) − ϕ(y)}{ŷ(T) − ẑ(T)} ≥ \frac{1}{A} \frac{ϕ(x) − ϕ(y)}{x̂(T) − ŷ(T)} ≥ \frac{1}{A} \frac{ϕ(x) − ϕ(y)}{−ϕ(y)} ≥ \frac{1}{A},

where Lemma 5 gives the last inequality. Therefore (7) holds for case 2.

While these first two cases required more work than in the situation in [11], the final case where x > 0 and z − x < x follows the arguments of Lemma 3.2 in [11] without any complications. The idea, similar to the strategy used in the previous case, is to let time run for S, where S is the first time that x(S)−ξ(S) = z(S) − x(S), and to show that the quotient q(t) is bounded on [0, S]. Thus, we end up in a setting similar to case 2. It remains then to verify that case 2 still applies and to run time backward from S to 0, again utilizing the boundedness of q(t).

We include the statement of Lemma 2.2 from [11] below, since we will use it in the proof of Theorem 3. It gives a condition in terms of the welding homeomorphism for when a slit-halfplane is a quasislit-halfplane.

Lemma 6. H\γ[0, T] is a quasislit-halfplane if and only if there is a constant 1 ≤ M < ∞ such that

\frac{1}{M} ≤ \frac{x − ξ(0)}{ξ(0) − ϕ(x)} ≤ M

for all x > ξ(0) and

\frac{1}{M} ≤ \frac{ϕ(x) − ϕ(y)}{ϕ(y) − ϕ(z)} ≤ M

whenever ξ(0) ≤ x < y < z with y − x = z − y. Furthermore, the quasislit constant K of H \ γ[0, T] depends on M only.

7. Proof of Theorem 3

By the scaling property, it suffices to show that f_{1}(H) is a quasislit plane. Let n ∈ N, and set t_{k} = k/n. Following the methods in [11], we wish to construct ξ_{n} ∈ Lip(\frac{1}{2}) so that ξ_{n}(t_{k}) = ξ(t_{k}) and ∥ξ_{n}∥_{1/2} ≤ c := ∥ξ∥_{1/2} . There are at least two ways to proceed. The first is by linear interpolation, and this is the method we will use. Alternatively, setting c_{k} = (ξ(t_{k}) − ξ(t_{k+1}))\sqrt{n} , we can define ξ̂_{n}(t) for t ∈ [0, 1] by ξ̂_{n}|_{[t_{k},t_{k+1}]}(t) = c_{k}\sqrt{t_{k+1} − t} + ξ(t_{k+1}). Although ξ̂_{n} ∈ Lip(\frac{1}{2}), it may not be true that ∥ξ̂_{n}∥_{1/2} ≤ c. However, it is possible to complete the proof using this construction for ξ̂_{n} by considering the larger space of locally Lip(\frac{1}{2}) functions and verifying that all the lemmas remain true for these functions as well. The benefit to using this construction is that we know slightly more about the generated domains. If ϕ̂_{t}^{k} is the map generated by

ξ̂_{n}(t_{k} + t) = c_{k}\sqrt{\frac{1}{n} − t} + α_{k+1}

for t ∈ [0, 1/n], then ϕ̂_{1/n}^{k} is a map from H onto the upper halfplane slit by a line segment whose angle with the real line is bounded away from 0 and π.

Using our first method of linear interpolation, we set m_{k} = n(ξ(t_{k+1})−ξ(t_{k})) and define ξ_{n}(t) for t ∈ [0, 1] by ξ_{n}|_{[t_{k},t_{k+1}]}(t) = m_{k}(t−t_{k})+ξ(t_{k}). First we check that ∥ξ_{n}∥_{1/2} ≤ c. Let x, y ∈ [0, 1]. If x, y ∈ [t_{k}, t_{k} + 1] for some k, then clearly |ξ_{n}(y) − ξ_{n}(x)| ≤ c\sqrt{|y − x|} . So assume that t_{j} ≤ x ≤ t_{j+1} ≤ t_{k} ≤ y ≤ t_{k+1} , and assume without loss of generality that ξ_{n}(y) ≥ ξ_{n}(x). If we maximize the function h(x, y) := ξ_{n}(y) − ξ_{n}(x) − c\sqrt{y − x} over (x, y) ∈ [t_{j}, t_{j+1}] × [t_{k}, t_{k+1}], we find that h(x, y) ≤ 0, as desired.

Let ϕ_{t}^{k} be the maps generated by ξ_{n}(t_{k} + t) = m_{k}t + ξ(t_{k}) for t ∈ [0, 1/n]. Then ϕ^{k} := ϕ_{1/n}^{k} is a map from H onto the upper halfplane slit by a smooth curve which makes an angle of \frac{1}{2}π with the real line. If f_{t}^{n} is the map generated by ξ_{n} for t ∈ [0, 1], we have that f_{1}^{n} = ϕ^{n} ∘ ϕ^{n−1} ∘ ⋯ ∘ ϕ^{2} ∘ ϕ^{1} . Hence, f_{1}^{n}(H) is a slit-halfplane. By Corollary 1, the first condition of Lemma 6 is satisfied, while the second condition is a result of Lemma 4. Therefore, we have that f_{1}^{n}(H) is a K -quasislit-halfplane, with K independent of n. By compactness of the space of K -quasislit-halfplanes, we have that f_{1}(H) is a quasislit-halfplane.

Appendix

In this appendix, we briefly describe some simplifications of the arguments in the fifth section of this paper, which were communicated to us by O. Schramm (private communication). The main idea is to make a reduction so that we only need to use properties of driving terms of the form c\sqrt{1 − t} or c\sqrt{t} which have been studied in [8].

This reduction is based on a simple observation. Let λ_{1} and λ_{2} be two driving terms defined on [0, T] with λ_{1}(t) ≤ λ_{2}(t) for every t ∈ [0, t]. For any x_{0} ≥ λ_{2}(0), let x_{1}(t) and x_{2}(t) be the corresponding solutions to (4) with initial point x_{0} . Notice then that we must have x_{1}(t) ≤ x_{2}(t) for all t for which x_{1} and x_{2} are defined.

To simplify the argument for Lemma 1, suppose that λ_{2} is a Lip(\frac{1}{2}) driving term with norm c that catches the solution x_{2}(t) to (4) at time 1. We shift our picture so that λ_{2}(1) = 0. Then take λ_{1}(t) = −c\sqrt{1 − t} , and note that λ_{1}(t) ≤ λ_{2}(t). Note also that it is not possible for −c\sqrt{1 − t} to catch a solution 158 Joan R. Lind

to (4) before time 1, since it has bounded derivative on the interval [0, 1 − ε]. Therefore by the observation in the previous paragraph, x_{1}(1) ≤ x_{2}(1). However, since x_{2}(t) is caught at time 1, x_{2}(1) = 0 = λ_{1}(1), and we have that x_{1}(t) is also caught at time 1. Therefore, we have reduced our problem to determining the values of c for which the driving term −c\sqrt{1 − t} catches a solution to (4) at time 1. In [8], L. Kadanoff, W. Kager, and B. Nienhuis have shown that this occurs precisely when c ≥ 4.

Corollary 1 can be proved from a study of driving terms of the form c\sqrt{t} , aided by the computations done in [8]. Then Lemma 2 is no longer needed.

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21 January 2004