TTangential Loewner hulls
Joan Lind
September 4, 2019
Abstract
Through the Loewner equation, real-valued driving functions generate sets calledLoewner hulls. We analyze driving functions that approach 0 at least as fast as a ( T − t ) r as t → T , where r ∈ (0 , / T . We also prove a result about trace existence and applyit to show that the Loewner hulls driven by a ( T − t ) r for r ∈ (0 , /
2) have a tangentialtrace curve.
Contents
The Loewner equation provides a correspondence between continuous functions (calleddriving functions) and certain families of growing sets (called hulls). We are interestedin the question of how analytic properties of the driving functions affect geometricproperties of the hulls, a question that has inspired much research (such as [MR], [Li],[LMR], [W], [LT], [KLS], [ZZ], among others.)In this paper, we examine the end behavior of Loewner hulls driven by functionsthat are bounded below by a ( T − t ) r , where r ∈ (0 , / T = 1). a r X i v : . [ m a t h . C V ] S e p igure 1: The trace curve driven by a (1 − t ) / hits back on itself tangentially when a = 2 . R tangentially when a = 4 (right). Theorem 1.1.
Assume that λ is a driving function defined on [0 , satisfying that λ (1) = 0 and λ ( t ) ≥ a (1 − t ) r for a ≥ and r ∈ (0 , / . Let K t be the Loewner hullgenerated by λ , and let p = inf { x ∈ K ∩ R } . Then near p , K is contained in theregion { x + iy : 0 ≤ x, ≤ y ≤ C ( x − p ) − r } for C = C ( a, r ) > . This is the counterpoint to a result in [KLS] which analyzes the initial behaviorof hulls driven by functions that begin faster than at r for r ∈ (0 , /
2) and shows thatthese hulls leave the real line tangentially. The end-hull question, however, is slightlyharder to analyze due to the influence of the past on hull growth.We view Theorem 1.1 as a partial extension of the following result from [LMR] tothe κ = ∞ case. Theorem 1.2 (Theorem 1.3 in [LMR]) . If λ : [0 , T ] → R is sufficiently regular on [0 , T ) and if lim t → T | λ ( T ) − λ ( t ) |√ T − t = κ > , then the trace γ driven by λ satisfies that γ ( T ) = lim t → T γ ( t ) exists, is real, and γ intersects R in the same angle as the trace for κ √ − t . Theorem 1.1 addresses the approach to R , but it does not address the question ofthe existence of a trace. To give a fuller extension, we address the existence of thetrace in the following result. Proposition 1.3. If λ : [0 , T ] → R is sufficiently nice on [0 , T ) with | λ ( T ) − λ ( t ) | ≥ √ T − t for all t ∈ [0 , T ) , then the trace γ driven by λ satisfies that γ ( T ) = lim t → T γ ( t ) exists and is real. The needed assumption of Proposition 1.3, which utilizes the notion of Loewnercurvature introduced in [LRoh], will be made explicit in Section 4. Taken together,Theorem 1.1 and Proposition 1.3 provide an understanding of the hulls driven byfunctions a ( T − t ) r , as illustrated in Figure 1. K π driven by Weierstrass function 4 ∞ (cid:88) k =0 − n/ cos(3 n t ). Corollary 1.4.
Let a (cid:54) = 0 and r ∈ (0 , / . The Loewner hulls generated by λ ( t ) = a ( T − t ) r have a trace curve for t ∈ [0 , T ] . This curve approaches the real line or itselftangentially as t → T . We have interest in applying Theorem 1.1 to some driving functions that lack theregularity of a ( T − t ) r . See Figure 2 for one such example. We will briefly discuss thisand other examples in the last section.Due to our desire to understand the hulls of less regular driving functions, onemight ask if there are weaker conditions than those of Proposition 1.3 that would stillgive the existence of a trace. In general, the question of the existence of the trace isdifficult and there has not been much progress on this front (as a notable exceptionto this statement, see the work in [ZZ]). We further discuss this question in the lastsection, and we give an example to show that monotonicity, while used in the proof ofProposition 1.3, is not enough to guarantee the trace existence.We end with a brief note about the organization of this paper. Section 2 containsbackground on the Loewner equation, and Section 3 contains the proof of Theorem1.1. In Section 4 we explore the trace existence question by proving Proposition 1.3and Corollary 1.4 and discussing some examples. This section briefly introduces the relevant background regarding the Loewner equa-tion. See [La] for a more detailed introduction.We work with the chordal Loewner equation in the setting of the upper halfplane H . In this context, the Loewner equation is the following initial value problem: ∂ t g t ( z ) = 2 g t ( z ) − λ ( t ) , g ( z ) = z (2.1)where λ is a continuous real-valued function and z ∈ H . For each initial value z ∈ H \ { λ (0) } , a unique solution to (2.1) exists as long as the denominator remains non-zero. We collect the initial values that lead to a zero in the denominator into sets calledhulls: K t = { z : g s ( z ) = λ ( s ) for some s ∈ [0 , t ] } . K t = [ a, a + i √ t ] driven by the constant driving function λ ≡ a .Top right: The hull driven by a linear driving function λ ( t ) = at with a >
0. Bottom left:The hull K driven by 3 √ − t . Bottom right: The hull K driven by 5 √ − t contains theblue trace curve γ and the points under γ in H . One can show that H \ K t is simply connected and g t is a conformal map from H \ K t onto H . Since the driving function λ determines the families of hulls K t , we say that λ generates K t or that K t is driven by λ .In many cases, there is a curve γ (called a trace) so that K t is the complement ofthe unbounded component of H \ γ [0 , t ] for all t . When the trace γ is a simple curvein H ∪ { λ (0) } , then the situation is especially nice and we have that K t = γ [0 , t ]. Inthis case, g t can be extended to the tip γ ( t ) and g t ( γ ( t )) = λ ( t ).See Figure 3 for some example Loewner hulls, which were computed in [KNK].Note that in the bottom right example, the hull K driven by 5 √ − t has a trace γ which is not a simple curve in H ∪ { λ (0) } . As a result, the hull also contains the pointsunder the curve and a real interval. The third and fourth hulls in Figure 3 are froman important family of driving functions k √ − t . We will use the following usefulinformation about this family: Lemma 2.1.
Let k ≥ . The Loewner hull K driven by k √ − t contains the realpoints in the interval [( k − √ k − / , k ] . Further, the Loewner hull K driven by λ with λ (1) = 0 and λ ( t ) ≥ k √ − t contains the real interval [( k − √ k − / , λ (0)] . The first statment can be established by a short computation (see, for instance,Lemma 2.3 in [LRob]). The second statement follows from a comparison between thedriving functions λ and k √ − t (see, for instance, the proof of Theorem 1.1 in [LRob]).While Lemma 2.1 can be used to determine when a Loewner hull is not a simplecurve, the following theorem gives a large class of hulls that are simple curves. We usethe notation || λ || / = sup t (cid:54) = s λ ( t ) − λ ( s ) (cid:112) | t − s | . heorem 2.2 (Theorem 2 in [Li]) . If || λ || / < , then the hulls K t driven by λ satisfy K t = γ [0 , t ] for a simple curve γ contained in H ∪ { λ (0) } . Additional driving function regularity provides additional regularity for the associ-ated trace curves.
Theorem 2.3 ([W], [LT]) . Let λ ∈ C β [0 , T ] for β > / with β + 1 / / ∈ N . Then theLoewner trace driven by λ is in C β + (0 , T ] . Loewner hulls satisfy some useful properties, which we will utilize frequently. If adriving function λ generates hulls K t , then the following hold: • Translation:
For a ∈ R , the driving function λ ( t ) + a generates hulls K t + a . • Scaling:
For k >
0, the driving function kλ ( t/k ) generates hulls kK t/k . • Reflection:
The driving function − λ ( t ) generates hulls R I ( K t ), where R I de-notes reflection about the imaginary axis. • Concatenation:
For s ∈ (0 , T ), the driving function λ ( s + t ) generates hulls g s ( K s + t ).There is an alternate flow that one can use to generate Loewner hulls. Setting ξ ( t ) = λ ( s − t ) for t ∈ [0 , s ], let f t satisfy the following initial value problem: ∂ t f t ( z ) = − f t ( z ) − ξ ( t ) , f ( z ) = z. (2.2)Then f s = g − s (where g t is the solution to (2.1) driven by λ ), and so the hull K s drivenby λ is the closure of H \ f s ( H ). We refer to (2.2) as the upward Loewner flow, since ∂ t Im( f t ( z )) > z ∈ H .For the convenience of the reader, we end this section with statements of resultsfrom other papers (possibly rewritten in our notation) that we will use. Lemma 2.4 (Lemma 3.3b in [CR]) . Let < (cid:15) < . If I ⊂ R is an interval of length √ T and I the concentric interval of size √ T , and if (cid:82) T { λ ( t ) ∈ I } d t ≤ (cid:15)T, then K T ∩ I × [4 √ (cid:15)T , ∞ ) = ∅ . Lemma 2.5 (Lemma 4.2 in [ZZ]) . Let I = { x ∈ R ∩ K \ ∪ t< K t } . If I is an intervaland there exists x ∈ I ◦ and c > so that | λ ( t ) − g t ( x ) |√ − t > c for all t ∈ [0 , , then there exists an open set B in C containing I ◦ so that B ∩ H ⊂ K \ ∪ t< K t . The last result uses the concept of Loewner curvature introduced in [LRoh]. For adriving function λ ∈ C [0 , T ) the Loewner curvature can be computed by LC λ ( t ) = (cid:40) λ (cid:48) ( t ) = 0 λ (cid:48) ( t ) λ (cid:48)(cid:48) ( t ) otherwise . (2.3) ote that driving functions α + c √ τ − t have constant Loewner curvature c /
2. TheLoewner curvature comparison principle (which is stated below in part) allows forcomparison with the hulls generated by constant curvature driving functions.
Theorem 2.6 (Theorem 15b in [LRoh]) . Let γ be the trace driven by λ ∈ C [0 , T ) .If ≤ c / ≤ LC λ ( t ) < ∞ , then γ [0 , T ) does not intersect the interior of the hull K ∗ τ driven by µ ( t ) = α + c √ τ − t , where α and τ are chosen so that λ (0) = µ (0) and λ (cid:48) (0) = µ (cid:48) (0) . In this section, we prove Theorem 1.1. Our first step is to consider the mapped downhull ˆ K s, := g s ( K \ K s ) and show that this hull must be low near 0 (see Lemma 3.2).Then, in the second step, we watch points from ˆ K s, under the upward Loewner flowto gain bounds on K near p . Lemma 3.1.
Suppose λ is defined on [0 , and satisfies that λ (1) = 0 and λ ( t ) ≥ k √ − t for t ∈ [0 , . Let K t be driven by λ . Then K ∩ (( −∞ , × [26 /k, ∞ )) = ∅ .Proof. When k <
13, the result is trivially true because any Loewner hull at time t = 1has its height bounded by 2, and so we assume k ≥
13. Let I ⊂ ( −∞ ,
2] be an intervalof length 1. The amount of time that λ spends in 10 I , the concentric interval of length10 which is contained in ( −∞ , . . /k ) . Therefore, applying Lemma2.4 with (cid:15) = (6 . /k ) , we conclude that K does not intersect I × [26 /k, ∞ ). Lemma 3.2.
Suppose that λ is defined on [0 , and satisfies that λ (1) = 0 and λ ( t ) ≥ a (1 − t ) r where a ≥ and r ∈ (0 , / . Then for s < , ˆ K s, = g s ( K \ K s ) satisfiesthat ˆ K s, ∩ (cid:0) ( −∞ , √ − s ] × [26 a − (1 − s ) − r , ∞ ) (cid:1) = ∅ . (3.1) Further, inf { x ∈ ˆ K s, ∩ R } ≤ a (1 − s ) − r .Proof. The rescaled hull √ − s ˆ K s, is generated by the driving functionˆ λ ( t ) = 1 √ − s λ ( s + t (1 − s )) , t ∈ [0 , λ (1) = 0 and ˆ λ ( t ) ≥ a (1 − s ) r − / √ − t . To obtain (3.1), we applyLemma 3.1 with k = a (1 − s ) r − / and then rescale by √ − s .To establish the second statement, we apply Lemma 2.1 to driving function ˆ λ . Thusfor k = a (1 − s ) r − / , the hull driven by ˆ λ contains the point k − √ k −
162 = 8 k + √ k − ≤ k . In other words, there is a real point ˆ p in the hull √ − s ˆ K s, with ˆ p ≤ a (1 − s ) / − r .Scaling then gives the desired result. ext we need to analyze the upward Loewner flow. For s fixed and t ∈ [0 , s ], weset ξ t = λ ( s − t ) and let f t = x t + iy t satisfy (2.2), which can be decomposed into thepair of equations ∂ t x t = 2 ξ t − x t ( ξ t − x t ) + y t and ∂ t y t = 2 y t ( ξ t − x t ) + y t . (3.2) Lemma 3.3.
Let s ∈ (0 , be fixed and let ξ t satisfy ξ t ≥ a (1 − s + t ) r for t ∈ [0 , s ] and r ∈ (0 , / . Let x t and y t be the solutions to (3.2) with initial values x and y ,respectively, and let p t = f t ( p s ) be the solution to (2.2) with initial value p s = inf { x ∈ ˆ K s, ∩ R } . There exists M = M ( r ) ≥ , so that when a ≥ M , the following hold:(i) If x ≤ √ − s , then x t ≤ √ − s + t for all t ∈ [0 , s ] .(ii) If x ≤ √ − s , then y t ≤ y for all t ∈ [0 , s ] .(iii) If x ∈ [ p s + √ − s, √ − s ] and y < a − (1 − s ) − r , then x t − p t ≥ x − p s for all t ∈ [0 , s ] .Proof. Assume x ≤ √ − s . Let τ ∈ [0 , s ] be a time when x τ = 2 √ − s + τ . Thensince ξ τ ≥ a (1 − s + τ ) r ≥ a √ − s + τ ,( ∂ t x t ) | t = τ ≤ ξ τ − x τ ≤ a − √ − s + τ ≤ √ − s + τ = (cid:0) ∂ t √ − s + t (cid:1)(cid:12)(cid:12) t = τ . This implies that x t can never surpass 2 √ − t + s and hence (i) holds.For (ii), we continue to assume that x ≤ √ − s . Then by (i), we have that ξ t − x t ≥ a (1 − s + t ) r − √ − s + t ≥ ( a − − s + t ) r . At times τ when y τ ≤ y , we have that( ∂ t y t ) | t = τ ≤ y ( a − (1 − s + τ ) r ≤ ∂ t (cid:18) y + 4 y ( a − (1 − r ) (1 − s + t ) − r (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) t = τ We choose a large enough so that 4( a − − (1 − r ) − ≤
1. Thus we can concludethat y t remains bounded by y + y (1 − s + t ) − r ≤ y .Lastly, assume that x ∈ [ p s + √ − s, √ − s ] and y < a − (1 − s ) − r , andassume that a is large enough for (ii) to hold. Then ∂ t ( x t − p t ) = 2 ξ t − x t ( ξ t − x t ) + y t − ξ t − p t = 2 ( ξ t − x t )( x t − p t ) − y t ( ξ t − p t )[( ξ t − x t ) + y t ] . (3.3)Let τ ∈ [0 , s ] be a time when x τ − p τ = x − p s . Our goal is to show that the numerator( ξ τ − x τ )( x τ − p τ ) − y τ >
0, meaning that x t − p t is increasing at time τ (and hence x t − p t ≥ x − p s for all t ∈ [0 , s ]). Now by applying (i) and (ii) and the fact that x − p s ≥ √ − s , we obtain( ξ τ − x τ )( x τ − p τ ) − y τ ≥ ( a − √ − s + τ √ − s − (cid:18) a (cid:19) (1 − s ) − r ≥ (1 − s ) (cid:34) a − − (cid:18) a (cid:19) (1 − s ) − r (cid:35) We can guarantee that this is positive by taking a ≥ roof of Theorem 1.1. We will first prove the case when a ≥ M (i.e. a is large enoughfor Lemma 3.3 to hold). Set K ∗ := K ∩ { z : Re( z ) ≤ } . We will show that K ∗ iscontained in the region { x + iy : 0 ≤ x, ≤ y ≤ a − ( x − p ) − r } .Let s ∈ [0 ,
1) and let z ∈ ˆ K s, with Re( z ) ∈ [ p s + √ − s, √ − s ] } . Then Lemma3.2 and Lemma 3.3 imply thatIm ( f s ( z )) ≤ z ) ≤ a − (1 − s ) − r and Re ( f s ( z )) − p ≥ Re( z ) − p s , where we used that f s ( p s ) = p . Thus f s ( z ) ∈ [ p + √ − s, ∞ ] × [0 , a − (1 − s ) − r ] . It remains to show that K ∗ ⊂ (cid:91) s ∈ [0 , f s (cid:16) ˆ K s, ∩ { z : Re( z ) ∈ [ p s + √ − s, √ − s ] } (cid:17) , (3.4)which will follow once we show the boundary ∂K in { x + iy : x ≤ , y > } iscontained in the right hand side of (3.4). Let z ∈ ∂K ∩ { x + iy : x ≤ , y > } . Then z is added to the hull K at some time T ( z ) <
1. (This follows from Proposition 4.27in [La], which says that there is at most one t -accessible point.) Therefore as s → T ( z ), f − s ( z ) → λ ( T ( z )) ≥ a (1 − T ( z )) r . Since p T ( z )0 ≤ a (1 − T ( z )) − r , there exists some T ( z ) < s < f − s ( z )) ∈ [ p s + √ − s, √ − s ].Now suppose that 4 ≤ a < M . Let s ∈ (0 ,
1) satisfy that a (1 − s ) r − / = M . Sincethe driving function ˆ λ of √ − s ˆ K s, satisfies ˆ λ ( t ) ≥ M (1 − t ) r , the previous case impliesthat (cid:18) √ − s ˆ K s, (cid:19) ∩ { x + iy : x ≤ } ⊂ { x + iy : 0 ≤ x, ≤ y ≤ M − ( x − ˆ p ) − r } for ˆ p = inf { x ∈ √ − s ˆ K s, ∩ R } . The desired result follows since f s ( √ − s z ) is con-formal in a neighborhood of ˆ p and takes √ − s ˆ K s, to K \ K s with ˆ p mapping to p . It remains to show that the constant C in the statement of Theorem 2 only dependson a and r . This will follow from showing that | f (cid:48) s ( p s ) | is bounded below, since when y = Cx b , then (ˆ x, ˆ y ) = ( kx, ky ) satisfy ˆ y = Ck − b ˆ x b . Note that by Schwarz reflection, f s can be extended to be conformal in C \ I for an interval I ∈ R . By the distortiontheorem | f (cid:48) s ( p s ) | ≥ dist( p, K s )dist( p s , I ) . Set d = min { ξ t : t ∈ [0 , s ] } . Then dist( p s , I ) ≤ d . We claim that dist( p, K s ) ≥ d .To prove the claim we will show that dist( p t , ˆ K s − t,s ) ≥ d for all t ∈ [0 , s ]. This holdsat time 0, since ˆ K s,s = { ξ } . For t close to 0, ˆ K s − t,s is near ξ t and ξ t − p t ≥ ξ t / ≥ d .Let τ be the first time t when dist( p t , ˆ K s − t,s ) = 2 d . Let z τ ∈ ˆ K s − τ,s with | p τ − z τ | = 2 d , nd for t ∈ [ τ, s ] let z t = x t + iy t satisfy (2.2). If y τ > d , then | p t − z t | > d for all t ∈ [ τ, s ] since z t is moving upwards. Suppose y τ ≤ d , which means that | x τ − p τ | ≥ d .We will consider the case that x τ − p τ ≥ d , as the other case is similar. Then by (3.3) ∂ t ( x t − p t ) | t = τ = 2 ( x τ − p τ )( ξ τ − p τ ) − d ( ξ τ − p τ )[( ξ τ − x τ ) + y τ ] . Note that the numerator satisfies( x τ − p τ )( ξ τ − p τ ) − d ≥ d ξ τ − d > . Therefore the distance between z t and p t is increasing at time τ , which shows thatdist( p, K s ) ≥ d . In this section we discuss the existence of a trace curve, especially in the context of thedriving functions that we are considering in this paper, i.e. those with end behaviorbounded by a ( T − t ) r for r ∈ (0 , / Question 4.1.
Let λ : [0 , T ] → R be a continuous function such that the correspondingLoewner hull K t has a trace curve for t ∈ [0 , T ) . What additional conditions are neededto guarantee that K t has a trace curve for t ∈ [0 , T ] ? Questions such as this about the existence of the trace have often proved to bedifficult to answer. When | λ ( T ) − λ ( t ) | / √ T − t is bounded as t → T , then Theorem1.2 in [ZZ] gives one possible answer to this question. Since this result does not applywhen the driving function is faster than a ( T − t ) r for r ∈ (0 , /
2) as t → T , we areinterested in other answers to Question 4.1, such as the following result. Proposition 1.3.
Let λ : [0 , T ] → R satisfy λ ∈ C [0 , T ) and | λ ( T ) − λ ( t ) | ≥ √ T − t for all t ∈ [0 , T ) . Assume the Loewner curvature satisfies ≤ LC λ ( t ) < ∞ for all t ∈ [0 , T ) , and there exists δ > so that for s ∈ (0 , , inf t ∈ [ s,T ) LC λ ( t ) ≥ δ √ T − s λ (cid:48) ( s ) . Then the trace γ driven by λ satisfies that γ ( T ) = lim t → T γ ( t ) exists and is real.Proof. From (2.3) and the bound on Loewner curvature, λ (cid:48) ( t ) (cid:54) = 0 for all t ∈ [0 , T ).Hence, λ must be monotone and λ (cid:48) does not change sign. We make the followingsimplifying normalizations: by the scaling property, we may assume that T = 1, bythe translation property, we may assume that λ (1) = 0, and by the reflection property,we may assume that λ (cid:48) ( t ) < q ˆ γ Figure 4: The Loewner curvature comparison principle implies that ˆ γ (shown in blue) doesnot intersect the region below the smaller curve (shown in red) that contains ˆ q . If t ∈ [0 , K t driven by λ satisfies K t = γ [0 , t ] for a simplecurve γ in H ∪ { λ (0) } , by Theorem 2.2. Lemma 2.1 guarantees that K ∩ R is a non-degenerate interval with right endpoint λ (0). Set p = inf { x ∈ K ∩ R } be the leftendpoint. We wish to show that lim t (cid:37) γ ( t ) = p. First we will rule out the case that there are additional limit points of γ in R . Byway of contradiction, we assume that there is q ∈ ( p, λ (0)) ⊂ R so that γ ( t n ) → q fora sequence t n increasing to 1. Let q t = g t ( q ) be the solution to (2.1). By Lemma 2.5,this implies that lim inf t → λ ( t ) − q t √ − t = 0 . Choose s so that λ ( s ) − q s √ − s < δ . Consider the mapped and rescaled hullsˆ K t = 1 √ − s g s ( K s + t (1 − s ) \ K s ) , which are generated by the driving functionˆ λ ( t ) = 1 √ − s λ ( s + t (1 − s )) , t ∈ [0 , . Note that for t <
1, ˆ K t = ˆ γ [0 , t ] for a simple curve ˆ γ and ˆ q = q s √ − s ∈ ˆ K is a limitpoint of ˆ γ ( t ) as t → λ (0) − ˆ q < δ/ B in H containing ˆ q so that B ∩ ˆ γ [0 , T ) = ∅ , as illustrated in Figure 4. We compare ˆ λ to µ ( t ) = α + c √ τ − t , withthe constants chosen as follows: c is chosen so that c / t ∈ [ s,T ) LC λ ( t ), τ is chosenso that ˆ λ (cid:48) (0) = µ (cid:48) (0), and α is chosen so that ˆ λ (0) = µ (0). Since LC ˆ λ ( t ) = LC λ ( s + t (1 − s )) ≥ c ≥ , Figure 5: The beginning of γ (shown in blue) when the set of limit points of γ as t → p ∈ R and points in H . the Loewner curvature comparison principle (Theorem 2.6) implies that ˆ γ stays aboveand never intersects the interior of the hull driven by µ . It remains to show ˆ q iscontained in this hull. Note by Lemma 2.1 the hull driven by c √ − t contains areal interval of length at least c/
2. Hence by scaling, the hull driven by µ ( t ) = α + c √ τ (cid:112) − t/τ contains an interval of length at least c √ τ c √ − s λ (cid:48) ( s ) ≥ δ λ (0). This implies that ˆ q is contained in the interior of the hullgenerated by µ (using the relative topology of H ), and hence ˆ q cannot be a limit pointof ˆ γ .To finish the proof, it remains to show that there cannot be a limit point of γ as t → H . If there were, the set limit points of γ as t → p ∈ R and γ would need to oscillate, such as in Figure 5. Since the setof limit points extends into H , γ must alternate between following its left side, i.e.the prime ends g − t ( x ) for x < λ ( t ), and its right side. This would require λ to benon-monotone.We now apply Proposition 1.3 and Theorem 1.1 to analyze the Loewner hulls drivenby a (1 − t ) r . Proof of Corollary 1.4.
By scaling, we may assume that T = 1, and by reflection wemay assume that a >
0. Thus we take λ ( t ) = a (1 − t ) r for a > r ∈ (0 , / K t be the Loewner hulls driven by λ . By Theorems 2.2 and 2.3, there is asimple curve γ ∈ C . (0 , T ) so that K t = γ [0 , t ] for t ∈ [0 , T ).We first assume that a ≥ √ − rr . Since √ − rr ≥ √ >
4, this assumption guaran-tees that | λ (1) − λ ( t ) | ≥ √ − t . Computing Loewner curvature gives LC λ ( t ) = λ (cid:48) ( t ) λ (cid:48)(cid:48) ( t ) = a r − r · − t ) − r ≥ δ = ar/ (1 − r ),inf t ∈ [ s, LC λ ( t ) √ − s λ (cid:48) ( s ) = ar − r − s ) / − r ≥ δ. λ of Proposition 4.2 is shown in solid blue.It is bounded by the dashed curve a (1 − t ) r . Thus Proposition 1.3 implies that γ ( T ) = lim t → T γ ( t ) exists and is real. Theorem 1.1implies that γ ( t ) approaches R tangentially as t → a < √ − rr follows from the large a case and the concatenationproperty.Since the monotonicity of the driving function played a role in the proof of Propo-sition 1.3, it is natural to ask whether this property is sufficient to answer Question4.1. The following example shows that monotonicty alone is not enough to guaranteethe existence of a trace on the full time interval. We also note that this example couldbe modified so that the driving function is in C [0 , Proposition 4.2.
Let r ∈ (0 , / . There exists a continuous monotone driving func-tion λ : [0 , → R with λ (1) = 0 and λ ( t ) ≥ a (1 − t ) r , where a > , such that thecorresponding Loewner hull K t is a simple curve for t ∈ [0 , , but K does not have atrace.Proof. The driving function λ will be constructed to alternate between constant andlinear portions, as pictured in Figure 6. In particular, each interval of the form I n =[1 − − n , − − ( n +1) ] is divided into two subintervals . On the first subinterval, λ isconstant, equal to 2 − nr a , where a satisfies a ≥ / (1 − − r ). On the second subinterval, λ is linear. Since we require that λ is continuous, choosing the slope of the linear piecewill uniquely identify λ on I n . For t <
1, this construction will give a simple curve γ in H ∪{ a } so that K t = γ [0 , t ]. Let β be the line segment { x + iy : x ∈ [ a − , a ] , y = a − x } .We will construct a nested sequence of subintervals β n converging to a and we willchoose the slopes of the linear portions to guarantee that γ intersects each β n . Thiswill show that the limit points of γ as t → R .We begin with the first interval I . Let s ∈ (0 , / λ ≡ a on [0 , s ], whichimplies γ [0 , s ] is the vertical slit from a to a + i √ s . Applying g s , the curve g s ( β )has an endpoint at a − √ s . For t ∈ [ s, / λ ( t ) = m s ( t − s ) + a , where √ − E t , where E t is an inverse α -stable subordinator with α = 0 . m s = a (2 − r − / − s and we let γ s be the Loewner curve generated by λ restricted to [ s, / a and moves to the left. Making s closer to 1 / m s , which in turn makes γ s [ s, /
2] closer to R . As s → / γ s [ s, /
2] convergesto the real interval [2 − r a, a ] of length a (1 − − r ) ≥
2. Since the distance from a to theendpoint of g s ( β ) is 2 √ s < √
2, we are able to choose s close enough to 1/2, so that γ s [ s, /
2] intersects g s ( β ). This gives us our definition of λ on [0 , /
2] and we set β tobe the connected component of β \ γ [0 , /
2] containing a . Note that we may assumethat g / ( β ) is as close to 2 − r a as we like (by simply taking s closer to 1 /
2, if needed.)The construction for subsequent intervals is similar.Despite the lack of trace, we note that Theorem 1.1 still applies to the aboveexample. We end this section by discussing two further examples where we can applyTheorem 1.1 but which lack the regularity of Proposition 1.3. It is currently unknownwhether either has a trace curve.The first example behaves similarly to the driving function of Proposition 4.2 in thatit is monotone and has periods where it is constant. In particular, we are interestedin the driving function k √ − E t , where E t is an inverse α -stable subordinator. SeeFigure 7. In [KLS] with Kobayashi and Starnes we looked at random time-changeddriving functions of the form φ ( E t ), and as an application of our results, we showedthat when α > /
2, then a.s. k √ E t generates a trace curve that leaves the real linetangentially. Analyzing k √ − E t is more difficult. When α > /
2, the work of [KLS]shows that a.s. k √ − E t generates a trace curve on [0 , T ), before the final time T .When the hull includes points from the real line, then Theorem 1.1 gives the tangentialbehavior of the final hull. However, the question remains open whether the trace existson the full time interval [0 , T ].The second example comes from the family of Weierstrass functions W ( t ) = W b,r,k ( t ) = k ∞ (cid:88) n =0 b − rn cos( b n t ) . The Loewner hulls driven by W have been studied in the r = 1 / G], [ZZ].) When r ∈ (0 , /
2) (and k large enough), then we enter the situation inwhich Theorem 1.1 applies. A simulation of one such example is shown in Figure 2.The tangential behavior on the left side of the hull is due to Theorem 1.1, whereasthe tangential behavior on the right side of the hull is due to Proposition 1.2 of [KLS].We also note that since the simulation that produced Figure 2 creates a trace thatapproximates the hull, this picture suggests that the hull may be a spacefilling curve(and the few white spots are most likely approximation error), but it is unknownwhether the trace exists for this example. Acknowledgement:
We thank Andrew Starnes for his comments.
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