A real variable characterization of Gromov hyperbolicity of flute surfaces
aa r X i v : . [ m a t h . C V ] M a y A REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OFFLUTE SURFACES
ANA PORTILLA (1) , JOS´E M. RODR´IGUEZ (1)
AND EVA TOUR´IS (1)
Departamento de Matem´aticasEscuela Polit´ecnica SuperiorUniversidad Carlos III de MadridAvenida de la Universidad, 3028911 Legan´es (Madrid), SPAINemails: [email protected], [email protected], [email protected]
Corresponding author: Jos´e M. Rodr´ıguez
Abstract.
In this paper we give a characterization of the Gromov hyperbolicity of trains (a large class ofDenjoy domains which contains the flute surfaces) in terms of the behavior of a real function. This functiondescribes somehow the distances between some remarkable geodesics in the train. This theorem has severalconsequences; in particular, it allows to deduce a result about stability of hyperbolicity, even though theoriginal surface and the modified one are not quasi-isometric.
Key words and phrases : Denjoy domain, flute surface, Gromov hyperbolicity, Riemann surface of infinitetype, train.2000 AMS Subject Classification: 41A10, 46E35, 46G10.1.
Introduction.
The theory of Gromov hyperbolic spaces is a useful tool in order to understand the connections betweengraphs and Potential Theory (see e.g. [4], [10], [13], [21], [22], [23], [24], [30], [31], [35]). Besides, the conceptof Gromov hyperbolicity grasps the essence of negatively curved spaces, and has been successfully used inthe theory of groups (see e.g. [15], [17], [18] and the references therein).A geodesic metric space is called hyperbolic (in the Gromov sense) if there exists an upper bound of thedistance of every point in a side of any geodesic triangle to the union of the two other sides (see Definition2.3). The latter condition is known as Rips condition.But, it is not easy to determine whether a given space is Gromov hyperbolic or not. Recently, there hasbeen some research aimed to show that metrics used in geometric function theory are Gromov hyperbolic.Some specific examples are showing that the Klein-Hilbert metric ([8], [25]) is Gromov hyperbolic (underparticular conditions on the domain of definition), that the Gehring-Osgood metric ([20]) is Gromov hyper-bolic, and that the Vuorinen metric ([20]) is not Gromov hyperbolic (except for a particular case). Recently,some interesting results by Balogh and Buckley [5] about the hyperbolicity of Euclidean bounded domainswith their quasihyperbolic metric have made significant progress in this direction (see also [9], [36] and thereferences therein). Another interesting instance is that of a Riemann surface endowed with the Poincar´emetric. With such metric structure a Riemann surface is always negatively curved, but not every Riemannsurface is Gromov hyperbolic, since topological obstacles may impede it: for instance, the two-dimensionaljungle-gym (a Z -covering of a torus with genus two) is not hyperbolic. Date : April 11, 2008.(1) Research partially supported by three grants from M.E.C. (MTM 2006-11976, MTM 2006-13000-C03-02 and MTM 2007-30904-E), Spain. (1) , JOS´E M. RODR´IGUEZ (1)
AND EVA TOUR´IS (1)
We are interested in studying when Riemann surfaces equipped with their Poincar´e metric are Gromovhyperbolic (see e.g. [32], [33], [34], [26], [27], [28], [3], [29]). To be more precise, in the current paper ourmain aim is to study the hyperbolicity of Denjoy domains, that is to say, plane domains Ω with ∂ Ω ⊂ R .This kind of surfaces are becoming more and more important in Geometric Theory of Functions, since, onthe one hand, they are a very general type of Riemann surfaces, and, on the other hand, they are moremanageable due to its symmetry. For instance, Garnett and Jones have proved the Corona Theorem forDenjoy domains ([14]), and in [2] the authors have got the characterization of Denjoy domains which satisfya linear isoperimetric inequiality.Denjoy domains are such a wide class of Riemann surfaces that characterization criteria are not straight-forward to apply. That is the main reason that led us to focus on a particular type of Denjoy domain, whichwe have called train . A train can be defined as the complement of a sequence of ordered closed intervals(see Definition 2.5). Trains do include a especially important case of surfaces which are the flute surfaces(see, e.g. [6], [7]). These ones are the simplest examples of infinite ends, and besides, in a flute surface itis possible to give a fairly precise description of the ending geometry (see, e.g. [19]). In [3] there are somepartial results on hyperbolicity of trains.This paper is a natural continuation of [3]. Although some of the theorems in the current work mightseem alike to some of the results in the preceding paper, the truth is that they are much more powerfuland the proofs developed are completely new. Without a doubt, the main contribution of this paper isTheorem 3.2, that provides a characterization of the hyperbolicity of trains in terms of the behavior of a realfunction with two integer parameters. (In [3] we give either necessary or sufficient conditions, but there areno characterizations). This function describes somehow the distances between some remarkable geodesics(called fundamental geodesics ) in the train. At first sight, Theorem 3.2 might not seem very user-friendly.However, in practice, this tool let us deduce a result about stability of hyperbolicity, even for cases whenthe original surface and the modified one are not quasi-isometric (see Theorem 3.8).Theorem 3.2 also allows to deduce both sufficient and necessary conditions that either guarantee ordiscard hyperbolicity (see Theorems 3.14, 3.16 and 3.17). Besides, these three theorems give a much simplercharacterization than Theorem 3.2 for an interesting case of trains: those for which the lengths of theirfundamental geodesics are a quasi-increasing sequence. We are talking about Theorem 3.18, another crucialresult in this paper.Theorem 3.22 gives some answers to the following question: how do some perturbations affect on thehyperbolicity of a flute surface?For the sake of clarity and readability, we have opted for moving all the technical lemmas to the lastsection of the paper. This makes the proof of Theorem 3.2, our main result, much more understandable. Notations.
We denote by X a geodesic metric space. By d X and L X we shall denote, respectively, thedistance and the length in the metric of X . From now on, when there is no possible confusion, we will notwrite the subindex X .We denote by Ω a train with its Poincar´e metric.Given a subset F of the complex plane, we define F + = F ∩ { z ∈ C : ℑ z ≥ } , where ℑ z is the imaginarypart of z .If E is either a function or a constant related to a domain Ω, we will denote by E ′ or E j the same functionor constant related to a domain Ω ′ or Ω j , respectively.Finally, we denote by c and c i , positive constants which can assume different values in different theorems. Acknowledgements.
We would like to thank Professor J. L. Fern´andez for some useful discussions.2.
Background in Gromov spaces and Riemann surfaces.
In our study of hyperbolic Gromov spaces we use the notations of [15]. We give now the basic facts aboutthese spaces. We refer to [15] for more background and further results.
REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OF FLUTE SURFACES 3
Definition 2.1.
Let us fix a point w in a metric space ( X, d ) . We define the Gromov product of x, y ∈ X with respect to the point w as ( x | y ) w := 12 (cid:0) d ( x, w ) + d ( y, w ) − d ( x, y ) (cid:1) ≥ . We say that the metric space ( X, d ) is δ - hyperbolic ( δ ≥ if ( x | z ) w ≥ min (cid:8) ( x | y ) w , ( y | z ) w (cid:9) − δ , for every x, y, z, w ∈ X . We say that X is hyperbolic (in the Gromov sense) if the value of δ is not important. It is convenient to remark that this definition of hyperbolicity is not universally accepted, since sometimesthe word hyperbolic refers to negative curvature or to the existence of Green’s function. However, in thispaper we only use the word hyperbolic in the sense of Definition 2.1.
Examples: (1) Every bounded metric space X is ( diamX )-hyperbolic (see e.g. [15, p. 29]).(2) Every complete simply connected Riemannian manifold with sectional curvature which is boundedfrom above by − k , with k >
0, is hyperbolic (see e.g. [15, p. 52]).(3) Every tree with edges of arbitrary length is 0-hyperbolic (see e.g. [15, p. 29]).
Definition 2.2. If γ : [ a, b ] −→ X is a continuous curve in a metric space ( X, d ) , the length of γ is L ( γ ) := sup n n X i =1 d ( γ ( t i − ) , γ ( t i )) : a = t < t < · · · < t n = b o . We say that γ is a geodesic if it is an isometry, i.e. L ( γ | [ t,s ] ) = d ( γ ( t ) , γ ( s )) = | t − s | for every s, t ∈ [ a, b ] .We say that X is a geodesic metric space if for every x, y ∈ X there exists a geodesic joining x and y ;we denote by [ x, y ] any of such geodesics (since we do not require uniqueness of geodesics, this notation isambiguous, but convenient as well). Definition 2.3.
Consider a geodesic metric space X . If x , x , x ∈ X , a geodesic triangle T = { x , x , x } is the union of three geodesics [ x , x ] , [ x , x ] and [ x , x ] . We say that T is δ - thin if for every x ∈ [ x i , x j ] we have that d ( x, [ x j , x k ] ∪ [ x k , x i ]) ≤ δ . The space X is δ - thin ( or satisfies the Rips condition with constant δ ) if every geodesic triangle in X is δ -thin. As the following basic result states, hyperbolicity is equivalent to Rips condition:
Theorem 2.4. ( [15, p. 41] ) Let us consider a geodesic metric space X . (1) If X is δ -hyperbolic, then it is δ -thin. (2) If X is δ -thin, then it is δ -hyperbolic. A non-exceptional Riemann surface S is a Riemann surface whose universal covering space is the unitdisk D = { z ∈ C : | z | < } , endowed with its Poincar´e metric, i.e. the metric obtained by projectingthe Poincar´e metric of the unit disk ds = 2 | dz | / (1 − | z | ). Therefore, any simply connected subset of S is isometric to a subset of D . With this metric, S is a geodesically complete Riemannian manifold withconstant curvature −
1, and therefore S is a geodesic metric space. The only Riemann surfaces which areleft out are the exceptional Riemann surfaces, that is to say, the sphere, the plane, the punctured plane andthe tori. It is easy to study the hyperbolicity of these particular cases. The Poincar´e metric is natural anduseful in Complex Analysis: for instance, any holomorphic function between two domains is Lipschitz withconstant 1, when we consider the respective Poincar´e metrics.A
Denjoy domain is a domain Ω in the Riemann sphere with ∂ Ω ⊂ R ∪ {∞} . As we mentioned in theintroduction of this paper, Denjoy domains are becoming more and more interesting in Geometric FunctionTheory (see e.g. [1], [2], [14], [16]).It is obvious that as we focus on more particular kind of surfaces, we can obtain more powerful results.For this reason we introduce now a new type of space.We have used the word geodesic in the sense of Definition 2.2, that is to say, as a global geodesic or aminimizing geodesic; however, we need now to deal with a special type of local geodesics: simple closed ANA PORTILLA (1) , JOS´E M. RODR´IGUEZ (1)
AND EVA TOUR´IS (1) geodesics, which obviously can not be minimizing geodesics. We will continue using the word geodesic withthe meaning of Definition 2.2, unless we are dealing with closed geodesics.
Definition 2.5. A train is a Denjoy domain Ω ⊂ C with Ω ∩ R = ∪ ∞ n =0 ( a n , b n ) , such that −∞ ≤ a and b n ≤ a n +1 for every n . A flute surface is a train with b n = a n +1 for every n .We say that a curve in a train Ω is a fundamental geodesic if it is a simple closed geodesic which justintersects R in ( a , b ) and ( a n , b n ) for some n > ; we denote by γ n the fundamental geodesic correspondingto n and l n := L Ω ( γ n ) . A curve in a train Ω is a second fundamental geodesic if it is a simple closedgeodesic which just intersects R in ( a n , b n ) and ( a n +1 , b n +1 ) for some n ≥ ; we denote by σ n the secondfundamental geodesic corresponding to n and r n := L Ω ( σ n ) (see figure below). If b n = a n +1 , we define σ n as the puncture at this point and r n = 0 . Given z ∈ Ω , we define the height of z as h ( z ) := d Ω ( z, ( a , b )) . PSfrag replacements a −∞ b a b a b a b a b γ γ σ (a) Train seen as a subset of the complex plane. PSfrag replacements a −∞ b a b a b a b a b γ γ σ (b) The same train seen with “Euclidean eyes”. Remark.
Recall that in every free homotopy class there exists a single simple closed geodesic, assumingthat punctures are simple closed geodesics with length equal to zero. That is why both the fundamentalgeodesic and the second fundamental geodesic are unique for every n .A train is a flute surface if and only if every second fundamental geodesic is a puncture.Flute surfaces are the simplest examples of infinite ends; furthermore, in a flute surface it is possible togive a fairly precise description of the ending geometry (see, e.g. [19]).3. The main results.
It is not difficult to see that the values of { l n } and { r n } determine a train, since for every n thereexists a single fundamental geodesic and a single second fundamental geodesic (see the Remark to Definition2.5). Then, there must exist a characterization of hyperbolicity in terms of the lengths of the fundamentalgeodesics. It would be desirable to obtain such a characterization, since these lengths describe the Denjoydomain from a simple geometric viewpoint.In order to obtain this characterization, we need to introduce the following functions. REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OF FLUTE SURFACES 5 (We refer to the next section for the details of the proofs of technical lemmas. We think that this structuremakes the paper more readable, because it shortens considerably the proof of Theorem 3.2).
Definition 3.1.
Let us consider a sequence of positive numbers { l n } ∞ n =1 and a sequence of non-negativenumbers { r n } ∞ n =1 . Consider n ≥ and ≤ h ≤ l n . We define A n ( h ) := max { m < n : l m ≤ h } if this setis non-empty and A n ( h ) := 1 in other case, B n ( h ) := min { m > n : l m ≤ h } if this set is non-empty and B n ( h ) := ∞ in other case, ∆( k ) := e − l k + e − l k +1 + e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + , and Γ nm ( h ) := (cid:0) r m + h − l m +1 (cid:1) + + e h n − X k = m +1 ∆( k ) , if m < n and l m ≤ h ,l m − h + e h n − X k = m ∆( k ) , if m < n and l m > h , min (cid:8) h, l n − h (cid:9) , if m = n ,l m − h + e h m − X k = n ∆( k ) , if m > n and l m > h , (cid:0) r m − + h − l m − (cid:1) + + e h m − X k = n ∆( k ) , if m > n and l m ≤ h . The functions Γ nm ( h ) are naturally associated to trains by taking { l n } ∞ n =1 and { r n } ∞ n =1 as the half-lengthsof their fundamental geodesics. Theorem 3.2.
A train Ω is hyperbolic if and only if K := sup n ≥ sup h ∈ [0 ,l n ] min m ∈ [ A n ( h ) ,B n ( h )] Γ nm ( h ) < ∞ . Furthermore, if Ω is δ -hyperbolic, then K is bounded by a constant which only depends on δ ; if K < ∞ , then Ω is δ -hyperbolic, with δ a constant which only depends on K . Remarks. (1) Notice that this is a real variable characterization of the hyperbolicity, although the hyperbolicity isa concept of complex geometry, since we consider the Poincar´e metric in each train.(2) Theorem 3.2 clearly improves [3, Theorem 5.3]: we need to know the lengths of the fundamentalgeodesics instead of the precise location of these geodesics and the distances to R from their points.(3) The proof of Theorem 3.2 gives that its conclusion also holds if we replace K by K ( l ) := sup n ≥ sup h ∈ [ l ,l n ] min m ∈ [ A n ( h ) ,B n ( h )] Γ nm ( h ) < ∞ , for any fixed l >
0. In this case, the constant δ depends on K ( l ) and l . Proof.
By [3, Theorem 5.3], Ω is δ -hyperbolic if and only if K := sup n ≥ sup z ∈ γ n inf m ≥ d Ω (cid:0) z, ( a m , b m ) (cid:1) < ∞ , with the appropriate dependence of the constants (if Ω is δ -hyperbolic, then K is bounded by a constantwhich only depends on δ ; if K < ∞ , then Ω is δ -hyperbolic, with δ a constant which only depends on K ).Fix any constant l >
0. Notice that:(1) d Ω (cid:0) z, ( a , b ) (cid:1) = h ( z ) and d Ω (cid:0) z, ( a n , b n ) (cid:1) = l n − h ( z ). Since any z with h ( z ) < l verifiesinf m ≥ d Ω (cid:0) z, ( a m , b m ) (cid:1) ≤ d Ω (cid:0) z, ( a , b ) (cid:1) = h ( z ) < l , we only need to consider z with l ≤ h ( z ) ≤ l n . ANA PORTILLA (1) , JOS´E M. RODR´IGUEZ (1)
AND EVA TOUR´IS (1)
From now on, let us fix n ≥ z ∈ γ n with l ≤ h ( z ) ≤ l n .(2) If k < m < n , with l m ≤ h ( z ), let us consider the geodesic σ which gives the minimum distancebetween z and ( a k , b k ). Define the point w := σ ∩ γ m ; hence d Ω (cid:0) z, w (cid:1) < d Ω (cid:0) z, ( a k , b k ) (cid:1) and Lemma 4.3 gives d Ω (cid:0) z, ( a m , b m ) (cid:1) ≤ d (cid:0) z, ( a m , b m ) ∩ γ m (cid:1) ≤ d (cid:0) z, w (cid:1) ≤ d Ω (cid:0) z, w (cid:1) < d Ω (cid:0) z, ( a k , b k ) (cid:1) . In a similar way, if k > m > n , with l m ≤ h ( z ), then d Ω (cid:0) z, ( a m , b m ) (cid:1) < d Ω (cid:0) z, ( a k , b k ) (cid:1) . Hence we onlyneed to consider d Ω (cid:0) z, ( a m , b m ) (cid:1) with m ∈ { } ∪ [ A n ( h ( z )) , B n ( h ( z ))], in order to study if K is finite.(3) If m ∈ ( A n ( h ( z )) , n ), then l ≤ h ( z ) < l m . By Lemma 4.4, we can replace d Ω (cid:0) z, ( a m , b m ) (cid:1) by d (cid:0) z, γ m ∩ ( a m , b m ) (cid:1) . If z m is the point in γ m with h ( z m ) = h ( z ), then d (cid:0) z, γ m ∩ ( a m , b m ) (cid:1) := d Ω ( z, z m ) + l m − h ( z ). Standard hyperbolic trigonometry in quadrilaterals (see e.g. [12, p. 88]) gives that d Ω ( z, z m ) = 2 Arcsinh (cid:16) sinh 12 d Ω ( γ m , γ n ) cosh h ( z ) (cid:17) . Recall that ( a , b ) contains the shortest geodesic joining γ m and γ n . By Corollary 4.7 we can replace d Ω ( z, z m ) by d Ω ( γ m , γ n ) e h ( z ) , and therefore d (cid:0) z, γ m ∩ ( a m , b m ) (cid:1) by d Ω ( γ m , γ n ) e h ( z ) + l m − h ( z ). Standardhyperbolic trigonometry in right-angled hexagons (see e.g. [12, p. 86]) gives that d Ω ( γ k , γ k +1 ) = Arccosh cosh r k + cosh l k cosh l k +1 sinh l k sinh l k +1 for every k ≥
1. Proposition 4.8 gives d Ω ( γ k , γ k +1 ) = f ( l k , l k +1 , r k ) ≍ e − l k + e − l k +1 + e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + = ∆( k ) , for every k ∈ ( A n ( h ( z )) , n ), since then l k , l k +1 ≥ h ( z ) ≥ l . Therefore we can replace d Ω (cid:0) z, ( a m , b m ) (cid:1) by l m − h ( z ) + e h ( z ) n − X k = m ∆( k ) . A symmetric argument gives that if m ∈ ( n, B n ( h ( z ))), then we can replace d Ω (cid:0) z, ( a m , b m ) (cid:1) by l m − h ( z ) + e h ( z ) m − X k = n ∆( k ) . (4) If m = A n ( h ( z )), then h ( z ) ≥ l m . If z m +1 is the point in γ m +1 with h ( z m +1 ) = h ( z ), by Lemma4.5, we can replace d Ω (cid:0) z, ( a m , b m ) (cid:1) by d Ω (cid:0) z, z m +1 (cid:1) + d Ω (cid:0) z m +1 , ( a m , b m ) (cid:1) . We have seen in (3) that we canreplace d Ω (cid:0) z, z m +1 (cid:1) by e h ( z ) n − X k = m +1 ∆( k ) . Standard hyperbolic trigonometry in pentagons (see e.g. [12, p. 87]) gives thatsinh d Ω (cid:0) z m +1 , ( a m , b m ) (cid:1) = − cosh l m sinh h ( z ) + sinh l m cosh h ( z ) cosh d Ω ( γ m , γ m +1 ) . Standard hyperbolic trigonometry in right-angled hexagons (see e.g. [12, p. 86]) gives thatcosh d Ω ( γ m , γ m +1 ) = cosh r m + cosh l m cosh l m +1 sinh l m sinh l m +1 , and hencesinh d Ω (cid:0) z m +1 , ( a m , b m ) (cid:1) = − cosh l m sinh h ( z ) + cosh h ( z ) cosh r m + cosh l m cosh l m +1 sinh l m +1 = cosh l m (cid:0) cosh l m +1 cosh h ( z ) − sinh l m +1 sinh h ( z ) (cid:1) + cosh r m cosh h ( z )sinh l m +1 = cosh l m cosh (cid:0) l m +1 − h ( z ) (cid:1) + cosh r m cosh h ( z )sinh l m +1 = sinh F (cid:0) l m , l m +1 , r m , h ( z ) (cid:1) , REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OF FLUTE SURFACES 7 where F is the function in Proposition 4.9. Therefore, Corollary 4.10 gives that we can replace d Ω (cid:0) z m +1 , ( a m , b m ) (cid:1) by (cid:0) r m + h ( z ) − l m +1 (cid:1) + . Consequently, we can substitute d Ω (cid:0) z, ( a m , b m ) (cid:1) by (cid:0) r m + h ( z ) − l m +1 (cid:1) + + e h ( z ) n − X k = m +1 ∆( k ) . A symmetric argument gives that if m = B n ( h ( z )), then we can replace d Ω (cid:0) z, ( a m , b m ) (cid:1) by (cid:0) r m − + h ( z ) − l m − (cid:1) + + e h ( z ) m − X k = n ∆( k ) . Notice that each time that we replace a quantity by another in this proof, the constants are undercontrol. Let us remark that (1), (2), (3) and (4) give the result, with inf m ∈ [ A n ( h ) ,B n ( h )] Γ nm ( h ) instead ofmin m ∈ [ A n ( h ) ,B n ( h )] Γ nm ( h ).Let us see now that this infimum is attained. Seeking for a contradiction, suppose that the latest statementis not true. Therefore, B n ( h ) = ∞ and l m > h for every m > n . Then, there exists an increasing sequenceof integer numbers { m j } with lim j →∞ Γ nm j ( h ) = inf m ∈ [ A n ( h ) , ∞ ) Γ nm ( h ). By choosing a subsequence if it isnecessary, we can assume that { Γ nm j ( h ) } j is a decreasing sequence. Hence,Γ nm j +1 ( h ) = l m j +1 − h + e h m j +1 − X k = n ∆( k ) < Γ nm j ( h ) = l m j − h + e h m j − X k = n ∆( k ) . Consequently, we have that l m j +1 < l m j < l m for every j , andΓ nm j ( h ) = l m j − h + e h m j − X k = n ∆( k ) ≥ e h m j X k = n e − l k ≥ e h j X k =1 e − l mk ≥ e h j e − l m . Hence, lim j →∞ Γ nm j ( h ) = lim j →∞ e h j e − l m = ∞ , which is a contradiction. This finishes the proof. (cid:3) Lemma 3.3.
For every r k ≥ and < l k ≤ h ≤ l k +1 , we have (cid:0) r k + h − l k +1 (cid:1) + < e h ∆( k ) . Proof.
Let us remark that it is sufficient to prove r k + h − l k +1 < e h (cid:0) e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + (cid:1) , for every r k ≥ < l k ≤ h ≤ l k +1 .Since the left hand side of the inequality does not depend on l k and the right hand side is a decreasingfunction on l k , it is sufficient to prove r k + h − l k +1 < e h (cid:0) e − ( h + l k +1 − r k ) + + ( r k − h − l k +1 ) + (cid:1) , for every r k ≥ < h ≤ l k +1 .If r k ≤ h + l k +1 , then the inequality is r k + h − l k +1 < e h e − ( h + l k +1 − r k ) = e ( r k + h − l k +1 ) , which trivially holds since t < e t/ for every real number t .If r k ≥ h + l k +1 , then the inequality is r k + h − l k +1 < e h (1 + r k − h − l k +1 ) . Since h >
1, it is clear that the function U ( r k ) := e h (1 + r k − h − l k +1 ) − r k − h + l k +1 is increasing in r k ∈ [ h + l k +1 , ∞ ). Then U ( r k ) ≥ U ( h + l k +1 ) = e h − h >
0, and the inequality holds. (cid:3)
ANA PORTILLA (1) , JOS´E M. RODR´IGUEZ (1)
AND EVA TOUR´IS (1)
Proposition 3.4.
In any train Ω we have min m ∈ [ A n ( h ) ,B n ( h )] Γ nm ( h ) = min m ≥ Γ nm ( h ) , for every n ≥ and ≤ h ≤ l n .Proof. Fix n ≥ ≤ h ≤ l n . If m < A n ( h ), then Lemma 3.3 gives Γ nm ( h ) > Γ nA n ( h ) ( h ):Γ nm ( h ) ≥ e h n − X k = m +1 ∆( k ) ≥ e h n − X k = A n ( h ) ∆( k ) = e h ∆( A n ( h )) + e h n − X k = A n ( h )+1 ∆( k ) > (cid:0) r A n ( h ) + h − l A n ( h )+1 (cid:1) + + e h n − X k = A n ( h )+1 ∆( k ) = Γ nA n ( h ) ( h ) . The case m > B n ( h ) is similar. (cid:3) Proposition 3.5.
If for some n we have l m ≥ l n for every m ≥ n , then the conclusion of Theorem 3.2 alsoholds if we replace [ A n ( h ) , B n ( h )] by [ A n ( h ) , n ] for this n .Proof. It suffices to remark that for every z ∈ γ n and m > n , we have d Ω ( z, ( a n , b n )) = l n − h ( z ) ≤ l m − h ( z ) Proposition 3.6. Let us consider a train Ω with l n ≤ c for every n . Then Ω is δ -hyperbolic, where δ is aconstant which only depends on c .Proof. For each positive integer n , we have Γ nn ( h ) := min (cid:8) h, l n − h (cid:9) ≤ l n ≤ c for every h ∈ [0 , l n ]. Hence, K ≤ c and Theorem 3.2 finishes the proof. (cid:3) One of the important problems in the study of any property is to obtain its stability under appropriatedeformations. Theorem 3.2 allows to prove a result which shows that hyperbolicity is stable under boundedperturbations of the lengths of the fundamental geodesics. Theorem 3.8 is particularly remarkable sincethere are very few results on hyperbolic stability which do not involve quasi-isometries. We need a previouslemma; it deals with some kind of reverse inequality to the one in Lemma 3.3. Lemma 3.7. For every r k , l k +1 ≥ and ≤ h ≤ l k , we have e h (cid:0) e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + (cid:1) ≤ (cid:0) r k + h − l k +1 ) + (cid:1) e ( r k + h − l k +1 ) + . Proof. Since the right hand side of the inequality does not depend on l k and the left hand side is a decreasingfunction on l k , it is sufficient to prove e h (cid:0) e − ( h + l k +1 − r k ) + + ( r k − h − l k +1 ) + (cid:1) ≤ (1 + (cid:0) r k + h − l k +1 ) + (cid:1) e ( r k + h − l k +1 ) + . for every r k , l k +1 , h ≥ h + l k +1 − r k ≥ 0, the inequality is direct since e h (cid:0) e − ( h + l k +1 − r k ) + + ( r k − h − l k +1 ) + (cid:1) = e h e − ( h + l k +1 − r k ) = e ( r k + h − l k +1 ) . If h + l k +1 − r k < 0, then r k − l k +1 > h and ( r k + h − l k +1 ) + > h ; consequently, e h (cid:0) e − ( h + l k +1 − r k ) + + ( r k − h − l k +1 ) + (cid:1) = e h (cid:0) r k − h − l k +1 (cid:1) < (1 + (cid:0) r k + h − l k +1 ) + (cid:1) e ( r k + h − l k +1 ) + . (cid:3) Next, the result about stability that we have talked about before Lemma 3.7. Theorem 3.8 is both aqualitative and a quantitative result. REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OF FLUTE SURFACES 9 Theorem 3.8. Let us consider two trains Ω , Ω ′ and a constant c such that | r ′ n − r n | ≤ c , and | l ′ n − l n | ≤ c for every n ≥ . Then Ω is hyperbolic if and only if Ω ′ is hyperbolic.Furthermore, if Ω is δ -hyperbolic, then Ω ′ is δ ′ -hyperbolic, with δ ′ a constant which only depends on δ and c . This result is a significant improvement with respect to [3, Theorem 5.33], since, in that paper, the lengths r n and r ′ n were required to be bounded, whereas Theorem 3.8 only requires r n − r ′ n to be bounded. Noticethat this is a much weaker condition. Furthermore, the argument in the proof is completely new. Remarks. (1) Notice that in many cases Ω and Ω ′ are not quasi-isometric (for example, if there exists a subsequence { n k } k with lim k →∞ l n k = 0 and l ′ n k ≥ c > { r n } is bounded and { r ′ n } is not bounded, then there exists { l n } = { l ′ n } with Ω hyperbolic and Ω ′ nothyperbolic. Proof. By symmetry, it is sufficient to prove that if Ω is δ -hyperbolic, then Ω ′ is δ ′ -hyperbolic, with δ ′ aconstant which only depends on δ and c . Therefore, let us assume that Ω is δ -hyperbolic.Notice that e − l k + e − l k +1 ≤ e c (cid:0) e − l ′ k + e − l ′ k +1 (cid:1) .If l k + l k +1 ≤ r k , then e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + = 1 + r k − l k − l k +1 and e − ( l ′ k + l ′ k +1 − r ′ k ) + + ( r ′ k − l ′ k − l ′ k +1 ) + ≤ c + r k − l k − l k +1 ≤ (1 + 3 c ) (cid:0) e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + (cid:1) . If l ′ k + l ′ k +1 ≥ r ′ k , then e − ( l ′ k + l ′ k +1 − r ′ k ) + + ( r ′ k − l ′ k − l ′ k +1 ) + = e − ( l ′ k + l ′ k +1 − r ′ k ) + ≤ e c/ (cid:0) e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + (cid:1) . If l k + l k +1 > r k and l ′ k + l ′ k +1 < r ′ k , then l k + l k +1 − r k ≤ l ′ k + l ′ k +1 − r ′ k + 3 c < c ,r ′ k − l ′ k − l ′ k +1 ≤ r k − l k − l k +1 + 3 c < c , and consequently e − ( l ′ k + l ′ k +1 − r ′ k ) + + ( r ′ k − l ′ k − l ′ k +1 ) + = 1 + r ′ k − l ′ k − l ′ k +1 < (1 + 3 c ) e c/ e − c/ < (1 + 3 c ) e c/ (cid:0) e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + (cid:1) . Therefore e − l ′ k + e − l ′ k +1 + e − ( l ′ k + l ′ k +1 − r ′ k ) + +( r ′ k − l ′ k − l ′ k +1 ) + ≤ (1+3 c ) e c/ (cid:0) e − l k + e − l k +1 + e − ( l k + l k +1 − r k ) + +( r k − l k − l k +1 ) + (cid:1) , i.e. ∆ ′ ( k ) ≤ (1 + 3 c ) e c/ ∆( k ). We also have( r ′ m + h − l ′ m +1 ) + ≤ c + ( r m + h − l m +1 ) + ,l ′ m − h ≤ c + l m − h , min (cid:8) h, l ′ n − h (cid:9) ≤ c + min (cid:8) h, l n − h (cid:9) . Hence, we conclude (cid:0) Γ nm (cid:1) ′ ( h ) ≤ (1 + 3 c ) e c/ Γ nm ( h ) + 2 c , for every n, m ≥ h ≥ m = n or l m , l ′ m ≤ h or l m , l ′ m > h .We deal now with the other cases. Let us assume that m ∈ [ A ′ n ( h ) , n ). The case m ∈ ( n, B ′ n ( h )] is similar.If l ′ m ≤ h < l m , then m = A ′ n ( h ) and l ′ m ≤ h < l ′ m +1 . Applying Lemma 3.3 we obtain (cid:0) Γ nm (cid:1) ′ ( h ) = (cid:0) r ′ m + h − l ′ m +1 (cid:1) + + e h n − X k = m +1 ∆ ′ ( k ) < e h n − X k = m ∆ ′ ( k ) ≤ l m − h + (1 + 3 c ) e c/ e h n − X k = m ∆( k ) ≤ (1 + 3 c ) e c/ Γ nm ( h ) . (1) , JOS´E M. RODR´IGUEZ (1) AND EVA TOUR´IS (1) If l m ≤ h < l ′ m , then m > A ′ n ( h ) and h < l ′ m +1 . We also have l ′ m − h ≤ l ′ m − l m ≤ c . Applying Lemma3.7 we obtain (cid:0) Γ nm (cid:1) ′ ( h ) = l ′ m − h + e h − l ′ m + e h − l ′ m +1 + e h (cid:0) e − ( l ′ m + l ′ m +1 − r ′ m ) + + ( r ′ m − l ′ m − l ′ m +1 ) + (cid:1) + e h n − X k = m +1 ∆ ′ ( k ) ≤ c + 2 + (cid:0) r ′ m + h − l ′ m +1 ) + (cid:1) e ( r ′ m + h − l ′ m +1 ) + + (1 + 3 c ) e c/ e h n − X k = m +1 ∆( k ) ≤ c + 2 + (cid:0) c + ( r m + h − l m +1 ) + (cid:1) e c e ( r m + h − l m +1 ) + + (1 + 3 c ) e c/ e h n − X k = m +1 ∆( k ) ≤ c + 2 + (cid:0) c + Γ nm ( h ) (cid:1) e c e Γ nm ( h ) + (1 + 3 c ) e c/ Γ nm ( h ) . We can conclude in any casesup h ∈ [0 , min { l n ,l ′ n } ] min m ∈ [ A ′ n ( h ) ,B ′ n ( h )] (cid:0) Γ nm (cid:1) ′ ( h ) = sup h ∈ [0 , min { l n ,l ′ n } ] min m ≥ (cid:0) Γ nm (cid:1) ′ ( h ) ≤ sup h ∈ [0 ,l n ] min m ≥ (cid:16) c + 2 + (cid:0) c + Γ nm ( h ) (cid:1) e c e Γ nm ( h ) + (1 + 3 c ) e c/ Γ nm ( h ) (cid:17) ≤ c + 2 + (cid:0) c + K (cid:1) e c e K + (1 + 3 c ) e c/ K, for every n ≥ 1, where K only depends on δ , by Theorem 3.2 and Proposition 3.4.If for some n we have l n < l ′ n and h ∈ [ l n , l ′ n ], then (cid:0) Γ nn (cid:1) ′ ( h ) ≤ l ′ n − h ≤ l ′ n − l n ≤ c andsup h ∈ [ l n ,l ′ n ] min m ∈ [ A ′ n ( h ) ,B ′ n ( h )] (cid:0) Γ nm (cid:1) ′ ( h ) ≤ c . Therefore, K ′ ≤ c + 2 + (cid:0) c + K (cid:1) e c e K + (1 + 3 c ) e c/ K , and the conclusion holds by Theorem 3.2. (cid:3) Theorem 3.8 has the following direct consequence. Corollary 3.9. Let us consider two trains Ω , Ω ′ such that r ′ n = r n , and l ′ n = l n for every n ≥ N . Then Ω is hyperbolic if and only if Ω ′ is hyperbolic. Theorems 3.11 and 3.12 are simpler versions of Theorem 3.2, which can be applied in many occasions,and are obtained by replacing Γ nm ( h ) for Γ ∗ nm ( h ) and Γ nm ( h ), respectively. We define now these functions. Definition 3.10. Let us consider a sequence of positive numbers { l n } ∞ n =1 and a sequence of non-negativenumbers { r n } ∞ n =1 . Consider n ≥ and ≤ h ≤ l n . We define Γ ∗ nm ( h ) := (cid:0) r m + h − l m +1 (cid:1) + + e h n X k = m +1 e − l k , if m < n and l m ≤ h ,l m − h + e h n X k = m e − l k , if m < n and l m > h , min (cid:8) h, l n − h (cid:9) , if m = n ,l m − h + e h m X k = n e − l k , if m > n and l m > h , (cid:0) r m − + h − l m − (cid:1) + + e h m − X k = n e − l k , if m > n and l m ≤ h , REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OF FLUTE SURFACES 11 and Γ nm ( h ) := e h n X k = m +1 e − l k , if m < n and l m ≤ h ,e h m − X k = n e − l k , if m > n and l m ≤ h , Γ ∗ nm ( h ) , if m > n in other case.The functions Γ ∗ nm ( h ) and Γ nm ( h ) are naturally associated to trains by taking { l n } ∞ n =1 and { r n } ∞ n =1 as thehalf-lengths of their fundamental geodesics. Theorem 3.11. Let us consider a train Ω such that there exists a constant c > with r n ≤ c + | l n − l n +1 | for every n ≥ . Then Ω is hyperbolic if and only if K ∗ := sup n ≥ sup h ∈ [0 ,l n ] min m ∈ [ A n ( h ) ,B n ( h )] Γ ∗ nm ( h ) < ∞ . Furthermore, if Ω is δ -hyperbolic, then K ∗ is bounded by a constant which only depends on δ and c ; if K ∗ < ∞ , then Ω is δ -hyperbolic, with δ a constant which only depends on K ∗ and c .Proof. First, let us consider the integer numbers k with l k + l k +1 ≥ r k . The inequality r k − l k − l k +1 ≤ c − { l k , l k +1 } (which is equivalent to r k ≤ c + | l k − l k +1 | ) gives e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + = e ( r k − l k − l k +1 ) ≤ e c − min { l k ,l k +1 } ≤ e c (cid:0) e − l k + e − l k +1 (cid:1) . And now, consider the integer numbers k with l k + l k +1 ≤ r k . The inequality 0 ≤ r k − l k − l k +1 ≤ c − { l k , l k +1 } gives min { l k , l k +1 } ≤ c , and consequently e − c ≤ e − min { l k ,l k +1 } , ≤ e c (cid:0) e − l k + e − l k +1 (cid:1) . Hence e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + = 1 + r k − l k − l k +1 ≤ c ≤ (1 + 2 c ) e c (cid:0) e − l k + e − l k +1 (cid:1) . Then e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + ≤ (1 + 2 c ) e c (cid:0) e − l k + e − l k +1 (cid:1) ,e − l k + e − l k +1 ≤ ∆( k ) ≤ (cid:0) c ) e c (cid:1)(cid:0) e − l k + e − l k +1 (cid:1) , for every k ≥ 1. Hence, if we apply Theorem 3.2 we obtain the conclusion, with inf m ∈ [ A n ( h ) ,B n ( h )] Γ ∗ nm ( h )instead of min m ∈ [ A n ( h ) ,B n ( h )] Γ ∗ nm ( h ). In order to see that the infimum is attained we can follow an argumentsimilar to the one at the end of the proof of Theorem 3.2. (cid:3) Theorem 3.12. Let us consider a train Ω such that there exists a constant c > with r n ≤ c for every n ≥ . Then Ω is hyperbolic if and only if K := sup n ≥ sup h ∈ [0 ,l n ] min m ∈ [ A n ( h ) ,B n ( h )] Γ nm ( h ) < ∞ . Furthermore, if Ω is δ -hyperbolic, then K is bounded by a constant which only depends on δ and c ; if K < ∞ , then Ω is δ -hyperbolic, with δ a constant which only depends on K and c . Remark. Notice that Γ nm is much simpler than Γ nm :Firstly, the four terms in the definition of ∆( k ) are replaced by its first term.Furthermore, in the first and fifth cases in the definition of Γ nm we remove the first term in the corre-sponding definition of Γ nm .In order to obtain these simplifications, we must pay with the hypothesis r n ≤ c , but this is a usualhypothesis: for instance, every flute surface satisfies it. Proof. Notice that (cid:0) r m + h − l m +1 (cid:1) + ≤ r m ≤ c if m = A n ( h ) (since l m +1 > h ) and (cid:0) r m − + h − l m − (cid:1) + ≤ r m − ≤ c if m = B n ( h ).Hence, if we apply Theorem 3.11 we obtain the conclusion, with inf m ∈ [ A n ( h ) ,B n ( h )] Γ nm ( h ) instead ofmin m ∈ [ A n ( h ) ,B n ( h )] Γ nm ( h ). (1) , JOS´E M. RODR´IGUEZ (1) AND EVA TOUR´IS (1) In order to see that the infimum is attained we can follow an argument similar to the one at the end ofthe proof of Theorem 3.2. (cid:3) Proposition 3.13. In any train Ω we have min m ∈ [ A n ( h ) ,B n ( h )] Γ nm ( h ) = min m ≥ Γ nm ( h ) , for every n ≥ and ≤ h ≤ l n .Proof. Fix n ≥ ≤ h ≤ l n . If m < A n ( h ), then Γ nm ( h ) > Γ nA n ( h ) ( h ):Γ nm ( h ) ≥ e h n X k = m +1 e − l k > e h n X k = A n ( h )+1 e − l k = Γ nA n ( h ) ( h ) . The case m > B n ( h ) is similar. (cid:3) Theorem 3.12 let us obtain an alternative proof of a result that appears in [3], but using now a completelynew argument. It is a simple sufficient condition for the hyperbolicity. Corollary 3.14. Let us consider a train Ω with l ≤ l , r n ≤ c for every n and (3.1) ∞ X k = n e − l k ≤ c e − l n , for every n > . Then Ω is δ -hyperbolic, where δ is a constant which only depends on c , c and l . Examples. Let us consider an increasing C function f with lim x →∞ f ( x ) = ∞ , and define l n := f ( n ) forevery n . A direct computation gives that { l n } satisfies (3.1) if and only if there exist constants c, M with f ′ ( x ) ≥ c > x ≥ M .Consequently, for a, b > c ∈ R , the sequence l n := an b + c satisfies (3.1) if and only if b ≥ Proof. Let us consider n ≥ h ∈ [ l , l n ]. Since l ≤ l ≤ h , we have that m = A n ( h ) satisfies l m ≤ h < l m +1 and Γ nm ( h ) = e h n X k = m +1 e − l k ≤ e h c e − l m +1 < c . If h ∈ [0 , l ], then Γ nn ( h ) ≤ h ≤ l . Hence, K ≤ max { c , l } , and Theorem 3.12 gives the result. (cid:3) Lemma 3.15. (1) Let us consider a sequence { l n } such that l m ≤ l n + c for every positive integer numbers m ≤ n .Then there exists a non-decreasing sequence { l ′ n } , such that | l n − l ′ n | ≤ c for every n . (2) Let us consider a non-decreasing sequence { l ′ n } . If { l n } is a sequence with | l n − l ′ n | ≤ c for every n ,then l m ≤ l n + 2 c for every positive integer numbers m ≤ n .Proof. We prove now the first part of the lemma. We define a sequence { l ′ n } in the following way: l ′ n :=max { l , l , . . . , l n } . It is clear that { l ′ n } is a non-decreasing sequence. Since l m ≤ l n + c for every m =1 , , . . . , n, we have l n ≤ l ′ n ≤ l n + c . Consequently, | l n − l ′ n | ≤ c for every n .In order to prove the second part, notice that if m ≤ n , then l m ≤ l ′ m + c ≤ l ′ n + c ≤ l n + 2 c . (cid:3) The two following theorems provide necessary conditions for hyperbolicity. Theorem 3.16. Let us consider an hyperbolic train Ω with l m ≤ l n + c for every positive integer numbers m ≤ n . If K is the constant defined in Theorem 3.2, then r n ≤ { K, } + 2 log max { K, } + 3 c , for every n with l n +1 > K + c ) . REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OF FLUTE SURFACES 13 Proof. Let us define M := max { K, } and fix n with l n +1 > K + c ).Let us assume that r n ≤ l n +1 . Consider ε ∈ (0 , / 2) and h n +1 := l n +1 − εr n . ThenΓ n +1 ,n +1 ( h n +1 ) = min { l n +1 − εr n , εr n } = εr n , Γ n +1 ,m ( h n +1 ) ≥ l m − h n +1 ≥ l n +1 − c − h n +1 = εr n − c , if m > n + 1 , Γ n +1 ,n ( h n +1 ) ≥ ( r n + h n +1 − l n +1 ) + = (1 − ε ) r n , if l n ≤ h n +1 , Γ n +1 ,m ( h n +1 ) ≥ e h n +1 ∆( n ) ≥ e l n +1 − εr n e − ( l n + l n +1 − r n ) ≥ e l n +1 − εr n e − ( l n +1 + l n +1 + c − r n ) = e − c +( − ε ) r n , if either m < n or m = n and l n > h n +1 . Since ε ∈ (0 , / M ≥ min (cid:8) εr n , εr n − c , (1 − ε ) r n , e − c +( − ε ) r n (cid:9) = min (cid:8) εr n − c , e − c +( − ε ) r n (cid:9) , and we deduce r n ≤ max n M + c ε , log M + c / / − ε o . Taking ε = ( M + c ) / (2 M + 2 log M + 3 c ) (notice that ε ∈ (0 , / M ≥ r n ≤ M + 2 log M + 3 c .We prove now that r n ≤ l n +1 . Seeking for a contradiction, assume that r n > l n +1 , and consider h n +1 := l n +1 . A similar argument, with h n +1 instead of h n +1 , gives:If l n + l n +1 < r n , since l n +1 > K + c ), K ≥ min n l n +1 , l n +1 − c , l n +1 , e l n +1 o = 14 l n +1 − c > K, since l n +1 > K + c ), and this is a contradiction. If l n + l n +1 ≥ r n , we obtain with a similar argument K ≥ min n l n +1 , l n +1 − c , l n +1 , e l n +1 − c o = min n l n +1 − c , e l n +1 − c o > K, since l n +1 > K + c ), and this is the contradiction we are looking for. (cid:3) Condition l m ≤ l n + c for every positive integer numbers m ≤ n in Theorem 3.16 can seem superfluous,but we have examples which prove that, in fact, if it is removed, then the conclusion of the theorem is nottrue.The following theorem obtains a similar inequality to (3.1) but with an explicit control of the constantsinvolved. Theorem 3.17. Let us consider an hyperbolic train Ω with l m ≤ l n + c for every positive integer numbers m ≤ n . If K is the constant defined in Theorem 3.2, then ∞ X k = n e − l k ≤ K e K + c e − l n , for every n with l n > K + c . Proof. Theorem 3.2 and Proposition 3.4 give thatmin m ≥ Γ nm ( h ) ≤ K , for every n ≥ h ∈ [0 , l n ] . Let us fix n with l n > K + c and n ≥ n . Consider ε > l n ≥ K + c + ε . If we define h := l n − K − c − ε/ ≥ K + ε/ > K , then for any m ≥ n we have l m − h ≥ l n − h − c = K + ε/ > K andΓ n m ( h ) ≥ Γ n m ( h ) ≥ K + ε/ > K . If m < n , we obtain Γ n m ( h ) ≥ Γ n m ( h ) ≥ e h n X k = n e − l k . Consequently, K ≥ min m ≥ Γ n m ( h ) = min ≤ m Let us consider a train Ω with l m ≤ l n + c for every positive integer numbers m ≤ n . (1) If { l n } is a bounded sequence, then Ω is hyperbolic. (2) If lim n →∞ l n = ∞ , then Ω is hyperbolic if and only if { r n } is a bounded sequence and ( ) holds forsome constant c . Remark. Note that Theorem 3.18 deals with every case under the hypothesis “ l m ≤ l n + c for m ≤ n ”: { l n } is either a bounded sequence or a sequence with limit ∞ .If we have an hyperbolic train, we want to study what kind of transformations in { l n } and { r n } allows toobtain another hyperbolic train. Theorem 3.19. Consider two trains Ω and Ω ′ . Let us assume that Ω is δ -hyperbolic. Then, Ω ′ is δ ′ -hyperbolic if we have either: (1) l ′ n = l n and r ′ n ≤ r n for every n ( and then K ′ ≤ K ) , or (2) l ′ n = λl n and r ′ n = λr n for every n ( λ ≥ 1) ( and then K ′ ≤ λK + (1 + λ ) K λ ) , or (3) l ′ n = λl n and r ′ n = µr n for every n ( λ ≥ , ≤ µ ≤ λ ) ( and then K ′ ≤ λK + (1 + λ ) K λ ) .Proof. In case (1), (cid:0) Γ nm (cid:1) ′ ( h ) ≤ Γ nm ( h ) for every n, m ≥ 1, since Γ nm ( h ) is a non-decreasing function ineach variable r k . This allows to deduce (1).In order to prove the second part, notice that (since λ ≥ e λh X k (cid:0) e − λl k + e − λl k +1 + e − ( λl k + λl k +1 − λr k ) + (cid:1) ≤ (cid:16) e h X k (cid:0) e − l k + e − l k +1 + e − ( l k + l k +1 − r k ) + (cid:1)(cid:17) λ . Notice that t ≤ (1 + t ) λ for every t ≥ λ ≥ 1. Hence, if r k − l k − l k +1 ≥ e λh X k ( λr k − λl k − λl k +1 ) + ≤ λ e λh X k (cid:0) r k − l k − l k +1 ) + (cid:1) λ ≤ λ (cid:16) e h X k (cid:0) e − ( l k + l k +1 − r k ) + + ( r k − l k − l k +1 ) + (cid:1)(cid:17) λ . We also have (cid:0) λr m + λh − λl m +1 (cid:1) + = λ (cid:0) r m + h − l m +1 (cid:1) + , λl m − λh = λ ( l m − h ) , min (cid:8) λh, λl n − λh (cid:9) = λ min (cid:8) h, l n − h (cid:9) . Consequently, (cid:0) Γ nm (cid:1) ′ ( λh ) ≤ λ Γ nm ( h ) + Γ nm ( h ) λ + λ Γ nm ( h ) λ for every n, m ≥ ≤ h ≤ l n , and then K ′ ≤ λK + (1 + λ ) K λ .Item (3) is a direct consequence of (1) and (2). (cid:3) We want to study now the following question: If we have an hyperbolic train with { r n } ∈ l ∞ , what kindof perturbations are allowed on { l n } so that the train is still hyperbolic? Theorem 3.22 answers this questionproviding a great deal of hyperbolic flute surfaces.We need the following definitions. Definition 3.20. We denote by H the following set of sequences: H : = (cid:8) { x n } : the train with l n = x n and r n = 0 for every n is hyperbolic (cid:9) = (cid:8) { x n } : every train with l n = x n for every n and { r n } ∈ l ∞ is hyperbolic (cid:9) . REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OF FLUTE SURFACES 15 The second equality is a direct consequence of Theorem 3.8. Definition 3.21. We say that the sequence { y n } is a union of the sequences { x n } , . . . , { x Nn } , if { x n } , . . . , { x Nn } are subsequences of { y n } , and { x n } , . . . , { x Nn } is a partition of { y n } . Theorem 3.22. Let us consider a sequence { l n } ∈ H . (1) If l ′ n = l n + x n with { x n } ∈ l ∞ , then { l ′ n } ∈ H . (2) Fix a positive integer N . Let us assume that { l n } is a subsequence { l ′ n k } of { l ′ n } such that n k +1 − n k ≤ N for every k , and max { l ′ n k , l ′ n k +1 } ≤ l ′ m + N for every m ∈ ( n k , n k +1 ) and every k . Then { l ′ n } ∈ H . (3) If { l ′ n } is any union of the sequences { l n } , . . . , { l Nn } ∈ H , then { l ′ n } ∈ H . (4) If { l ′ n } is a union of { l n } and a sequence { x n } ∈ l ∞ , then { l ′ n } ∈ H . (5) Let us assume that { l ′ n } is any union of the sequences { l n } , . . . , { l Nn } which verify ∞ X k = n e − l jk ≤ c e − l jn , for every n > and j = 1 , . . . , N . Then { l ′ n } ∈ H . (6) Fix a positive integer N . Let us assume that { x n } is a subsequence { l ′ n k } of { l ′ n } such that max { l ′ n k , l ′ n k +1 } ≤ l ′ m + N for every m ∈ ( n k , n k +1 ) and every k . If { x n } / ∈ H , then { l ′ n } / ∈ H . (7) Fix a positive integer N . Let σ be a permutation of the positive integer numbers such that | σ ( n ) − n | ≤ N for every n , and consider l ′ n := l σ ( n ) . Then { l ′ n } ∈ H . Remarks. (1) In fact, (7) gives the following stronger statement: If σ is a permutation of the positive integer numberssuch that | σ ( n ) − n | ≤ N for every n , then { l σ ( n ) } ∈ H if and only if { l n } ∈ H (since σ − also satisfies | σ − ( n ) − n | ≤ N for every n ).(2) We have examples showing that the conclusions of Theorem 3.22 do not hold if we remove any of thehypothesis. Proof. (1) is a direct consequence of Theorem 3.8.(2) Fix n ≥ h ∈ [0 , l ′ n ].Let us consider the maximum integer k such that n k ≤ n < n k +1 .If l ′ s ≤ h for some s ∈ [ n k , n k +1 ], by symmetry, without loss of generality we can assume that there existssome s ∈ [ n k , n ) with l ′ s ≤ h (the case s = n is trivial: if l ′ n ≤ h , then h = l ′ n and (cid:0) Γ nn (cid:1) ′ ( h ) = 0). Hence A ′ n ( h ) ∈ [ n k , n ) and then l ′ k ≥ h for every k ∈ ( A ′ n ( h ) , n ] and n − A ′ n ( h ) ≤ n − n k ≤ N − 1; consequently, (cid:0) Γ nA ′ n ( h ) (cid:1) ′ ( h ) = n X k = A ′ n ( h )+1 e h − l ′ k ≤ n X k = A ′ n ( h )+1 n − A ′ n ( h ) ≤ N − . Let us assume now that l ′ s > h for every s ∈ [ n k , n k +1 ]. There exists some integer m with Γ k m ( h ) ≤ K .By symmetry, without loss of generality we can assume that m ≤ k .If m = k , then min { h, l k − h } ≤ K . If min { h, l k − h } = h , then h ≤ K and we can deduce (cid:0) Γ nn (cid:1) ′ ( h ) = min { h, l ′ n − h } ≤ h ≤ K . If min { h, l k − h } = l k − h , then l k − h ≤ K and (cid:0) Γ nn k (cid:1) ′ ( h ) = l ′ n k − h + n X k = n k e h − l ′ k ≤ l k − h + n X k = n k ≤ K + N. (1) , JOS´E M. RODR´IGUEZ (1) AND EVA TOUR´IS (1) If m < k and l m > h , then Γ k m ( h ) = l m − h + e h P k k = m e − l k ≤ K . Hence (cid:0) Γ nn m (cid:1) ′ ( h ) = l ′ n m − h + e h n k X k = n m e − l ′ k + n X k = n k +1 e h − l ′ k ≤ l ′ n m − h + e h (cid:16) e − l ′ nm + k X j = m +1 n j X k = n j − +1 e − l ′ k (cid:17) + n X k = n k +1 ≤ l ′ n m − h + e h (cid:16) e − l ′ nm + k X j = m +1 N e N − l ′ nj (cid:17) + N − ≤ N e N (cid:16) l m − h + e h k X j = m e − l j (cid:17) + N − ≤ N e N K + N − . If m < k and l m ≤ h , a similar argument gives the same bound for (cid:0) Γ nn m (cid:1) ′ ( h ).Then, (cid:0) K (cid:1) ′ ≤ N e N K + N and Theorem 3.12 implies (2).(3) Assume first that N = 2; then { l ′ n } is the union of { l n } and { l n } . We denote by { l ′ n ik } the subsequence { l in } in { l ′ n } , for i = 1 , 2. Fix n ≥ h ∈ [0 , l ′ n ]. By symmetry, without loss of generality we can assumethat there exist k with n k = n and m ≤ k with (cid:0) Γ k m (cid:1) ( h ) ≤ ( K ) .We can assume that l ′ s > h for every s ∈ ( n m , n k ), since the other case is similar.If there is no k with n k ∈ [ n m , n k ], then (cid:0) Γ n k n m (cid:1) ′ ( h ) = (cid:0) Γ k m (cid:1) ( h ) ≤ ( K ) .Assume now that there exists k with n k ∈ ( n m , n k ). Let us define k := max { k : n k ∈ ( n m , n k ) } .If there exists m ≤ k such that (cid:0) Γ k m (cid:1) ( h ) ≤ ( K ) , then (cid:0) Γ n k , max { n m , n m } (cid:1) ′ ( h ) ≤ (cid:0) Γ k m (cid:1) ( h ) + (cid:0) Γ k m (cid:1) ( h ) ≤ ( K ) + ( K ) . If there exists k verifying the next three conditions simultaneously:( a ) n k ∈ ( n m , n k ),( b ) there exists m ≤ k such that (cid:0) Γ k m (cid:1) ( h ) ≤ ( K ) ,( c ) for every k ∈ ( k , k ] we have (cid:0) Γ km (cid:1) ( h ) > ( K ) for every m ≤ k ,then there exists m > k such that (cid:0) Γ k +1 ,m (cid:1) ( h ) ≤ ( K ) : In fact, seeking for a contradiction, let us as-sume that there exists m ∈ ( k +1 , k ] with (cid:0) Γ k +1 ,m (cid:1) ( h ) ≤ ( K ) ; then (cid:0) Γ m m (cid:1) ( h ) ≤ (cid:0) Γ k +1 ,m (cid:1) ( h ) ≤ ( K ) (recall that l ′ s > h for every s ∈ ( n m , n k )), which is actually a contradiction with ( c ). Hence, (cid:0) Γ n k , max { n m , n m } (cid:1) ′ ( h ) ≤ (cid:0) Γ k m (cid:1) ( h ) + (cid:0) Γ k m (cid:1) ( h ) + (cid:0) Γ k +1 ,m (cid:1) ( h ) ≤ ( K ) + 2( K ) . If for any k with n k ∈ ( n m n k ) we have (cid:0) Γ km (cid:1) ( h ) > ( K ) for every m ≤ k , let us define k := min { k : n k ∈ ( n m , n k ) } . As in the last case, then there exists m > k such that (cid:0) Γ k m (cid:1) ( h ) ≤ ( K ) , and hence (cid:0) Γ n k n m (cid:1) ′ ( h ) ≤ (cid:0) Γ k m (cid:1) ( h ) + (cid:0) Γ k m (cid:1) ( h ) ≤ ( K ) + ( K ) . Consequently, (cid:0) K (cid:1) ′ ≤ K ) + 2( K ) and Theorem 3.12 implies (3) with N = 2. The result for N sequences is obtained by applying N − { x n } / ∈ H , by Theorem 3.12 and Proposition 3.13, for each M > N there exist k and h ∈ (0 , x k )with Γ k m ( h ) ≥ M , for every m ≥ m ≥ 1. By symmetry, without loss of generality we can assume that m ≤ n k . If m = n k , then (cid:0) Γ n k n k (cid:1) ′ ( h ) = min (cid:8) h, l ′ n k − h (cid:9) = min (cid:8) h, x k − h (cid:9) = Γ k k ( h ) ≥ M. REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OF FLUTE SURFACES 17 Notice that if m ∈ ( n k − , n k ), then l ′ m − h ≥ l ′ n k − h − N = x k − h − N ≥ Γ k k ( h ) − N ≥ M − N > , and l ′ m > h . Hence (cid:0) Γ n k m (cid:1) ′ ( h ) ≥ l ′ m − h ≥ M − N .In the case m ≤ n k − , we have n k − < m ≤ n k for some k < k .If x k ≤ h , then (cid:0) Γ n k m (cid:1) ′ ( h ) ≥ e h n k X k = m +1 e − l ′ k ≥ e h k X k = k +1 e − x k = Γ k k ( h ) ≥ M. If x k > h and l ′ m > h , then (cid:0) Γ n k m (cid:1) ′ ( h ) = l ′ m − h + e h n k X k = m e − l ′ k ≥ l ′ n k − h − N + e h n k X k = m e − l ′ k ≥ x k − h − N + e h k X k = k e − x k = Γ k k ( h ) − N ≥ M − N. If x k > h and l ′ m ≤ h , then x k − N = l ′ n k − N ≤ l ′ m ≤ h and 0 ≥ x k − h − N ; therefore (cid:0) Γ n k m (cid:1) ′ ( h ) = e h n k X k = m +1 e − l ′ k ≥ x k − h − N + e h e − x k − e h k X k = k +1 e − x k = Γ k k ( h ) − N − ≥ M − N − . Consequently, (cid:0) K (cid:1) ′ ≥ M − N − M > N , and hence (cid:0) K (cid:1) ′ = ∞ . Then { l ′ n } / ∈ H by Theorem3.12.(7) First, we want to remark the following elementary fact: If i < j and σ ( i ) > σ ( j ), then | i − j | < N : | i − j | = j − i < j − σ ( j ) + σ ( i ) − i ≤ N .Fix n ≥ h ∈ [0 , l ′ n ]. There exists σ ( m ) with Γ σ ( n ) σ ( m ) ( h ) ≤ K . By symmetry, without loss ofgenerality we can assume that σ ( m ) ≤ σ ( n ).If m = n , then σ ( m ) = σ ( n ) and (cid:0) Γ nn (cid:1) ′ ( h ) = Γ σ ( n ) σ ( n ) ( h ) ≤ K .We consider now the case σ ( m ) < σ ( n ).If m > n , then m − n < N .If B ′ n ( h ) > m , then l ′ k > h for every k ∈ ( n, m ] and (cid:0) Γ nm (cid:1) ′ ( h ) = l ′ m − h + m X k = n e h − l ′ k ≤ l σ ( m ) − h + 2 N ≤ Γ σ ( n ) σ ( m ) ( h ) + 2 N ≤ K + 2 N. If B ′ n ( h ) ≤ m , then l ′ k > h for every k ∈ ( n, B ′ n ( h )) and (cid:0) Γ nB ′ n ( h ) (cid:1) ′ ( h ) = B ′ n ( h ) − X k = n e h − l ′ k ≤ N. We deal now with the case m < n . Notice first that σ ([ m, n ]) ⊂ [ m − N, n + N ] and [ m + N, n − N ] ⊂ [ σ ( m ) , σ ( n )]; then, in σ ([ m, n ]) \ [ σ ( m ) , σ ( n )] there are at most 4 N integers.If A ′ n ( h ) ≥ m , then l ′ k > h for every k ∈ ( A ′ n ( h ) , n ), and (cid:0) Γ nA ′ n ( h ) (cid:1) ′ ( h ) = e h n X k = A ′ n ( h )+1 e − l ′ k ≤ e h X k ∈ [ m,n ] l σ ( k ) ≥ h e − l σ ( k ) = e h X j ∈ σ ([ m,n ]) l j ≥ h e − l j ≤ X j ∈ σ ([ m,n ]) \ [ σ ( m ) ,σ ( n )] l j ≥ h e h − l j + e h σ ( n ) X j = σ ( m ) l j ≥ h e − l j ≤ N + 1 + e h σ ( n ) X j = σ ( m )+1 e − l j ≤ N + 1 + Γ σ ( n ) σ ( m ) ( h ) ≤ N + 1 + K . (1) , JOS´E M. RODR´IGUEZ (1) AND EVA TOUR´IS (1) If A ′ n ( h ) < m , then l ′ k > h for every k ∈ [ m, n ), and (cid:0) Γ nm (cid:1) ′ ( h ) = l ′ m − h + e h n X k = m e − l ′ k = l σ ( m ) − h + e h X k ∈ [ m,n ] e − l σ ( k ) = l σ ( m ) − h + e h X j ∈ σ ([ m,n ]) e − l j ≤ X j ∈ σ ([ m,n ]) \ [ σ ( m ) ,σ ( n )] e h − l j + l σ ( m ) − h + e h σ ( n ) X j = σ ( m ) e − l j ≤ N + Γ σ ( n ) σ ( m ) ( h ) ≤ N + K . Hence, (cid:0) K (cid:1) ′ ≤ N + 1 + K , and Theorem 3.12 gives (7). (cid:3) Trigonometric lemmas. In this section some technical lemmas are collected. All of them have been used in Section 3 in order tosimplify the proof of Theorem 3.2. Definition 4.1. Given a surface M , a geodesic γ in M , and a continuous unit vector field ξ along γ ,orthogonal to γ , we define the Fermi coordinates based on γ as the map E ( u, v ) := exp γ ( u ) vξ ( u ) . It is well known that the Riemannian metric can be expressed in Fermi coordinates as ds = dv + η ( u, v ) du , where η ( u, v ) is the solution of the scalar equation ∂ η/∂v + Kη = 0, η ( u, 0) = 1, ∂η/∂v ( u, 0) =0, and K is the curvature of M (see e.g. [11, p. 247]). Consequently, if M is a non-exceptional Riemannsurface, the Poincar´e metric in Fermi coordinates (based on any geodesic γ ) is ds = dv + cosh v du , since K = − a , b ). Definition 4.2. Let us consider Fermi coordinates ( u, v ) in D . We define the distances d (cid:0) ( u , v ) , ( u , v ) (cid:1) , d (cid:0) ( u , v ) , ( u , v ) (cid:1) as follows: without loss of generality we can assume that v ≥ v ; then d (cid:0) ( u , v ) , ( u , v ) (cid:1) : = d (cid:0) ( u , v ) , ( u , v ) (cid:1) + d (cid:0) ( u , v ) , ( u , v ) (cid:1) = v − v + d (cid:0) ( u , v ) , ( u , v ) (cid:1) ,d (cid:0) ( u , v ) , ( u , v ) (cid:1) : = d (cid:0) ( u , v ) , ( u , v ) (cid:1) + d (cid:0) ( u , v ) , ( u , v ) (cid:1) = d (cid:0) ( u , v ) , ( u , v ) (cid:1) + v − v . The following lemma shows that the “cartesian distances” d and d are comparable to d . Lemma 4.3. Let us consider Fermi coordinates ( u, v ) in D and the distances d and d . Then d ≤ d ≤ d , d ≤ d ≤ d . Proof. Triangle inequality gives directly d ≤ d and d ≤ d . Let us consider v ≥ v . It is easy to check that d (cid:0) ( u , v ) , ( u , v ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , v ) (cid:1) , d (cid:0) ( u , v ) , ( u , v ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , v ) (cid:1) and this implies d ≤ d .We also have d (cid:0) ( u , v ) , ( u , v ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , v ) (cid:1) , and then d (cid:0) ( u , v ) , ( u , v ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , v ) (cid:1) + d (cid:0) ( u , v ) , ( u , v ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , v ) (cid:1) ,d (cid:0) ( u , v ) , ( u , v ) (cid:1) = d (cid:0) ( u , v ) , ( u , v ) (cid:1) + d (cid:0) ( u , v ) , ( u , v ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , v ) (cid:1) . (cid:3) Lemma 4.4. Let Ω be a train and l any positive constant. We have d ( z, γ n ∩ ( a n , b n )) ≤ d Ω ( z, ( a n , b n )) + 2 Arcsinh 1 √ l , for every n > and z ∈ Ω with l ≤ h ( z ) ≤ l n . REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OF FLUTE SURFACES 19 Proof. Let w be the nearest point in ( a n , b n ) to z , and define v := γ n ∩ ( a n , b n ), let v be the nearest pointin ( a , b ) to v and w the nearest point in ( a , b ) to w . Consider the geodesic quadrilateral in Ω + withvertices v , w , w and v . Standard hyperbolic trigonometry gives thattanh d Ω ( w, w ) = tanh d Ω ( v, v ) cosh d Ω ( v , w ) = tanh l n cosh d Ω ( v , w ) . Denote by v ′ (respectively w ′ ) the point in γ + n = [ v, v ] ⊂ Ω + (respectively in [ w, w ] ⊂ Ω + ) with h ( v ′ ) = h ( z )(respectively h ( w ′ ) = h ( z )). Consider the geodesic quadrilateral in Ω with vertices v ′ , w ′ , w and v .Standard hyperbolic trigonometry (see e.g. [12, p. 88]) gives thatsinh d Ω ( v ′ , w ′ )2 = sinh d Ω ( v , w )2 cosh h ( z ) = cosh h ( z ) r cosh d Ω ( v , w ) − 12= 1 √ h ( z ) s tanh d Ω ( w, w )tanh l n − ≤ √ h ( z ) s h ( z ) − 1= 1 √ h ( z ) s − tanh h ( z )tanh h ( z ) = 1 p h ( z ) ≤ √ l . This fact and Lemma 4.3 imply d ( z, v ) = d Ω ( z, v ′ ) + d Ω ( v ′ , v ) ≤ d Ω ( v ′ , w ′ ) + d Ω ( z, w ′ ) + d Ω ( w ′ , w ) ≤ √ l + d ( z, w ) ≤ d Ω ( z, w ) + 2 Arcsinh 1 √ l . (cid:3) Lemma 4.5. Let us consider Fermi coordinates ( u, v ) in D . Fix u < u , g := { ( u, v ) : u = u , ≤ v ≤ x } , g := { ( u, v ) : u = u , v ≥ } , and g the (infinite) geodesic orthogonal to g in ( u , x ) . We assume that g does not intersects g . Consider ( u , h ) ∈ g , with h ≥ x , and ( u , v ) ∈ g , with d (cid:0) ( u , v ) , ( u , h ) (cid:1) = d (cid:0) g , ( u , h ) (cid:1) . Then d (cid:0) g , ( u , h ) (cid:1) ≤ d (cid:0) g , ( u , h ) (cid:1) + d (cid:0) ( u , h ) , ( u , h ) (cid:1) ≤ d (cid:0) g , ( u , h ) (cid:1) , for every u ≤ u ≤ u .Proof. We only need to prove the second inequality. Fix u ∈ [ u , u ].Let us assume that v ≤ h . Then Lemma 4.3 implies d (cid:0) g , ( u , h ) (cid:1) + d (cid:0) ( u , h ) , ( u , h ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , h ) (cid:1) + d (cid:0) ( u , h ) , ( u , h ) (cid:1) + d (cid:0) ( u , h ) , ( u , h ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , h ) (cid:1) + 2 d (cid:0) ( u , h ) , ( u , h ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , h ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , h ) (cid:1) = 6 d (cid:0) g , ( u , h ) (cid:1) . Let us assume now that v ≥ h . Lemma 4.3 also implies d (cid:0) g , ( u , h ) (cid:1) + d (cid:0) ( u , h ) , ( u , h ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , h ) (cid:1) + d (cid:0) ( u , h ) , ( u , h ) (cid:1) + d (cid:0) ( u , h ) , ( u , h ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , h ) (cid:1) + 2 d (cid:0) ( u , h ) , ( u , h ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , h ) (cid:1) ≤ d (cid:0) ( u , v ) , ( u , h ) (cid:1) = 4 d (cid:0) g , ( u , h ) (cid:1) . (cid:3) Lemma 4.6. Let us define F as F ( a, x ) := a cosh x , if ≤ a ≤ , log (cid:0) sinh a cosh x (cid:1) , if a ≥ . Then F ( a, x ) ≤ a e x ≤ a cosh x , for every a, x ≥ . (1) , JOS´E M. RODR´IGUEZ (1) AND EVA TOUR´IS (1) Proof. The last inequality is a direct consequence of a ≤ sinh a and e x ≤ x .If a ≥ 1, the function h ( x ) := a e x − a − x satisfies h ′ ( x ) = a e x − ≥ a − ≥ x ≥ 0. Hence, h ( x ) ≥ h (0) = 0 for every x ≥ 0, and we conclude a e x ≥ a + x = log (cid:0) e a e x (cid:1) ≥ log (cid:0) sinh a cosh x (cid:1) , for a ≥ x ≥ H ( a ) := sinh a − a sinh 1 is convex in [0 , H ( a ) ≤ max { H (0) , H (1) } = 0for every 0 ≤ a ≤ 1. Hence, a e x ≥ a e x ≥ a cosh x , for 0 ≤ a ≤ x ≥ (cid:3) This result has the following direct corollary. Corollary 4.7. For a set E ⊂ { ( a, x ) : a, x ≥ } , we have Arcsinh (cid:0) sinh a cosh x (cid:1) ≤ c , for every ( a, x ) ∈ E and some constant c , if and only if a e x ≤ c , for every ( a, x ) ∈ E and some constant c .Furthermore, if one of the inequalities holds, the constant in the other inequality only depends on the firstconstant. As usual, we denote by x + the positive part of x : x + := x if x ≥ x + := 0 if x < Proposition 4.8. (1) There exists a universal constant c such that f ( x, y, t ) := Arccosh cosh t + cosh x cosh y sinh x sinh y ≥ c (cid:0) e − x + e − y + e − ( x + y − t ) + + ( t − x − y ) + (cid:1) , for every x, y, t ≥ . (2) For each l > , there exists a constant c , which only depends on l , such that Arccosh cosh t + cosh x cosh y sinh x sinh y ≤ c (cid:0) e − x + e − y + e − ( x + y − t ) + + ( t − x − y ) + (cid:1) , for every t ≥ and x, y ≥ l . Remark. This result is interesting by itself: if H is a right-angled hexagon in the unit disk for which threepairwise non-adjacent sides X , Y , T are given (with respective lengths x , y , t ), then the opposite side of T in H has length f ( x, y, t ) (see e.g. [12, p. 86], or the proof of Theorem 3.2). Proof. First, we remark that if x ≥ l , then e − l e x ≥ e x − ≥ (1 − e − l ) e x . Therefore, if wedefine c − := (1 − e − l ) / 2, we have e x − ≥ c − e x , sinh x ≥ c − e x , coth x = 1 + 2 e x − ≤ c e − x , for every x ≥ l . We also have coth x = 1 + 2 e x − ≥ e − x , for every x ≥ . Let us start with the proof of item (1).If f ≥ 3, then f ≥ e − x + e − y + e − ( x + y − t ) + . If f ≤ 3, then 1 + c − f ≥ cosh f , for some universalconstant c ≤ 1, and1 + 23 c − f ≥ cosh f ≥ e t − x − y + coth x coth y ≥ e − ( x + y − t ) + (cid:0) e − x (cid:1)(cid:0) e − y (cid:1) , c − f ≥ (cid:0) e − x + e − y + e − ( x + y − t ) + (cid:1) ,c − f ≥ √ p e − x + e − y + e − ( x + y − t ) + ≥ e − x + e − y + e − ( x + y − t ) + ,f ≥ c (cid:0) e − x + e − y + e − ( x + y − t ) + (cid:1) , REAL VARIABLE CHARACTERIZATION OF GROMOV HYPERBOLICITY OF FLUTE SURFACES 21 where we have used the inequality √ √ a + b + c ≥ √ a + √ b + √ c , for every a, b, c ≥ 0. This inequality is(1) if t ≤ x + y . If t ≥ x + y , thencosh f > cosh t sinh x sinh y + 1 ≥ e t − x − y + 1 > e t − x − y + 14 · e − ( t − x − y ) = cosh (cid:0) t − x − y + log 4 (cid:1) and f > t − x − y + log 4 > ( t − x − y ) + + e − ( x + y − t ) + .Consequently we have f ≥ c (cid:0) e − x + e − y + e − ( x + y − t ) + + ( t − x − y ) + (cid:1) , for every x, y, t ≥ 0, with c := c / 2, since c ≤ l > 0. We have seen that sinh x ≥ c − e x and coth x ≤ c e − x , forevery x ≥ l .Let us assume t ≥ x + y . If x, y ≥ l , then12 e f ≤ cosh f = cosh t + cosh x cosh y sinh x sinh y ≤ c e t − x − y + cotanh l . Consequently, e f ≤ c e t − x − y + 2 cotanh l ≤ e t − x − y + c , with c := log (cid:0) c +2 cotanh l (cid:1) , since t − x − y ≥ 0. Hence, f ≤ t − x − y + c = ( t − x − y ) + + c e − ( x + y − t ) + ,for every t ≥ x, y ≥ l with t ≥ x + y .Let us assume t ≤ x + y . If x, y ≥ l , then1 + 12 f ≤ cosh f ≤ c e t − x − y + cotanh x cotanh y ≤ c e t − x − y + (cid:0) c e − x (cid:1)(cid:0) c e − y (cid:1) , f ≤ c e t − x − y + c e − x + c e − y + c e − x − y , f ≤ c e t − x − y + c e − x + c e − y + 12 c (cid:0) e − x + e − y (cid:1) ,f ≤ c e − ( x − y − t ) + (cid:0) c + c (cid:1) e − x + (cid:0) c + c (cid:1) e − y ,f ≤ c (cid:0) e − x + e − y + e − ( x + y − t ) + (cid:1) ,f ≤ c (cid:0) e − x + e − y + e − ( x + y − t ) + + ( t − x − y ) + (cid:1) , where c := max (cid:8) c , c + c (cid:9) , for every t ≥ x, y ≥ l with t ≤ x + y . Then we have (2) with c := max { , c , c } . (cid:3) Proposition 4.9. For each l > , we have F ( x, y, t, h ) := Arcsinh cosh x cosh( y − h ) + cosh t cosh h sinh y ≍ e − h + x + e − ( y − h − t ) + + ( t + h − y ) + , for every x, y, t, h ≥ , verifying y ≥ h ≥ x and y ≥ l . Furthermore, the constants in the inequalities onlydepend on l . Remark. This result is interesting by itself: if H is a right-angled hexagon in the unit disk for which threepairwise non-adjacent sides X , Y , T are given (with respective lengths x , y , t ), P is the nearest point to X in Y , and P h is the point in Y with d ( P h , P ) = h , then F ( x, y, t, h ) is the distance between P h and theopposite side of Y in H (see the proof of Theorem 3.2). Proof. We have seen that if y ≥ l , and c − := (1 − e − l ) / 2, we have c − e y ≤ sinh y ≤ e y / 2. We also have e z / ≤ cosh z ≤ e z , for every z ≥ F ≍ e − h + x + e − y + h + t , since y ≥ l and y ≥ h , and the constants in the inequalities only dependon l .If h + t ≤ y , then e − h + x + e − y + h + t ≤ 2, and F ≍ sinh F ≍ e − h + x + e − ( y − h − t ) = e − h + x + e − ( y − h − t ) + + ( t + h − y ) + . (1) , JOS´E M. RODR´IGUEZ (1) AND EVA TOUR´IS (1) If h + t ≥ y , then e − h + x + e − y + h + t ≥ 1, and e F ≍ sinh F ≍ e − h + x + e − y + h + t ≍ e t + h − y = e − e t + h − y ) + . Since F ≥ Arcsinh (cid:0) e x e y − h + e t e h (cid:1) / e y / ≥ Arcsinh 12 (cid:0) e − h + x + e − y + h + t (cid:1) ≥ Arcsinh 12 > , and 1 + ( t + h − y ) + ≥ > x, y, t, h ≥ 0, and e F ≍ e t + h − y ) + for every x, y, t, h ≥ 0, verifying h + t ≥ y ≥ h ≥ x and y ≥ l , we obtain that F ≍ t + h − y ) + . Since 1 ≤ e − h + x +1 = e − h + x + e − ( y − h − t ) + ≤ 2, we also conclude that F ≍ e − h + x + e − ( y − h − t ) + + ( t + h − y ) + , if h + t ≥ y . (cid:3) The following corollary can be directly deduced from this result. Corollary 4.10. For each l > , let us consider a set E ⊂ { ( x, y, t, h ) : x, y, t, h ≥ , y ≥ h ≥ x, y ≥ l } .We have F ( x, y, t, h ) ≤ c , for every ( x, y, t, h ) ∈ E and some constant c , if and only if ( t + h − y ) + ≤ c ,for every ( x, y, t, h ) ∈ E and some constant c .Furthermore, if one of the inequalities holds, the constant in the other inequality only depends on the firstconstant and l . 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