Harmonic quasi-isometric maps II : negatively curved manifolds

We prove that a quasi-isometric map, and more generally a coarse embedding, between pinched Hadamard manifolds is within bounded distance from a unique harmonic map.


Introduction
The aim of this article, which is a sequel to [4], is the following theorem.
Theorem 1.1. Let f : X → Y be a quasi-isometric map between two pinched Hadamard manifolds. Then there exists a unique harmonic map h : X → Y which stays within bounded distance from f i.e. such that sup x∈X d(h(x), f (x)) < ∞ .
We first recall a few definitions. A pinched Hadamard manifold X is a complete simply-connected Riemannian manifold of dimension at least 2 whose sectional curvature is pinched between two negative constants: −b 2 ≤ K X ≤ −a 2 < 0. A map f : X → Y between two metric spaces X and Y is said to be quasi-isometric if there exist constants c ≥ 1 and C ≥ 0 such that f is (c, C)-quasi-isometric. This means that one has for all x, x ′ in X. A C 2 map h : X → Y between two Riemannian manifolds X and Y is said to be harmonic if it satisfies the elliptic nonlinear partial differential equation tr(D 2 h) = 0 where D 2 h is the second covariant derivative of h. Partial results towards the existence statement were obtained in [28], [38], [15], [24], [5]. A major breakthrough was achieved by Markovic who solved the Schoen conjecture, i.e. the case where X = Y is the hyperbolic plane H 2 R , and by Lemm-Markovic who proved the existence for the case X = Y = H k R in [26], [25] and [20]. The existence when both X and Y are rank one symmetric spaces, which was conjectured by Li and Wang in [22,Introduction], was proved in our paper [4]. We refer to [4, Section 1.2] for more motivations and a precise historical perspective on this result.
Partial results towards the uniqueness statement were obtained by Li and Tam in [21], and by Li and Wang in [22]. All these papers were dealing with rank one symmetric spaces.
Note that Theorem 1.1 was conjectured by Markovic at the end of the conference talk www.youtube.com/watch?v=A5Yt83I1FrY, during a 2016 Summer School in Grenoble. According to our knowledge, Theorem 1.1 is new even in the case where both X and Y are assumed to be surfaces.
The strategy of the proof of the existence follows the lines of the proof in [4]. As in [4], we replace the quasi-isometric map f by a C ∞ -map whose first two covariant derivatives are bounded. But we need to modify the barycenter argument we used in [4] for this smoothing step. See Subsection 2.2.1 for more details on this step. As in [4], we then introduce the harmonic maps h R that coincide with f on a sphere of X with large radius R and we need a uniform bound for the distances between the maps h R and f . The heart of our argument is in Chapter 3 which contains the boundary estimates, and in Chapter 4 which contains the interior estimates, for d(h R , f ). The proof of these interior estimates is based on a new simplification of an idea by Markovic in [25]. Indeed we will introduce a point x where d(h R (x), f (x)) is maximal and focus on a subset U ℓ 0 of a sphere S(x, ℓ 0 ) whose definition (4.10) is much simpler than in [25] or [4]. This simplification is the key point which allows us to extend the arguments of [4] to pinched Hadamard manifolds. In this proof we use a uniform control on the harmonic measures on all the spheres of X, which is given in Proposition 4.9. We refer to Section 4.1 for more details on our strategy of proof of the existence.
In order to prove the uniqueness, we need to introduce Gromov-Hausdorff limits of the pointed metric spaces X and Y with respect to base points going to infinity and therefore to deal with C 2 -Riemannian manifolds with C 1 -metrics. This will be done in Chapter 5. We refer to Section 5.1 for more details on our strategy of proof of the uniqueness.
In Chapter 7, we extend Theorem 1.1 to coarse embeddings (see Definition 6.2 and Theorem 7.1). The proof is similar but relies on the existence of a boundary map for coarse embeddings. We also show that Theorem 1.1 can not be extended to Lipschitz maps (Example 7.3) Chapter 6 is dedicated to the existence of this boundary map which, for a coarse embedding, is well-defined outside a set of zero Hausdorff dimension (Theorem 6.5). The existence of such a boundary map seems to be new.
We thank the MSRI for its hospitality during the Fall 2016 where this project was developed. We are also very grateful to A. Ancona, U. Hamenstadt, M. Kapovich and F. Ledrappier for sharing their insight with us.

Smoothing
In this chapter, we recall a few basic facts on Hadamard manifolds, and we explain how to replace our quasi-isometric map f by a C ∞ map whose first two covariant derivatives are bounded.

The geometry of Hadamard manifolds
We first recall basic estimates on Hadamard manifolds for triangles, for images of triangles under quasi-isometric maps, and for the Hessian of the distance function.
All the Riemannian manifolds will be assumed to be connected. We will denote by d their distance function.
A Hadamard manifold is a complete simply connected Riemannian manifold X of dimension k ≥ 2 whose curvature is non positive K X ≤ 0. For instance, the Euclidean space R k is a Hadamard manifold with zero curvature K X = 0, and the real hyperbolic space H k R is a Hadamard manifold with constant curvature K X = −1. We will say that X is pinched if there exist constants a, b > 0 such that For instance, the non-compact rank one symmetric spaces are pinched Hadamard manifolds.
Let x 0 , x 1 , x 2 be three points on a Hadamard manifold X. The Gromov product of the points x 1 and x 2 seen from x 0 is defined as (x 1 |x 2 ) x 0 := (d(x 0 , x 1 ) + d(x 0 , x 2 ) − d(x 1 , x 2 ))/2. (2.1) We recall the basic comparison lemma which is one of the motivations for introducing the Gromov product.
Proof. This is a general property of quasi-isometric maps between Gromov δ-hyperbolic spaces which is due to M. Burger. See [12,Prop. 5.15].
When x 0 is a point in a Riemannian manifold X, we denote by d x 0 the distance function defined by d x 0 (x) = d(x 0 , x) for x in X. We denote by d 2 x 0 the square of this function. When F : X → R is a C 2 function, we denote by DF its differential and by D 2 F its second covariant derivative. Lemma 2.3. Let X be a Hadamard manifold and x 0 ∈ X. Assume that −b 2 ≤ K X ≤ −a 2 ≤ 0. The Hessian of the distance function

3)
where g 0 := g X − Dd x 0 ⊗ Dd x 0 and g X is the Riemannian metric on X.
When a = 0 the left-hand side of (2.3) must be interpreted as d −1 x 0 g 0 .

Smoothing rough Lipschitz maps
The following proposition will allow us to assume in Theorem 1.1 that the quasi-isometric map f we start with is C ∞ with bounded derivative and bounded second covariant derivative.

Rough Lipschitz maps
A map f : X → Y between two metric spaces X and Y is said to be rough Lipschitz if there exist constants c ≥ 1 and C ≥ 0 such that, for all x, x ′ in X, one has d(f (x), f (x ′ )) ≤ c d(x, x ′ ) + C .
(2.4) Proposition 2.4. Let X, Y be two Hadamard manifolds with bounded curvatures −b 2 ≤ K X ≤ 0 and −b 2 ≤ K Y ≤ 0. Let f : X → Y be a rough Lipschitz map. Then there exists a C ∞ map f : X → Y within bounded distance from f and whose first two covariant derivatives D f and D 2 f are bounded on X.
We denote k = dim X and k ′ = dim Y . We will first construct in 2.2.2 a regularized map f : X → Y which is Lipschitz continuous. This first construction is the same as for rank one symmetric spaces in [4,Proposition 2.4]. This construction will not allow us to control the second covariant derivative, hence we will have to combine this first construction with an iterative smoothing process in local charts that we will explain in 2.2.3.

Lipschitz continuity
The first part of the proof of Proposition 2.4 relies on the following lemma. Lemma 2.5. Let Y be a Hadamard manifold. a) Let µ be a positive finite Borel measure on Y supported by a ball B(y 0 , R). The function Q µ on Y defined by has a unique minimum y µ in Y called the center of mass of µ. This center of mass y µ belongs to the ball B(y 0 , R). b) Let µ 1 , µ 2 be two positive finite Borel measures on Y . Assume that (i) µ 1 (Y ) ≥ m and µ 2 (Y ) ≥ m for some m > 0, (ii) both µ 1 and µ 2 are supported on the same ball B(y 0 , R), (iii) the norm of µ 1 − µ 2 ≤ ε. Then, the distance between their centers of mass y µ 1 and y µ 2 is bounded by d(y µ 1 , y µ 2 ) ≤ 4εR/m . (2.5) Proof of Lemma 2.5. a) Since the space Y is a proper space, i.e. its balls are compact, the function Q µ is proper and admits a minimum y µ . Since Y has non-positive curvature the median inequality holds, namely one has for all y, y 1 , y 2 , y 3 in Y where y 3 is the midpoint of y 1 and y 2 : 1 2 d(y 1 , y 2 ) 2 ≤ d(y, y 1 ) 2 + d(y, y 2 ) 2 − 2d(y, y 3 ) 2 . (2.6) Integrating (2.6) with respect to µ, one checks that the function Q µ satisfies the following uniform convexity property : if y 3 is the midpoint of y 1 and y 2 m 2 d(y 1 , y 2 ) 2 ≤ Q µ (y 1 ) + Q µ (y 2 ) − 2Q µ (y 3 ) .
Applying this inequality with y 1 = y µ and y 2 = y, one gets for each y in Y m 2 d(y µ , y) 2 ≤ Q µ (y) − Q µ (y µ ) , so that y µ is the unique minimum of Q µ . We now check that y µ ∈ B(y 0 , R). By the median inequality (2.6), the ball B(y 0 , R) is convex, every point y in Y admits a unique nearest point y ′ on B(y 0 , R), and this point y ′ also satisfies the inequality d(y ′ , w) ≤ d(y, w) for all w in B(y 0 , R).
Therefore, one has Q µ (y ′ ) ≤ Q µ (y). This proves that the center of mass y µ belongs to the ball B(y 0 , R). b) Applying twice Inequality (2.7), one gets the two inequalities Summing these two inequalities yields which proves (2.5).
We now choose a non-negative C ∞ function χ : R → R with support included in ] − 1, 1[, which is equal to 1 on a neighborhood of [− 1 2 , 1 2 ] and whose first derivative is bounded by |χ ′ | ≤ 4.
Proof of Proposition 2.4. First step: Lipschitz continuity. We now explain this first construction. We can assume b = 1. Since a rough Lipschitz map f : X → Y is always within bounded distance from a Borel measurable map, we can assume that f itself is Borel measurable. For x in X, we introduce the positive finite measure µ x on Y such that the equality holds for any positive function ϕ on Y . The measure µ x is the image by f of a measure supported in the ball B(x, 1). We define f (x) ∈ Y to be the center of mass of this measure µ x . Lemma 2.5.a tells us that the map x → f (x) is well-defined. The Lipschitz continuity of f will follow from Lemma 2.5.b applied to two measures µ 1 := µ x 1 and µ 2 := µ x 2 with x 1 , x 2 in X. Let us check that the three assumptions in Lemma 2.5.b are satisfied.
(i) Because of the pinching of the curvature on X, the Bishop volume estimates tell us that there exist positive constants 0 < m 0 < M 0 such that one has, for all x : ensures that both µ x 1 and µ x 2 are supported on the ball B(f (x 1 ), 2c + C).
(iii) The norm of the difference of these measures satisfies Thus Lemma 2.5 applies and yields a bound on the Lipschitz constant of f , namely Lip( f ) := sup

Bound on the second derivative
The second step of the proof of Proposition 2.4 relies on three lemmas. The first lemma provides a nice system of charts on Y .
Proof of Lemma 2.6. This is classical. One can for instance choose the socalled almost linear coordinates, as in [17,Section 2] or [29,Section 3]. They are defined in the following way. We fix an orthonormal basis (e i ) 1≤i≤k ′ for the tangent space T y Y and introduce the points y i := exp y (−e i ) in Y . The map Φ y is defined by the formula where z belongs to a sufficiently small ballB(y, r 0 ). See [17, p. 43 and 58] for a detailed proof.
There exist better systems of coordinates, the so-called harmonic coordinates. We will not need them in this chapter, but we will need them in Chapter 5 to prove uniqueness (see Lemma 5.2).
The second lemma explains how to modify a Lipschitz map g inside a tiny ball B(x, r) of X so that the new map g x,r is constant on the ball B(x, r 2 ) and the first two derivatives of g x,r are controlled by those of g. We recall that χ : R → R is a non-negative C ∞ function with support included in ] − 1, 1[, which is equal to 1 on a neighborhood of [− 1 2 , 1 2 ] and which is 4-Lipschitz, i.e. |χ ′ | ≤ 4. Lemma 2.7. Let X and Y be two Hadamard manifolds with bounded cur- Let r 0 > 0 and c 0 ≥ 1 be as in Lemma 2.6. Let g : X → Y be a Lipschitz map, x be a point in X, y = g(x) and let 0 < r < r 0 . Assume that Then the following formulas define a Lipschitz map g r,x : when d(z, x) ≥ r .
One has the inequality between the Lipschitz constants on the ball B(x, r) : In particular, one has Moreover, if g is C 2 in a neighborhood of a point z in X, then g r,x is also C 2 in this neighborhood and one has where the constant M r ≥ 1 depends only on r, b, k, k ′ and χ.
Proof of Lemma 2.7. Condition (2.11) ensures that, for any point z in the ball B(x, r), the image g(z) belongs to the ballB(y, c −2 0 r 0 ). Therefore, by (2.10), the vector Φ y (g(z)) belongs to the ballB(0, c −1 0 r 0 ) ⊂ R k ′ . When we multiply this vector by the scalar 1−χ(.), the new vector is still in the same ball. This is why, using again (2.10), the element g r,x (z) is well-defined and belongs to B(y, r 0 ).
The upper bound (2.12) follows from the chain rule. Indeed, when z is a point in B(x, r) where g is differentiable, the bound (2.9) yields The upper bound (2.14) follows from similar and longer computations left to the reader, which also use the bounds (2.3) for D 2 d x .
We will also need a third lemma. We recall that a subset X 0 of a metric space X is said to be r-separated if the distance between two distinct points of X 0 is at least r. Lemma 2.8. Let X be a Hadamard manifold with −b 2 ≤ K X ≤ 0. Let k = dim X and N 0 := 100 k . There exists a radius r 0 = r 0 (k, b) > 0 such that, for any r < r 0 , every r 2 -separated subset X 0 of X can be decomposed as a union of at most N 0 subsets which are 2r-separated.
Proof of Lemma 2.8. The bound on the curvature of X and the Bishop volume estimates ensure that we can choose r 0 > 0 such that volB(x, 4r) ≤ N 0 volB(x, r 4 ) for all r < r 0 and x in X. (2.15) This r 0 works. Indeed, let X 1 , X 2 , . . ., X N 0 be a sequence of disjoint 2rseparated subsets of X 0 with X 1 maximal in X 0 , X 2 maximal in X 0 X 1 , and so on. Every point x of X 0 must be in one of the X i 's with i ≤ N 0 because, if it is not the case, each X i contains a point in B(x, 2r), contradicting (2.15).
Proof of Proposition 2.4. Second step: bound on D 2 f . According to the first step of this proof, we can now assume that the map f : X → Y is c-Lipschitz with c ≥ 1. We can choose a new radius r 0 = r 0 (k, k ′ , b) that satisfies both conclusions of Lemma 2.8 for X and of Lemma 2.6 for Y . We will use freely the notations of these two lemmas. Now let and pick a maximal r 1 4 -separated subset X 0 of X. Thanks to Lemma 2.8, we write this set X 0 as a union X 0 = X 1 ∪ · · · ∪ X N 0 of N 0 subsets X i which are 2r 1 -separated.
In order to construct f from f , we will use a finite iterative process based on Lemma 2.7. Starting with f 0 = f , we construct by induction a finite sequence of maps f i for i ≤ N 0 and we set f := f N 0 . Using the notations of Lemma 2.7, the map f i is defined from f i−1 by letting so that the Lipschitz constants of these maps satisfy Indeed, once f i is known to be well defined and to satisfy (2.16), it also satisfies the bound (2.11) : Lip(f i ) < r 0 c 2 0 r 1 . Therefore Lemma 2.7 ensures that f i+1 is well defined and, using (2.12), that f i+1 also satisfies (2.16) : Let Λ := M r 1 + 25c 4 0 + 1. By (2.14) and (2.16), one has for any i ≤ N 0 and z in X : Since X 0 is a maximal r 1 4 -separated subset of X, every z in X belongs to at least one ballB(x, r 2 ) where x is in one of the sets X i 0 . But then the function f i 0 is constant in a neighborhood of z. Therefore, using (2.16) and applying (N 0 −i 0 ) times the bound (2.17) one deduces that f is a C 2 -map that satisfies the uniform upper bound

Harmonic maps
In this chapter we begin the proof of the existence part in Theorem 1.1. We first recall basic facts satisfied by harmonic maps. We explain why a standard compactness argument reduces this existence part to proving a uniform upper bound on the distance between f and the harmonic map h R which is equal to f on the sphere S(O, R). Then we provide this upper bound near this sphere S(O, R).

Harmonic functions and the distance function
We recall basic facts on the Laplace operator on Hadamard manifolds.
The Laplace-Beltrami operator ∆ on a Riemannian manifold X is defined as the trace of the Hessian. In local coordinates, the Laplacian of a function ϕ is ∆ϕ = tr( where v = det(g X ij ) is the volume density. The function ϕ is said to be harmonic if ∆ϕ = 0 and subharmonic if ∆ϕ ≥ 0. We will need the following basic lemma.
Lemma 3.1. Let X be a Hadamard manifold with K X ≤ −a 2 ≤ 0 and x 0 be a point in X. Then, the function d x 0 is subharmonic. More precisely, the distribution ∆ d x 0 −a is non-negative.

Harmonic maps and the distance function
In this section, we recall two useful facts satisfied by a harmonic map h : the subharmonicity of the functions d y 0 • h, and Cheng's estimate for the differential Dh. Note that the tension field τ (h) is a Y -valued vector field on X, i.e. it is a section of the pulled-back of the tangent bundle T Y → Y under the map h : X → Y .
For instance, an isometric immersion with minimal image is always harmonic. The problem of the existence, regularity and uniqueness of harmonic maps under various boundary conditions is a very classical topic (see [10], [32], [17], [9], [37], [35] or [23]). In particular, when Y is simply connected and has non positive curvature, a harmonic map is always C ∞ i.e. it is indefinitely differentiable, and is a minimum of the energy functional among maps that agree with h outside a compact subset of X. Lemma 3.3. Let h : X → Y be a harmonic C ∞ map between Riemannian manifolds. Let y 0 ∈ Y and let ρ h : X → R be the function ρ h := d y 0 • h. If Y is Hadamard, the continuous function ρ h is subharmonic on X.
Another crucial property of harmonic maps is the following bound for their differential due to Cheng. Lemma 3.4. Let X, Y be two Hadamard manifolds with −b 2 ≤ K X ≤ 0. Let k = dim X, z be a point in X, r > 0 and let h : B(z, r) → Y be a harmonic C ∞ map such that the image h(B(z, r)) lies in a ball of radius R 0 . Then one has the bound Dh(z) ≤ 2 5 k 1+br r R 0 .
In the applications, we will use this inequality with r = b −1 .
Proof. This is a simplified version of [8, Formula 2.9].

Existence of harmonic maps
In this section we prove Theorem 1.1, taking for granted Proposition 3.5 below.
Let X and Y be two Hadamard manifolds whose curvatures are pinched −b 2 ≤ K X ≤ −a 2 < 0 and −b 2 ≤ K Y ≤ −a 2 < 0. Let k = dim X and k ′ = dim Y . Let f : X → Y be a (c, C)-quasi-isometric C ∞ map whose first two covariant derivatives are bounded.
We fix a point O in X. For R > 0, we denote by B R := B(O, R) the closed ball in X with center O and radius R and by ∂B R the sphere that bounds B R . Since the manifold Y is a Hadamard manifold, there exists a unique harmonic map h R : B R → Y satisfying the Dirichlet condition h R = f on the sphere ∂B R . Thanks to Schoen and Uhlenbeck in [33] and [34], the harmonic map h R is known to be C ∞ on the closed ball B R . We denote by the distance between these two maps. The main step for proving existence in Theorem 1.1 is the following uniform estimate.
Proposition 3.5. There exists a constant ρ ≥ 1 such that, for any R ≥ 1, The constant ρ is a function of a, b, c, C, k and k ′ . More precisely, when f satisfies (4.1), ρ only needs to satisfy Conditions (4.6), (4.7) and (4.8).
We briefly recall the classical argument used to deduce Theorem 1.1 from this Proposition.
Proof of Theorem 1.1. As explained in Proposition 2.4, we may assume that the (c, C)-quasi-isometric map f is C ∞ with bounded first two covariant derivatives. Pick an unbounded increasing sequence of radii R n and let h Rn : B Rn → Y be the harmonic C ∞ map that agrees with f on the sphere ∂B Rn . Proposition 3.5 ensures that the sequence of maps (h Rn ) is locally uniformly bounded. Using the Cheng Lemma 3.4 it follows that the first derivatives are also locally uniformly bounded. The Ascoli-Arzela theorem implies that, after extracting a subsequence, the sequence (h Rn ) converges uniformly on every ball B S towards a continuous map h : X → Y . Using the Schauder's estimates, one also gets a uniform bound for the C 2,α -norms of h Rn on B S . If needed, note that these classical estimates will be recalled in Formulas (5.29), (5.32) and (5.33) in Section 5.6. Therefore, using the Ascoli-Arzela theorem again, the sequence (h Rn ) converges in the C 2 -norm and the limit map h is C 2 and harmonic. By construction, this limit harmonic map h stays within bounded distance from the quasi-isometric map f . Remark 3.6. By the uniqueness part of our Theorem 1.1 that we will prove in Chapter 5, the harmonic map h which stays within bounded distance from f is unique. Hence the above argument also proves that the whole family of harmonic maps h R converges to h uniformly on the compact subsets of X when R goes to infinity.

Boundary estimate
In this section, we begin the proof of Proposition 3.5 : we bound the distance between h R and f near the sphere ∂B R .
Proposition 3.7. Let X, Y be Hadamard manifolds and k = dim X. Assume moreover that K X ≤ −a 2 < 0 and −b 2 ≤ K Y ≤ 0. Let c ≥ 1 and f : X → Y be a C ∞ map with Df (x) ≤ c and D 2 f (x) ≤ bc 2 . Let O ∈ X, R > 0 and set B R := B(O, R).
Let h R : B R → Y be the harmonic C ∞ map whose restriction to the sphere ∂B R is equal to f . Then, one has for every x in B R : An important feature of this upper bound is that it does not depend on the radius R, provided the distance d(x, ∂B R ) remains bounded. This is why we call (3.2) the boundary estimate. The proof relies on an idea of Jost in [17,Section 4].
Proof. This proposition is already in [4,Proposition 3.8]. We give here a slightly shorter proof. Let x be a point in B R and y be a point in Y chosen so that d(y, f (B R )) ≥ b −1 and This point y is far away on the geodesic ray starting at h R (x) and containing f (x). Let ϕ be the C ∞ function on the ball B R defined by This function is the sum of three functions ϕ = ϕ 1 + ϕ 2 + ϕ 3 . The first function ϕ 1 : z → d y (h R (z)) is subharmonic on B R i.e. one has ∆ϕ 1 ≥ 0. This follows from Lemma 3.3 and the harmonicity of the map h R .

Interior estimate
In this chapter we complete the proof of Proposition 3.5.

Strategy
We first explain more precisely the notations and the assumptions that we will use in the whole chapter.
Let X and Y be Hadamard manifolds whose curvatures are pinched −b 2 ≤ K X ≤ −a 2 < 0 and −b 2 ≤ K Y ≤ −a 2 < 0. Let k = dim X and k ′ = dim Y . We start with a C ∞ quasi-isometric map f : X → Y whose first and second covariant derivatives are bounded. We fix constants c ≥ 1 and C > 0 such that one has, for all x, x ′ in X : Note that the additive constant C on the right-hand side term of (1.1) has been removed since the derivative of f is now bounded by c.

Choosing the radius ℓ 0
We fix a point O in X. We introduce a fixed radius ℓ 0 depending only on a, b, k, k ′ , c and C. This radius ℓ 0 is only required to satisfy the following three inequalities (4.3), (4.4) and (4.5) that will be needed later on.
The first condition we impose on the radius ℓ 0 is The second condition we impose on the radius ℓ 0 is where A is the constant given by Lemma

Assuming ρ to be large
We want to prove Proposition 3.5. For R > 0, recall that h R : B(O, R) → Y is the harmonic C ∞ map whose restriction to the sphere ∂B(O, R) is equal to f . We let ρ := sup We argue by contradiction. If this supremum ρ is not uniformly bounded with respect to R, we can fix a radius R such that ρ satisfies the following three inequalities (4.6), (4.7) and (4.8) that we will use later on.
The first condition we impose on the radius ρ is The second condition we impose on the radius ρ is The third condition we impose on the radius ρ is where M , N are the constants given by Proposition 4.9.
We denote by x a point of According to the boundary estimate in Proposition 3.7, Condition (4.6) yields . This inclusion will allow us to apply the Cheng 's lemma 3.4 at each point z of the ball B(x, ℓ 0 ).

Getting a contradiction
We will focus on the restrictions of both maps f and h R to this ball B(x, ℓ 0 ). We introduce the point y := f (x). For y 1 , y 2 in Y {y}, we denote by θ y (y 1 , y 2 ) the angle at y of the geodesic triangle with vertices y, y 1 , y 2 . For z on the sphere S(x, ℓ 0 ), we will analyze the triangle inequality : and prove that on a subset U ℓ 0 of the sphere, each term on the right-hand side is small (Lemmas 4.5 and 4.6) while the measure of U ℓ 0 is large enough (Lemma 4.4) to ensure that the left-hand side is not that small (Lemma 4.8), giving rise to the contradiction. These arguments rely on uniform lower and upper bounds for the harmonic measures on the spheres of X that will be given in Proposition 4.9.
We denote by ρ h the function on B(x, ℓ 0 ) given by ρ h (z) = d(y, h R (z)) where again y = f (x). By Lemma 3.3, this function is subharmonic.
Definition 4.1. The subset U ℓ 0 of the sphere S(x, ℓ 0 ) is given by

Measure estimate
We first observe that one can control the size of ρ h (z) and of Dh R (z) on the ball B(x, ℓ 0 ). We then derive a lower bound for the measure of U ℓ 0 .
Proof. The triangle inequality and (4.2) give, for any z in B(x, ℓ 0 ) : For these last two inequalities, we used Conditions (4.3) and (4.6). Applying the Cheng's lemma 3.4 with R 0 = 3ρ and r = b −1 , one then gets for all z in We now give a lower bound for the measure of U ℓ 0 . Lemma 4.4. Let σ = σ x,ℓ 0 be the harmonic measure on the sphere S(x, ℓ 0 ) at the center point x. Then one has (4.11) Proof. By Lemma 3.3, the function ρ h is subharmonic on the ball B(x, ℓ 0 ). Hence this function ρ h is not larger than the harmonic function on the ball with same boundary values on the sphere S(x, ℓ 0 ). Comparing these functions at the center x, one gets

Upper bound for
For all z in U ℓ 0 , we give an upper bound for the the angle between f (z) and h R (z) seen from the point y = f (x).
Proof. For z in U ℓ 0 , we consider the triangle with vertices y, f (z) and h R (z). Its side lengths satisfy where we used successively the definition of ρ, the quasi-isometry lower bound (4.2) and the definition of U ℓ 0 . Hence, one gets the following lower bound for the Gromov product

Upper bound for
For all z in S(x, ℓ 0 ), we give an upper bound for the angle between h R (z) and h R (x) seen from the point y = f (x). Lemma 4.6. For all z in the sphere S(x, ℓ 0 ), one has . (4.14) The proof will rely on the following lemma which also ensures that this Proof of Lemma 4.7. Assume by contradiction that there exists a point z 1 in the ball B(x, ℓ 0 ) such that ρ h (z 1 ) = ρ/2. Set r 1 := d(x, z 1 ). One has 0 < r 1 ≤ ℓ 0 . According to Lemma 4.3, one can bound the differential of h R on the ball B(x, ℓ 0 ), namely

Hence one has
By comparison with the hyperbolic plane with curvature −b 2 , this intersection contains the trace on the sphere S(x, r 1 ) of a cone C α with vertex x and angle α as soon as sin α 2 ≤ sinh(2 −11 /k) sinh(br 1 ) . For instance we will choose for α the angle α := e −bℓ 0 2 −10 /k.
Let σ ′ = σ x,r 1 be the harmonic measure on the sphere S(x, r 1 ) for the center point x. Using the subharmonicity of the function ρ h as in the proof of Lemma 3.3, one gets the inequality Using the uniform lower bounds for the harmonic measures on the spheres of X in Proposition 4.9, one gets which contradicts Condition (4.8).
Proof of Lemma 4.6. Let us first sketch the proof. Let z be a point on the sphere S(x, ℓ 0 ). We denote by t → z t , for 0 ≤ t ≤ ℓ 0 , the geodesic segment between x and z. By Lemma 4.7, the curve t → h R (z t ) lies outside of the ball B(y, ρ/2) and by Cheng's bound on Dh R (z t ) one controls the length of this curve. We now detail the argument. We denote by (ρ(y ′ ), v(y ′ )) ∈ ]0, ∞[×T 1 y Y the polar exponential coordinates centered at y. For a point y ′ in Y {y}, they are defined by the equality y ′ = exp y (ρ(y ′ )v ρ (y ′ )). Since K Y ≤ −a 2 the Alexandrov comparison theorem for infinitesimal triangles and the Gauss lemma ( [11, 2.93 Hence, Lemma 4.7 yields the inequality Therefore, using Lemma 4.3 and Condition (4.6), one gets We find a point z in U ℓ 0 for which the angle between f (z) and h(x) seen from y = f (x) has an explicit lower bound.
where θ 0 is the angle given by (4.5).
Proof of Lemma 4.8. Let σ 0 := 1 3c 2 . According to Lemma 4.4, one has σ(U ℓ 0 ) ≥ σ 0 > 0 Thus, using the uniform upper bounds for the harmonic measures on the spheres of X in Proposition 4.9, one can find z 1 , z 2 in U ℓ 0 such that the angle θ x (z 1 , z 2 ) between z 1 and z 2 seen from x satisfies This can be rewritten as where ε 0 is the angle introduced in (4.4) by the equality σ 0 = M ε 1 N 0 . Therefore, using Lemma 2.1.a and Condition (4.4), we get the following lower bound on the Gromov products Using then Lemma 2.2, one gets This inequality (4.17) allows us to apply Lemma 2.1.c, which gives Therefore, by Lemma 2.2, one has Using Lemma 2.1.b and Condition (4.16), one gets according to the definition (4.5) of θ 0 .
End of the proof of Proposition 3.5. Using Lemmas 4.5 and 4.6 and the triangle inequality (4.9), one gets for any two points z i = z 1 or z 2 in U ℓ 0 : by Conditions (4.5) and (4.7).
Therefore, using again a triangle inequality, one has which contradicts Lemma 4.8.

Harmonic measures
The following proposition gives the uniform lower and upper bounds for the harmonic measure on a sphere for the center which were used in the proof of Lemmas 4.7 and 4.8.
Proposition 4.9. Let 0 < a < b and k ≥ 2 be an integer. There exist positive constants M , N depending only on a, b, k such that for every kdimensional Hadamard manifold X with pinched curvature −b 2 ≤ K X ≤ −a 2 , for every point x in X, every radius r > 0 and every angle θ ∈ [0, π] one has where σ x,r denotes the harmonic measure on the sphere S(x, r) at the point x and where C x,θ stands for any cone with vertex x and angle θ.
We recall that, by definition, σ x,r is the unique probability measure on the sphere S(x, r) such that, for every continuous function h on the ball B(x, r) which is harmonic in the interiorB(x, r), one has the equality A proof of Proposition 4.9 is given in [3]. It relies on various technical tools of the potential theory on pinched Hadamard manifolds : the Harnack inequality, the barrier functions constructed by Anderson [19,Proposition 3.9] where the Hölder regularity of the Martin kernel is proven.

Uniqueness of harmonic maps
In this chapter we prove the uniqueness part in Theorem 1.1.

Strategy
In other words we will prove the following proposition.
Proposition 5.1. Let X, Y be two pinched Hadamard manifolds and let h 0 , h 1 : X → Y be two quasi-isometric harmonic maps that stay within bounded distance of one another: Then one has h 0 = h 1 .
When X = Y = H 2 , this proposition was first proven by Li and Tam in [21]. When both X and Y admit a cocompact group of isometries, this proposition was then proven by Li and Wang in [22,Theorem 2.3]. The aim of this chapter is to explain how to get rid of these extra assumptions.
Note that the assumption that the h i are quasi-isometric is useful. Indeed there does exist non constant bounded harmonic functions on X. Note that there also exist bounded harmonic maps with open images. Here is a very simple example. Let 0 < λ < 1. The map h λ from the Poincaré unit disk D of C into itself given by z → λz is harmonic. More generally, for any harmonic map h : D → D, the map h λ : D → D : z → h(λz) is a harmonic map with bounded image.
Before going into the technical details, we first explain the strategy of the proof of this uniqueness.
Strategy of proof of Proposition 5.1. We recall that the distance function x → d(h 0 (x), h 1 (x)) is a subharmonic function on X and that, by the maximum principle, a subharmonic function that achieves its maximum value is constant. Unfortunately since X is non-compact we can not a priori ensure that this bounded function achieves its maximum. This is why we will use a recentering argument.
We assume, by contradiction, that h 0 = h 1 and we choose a sequence of points p n in X for which the distances and we set q n := h 0 (p n ). The pinching condition on X and Y ensure that, after extracting a subsequence, the pointed metric spaces (X, p n ) and (Y, q n ) converge in the Gromov-Hausdorff topology to pointed metric spaces (X ∞ , p ∞ ) and (Y ∞ , q ∞ ) which are C 2 Hadamard manifolds with C 1 Riemannian metrics satisfying the same pinching conditions (Proposition 5.14). Moreover, extracting again a subsequence, the harmonic map h 0 (resp. h 1 ) seen as a sequence of maps between the pointed Hadamard manifolds (X, p n ) and (Y, q n ) converges locally uniformly to a map h 0,∞ (resp h 1,∞ ) between the pointed C 2 Hadamard manifolds (X ∞ , p ∞ ) and (Y ∞ , q ∞ ). These harmonic maps h 0,∞ and h 1,∞ are still harmonic quasi-isometric maps (Lemma 5.15).
) is a subharmonic function on X ∞ that now achieves its maximum δ > 0 at the point p ∞ . Hence, by the maximum principle, this distance function is constant and equal to δ (Lemma 5.16). Generalizing [22, Lemma 2.2], we will see in Corollary 5.19 that this equidistance property implies that both h 0,∞ and h 1,∞ take their values in a geodesic of Y ∞ . This contradicts the fact that h 0,∞ and h 1,∞ are quasi-isometric maps, and concludes this strategy of proof.
In the following sections of Chapter 5, we fill in the details of the proof.

Harmonic coordinates
We first introduce the so-called harmonic coordinates, which improve the quasilinear coordinates introduced in Lemma 2.6. We refer to [14] or [17] for more details.
The harmonic coordinates have been introduced by DeTurk and Kazdan and extensively used by Cheeger, Jost, Karcher, Petersen... to prove various compactness results for compact Riemannian manifolds. Beyond being harmonic, the main advantage of these coordinates is that, for every α ∈]0, 1[, they are uniformly bounded in C 2,α -norm, i.e. they are uniformly bounded in C 2 -norm and one also has a uniform control of the α-Hölder norm of their second covariant derivatives. Moreover, one has a uniform control on the size of the balls on which these harmonic charts are defined. This is what the following lemma tells us.
We endow R k with the standard Euclidean structure.
and such that each component z 1 , . . . , z k of Ψ x is a harmonic function.
In particular, one has for all r < r 0 : The second covariant derivatives of all Ψ x are also uniformly α-Hölder : This α-Hölder semi-norm D 2 Ψ x C α is defined as follows. Using the vector fields ∂ ∂z 1 , . . . , ∂ ∂z k onB(x, r 0 ) associated to our coordinate system Ψ x = (z 1 , . . . , z k ), we reinterpret the tensor D 2 Ψ x as a family of vector valued functions onB(x, r 0 ). Indeed, we set and the bound (5.5) means that These uniform bounds (5.3) and (5.5) have three consequences. First, in the harmonic coordinate systems Ψ x = (z 1 , . . . , z k ), the Christoffel coefficients Γ ℓ ij are uniformly bounded in C α -norm. Indeed, these coef- Third, in the coordinate systems Ψ x = (z 1 , . . . , z k ), the coefficients of the metric tensor

Gromov-Hausdorff convergence
In this section, we recall the definition of Gromov-Hausdorff convergence for pointed metric spaces and some of its key properties. We refer to [7] for more details.

Definition
When X is a metric space, we will denote by d or d X the distance on X. We denote by B(x, R) the closed ball with center x and radius R, and bẙ B(x, R) the open ball. We recall that a metric space X is proper if all its balls are compact or, equivalently, if X is complete and for all R > 0 and ε > 0 every ball of radius R can be covered by finitely many balls with radius ε.
We also recall the notion of Gromov-Hausdorff distance between two (isometry class of proper) pointed metric spaces. Definition 5.3. The Gromov-Hausdorff distance between two pointed metric spaces (X, p) and (Y, q) is the infimum of the ε > 0 for which there exists a subset R of X × Y , called a correspondence, such that : Heuristically, this correspondence R must be thought as an ε-rough isometry between these two balls with radius ε −1 .
Based on this definition, a sequence (X n , p n ) of pointed metric spaces converges to a pointed metric space (X ∞ , p ∞ ) if, for all ε > 0, there exists n 0 such that for n ≥ n 0 , there exists a map f n : This definition 5.3 is only useful for complete metric spaces. Indeed, the Gromov-Hausdorff topology does not distinguish between a metric space and its completion. It does not distinguish either between two pointed metric spaces that are isometric : the Gromov-Hausdorff distance is a distance on the set of isometry classes of proper pointed metric spaces. See [7,Theorem 8.1.7] The following equivalent definition of Gromov-Hausdorff convergence is useful when one wants to get rid of the ambiguity coming from the group of isometries of (X ∞ , p ∞ ).
Fact 5.4. Let (X n , p n ), for n ≥ 1, and (X ∞ , p ∞ ) be pointed proper metric spaces. The sequence (X n , p n ) converges to (X ∞ , p ∞ ) if and only if there exists a complete metric space Z containing isometrically all the metric spaces X n and X ∞ as disjoint closed subsets, and such that (a) the sequence of points p n converges to p ∞ in Z, (b) the sequence of closed subsets X n converges to X ∞ for the Hausdorff topology.
Statement (b) means that -every point z of X ∞ is the limit of a sequence (x n ) n≥1 with x n ∈ X n , -every cluster point z ∈ Z of a sequence (x n ) n≥1 with x n ∈ X n belongs to X ∞ .
Sketch of proof of Fact 5.4. Assume that the sequence (X n , p n ) converges to (X ∞ , p ∞ ). We want to construct the metric space Z. We can choose a sequence ε n ց 0, and correspondences R n on X n × X ∞ as in Definition 5.3 with p = p n , q = p ∞ and ε = ε n . This allows us to construct, for every n ≥ 1, a metric space Y n which is the disjoint union of X n and X ∞ , which contains isometrically both X n and X ∞ and such that the distance between two points x in X n and y in X ∞ is given by where the infimum is over all the pairs (x ′ , y ′ ) which belong to R n . The space Z is defined as the disjoint union of all the X n and of X ∞ . The distance on Z is given on each union Y n := X n ∪ X ∞ by (5.9) and the distance between points x in X m and z in X n with m = n is given by where the infimum is over all the points y in X ∞ . Then (a) follows from (i) and (b) follows from (ii), (iii) and (iv).
The choice of such isometric embeddings of all X n and X ∞ in a fixed metric space Z will be called a realization of the Gromov-Hausdorff convergence. Such a realization is not unique. It is useful since it allows us to define the notion of a converging sequence of points x n in X n to a limit x ∞ in X ∞ by the condition d Z (x n , x ∞ ) −−−→ n→∞ 0.

Compactness criterion
A fundamental tool in this topic is the following compactness result for uniformly proper pointed metric spaces due to Cheeger-Gromov : Fact 5.5. Let (X n , p n ) n≥1 be a sequence of pointed proper metric spaces. Suppose that, for all R > 0 and ε > 0, there exists an integer N = N (R, ε) such that, for all n ≥ 1, the ball B(p n , R) of X n can be covered by N balls with radius ε. Then there exists a subsequence of (X n , p n ) which converges to a proper pointed metric space (X ∞ , p ∞ ).
For a proof see [7, Theorem 8.1.10]. The following lemma gives us a compactness property for sequences of Lipschitz functions between pointed metric spaces. Lemma 5.6. Let (X n , p n ) n≥1 and (Y n , q n ) n≥1 be sequences of pointed proper metric spaces which converge respectively to proper pointed metric spaces (X ∞ , p ∞ ) and (Y ∞ , q ∞ ). As in Fact 5.4, we choose metric spaces Z X and Z Y which realize these Gromov-Hausdorff convergences as Hausdorff convergences.
Let c > 1 and let (f n : X n → Y n ) n≥1 be a sequence of c-Lipschitz maps such that f n (p n ) = q n . Then there exists a c-Lipschitz map f ∞ : X ∞ → Y ∞ such that, after extracting a subsequence, the sequence of maps f n converges to f ∞ . This means that for each sequence x n ∈ X n which converges to Proof. This follows from basic topology arguments.
First step. We first choose a point x ∞ in X ∞ and a sequence x n in X n converging to x ∞ . Since the metric space Z Y is proper and the sequence f n (x n ) is bounded in Z Y we can assume after extracting a subsequence that the sequence f n (x n ) converges to a point y ∞ ∈ Y ∞ . Since the f n are c-Lipschitz, this limit y ∞ does not depend on the choice of the sequence x n converging to x ∞ . We define f ∞ (x ∞ ) := y ∞ .
Second step. We choose a countable dense subset S ∞ ⊂ X ∞ and use Cantor's diagonal argument to ensure that the first step is valid simultaneously for all points x ∞ in S ∞ .
Last step. One checks that the limit map f ∞ : S ∞ → Y ∞ is c-Lipschitz. Hence it extends uniquely as a c-Lipschitz map f ∞ : X ∞ → Y ∞ and the sequence f n converges locally uniformly to f ∞ .

Length spaces and Alexandrov spaces
We recall a few well-known definitions (see [7]).
A length space is a complete metric space for which the distance δ between two points is the infimum of the length of the curves joining them. When X is proper, any two points at distance δ can be joined by a curve of length δ. Such a curve is called a geodesic segment.
Let K ≤ 0. A CAT(K)-space or CAT-space with curvature at most K is a length space in which any geodesic triangle (P, Q, R) is thinner than a comparison triangle (P , Q, R) in the plane X of constant curvature K. Let us explain what this means. A comparison triangle is a triangle in X with the same side lengths. For every point P ′ on the geodesic segment [P, Q] we denote by P ′ the corresponding point on the geodesic segment [P , Q] i.e.
the point such that d(P, P ′ ) = d(P , P ′ ). Thinner means that one always has d(P ′ , R) ≤ d(P ′ , R). Note that a CAT(0)-space is always simply connected (See [6, Corollary II.1.5]). We also recall that in a proper CAT(0)-space, any two points can be joined by a geodesic and that this geodesic is unique. Similarly, a metric space with curvature at least K is a length space in which any geodesic triangle (P, Q, R) is thicker than a comparison triangle (P , Q, R) in the plane X of constant curvature K. Thicker means that one always has d(P ′ , R) ≥ d(P ′ , R).
The following proposition tells us that these properties are closed for the Gromov-Hausdorff topology.
Fact 5.7. Let (X n , p n ) n≥1 and (X ∞ , p ∞ ) be pointed proper metric spaces. Let K ≤ 0. Assume that the sequence (X n , p n ) converges to (X ∞ , p ∞ ). (i) If the X n 's are length spaces, then X ∞ is also a length space.
(ii) If the X n 's are CAT(K) spaces, then X ∞ is also a CAT(K) space.
(iii) If moreover the X n 's have curvature at least K, then X ∞ too.

Hadamard manifolds with C 1 metrics
In this section we focus on C 2 Hadamard manifolds when the Riemannian metric is only assumed to be C 1 . These Hadamard manifolds will occur in Section 5.5 as Gromov-Hausdorff limits of pinched C ∞ Hadamard manifolds.

Definition
We need first to clarify the definitions. We will deal with C 2 manifolds X. This means that X has a system of charts x → (x 1 , . . . , x k ) into R k for which the transition functions are of class C 2 . These manifolds will be endowed with a C 1 Riemannian metric g. This means that in any C 2 chart, the functions g( ∂ ∂x i , ∂ ∂x j ) are continuously differentiable. In general, on such a Riemannian manifold, there might exist two different geodesics which are tangent at the same point (see [16] for an example with a C 1,α -Riemannian metric). The following lemma tells us that this kind of examples will not occur here since we are dealing only with CAT(0)spaces whose curvature is bounded below. Note that, since the metric tensor is not assumed to be twice differentiable, the expression "curvature bounded below" refers to the definitions in Section 5.3.
Definition 5.8. By a C 2 Hadamard manifold with a C 1 metric, we mean a C 2 manifold endowed with a C 1 Riemannian metric which is CAT(0) and complete. Lemma 5.9. Let X be a C 2 Hadamard manifold with a C 1 metric of bounded curvature. a) For all x in X and v in T x X there is a unique geodesic t → exp x (tv) starting from x at speed v. This geodesic is of class C 2 . b) This exponential map induces an homeomorphism Ψ : T X

Exponential map
Proof. This lemma looks very familiar. But, since the Christoffel coefficients might not be Lipschitz continuous, we cannot apply Cauchy-Lipschitz theorem on Existence and Uniqueness of solutions of differential equations.
a) Since the Christoffel coefficients are continous, we can apply Peano-Arzela theorem. It tells us that there exists at least one geodesic of class C 2 starting from x at speed v. Uniqueness follows from the lower bound on the curvature. b) Since X is CAT(0), the map Ψ is a bijection. Since a uniform limit of geodesic on X is also a geodesic, the map Ψ is continuous. This map Ψ is also proper, therefore it is an homeomorphism.

Geodesic interpolation of h 0 and h 1
In the sequel of this section we prove a few technical properties of the interpolation h t of two equidistant Lipschitz maps h 0 and h 1 with values in a Hadamard manifold (lemmas 5.10). In Section 5.8, we will apply this lemma to two equidistant harmonic maps h 0 and h 1 obtained by a limit process. This lemma 5.10 will be used to compare the energy of h 0 and h 1 with the energy of some small perturbations of h 0 and h 1 . However, in this section 5.4, we do not need to assume h 0 and h 1 to be harmonic. Here are the precise assumptions and notations for Lemma 5.10.
Let X be a C 2 Riemannian manifold with a C 1 metric and Y be a C 2 Hadamard manifold with C 1 metric. Let h 0 , h 1 : X → Y be two C 1 maps such that one has Since Y is a Hadamard manifold, there exists a unique map such that, for all x in X, the path t → h t (x) is the unit speed geodesic joining h 0 (x) and h 1 (x). This map h is called the geodesic interpolation of h 0 and h 1 . By convexity of the distance function, h is Lipschitz continuous. Therefore, by Rademacher's theorem, the map h is differentiable on a subset of full measure (with respect to the Riemannian measure on X). In particular, there exists a subset X ′ ⊂ X of full measure such that, for all x in X ′ , the map h is differentiable at (x, t) for almost all t in [0, 1]. In particular, for all tangent vector V ∈ T x X at a point x ∈ X ′ , the following derivative is well-defined for almost all t in [0, 1]. Such a measurable vector field J V on the geodesic t → h t (x) will be called a Jacobi field. We denote by the unit tangent vector to the geodesic t → h t (x).
Lemma 5.10. We keep these assumptions and notations. Let x be a point in X ′ and V ∈ T x X. a) There exists a constant α V ∈ R such that Proof. When Y is a C ∞ Hadamard manifold, the vector field J V is a classical Jacobi field and this lemma is well known. Indeed, the function ψ V is the norm of the orthogonal component K V of the Jacobi field J V , and Inequality (5.12) follows from the Jacobi equation satisfied by this Jacobi field K V . We now explain how to adapt the classical proof when Y is only assumed to be a C 2 Hadamard manifold with a C 1 metric. a) Since the path t → h t (x) is a unit speed geodesic, one has the equality d(h s (x), h t (x)) = |t−s| for all s, t in [0, 1]. Differentiating this equality gives, when J V (s) and J V (t) are defined, Since these functions ϕ s are convex, the limit ϕ V (t) := lim s→0 ϕ s (t) exists for all t in [0, 1] and is a convex function. c) We slightly change the parametrization of the geodesic interpolation : the function k : (t, s) → k t (s) := h t−sα V (c(s)) is well defined when t − sα V is in [0, 1], and the paths t → k t (s) are also unit speed geodesics. Hence, for almost all t in [0, 1], the vector field is well-defined and one has the orthogonal decomposition In particular, one has the equality, The same argument as in b) with the Jacobi field K V proves that the function ψ V is also convex.

Geodesic interpolation in negative curvature
The following Lemma 5.11 improves Lemma 5.10 when the curvature of Y is uniformly negative. Indeed, it tells us that the norm t → ψ V (t) of the Jacobi field K V is uniformly convex. Lemma 5.11. We keep the assumptions and notations of Lemma 5.10. Moreover we assume that Y is a CAT(−a 2 )-space with a > 0. Then the function ψ V satisfies the following uniform convexity property, Proof. This inequality (5.19) will follow from an upper bound for the norm of the Jacobi field t → K V (t) by the norm of a well chosen Jacobi field t → K(t) along a geodesic segment in the hyperbolic plane of curvature −a 2 . Here are the details of the construction of this Jacobi field t → K(t).
Using a slight rescaling, we can assume without loss of generality that the geodesics t → k t (s) are defined for t in [0, 1] and that the Jacobi field K V (t) is well defined for t = 0 and for t = 1. We choose s > 0. Later on we will let s go to 0. We set P t := k t (0) and Q s,t := k t (s), and we apply Reshetnyak Lemma 5.13 to the four points P 0 , P 1 , Q s,1 , Q s,0 . According to this lemma, there exists a convex quadrilateral C s in the hyperbolic plane Y of curvature −a 2 with vertices P 0 , P 1 , Q s,1 , Q s,0 , and a 1-Lipschitz map j : C s → Y whose restriction to each of the four geodesic sides P 0 P 1 , P 1 Q s,1 , Q s,1 Q s,0 , Q s,0 P 0 is an isometry onto each of the four geodesic segments P 0 P 1 , P 1 Q s,1 , Q s,1 Q s,0 , Q s,0 P 0 . Indeed, since d(P 0 , P 1 ) = 1, we can assume that the two vertices P 0 and P 1 do not depend on s and that the quadrilateral C s is positively oriented.
Since the vectors K V (0) and K V (1) are orthogonal to the geodesic segment t → k t (0), by Lemma 5.9, each of the four successive angles θ i (for i = 1, . . . , 4 ) between the four successive geodesic segments P 0 P 1 , P 1 Q s,1 , Q s,1 Q s,0 , Q s,0 P 0 in Y is equal to π 2 +o(1), where o(1) denotes a quantity that goes to 0 when s goes to 0. Since j is 1-Lipschitz, each of the corresponding four successive angles θ i between the four successive geodesic sides P 0 P 1 , P 1 Q s,1 , Q s,1 Q s,0 , Q s,0 P 0 in the hyperbolic plane Y is not smaller than θ i . Since the sum of these four angles θ i is bounded above by 2π, each of these four angles θ i also satisfies when s goes to 0 : Denote by t → P t and t → Q s,t the unit speed parametrizations of the sides P 0 P 1 and Q 0 Q 1 . For t in [0, 1], one has j(P t ) = P t and j(Q s,t ) = Q s,t , and also with equality when t = 0 or 1 : d(P 0 , Q s,0 ) = d(P 0 , Q s,0 ) and d(P 1 , Q s,1 ) = d(P 1 , Q s,1 ) .

(5.22)
We now focus on these convex quadrilaterals C s in the hyperbolic plane Y of curvature −a 2 . We write Q s,t = exp P t (sK s,t ) where K s,t belongs to T P t Y .
Since Since the vector field t → K(t) is a Jacobi field along the geodesic segment t → P t , which is orthogonal to the tangent vector, its norm ψ(t) := K(t) satisfies the Jacobi differential equation We have used the following existence result for a majorizing quadrilateral due to Reshetnyak in [31]. More precisely we have used the boundary of this majorizing quadrilateral C. Lemma 5.13. Let Y be a CAT(−a 2 ) metric space and Y be the hyperbolic plane of curvature −a 2 . Then, for every four points P 0 , P 1 , Q 1 Q 0 in Y there exists a convex quadrilateral C in Y with vertices P 0 ,P 1 , Q 1 ,Q 0 and a 1-Lipschitz map j : C → Y which is an isometry on each of the four geodesic sides of C, and which sends each of these four vertices R i on the corresponding given point R i in Y .

Limits of Hadamard manifolds
In this section we describe the Gromov-Hausdorff limits of pinched Hadamard manifolds.
The following proposition is a variation on the Cheeger compactness theorem.
Proposition 5.14. Let (X n , p n ) n≥1 be a sequence of k-dimensional pointed Hadamard manifolds with pinched curvature −1 ≤ K X n ≤ −a 2 ≤ 0. a) There exists a subsequence of (X n , p n ) which converges to a pointed proper CAT-space (X ∞ , p ∞ ) with curvature between −1 and −a 2 . b) This space X ∞ has a structure of a C 2 Hadamard manifold such that the distance on X ∞ comes from a C 1 Riemannian metric.
The same proof shows that X ∞ has a structure of a C 2,α Hadamard manifold with a C 1,α Riemannian metric, for every 0 < α < 1. We will not use this improvement.
Even though this proposition follows from [30, Theorem 72 p. 311], we give a sketch of proof below.

Proof. a)
The assumption on the curvature of X n ensures that for each R > 0, one has uniform estimates for the volumes of balls with radius R in X n : for all n ≥ 1 and x in X n , one has Therefore, for each 0 < ε < R, there exists an integer N = N (R, ε) such that every ball B Xn (p n , R) of X n can be covered by N balls of radius ε. Hence, according to Fact 5.5, there exists a subsequence of (X n , p n ) which converges to a proper pointed metric space (X ∞ , p ∞ ). According to Fact 5.7, X ∞ is a CAT-space with curvature between −1 and −a 2 .
b) It remains to check that X ∞ is a C 2 manifold with a C 1 Riemannian metric. We isometrically imbed the converging sequence (X n , p n ) in a proper metric space Z as in Fact 5.4. We fix r 0 > 0 and c 0 > 0 as in Lemma 5.2 where we introduced the harmonic coordinates, and we choose a maximal r 0 2c 0 -separated subset S ∞ of X ∞ . For each x ∞ in S ∞ , we choose a sequence x n of points in X n that converges to x ∞ . By (5.3), the harmonic charts are uniformly bi-lipschitz. More precisely, for all z, z ′ inB(x n , r 0 c 0 ), one has . Hence after extracting a subsequence, this sequence of charts Ψ xn converges towards a bi-lipschitz map The extraction can be chosen simultaneously for all the points x ∞ in the countable set S ∞ . This collection of maps Ψ x∞ endows X ∞ with a structure of a Lipschitz manifold. We now want to prove that this manifold X ∞ is a C 2 manifold. Indeed we will check that, for any are of class C 2 . This just follows from the fact that these transition functions are uniform limit on compact sets of the transition functions Φ x ′ n •Φ −1 xn which are, by (5.7), uniformly bounded in the C 2,α -norm.
Finally, we check that the distance d on X ∞ comes from a C 1 Riemannian metric on X ∞ . By (5.8), the Riemannian metrics (g n ) ij on X n , seen as functions in the charts Ψ xn of X n , are uniformly bounded in the C 1,α -norm. Extracting again a subsequence, there exists a C 1 Riemannian metric (g ∞ ) ij in the charts Ψ x∞ of X ∞ such that the sequence of metrics (g n ) ij converges to (g ∞ ) ij in the C 1 topology. (5.28) Let d ∞ be the distance on X ∞ associated with g ∞ . We check that d ∞ = d on X ∞ . Let x ′ ∞ and x ′′ ∞ be points in X ∞ . They are limits of points x ′ n and x ′′ n in X n . Let c n be the geodesic segment joining x ′ n to x ′′ n . Extracting once more a subsequence, the curves c n converge uniformly to a curve joining x ′ ∞ and x ′′ ∞ . This curve must be a geodesic for g ∞ . This proves that

Convergence of harmonic maps
We now explain how to obtain the limit harmonic maps.
We first notice that we can extend Definition 3.2 : A C 2 map h : X → Y between two C 2 Riemannian manifolds with C 1 metrics X and Y is said to be harmonic if its tension field is zero, namely τ (h) := trD 2 h = 0. Indeed, the tension field of a C 2 map h at a point x depends only on the 2-jet of h and on the 1-jet of the metrics of X and Y at the points x and h(x). More precisely, writing h in a coordinate system h : (x 1 , . . . , x k ) → (h 1 , . . . , h ′ k ), the equation trD 2 h = 0 reads as where Γ λ µν are the Christoffel coefficients on Y and where ∆ is the Laplace operator on X defined as in (3.1) : where v = det(g ij ) denotes the volume density on X. See [17, Section 1.3] for more details. Lemma 5.15. Let (X n , p n ) n≥1 and (Y n , q n ) n≥1 be two sequences of equidimensional pointed Hadamard manifolds with curvature between −1 and 0.
Let c, C > 0 and let h n : X n → Y n be a sequence of (c, C)-quasi-isometric harmonic maps such that sup n d(h n (p n ), q n ) < ∞. After extracting a subsequence, the sequences of pointed metric spaces (X n , p n ) and (Y n , q n ) converge respectively to pointed C 2 manifolds with C 1 Riemannian metrics (X ∞ , p ∞ ) and (Y ∞ , q ∞ ), and the sequence of maps h n converges to a c-quasi-isometric map h ∞ : X ∞ → Y ∞ . This map h ∞ is of class C 2 and is harmonic.
Proof. Since they are harmonic, the maps h n are C ∞ . Since these maps are also (c, C)-quasi-isometric, according to Cheng's Lemma 3.4, there exists some constant C ′ > 0 such that the maps h n are C ′ -Lipschitz. The first two statements then follow from Proposition 5.14 and Lemma 5.6. It remains to show that the limit map h ∞ is of class C 2 and harmonic. The key point will be a uniform bound for the C 2,α -norm of h n in suitable harmonic coordinates. Let k := dim X n and k ′ := dim Y n . Let x ∞ be a point in X ∞ and y ∞ := h ∞ (x ∞ ). Let x n be a sequence in X n converging to x ∞ and let y n := h n (x n ).
We look at the maps h n through the harmonic charts Ψ xn of X n and Ψ yn of Y n as in (5.26). By (5.27), these charts converge respectively to charts Ψ x∞ of X ∞ and Ψ y∞ of Y ∞ . By (5.28), in these charts, the Riemannian metrics of X n and Y n converge to the Riemannian metrics of X ∞ and Y ∞ in the C 1,α -norm.
Let 0 < α < 1. When one writes Equation (5.29) for h = h n in these harmonic coordinates on a small open ball Ω :=B(0, r 0 c 0 C ′ ) of R k that does not depend on n, one gets The coefficients of this equation depend on n, but Lemma 5.2 ensures that they are uniformly bounded in the C α -norm. The Schauder estimates for functions u on Ω and compact sets K of Ω as in [30, Theorem 70 p. 303] thus tell us that for some constant M = M (k, Ω, K). Therefore, since the maps h n are C ′ -Lipschitz, combining (5.29), (5.32) and (5.33) yields a uniform bound for the C 2,α -norm of the maps h n . Hence the Ascoli Lemma ensures that, after extracting a subsequence, the sequence of maps h n converges towards a C 2 map in the C 2 topology. This proves that the limit map h ∞ is C 2 and is harmonic.

Construction of the limit equidistant harmonic maps
We now explain why the limit harmonic maps h 0,∞ and h 1,∞ constructed in the strategy of Proposition 5.1 are equidistant.
We first sum up the construction of these limit maps. We start with two Hadamard manifolds X, Y of bounded curvatures, and with two distinct quasi-isometric harmonic maps h 0 , h 1 : X → Y such that δ := d(h 0 , h 1 ) is finite and non-zero. We choose a sequence of points p n in X such that d(h 0 (p n ), h 1 (p n )) converges to δ and we set q 0,n := h 0 (p n ) and q 1,n := h 1 (p n ). We will frequently replace this sequence by subsequences without mentioning it. By Proposition 5.14, there exist two C 2 Hadamard manifolds with C 1 metrics (X ∞ , p ∞ ) and (Y ∞ , q 0,∞ ) which are the Gromov-Hausdorff limits of the pointed metric spaces (X, p n ) and (Y, q 0,n ). These limit Hadamard manifolds also have bounded curvature. We denote by q 1,∞ the limit in Y ∞ of the sequence q 1,n . By the Cheng Lemma 3.4, the harmonic quasi-isometric maps h 0 and h 1 are Lipschitz continuous. By Lemma 5.6, there exists a limit map h 0,∞ : (X ∞ , p ∞ ) → (Y ∞ , q 0,∞ ) of the sequence of Lipschitz continuous maps h 0 : (X, p n ) → (Y, q 0,n ). There exists also a limit map h 1,∞ : (X ∞ , p ∞ ) → (Y ∞ , q 1,∞ ) of the sequence of Lipschitz continuous maps h 1 : (X, p n ) → (Y, q 1,n ). By Lemma 5.15, these limit maps h 0,∞ and h 1,∞ are still harmonic quasi-isometric maps.
We will apply this lemma to two pinched Hadamard manifolds X, Y . In this case, the limit C 2 Hadamard manifolds X ∞ , Y ∞ will also be pinched.
Proof. Let ∆ ∞ be the Laplace operator on X ∞ defined as in (5.30). We first check that the function ϕ ∞ : x → d(h 0,∞ (x), h 1,∞ (x)) is subharmonic on X ∞ . This means that ∆ ∞ ϕ ∞ is a positive measure on X ∞ . Assume first that the Riemannian metric on Y ∞ is C ∞ . In this case, ϕ ∞ is the composition of a harmonic map h = (h 0 , h 1 ) : where (e i ) 1≤i≤k is an orthonormal basis of the tangent space to X.
Since the Riemannian metric on Y might not be of class C ∞ , we will use instead a limit argument. We fix a point x ∞ in X ∞ . In a chart (x 1 , . . . , x k ), the Laplace operator ∆ ∞ of the Riemannian metric (g ∞ ) ij of X ∞ reads as where v ∞ is the volume density. We want to prove that for every C 2 function ψ ≥ 0 with compact support in a small neighborhood of x ∞ , one has This function ϕ ∞ on the pointed metric space (X ∞ , p ∞ ) is equal to the limit of the sequence of functions ϕ n : x → d(h 0 (x), h 1 (x)) on the pointed metric spaces (X, p n ), as defined in Lemma 5.6. Note that the dependence on n comes from the base point p n which moves with n. We choose a sequence x n in X n converging to x ∞ . As in the proof of Lemma 5.15, we look at the functions ϕ n through the harmonic charts Ψ xn of X n . By (5.27), these charts converge to a chart Ψ x∞ of X ∞ . By (5.28), in these charts (x 1 , . . . , x k ) the Riemannian metric (g n ) ij of X n converge to the Riemannian metric (g ∞ ) ij of X ∞ in C 1 topology.
Since, by the above argument, the functions ϕ n are subharmonic for the metric (g n ) ij , for every C 2 function ψ ≥ 0 with compact support in these charts, one has at each step n ϕ n ∆ n ψ v n dx ≥ 0 (5.36) where ∆ n and v n are the Laplace operator and the volume density of the metric (g n ) ij . Letting n go to ∞ in (5.36) gives (5.35). This proves that the function ϕ ∞ is subharmonic. By construction, this subharmonic function ϕ ∞ on X ∞ achieves its maximum δ > 0 at the point p ∞ . By (5.34), the Laplace operator is an elliptic linear differential operator with continuous coefficients. Hence, by the strong maximum principle in [13,Theorem 8.19 p.198], this function ϕ ∞ is constant and equal to δ.
The aim of Sections 5.8 and 5.9 is to prove that such equidistant harmonic quasi-isometric maps h 0,∞ and h 1,∞ can not exist (Corollary 5.19) when Y ∞ is pinched. This will conclude the proof of Proposition 5.1.

Equidistant harmonic maps
We first study equidistant harmonic maps without any pinching assumption.
The following lemma 5.17 extends [22, Lemma 2.2] to the case where the source space X is only assumed to be a C 2 -Hadamard manifold. Lemma 5.17. Let X, Y be two C 2 Hadamard manifolds with C 1 Riemannian metrics of bounded curvature. Let h 0 , h 1 : X → Y be two harmonic maps such that the distance function x → d(h 0 (x), h 1 (x)) is constant. For t in [0, 1], let h t be the geodesic interpolation of h 0 and h 1 as in (5.12). Then for almost all x in X, t in [0, 1] and V in T x X, one has Note that we can not conclude that Equality (5.37) is valid for all x and t since the interpolation h t might not be of class C 1 .
We will use the following straightforward inequality for convex functions.
Proof of Lemma 5.18. We just add the following two convexity inequalities Proof of Lemma 5.17. The idea is to construct two small perturbations f ε and g ε of the harmonic maps h 0 and h 1 with support in a compact set K of X, and to compare the sum of the energies of f ε and g ε with the sum of the energies of h 0 and h 1 .
Here is the definition of the two maps f ε : X → Y and g ε : X → Y . We fix a C 1 cut-off function η : X → [0, 1]; x → η x with compact support K, and we let for all x in X : These functions are Lipschitz continuous, so that they are almost everywhere differentiable. In order to compute their differentials, we use the notations (5.13) and (5.14) : for all x in a subset X ′ ⊂ X of full measure, for all V in T x X, for almost all t in [0, 1], we let For such a tangent vector V , it follows from Lemma 5.10.b that there exists a convex function t → ϕ V (t) such that ϕ V (t) = J V (t) for all t where the derivative J V (t) exists. By the chain rule, for almost all ε in [0, 1], the differentials of f ε and g ε are given, for almost all x in X and all V in T x X, by where V.η = dη(V ) is the derivative of the function η in the direction V . According to Lemma 5.10.a, for almost all x in X and all V in T x X, the scalar product J V (t), τ x (t) is almost surely constant. Therefore, for almost all ε in [0, 1], x in X and V in the unit tangent bundle T 1 x X, one has the equality We introduce the convex function t → Φ V t := ϕ V (t) 2 . Using Inequality (5.38), one gets for almost all ε in [0, 1], x in X and V in T 1 x X the bound We recall that the energy over K of a Lipschitz map h : X → Y is where dx is the Riemannian measure on X and dV the Riemannian measure on T 1 X. Integrating the previous inequality on the unit tangent bundle of K, one gets the following inequality relating the energy over K of f ε , g ε , h 0 and h 1 : where A is the function on T 1 X defined for almost all x in X and V in T 1 x X, by ). Since the harmonic maps h 0 and h 1 are critical points for the energy functional, Inequality (5.43) implies that Since the function Φ V is convex, the function A is non-negative. Therefore Inequality (5.44) implies that the function A is almost surely zero. Since the function η was arbitrary, this tells us that, for almost all V in T 1 X, one has Since Φ V is the square of the convex function ϕ V , this tells us that for almost all V in T X, the function ϕ V is constant. This proves (5.37).

Equidistant harmonic maps in negative curvature
The following Corollary 5.19 improves the conclusion of Lemma 5.17 when the curvature of Y is uniformly negative.
Corollary 5.19. Let a > 0. Let X, Y be two C 2 Hadamard manifolds with C 1 Riemannian metrics. Assume moreover that Y is CAT(−a 2 ). Let h 0 , h 1 : X → Y be two harmonic maps such that the distance function This means that, when h 0 = h 1 , there exists a geodesic t → γ(t) in Y and two harmonic functions u 0 , u 1 on X such that h 0 = γ • u 0 , h 1 = γ • u 1 and the difference u 1 −u 0 is a bounded harmonic function on X.
Note that this case is ruled out when h 0 and h 1 are within bounded distance from a quasi-isometric map f : X → Y since X has dimension k ≥ 2.
Proof of Corollary 5.19. We can assume that the distance between h 0 and h 1 is equal to 1. We recall a few notations that we have already used. For t in [0, 1], let h t be the geodesic interpolation of h 0 and h 1 . For By Lemmas 5.10.a and 5.17, one has the equalities for almost all t in [0, 1] and almost all V in T X, so that Comparing these equalities with the uniform convexity of the function ψ V in (5.19), one infers that ψ V (t) = 0. Hence, when J V (t) is defined, one has Since the speed of c is bounded by 1/3, the curve C is well-defined. By construction C is a Lipschitz continuous path, and by (5.46) and (5.47), for almost all s, its derivative is, Therefore, one has C(s) = C(0) for all s in [0, 1], that is Using this equality for two distinct values of t 0 , we deduce that the geodesic segments h This also ends the proof of Proposition 5.1.

Boundary maps for weakly coarse embeddings
This chapter is independent of the previous ones. We prove that a weakly coarse embedding between pinched Hadamard manifolds admits a boundary map which is well-defined outside a set of zero Hausdorff dimension. We prove that the fibers of this boundary map also have zero Hausdorff dimension (Theorem 6.5). More precisely, we will prove quantitative versions of these facts (Propositions 6.13 and 6.15) that we will use in Chapter 7.

Weakly coarse embeddings
In this section, we introduce various classes of rough Lipschitz maps f : X → Y between pinched Hadamard manifolds generalizing the quasi-isometric maps.
Let X and Y be Hadamard manifolds with pinched sectional curvatures When f : X → Y is a rough c-Lipschitz map, one has for all x, x ′ in X : Definition 6.2. A map f : X → Y is a coarse embedding if there exist two non-decreasing unbounded functions ϕ 1 , ϕ 2 such that, for all x, x ′ ∈ X : Note that a map which is within bounded distance from a coarse embedding is also a coarse embedding. In Definition 6.2 one may always assume that the function ϕ 2 is an affine function, that is, f is rough Lipschitz. A quasi-isometric map is a special case of a coarse embedding, where ϕ 1 is also an affine function. Definition 6.3. A weakly coarse embedding f : X → Y is a rough Lipschitz map for which there exist c 0 , C 0 > 0 such that, for all x, x ′ in X : Equivalently, this means that there exist two non-decreasing non-negative and non-zero functions ϕ 1 , ϕ 2 such that (6.1) holds. Of course, any coarse embedding f : X → Y is a weakly coarse embedding. Proof. Indeed, one first constructs a unit-speed C 1 curve f 0 : R → H 2 such that ϕ 1 (t) = min s∈R d(f 0 (s + t), f 0 (s)) for every t ≥ 0. We set H 1 := R and, for k ≥ 1, we embed each space H k as a totally geodesic hyperplane in H k+1 and denote by x → n x a unit normal vector field to H k in H k+1 . We now define the Lipschitz map f : H 2 → H 3 as f (exp(tn s )) := exp(tn f 0 (s) ) for all s in H 1 and t ∈ R.
For any point x 0 ∈ X and r > 0, we identify through the exponential map each sphere S(x 0 , r) with the unit tangent sphere More precisely, when ξ ∈ S x 0 is a unit tangent vector at the point x 0 , we denote by r → ξ r := exp x 0 (rξ) the corresponding unit-speed geodesic ray (so that ξ 0 = x 0 ).
We denote by X = X ∪ ∂X the visual compactification of X. The boundary ∂X is the set of (equivalence classes of) rays in X. The map ψ x 0 : ξ → lim r→∞ ξ r gives an homeomorphism of the unit tangent sphere S x 0 with the sphere at infinity ∂X. We say that a subset A of ∂X has zero Hausdorff dimension if, seen in S x 0 , it has zero Hausdorff dimension. One can check that this property does not depend on the choice of x 0 , because for any another point x 1 ∈ X, the homeomorphism ψ x 1 −1 • ψ x 0 is bi-Hölder. In this Chapter 6 we will prove the following theorem. b) For every ξ ∈ ∂X A, the fiber {η ∈ ∂X A | ∂f (η) = ∂f (ξ)} has zero Hausdorff dimension.
The map ∂f : ∂X A → ∂Y is called the boundary map of f . The proof of Theorem 6.5 will last up to the end of this Chapter. The quantitative estimates (6.7) and (6.9) that we will obtain during this proof will be reused in Chapter 7.

Hausdorff dimension and Frostman measures
In this section we introduce classical notations and definitions from geometric measure theory. Definition 6.6. Let M, ν > 0. A Borel probability measure σ on a compact metric space S is said to be (M, ν)-Frostman if, for all ξ ∈ S and all r > 0, one has the following bound for the measures of the balls : Let ν > 0 and δ > 0. For a subset A ⊂ S, we denote When δ = ∞, we denote similarly We recall that the ν-dimensional Hausdorff measure of A is defined as Observe that one also has The following easy lemma relates H ν ∞ (A) with Frostman measures.

Image of a large sphere
In this section we focus on those points of a sphere S(x 0 , r) whose images under a weakly coarse embedding are too close from a given point.
The following definition plays a key role in the proof of Theorem 6.5.
can be covered by at most C 1 e bk ′ αr balls of radius C 2 e −ar , where k ′ = dim Y .
If such constants C 1 , C 2 exist, we say that f satisfies property C.
In this definition the unit sphere S x 0 is endowed with the distance induced by the Riemannian norm on T x 0 X.
The bound on the size of a covering of the set (6.6) will be very useful for Hausdorff dimension estimations. The precise value bk ′ α for the exponential growth in Definition 6.8 is not very important. It is the one obtained in the next proposition and it just avoids the introduction of another parameter. Proposition 6.9. Every weakly coarse embedding f : X → Y satisfies property C.
In particular, Propositions 6.13 and 6.15 below apply to all weakly coarse embeddings f .
We will use the Bishop volume estimates (see for example [11]) which compare the volume of balls in X and in the hyperbolic space H k . Lemma 6.10. Let X be a pinched Hadamard manifold with dimension k and sectional curvature −b 2 ≤ K X ≤ −a 2 < 0. Then, for R > 0, one has We will also need to bound angles by Gromov products as in Lemma 2.1. Lemma 6.11. Let Y be a Hadamard manifold with curvature K Y ≤ −a 2 < 0. Then, for all y 0 ∈ Y and y 1 , y 2 ∈ Y {y 0 } one has the bound θ y 0 (y 1 , y 2 ) ≤ 4 e −a(y 1 ,y 2 )y 0 , where θ y 0 (y 1 , y 2 ) is the angle at y 0 of the geodesic triangle (y 0 , y 1 , y 2 ) and (y 1 , y 2 ) y 0 := 1 2 (d(y 0 , y 1 ) + d(y 0 , y 2 ) − d(y 1 , y 2 )) is the Gromov product.
Proof of Proposition 6.9. We will see that f satisfies property C C 1 ,C 2 where the constants C 1 , C 2 depend only on a, b, k ′ , and on c 0 , C 0 from (6.2). It follows from the volume estimates of Lemma 6.10 that there exists a constant C 1 > 0 such that for each ball B(y 0 , αr) ⊂ Y (r > 0) and each covering of minimal cardinality of this ball by balls with radii c 0 /2 B(y 0 , αr) ⊂ ∪ i∈I B(y i , c 0 /2), this cardinality is at most C 1 e bk ′ αr .
Since f is a (c 0 , C 0 )-weakly coarse embedding, for each i ∈ I, the inverse image of this ball f −1 (B(y i , c 0 /2)) is either empty or lies in a ball B(x i , C 0 ) ⊂ X. By Lemma 6.11, the intersection B(x, C 0 ) ∩ S(x 0 , r) lies in a cone with vertex x 0 and angle θ r = C 2 e −ar .
Remark 6.12. Any map f : X → Y within bounded distance from a map f : X → Y satisfying Property C also satisfies Property C.

Construction of the boundary map
We now investigate the long-term behavior of the images of geodesic rays under a rough Lipschitz map satisfying Property C.
Let X, Y be pinched Hadamard manifolds and f : X → Y be a rough Lipschitz map satisfying Property C. The following proposition 6.13 tells us that, except for a set of rays of zero Hausdorff dimension, the image under f of a ray goes to infinity in Y at positive speed and this image converges to a point in ∂Y .
We need some notations. For x 0 ∈ X, let A x 0 be the set of rays whose image do not go to infinity at positive speed, namely
The bound (6.8) can be interpreted as a large deviation inequality for the random path f (ξ t ) when the ray ξ is chosen randomly with law σ. We will apply it later on to various harmonic measures σ = σ x 0 ,r . A key point in this bound is that the constants involved in (6.8) do not depend on the (M, ν)-Frostman measure σ.
Proof of Proposition 6.13. a) Since f has Property C C 1 ,C 2 , one has b) This follows from a) and Lemma 6.7. c) Letting n 0 go to infinity in (6.7), one gets H ν ∞ (A x 0 ,α ) = 0 for all ν > ν α . Therefore, (6.5) e) Since f is rough Lipschitz, one can assume that the parameters r are integers and applies the next lemma 6.14 to the sequence y n = f (ξ n ). Lemma 6.14. Let Y be a Hadamard manifold with sectional curvature K Y ≤ −a 2 < 0. Let (y n ) n∈N be a sequence in Y such that sup n≥0 d(y n , y n+1 ) < ∞ and lim inf n→∞ 1 n d(y 0 , y n ) > 0.
Then, the limit y ∞ := lim n→∞ y n exists in the visual boundary ∂Y .
Unlike quasi-isometric maps, coarse embedding may not have boundary values in every direction. See Example 6.4 where we could begin with a curve f 0 that spirals away in H 2 .

The fibers of the boundary maps
We now investigate the fibers of the boundary map ∂f of a rough Lipschitz map satisfying property C.
The following proposition 6.15 tells us that the fibers of the boundary map have zero Hausdorff dimension.
We begin with a technical covering lemma. Lemma 6.16. Notations as in Proposition 6.15. For ξ ∈ S x 0 and p ≥ n 0 , let If ξ is not in A x 0 ,α (n 0 ), the set B ξ x 0 ,α,p can be covered by at most C 1 e αbk ′ p 1 − e −βαbk ′ balls of radius C 2 e −ap .
The value β α has been chosen in order to obtain the same exponential growth αbk ′ for these coverings as in Definition 6.8.
Proof of Lemma 6.16. Using the notation (6.6), we have the equality The key point is that, since f is rough c-Lipschitz and ξ ∈ A x 0 ,α (n 0 ), this union is a finite union. Indeed, assume that an integer n ≥ n 0 satisfies By our choice of β α , this inequality is equivalent to (n+p)β α ≤ p α.
Therefore, using Definition 6.8, one can cover the set B ξ x 0 ,α,p by at most C 1 n e (n+p)bk ′ βα balls of radius C 2 e −ap , where this sum runs over the integers n ≥ n 0 such that (n+p)β α ≤ pα. Computing this sum, one deduces that the set B ξ x 0 ,α,p can be covered by at most C 1 e αpbk ′ 1 − e −βαbk ′ balls of radius C 2 e −ap .
Proof of Proposition 6.15. a) Using Lemma 6.16, one has b) This follows from a) and Lemma 6.7. c) Letting n 0 go to infinity in (6.9) one gets H ν ∞ (B ξ x 0 ,α ) = 0, for all ν > ν α . Therefore, using (6.5), it follows that dim This is a consequence of the following lemma 6.17 applied to the sequences y n = f (ξ n ) and z p = f (η p ).
f ) This follows from e).

Two sequences going away from one another
The aim of this section is to prove the following lemma which provides, in a pinched Hadamard manifold, a lower bound for the angle between points in two sequences with bounded speed that "go away from one another at positive speed". Lemma 6.17. Let Y be a Hadamard manifold with sectional curvature 1−e −aβ . Let (y n ) n∈N and (z p ) p∈N be two sequences of points in Y with y 0 = z 0 , such that d(y n , y n+1 ) ≤ c and d(z p , z p+1 ) ≤ c for any integers n, p ≥ 0, (6.12) d(y 0 , y n ) ≥ nα, d(y 0 , z p ) ≥ pα and d(y n , z p ) ≥ (n+p)β for n, p ≥ n 0 . (6.13) Then, for any integer n, p ≥ ℓ 0 := 4n 0 c α , one has θ y 0 (y n , z p ) ≥ 1 2 e −2n 0 bc . (6.14) We will need two geometric lemmas. We know that the orthogonal projection from a Hadamard manifold on a geodesic is a 1-Lipschitz map. The following lemma gives us a more precise information when the curvature is bounded from above. Lemma 6.18. Let Y be a Hadamard manifold with sectional curvature K Y ≤ −a 2 < 0. Let γ ⊂ Y be a geodesic. Then, the orthogonal projection π : Y → γ is smooth and, for y ∈ Y , the norm of its differential satisfies D y π ≤ 1 cosh(a d(y, γ)) ≤ 2 e −a d(y,γ) .
Let v ∈ T y Y . We want to bound the ratio D y π(v) / v . We may thus assume that v is orthogonal to KerD y π, i.e. that v is orthogonal to the geodesic c at the point y.
The map (s, t) → c t (s) is a variation of geodesics, so that J : s ∈ [0, ℓ] → d dt c t (s) |t=0 ∈ T c(s) Y is a Jacobi field along the geodesic c. We have J(0) = D y π(v) and J(ℓ) = v. Since both J(0) and J(ℓ) are normal to c, it follows that J is a normal Jacobi field. Since γ is a geodesic and each u(t) is normal to γ, we infer from the equality J ′ (0) = u ′ (0) that J ′ (0) is normal to γ, i.e. orthogonal to J(0). The Jacobi field equation J ′′ + R(c ′ , J)c ′ = 0 and the hypothesis on the curvature now yield Since J ′ (0) = J(0),J ′ (0) = 0, one deduces that J(t) ≥ J(0) cosh(at), for all t ≥ 0. In particular, one has D y π(v) ≤ v / cosh(aℓ).
Proof of Lemma 6.19. The angles of a triangle in H 2 (−b 2 ) with same sidelengths are smaller than the angles of the triangle (y 0 yȳ). It follows that θ y 0 (y,ȳ) ≥ ϕ, where ϕ is the angle of an isosceles right triangle in H 2 (−b 2 ) with adjacent sides R, which is ϕ = arctan( 1 cosh(bR) ) ≥ 1 2 e −bR .
Hence by Lemma 6.19, one has either θ y 0 (y n , γ + ) ≥ 1 2 e −bR or θ y 0 (z p , γ + ) ≥ 1 2 e −bR . Since this is true for any ray γ + based at y 0 , one gets θ y 0 (y n , z p ) ≥ 1 2 e −bR . Proof of Theorem 6.5. Point a) follows from Propositions 6.13.d and 6.13.e. Point b) follows from Propositions 6.15.d and 6.15.f . Remark 6.20. It follows from the proof that Theorem 6.5 also holds true for any rough Lipschitz map f : X → Y between pinched Hadamard manifolds that satisfies property C.

Beyond quasi-isometric maps
The aim of this Chapter 7 is the following extension of Theorem 1.1 to all weakly coarse embeddings f , and in particular to all coarse embeddings f (see Definitions 6.2 and 6.3).

Weakly coarse embeddings and harmonic maps
Theorem 7.1. Every weakly coarse embedding f : X → Y between two pinched Hadamard manifolds is within bounded distance from a unique harmonic map h : X → Y .
Indeed we will prove a more general proposition using Definition 6.8.
Proposition 7.2. Every rough Lipschitz map f : X → Y satisfying property C between two pinched Hadamard manifolds is within bounded distance from a unique harmonic map h : X → Y .
The main new ingredients in the proof are the construction and the properties of a boundary map of f . These new ingredients which do not involve harmonic maps were explained in Chapter 6. We now explain how to adapt the proof of Theorem 1.1 using these new ingredients.

Rough Lipschitz harmonic maps
We first want to point out that Theorem 7.1 can not be extended to all rough Lipschitz maps. Example 7.3. There exists an injective Lipschitz map f : H 2 → H 2 from the hyperbolic plane to itself, that extends continuously to the visual boundary as the identity map, and which is not within bounded distance from any harmonic map.
Proof. We will consider a map f : H 2 → H 2 that commutes to a parabolic subgroup of Isom(H 2 ). Let us work in the upper half-plane model. The map f is defined by for any t ∈ R. Observe that f extends continuously to the visual compactification of H 2 by the identity, and that f is 2-Lipschitz. Assume by contradiction that there exists a harmonic map h : H 2 → H 2 within bounded distance from f .
First case: the map h is unique. In this case h also commutes to the isometries s t , so that there exists a continuous function g : [0, ∞] → [0, ∞] such that h(u, v) = (u, g(v)) u ∈ R , v > 0 , and with g(0) = 0, g(∞) = ∞. Saying that h is harmonic is equivalent to requiring the function g to satisfy the differential equation g g ′′ = (g ′ ) 2 − 1 .
It follows that the harmonic map h coincides with one of the maps h a : H 2 → H 2 defined by h a (u, v) = (u, 1 a sinh(av)) for some constant a ≥ 0. Observe that none of the maps h a is within bounded distance from f , hence the contradiction.
Second case: the map h is not unique. Let h 0 , h 1 be two distinct harmonic maps within bounded distance from f . We want again to find a contradiction. We will use arguments similar to those in Chapter 5. Let x 0 := (0, 1) ∈ H 2 . We choose a sequence of points x n in H 2 for which the distances d(h 0 (x n ), h 1 (x n )) converge to δ := sup x∈H 2 d(h 0 (x), h 1 (x)) > 0 and we set y n := f (x n ). Let ϕ n and ψ n be the isometries of H 2 fixing the point ∞ ∈ ∂H 2 and such that ϕ n (x 0 ) = x n and ψ n (x 0 ) = y n . After extraction, the sequence of maps ψ −1 n • f • ϕ n converges to one of the maps f β : For i = 0 and 1, the sequence of harmonic maps h i,n := ψ −1 n • h i • ϕ n converges, after extraction, to a harmonic map h i,∞ : H 2 → H 2 within bounded distance to f β . The subharmonic function x → d(h 0,∞ (x), h 1,∞ (x)) achieves its maximum value at x = x 0 , hence is a constant function equal to δ. Therefore, by Corollary 5.19, the harmonic maps h 0,∞ and h 1,∞ take their values in the same geodesic Γ. This forces the equality β = ∞ and the geodesic Γ is the image of f ∞ . Now we write f ∞ (u, v) = (0, e 2F∞(u,v) ) and h 0,∞ (u, v) = (0, e 2H 0,∞ (u,v) ), where F ∞ (u, v) = log v and where H 0,∞ is a harmonic function.
The function G ∞ := F ∞ − H 0,∞ is then a bounded function on H 2 such that ∆G ∞ = 1. Such a function G ∞ does not exist. Indeed the function G : x → 2 log(cosh(d(x 0 , x)/2)) also satisfies ∆G = 1 and the function G−G ∞ would be proper and harmonic, contradicting the maximum principle.

An overview of the proof of Proposition 7.2
Proof of Proposition 7.2. The strategy is the same as for Theorem 1.1: Step 1 : smoothing f out. By Proposition 2.4 there exists a smooth map f : X → Y within bounded distance from f and whose first and second covariant derivatives are bounded on X. This function f is Lipschitz and still satisfies property C. Hence we can assume that f = f .
Step 2 : solving a bounded Dirichlet problem. We fix an origin O ∈ X. For any radius R we consider the unique harmonic map h R : B(O, R) → Y satisfying the Dirichlet condition h R = f on the sphere S(O, R).
Step 3 : estimating the distance d(h R , f ). We will check in Section 7.4: Proposition 7.4. There exists a constant ρ ≥ 1 such that, for any R ≥ 1, one has d(h R , f ) ≤ ρ.
Step 4 : letting the sequence h R converge to h. We prove this convergence as in Section 3.3.
The proofs of Steps 1, 2 and 4, as well as the proof of uniqueness, require only minor modifications from the ones for quasi-isometric maps. Thus, the remaining of this paper will be devoted to the proof of Step 3.

Interior estimate for rough Lipschitz
In this section we complete the proof of Proposition 7.4 whose structure is exactly the same as the proof of Proposition 3.5. We will just repeat quickly the arguments of Section 4 pointing out the changes in the choice of the many constants involved in the proof.

Strategy
Let X and Y be two Hadamard manifolds whose curvatures are pinched −b 2 ≤ K ≤ −a 2 < 0. Let k = dim X and k ′ = dim Y . We fix two constants M, N > 0 as in Proposition 4.9. We set α = a 2bk ′ N so that, with the notation of Propositions 6.13 and 6.15, one has ν α = 1 2N . We set ν = 2ν α = 1 N . We start with a C ∞ Lipschitz map f : X → Y whose first and second covariant derivatives are bounded. We fix constants c, C 1 , C 2 ≥ 1 such that, f satisfies property C C 1 ,C 2 as in Definition 6.8 and such that, for all x in X, one has Df (x) ≤ c , D 2 f (x) ≤ bc 2 . (7.1) We let C 3 = C 3,α,ν ≤ C 4 = C 4,α,ν be the two constants as in Proposition 6.13 and 6.15. (1 − e −βbk ′ )(1 − e −a/(2N ) ) where β = α 2 2α + c .
The first version of this paper containing Chapters 1 to 5 was released in February 2017. In this second version, Chapters 6 and 7 were added. In between, two related preprints were posted in the ArXiv: [27] and [36].