The Spherical Boundary and Volume Growth

We consider the spherical boundary, a conformal boundary using a special class of conformal distortions. We prove that certain bounds on volume growth of suitable metric measure spaces imply that the spherical boundary is “small” in cardinality or dimension and give examples to show that the reverse implications fail. We also show that the spherical boundary of an annular convex proper length space consists of a single point. This result applies to l2-products of length spaces, since we prove that a natural metric, generalizing such “norm-like” product metrics on a possibly infinite product of unbounded length spaces, is annular convex.


Introduction
There are various notions of "boundaries at infinity" of metric spaces in the literature.One of these is the spherical boundary ∂ S X of certain unbounded metric spaces, as introduced in 1 .This is defined in detail in Section 2, but let us mention here that it is a byproduct of the concept of sphericalization, which replaces an unbounded metric l by a conformally distorted bounded metric σ.This allows one to interpret results in 2 concerning the quasihyperbolizations of bounded length spaces in the context of certain unbounded spaces.To this end, the relationship between l and σ, together with the relationship between their associated quasihyperbolizations, was studied in 1 .
The spherical boundary ∂ S X is a key ingredient in studying the invertibility of the sphericalization process as is clear from the results in 1 , but no detailed study of the links between features of X and features of ∂ S X was carried out there.The current paper aims to throw more light on one such link by proving results of the following type: if the spherical boundary of a suitable metric measure space is sufficiently large, then X has rapid volume growth.For instance, contrast Euclidean space R n whose spherical boundary is a singleton set if n > 1 with the hyperbolic plane whose spherical boundary has infinite Minkowski dimension.
Suppose X, d, μ is a metric measure space and C ≥ 1.We say that X, d, μ is Cdoubling if μ 2B ≤ Cμ B whenever tB B x, tr for fixed but arbitrary x ∈ X, r > 0. We say that X, d, μ is C-translate doubling if instead μ B ≤ Cμ B whenever B, B are overlapping balls of the same radius or weak C-translate doubling if we merely have μ B ≤ C 1 r μ B whenever B, B are overlapping balls of radius r.
A measure is doubling if and only if it is translate doubling and the underlying space has finite Assouad dimension equivalently, all balls can be covered by a bounded number of balls of half the radius .Thus, doubling and translate doubling are quantitatively equivalent in Euclidean space and examples of length spaces with translate doubling measures that fail to be doubling including Hausdorff n-measure on hyperbolic n-space and arclength measure on the Cayley graph of an n-generator free-group.
The lower Minkowski dimension dim M E of a subset E of a metric space X, d is defined by where N E, is the maximum cardinality of a collection of disjoint open balls of radius and centers in E.

The Spherical Boundary
A Borel function g : 0, ∞ → 0, ∞ is said to be a C-sphericalizing function, C > 2, if it has the following properties: S1 g r ≤ Cg s whenever r, s ≥ 0, r ≤ 2s 1, and s ≤ 2r 1; S2 ∞ r g t dt ≤ Crg r , r ≥ 1.We recall the following property of a sphericalizing function, taken from 1 .
In particular, tg t → 0 as t → ∞.Suppose X, l, o is an unbounded pointed local length space, and let us write |x| l x, o , x ∈ X l .Given a sphericalizing function g : 0, ∞ → 0, ∞ , we define a new metric S l, o, g on X by the equation: We usually write σ in place of S l, o, g .If γ ∈ Γ l x, y , x ∈ X, y ∈ ∂X l , then it is clear that γ is also of finite σ-length, so the length boundary ∂ 0 X l can be viewed as a subset of ∂X σ .We define the spherical boundary of X, ∂ S X to be ∂X σ \ ∂ 0 X l , and the spherical closure of X to be X σ .Since any point in x ∈ X ∪ ∂ 0 X l is at a finite l-distance from o, and since g is bounded away from zero on bounded intervals, it follows that a sequence in X is σ-convergent to a point in X ∪ ∂ 0 X l if and only if it is l-convergent.It also follows that if x ∈ X, y ∈ ∂ S X, and γ ∈ Γ σ x, y , then γ cannot be contained in any ball B l o, r .
We record some useful elementary estimates involving the two metrics l and σ S l, o, g .Below, G t ≡ 1 t g t , |x| ≡ l x, o , δ ∞ x σ x, ∂ S X , and C g is the sphericalization constant of g.It follows by standard analysis as in 1, Proposition 2.14 that δ ∞ x ≥ ∞ |x| g t dt, for all x ∈ X.Using S2 and S3 , we readily deduce that Suppose γ ∈ Γ l o, x for some x ∈ X, with len l γ L < |x| 1, and let ν γ| s,L , where γ s v, |v| ≥ 1.Then |γ t | ∈ t − 1, t , 0 ≤ t ≤ L, and so using S1 and S2 , we deduce that 2.4

Spherical Boundaries and Annular Convexity
A metric space X, d is said to be C-annular convex for some C ≥ 2 if for every o ∈ X, r > 0, and every pair of points Annular convex spaces, introduced by Herron et al. in 3 , form a large class of spaces that include all Banach spaces with the exception of one-dimensional real Banach spaces and most spaces equipped with a doubling measure that supports a Poincaré inequality as follows from the results in 4 .
In this section, we show that finite and countable l 2 products of metric spaces are annular convex.This is of interest to us because of the following simple result.Proposition 3.1.The spherical boundary of an unbounded proper annular convex pointed length space X, l, o is a singleton set.
Proof.Since X is proper, it follows from 5, Theorem 2.4 that ∂ S X is nonempty.We write |a| l a, o , a ∈ X, and let σ S l, o, g , where g is a given C g -sphericalizing function.
Fixing a fix a pair of points z, w ∈ ∂ S X, we pick sequences z n and w n in X converging to z and w, respectively.Since necessarily Joining z n to o by a path of length at most |z n | 1, and picking a point z n on this path with the property that |z n | n, it follows from 2.4 that σ z n , z n ≤ C 3 g ng n .Using Lemma 2.1, we deduce that σ z n , z n → 0 as n → ∞.We similarly find points w n such that |w n | n and σ w n , w n → 0 as n → ∞.
But by annular convexity and the properties of sphericalizing functions, it is easy to see that σ z n , w n ≤ C ng n , where C C C, C g and so σ z n , w n → 0. Thus σ z n , w n → 0 and so z w, as required.
Note that the assumption that X is proper in Proposition 3.1 was needed only to show that the spherical boundary is nonempty.Some such condition is needed to deduce this fact: for instance if X is a bouquet of line segments of length n for each n ∈ N, joined together by identifying with each other the left endpoints of all such intervals, then it is easy to show that ∂ S X is empty.
It is clear from the proof of Proposition 3.1 that C-annular convexity can be replaced by the following formally weaker condition: a metric space X, d is weakly C-annular convex, where C ≥ 1, if for every o ∈ X, r > 0, and every pair of points x 1 , x 2 ∈ X such that d o, x i r, i 1, 2, there exists a path γ from x 1 to x 2 in X \ B o, r/C of length at most 2Cr.However, replacing annular convexity by weak annular convexity is of no real benefit in Proposition 3.1, since for length spaces the two conditions are quantitatively equivalent.We record the simple argument for completeness., where > 0 is so small that d x 1 , x 2 < 2d x 1 , x 2 .Clearly λ remains inside B o, 3Cr , so it certainly verifies the 3C -annular convexity condition if it remains outside B o, r/3 .Assume therefore that λ ventures inside this ball.Let z 1 , z 2 be the first and last points z on λ such that d z, o r/3, let γ 1 be the initial segment of λ from x 1 to z 1 , let γ 2 be the final segment of λ from z 2 to x 2 , and let γ 3 be a path given by weak annular convexity for the pair z 1 , z 2 and center point o.Let γ be the concatenation of γ 1 , γ 3 , and γ 2 .
Then len d γ 1 len d γ 2 ≤ d x 1 , x 2 and weak annular convexity gives len when > 0 is sufficiently small.By construction, γ remains outside B o, r/3C and, since λ ⊂ B o, 3Cr , it suffices to verify that γ 3 also lies in this ball.But len γ 3 ≤ 2Cr/3, and the endpoints of γ 3 are a distance r/3 from o, so It turns out that product spaces are annular convex.Rather than proving Proposition 3.3, we prove a much more general result that is modeled on the previously mentioned fact that Banach spaces of dimension at least 2 are annular convex.We will generalize this to a large class of what could roughly be termed ISRN Geometry normed spaces with values in unbounded length spaces which can vary from point to point .More precisely, we look at metrics constructed in the following manner.
We begin with a real normed vector space V, • which we assume to consist of functions defined on an index set I. We say that the norm Assume now that X i , d i , i ∈ I are metric spaces and let P : i∈I X i .We define as follows the metric subproduct X, d of X i , d i , i ∈ I, relative to some fixed a ∈ P and the norm • , which is monotonic on I: if x x i i∈I ∈ P , then x ∈ X whenever f i : d i a i , x i defines a function f ∈ V , and we define d x, y i → d i x i , y i , which makes sense by monotonicity of the norm.We write X, d V ;a i∈I X i , d i .
Theorem 3.5.Suppose V is a normed vector space of dimension at least 2 which is monotonic on I, and suppose X, d V ;a i∈I X i , d i , where each X i , d i is an unbounded length space.Then X, d is 4-annular convex.
Before proving Theorem 3.5, we discuss metric subproducts X and give some examples.If x, x ∈ X, it is clear that d x, x ≤ d x, a d a, x , so certainly d x, x < ∞.It is routine to verify that d is a metric on X.Although X is defined with respect to some a ∈ P , it is clear that we get the same metric space X, d if a is replaced by any b ∈ X.However, we get a metric subproduct disjoint from our original X if we replace a by any b ∈ P \ X. Lemma 3.6.Suppose V is a normed vector space of dimension at least 2 which is monotonic on I, and suppose X i , d i is a length space for all i ∈ I. Then X, d V ;a i∈I X i , d i is also a length space.
Proof.Suppose x, x ∈ X and let > 0 be fixed but arbitrary.We define a path λ : 0, 1 → X such that λ t λ i t i∈I , 0 ≤ t ≤ 1, with the following important properties which we record for later reference: In fact this is quite easy to do: since X i is a length space, we can certainly pick λ i satisfying 3.3 .Then for all i ∈ I and 0 ≤ s ≤ t ≤ 1. Assembling together these paths λ i to get a path λ : 0, 1 → X, 3.4 follows from the last estimate and the fact that d is defined via a monotone norm.In particular, 3.4 implies that len d λ ≤ 1 d x, x .Since x, x ∈ X and > 0 are all arbitrary, the result follows.
Suppose we fix o ∈ X, where o o i i∈I .For every i ∈ I and R > 0 we can find a point v .We use function notation f i for V versus subscript notation x i for X to emphasize the difference between the normed space V, • and the subproduct X, d .
The simplest examples of monotonic normed spaces are l p spaces associated with finite or countably infinite I, for 1 ≤ p ≤ ∞.In the case of finite I, the subproduct X coincides as a set with the full Cartesian product i∈I X i .In the special case where I has cardinality 2 and p 2, we deduce Proposition 3.3.
Beyond the above l p spaces, other examples of monotonic-normed spaces include l p sums over uncountable index sets, but more interesting examples are normed sequence spaces of Orlicz or variable exponent l p • type.
Note that if I N and each X i is the real line, then the subproduct is merely the normed space V translated by a sequence a ∈ R N : thus these subproducts are all cosets of V , and any two such subproducts for different choices of a can be put into a natural 1-1 correspondence.
However, there is not always such a natural 1-1 correspondence.Consider for instance the case where X i is the metric subspace of the real line given by X i {0}∪ n∈N {n, 1/n} for each i ∈ N and V l 2 .If a 0 i∈N , then it is readily verified that the metric subproduct X : V ;a n∈N X i has the cardinality of the continuum, whereas if a c i∈N for any fixed c ∈ X i \ {0}, then X is a countable space.
It can be shown that this dependence of the cardinality of X on our choice of a does not occur when the spaces X i , d i , i ∈ N are length spaces essentially because its cardinality is at least that of the continuum if V is nontrivial .However, the above example suggests that there is in general no natural map from one metric subproduct to another.
Example 3.7.The constant 4 cannot be improved in Theorem 3.5.For instance if X is the closed first quadrant of the l 1 -plane; this choice of X corresponds to taking V to be the l 1 plane, with X 1 X 2 being the Euclidean half line 0, ∞ .Let r 4, x 0, 0 , y 2, 0 , and o 1 δ, 1 δ where 0 < δ < 1.Then every path from x to y intersects B o, cr for any c > 1/4, as long as δ < 4c − 1.
We now move on to the proof of Theorem 3.5.
Proof of Theorem 3.5.Let x, y ∈ B o, r \ B o, r/2 be the pair of points for which we want to verify the 4-annular convexity condition with other data r, o as usual , and let S : d x, o d y, o , so that S > r.As in the proof of Proposition 3.2, a path γ connecting x and y of length at most 1 d x, y , where > 0, verifies the 4-annular convexity condition for data x, y, o, r unless γ intersects B o, r/4 .We may therefore assume that this intersection occurs and so

3.6
Taking a limit as → 0 we get d x, y ≥ S − r/2.In particular, d x, y > S/2.Let us write e A ∈ {0, 1} I for the characteristic function of any A ⊂ I: thus e A i 1 if and only if i ∈ A. For x ∈ X and A ⊂ I, let x A ∈ X be defined by For convenience, we write e i : e {i} and x i : x {i} for any i ∈ I. Since V has dimension at least 2, monotonicity readily implies that there are distinct indices j, k ∈ I such that the basic functions e j , e k lie in V .

ISRN Geometry
We now define "scalar multiplication" on X, restricted to scalar values 0 ≤ α ≤ 1. Choosing λ : 0, 1 → X to be as in the proof of Lemma 3.6, with x, x , x, o, and 0 < ≤ 1/3, we let α • x : λ s , where s is the minimal s ∈ 0, 1 such that d λ s , o αd x, o .This definition is typically not unique since λ is not unique, but note that if Suppose first that we can find some such as A such that d x A , o ≥ r/4 and d y B , o ≥ r/4, where B : I \ A. Let z x : α • x A and z y : β • y B for α r/4d x A , o and β r/4d y B , o , so that d z x , o d z y , o r/4.Let λ i , i ∈ A, and s be the associated coordinate paths and argument for α • x A , as in the last paragraph.Now join x and y by a path γ defined in the following piecewise manner by concatenating, in the natural order, paths γ m : 0, 1 → X, 1 ≤ m ≤ 4. First γ 1 is a path from x to z x which has component paths γ for i ∈ B. Since • is a monotonic norm, we deduce that len d γ 2 ≤ 1 r/4.Finally γ 3 and γ 4 are analogues of γ 2 and γ 1 , respectively, but with x, A, B replaced by y, B, A , and with the directions of the paths reversed.It follows that the d-length of our concatenated path γ is at most 1 S r/2 ≤ 3 1 S/2.Since d x, y ≥ S/2 and ≤ 1/3, we deduce that len γ ≤ 4d x, y , as required.
We next need to show that γ m , 1 ≤ m ≤ 4, stays outside B o, r/4 .It follows from the definition of α • x that d λ 1 t , o ≥ r/4 for all 0 ≤ t ≤ 1, the same estimate follows for γ 2 by monotonicity of the norm and symmetry with γ 2 and γ 1 then gives the same estimates for γ 3 and γ 4 , respectively.
Finally we need to show that each γ m is contained in B o, 4r .The triangle inequality ensures that d γ 1 t , o < r /2 r < 4r and d γ 2 t , o < r/4 1 r/4 < 4r for all 0 ≤ t ≤ 1.The same estimates for γ 3 and γ 4 follow by symmetry.
We may therefore make the added assumption that there is no way to split I into complementary subsets A and B such that In particular, one of these last pairs of conditions holds for a set A that contains j but not k.By switching the definitions of A, j and B, k if necessary, we assume that d w w k .We now join x and y by a path γ defined in the following piecewise manner by concatenating, in the natural order, paths γ m : 0, 1 → X, 1 ≤ m ≤ 5, where γ 1 is a path from x to z x defined as in the previous case, γ 2 is a path of length at most 1 r/4 from z x to w x which is stationary except in coordinate k, γ 3 is a path of length at most 1 r/2 from w x to w y which is stationary in coordinate k, γ 4 is a path of length at most 1 r/4 from w y to z y which is stationary except in coordinate k, and γ 5 is analogous to γ 1 in reverse, but from z y to y.As in the previous case, we see that len γ ≤ 1 S r.Since d x, y > S/2 > r/2, by taking to be sufficiently small we get len γ ≤ 4d x, y , as required.The fact that γ ⊂ B o, 4r \ B o, r/4 can be verified as before, so we leave it to the reader.
The examples that we have so far include the cases where V is an l p or related space, but we cannot handle general L p spaces because the requirement that the norm is monotonic restricts us to spaces V where nonnegative functions that are pointwise less than a given function in V must also lie in V .This is incompatible with spaces of measurable functions unless the sigma algebra is the power set , let alone spaces of continuous or smooth functions.To get similar results for such spaces, the basic problem is getting fine control over the relationship between d i λ i t , o i for different values of i ∈ I and fixed 0 < t < 1, where λ λ i is as in Proposition 3.2.One way to get such control is to assume that each X i , d i is a geodesic space, so that we can assume that λ i : 0, 1 → X i is a constant speed geodesic.Then λ λ i is also a constant speed geodesic and d λ t , o 1 − t d x, o , allowing us to get analogues of Theorem 3.5 for more general spaces.The assumptions of monotonicity and dimension at least 2 would need to be replaced by assumptions appropriate to the context.

Large Spherical Boundary and Fast Volume Growth
A metric measure space, even a proper one, can have very fast volume growth and small spherical boundary, in the sense that its spherical boundary is a singleton set.For instance the product Riemannian manifold X H 2 × H 2 has exponential volume growth and constant negative Ricci curvature, but Corollary 3.4 implies that ∂ S X is a one-point space.
However, implications in the reverse direction are possible.Our Guiding Principle is that for reasonably general classes of pointed length measure spaces X, l, o, μ , a large spherical boundary forces X, l to have rapid volume growth.By making appropriate choices for the vague italicized phrases in our Guiding Principle, we get some theorems.We state and prove three such results in this section and discuss some relevant examples.In all instances, the reasonably large class of spaces consists of spaces satisfying a doubling condition or some weak variant thereof.
Throughout this section, X, l, o is a pointed length space, g is a C g -sphericalizing function, with associated spherical metric σ S l, o, g and spherical boundary ∂ S X.Also G t 1 t g t , |x| l x, o , and δ ∞ x dist σ x, ∂ S X .In our first result, we assume that our metric measure space is doubling.This is a rather strong condition and it implies slow meaning polynomial rate volume growth so, without any explicit mention of volume growth, we deduce that the spherical boundary is quite small in the sense of having finite cardinality.Theorem 4.1.Suppose X, l, μ is C μ -doubling.Then ∂ S X is a finite set whose cardinality is bounded by a number dependent only on C g and C μ .
In our other results, we replace doubling by translate doubling or weak translate doubling.Unlike doubling, weak translate doubling puts no real constraint on volume growth, so volume growth enters the statements of our results explicitly.Theorem 4.2.Suppose X, l, μ is C μ -translate doubling.If dim M ∂ S X > 0, then f r ≡ μ B l o, r grows faster than any polynomial.In fact, , r grows at a polynomial rate or faster, that is, Proof of Theorem 4.2.Part of the proof is similar to that of Theorem 4.1, so we will be sketchy.Since dim M ∂ S X > 0, there are constants c, Q > 0 such that ∂ S X contains c −Q disjoint σ-balls B ,i of radius for all 0 < ≤ dia σ X .Taking j A −j dia σ X for a fixed number A > 1, we can associate radii t j such that each of the c −Q j balls B j ,i contains an l-ball B j,i whose radius is t j and whose distance from the origin is contained in the interval r j , R j for some numbers r j , R j are comparable with t j ; the constants of comparability can be taken to depend only C g .We assume, as we may, that A is chosen so large that c −Q j ≥ 2 j , r 1 ≥ 1, and R j 2t j < r j 1 , j ∈ N. Note also that the ratios r j 1 /r j are uniformly bounded by a constant dependent only on A and C g , so that log r j ≈ j.
Translate doubling ensures that the balls B j,i are of comparable measure with B j ≡ B l o, r j , so there exists a constant C > 0 such that f r j 1 ≥ 2 j f r j /C for each j ∈ N. Iterating this, we get that f r j ≥ 2 j j−1 /2 f r 1 /C j .Since log r j ≈ j, the result follows.
We omit the proof of Theorem 4.3 as it is so similar to that of Theorem 4.2.In fact it differs from it only in the last paragraph above, and the required modifications are straightforward.
We now consider some examples.All of our examples are either n-dimensional Riemannian manifolds or one-point joins of a finite number of n-dimensional Riemannian manifolds meaning that the distinguished points o in these manifolds are all identified with each other .In all these cases, the associate measure is the usual measure on a Riemannian manifold or equivalently Hausdorff n-measure .
It is easy to give examples relevant to Theorem 4.1.Euclidean space R n has spherical boundary of cardinality 2 for n 1, and 1 for n > 1: the n 1 case follows easily from the definition, while the n > 1 case follows for instance from Corollary 3.4.The one-point join at 0 of k ∈ N copies of the half-line 0, ∞ is a doubling space whose spherical boundary has cardinality k.
We do not know whether or not there exists a space X, l, μ that satisfies the assumptions of Theorem 4.2 and has sharp volume growth rate lim inf However, hyperbolic space H n , l, μ is an example of a translate doubling space with much faster volume growth whose spherical boundary has infinite Minkowski dimension, as follows from the following more precise result.
Proposition 4.4.Let σ S l, o, g be the sphericalized metric on H n for the standard sphericalizing function g t 1 t −2 .The minimum number N r of σ-balls of radius r > 0 required to cover ∂ S H n satisfies where C n depends only on n.For a general sphericalizing function g, N r grows faster than c g • exp n − 1 /r c g /C n , where c g , C n > 0 depend only on their subscripted parameters.
Proof.We assume that n 2: this does not change anything essential in the proof but it simplifies the notation.It suits us to think of H 2 as the warped product B × f F, where B 0, ∞ , F S 1 , and the warping function is f t sinh t.We identify ∂ S H 2 with F as a set and view F as the set of points in the complex plane of the form exp iθ , θ ∈ R. For the moment, assume that g t 1 t −2 .Due to the symmetry of H 2 , to get the lower bound on N r , it suffices to show that σ exp iθ , 1 ≥ 2r, whenever θ ≈ exp −1/r /r, is sufficiently small.
Suppose we join exp iθ with 1 via a path γ whose B-coordinate achieves a minimum value s ≥ 0. Considering only the horizontal component of arclength, we deduce from that len σ γ ≥ 2 ∞ s g t dt 2/ 1 s .Considering only the vertical component of arclength, we have len σ γ ≥ H s : θg s sinh s.Thus σ exp iθ , 0 ≥ m, where m is the minimum over all s ≥ 0 of M s : 2/ 1 s ∨H s .Since we may take θ to be less than θ 0 for any θ 0 > 0 of our choice, we may assume that the minimum of M s occurs when s > 1.But then m equals the minimum over all s ≥ 1 of 2 1 s −1 ∨ θ 1 s −2 sinh s , which occurs when 1 s θ sinh s /2.Taking r 1/ 1 s in this last equation gives θ 2 sinh 1 − 1/r /r, as required.
To obtain an upper bound on N r , it suffices to consider the path γ consisting of a horizontal segment from exp iθ to the point with first coordinate s where s θ sinh s , then the shorter vertical segment to the point with second coordinate 1, and finally a horizontal segment to 1 ∈ F ∂ S H 2 .Then σ exp iθ , 1 ≤ len σ γ 2/ 1 s H s , where H s is as above.Taking s as above gives the required upper bound for N r .
For a general sphericalizing function g, we obtain as above that len σ γ G s ∨ θg s sinh s , 4.6 where G s 1 s g s .This lower bound is minimal when 1 s θ sinh s.Using the fact that G s decays at a polynomial rate as s → ∞, it is a routine matter to obtain the desired conclusion.
Finally we show that Theorem 4.3 is sharp by considering the warped product X, l , where X B × f F, B 0, ∞ , F S 1 , the warping function is f t t 2 , and the sphericalizing function is g t 1 t −2 .We take o ∈ X to be the unique point with first coordinate 0. Proposition 4.5.If X, l, o is as above then ∂ S X, σ is bilipschitz equivalent to the arclength metric on S 1 .Moreover, X, l, μ is weak translate doubling, where μ is Hausdorff 2-measure.
Proof.As in the proof of Proposition 4.4, a lower bound on σ exp iθ , 0 is given by the minimum m over all s ≥ 0 of 2/ 1 s ∨ H s , where H s θf s g s .Taking θ > 0 to be small, we may assume that the minimum occurs when s > 1.But then m is comparable with the minimum over all s ≥ 1 of s −1 ∨ θ, which occurs when s 1/θ and equals θ.Thus σ exp iθ , 0 θ.On the other hand, as in Proposition 4.4, we see that The fact that X is weak translate doubling follows from the fact that there exists constants c, C > 0 such that c r 2 ∧ r 3 ≤ μ B r ≤ C r 2 ∨ r 3 whenever B r is a ball of radius r > 0. We leave this as an exercise to the reader.
Note that the space X in Proposition 4.5 and Euclidean 3-space have the same volume growth rate, but ∂ S X is topologically S 1 at least for g decaying no faster than the standard sphericalizing function whereas ∂ S R 3 is a one-point space.This again emphasizes that although the size of the spherical boundary constrains volume growth for a large class of spaces , volume growth does not determine the size of the spherical boundary.We have also seen that restrictions such as negative Ricci curvature in the case of Riemannian manifolds is also not sufficient to ensure a nontrivial boundary.We would need more detailed curvature conditions, like an upper bound on the decay rate of Alexandrov curvature, in order to obtain results in that direction.

Proposition 3 . 3 . 3 . 3 . 4 .
Let X, d be the Cartesian product of unbounded length spaces X 1 , d 1 and X 2 , d 2 , with d being the l 2 product of d 1 and d 2 .Then X, d is a 4-annular convex length space.We get the following immediate corollary of Propositions 3.1 and 3.Corollary Let X, d be the Cartesian product of unbounded length spaces X 1 , d 1 and X 2 , d 2 , with d being the l 2 product of d 1 and d 2 .Then the spherical boundary of X, d is a singleton set.
d x A , o ≥ r/4 and d y B , o ≥ r/4.Note though that for any w ∈ X and A ⊂ I, we either have d w A , o ≥ d w, o − r/4 > r/4 or d w I\A , o ≥ r/4.By our added assumption, it follows that for every A ⊂ I and B : I \A, either d w A , o ≥ d w, o − r/4 and d w B , o < r/4 both hold for w ∈ {x, y} or d w B , o ≥ d w, o − r/4 and d w A , o < r/4 both hold for w ∈ {x, y}.

Proposition 3.2. If
a length space is weak C-annular convex, then it is 3C -annular convex.Proof.Consider distinct points x 1 , x 2 ∈ B o, r \ B o, r/2 .Let d i : d o, x i , i 1, 2. Join x 1 , x 2 by a path λ of length less than d x 1 , x 2 X i such that d i o i , z ≥ R, and then we use continuity to pick the required z i R on a path from z to o i .It follows that if v ∈ V , we can find a point x v the unboundedness of X i ensures that there ISRN Geometry 7 exists z ∈ i | and so d x v , o where γ 1 i is a rescaled copy of λ i | 0,s if i ∈ A and λ λ i , s are as above , while γ 1 i is a constant speed path from x i to o i of length at most 1d i x i , o i if i ∈ B.Thus, each γ i is a constant speed path of length at most 1 d i x i , o i , and so len d γ 1 ≤ 1 Let γ 2 : 0, 1 → X be any path z x to z such that the coordinate paths γ 2 i are stationary paths for i ∈ A and are of d i -length at most 1 A, o ≥ d w, o − r/4 and d w B , o < r/4 both hold for w ∈ {x, y}, and that j ∈ A. We choose w ∈ X such that w i o i for i / k and d w, o r/4.As in the previous case, we let z x : α • x A for α r/4d x A , o , but now we choose z y : β • y A for β r/4d y A , o .
3g so small that the balls 4B i ≡ B σ z i , 4 are all disjoint and choose points u i ∈ B i .Suppose δ ∞ z ≤ 3 .Using 2.3 , we get It now follows from S1 and the definition of G that |z| ≥ 2.We carry out the following construction for each index i,1 ≤ i ≤ N. Choose γ i ∈ Γ l o, u i with len l γ L < |u i | 1, and let B i : B σ v i ,, where v i is the first point at which γ i meets 3B i .Now d ∞ v i ≤ 3 , and so |v i | ≥ 2. Using 2.3 and 2.4 , we see that G |v i | ≈ .In view of S3 , we see that the distances |v i | are mutually comparable, so let us choose a pair of mutually comparable radii r, R such that 2 ≤ r ≤ |v i | ≤ R. By S1 , B l v i , t ⊂ B σ v i , C g tg t for every 0 < t ≤ r.We can therefore fix t ≤ r, t ≈ r, so that B i :B l v i , t ⊂ B i .Every B i is contained in the single ball B 0 B l o, R t and in turn B 0 is contained in each of the balls sB i , s 2R t /t.Since t and R are comparable, doubling ensures that μ B 0 ≤ C 1 μ B i , where C 1 depends only on C μ and 2R t /t 1.Since B 0 contains N disjoint balls of measure at least μ B 0 /C 1 , it follows that N ≤ C 1 , as required.