Isoperimetric Regions in $\mathbb{R}^n$ with density $r^p$

We show that the unique isoperimetric hypersurfaces in $\mathbb{R}^n$ with density $r^p$ for $n \ge 3$ and $p>0$ are spheres that pass through the origin.


INTRODUCTION
Recently, there has been a surge of interest in manifolds with density, partly because of their role in Perelman's proof of the Poincaré Conjecture. We consider the isoperimetric problem when volume and perimeter are weighted by the density function r p and prove the following theorem: Theorem 3.3. In R n with density r p , where n ≥ 3 and p > 0, spheres that pass through the origin are uniquely isoperimetric with respect to weighted perimeter and volume.
The density r p is one of the simplest radial density functions, but it has some interesting properties. First, r p is homogeneous in degree p, which means that given an isoperimetric region of one volume, we can scale it to get an isoperimetric region of a different volume. Second, r p (or a constant multiple) is the only density for which spheres through the origin could be isoperimetric (see e.g. Rmk. 4.4). We can view our present problem as a venture either to prove a partial converse of this statement in the case that p > 0 or to extend the work of Dahlberg et al., who proved the result in R 2 [DDNT,Thm. 3.16]. Díaz et al. [DHHT,Conj. 7.6] conjectured the generalization to R n and reduced the problem to analyzing planar curves. Recently, Chambers [C,Thm. 1.1] proved that spheres centered at the origin are isoperimetric in R n with any log-convex density.
We adapt Chambers' proof to density r p . Like Chambers, we first consider an isoperimetric region that is spherically symmetric (see Defn. 2.6), then prove the result in the general case. Given a spherically symmetric isoperimetric region, we prove that the generating curve for the boundary is a circle through the origin. The behavior of this curve is determined by a differential equation corresponding to the fact that isoperimetric hypersurfaces have constant generalized mean curvature [MP,Defn. 2.3]. By spherical symmetry and regularity, the rightmost point of the curve is on the e 1 -axis, and the tangent vector at this point is vertical. Our Lemmas 4.5 and 4.7 show that if the osculating circle at the rightmost point of the curve, which we may assume to be (1,0), goes through the origin, then the curve is a circle through the origin.
We suppose for contradiction that the initial osculating circle does not pass through the origin, then  take two cases according to whether its center is right or left of (1/2, 0). In the case that the center is right of (1/2, 0), the curve is like that in Chambers' proof in that the curvature is greater at a point above the e 1 -axis with tangent vector in the third quadrant than at the point of the same height with tangent vector in the second quadrant. As a result, the curve has a vertical tangent before it meets the e 1 -axis again and then curves in to meet the axis at an angle (Fig. 1,right). In the left case, the opposite inequality regarding curvatures holds, and, as a result, the curve never returns to vertical before reaching the axis ( Fig. 1, left). The left case presents the new challenge of showing that there is only one point on the upper half of the curve where the tangent vector is horizontal (Prop. 7.21). Additionally, although the curve in the right case is similarly to that in Chambers, the proof is different in that we do not have the hypothesis that an isoperimetric hypersurface is mean convex, which is what Chambers used to prove that curvature was positive on the final segment of the curve. We achieve the same result by computations that depend on the fact that our curve ends right of the e 2 -axis (Lemma 6.12), which is a property that may not hold for the generating curve in Chambers. Proof. Since E is an isoperimetric minimizer and the oriented tangent cone at P lies in a halfspace, the oriented tangent cone is a hyperplane. The result follows by [M,Prop. 3.5,Rmk. 3.10].
Corollary 2.5. All points in ∂ E of maximal distance from the origin are regular.
Definition 2.6. (Spherical Symmetrization) Given a region E ⊂ R n , let A E (r) denote the area of the intersection of E with S r , the sphere through the origin of radius r. We define the spherical symmetrization of E to be the unique set E * such that for all r ≥ 0, A E (r) = A E * (r) and E * ∩ S r is a closed spherical cap centered at (r, 0, ..., 0).
Remark 2.7. Since the set of singularities on the boundary of an isoperimetric set E ⊂ R n has dimension at most (n − 8), it follows that if E is spherically symmetric about the e 1 -axis, then all points in ∂ E that are not on the e 1 -axis are regular.
The following theorem demonstrates that for a radial density, spherical symmetrization preserves weighted volume but does not increase weighted perimeter. Moreover there are certain conditions under which the perimeter of a region remains the same after symmetrization only if the original region was spherically symmetric about some (oriented) line through the origin.
Theorem 2.8. [MP,Thm. 6.2] Let f be a radial density on R n and let E be a set of finite perimeter. Then the spherical symmetrization E * satisfies |E * | = |E| and P(E * ) ≤ P(E). Suppose further that E is an open set of finite perimeter, and let ν(x) denote the normal vector at any x ∈ It is immediate that if E is an isoperimetric region in Euclidean space with a radial density, then E * is also isoperimetric.

SPHERES THROUGH THE ORIGIN ARE UNIQUELY MINIMIZING
To prove our main result Theorem 3.3, we begin by showing that that any spherically symmetric isoperimetric region is a ball whose boundary is a sphere through the origin (Prop. 3.1). The proof of Proposition 3.1 comprises most of the paper, but we provide a sketch below. We apply this proposition to the symmetrized version of an arbitrary isoperimetric region to show that, in fact, any isoperimetric region is spherically symmetric about some oriented line through the origin (Prop. 3.2).
Proposition 3.1. Suppose that E ⊂ R n is a spherically symmetric isoperimetric region in R n with density r p . Then E is a ball whose boundary goes through the origin.
Proof. Assume without loss of generality that E is spherically symmetric about the positive e 1 -axis. Then E can be generated by rotating a planar set A about the e 1 -axis. By regularity of E (Defn. 2.3), we are assuming that A is open and that its boundary is a curve (possibly having multiple connected components). Since E is spherically symmetric about the positive e 1 -axis, A is also spherically symmetric about the positive e 1 -axis. We define γ ⊂ ∂ A by beginning at the rightmost point on ∂ A and following the curve through this point in both directions until it intersects the e 1 -axis again. This definition relies on regularity properties of ∂ E; see the beginning of Section 4 for more details.
We assume that γ : [−β , β ] is an arclength parameterization so that γ(0) is the rightmost point on ∂ A and γ(±β ) is the other intersection of γ with the e 1 -axis. Since r p is homogeneous, all isoperimetric regions are similar, and we can assume without loss of generality that γ(0) = (1, 0). We will show that γ is a circle through the origin. Given that γ is a circle through the origin, γ must comprise all of ∂ A by spherical symmetrization. By Lemma 4.5, to prove that γ is a circle through the origin, it suffices to prove that there exists an s so that the associated canonical circle C s (see Defn. 4.2) has the same curvature as γ at γ(s) and C s goes through the origin. By Lemma 4.7 the canonical circle C 0 at the rightmost point has the same curvature as γ at γ(0). Therefore, it suffices to prove that C 0 passes through the origin, which occurs if and only if the center of C 0 is (1/2, 0).
The only remaining possibility is that γ is a circle through the origin. Thus, γ = ∂ A and, when rotated, γ generates a sphere through the origin.
Given Proposition 3.1, we can prove our claim that any isoperimetric region in R n with density r p is spherically symmetric.
Proposition 3.2. If E is an isoperimetric region in R n with density r p , then E = E * , up to a rotation about the origin.
Proof. By regularity (Defn. 2.3), we are assuming E is open. By Theorem 2.8, it suffices to show that I E is an interval and that We call a point x with ν(x) = ± x/|x| tangential. Since symmetrization (Defn. 2.6) preserves weighted volume without increasing weighted perimeter, E * is also isoperimetric. Applying Proposition 3.1, we conclude that E * is a ball with boundary through the origin. It follows that I E is an interval. Moreover, there exists no r ∈ I E such that the spherical cap S r ∩ E is a full sphere. This will be important in our proof of (1). Suppose for contradiction that there exists a positive area subset of ∂ E that is tangential. As in Morgan-Pratelli [MP,Pf. of Cor. 6.4], at any smooth point of density of this tangential subset of ∂ E, ∂ E has the same generalized mean curvature as a sphere centered at the origin. It follows by uniqueness of solutions to elliptic partial differential equations that a component of ∂ E is a sphere centered at the origin. E must contain an annular region centered at the origin with this spherical component as one of its bounding components. Thus, there exists an interval (r 0 , r 1 ) such that for any r in (r 0 , r 1 ), S r ∩ E is a full sphere, contradicting the fact that the boundary of E * is a sphere through the origin.
Combining Propositions 3.1 and 3.2 along with Theorem 2.2 we have proved: Theorem 3.3. In R n with density r p , where n ≥ 3 and p > 0, spheres that pass through the origin are uniquely isoperimetric with respect to weighted perimeter and volume.

STRUCTURE OF PROOF
Sections 5, 6, and 7 are devoted to filling in the details of the proof of Proposition 3.1. Throughout these sections, we work within the following framework: Let E be a spherically symmetric isoperimetric region. Then there is a set A ⊂ R 2 such that E is the rotation of A about the e 1 -axis. We will analyze a certain curve on the boundary of A. We begin at the point P on the e 1 -axis that is the rightmost point on ∂ A. By spherical symmetry, P is a point of E farthest from the origin, so ∂ E is regular at P by Corollary 2.5. The tangent space to ∂ A at P is spanned by e 2 . We follow ∂ A, which has finite length, in both directions until it intersects the e 1 -axis at another point. The result is a Jordan curve γ(s) : [−β , β ] → R 2 such that γ(0) = P and γ(±β ) is the other intersection of the curve with the e 1 -axis (Fig. 2). Since r p is homogeneous, all isoperimetric regions are similar to each other. Therefore, we may assume without loss of generality that P = (1, 0). We assume that γ is a counterclockwise arclength parameterization. Let γ 1 and γ 2 denote the coordinates of γ. Then γ 1 (−s) = γ 1 (s) and γ 2 (−s) = −γ 2 (s) for all s.
By Corollary 2.5, γ is smooth at 0. By Remark 2.7, γ is smooth at all remaining points in (−β , β ). Since γ is smooth at 0 and 0 is the global maximum point of γ 1 , γ (0) = (0, 1). We let κ(s) denote the curvature of γ at γ (s). Then κ(0) > 0; otherwise a contradiction to spherical symmetry would result. We will also consider the generalized mean curvature of the surface generated by ∂ A at a point γ (s). As in where H 0 is the unaveraged Riemannian mean curvature and ν is the outward unit normal vector. If ψ(x) = g(|x|) for some smooth function g and x = 0, we have In R n with density r p , g(r) = log (r p ). Henceforth, we will denote by H 1 (x). For concision, given a point γ(s), we refer to H 1 (γ(s)) as H 1 (s) with analogous notation for the values of H 0 and H f at γ(s).
The following lemma of Chambers gives a trivial but useful result of spherical symmetrization. At each point on γ, we define a related circle that we call the canonical circle. We show in Proposition 5.1 that the curvature of the canonical circle is one of two terms in a formula for the mean curvature of the surface of revolution.
Definition 4.2. [C,Defns. 3.1,3.2] For s ∈ (0, β ), let the canonical circle at s, denoted C s , be the unique oriented circle centered on the e 1 -axis that passes through γ(s) and has unit tangent vector at γ(s) equal to γ (s). If γ (s) is a multiple of e 2 , then C s is an oriented vertical line. We define C 0 to be lim s→0 + C s . The regularity of the surface at γ(0) guarantees the existence of this limit. We let R(s) denote the radius of C s and let λ (s) denote its signed curvature. Then λ (s) = 1/R(s) if C s is counterclockwise oriented, and λ (s) = −1/R(s) if C s is clockwise oriented. Finally, we let F(s) denote the abscissa of the center of C s .
The following lemma shows that spheres through the origin have constant generalized mean curvature. We apply this result to prove Lemmas 4.5 and 4.7, which imply that γ is a sphere through the origin, given that the curvature at the rightmost point is the same as the curvature of the circle through that point and the origin.
Proposition 4.3. In R n with density r p , hyperspheres through the origin have constant generalized mean curvature.
Proof. Let S be a hypersphere through the origin and assume without loss of generality that S can be obtained by rotating a circle C in the plane about the e 1 -axis. It suffices to prove that all points on C have the same generalized mean curvature. H 0 is constant on C since it is constant on S. It remains to prove that H 1 is constant on C.
Let the center of C be (a, 0) with a > 0. Then the polar coordinates equation for C is r = 2a cos θ . At a point (r(θ ), θ ), the unit outward normal vector makes angle 2θ to the positive e 1 -axis, and the angle between the position vector and the unit outward normal vector is θ . Supposing that x has polar coordinates (r, θ ), we have Therefore, H 1 is constant on C, as required.
Remark 4.4. These computations show that the only density on R 2 − {0} (R n − {0}) for which circles (spheres) through the origin are isoperimetric is r p , or a constant multiple thereof. On a circle C through the origin, parameterized by α, the quantity α(t)/|α(t)| · ν(t) is a constant multiple of the magnitude of the position vector. Hence, for H 1 to be constant it must be the case that g (r) is inversely proportional to r. This occurs only if g = log(r p ) + c for some p and some constant c. Proof. Supposing that C s is arclength parameterized, to prove that C s agrees with γ locally, it suffices by uniqueness theorems concerning solutions of ODEs to prove that both satisfy the differential equation H f = c. This is clearly true since the tangent vectors of the two curves agree at γ(s) and the generalized mean curvature of the surfaces generated by these curves is the same at γ(s). To prove that H f = c at all points on C s , it suffices to show that H f is constant on C s . This follows from the computations in Lemma 4.3. Having proved that γ and C s coincide locally, we claim that, in fact, γ and C s must coincide everywhere.
Since γ and C s agree near γ(s), S is nonempty and therefore has a least upper bound m. Letting α be an arclength parameterization of C s , it follows by smoothness of α and of γ that m ∈ S, that C s is tangent to γ at γ(m), and that κ(m) = λ (s). (To conclude smoothness of γ at m, we are using our assumption that m < β .) By an identical argument to that in the first paragraph, there exists an open interval I containing m such that γ(I) ⊂ C s , contradicting the fact that m = sup S. We conclude that m = β . A similar argument shows that γ coincides with C s on [−β , s].
Remark 4.6. By radial symmetry, spheres centered at the origin also have constant generalized mean curvature. Thus, if C s is centered at the origin and κ(s) = λ (s), then γ is a circle that is centered at the origin. We use this result to obtain contradictions in several places.
Since κ(0) > 0 and κ is continuous at 0, there is a neighborhood of 0 on which γ 1 (s) = 0 except when s = 0. By definition, By Lemmas 4.5 and 4.7, if C 0 is a circle through the origin, then γ is a circle through the origin. This means that if F(0) = 1/2, then γ is a circle through the origin. We will argue by contradiction, taking cases according to whether F(0) > 1/2 or F(0) < 1/2. We call these cases the right case and the left case, respectively. In each case, we can obtain a result that contradicts spherical symmetry. We state these results as the Right Tangent Lemma and the Left Tangent Lemma, and we will devote a section to proving each.

PRELIMINARY LEMMAS
This section contains results relevant to both cases. Proposition 5.1 and Corollary 5.2 give expressions for the mean curvature and generalized mean curvature at a point on the hypersurface generated by γ in terms of the curvature of γ, the curvature of the canonical circle, and the normal derivative of the log of the density at that point. We then discuss computational techniques that we use to determine how these functions (and others) vary with arclength. Finally, Proposition 5.4 is used in both cases to compare curvatures at pairs of points on the curve that are at the same height.
Proof. We consider the principal curvatures of the surface at a point P = γ(s). We treat the case that y = γ 2 (s) > 0 and that γ (s) = (0, ±1). A similar argument shows that the equation holds if γ 2 (s) < 0 and γ (s) = (0, ±1). We claim that there exists no interval on which γ 2 = 0 or γ is vertical; then it will follow by smoothness of γ that the equation holds at the remaining points.
To prove the claim, recall that γ is smooth at 0 and that, as a consequence of spherical symmetry, κ(0) > 0. Thus, γ 2 cannot be 0 on an interval including 0. On the other side of the curve, γ(β ) is defined to be the first point where the curve intersects the axis again, so even if a portion of the curve were a line segment along the e 1 -axis, that segment would not be parameterized by the function γ. The curve cannot have vertical tangent vector on an interval either. If a portion of the curve were a vertical line segment, then this vertical line segment, when rotated, would generate a portion of a hyperplane, which would have zero mean curvature. However, H 1 (the normal derivative of the log of the density) would vary as one moved up or down along the line segment, contradicting the fact that the surface has constant generalized mean curvature.
With this technical point out of the way, we proceed in the case that y = γ 2 (s) > 0 and that γ (s) = (0, ±1). One of the principal curvatures at P is the the curvature of γ at this point. The cross section of the surface obtained by fixing the first coordinate is an (n − 2)-dimensional sphere of revolution. The remaining principal curvatures of the surface are the principal curvatures of the sphere, which are equal. Thus, to compute one of the principal curvatures of the sphere, it is sufficient to compute the second principal curvature of a 2-dimensional surface in the n = 3 case. This second principal curvature is the curvature of a circle of revolution C.
By assumption that y = γ 2 (s) > 0, the curvature of the circle C is 1/y. We let n denote the inward normal vector to the surface and N denote the normal vector to the circle of revolution. Since y > 0, C s is counterclockwise oriented if and only if n is downward (i.e. n has a negative e 2 -component). Thus, Meanwhile, by Meusnier's formula, the second principal curvature is given by where φ is the angle between n and N. Again, since y > 0, . A cross section of the surface in the xy-plane and the unit inward normal vectors to the surface and to a circle of revolution The first of these cases is depicted in Figure 3. In both cases, the second principal curvature is the curvature of the canonical circle.
Since one principal curvature of the surface at γ(s) equals κ(s) and all of the others equal λ (s), the mean curvature of the surface at γ(s) is given by Corollary 5.2. The generalized mean curvature of the surface at a point γ(s) is given by In the left and right cases delineated on p. 7, for any s ∈ [0, β ) we can analyze how γ and related functions are instantaneously changing at γ(s) by computing the requisite derivatives on the osculating circle to γ at γ(s). A justification for this procedure follows.
For a given s in [0, β ), let A s denote the unique oriented circle that is tangent to γ at γ(s) and has curvature κ (s). Note that if κ(s) = 0, then A s is an oriented line with direction vector γ (s). For a fixed s, let α be a arclength parameterization of A s such that α(s) = γ(s). Since A s is tangent to γ at γ(s) and has curvature κ(s), we have α (s) = γ (s) and α (s) = γ (s).
We letκ(t) denote the signed curvature of α at t andH 1 (t) denote ∂ g/∂ ν at t, where ν is the unit outward normal to α at t. Bothκ andH 1 are smooth functions on their domains. Moreover, we consider the canonical circles to α and their related functions. For a fixed t we denote the canonical circle to A s at α(t) byC t . If α 2 (t) = 0, then we defineC t as before. If α 2 (t) = 0 and α (t) = (0, ±1), then we definẽ C t to be A s . If α 2 (t) = 0 and α (t) = (0, ±1), thenC t is undefined. For each t so thatC t is defined, the canonical circle is defined on a neighborhood of t. Finally, we define the functionsλ ,R, andF by lettingλ (t),R(t), andF(t) be the curvature ofC t , the radius ofC t , and the abscissa of the center ofC t , respectively.
We use these functions to approximate their counterparts on γ. Fix s in [0, β ), and let α be a arclength parameterization of A s such that α(s) = γ(s). For a given t, the canonical circleC t depends only on α(t) and on the inward normal at α(t), which is a rotation of α (t) by π/2 (with the direction depending on the orientation of α). It follows thatF,R, andλ can be computed as in terms of α and α and that their derivatives depend on α and its first two derivatives. In particular, since α (s) = γ (s) and α (s) = γ (s), we haveF (s) = F (s),R (s) = R (s), andλ (s) = λ (s).
As well as analyzing these functions, we also consider the angle the tangent vector makes to the horizontal.
Definition 5.3. We define the function θ : S 1 → (0, 2π] by letting θ (v) be the angle in the specified interval that v makes to the positive e 1 -axis.
The next proposition of Chambers concerns two C 2 functions on an interval (a, b).
and we let κ h (x) denote the upward curvature of the graph of h at x.
Proposition 5.4. [C,Prop 3.8] Consider two C 2 functions f , g : (a, b) → R ≥ 0 with b > a that satisfy the following: for all x ∈ (a, b).

PROOF OF RIGHT TANGENT LEMMA
To prove Proposition 4.8, we assume that F(0) > 1/2, then investigate three portions of the curve, beginning at the rightmost point and traveling counterclockwise to its other intersection point with the e 1 -axis. One may think of the upper curve as the portion of [0, β ) on which the tangent vector γ (s) is in the second quadrant and the lower curve as the portion of [0, β ) on which the tangent vector is in the third quadrant. We will prove that the lower curve ends in a vertical tangent at a point right of the e 2 -axis (Lemma 6.12) and that, past this point, curvature is positive and the tangent vector is strictly in the fourth quadrant (Lemma 6.14). The end behavior of the curve is similar to that of the generating curve in [C] except that our curve must terminate right of the y-axis, an additional feature which allows us to achieve a contradiction to spherical symmetry without an analogue of Chambers' Second Tangent Lemma [C,Lemma 2.5]. As such, many intermediate results are also similar to results in [C] and are The vectors v 1 and v 2 are admissible with respect to (x 1 , y) and (x 2 , y).

cross-referenced.
Our analysis requires comparing curvatures at points of the same height on the upper and lower curves (technically, the portions of the curve parameterized by these intervals in [0, β )). Specifically, we show that the curvature at the point on the left is strictly greater than the curvature at the corresponding point on the right (Proposition 6.11). By Corollary 5.2, it suffices to prove that λ , the canonical circle curvature, is less at the point on the left and H 1 , the normal derivative of the log of the density, is strictly less at the point on the left. For any s ∈ [0, β ), γ(s) = (0, 0), so the normal derivative of log(r p ) at γ(s) is given by More generally, given points (x 1 , y), (x 2 , y) ∈ R 2 − {0}, and unit vectors v 1 and v 2 , one can compare the quantities where v ⊥ 1 and v ⊥ 2 denote clockwise rotations of v 1 and v 2 by π/2 radians. (In our context, v 1 and v 2 will be tangent vectors to the curve at two points, so v ⊥ 1 and v ⊥ 2 will be the unit outward normal vectors.) We have discovered a set of sufficient conditions for the points (x 1 , y) and (x 2 , y) and the vectors v 1 and v 2 to satisfy the inequality In Definition 6.1, we define two unit vectors v 1 and v 2 to be admissible with respect to (x 1 , y) and (x 2 , y) if they satisfy these conditions. Definition 6.1. Consider a pair of points (x 1 , y) and (x 2 , y) with y > 0, and a pair of unit vectors, v 1 and v 2 , which lie strictly in the second and third quadrants, respectively. Let v 1 denote the reflection of v 1 over the e 1 -axis. Let C i denote the canonical circle with respect to v i at (x i , y), with center (a i , 0) and radius R i . As depicted in Figure 4, v 1 and v 2 are admissible with respect to (x 1 , y) and (x 2 , y) if the following occur: (1) a 1 > R 1 , Proposition 6.2. Consider a pair of points (x 1 , y) and (x 2 , y) in the upper half plane with x 1 ≥ x 2 . Let v 1 and v 2 be two unit vectors, and v ⊥ 1 and v ⊥ 2 denote the clockwise rotations of these respective vectors through π/2 radians. If v 1 and v 2 are admissible with respect to (x 1 , y) and (x 2 , y), then Proof. Let (x 1 , y) be the reflection of (x 1 , y) over the vertical line x = a 1 . It follows that x 1 = a 1 − (x 1 − a 1 ). By symmetry, C 1 is also the canonical circle with respect to v 1 at (x 1 , y). We will show that and that To prove (5), we parameterize C 1 by α(t) = (a 1 + R 1 cost, R 1 sint) for t in [0, 2π). Taking t 1 ∈ (0, π/2) so that α(t 1 ) = (x 1 , y), we have by symmetry that (x 1 ,t) = α(π −t 1 ). Using this parametrization to simplify the quantities in (5), we have The denominator is positive, so we need only show that the numerator is positive to conclude that (5) holds. A short computation reveals that (a 1 cost 1 + R 1 )|α(π − t 1 )| 2 − (−a 1 cost 1 + R 1 )|α(t 1 )| 2 = 2a 1 (a 2 1 − R 2 1 ) cost 1 > 0. To prove (6), we first note that since v 1 and v 2 are admissible with respect to (x 1 , y) and (x 2 , y), we have that x 2 ≥ a 1 − (x 1 − a 1 ) = x 1 . Moreover, x 1 must be positive, as Therefore, to prove (6), it suffices to show that We note that the left-hand side of (7) is cos(θ (v ⊥ 1 ) − θ ((x 1 , y))) and the right-hand side is equal to cos(θ (v ⊥ 2 ) − θ ((x 2 , y))). Since v 1 is strictly in the second quadrant, v 2 is strictly in the third, and As cosine is decreasing on (0, π), it suffices to show that As noted above, x 2 ≥ x 1 , so θ ((x 2 , y)) ≤ θ ((x 1 , y)). By the admissibility of v 1 and v 2 , we have that θ (v ⊥ 2 ) ≥ θ (v ⊥ 1 ). Combining these inequalities establishes (8), completing our proof of (6). Having proved Proposition 6.2, we define the upper and lower curves and prove various properties that hold on these intervals. Our definition of the upper curve is motivated by the following observation. Lemma 6.3. (cf. [C,Lemma 3.5]) Given that F(0) > 1/2, we have κ (0) > 0.
We will soon prove several properties of δ , but first we require one more lemma.
Given this result, to prove that κ (s) > 0, it suffices to prove that H 1 (s) < 0. Parameterizing A s by we can compute thatH Since A s = C s , we have that b = 0, a = F(s), and r = R(s). By Lemma 6.5, a > r > 0. Thus, we have Proposition 6.7. (cf. [C,Prop. 3.12]) The following properties of δ hold: Proof. The proofs of (1)-(3) are identical to their counterparts in [C,Prop. 3.12]. Setting s = 0 in the inequality γ 1 (δ ) ≥ F(s) we have The proof that γ (δ ) = (−1, 0) is similar to that in [C,Prop. 3.12]; however, we do not require a second sub-case, and we apply our Lemma 6.6 in place of [C,Lemma 3.4].
Looking to (10), we claim that a 2 + b 2 > r 2 . To prove so, let R = R(δ ) be the radius of C δ . Since γ (δ ) = (−1, 0), we have that R = r + b and a = F(δ ). We apply Lemma 6.5 to give a > R = r + b. Since both sides of a − b > r are positive, we may square to give (a − b) 2 > r 2 . Since b ≥ 0, this implies that a 2 + b 2 > r 2 . Therefore, we have that Definition 6.9. (cf. [C,Defn. 3.5]) Let the lower curve L be defined as the set of all t in [δ , β ) such that for all s ∈ [δ ,t] the following hold: (2) If s is the unique point in K with γ 2 (s) = γ 2 (s), then κ(s) ≤ κ(s).
(12) The function g gives the e 1 -coordinate of a point in the lower curve with a given e 2 -coordinate. If we begin with the point in the upper curve with a given e 2 -coordinate, then f gives the e 1 -coordinate of the reflection of this point over the line x = δ . We can use these functions to prove two properties of the lower curve.
Proof. Both inequalities are trivially true if s = δ . Now let s ∈ (δ , η) be fixed, and let y = γ 2 (s). By the definition of L (Def. 6.9), f and g satisfy the hypotheses of Proposition 5.4. From the inequality f ≤ g in Proposition 5.4, (13) above is immediate. To arrive at (14), let t f (y) and t g (y) denote the unit tangent vectors to the graphs of f and g at y. Note that γ( (δ , η) ) is the set {(g(y), y) : y ∈ (γ 2 (η), γ 2 (δ )}, and the reflection of γ((0, δ )) over the line x = δ is the set {( f (y), y) : y ∈ (γ 2 (η), γ 2 (δ )}. Let y = γ 2 (s). Then we obtain the tangent vector t g (y) from γ (s) by rotating γ (s) clockwise through π radians and reflecting the resulting vector in the first quadrant over the line y = x. Therefore, we have Similarly, we obtain t f (y) from γ (s) by reflecting over the line x = δ and reflecting over the line y = x. Thus, Substituting these results into the second inequality in 5.4 completes the proof.
Because γ (s) and γ (s) are admissible with respect to γ(s) and γ(s), we conclude by Proposition 6.2 that By a similar argument to that in [C,Lemma 3.14] along with Proposition 6.11, η < β , η ∈ L, and γ (η) = (0, −1). In addition to these properties of η, we can also show using the curvature comparison that γ 1 (η) > 0. Then proving that γ 1 (β ) > 0 is a matter of showing that γ 1 is increasing on (η, β ). To establish the second claim of the Right Tangent Lemma, we consider the functions κ and γ on (η, β ). Lemma 6.13 gives a computational result regarding κ, whereas Lemma 6.14 extends this result as well as showing that γ is strictly in the fourth quadrant on (η, β ).

PROOF OF LEFT TANGENT LEMMA
In the previous section, the key to proving the Right Tangent Lemma was to show that the curvature was greater at a point on the lower curve than at its corresponding point on the upper curve, allowing us to find η < β where γ (η) = (0, −1). Now, in the left case (Prop. 4.9), we will prove the opposite inequality concerning curvatures at corresponding points, with the aim of showing that lim s→β − γ (s) is strictly in the third quadrant. This case, however, presents new obstacles. One difficulty we eliminate is the possibility that there are multiple points on the portion of γ parameterized by [0, β ) where the tangent vector is (−1, 0). In the right case, the lower curve naturally terminated at a point where the tangent vector was (0, −1). However, the goal in the left case will be to show that the lower curve does not terminate before β , allowing us to apply the curvature comparison all the way up to β . We begin with a new definition of admissibility for the left case and an analogue of Proposition 6.2.
Definition 7.1. Consider two points (x 1 , y) and (x 2 , y) and two unit vectors v 1 and v 2 , strictly in the second and third quadrants, respectively. Let C i , a i , R i , x 1 , and v 1 be as in Definition 6.1. Finally, let (x * , 0) be the unique point on the e 1 -axis so that v 2 is tangent at (x * , y) to the circle centered at the origin that passes through (x * , y). We say that v 1 and v 2 are admissible with respect to (x 1 , y) and (x 2 , y) if the following hold: ( Figures 5 and 6 depict vectors v 1 and v 2 that are admissible with respect to (x 1 , y) and (x 2 , y) when x 2 ≥ 0 and when x 2 < 0. Proposition 7.2. If v 1 and v 2 are admissible with respect to (x 1 , y) and (x 2 , y), then H 1 is larger at (x 2 , y) with respect to v 2 than at (x 1 , y) with respect to v 1 .
Hence, we have By (4) in Definition 7.1 and the assumption that x 2 < 0, we have x * ≤ x 2 < 0. We multiply through by x 2 to obtain x * x 2 ≥ x 2 2 > 0. Substituting this into the above equation, we have (x 2 , y) completing the second case.
Before we define the upper and lower curves, we require several lemmas. Proposition 7.7 and Proposition 7.8, which concern points where the unit tangent vector is in the second quadrant, are later used to check the conditions for admissibility. Meanwhile, we determine some properties that hold at points on the curve with positive first coordinates.
Proof. We take three cases according to whether κ(s) > 0, κ(s) = 0, or κ(s) < 0. If κ(s) = 0, then A s is the oriented line through γ(s) that has direction vector γ (s). By Lemma 7.3, γ 1 (s) < 0. We parameterize A s by α(t) = γ(s) + tγ (s). LetF(t) and denote the e 1 -coordinate of the canonical circle to A s at α(t), letR(t) denote its radius, and letG =F −R. Then we can compute that Since γ is an arclength parameterization, the numerator ofG is necessarily nonnegative. Thus,G ≤ 0. If κ = 0, let (a, b) be the center of A s , and let r be the radius. If κ > 0, then b < 0, whereas if κ < 0, then b > 0. We parameterize in arclength on [0, 2πr) so that α(s) = γ (s). Then, as in [C,Lemma 5.3], we obtain expressions forF andR and differentiate. In each case, the signs ofF (s) andG (s) are determined by the sign of b.
Although we used both hypotheses of Proposition 7.5 in the proof, it is actually the case that the first hypothesis implies the second, as we prove below.
Suppose for contradiction that γ 1 (u) > 0. We will show that u is not an upper bound for S, but, instead, that there exists ε > 0 so that [u, u + ε) ⊂ S. We can obviously find ε 1 > 0 so that γ 1 ≥ 0 on [u, u + ε 1 ). It remains to show that there exists ε 2 > 0 so that κ ≤ λ on [u, u + ε 2 ). The proof will be similar to that of Lemma 3.4 in [C].
First, we can prove by contradiction that κ(u) = λ (u). Given this assumption, we have that C u = A u , so λ (u) =λ (ũ) = 0. Thus, to guarantee the existence of a ε 2 > 0 so that κ ≤ λ on [u, u + ε 2 ), it suffices to show that κ (u) < 0. Since κ ≤ λ and γ 1 ≥ 0 on [0, u], it follows from Proposition 7.5 that G ≤ 0 on [0, u]. Therefore, G(u) ≤ G(0) < 0 by assumption that F(0) < R(0), and it follows by a similar argument to that in the proof of Lemma 6.6 that κ (u) < 0. While the inequality a > 0 was immediate in the case that a > r, here it is more subtle. The fact that a ≥ 0 follows from a similar argument to that in [C,Lemma 3.3]. To prove strict inequality, note that if a = 0, then γ is a circle centered at the origin, which contradicts the fact that balls centered at the origin are not stable ( [RCBM,Thm. 3.10]).
We use Lemmas 7.5 and 7.6 to prove two propositions used in checking the conditions for admissibility (Props. 7.8 and 7.9), but first we require one additional lemma. Lemma 7.7. Suppose that s ∈ (0, β ) and that γ (s) is in the second quadrant. Then γ 1 (s) > 0.
Having proved the propositions necessary for checking the definitions of admissibility, we define the upper and lower curves and prove that the curvature at a point on the lower curve is less than the curvature at its counterpart on the upper curve (Prop. 7.15). Definition 7.10. A point s is in the upper curve K ⊂ (0, β ) if and only if for all t ∈ (0, s), γ (t) is strictly in the second quadrant.