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d ρ S ( ( ,y x, ρ t 62 piay.5F9 60 (secondary). 46B07 51F99, (primary). 46B20 fara omdsae h egho path a of length the space, normed real a of j +1 [0 : ) 1 ρ ≤ S ) 0 = (0) , k . ,y x, √ 1]

2 → π ∈ = 0 k this-really-should-exist-somewhere- ,ρ x, x S S nlsosarvda r hs fthe of those are at arrived onclusions − edfiethe define we , continuous t y 1 = (1) 0 aSS. k t < . . . < , n ∈ r . ttdbelow). stated 1.2 ure y lnei ahmtcland Mathematical in llence N , i ntsphere”. unit ric , sudrinvestiga- under ns n stu,wt opti- with true, is t  mpsdi this in posed em 1 = . nrni metric intrinsic tdotthe out nted    , ti now is lt strongly THEINTRINSICMETRICONUNITSPHERES 2

The above question is then rephrased as: For the unit sphere S in a normed space, do there exist constants A,B > 0 such that, for all x, y S, ∈ A x y d(x, y) B x y ? k − k≤ ≤ k − k To the authors’ knowledge, the answer to this question does not appear in the literature1, a fact that is perhaps more surprising than the positive solution to the problem which will be presented in this paper. The problem can essentially be reduced to one in two dimensions and apart from relying on John Ellipsoids—a fundamental structure from local theory—the result follows from entirely elementary (albeit somewhat technical) arguments. We will now describe the structure of the paper. After introducing the needed notation, definitions and preliminary results in Section 2, in Section 3 our goal will be to prove Theorem 3.6: Theorem 3.6. For any norm on a real vector space V , let d denote the intrinsic metric on the unit sphere S ofk·kV . For all x, y S, ∈ x y d(x, y) √2π x y . k − k≤ ≤ k − k A crucial ingredient is that of the John Ellipsoid: The largest ellipsoid (Euclidean ball of largest volume) that can be contained inside a unit ball of a finite dimensional normed space. Theorem 1.1 (John’s Theorem [1, Theorem 12.1.4]). Let W be any normed space Rn of dimension n > 1. With E denoting the Euclidean norm on , there exists k·kRn −1 Rn a norm one isomorphism T : ( , E) W with inverse T : W ( , E) whose norm is at most √n. k·k → → k·k Specifically, all two-dimensional subspaces of a normed space has a Banach- Mazur distance of at most √2 from two-dimensional . Of course, the intrinsic metric on a normed space V ’s unit sphere is bounded above by the “planar” intrinsic metric: where all paths in the defining infimum are taken to live in any two-dimensional subspace of V . This allows us to reduce the question to R2, where S lies between a Euclidian unit sphere SE and √2SE. In this setting, the crucial ingredients are Lemmas 3.3 and 3.4, which allows us to conclude the local bi-Lipschitzness of the map σ : S SE defined by σ(x) := x/ x where both S and S are endowed with the Euclidean→ induced metric. k kE E Since the Euclidean induced– and intrinsic metrics on SE are easily calculated and related (Lemma 2.1), our main results, Theorems 3.5 and 3.6, then easily follow. We note that the constant √2π obtained in our main result (Theorem 3.6) is likely not optimal. In two dimensions, the -norm provides a worst case for the k·k∞ John Ellipsoid. Let S∞ be the unit sphere of this norm, with intrinsic metric d∞, then for x, y S∞, it is easily seen that x y ∞ d∞(x, y) 2 x y ∞. This prompts the∈ following conjecture: k − k ≤ ≤ k − k Conjecture 1.2. 2 For any norm on a real vector space V , let d denote the intrinsic metric on the unit sphere Sk·kof V . For all x, y S, ∈ x y d(x, y) 2 x y . k − k≤ ≤ k − k 1False! See the correction above! 2This conjecture is proven true in [2, Theorem 3.5]. THEINTRINSICMETRICONUNITSPHERES 3

2. Definitions, notation and preliminary results This section will explicitly define all notation used in this paper. Since we will translate between many different metrics on many different sets, we will take extreme care to make our notation as explicit as possible.

Let V be a real vector space. Let A be an arbitrary index symbol and A any norm on V . We will denote unit sphere, closed unit ball and open unit bak·kll with S B B respect to A respectively by A, A and A. For anyk·k subset M V , we define the induced metric d : M R on M by ⊆ A → ≥0 dA(x, y) := x y A for v, w M. We definek the− Ak-M-path-space∈ by (M) := ρ ρ : [0, 1] M, continuous , PA { | → k·kA − } and the planar A-M-path-space by planar(M) := ρ (M) dim (span (Imρ))=2 . PA { ∈PA | } We define the A-path-length operator L : (V ) R by A PA → ≥0 ∪ {∞} n−1 n N, LA(ρ) := sup ρ(tj ) ρ(tj+1) A ∈ (ρ A(V )).  k − k 0= t0 <...

We define the (extended) A-M-intrinsic metric dA,M : M R≥0 , by  → ∪ {∞} d (v, w) := inf L (ρ) ρ A(M), ρ(0) = v, ρ(1) = w (v, w M) A,M A ∈P ∈ and the (extended) A-M-planar-intrinsic metric dplanar : M R by A,M → ≥0 ∪ {∞} planar planar d (v, w) := inf LA(ρ) ρ (M), ρ(0) = v, ρ(1) = w (v, w M). A,M ∈PA ∈ We introduce the following abbreviated notation that will aid in readability of the paper: For (extended) metrics d and d′ on some set D, subset M D, and constant K > 0, by ⊆ “d Kd′ on M” ≤ we will mean d(a,b) Kd′(a,b) for all a,b M. ≤ ∈ planar For any real normed space (V, A) and subset M V , since A (M) (M) (V ), the chain of inequalitiesk·k ⊆ P ⊆ PA ⊆PA d = d d dplanar on M A A,V ≤ A,M ≤ A,M ≤ ∞ is easy to verify. An elementary calculation establishes the following lemma: Lemma 2.1. With denoting the Euclidean norm on R2, k·kE π d d SE d on S . E ≤ E, ≤ 2 E E R2 If is endowed with the euclidean norm E arising from an inner product 2 k·k , for elements x, y R the ray from x through y is denoted by rx,y and defined h·by |r ·i := (1 t)x + ty∈ t 0 . For a point x and points y,z R2 distinct from x,y { − | ≥ } ∈ x, when referring to the size of the between rx,y and rx,z we will mean the quantity y x z x arccos h − | − i [0, π]. y x z x ∈ k − kE k − kE  THEINTRINSICMETRICONUNITSPHERES 4

2 For points v,w,x,y R , we will say the ray rx,y lies between the rays rx,v and r if v, w and x are∈ in general position and r (1 t)v + tw t [0, 1] = , x,w x,y ∩{ − | ∈ } 6 ∅ or, rx,y = rx,v = rx,w.

3. The intrinsic metric on unit in R2 In this section we will prove our main results. Although somewhat technical, our results follow mostly from elementary trigonometry and Euclidian plane geometry. Let denote the Euclidean norm on R2 and let be any norm on R2 k·kE k·kX satisfying BE BX KBE for some K 1. A large part of our attention will be devoted to⊆ proving⊆ that the map σ : S ≥ S defined by σ(x) := x/ x is X → E k kE locally bi-Lipschitz when both SX and SE are both endowed with the Euclidean induced metrics. Once this has been achieved through Lemmas 3.3 and 3.4, a straightforward calculation will prove our main results Theorems 3.5 and 3.6. R2 Let E denote the Euclidean norm on and let X be any other norm on 2 k·k k·k R satisfying BE BX . We will first relate points on SX to lines tangent to SE. Specifically, for any⊆ point x S that is not in S , the two lines through x that ∈ X E are tangent to SE are such that points in SX “close to” x are “wedged between” the tangent lines. Also, if x SX SE, then the whole of SX lies on the same side of the line y R2 x y ∈=1 ∩. ∈ h | i Let x R2 B , and let τ(x) S be a point on a tangent line to S through x. ∈ \ E ∈ E X Then the angle between r and r equals arccos( x −1). Let x⊥ S x ⊥ 0,x 0,τ(x) k kE ∈ X ∩{ } be such that τ(x) x⊥ 0. If we now define ≥ 1 x x a (x) := cos arccos = x x x 2  k kE  k kE k kE and 1 1 b(x) := sin arccos x⊥ = 1 x⊥, x − x 2  k kE  s k kE then τ(x)= a(x)+ b(x).

Lemma 3.1. Let be an inner product and E be the associated Euclidean R2 h· | ·i Rk·k2 B B norm on . Let X be any other norm on such that E X . For all x S , k·k ⊆ ∈ X (1) For all t R, tx + (1 t)τ(x) τ(x) =1. (2) For all t ∈ [0, 1]h , tx +− (1 t)τ(|x) i 1. ∈ k − kX ≤ (3) For all t> 1, tx + (1 t)τ(x) X 1. (4) If x S S k and y −S , thenk ≥x y 1. ∈ X ∩ E ∈ X h | i≤ Proof. We prove (1). Let x S . For all t R, ∈ X ∈ tx + (1 t)τ(x) τ(x) = t x τ(x) + (1 t) τ(x) τ(x) h − | i h | i − h | i x x = th | i + (1 t) x 2 − k kE = t + (1 t) − = 1.

We prove (2). Let x SX . Since τ(x) SE BX , and BX is convex, the result follows. ∈ ∈ ⊆ THEINTRINSICMETRICONUNITSPHERES 5

We prove (3). Let x S . Since x = 1, if τ(x) = x, then, tx + (1 ∈ X k kX k − t)τ(x) X = 1, and the result is trivial. We therefore assume τ(x) = x. Since kS 6 τ(x) E, so that τ(x) X 1, by the reverse triangle inequality and intermediate value∈ theorem therek existsk ≤ some t 0 such that 1 = t x + (1 t )τ(x) = 0 ≤ k 0 − 0 kX 1x+(1 1)τ(x) X (here we used τ(x) = x). Since the map t tx+(1 t)τ(x) X isk convex,− we cannotk have that tx + (16 t)τ(x) < 1 for any7→t> k 1, as− this wouldk k − kX contradict 1 = t0x + (1 t0)τ(x) X = 1x + (1 1)τ(x) X . We conclude that tx + (1 t)τ(xk) 1 for− all t> k1. k − k k − kX ≥ We prove (4). Let x SX SE and y SX , but suppose x y > 1. ∈ ∩ ∈ h | i S If y and x are linearly dependent, then y X > 1, contradicting y X , and we therefore may assume that y and x are linearlyk k independent. ∈ Let L denote the line through x and y, parameterized by the affine map η(t) := (1 t)y + tx for t R. The line L is not tangent to SE (else we would have x −y = 1). Therefore∈ L intersects S in two distinct points, one being x; let h | i E t0 R be such that η(t0) SE L is the other. We must have t0 > 1, since ∈ ∈ ∩ R2 1 < x η(t) for t [0, 1). Since η is an affine map and ( , E) is a strictly convexh space,| i ∈ k·k 1+ t η(t )+ η(1) η 0 = 0 2 2   X X

η(t0)+ η(1) ≤ 2 E

< 1. −1 1+ t0 Let λ := 2(1 + t0) (0, 1), so that λ 2 + (1 λ)0=1. Then, again since η is affine, ∈ −  1 = x k kX = η(1) k kX 1+ t = η λ 0 + (1 λ)0 2 −     X

1+ t0 = λη + (1 λ)η(0) 2 −   X

1+ t0 λ η + (1 λ) η(0) ≤ 2 − k kX   X 1+ t = λ η 0 + (1 λ) y 2 − k kX   X

< λ 1+(1 λ) 1 · − · = 1, which is absurd. We conclude that x y 1 for all x SX SE and all y SX . h | i≤ ∈ ∩ ∈ 

Next, we show that for points x, y S that are “sufficiently close”, the size of ∈ X the angle formed by the rays r0,x and r0,y bounds the size of the acute angle formed by the ray rx,y and the perpendicular line to r0,x through x. Lemma 3.2. Let denote the Euclidean norm on R2 and be any norm k·kE k·kX on R2 such that B B . Let x, y S and let x⊥ S x ⊥ be such E ⊆ X ∈ X ∈ E ∩ { } THEINTRINSICMETRICONUNITSPHERES 6 that x⊥ y 0 and define v := x + x⊥. If K 1 and x, y S are such | i ≥ ≥ ∈ X that x E K and the size of the angle between the rays r0,x and r0,y is at most k k −1≤ arccos(K ), then α, the size of the angle between the rays rx,v and rx,y, is also at most arccos(K−1). Proof. As a visual aid, the reader is referred to Figure 3.1.

w

x v β α

τ2(x) y τ1(x)

β S u X Pux 0 SE Figure 3.1.

−1 ⊥ Let β := arccos(K ) and u SE be such that x u > 0 and that the size of the angle formed by the rays ∈r and r equals β (i.e.,| i x u = x cos β). 0,x 0,u h | i k kE Let τ1(x), τ2(x) SE be the point(s) on the lines through x that are tangent to S , such that x∈⊥ τ (x) 0. Let E | 1 i≥ v x = τ (x)= τ (x) w := 1 2 2x τ (x) otherwise, ( − 2 so that w rτ2(x),x is distinct from x, and is such that rx,w rτ2(x),x. Let Pu denote the∈ orthogonal projection onto the span of u. Then size of⊆ the angle formed between the rays rx,Pux and rx,v is exactly β. Since x E K, the point Pux lies on the line segment tu t (0, 1] (if x = K,k thenk P≤ x = u = τ (x)), and { | ∈ } k kE u 1 therefore the size of angle between rays rx,u and rx,v is at most β. Since rx,τ1(x) is between the rays rx,u and rx,v, and since rx,v bisects the angle formed by the rays rx,τ1(x) and rx,w, the size of the angle formed by rx,v and rx,w is also at most β. Finally, by Lemma 3.1 (2),(3) and (4) and the fact that B B , the ray r lies E ⊆ X x,y either between the rays rx,u and rx,v or the rays rx,v and rx,w (The point y can only lie in the shaded area in Figure 3.1). We conclude that α, the size of the angle −1 between rays rx,v and rx,y, is at most β = arccos(K ). 

Lemma 3.3. Let denote the Euclidean norm on R2 and be any norm k·kE k·kX on R2 such that B B KB for some K 1. If x, y S is such that θ, E ⊆ X ⊆ E ≥ ∈ X THEINTRINSICMETRICONUNITSPHERES 7

−1 the size of angle between the rays r0,x and r0,y, is at most arccos(K ), then x y x y K2 . k − kE ≤ x − y E E E k k k k

Proof. As a visual aid, the reader is referred to Figure 3.2.

x v α

α y Pxy x/ x k k λy S y/ y X k k KSE θ u

SE 0

Figure 3.2.

⊥ ⊥ ⊥ ⊥ Let x SE x be such that x y > 0 and define v := x + x . Let Px and P be∈ the orthogonal∩{ } projections onto| thei span of x and y respectively. Define y u := P (x/ x ), and λ := P y −1 so that P (λy)= x/ x . Let α denote the y k kE k x kE x k kE size of the angle formed by the rays rx,y and rx,v. We note that then size of the angle formed between the rays rx/kxkE ,λy and rx/kxkE ,u also equals θ. Elementary trigonometry will establish 1 x y = y P y k − kE cos α k − x kE 1 = λy P (λy) λ cos α k − x kE 1 x = λy λ cos α − x E E k k 1 x = u . λ cos α cos θ − x E E k k x y x Now we note that u k k k k k k , since u the is the closest point − x E ≤ y E − x E (with respect to ) in the span of y to the point x/ x . Also, by Lemma 3.2 k·kE k kE we have α arccos(K −1), so that cos α K− 1. Furthermore λ−1 = P y = ≤ ≥ k x kE THEINTRINSICMETRICONUNITSPHERES 8

y cos θ K cos θ. Finally we conclude k kE ≤ 1 x x y = u k − kE λ cos α cos θ − x E E k k K cos θ y x

≤ cos α cos θ y − x E E k k k k 2 y x = K . y − x  E E k k k k Lemma 3.4. Let denote the Euclidean norm on R2 and be any norm k·kE k·kX on R2 such that B B . If x, y S , then E ⊆ X ∈ X x y x y . x − y ≤k − kE E E E k k k k Proof. As a visual aid we refer the reader to Figure 3.3.

x y

x/ x y/ x k k k k y/ y k k u SX

SE

0 Figure 3.3.

Let x, y SX . By exchanging the roles of x and y if necessary, we may assume y x ∈ 1. Let P be the orthogonal projection onto the span of y and let k kE ≥k kE ≥ y x u := Py . Then u 1 and y/ x 1= y/ y . Then, by the kxkE k kE ≤ k k kEkE ≥ k k kEkE Pythagorean theorem, x y 2 x 2 y 2 = u + u x − y x − − y k kE k kE E k kE E k kE E 2 2 x y u + u ≤ x − − x k kE E k kE E 2 x y = x − x E E E k k k k 1 2 = 2 x y E x k − k k kE x y 2 .  ≤ k − kE THEINTRINSICMETRICONUNITSPHERES 9

In essence, the previous two Lemmas together establish that the local bi-Lip- schitzness of the map σ : S S defined by σ(x) := x/ x when B B X → E k kE E ⊆ X ⊆ KBE. We will now use the previous results to prove one of our main results which relates the intrinsic metric on SX to the induced metric on SX when BE BX KBE for some K 1. ⊆ ⊆ ≥ Theorem 3.5. Let denote the Euclidean norm on R2 and be any norm k·kE k·kX on R2 such that B B KB for some K 1. Then E ⊆ X ⊆ E ≥ 3 π d d SX K d on S . X ≤ X, ≤ 2 X X S Proof. We have already noted in Section 2 that dX dX,SX on X . 2 ≤ Let x, y SX be arbitrary. Let c : R R be the map defined by c(θ) := ∈ R R → (cos(θ), sin(θ)) for θ . Let θx,θy be such that c(θx)/ c(θx) X = x and c(θ )/ c(θ ) = y.∈ By switching the∈ roles of x and y, if necessary,k k we may y k y kX assume that 0 θy θx π. Consider the path ρ :[θx,θy] SX defined by ρ(θ) := c(θ)/ c(≤θ) −for θ≤ [θ ,θ ]. → k kX ∈ x y Let ε> 0 be arbitrary and θx = θ0 <θ1 <...<θn = θy be a partition of [θx,θy] such that n−1 ρ(θ ) ρ(θ ) L (ρ) ε. k j − j+1 kX ≥ X − j=0 X −1 We may assume that 0 < θj+1 θj arccos K , since the triangle inequality ensures that every refinement of −θ n≤ still satisfies the above inequality. { j }j=1 We note that B B KB implies w w K w for all E ⊆ X ⊆ E k kX ≤ k kE ≤ k kX w R2. Then, by Lemmas 3.3, 2.1 and 3.4, we obtain ∈ dX,SX (x, y) L (ρ) ≤ X n−1 ρ(θ ) ρ(θ ) + ε ≤ k j − j+1 kX j=0 X n−1 ρ(θ ) ρ(θ ) + ε ≤ k j − j+1 kE j=0 X n−1 K2 c(θ ) c(θ ) + ε ≤ k j − j+1 kE j=0 X m−1 N 2 m K sup c(φj ) c(φj+1) E ∈ + ε ≤  k − k θx = φ0 <...<φm = θy  j=0  X 

2 x y K d SE , + ε  ≤ E, x y k kE k kE  π x y K2 + ε ≤ 2 x − y E E E π k k k k K2 x y + ε ≤ 2 k − kE π K3 x y + ε. ≤ 2 k − kX THEINTRINSICMETRICONUNITSPHERES 10

Since ε> 0 was chosen arbitrarily, the result follows.  Our final result now follows through an easy application of the previous result and John’s Theorem (Theorem 1.1): Theorem 3.6. For any norm on a real vector space V , k·kX

d d SX √2π d on S . X ≤ X, ≤ X X planar S Proof. We have already noted in Section 2 that d d SX d S on . X ≤ X, ≤ X, X X Let x, y SX be arbitrary and let W V be any two-dimensional subspace ∈ planar⊆ containing x and y, noting that then d S (x, y) d SX (x, y). By John’s X, X ≤ X, ∩W Theorem (Theorem 1.1), there exists a Euclidean norm on W such that k·kE w w √2 w for all w W , i.e., B B W √2B . Then, by k kX ≤k kE ≤ k kX ∈ E ⊆ X ∩ ⊆ E Theorem 3.5, we may conclude that d d SX √2π d on S .  X ≤ X, ≤ X X References

1. F. Albiac and N.J. Kalton, Topics in Banach space theory, Graduate Texts in Mathematics, Springer, New York, 2006. 2. J.J. Schäffer, Inner diameter, perimeter, and girth of spheres, Math. Ann. 173 (1967), 59–79; addendum, ibid. 173 (1967), 79–82.

Miek Messerschmidt; Unit for BMI; North-West University; Private Bag X6001; Potchefstroom; South Africa; 2520 E-mail address: [email protected]

Marten Wortel; Unit for BMI; North-West University; Private Bag X6001; Potchef- stroom; South Africa; 2520 E-mail address: [email protected]