Linearity of isometries between convex Jordan curves
aa r X i v : . [ m a t h . M G ] F e b Linearity of isometries between convex Jordan curves
Javier Cabello S´anchez
Departamento de Matem´aticas, Universidad de Extremadura, Avda. de Elvas s/n, 06006 Badajoz.Spain. [email protected]
Supported in part by DGICYT projects MTM2016-76958-C2-1-P and PID2019-103961GB-C21(Spain) and Junta de Extremadura programs GR-15152 and IB-16056.Keywords: Tingley’s Problem; differentiability; finite-dimensional spaces; metric invariants.2020 Mathematics Subject Classification: 46B04, 15A03, 52A10.
Abstract
In this paper, we show that the C -differentiability of the norm of a two-dimensionalnormed space depends only on distances between points of the unit sphere in two differentways.As a consequence, we see that any isometry between the spheres of normed planes τ : S X → S Y is linear, provided that there exist linearly independent x, x ∈ S X where S X is not differentiable and that S X is piecewise differentiable.We end this work by showing that the isometry τ : C X → C Y is linear even if itis not an isometry between spheres: every isometry between (planar) Jordan piecewise C -differentiable convex curves extends to X whenever X and Y are strictly convex andthe amount of non-differentiability points of S X and S Y is finite and greater than 2.
1. Introduction
The study of isometries between Banach spaces led, back in the 30’s, to one ofthe best known results in Functional Analysis, the Mazur–Ulam Theorem. This result,see [16], states that every onto isometry between two Banach spaces is affine. So, ifan onto isometry preserves the origin, then the isometry is linear. Forty years later,P. Mankiewicz ([15]) proved that every onto isometry between convex bodies in twoBanach spaces is also affine. The foreseeable generalisation of these results is
Every ontoisometry between the spheres of two Banach spaces is linear, and this could be ultimatelygeneralised as
Every onto isometry between the boundaries of convex bodies of two Banachspaces is affine, but, up to now, no-one has been able to prove or disprove this statement.This innocent-looking problem was stated in 1987 by D. Tingley ([27]), but it turnsout to be way more challenging than it could seem at first glance. Tingley’s Problem
Preprint submitted to Elsevier February 23, 2021 as evolved to that of extending isometries between spheres to isometries between thewhole spaces, and the greatest advances have been achieved when both spaces have somecommon structure, such as von Neumann algebras, trace class operators spaces, sumsof strictly convex spaces. . . This extension of isometries problem has experienced a rapiddevelopment in the last few years, and there are lots of kinds of spaces where Tingley’sProblem has a positive answer, see [2, 5, 6, 7, 8, 9, 10, 11, 13, 14, 17, 18, 19, 20, 21, 23,24, 25, 26, 28].Nevertheless, there is another way to look at this Problem. Instead of extending anisometry, one can rule out the existence of an isometry between two spheres –a trivialexample: there is no isometry between the spheres of ( R , k · k ) and ( R , k · k ∞ ), say S and S ∞ , because there exist x, y, z ∈ S ∞ such that k x − y k ∞ = k x − z k ∞ = k y − z k ∞ = 2but this cannot happen in S . To the best of our knowledge, the first great achievementin this setting can be found in [13], where the authors prove that, in finite-dimensionalspaces, no sphere can be isometric to a polyhedral sphere unless it is polyhedral, too.Actually, they also extend the isometry between the spheres, so the main result in [13]is If X is finite-dimensional and polyhedral and there is an onto isometry τ : S X → S Y ,then Y is also polyhedral and X and Y are linearly isometric. In the same spirit, a few years later appeared this result: If X is an inner product space and there is an onto isometry τ : S X → S Y , then Y is also an inner product space and X and Y are linearly isometric, see [3, 4, 18].So, motivated by the huge advance that can be seen at a recent paper by Tar´asBanakh, [2], whose main result is Every isometry τ : S X → S Y between the spheres of absolutely smooth two-dimensionalspaces is linear ,we began to study whether the C -differentiability (that implies absolute differentia-bility) of some two-dimensional normed space ( X, k · k X ) can be expressed in terms of( S X , k · k X ) –unsuccessfully.However, we have been able to determine C -differentiability in two independentways. The first way is easily seen to hold in finite-dimensional spaces, whereas we havenot been able to prove whether the second one works in dimensions higher than 2 or not.These two ways are the following:1. Given a finite-dimensional normed space ( X, k · k X ), the norm k · k X fails to bedifferentiable at x if and only if there exist α, ε > ε < ε there exist u, v ∈ S X \{± x } such thatmax {k u − x k X , k v + x k X } ≤ ε, k u − v k X ≤ − αε.
2. Given x, y, z ∈ S X such that k · k X is differentiable at z and z − y = λx for some2 >
0, the norm k · k X is differentiable at x ∈ S X if and only if G ( t ) = k γ z ( t ) − y k is differentiable at 0, where γ z : R → S X is an arc-length parameterization suchthat γ z (0) = z .With these two facts in mind, it is not too difficult to show the Tingley-type resultin this paper: Theorem 1.1.
Let ( X, k · k X ) and ( Y, k · k Y ) be two-dimensional normed spaces andlet τ : S X → S Y be an isometry. If the amount of points in S X where k · k X is notdifferentiable is finite and greater than 3, then τ is linear. With the same ideas as in the proof of Theorem 1.1, we have also been able to showthis result about, so to say, boundaries of convex open subsets of R : Theorem 1.2.
Let ( X, k · k X ) be a strictly convex normed plane such that the amount ofpoints in S X where k · k X is not differentiable is finite and greater than 3. Let C X ⊂ X be a piecewise C Jordan curve that encloses a convex set. If there is some isometry C X → C Y for some piecewise C curve C Y ⊂ Y , being ( Y, k · k Y ) another normed planethat fulfils the same that ( X, k · k X ) , then τ is affine and, so, X and Y are isometric.1.1. Notations and backgroundRemark . Given some Jordan curve C ⊂ R , we will say that C is a convex curvewhen it encloses a convex region.For any convex piecewise smooth curve C ⊂ R , it is known that there are parame-terizations γ : R → C that are smooth at t if and only if C is smooth at γ ( t ) and haveone-sided derivatives γ ′− ( t ) and γ ′ + ( t ) at every other t ∈ R . We will only consider theseparameterizations.As we will heavily use the arc-length anticlockwise parameterization of S X beginningat some point, we will denote this curve in a special way:If z belongs to some piecewise C -differentiable, convex, Jordan curve C ⊂ R , then γ z : R → C will denote the only anticlockwise parameterization of C that fulfils γ z (0) = γ z ( L ) = z , is injective when restricted to [0 , L ), is L -periodic and has k γ ′ z, − ( t ) k X = k γ ′ z, + ( t ) k X = 1 for every t ∈ R . This parameterization is also known as the naturalparameterization of C , see [2]. Definition 1.4.
Let ( X, k · k X ) be a normed space. We say that x is Birkhoff orthogonalto y , denoted as x ⊥ B y , if k x + λy k X ≥ k x k X for every λ ∈ R . We will denote x ⊥ = { y ∈ X : x ⊥ B y } . Proposition 1.5. [12, Theorem 2.2], [1, Theorem 4.12]
For any vector x in a normedlinear space X there exists a hyperplane H ⊂ X such that x ⊥ B H . Proposition 1.6. [12, Theorem 4.2], [1, Theorem 4.15]
The norm of a normed linearspace X is Gˆateaux differentiable at x ∈ X \ { } if and only if x ⊥ is a hyperplane. As for the differentiability of finite-dimensional norms, joining [22, Theorem 25.2 andCorollary 25.5.1] we obtain:
Proposition 1.7.
Let f be a convex function on an open convex set A ⊂ R d . If f hasall partial derivatives at each point of A , then f ∈ C ( A ) . Proposition 1.7 implies that the usual differences between the various kinds of differ-entiability do not exist when we deal with a convex function like k · k X : R n → R . Inparticular, Lemma 1.8.
Let ( X, k · k X ) be a finite-dimensional normed space. Then, the followingconditions are equivalent to one another: • k · k X is C -differentiable. • k · k X is Fr´echet differentiable. • k · k X is Gˆateaux differentiable. • S X is a differentiable manifold. • For each x ∈ X, x = 0 , x ⊥ is a hyperplane. • If, in addition, X is two-dimensional, then the above conditions are equivalent tothe fact that for every x ∈ S X , t ∈ R , the equality γ ′ x, − ( t ) = γ ′ x, + ( t ) holds. We will also use this Lemma that Professor Javier Alonso gifted me some years ago:
Lemma 1.9 (J. Alonso) . Let ( R , k·k X ) be a two-dimensional normed space and x ∈ S X .If y ∈ S X is a side derivative of the natural parameterization of S X at x , then x ⊥ B y . roof. We need to show that k x + λy k X ≥ λ ∈ R , with y = lim t → + γ x ( t ) − xt . We have the following: k x + λy k X = (cid:13)(cid:13)(cid:13)(cid:13) x + λ lim t → + γ x ( t ) − xt (cid:13)(cid:13)(cid:13)(cid:13) X = lim t → + (cid:13)(cid:13)(cid:13)(cid:13) x + λ γ x ( t ) − xt (cid:13)(cid:13)(cid:13)(cid:13) X =lim t → + (cid:13)(cid:13)(cid:13)(cid:13) λt γ x ( t ) + (cid:18) − λt (cid:19) x (cid:13)(cid:13)(cid:13)(cid:13) X ≥ lim t → + (cid:12)(cid:12)(cid:12)(cid:12)(cid:13)(cid:13)(cid:13)(cid:13) λt γ x ( t ) (cid:13)(cid:13)(cid:13)(cid:13) X − (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) − λt (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) X (cid:12)(cid:12)(cid:12)(cid:12) =lim t → + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) λt (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) − λt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1 , for every λ ∈ R . (1)
2. Main results
We will prove that the differentiability of k· k X at some x depends on the infinitesimalmetric structure of S X around x . Later, we will show that it can be determined by meansof computations carried away far from x . Joining both facts we will arrive at our mainresults after some extra work. Proposition 2.1.
Let ( X, k · k X ) be a finite-dimensional space. The differentiability of k · k X at any x ∈ S X depends only on the metric structure of the unit sphere ( S X , k · k X ) near x and − x . Namely, k · k X fails to be differentiable at x if and only if there exist δ, ε > such that for every ε < ε there exist u, v ∈ S X \{± x } such that max {k u − x k X , k v + x k X } ≤ ε, k u − v k X ≤ − δε. (2) Proof.
Let x ∈ S X .It is clear that if k · k X is differentiable at x then for every δ, ε > < ε < ε such that (2) cannot hold for every u, v ∈ S X \{± x } . If k · k X is not differentiable at x , then Proposition 1.5 and Lemma 1.8 imply that x ⊥ contains strictly a hyperplane, so there is some pair of independent vectors y, z ∈ S X ∩ x ⊥ such that x ∈ span { y, z } . We are going to show that there are u, v ∈ S X ∩ span { y, z } that fulfil (2), so we may suppose that X is two-dimensional and X = span { y, z } . Takingany orientation on X , we may define γ x . As the side derivatives of γ x at 0 are differentand fulfil x ⊥ B γ ′ x, − (0) and x ⊥ B γ ′ x, + (0) (Lemma 1.9), we may suppose y = γ ′ x, + (0), z = γ ′ x, − (0). 5aking into account that x ⊥ B y if and only if x ⊥ B − y , we may suppose thatthere exist λ, µ > x = − λy + µz . Consider the basis B = { x, y } . Takingcoordinates with respect to B , we have z = ( z , z ) and z = 1 /µ, z = λ/µ >
0. By thevery definition of Birkhoff orthogonality, x ⊥ B y implies k (1 , t ) k X = k x + ty k X ≥ − x ⊥ B z implies k ( − tz /z , t ) k X = k − x + tz/z k X ≥ t ∈ R . It is clearthat, moreover, k ( α, t ) k X ≥ α and k ( − α + tz /z , t ) k X ≥ α for every α >
0, so we have B X ⊂ { ( α, β ) ∈ R : α ≤ , βz /z ≤ α + 1 } . Furthermore, B X contains the convex hull of { ( − , , (0 , , (1 , } . On the one hand,this means that k · k X ≤ k · k . On the other hand, this implies that for each t ∈ ]0 ,
1[ theline { ( α, t ) : α ∈ R } intersects with S X at exactly two points u = ( a + ( t ) , t ) , v = ( a − ( t ) , t ),with − tz /z ≤ a − ( t ) ≤ − t, − t ≤ a + ( t ) ≤ . Thus, we obtain k u − v k X = k ( a + ( t ) − a − ( t ) , k X ≤ − tz /z and k u − x k X = k ( a + ( t ) − , t ) k X ≤ | a + ( t ) − | + | t | = 1 − a + ( t ) + t ≤ t, k v + x k X = k ( a − ( t ) + 1 , t ) k X ≤ | a − ( t ) + 1 | + | t | = 1 + a − ( t ) + t ≤ t. The last three inequalities end the proof, taking ε = 2 t and δ = z / (2 z ). Corollary 2.2.
Let ( X, k · k X ) , ( Y, k · k Y ) be finite-dimensional normed spaces whosespheres are isometric. Then, k · k Y is differentiable if and only if k · k X is also differen-tiable.Proof. It is straightforward from Proposition 2.1 and the fact that every onto isometrybetween finite-dimensional spheres preserves antipodes ([27, Theorem, p. 377]).
Corollary 2.3.
Let ( X, k · k X ) , ( Y, k · k Y ) be two-dimensional normed spaces and τ : S X → S Y a surjective isometry between their spheres. Then, either both spaces arepiecewise C -differentiable or none of them is. Moreover, if k · k X is piecewise C thenit is C -differentiable at x if and only if k · k Y is C -differentiable at τ ( x ) .Remark . To avoid confusion, we will use the notation ] α, β [ to denote the openinterval whose endpoints are α and β . Thus, ( α, β ) will always be a vector in R . Proposition 2.5.
Let k · k X be a strictly convex norm defined on X = R . Consider x, y, z ∈ S X , and λ ∈ ]0 , such that z = y + λx and k · k X is differentiable at z . In theseconditions, k · k X is differentiable at x if and only if G ( t ) = k γ z ( t ) − y k X is differentiable at t = 0 . In particular, the differentiability at x depends on the metricat y and around z . roof. It is clear that if k · k X is differentiable at x and z , then G is differentiable at 0because it is the composition of differentiable functions.Suppose, on the other hand, that k·k X is not differentiable at x , i.e., γ ′ x, + (0) = γ ′ x, − (0).For the sake of clarity, we will consider the basis B = {− γ ′ x, − (0) , x } so the position of x, y, z is like in Figure 1, i.e., x = (0 ,
1) and z − y = (0 , λ ). x yz xyz Figure 1: With the basis B , y and z have the same first coordinate and S X arrives at x horizontally. Let us show that G is not differentiable at 0. To this end, let z ′ and z ′ be thecoordinates of γ ′ z (0) in the basis B . With these assumptions, z ′ is negative (as in thefigure). As G (0) = λ, we have G ′− (0) = lim ε → − k γ z ( ε ) − y k X − λε · As S X is differentiable at z , k γ z ( ε ) − ( z + εγ ′ z (0)) k X = o ( ε ) , so we have G ′− (0) = lim ε → − k z + εγ ′ z (0) − y k X − λε = lim ε → − k λx + εγ ′ z (0) k X − λε =lim ε → − k (0 , λ ) + ε ( z ′ , z ′ ) k X − λε = z ′ , (3)where the last equality holds because (0 , λ ) + ε ( z ′ , z ′ ) lies in the first quadrant and isclose to (0 , λ ). In this situation, k (0 , λ ) + ε ( z ′ , z ′ ) k X = k (0 , λ + εz ′ ) k X + o ( ε ) = λ + εz ′ + o ( ε ) . So, G ′− (0) = z ′ . 7he choice of the basis has nothing to do with the value of G ′− (0). Had we computed G ′ + (0) by using the basis B = {− γ ′ x, + (0) , x } , we would have arrived at G ′ + (0) = z ′ ,where ( z ′ , z ′ ) are the coordinates of γ ′ z (0) with respect to B . What we need to see isthat z ′ = z ′ . So, consider the linear automorphism of R given by T ( a, b ) = ( a, b ) when aγ ′ x, − (0) + bx = aγ ′ x, + (0) + bx. If we had z ′ = z , then T and the map ( a, b ) ( z ′ a/z , b ) would agree at (0 ,
1) and( z , z ). Both maps are linear and these vectors form a basis, so they must be the samemap. This readily implies that S X is differentiable at x , a contradiction that ends theproof. Remark . Consider R endowed with the hexagonal norm k · k X defined as k ( a, b ) k X = (cid:26) max {| a | , | b |} if ab ≥ , | a | + | b | if ab < k · k X is not differentiable at x = (0 , y = (1 , /
3) and z = (1 , /
3) then k γ z ( t ) − y k X = 1 / t for t ∈ [ − / , / k γ z ( t ) − y k X is differentiableat t = 0. So, if k · k X is not strictly convex then Proposition 2.5 does not need to hold. Questions . Can Proposition 2.5 be generalised to finite-dimensional spaces or arbi-trary dimension?Is Proposition 2.5 true if we replace differentiability by C -differentiability?2.8 . As we will often need to refer to differentiability and non-differentiability points, forthe sake of readability we will denote D( C X ) (resp., ND( C X )) the sets where a curve C X is differentiable (resp., where it is not differentiable) so we will write x ∈ D( C X ) (resp., x ∈ ND( C X )) instead of x is a differentiability (resp., non-differentiability ) point of C X . Before we proceed with our main results, we need these technical Lemmas:
Lemma 2.9.
Let C ⊂ R be a Jordan curve that encloses a convex region and supposethat the extreme points are all different –that is, there are some c = ( c , c ) , c = ( c , c ) ,c = ( c , c ) , c = ( c , c ) ∈ C such that c = min { c : ( c , c ) ∈ C } , c = min { c : ( c , c ) ∈ C } ; c = max { c : ( c , c ) ∈ C } , c = max { c : ( c , c ) ∈ C } ; c < min { c , c , c } , c < min { c , c , c } , c > max { c , c , c } , c > max { c , c , c } . Then, for any x = ( x , x ) ∈ R there are u = ( u , u ) , v = ( v , v ) , w = ( w , w ) ∈ C,t = 0 such that u − v = tx, w = u and w = v . roof. If x = 0 then we may take any u ∈ C and w = v = u, t = 1 . So, we only need toshow that the result holds for x = 0. It is clear that we may suppose k x k ∞ = 1, so wewill show that for every x ∈ S ∞ there exist t, u, v, w as in the statement. If x = (1 ,
0) wejust need to find u, v ∈ C that belong to the same horizontal line (in this case, w = u )and if x = (0 ,
1) it suffices to find u, v ∈ C in the same vertical line and take w = v , sowe may suppose x x = 0. Suppose that x , x >
0, the other cases are analogous.As C is convex, there is exactly one point or segment at the undermost end of C .Suppose it is just one point, say w = c , and analyse what happens when we move alongthe curve anticlockwise until we reach the rightmost point or segment of C , suppose againthat it is a singleton, say w = c . As w = w , we may consider some anticlockwiseparameterization of C that has γ (0) = w and γ (1) = w , we will denote w s = γ ( s ).It is clear that, for any s ∈ ]0 , w s is the undermost point of the intersection of C with the vertical line where it lies. Denote u s the uppermost point of this intersection.Analogously, w s is the rightmost intersection of C with the horizontal line where it lies,we will denote its leftmost point as v s . What we need to see is that for every proportion x /x there is some w s such that ( w s − v s ) / ( u s − w s ) = x /x .But the map s ∈ ]0 , ( w s − v s ) / ( u s − w s ) is continuous and its limits are 0 at 0and ∞ at 1. So, at some s ∈ ]0 ,
1[ we get the desired equality.If instead of one point there is a segment at the bottom of C then we take w as theleftmost point of this segment, if there is one segment at the rightmost end of C then w is at the top of the segment and everything goes undisturbed. Lemma 2.10.
Consider R endowed with the norm k ( λ, µ ) k = | λ | + | µ | . Let C ⊂ R be aconvex Jordan curve that does not fulfil the conditions of Lemma 2.9 because some point,say c , is extreme in two directions. For each a ∈ C , consider the sequence ( a n ) n ⊂ C defined as a = a and, for n ≥ , a n +1 is the closest point from c that shares somecoordinate with a n . In these conditions, ( a n ) n → c unless a is the strict extreme in thetwo other directions, in which case a n = a, ∀ n ∈ N . . If there are two possible choices for a given a n +1 then we choose the point lying inthe same vertical line as a n . Proof.
If the conditions in the statement are fulfilled, then it is clear that ( a n ) n has someaccumulation point because C is compact and the accumulation point must be its limitbecause ( a n ) n is monotonic in both coordinates. The only possible limit is c , so we aredone. Theorem 2.12.
Let ( X, k· k X ) be a two-dimensional normed space whose unit sphere S X is piecewise C -differentiable and has at least two points x = ± x, with x, x ∈ ND( S X ) .For any normed plane ( Y, k · k Y ) , every isometry τ : S X → S Y is linear. roof. If X is not strictly convex then the result holds by [4, Corollary 3.8], so we maysuppose that k · k X is strictly convex.Suppose there are linearly independent x, x ∈ ND( S X ), consider the basis B X = { x, x } , and let τ : S X → S Y be an onto isometry. The only point in S X at distance2 from x is − x , so it is obvious that τ ( − x ) = − τ ( x ) and we obtain that τ ( x ) and τ ( x ) are linearly independent so we may consider the basis B Y = { τ ( x ) , τ ( x ) } . Takingcoordinates with respect to B X and B Y , we have x = (1 , X , τ ( x ) = (1 , Y , so both x and − x lie on the same horizontal line and τ ( x ) and τ ( − x ) do, too. We are going tosee that this happens to τ ( u ) , τ ( v ) ∈ S Y for any couple u = ( u , u ) , v = ( v , v ) ∈ S X such that u > v and v = u , i.e., such that u − v = λx for some λ > λ = k u − v k X ∈ ]0 , ⊥ x be the only point in S X such that ⊥ x ⊥ B x andwhose second coordinate is negative ( ⊥ x is unique because k · k X is strictly convex, see[1, Theorem 4.15]). We will denote as C the (relative) interior of the arc of S X that joins ⊥ x with − ⊥ x and contains x , observe that S X = C ∪ − C ∪ {± ⊥ x } . Consider C x = { u ∈ C : τ ( u ) − τ ( v ) = λτ ( x ) if u − v = λx, λ ∈ ]0 , } We are going to show that C x = C , so we will have the equivalence u − v = λx ⇔ τ ( u ) − τ ( v ) = λτ ( x ) . If ( u n ) n → u ∈ C and u n ∈ C x for every n ∈ N , then consider the correspondingsequences ( v n ) n , ( λ n ) n . We have, for every n ∈ N , u n − v n = λ n x and τ ( u n ) − τ ( v n ) = λ n τ ( x ) . (4)It is clear that both ( v n ) n , ( λ n ) n must converge and that u − v = λx and τ ( u ) − τ ( v ) = λτ ( x ) , where λ = lim( λ n ) , v = lim( v n ). So, u ∈ C x and C x is, therefore, closed in C .Suppose now that ( u n ) n → u and u ∈ C x . Take λ, v and ( λ n ) n , ( v n ) n such that u − v = λx, u n − v n = λ n x . As there are finitely many points in ND( S X ), we maysuppose that none of them is u n or v n . In this situation, Proposition 2.1 implies that S Y is differentiable at τ ( u n ) and so, Proposition 2.5 implies that τ ( u n ) − τ ( v n ) = λ n y n , with y n ∈ ND( S Y ). As ND( S Y ) is finite, there is some y that appears infinitely many times in( y n ) n , so passing to a subsequence we may suppose that τ ( u n ) − τ ( v n ) = λ n y for every n ∈ N . Of course, lim( τ ( u n )) n = τ ( u ), lim( τ ( v n )) n = τ ( v ) and lim( λ n ) n = λ , so λy = lim( τ ( u n ) − τ ( v n )) n = τ ( u ) − τ ( v ) = λτ ( x ) , we obtain that y = τ ( x ) and this means that C x is open.10e have seen that C x is non-empty –because x ∈ C x –, closed and open, so theconnectedness of C shows that C x = C .So, u − v = λx implies τ ( u ) − τ ( v ) = λτ ( x ). Of course, the same applies to x , so whatwe actually have is that u − v = λx + µx implies τ ( u ) − τ ( v ) = λτ ( x ) + µτ ( x ) wheneverthere exists w ∈ S X such that either w = v + µx = u − λx or w = v + λx = u − µx .Now we have two options. If we are in the hypotheses of Lemma 2.9, then w existsfor every possible direction, and from the fact that τ is an isometry, we get k λτ ( x ) + µτ ( x ) k Y = k λx + µx k X . As we have taken coordinates with respect to { x, x } and { τ ( x ) , τ ( x ) } , we get k ( λ, µ ) k Y = k ( λ, µ ) k X . This means that in these coordinates we have k·k Y = k·k X , so [4, Theorem 2.3]implies that τ is linear.If we cannot apply Lemma 2.9, then there is some c ∈ S X that is strictly extremalin two different directions. So, we may apply Lemma 2.10 to show that, given any a = ± c ∈ S X , the sequence ( τ ( a n )) n is the same as the sequence (( τ ( a )) n ) n , i.e., thesequence originated in τ ( a ). This implies that for every n ∈ N , and with the coordinatestaken again with respect to { x, x } and { τ ( x ) , τ ( x ) } we have τ ( a n ) − τ ( a ) = a n − a . As( a n ) n → c , this means that for every a ∈ C, we have τ ( a ) − τ ( c ) = a − c . From here wereadily see that τ is linear in S X and this completes the proof. Remark . If the only non-differentiability points in S X are ± x , then it is clear fromthe previous proof that u − v = λx implies τ ( u ) − τ ( v ) = λτ ( x ), but we have notbeen able to infer from here that τ must be linear. Taking any basis B X = { x, x } andconsidering B y = { τ ( x ) , τ ( x ) } we have, in coordinates, τ ( α, β ) − τ ( α ′ , β ) = ( α − α ′ , α, β ) , ( α ′ , β ) ∈ S X but we have not been able to deduce anything for pointswith different second coordinates. Theorem 2.14 (Mankiewicz Property) . Let ( X, k · k X ) , ( Y, k · k Y ) be strictly convexnormed planes such that both ND( S X ) and ND( S Y ) are finite and contain more thanthree points. Let C X ⊂ X be a piecewise C Jordan curve that encloses a convex set. Ifthere is some isometry C X → C Y for some piecewise C curve C Y ⊂ Y , then τ is affineand X and Y are isometric.Proof. First of all, we hasten to remark that in strictly convex spaces, if three points z , z , z fulfil k z − z k = k z − z k + k z − z k , then z belongs to the segment whoseendpoints are z and z , we will denote this segment as [ z , z ]. With this, it is not hardto see that a curve C encloses a convex region if and only if for every triple of collinearpoints z , z , z ∈ C , the curve C contains the segment [ z , z ]. So, our hypotheses implythat C Y is convex, too. 11he first we need to show is that Corollary 2.3 and Proposition 2.5 still apply inthis situation, i.e., that if u , v ∈ C X fulfil that ( u − v ) / k u − v k X ∈ ND( S X ),then for every u, v ∈ C X we have the equivalence u − v = λ ( u − v ) if and only if τ ( u ) − τ ( v ) = λ ( τ ( u ) − τ ( v )).For the equivalent of Corollary 2.3, we have to make do without − x , but the onlything really useful of having − x was that x ∈ D( S X ) if and only if − x ∈ D( S X ). In anycase, we are going to show that x ∈ D( C X ) is equivalent to τ ( x ) ∈ D( C Y ). As the proofis going to be quite different, we will denote the point as a instead of x .Let a ∈ C X and let us analyse the setNDif( a ) = { b ∈ C X : k γ a ( t ) − b k X is not differentiable at t = 0 } , observe that one has a ∈ NDif( a ) for every a ∈ C X .If a ∈ D( C X ), then it is clear that a = b ∈ NDif( a ) implies ( a − b ) / k a − b k X ∈ ND( S X )no matter whether b ∈ D( C X ) or not. As there are only finitely many points in ND( S X ),say ND( S X ) = { x , . . . , x m } , when b ∈ NDif( a ) one has b − a = k b − a k X x i for some i ∈ { , . . . , m } –we are considering as unrelated points x i and − x i . Claim . For any a ∈ D( C X ), NDif( a ) contains, at most, one segment and finitely manyisolated points. If it contains one segment, then one of its endpoints is a . Proof.
We need to show that for every i ∈ { , . . . , m } , there is at most one point inNDif( a ) that can be written as b − a = k b − a k X x i unless there is a segment thatfulfils it. Indeed, if b , b fulfil b − a = k b − a k X x i and b − a = k b − a k X x i with k b − a k X < k b − a k X , then b lies in the interior of the segment [ a, b ] –i.e., the closedsegment whose endpoints are a and b . As a, b , b ∈ C X and C X encloses a convexregion, the segment [ a, b ] is included in C X and it is obvious that there is only onesegment included in C X that has a as its endpoint –recall that C X is differentiable at a .If a is interior to some segment, then no more segments can arrive to a and it is clearthat k γ a ( t ) − b k X is differentiable at 0 for any point in the same segment.If we have, instead, a ∈ ND( C X ) , then NDif( a ) includes every b ∈ C X such that( a − b ) / k a − b k X ∈ D( S X ). Indeed, let x = ( a − b ) / k a − b k X ∈ D( S X ) and consider ¯ x asany of the two opposite vectors in S X such that x ⊥ B ¯ x –i.e, ¯ x = ± γ ′ x (0) ∈ S X . Withthe basis B X = { ¯ x, x } , the sphere S X and the line { ( λ,
1) : λ ∈ R } are tangent. Thisimplies that, for every µ ∈ ] − , { ( λ, µ ) : λ ∈ R } meets S X in two points,say b = ( b , b ) , c = ( c , c ), and the first coordinates of these points have different sign.Moreover, as k · k X is strictly convex, { ( λ,
1) : λ ∈ R } ∩ S X = { x } . Both these facts willbe important later.Denote a ′− , a ′ + the (different) side derivatives of C X in a . In the basis B X = { ¯ x, x } ,we have a ′− = ( a ′− , , a ′− , ) , a ′ + = ( a ′ + , , a ′ + , ) . The speed of growing of k γ a ( t ) − b k X as we12re arriving at a in the direction of a ′− is a ′− , and the speed of growing in the directionof a ′ + is a ′ + , . This can be seen as in Proposition 2.5 or by thinking this situation as ifwe had partial derivatives: k γ a ( t ) − b k X would grow at speed 1 if γ ′ a (0) = x = (0 ,
1) andthe speed would be 0 if γ ′ a (0) = ¯ x = (1 , a , a ) we havespeed a . So, we need to show that a ′− , = a ′ + , . As C X encloses a convex region, thesigns of a ′− , and a ′ + , are the same –maybe one of them is zero. So, if a ′− , = a ′ + , , thenwe would have two points in S X with the same second coordinate in the same quadrant,but we have just seen that this cannot happen.This means that NDif( a ) contains every point in C X but, at most, two segments thatinclude a and finitely many other points. In particular, there exist some open U ⊂ C X such that U ⊂ NDif( a ) and U ∪ { a } is not contained in a metric segment.Gathering all these facts, we obtain that the metric structure of NDif( a ) determinesthe differentiability of C X at a –and this implies that τ ( a ) ∈ D( C Y ) if and only if a ∈ D( C X ).As for the analogous of Proposition 2.5, we need to show that ( u − v ) / k u − v k X ∈ ND( S X ) implies ( τ ( u ) − τ ( v )) / k τ ( u ) − τ ( v ) k Y ∈ ND( S Y ).So, let x = ( u − v ) / k u − v k X ∈ ND( S X ) and consider, as in the proof of Proposition 2.5,the basis B = {− γ ′ x, − (0) , x } . We have u and v in the same vertical line and u is over v .In coordinates, u = v , u > v .Suppose that u ∈ D( C X ) and that there is no segment in C X that contains u and v .Then, the map G ( t ) = k γ u ( t ) − v k X is not differentiable at t = 0, the proof is the same as the one in Proposition 2.5.In the proof of Theorem 2.12, we defined C as the interior of one of the arcs that join ⊥ x and − ⊥ x . The analogous way to define this is by taking C as the interior of the arcthat joins the lowermost point or segment in C X with its uppermost point or segmentpassing through the right part of C X –so, C does not include any of its endpoints.Later, we defined C x = { u ∈ C : τ ( u ) − τ ( v ) = λτ ( x ) if u − v = λx, λ > } , but now we do not have τ ( x ), so we need to modify the definition of the set C x . Forthis, there is an equivalent way to state u − v = λx and τ ( u ) − τ ( v ) = λτ ( x ). We cantake u ∈ C, v ∈ C X such that u − v = λ x with λ >
0, eliminate the condition u − v = λx by writing u − λx instead of v and define our new subset as any of thefollowing equivalent ways: C x = { u ∈ C : τ ( u ) − τ ( u − λx ) = λ ( τ ( u ) − τ ( v )) /λ , λ > } , C x = { u ∈ C : τ ( u ) − τ ( u − λx ) = λ ( τ ( u ) − τ ( u − λ x )) /λ , λ > } . C x is open and closed in C . Again, C x is notempty because u ∈ C x , so C x = C . It is clear that if for every u in C X such that there isexactly one v ∈ C X such that u − v = k u − v k X x one has τ ( u ) − τ ( v ) = λ ( τ ( u ) − τ ( v ))for some λ >
0, we have the same when u belongs to a segment whose direction is x . So,denoting y = ( τ ( u ) − τ ( v )) / k τ ( u ) − τ ( v ) k Y we have u − v = k u − v k X x if and onlyif τ ( u ) − τ ( v ) = k τ ( u ) − τ ( v ) k Y y .If we consider x ∈ ND( S X ) , x = ± x and u , v ∈ C X such that u − v = k u − v k X x ,then the same argument as before shows that u − v = k u − v k X x is equivalent to τ ( u ) − τ ( v ) = k τ ( u ) − τ ( v ) k Y y , with y = ( τ ( u ) − τ ( v )) / k τ ( u ) − τ ( v ) k Y .With this, if we consider the bases { x, x } and { y, y } , we have the equivalences u − v = ( k u − v k X ,
0) if and only if τ ( u ) − τ ( v ) = ( k u − v k X , ,u − v = (0 , k u − v k X ) if and only if τ ( u ) − τ ( v ) = (0 , k u − v k X ) . Now we have two options: if we can apply Lemma 2.10 then the remainder of theproof goes as the last part of the proof of Theorem 2.12. Otherwise, we can applyLemma 2.9 to obtain that, in the bases { x, x } and { y, y } , we have k · k X = k · k Y . Itremains to show that τ is affine.Now, we may suppose that Y = X and we need to show that τ ( a ) − τ ( b ) = a − b forevery a, b ∈ C X .In what followswe suppose that C X has no horizontal nor vertical segment. Ananalogous idea gives a proof for the other cases. Observe that if a and b belong to thesame horizontal or vertical segment, then we have τ ( a ) − τ ( b ) = a − b . We will denoteby W, S, E and N respectively the leftmost, the undermost, rightmost and uppermostpoints in C X . We will also denote as SW, SE, N E and
N W the closed arcs that joineach pair of consecutive extremal points.We will denote E a = { b ∈ C X : τ ( a ) − τ ( b ) = a − b } . So, we need to show that E a = C X for some (every) a ∈ C X .For any a ∈ C X , denote a = a ; a ∈ C X is the other point that lies in the samehorizontal line as a , a ∈ C X lies in the same vertical line as a and so on. Furthermore,let a − ∈ C X be the point that lies in the same vertical line as a , a − ∈ C X is in thesame horizontal line as a − . . . This bi-infinite sequence may hit some extremal point and,so to say, get stuck –but this changes nothing. Moreover, if C X has some vertical orhorizontal symmetry, then a = a = a − . Still, everything goes fine. It is clear that,given n, m ∈ Z , one has τ ( a n ) − τ ( a m ) = a n − a m . The continuity of the isometry isenough to ensure τ (lim( a n k )) − τ (lim( a m k )) = lim( a n k ) − lim( a m k )for any convergent subsequences ( a n k ) , ( a m k ). So, τ ( c ) − τ ( c ) = c − c whenever c , c ∈ { a n : n ∈ Z } , i.e., { a n : n ∈ Z } ⊂ E a .14o end the proof we need three more facts. Claim . For a, b ∈ C X , if there are i , i , i , i , j , j , j , j ∈ Z such that each a i k − b j k lies in the k -th quadrant, then b ∈ E a or, equivalently, E b = E a . Proof.
We may suppose, after composition with some translation, that τ ( a ) = a . If τ ( a ) − τ ( b ) = a − b , then there is some v = 0 such that τ ( b n ) = b n + v for every n ∈ Z .As k · k X is strictly convex, there is a half-plane H such that k u − v k X > k u k X whenever u ∈ H . Namely, with w = ⊥ v , H = { αv + βw : α ∈ ] − ∞ , , β ∈ R } . In particular, thereis a whole quadrant included in H , so for some k ∈ { , , , } one has k τ ( a i k ) − τ ( b j k ) k X = k a i k − b j k − v k X > k a i k − b j k k X , a contradiction with the fact that τ is an isometry. Claim . For every a ∈ C X , the intersection of { a n : n ∈ Z } with every arc SW, SE, N E and
N W is nonempty.
Proof.
Let us see that SW ∩ { a n : n ∈ Z } 6 = ∅ , the other cases follow by symmetry. As C X is convex, some point b = ( b , b ) belongs to the arc SW if and only if there is nopoint c = ( c , c ) ∈ C X such that c ≤ b , c ≤ b and c = b . Given a ∈ C X \ SW , either a or a − –maybe both– has a coordinate that is smaller than that of a , say a < a .If a SW , then a < a and so on. If there is no n such that a n ∈ SW , then thesequence has an accumulation point, but this accumulation point must be the limit ofthe sequence because the sequence is bounded and nonincreasing in both coordinates.So, we are in the conditions of Lemma 2.10, a contradiction. Claim . For S = ( S , S ) –the undermost point of C X – we have either SW ⊂ E S or SE ⊂ E S . Proof.
Let b = ( S , b ) ∈ C X be the other point with the same first coordinate as S . Itis clear that b ∈ E S . Given c = ( c , c ) ∈ C X with c < b , take c = ( c , c ) ∈ C X with,say, c < c . We have c ∈ E c and moreover b − c , b − c , S − c , S − c are, respectivelyin the first, second, third and fourth quadrants. Claim 2 implies that c , c ∈ E S so, forevery c = ( c , c ) ∈ C X such that c ≤ b one has c ∈ E S . As C X encloses a convexregion, E < b < W implies b ≥ min { E , W } . Now we may suppose E ≤ b , thatimplies u ≤ b for every ( u , u ) ∈ SE , so SE ⊂ E S .Now we just need to use Claims 3 and 4 to see that E S = C X , so we have finishedthe proof. Remark . After Theorem 2.12, [2, Theorem 1.5] and [4, Corollary 3.8], the onlypossibility for the existence of a nonlinear isometry between two-dimensional spheres isthat both of them are strictly convex and one of the following holds:15
Both spheres are C -differentiable and at least one of them is not absolutelysmooth. • None of the spheres is piecewise differentiable, i.e., there are infinitely many pointsof non-differentiability in each sphere. • Both spheres have exactly two points of non-differentiability, say, x and − x . Acknowledgements
I would like to thank Professor Tar´as Banakh and my colleagues Daniel Moralesand Jos´e Navarro for some valuable discussions regarding the two-dimensional Tingley’sProblem.It is a pleasure to thank Professor Javier Alonso for the Gift 1.9.I absolutely need to thank the anonymous referee for their fantastic reports. Thework reads much better because of these reports and, in particular, the proofs of Propo-sition 2.1, Proposition 2.5 and Theorem 2.14 owe this referee a great debt.Supported in part by Junta de Extremadura programs GR-15152 and IB-16056 andDGICYT projects MTM2016-76958-C2-1-P and PID2019-103961GB-C21 (Spain).
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