Angular measures and Birkhoff orthogonality in Minkowski planes
aa r X i v : . [ m a t h . M G ] S e p Angular measures and Birkhoff orthogonality inMinkowski planes ∗ Márton Naszódi † Vilmos Prokaj ‡ Konrad Swanepoel § Abstract
Let x and y be two unit vectors in a normed plane R . We say that x is Birkhoff orthogonal to y if the line through x in the direction y supports the unit disc. A B-measure (Fankhänel 2011) is an angularmeasure µ on the unit circle for which µ ( C ) = π/ C is ashorter arc of the unit circle connecting two Birkhoff orthogonal points.We present a characterization of the normed planes that admit a B-measure. Let K be an o -symmetric convex body on the plane, and consider the normedplane ( R , k·k K ), where k x k K = min { λ > x ∈ λK } for any x ∈ R .Let x and y be two non-zero vectors in R . We say that x is Birkhofforthogonal to y , and denote it by x ⊣ y , if k x k K ≤ k x + ty k K for all t ∈ R .Geometrically, it means that the line through x in the direction y supports k x k K K . In general, Birkhoff orthogonality is not a symmetric relation.Normed planes where Birkhoff orthogonality is symmetric are called Radonplanes and the boundaries of their unit balls
Radon curves (see the survey[MS06]).A Borel measure µ on bd K is called an angular measure , if µ (bd K ) = 2 π , µ ( X ) = µ ( − X ) for every Borel subset X of bd K , and µ is continuous , thatis, µ ( { x } ) = 0 for every x ∈ bd K . Clearly, for any K , there is an angu-lar measure, as the one-dimensional Lebesgue measure on bd K normalizedto 2 π is one, but has no relation to the geometry of ( R , k·k K ). A natural ∗ Part of the research was carried out while MN was a member of János Pach’s chairof DCG at EPFL, Lausanne, which was supported by Swiss National Science FoundationGrants 200020-162884 and 200021-175977, and while MN and KS visited the Mathemat-ical Research Institute Oberwolfach in their Research in Pairs programme. MN also ac-knowledges the support of the National Research, Development and Innovation Fund grantK119670. † Alfréd Rényi Inst. of Math. and Dept. of Geometry, Eötvös Loránd University,Budapest ‡ Department of Probability Theory and Statistics, Eötvös Loránd University, Budapest § Department of Mathematics, London School of Economics µ is called a B-measure [Fan11] if µ ( C ) = π/ C of bd K that contains no opposite points of bd K , andwhose end points x and y satisfy x ⊣ y .The main result, Theorem 1, of this note is a characterization of thenormed planes ( R , k·k K ) which admit a B-measure.We call a point x in bd K an Auerbach point , if there is a y ∈ bd K suchthat x ⊣ y and y ⊣ x . It is well known that Auerbach points exist for anynorm [Tho96, Section 3.2]. We denote the set of Auerbach points of K by A ( K ). Note that A ( K ) is a closed subset of bd K . We denote the union ofopen non-degenerate line segments contained in bd K by E ( K ). Theorem 1.
Let K be an origin-symmetric convex body in R . Then thereis a B-measure on bd K if, and only if, the set A ( K ) \ E ( K ) is uncountable. This result strengthens Fankhänel’s [Fan11, Theorem 1], where the exis-tence of a B-measure is shown under the condition that A ( K ) \ E ( K ) containsan arc. (Fankhänel does not explicitly exclude line segments, but it is clearthat they have to be excluded, as line segments in A ( K ) necessarily havemeasure 0 for any B-measure; see Lemma 3.) We prove Theorem 1 in Sec-tion 2, where we also present a smooth, strictly convex, centrally symmetricplanar body K such that A ( K ) is the union of two disjoint copies of theCantor set and a countable set of isolated points (Example 4). Thus, A ( K )is of Lebesgue measure zero and yet, by Theorem 1, there is a B-measure onbd K .We recall that a subset of a topological space is called perfect , if it isclosed and has no isolated point. Recall that the support supp( µ ) of a Borelmeasure µ on a topological space X is the set of all x ∈ X such that all opensets containing x have positive µ -measure. In the proof of Theorem 1, werely on the following result. Proposition 2.
Let H ⊂ [0 , be a non-empty, closed, perfect set. Thenthere is a continuous probability measure on [0 , whose support is H . This is a well-known result holding more generally for any separable com-plete metric space [Par05, Chapter II, Theorem 8.1], but for the convenienceof the reader we present an explicit construction in Section 3. It is easyto see that the converse of Proposition 2 holds, namely that the support ofany continuous measure on [0 ,
1] is a perfect set. It is well known that ev-ery non-empty perfect set is uncountable [Rud76, Theorem 2.43] and everyuncountable closed set contains a perfect set [Kec95, Section 6B]. (More gen-2rally, all Borel sets and analytic sets have this property [Kec95], but we willonly need it for F σ sets). Given two non-opposite points a, b ∈ bd K , we denote by ∢ ( a, b ) the closedarc from a to b that does not contain any opposite pairs of points. We denotethe closed line segment with endpoints a, b ∈ R by [ a, b ]. Lemma 3.
Let K be an origin-symmetric convex body in R and µ be aB-measure on bd K . Then supp( µ ) ⊆ A ( K ) \ E ( K ) .Proof. Consider a non-degenerate line segment [ x − , x + ] ⊂ bd K . Let y ∈ bd K parallel to [ x − , x + ]. Since x − , x + ⊣ y , we have µ ([ x + , y ]) = µ ([ x − , y ]) = π/
2. Thus, µ ([ x − , x + ]) = 0, and hence, no x strictly between x − and x + isin supp( µ ).Next, let x be a point in bd K \ A ( K ). Let y , y ∈ bd K such that x ⊣ y and y ⊣ x . Then y = y . By possibly replacing y by − y , we assumewithout loss of generality that y and y are in the same open half planebounded by the line ox . By possibly replacing x by − x , we may also assumewithout loss that y and x are in the same open half plane bounded by oy .Let x and x be points on the same side of oy as x such that y ⊣ x and x ⊣ y . Then x , x = x . Because y is between x and y , we havethat x and x are in opposite open half planes bounded by ox . Since µ isa B-measure, µ ( ∢ ( x, x )) = µ ( ∢ ( x, x )) = 0. Thus, µ ( ∢ ( x , x )) = 0, hence x / ∈ supp( µ ). Proof of Theorem 1.
Let µ be a B-measure. Since supp( µ ) is a perfect set,hence uncountable, Lemma 3 gives that A ( K ) \ E ( K ) is uncountable.Conversely, assume that A ( K ) \ E ( K ) is uncountable. Define a map φ : A ( K ) \ E ( K ) → bd K by setting φ ( x ) to be the first y ∈ A ( K ) in thepositive direction along bd K from x such that x ⊣ y and y ⊣ x . Then φ is monotone, but not necessarily injective. However, if φ ( x ) = φ ( x ), then x ⊣ y and x ⊣ y , hence [ x , x ] is a line segment on bd K . It follows that forany given y ∈ bd K , there are at most two values of x ∈ A ( K ) \ E ( K ) suchthat φ ( x ) = y , and there are at most countably many y ∈ bd K for whichthere is more than one x ∈ A ( K ) \ E ( K ) such that φ ( x ) = y . Thus, φ is aBorel measurable map.For any x ∈ bd K , let x + denote the first element of A ( K ) \ E ( K ) inthe positive direction from x , and similarly, let x − be the first element of A ( K ) \ E ( K ) in the negative direction from x .Fix a ∈ A ( K ) \ E ( K ). Let b = φ ( a ) and a ′ = φ ( b + ). If b + = b then a ′ is not strictly between − a and b . If b + = b and a ′ is strictly between − a and b , then [ − a, a ′ ] is a line segment on bd K . In either case, we obtainthat ∢ ( a, b ) ∩ A ( K ) \ E ( K ) or ∢ ( b + , a ) ∩ A ( K ) \ E ( K ) is uncountable. Thuswe may assume without loss of generality that ∢ ( a, b ) ∩ A ( K ) \ E ( K ) is3ncountable. It then contains a perfect set H and there is a continuousprobability measure ν on the Borel sets of bd K with support H , of whichthe existence is guaranteed by Proposition 2. We use ν to define the B-measure as follows. For any Borel set A ⊂ bd K , let µ ( A ) = π (cid:2) ν ( A ) + ν ( − A ) + ν ( φ − ( A )) + ν ( φ − ( − A )) (cid:3) . (1)Then µ is clearly an angular measure. To see that µ is a B-measure, let x, y ∈ bd K with x ⊣ y . By possibly replacing x by − x and y by − y , we mayassume that x ∈ ∢ ( a, b ) or x ∈ ∢ ( b, − a ), and that y ∈ ∢ ( a, b ) ∪ ∢ ( b, − a ). Case 1: x ∈ ∢ ( a, b ). Then either y ∈ ∢ ( a, b ) or y ∈ ∢ ( b, − a ) \ { b } . Case 1.1: y ∈ ∢ ( a, b ). Then there are two cases depending on the relativeposition of x and y . Case 1.1.1: x ∈ ∢ ( a, y ). In this case necessarily y = b or x = a . If y = b then [ a, x ] is a segment on bd K , hence µ ( ∢ ( a, x )) = 0 and µ ( ∢ ( x, y )) = π/ x = a and y = b , then we show that ∢ ( y, b ) ∩ A ( K ) is finite. Suppose thereexist y ′ ∈ ∢ ( y, b ) ∩ A ( K ) \ { b, y } . Then there exist a ′ ∈ ∢ ( a, − b ) \ { a } suchthat a ′ ⊣ y ′ , y ′ ⊣ a ′ . However, since a ⊣ y and a ⊣ b , we obtain a ⊣ y ′ , hence[ a ′ , a ] is a line segment on bd K and y ′ = y . Therefore, µ ( ∢ ( y, b )) = 0. Case 1.1.2: x ∈ ∢ ( y, b ). In this case necessarily x = b or y = a . This caseis finished in a similar way as Case 1.1.1. Case 1.2: y ∈ ∢ ( b, − a ) \ { b } . We have to show that φ − ( ∢ ( b, y )) = ∢ ( a, x ) ∩ A ( K ) \ E ( K ) up to measure 0. Case 1.2.1: x / ∈ A ( K ) \ E ( K ). Then φ ( x − ) ∈ ∢ ( b, y ) and φ ( x + ) ∈ ∢ ( y, − a ).Note that φ − ( ∢ ( b, φ ( x − ))) = ∢ ( a, x ) ∩ A ( K ) \ E ( K ) ∪ φ − ( { φ ( x − ) , b } ), where φ − ( { φ ( x − ) , b } ) contains at most 4 points. Similarly, φ − ( ∢ ( b, φ ( x + ))) = ∢ ( a, x ) ∩ A ( K ) \ E ( K ) ∪ φ − ( { φ ( x + ) , b } ), and it follows that φ − ( ∢ ( b, y )) = ∢ ( a, x ) ∩ A ( K ) \ E ( K ) up to 2 points. Case 1.2.2: x ∈ A ( K ) \ E ( K ). We show that ∢ ( φ ( x ) , y ) ∩ A ( K ) is finite,from which follows that φ − ( ∢ ( b, y )) equals ∢ ( a, x ) up to finitely many points.Suppose that there exists y ′ ∈ ∢ ( φ ( x ) , y ) ∩ A ( K ) \ { φ ( x ) , y } . Case 1.2.2.1: y ∈ ∢ ( b, φ ( x )). Then there exists x ′ ∈ ∢ ( a, x ) such that x ′ ⊣ y ′ , y ′ ⊣ x ′ . Since x ⊣ φ ( x ) and x ⊣ y , we have x ⊣ y ′ . It follows that[ x, x ′ ] is a segment on bd K parallel to y ′ , but then y = y ′ unless x = x ′ , butthen φ ( x ) = y ′ . Case 1.2.2.2: φ ( x ) ∈ ∢ ( b, y ). Similar to Case 1.2.2.1 we obtain that x = x ′ ,and then φ ( x ) ⊣ x and y ′ ⊣ x , hence [ φ ( x ) , y ′ ] is a segment on bd K parallelto x . This holds for all y ′ ∈ A ( K ) ∩ ∢ ( φ ( x ) , y ) \ { φ ( x ) , y } . It follows that φ − ( ∢ ( φ ( x ) , y )) ⊆ φ − ( { φ ( x ) , y } ), which consists of at most 4 points. Case 2: x ∈ ∢ ( b, − a ) \ { b, − a } . If y ∈ ∢ ( b, − a ), then as in Case 1.1, y = b or y = − a . But then, by possibly replacing y with − y , we can assume that y ∈ ∢ ( a, b ). Thus without loss of generality, y ∈ ∢ ( a, b ). Then φ ( y + ) ∈ ∢ ( x, − a )and φ ( y − ) ∈ ∢ ( b, x ). Then φ − ( ∢ ( b, φ ( y )), which equals ∢ ( a, y − ) ∩ A ( K ) \ ( K ) up to 2 points, is contained in φ − ∢ ( b, x )), which is in turn containedin φ − ( ∢ ( b, φ ( y + ))), which equals ∢ ( a, y + ) ∩ A ( K ) \ E ( K ) up to 2 points. Itfollows that ν ( φ − ( ∢ ( b, x )) = ν ( ∢ ( a, y )).This completes the proof of Theorem 1. Example 4.
We present a smooth, strictly convex, centrally symmetric pla-nar body K such that A ( K ) is the union of two disjoint copies of the Cantorset and a countable set of isolated points.First, let K be the Euclidean unit disk centered at the origin, and let C denote the shorter arc connecting the two points whose angles with thepositive x axis are − π/ π/
4. Let C denote the Cantor set in C . Now, C can be written as C = C \ ∞ [ n =1 I n , where the I n are disjoint open arcs in C .For each n ∈ Z + , we construct a smooth and strictly convex curve C n connecting the two endpoints of I n with the following properties.1. C n has the same tangents at the endpoints as K ;2. C n is contained in conv I n ;3. for any point x of C n which is neither an end point, nor the midpointof C n , the tangent of C n at x is not orthogonal to x .Let Ψ ( x ) = exp (cid:16) − − x (cid:17) for x ∈ ( − , bump function . It is well known that Ψ is non-negative, smooth, itssupport is [ − , − , / α < α denote the angles for which the two end points of I n havepolar coordinates ( α ,
1) and ( α , C n be the curve whose polar coor-dinates are given for each ϕ ∈ [ α , α ] by (cid:18) ϕ, − εΨ (cid:18) α − α (cid:20) ϕ − α + α (cid:21)(cid:19) (cid:19) , where ε > C n is a smooth curve, and if ε is small, then it is also strictlyconvex. Moreover, C n has the first property, as Ψ ′ ( −
1) = Ψ ′ (1) = 0. If ε issmall, then C n has the second property as well. Finally, to verify the thirdproperty, observe that the tangent of C n is orthogonal to x at a point x of C n if and only if, the derivative of1 − εΨ (cid:18) α − α (cid:20) ϕ − α + α (cid:21)(cid:19)
5s a function of ϕ is zero at the polar angle ϕ of x . However, that is only thecase at the two end points and at the mid point of C n .Consider the following closed curve. L = ( C ∪ − C ) ∪ ∞ [ n =1 ( C n ∪ − C n ) . Clearly, L is the boundary of a smooth, strictly convex, centrally symmetricplanar body, call it K . By construction, for each n ∈ Z + , only the mid-pointof the relative interior of the arc C n is an Auerbach point. The same holdsfor − C n . On the other hand, all points of C ∪ − C are Auerbach points.Thus, K is as promised. We may assume that 0 , ∈ H . Enumerate the components of R \ H as I , I , . . . , where I = ( −∞ ,
0) and I = (1 , ∞ ). We will assign recursively areal number a n to each open interval I n . Let a = 0 and a = 1.If a k is defined for k < n , then let a n = 12 (cid:18) max ℓ
0. Otherwise, I is intersected by at least two I k . Indeed, if only one I k would intersect I , then I ∩ H would be the union of at most two intervalscontradicting that H is perfect and of Lebesgue measure zero.Since the values of f on distinct intervals I k are distinct, f is not constanton I , and hence, µ ( I ) > µ ( I ) >
0, completing the proof of Proposition 2.6 cknowledgement
KS thanks Adam Ostaszewski for enlightening conversations and for drawinghis attention to [Par05].
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