The (B) conjecture for uniform measures in the plane
aa r X i v : . [ m a t h . F A ] N ov THE (B) CONJECTURE FOR UNIFORM MEASURES INTHE PLANE
AMIR LIVNE BAR-ON
Abstract.
We prove that for any two centrally-symmetric convex shapes
K, L ⊂ R , the function t
7→ | e t K ∩ L | is log-concave. This extends a re-sult of Cordero-Erausquin, Fradelizi and Maurey in the two dimensional case.Possible relaxations of the condition of symmetry are discussed. Introduction
It was conjectured by Banaszcyk (see Lata la [1]) that for any convex set K ⊂ R n that is centrally-symmetric (i.e., K = − K ) and for a centered Gaussian measure γ , γ ( s − λ t λ K ) ≥ γ ( sK ) − λ γ ( tK ) λ (1)for any λ ∈ [0 ,
1] and s, t > t γ ( e t K ) is log-concave. The same paper raises the question whether (1) re-mains valid when γ is replaced by other log-concave measures. The proof of (1) for unconditional sets and log-concave measures was given in [2] as well: Theorem 1 ([2], Proposition 9) . Let K ⊂ R d be a convex set and let µ be alog-concave measure on R d , and assume that both are invariant under coordinatereflections. Then t µ ( e t K ) is a log-concave function. This paper explores the situation in R . To distinguish this special case, we calla convex set K ⊂ R , which is compact and has a non-empty interior, a shape . Themain result is Theorem 2.
Let
K, L ⊂ R be centrally-symmetric convex shapes. Then t
7→ | e t K ∩ L | is a log-concave function. Here | · | is the Lebesgue measure, so Theorem 2 is an analog of (1) for uniform measures – with density dµ ( x ) = L ( x ) dx . Note that a uniform measure on a setis log-concave if and only if the set is convex.The condition of central symmetry in Theorem 2 can be replaced by dihedralsymmetry. For an integer n ≥
2, let D n be the group of symmetries of R that isgenerated by two reflections, one across the axis Span { (1 , } and the other acrossthe axis Span { (cos πn , sin πn ) } . The dihedral group D n contains 2 n transformations.A D n -symmetric shape A ⊂ R is one invariant under the action of D n . Theorem 3.
Let n ≥ be an integer, and let K, L ⊂ R be D n -symmetric convexshapes. Then t
7→ | e t K ∩ L | is a log-concave function. Key words and phrases.
B conjecture, uniform measure.Supported in part by a grant from the European Research Council.
Examples and open questions.
For what sets and measures is (1) valid?The (B)-conjecture, or (1), is not necessarily true for measures and sets with justone axis of symmetry in R . An example with a log-concave uniform measure is L = conv { ( − , − , (0 , , (5 , − } K = [ − , × [ − , t
7→ | e t K ∩ L | is not log-concave in a neighbourhood of t = 0.Another negative result is for quasi-concave measures. These are measures withdensity dµ ( x ) = ϕ ( x ) dx satisfying ϕ ((1 − λ ) x + λy ) ≥ max { ϕ ( x ) , ϕ ( y ) } for all0 ≤ λ ≤
1. If µ ( A ) = | A ∩ Q | + | A | , Q = [ − , × [ − , t µ ( e t Q ) is not log-concave in a neighbourhoodof t = 1.The (B)-conjecture for general centrally-symmetric log-concave measures is notsettled yet, even in two dimensions. It is also of interest to generalize the methodof this paper to higher dimensions. Notation.
For a convex shape K ⊂ R , its boundary is denoted by ∂K . Thesupport function is denoted h K ( x ) = sup y ∈ K h x, y i . The normal map ν K : ∂K → S is defined for all smooth points on the boundary, and ν K ( p ) is the uniquedirection that satisfies h ν K ( p ) , x i = h K ( x ). We denote the unit square by Q =[ − , × [ − ,
1] = B ∞ . The Hausdorff distance between sets A, B ⊂ R n is defined as d H ( A, B ) = max { sup a ∈ A d ( a, B ) , sup b ∈ B d ( b, A ) } . The radial function ρ K : R → R of a convex shape K ⊂ R is ρ K ( θ ) = max { r ∈ R : ( r cos θ, r sin θ ) ∈ K } , withperiod 2 π . 2. Main result
This section proves Theorem 2:
Theorem.
Let
K, L ⊂ R be centrally-symmetric convex shapes. Then the function f K,L ( t ) = | e t K ∩ L | is log-concave. Obviously, it suffices to show log-concavity around t = 0.If we consider the space of centrally-symmetric convex shapes in the plane,equipped with the Hausdorff metric d H , then the operations K, L K ∩ L and K
7→ | K | are continuous. This means that the correspondence K, L f K,L iscontinuous as well. Since the condition of log-concavity in the vicinity of a pointis a closed condition in the space C ( R ) of bounded continuous functions, the classof pairs of centrally-symmetric shapes K, L ⊂ R for which f K,L ( t ) is log-concavenear t = 0 is closed w.r.t Hausdorff distance. Thus in order to prove Theorem 2 itsuffices to prove that f K,L ( t ) is a log-concave function near t = 0 for a dense set inthe space of pairs of centrally-symmetric convex shapes.As a dense subset, we shall pick the class of transversely-intersecting convexpolygons. This class will be denoted by F . The elements of F are pairs ( K, L ) ofshapes
K, L ⊂ R that satisfy: • The sets (
K, L ) are centrally-symmetric convex polygons in R . • The intersection ∂K ∩ ∂L is finite. • None of the points x ∈ ∂K ∩ ∂L are vertices of K or of L . That is, there issome ε > B ( x, ε ) ∩ ∂K and B ( x, ε ) ∩ ∂L are line segments. HE (B) CONJECTURE FOR UNIFORM MEASURES IN THE PLANE 3 • For every x ∈ ∂K ∩ ∂L , ν K ( x ) = ν L ( x ). Claim.
The class F is dense in the space of pairs of centrally-symmetric convexshapes (with respect to the Hausdorff metric). Hence, in order to prove Theorem 2, it is enough to consider polygons withtransversal intersection.
Deriving a concrete inequality.Lemma. If ( K, L ) ∈ F , then f K,L ( t ) is twice differentiable in some neighbourhoodof t = 0 . Remark.
In this case, log-concavity around t = 0 amounts to the inequality d dt log f ( t ) (cid:12)(cid:12)(cid:12)(cid:12) t =0 ≤ f (0) · f ′′ (0) ≤ f ′ (0) . (2) Proof.
The area of the intersection is | aK ∩ L | = a Z dr Z x ∈ r∂K ∩ L h K ( ν K ( xr )) dℓ, where dℓ is the length element.Denote g K,L ( r ) = Z x ∈ r∂K ∩ L h K ( ν K ( xr )) dℓ. The transversality of the intersection implies that g K,L ( r ) is continuous near r = 1.Therefore a
7→ | aK ∩ L | is continuously differentiable near a = 1.The contour r∂K ∩ L is a finite union of segments in R . Transversality im-plies that the number of connected components does not change with r in a smallneighbourhood of r = 1. The beginning and end points of each component aresmooth functions of r , also in some neighbourhood of r = 1. Therefore g K,L ( r ) isdifferentiable as claimed. (cid:3) Note that in such a neighbourhood of r = 1, the function g K,L ( r ) only depends onthe parts of K and L that are close to ∂K ∩ L , and is in fact a sum of contributionsfrom each of the connected components.Writing (2) in terms of g ( r ), we get the following condition: Definition.
For convex shapes ( K, L ) ∈ F , we say that K and L satisfy property B , or that B ( K, L ) , if | K ∩ L | · [ g K,L (1) + g ′ K,L (1)] ≤ g K,L (1) . (3)The set F is open with respect to the Hausdorff metric, and in particular, if( K, L ) ∈ F then ( K, rL ) ∈ F for every r in some neighbourhood of r = 1. If B ( K, rL ) holds for every r in such a neighbourhood, then f K,L ( t ) is log-concave insome neighbourhood of t = 0, as f K,L ( t + t ) = e t f K,e − t L ( t ) . Therefore verifying (3) for all pairs (
K, L ) ∈ F will prove Theorem 2. AMIR LIVNE BAR-ON
Reduction to parallelograms.
Given two polygons (
K, L ) ∈ F , the intersection ∂K ∩ L consists of a finite number of connected components. Due to central sym-metry, they come in opposite pairs. We denote these components by S , . . . , S n ,and S i + n = {− x : x ∈ S i } .We define a pair of convex shapes K ( i ) , L ( i ) for each 1 ≤ i ≤ n via the followingproperties. • The shape K ( i ) is the largest convex set whose boundary contains S i ∪ S i + n .Equivalently, denoting by x , x the endpoints of S i , and by x the solutionof the equations ( h ν K ( x ) , x i = h K ( ν K ( x )) h ν K ( x ) , x i = − h K ( ν K ( x )) K ( i ) = conv ( S i ∪ S i + n ∪ { x, − x } ). • The shape L ( i ) is the parallelogram defined by the four lines h ν L ( x ) , x i = ± h L ( ν L ( x )) , h ν L ( x ) , x i = ± h L ( ν L ( x ))See Figure 1 for examples.If S i is a segment then K ( i ) described above is an infinite strip, and if ν L ( x ) = ν L ( x ) then L ( i ) is an infinite strip. We would like to work with compact shapes,thus we apply a procedure to modify K ( i ) , L ( i ) to become bounded without changingtheir significant properties. Transversality implies that the intersection K ( i ) ∩ L ( i ) is bounded, even if both sets are strips. For each 1 ≤ i ≤ n we pick a centrally-symmetric strip A ⊂ R such that A ∩ K ( i ) and A ∩ L ( i ) are both bounded, andwhich contains K and L , and whichever of K ( i ) , L ( i ) that is bounded. From nowon we replace K ( i ) and L ( i ) by their intersection with A . Figure 1.
Two examples of the extension
K, L = ⇒ K (1) , L (1) . The shaded shape in each diagram is K and the white shape with a solidboundary line is the corresponding L . Remark.
Note that the sets grow in the process: K ⊂ K ( i ) and L ⊂ L ( i ) for all i =1 . . . n . They satisfy ∂K ( i ) ∩ L ( i ) = S i ∪ S i + n . Also note that if K is a parallelogramthen so are the K ( i ) , for every i . It is trivial to check that ( K ( i ) , L ( i ) ) ∈ F when ( K, L ) ∈ F . Lemma. If B ( K ( i ) , L ( i ) ) for all i = 1 . . . n , then B ( K, L ) .Proof. The function g K,L ( r ) takes non-negative values for r >
0. In addition,its value is the sum of contributions from the different connected components of r∂K ∩ L . From transversality, these components vary continuously around r = 1, HE (B) CONJECTURE FOR UNIFORM MEASURES IN THE PLANE 5 hence g ′ K,L (1) is also a sum of values coming from the different components. There-fore we can write | K ∩ L |· [ g K,L (1) + g ′ K,L (1)] = | K ∩ L | · n X i =1 h g K ( i ) ,L ( i ) (1) + g ′ K ( i ) ,L ( i ) (1) i ≤ n X i =1 | K ( i ) ∩ L ( i ) | · h g K ( i ) ,L ( i ) (1) + g ′ K ( i ) ,L ( i ) (1) i ≤ by B ( K ( i ) ,L ( i ) ) n X i =1 g K ( i ) ,L ( i ) (1) ≤ n X i =1 g K ( i ) ,L ( i ) (1) ! = g K,L (1) (cid:3) Lemma. If B ( K, L ) holds for all pairs of parallelograms ( K, L ) ∈ F , then Theo-rem 2 follows.Proof. Let (
K, L ) ∈ F be any polygons. Construct the sequence of pairs K ( i ) , L ( i ) from K, L . The shape L ( i ) is a parallelogram for every i . Then construct the pairs (cid:0) L ( i ) (cid:1) ( j ) , (cid:0) K ( i ) (cid:1) ( j ) from L ( i ) , K ( i ) , for all i . The shapes (cid:0) L ( i ) (cid:1) ( j ) and (cid:0) K ( i ) (cid:1) ( j ) willbe parallelograms for every i, j . Under our assumption, we have B (cid:16)(cid:0) L ( i ) (cid:1) ( j ) , (cid:0) K ( i ) (cid:1) ( j ) (cid:17) .From this and the previous lemma, B ( L ( i ) , K ( i ) ) follows.The property B is symmetric in the shapes. That is, B ( S, T ) ⇐⇒ B ( T, S ) forall (
S, T ) ∈ F . This is since f S,T and f T,S differ by a log-linear factor: f S,T ( t ) = | e t S ∩ T | = e t f T,S ( − t )This means that we have B ( K ( i ) , L ( i ) ) as well. Applying the previous lemmaagain gives B ( K, L ). (cid:3) All that remains in order to deduce Theorem 2 is to analyse the case of centrally-symmetric parallelograms.If
K, L are parallelograms and K = T Q where T is an invertible linear map and Q = [ − , × [ − , f K,L = det T · f Q,T − L . Therefore we can take one of the parallelograms to be a square. In other words,establishing B ( Q, L ) where Q is the unit square and L is a parallelogram, and( Q, L ) ∈ F , will imply Theorem 2.In fact, we may place additional geometric constraints on the square and theparallelogram.If neither Q nor L contains a vertex of the other quadrilateral in its interior,then ∂Q ∩ L has 4 connected components. Applying the reduction above to Q, L gives Q ( i ) , L ( i ) with i = 1 ,
2, and the intersection ∂Q ( i ) ∩ L ( i ) has only 2 connectedcomponents, as remarked above.Since the shapes are convex, if all the vertices of one shape are contained in theother, we have Q ⊂ L or L ⊂ Q , and then (3) holds trivially. If L contains cornersof Q but Q does not contain vertices of L , we shall swap them.These arguments leave two cases to be considered: • Q contains 2 vertices of L , and L does not contain corners of Q . In thiscase the intersection ∂Q ∩ L is contained in two opposite edges of Q . • Q contains 2 vertices of L , and L contains 2 corners of Q . In this case theintersection ∂Q ∩ L is a subset of the edges around these corners of Q . AMIR LIVNE BAR-ON
Computation of the special cases.
These cases are defined by 4 real parame-ters – the coordinates of the vertices of L . A symbolic expression for f ( t ) can bederived, and (3) will be a polynomial inequality in these parameters. The geometricconditions given above are also polynomial inequalities in these parameters. Thuseach of the two cases can each be expressed by a universally-quantified formula inthe language of real closed fields. By Tarski’s theorem [3], this first order theoryhas a decision procedure. This is implemented in the QEPCAD B computer pro-gram [4]. Relevant computer files, for generation of the symbolic condition and forrunning the logic solver, for one of the two cases above, are available at A human-readable proof of both cases is included here as well.
Lemma. If L is a centrally-symmetric parallelogram that satisfies ( Q, L ) ∈ F , andif L crosses Q only inside the vertical edges of Q , then B ( Q, L ) .Proof. Let α, β, c, d be as in Figure 2.
K = QL (c,d) β α
Figure 2.
The equations for the edges of L are x cos α + y sin α = ± ( c cos α + d sin α ) x cos β + y sin β = ± ( c cos β + d sin β )Relevant parameters are computed as follows: ∂Q ∩ ∂L = (cid:8) ± (cid:0) , ( c −
1) cot α + d (cid:1) , ± (cid:0) , ( c −
1) cot β + d (cid:1)(cid:9) g Q,L (1) = 2( c − α − cot β ) g ′ Q,L (1) = − α − cot β ) g Q,L (1) + g ′ Q,L (1) = (2 c − α − cot β )The area of L is comprised of Q ∩ L and of two triangles. The area of the trianglesis g (1) · ( c −
1) so | Q ∩ L | = | L | − ( c − (cot α − cot β ) . Note that 0 < α < π < β < π so cot α − cot β is a positive quantity, and that if c < g (1) + g ′ (1) is negative so inequality (3) is satisfied immediately. HE (B) CONJECTURE FOR UNIFORM MEASURES IN THE PLANE 7
Assume c > c − α − cot β ) · (cid:2) | L | − ( c − (cot α − cot β ) (cid:3) ≤ c − (cot α − cot β ) . Or equivalently (2 c − | L | ≤ ( c − (cot α − cot β ) · (4 + 2 c − , or still | L | ≤ (cid:16) c − (cid:17) · ( c − g (1) . The amount ( c − g (1) is the area of the triangles L \ Q . By convexity the areaof L cannot be larger than that times (cid:16) cc − (cid:17) . It remains to verify that for c > c ( c − < c − . This is a simple exercise in algebra: c ( c − = 1+ 2 c − c − = 1+ c − · ( c − )( c − c − = 1+ c − (cid:20) − c/ c − (cid:21) ≤ c − (cid:3) Lemma. If L is a centrally-symmetric parallelogram that satisfies ( Q, L ) ∈ F , andeach of Q, L contains two vertices of the other, then B ( Q, L ) .Proof. Let a and b be as in Figure 3, and let S stand for the area S = | Q ∩ L | . Thenumbers a and b are in the range 0 < a, b <
2, and α and β satisfy π < α < β < π .The area S is in the range 4 − ab < S < K = QL ab S αβ p Figure 3.
The quantity g (1) is simply 8 − a − b , and g ′ (1) will soon be shown to bebounded by g ′ (1) ≤ − S − (4 − ab )(4 − S ) + ( a − b ) . This gives an inequality in the 3 variables a, b, S , which will be proved for valuesin the prescribed ranges.
AMIR LIVNE BAR-ON
The length of each dotted line in Figure 3 is ( a + b ) / . Denoting the height ofthe triangle (the distance between p and the closest dotted line) by h , the area is S = (4 − ab ) + 2 · h · ( a + b ) / , so h = S − (4 − ab )( a + b ) / . The formula for g ′ (1) in terms of the angles α, β is g ′ (1) = 4 + 2 tan α + 2 cot β. Denote c = β − α . Holding c fixed, the function α g ′ (1) = 4 + 2 tan α + 2 cot( α + c )is concave and takes the same value for α as for π − c − α . Therefore its maximumis attained at α = π − c . This gives a bound for g ′ (1) for a given c = β − α : g ′ (1) ≤ (cid:0) π − c (cid:1) + 2 cot (cid:0) π + c (cid:1) . This bound is stronger for higher values of c , since tan is an increasing functionand cot is a decreasing function.The angle between the edges of L meeting at p is π − ( β − α ) = π − c . When a , b , and h are kept fixed, the position of p gives a bound for g ′ (1). This bound is theweakest when the angle π − c is largest. Simple geometric considerations show thatin a family of triangles with the same base and height, the apex angle is largestwhen the triangle is isosceles, so we will pursue the case where the triangle formedby p and the nearest dotted line is isosceles.The value of c in this case is c = 2 tan − ( a + b ) / h , and we get g ′ (1) ≤ π − π + tan − ( a + b ) / h ! + 2 cot π + π − tan − ( a + b ) / h ! = 4 + 4 tan (cid:18) π + tan − a + b S − (4 − ab ) (cid:19) = 4 + 4 · a + b S − (4 − ab ) − a + b S − (4 − ab ) = 81 − a + b S − (4 − ab ) = − S − (4 − ab )(4 − S ) + ( a − b ) , which proves the forementioned bound for g ′ (1).Therefore, to prove (3) it is enough to show S · (cid:18) − a − b − S − (4 − ab )(4 − S ) + ( a − b ) (cid:19) ≤ (8 − a − b ) Rearranging and taking into account that
S <
4, this is equivalent to(8 − a − b )(8 − a − b − S ) (cid:0) (4 − S ) + ( a − b ) (cid:1) + 8 S ( S − (4 − ab )) | {z } E ≥ HE (B) CONJECTURE FOR UNIFORM MEASURES IN THE PLANE 9
When a and b are held fixed, this is a 2 nd degree condition on S . Since 0 < a, b < S = 4 − ab are positive: E | S =4 − ab = (8 − a − b )(2 − a )(2 − b ) · ( a + b ) > ∂E∂S (cid:12)(cid:12)(cid:12)(cid:12) S =4 − ab = ( a + b ) (cid:0) (5 − a − b ) − (cid:1) + 2( a − b ) > ∂ E∂S (cid:12)(cid:12)(cid:12)(cid:12) S =4 − ab = 18(4 − ab ) > S > − ab , as required. (cid:3) Dihedral symmetry
This section deals with dihedrally symmetric sets. The group D n is defined inthe introduction. Theorem (3) . Let n ≥ be an integer, and let K, L ⊂ R be D n -symmetric convexshapes. Then t
7→ | e t K ∩ L | is a log-concave function. For n = 2 the group D n is generated by reflections across the standard axes. Thiscorresponds to unconditional sets and functions, and Theorem 1 from [2] solves thiscase.The proof for n ≥ smooth strongly-convex shape K ⊂ R is one whose boundary is a smoothcurve with strictly positive curvature everywhere. The radial function ρ K of asmooth strongly-convex shape K ⊂ R is a smooth function. The boundary ∂K isthe curve γ K ( θ ) = ( ρ K ( θ ) cos θ, ρ K ( θ ) sin θ ) . The convexity of K is reflected in the sign of the curvature of γ K . Positive curvaturecan be written as a condition on the radial function: ρ ( θ ) + 2 ρ ′ ( θ ) − ρ ( θ ) ρ ′′ ( θ ) > . (4) Proof of theorem 3.
For any D n -symmetric convex shape K ⊂ R there is a seriesof D n -symmetric convex shapes whose boundaries are smooth and strongly convexcurves, and whose Hausdorff limit is K . By the continuity argument from theprevious section, the general case follows from the smooth case. From here on, K and L are smooth D n -symmetric shapes. D n -symmetric shapes correspond to radial functions that are even and haveperiod πn . These shapes are completely determined by their intersection with thesector G n = (cid:8) ( r cos θ, r sin θ ) : r ≥ , θ ∈ [0 , πn ] (cid:9) . Given two such shapes
K, L the area function is f K,L ( t ) = | e t K ∩ L | = 2 nf K ∩ G n ,L ∩ G n ( t ) . Let K ⊂ R be a D n -symmetric strongly convex shape, and consider the function˜ ρ ( θ ) = ρ K ( n θ ). This is an even function with period π . The function ˜ ρ ( θ ) alsosatisfies (4):˜ ρ ( θ ) + ˜ ρ ′ ( θ ) − ˜ ρ ( θ )˜ ρ ′′ ( θ ) = n (cid:0) ρ K ( n θ ) + 2 ρ ′ K ( n θ ) − ρ K ( n θ ) ρ ′′ K ( n θ ) (cid:1) + (1 − n ) ρ K ( n θ ) > . This means that ˜ ρ ( θ ) is the radial function of some D -symmetric (unconditional)strongly convex shape. We denote this w ( K ): the unique shape that satisfies ρ w ( K ) ( θ ) = ρ K ( n θ ).The following function, also named w , is defined on G n : w (cid:18) r cos θr sin θ (cid:19) = (cid:18) r cos n θr sin n θ (cid:19) . (cid:0) for r ≥ , θ ∈ [0 , πn ] (cid:1) The point function w is an bijection between G n and G . It relates to the shapefunction w by the formula { w ( x ) : x ∈ K ∩ G n } = w ( K ) ∩ G . The point function w is differentiable inside G n , and has a constant Jacobian de-terminant n .Hence f K,L ( t ) = 2 nf K ∩ G n ,L ∩ G n ( t ) = 4 f w ( K ) ∩ G ,w ( L ) ∩ G ( t ) = f w ( K ) ,w ( L ) ( t ) , and the theorem follows from the result in [2]. (cid:3) References [1] R. Lata la. On some inequalities for Gaussian measures.
Proceedings of the InternationalCongress of Mathematicians , vol II, 813–822, 2002.[2] D. Cordero-Erausquin, M. Fradelizi and B. Maurey. The (B) conjecture for the Gaussianmeasure of dilation of symmetric convex sets and related problems.
Journal of FunctionalAnalysis , 214(2):410–427, 2004.[3] A. Tarski. A Decision Method for Elementary Algebra and Geometry.
University of Cali-fornia Press , 1951.[4] C. W. Brown. QEPCAD B: a program for computing with semi-algebraic sets using CADs.
SIGSAM Bulletin , 37(4): 97–108, 2003.
Tel Aviv University, Tel Aviv 69978, Israel
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