Convex bodies with minimal volume product in R

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Discrete Mathematics

journal homepage:www.elsevier.com/locate/disc

Convex bodies with minimal volume product in R

²

— a new proof

Youjiang Lin

^∗

, Gangsong Leng

Department of Mathematics, Shanghai University, Shanghai, 200444, People’s Republic of China

a r t i c l e i n f o

Article history:

Received 6 March 2010

Received in revised form 2 July 2010 Accepted 13 July 2010

Available online 5 August 2010 Keywords:

Convex body Polar body Mahler conjecture Polytope

a b s t r a c t

A new proof of the Mahler conjecture inR²is given. In order to prove the result, we introduce a new method — the vertex removal method; i.e., for any origin-symmetric polygonP, there exists a linear imageφ^Pcontained in the unit diskB², and there exist three contiguous vertices ofφ^Plying on the boundary ofB². We can show that the volume- product ofPdecreases when we remove the middle vertex of the three vertices.

©2010 Elsevier B.V. All rights reserved.

1. Introduction

IfK

⊂

_Rⁿis an origin-symmetric convex body, letK^∗denote its polar body which is defined by K^∗

= {

_x

∈

_Rⁿ

:

_x

·

_y

≤

₁

, ∀

_y

∈

_K

} .

Define the volume-productP

(

^K

)

^ofKas P

(

^K

) =

Vol

(

^K

)

^Vol

(

^K∗

).

The famous Mahler conjecture [6] is to find a lower bound ofP

(

^K

)

. Is it true that we always have

P

(

^K

) ≥

_P

(

^Bn∞

),

^(1.1)

whereBⁿ_∞

= {

x

∈

_Rⁿ

: |

x_i

| ≤

1

,

¹

≤

i

≤

n

}

?

Forn

=

2, Mahler [5] proved that the answer is affirmative, and in 1986 Reisner [9] showed that equality holds only for parallelograms. Forn

=

2, a new proof of inequality(1.1)was obtained by Campi and Gronchi [2]. In then-dimensional case, the conjecture has been verified for some special classes of bodies, namely, for 1-unconditional bodies, [7,10,11], and for zonoids, [3,9].

In 1987, Bourgain and Milman [1] proved that there exists a universal constantc

>

0 such thatP

(

^K

) ≥

cⁿP

(

^Bn∞

)

^, which is now known as the Bourgain–Milman inequality. Very recently, Kuperberg [4] found a beautiful new approach to the Bourgain–Milman inequality. What’s especially remarkable about Kuperberg’s inequality is that it provides an explicit value forc.

Forn

=

2, Mahler [5] proved that the volume-product of a polygon is concave down as a pair of opposite sides are pivoted and the volume-product can be maximized as all pairs of sides are repeatedly pivoted, at the same time the polygon converges to a regular polygon. In this paper, to prove the Mahler conjecture forn

=

2, we provide a new method — thevertex removal method, which is illustrated as follows: Firstly, we prove that any origin-symmetric polygonPcan be transformed

∗Corresponding author.

E-mail addresses:linyoujiang@shu.edu.cn(Y. Lin),gleng@staff.shu.edu.cn(G. Leng).

0012-365X/$ – see front matter©2010 Elsevier B.V. All rights reserved.

doi:10.1016/j.disc.2010.07.008

Y. Lin, G. Leng / Discrete Mathematics 310 (2010) 3018–3025 3019

into a new polygon contained in the unit diskB², in which there are three contiguous vertices lying on the boundary ofB² (seeTheorem 3.1). Next, we prove that the volume-product of the new polygon decreases as the middle vertex of the three contiguous vertices moves toward its adjacent vertices on the boundary ofB²(seeTheorem 3.3). Therefore, we obtain that the volume-product ofPdecreases when we remove the middle vertex of the three vertices. Compared with Mahler’ proof, thevertex removal methodprovides a specific downward path to a square from any origin-symmetric polygon. Forn

=

3, the conjecture could probably be solved following the same idea.

It is worth mentioning that Meyer [8] proved the Mahler conjecture for the general convex bodies inR². He showed that the volume-product of a convex body is always bigger than that of a triangle and established the case of equality.

2. Definition, notation and preliminaries

As usual,Sⁿ⁻¹denotes the unit sphere,Bⁿthe unit ball centered at the origin,othe origin and

k · k

the norm in Euclidean n-spaceRⁿ. Ifx

,

^y

∈

_Rⁿ, thenx

·

yis the inner product ofxandy.

IfKis a set,

∂

^Kis its boundary, intKis its interior, and convKdenotes its convex hull. LetRⁿ

\

Kdenote the complement ofK; i.e.,

Rⁿ

\

K

= {

x

∈

_Rⁿ

:

x

6∈

K

} .

^(2.1)

IfKis ann-dimensional convex subset ofRⁿ, thenV

(

^K

)

denotes Vol_n

(

^K

)

^.

LetKⁿdenote the set of convex bodies (compact, convex subsets with non-empty interiors) inRⁿ. LetK_oⁿdenote the subset ofKⁿthat contains the origin in its interior. Leth

(

^K

, · ) :

Sⁿ⁻¹

→

_R, denote the support function ofK

∈

_Koⁿ; i.e.,

h

(

^K

,

^u

) =

max

{

u

·

x

|

x

∈

K

} ,

^u

∈

Sⁿ⁻¹

.

^(2.2)

Let

ρ(

^K

, · ) :

_Sⁿ⁻¹

→

_R, denote the radial function ofK

∈

_Koⁿ_{; i.e.,}

ρ(

^K

,

^u

) =

max

{ λ ≥

0

| λ

^u

∈

K

} ,

^u

∈

Sⁿ⁻¹

.

^(2.3)

A linear transformation (or affine transformation) ofRⁿis a map

φ

^fromRⁿto itself such that

φ

^x

=

Ax(or

φ

^x

=

Ax

+

t, respectively), whereAis ann

×

nmatrix andt

∈

_Rⁿ. For a convex bodyPand a linear transformation

φ

, we define

φ

^Pas the linear image ofPabout

φ

^.

Geometrically, an affine transformation in Euclidean space is one that preserves:

(i) The collinearity relation between points; i.e., three points which lie on a line continue to be collinear after the transformation.

(ii) Ratios of distances along a line; i.e., for distinct collinear pointsP₁

,

^P2

,

^P3, the ratio

|

P₂

−

P₁

| / |

P₃

−

P₂

|

is preserved.

IfK

∈

_Koⁿ, it is easy to verify that (see p. 44 in [12]) h

(

^K∗

,

^u

) =

¹

ρ(

^K

,

^u

)

^and

ρ(

^K∗

,

^u

) =

¹

h

(

^K

,

^u

) .

^(2.4)

IfPis a polygon; i.e.,P

=

conv

{

A₁

, . . . ,

^Am

}

, whereA_i

(

ⁱ

=

1

, . . . ,

^m

)

are vertices ofP. Leta_idenote the vector ofA_i. By the definition of the polar body, we have

P^∗

=

m

\

i=1

{

x

∈

_R²

:

x

·

a_i

≤

1

} ,

^(2.5)

which implies thatP^∗is the intersection ofmclosed half-planes with exterior normal vectora_i. The distance of the straight line

{

x

∈

_R²

:

x

·

a_i

=

1

}

from the origin is 1

/ k

a_i

k

. Thus, ifPis an inscribed polygon in a unit circle, thenP^∗is a polygon circumscribed around the unit circle. In the proof ofTheorem 3.3, we shall use these properties.

ForK

,

^L

∈

_Kⁿ, the Hausdorff distance is defined by

d

(

^K

,

^L

) =

min

{ λ ≥

0

:

K

⊂

L

+ λ

^Bn

,

^L

⊂

K

+ λ

^Bn

} ,

^(2.6)

which can be conveniently defined by (see p. 53 in [12]) d

(

^K

,

^L

) =

_max

u∈Sⁿ−1

|

_h

(

^K

,

^u

) −

_h

(

^L

,

^u

) | .

^(2.7)

Therefore, a sequence of convex bodiesK_iconverges toKif and only if the sequence of support functionsh

(

^Ki

, · )

^converges uniformly toh

(

^K

, · )

^.

The following lemmas are well-known and important for our proof:

Lemma 2.1. For any origin-symmetric convex body K

⊂

_Rⁿ

,

^P

(

^K

)

is linear invariant, that is, for every linear transformation

φ :

_Rⁿ

→

_Rⁿ, we haveP

(φ

^K

) =

_P

(

^K

)

^.

Lemma 2.2. The volume-productP

(

^K

)

is continuous under the Hausdorff metric.

Fig. 3.1. Linear transformationφsuch thatA⁰=φA.

3. Main result and its proof

Theorem 3.1. InR², for any origin-symmetric polygon P, there exists a linear image P⁰

= φ

P which satisfies that P⁰

⊂

B²and there exist three contiguous vertices of P⁰contained in

∂

^B2.

To proveTheorem 3.1, we need the following lemma.

Lemma 3.2. InFig.3.1, A

(

^x1

,

^y1

)

^{and A0}

(

^x2

,

^y2

)

^{are on}

∂

^B2.

φ

is a linear transformation fromR²to itself such that A⁰

= φ

^{A. If} B

(

^x

,

^y

) ∈

B²and0

<

^x2

<

^x1

<

^{x, then}

φ

^B

∈

B².

Proof. LetB⁰

= φ

^BandB0

= (

^x0

,

^y0

)

^{. Since}

φ

is a linear transformation, we get x

x₁

=

^x

0

x₂ and y y₁

=

^y

0

y₂

.

Therefore,

x⁰

=

^xx2

x₁ and y⁰

=

^yy2 y₁

.

Noting thatx²₁

+

y²₁

=

1

,

^x22

+

y²₂

=

1 andx²

+

y²

≤

1, we have x⁰²

+

y⁰²

=

^x

2x²₂ x²₁

+

^y

2y²₂ y²₁

≤

^x

2x²₂

x²₁

+ (

¹

−

x²

)(

¹

−

x²₂

)

1

−

x²₁

=

1

− (

^x21

−

x²₂

)(

^x2

−

x²₁

)

x²₁

(

¹

−

x²₁

)

≤

1

.

^(3.1)

Hence,

φ

^B

∈

B².

Now we proveTheorem 3.1.

Proof. SincePis an origin-symmetric polygon, the number of sides ofPis even and the opposite sides ofPare parallel. Let A₁

, . . . ,

^An

,

^B1

, . . . ,

^Bndenote all vertices ofP, whereB_iis the symmetric point ofA_iabout the origin. Our proof is in three steps.

The first step, transform the parallelogramA₁A₂B₁B₂into the rectangularA⁰₁A⁰₂B⁰₁B⁰₂inscribed inB². NowPis transformed intoP₁(see (2) or

(

²

)

⁰ⁱⁿFigs. 3.2.1and3.2.2).

Y. Lin, G. Leng / Discrete Mathematics 310 (2010) 3018–3025 3021

Fig. 3.2.1. Transforms (1) into (2) or(²)^0.

Fig. 3.2.2. Transforms (2) or(²)^{0into (3).}

The second step, transformP₁intoP₂(see (3) inFig. 3.2.2). Consider the following two cases for the polygonP₁, (i) If there exist some vertices ofP₁satisfying

{

A⁰_i

:

i

∈

I

⊂ {

3

, . . . ,

ⁿ

}} ⊂

_R²

\

B²

,

then we transformP₁intoP₂

⊂

B². The transformation shortens the segmentA⁰₁A⁰₂intoA⁰⁰₁A⁰⁰₂and satisfies that some vertices

{

A⁰⁰_i

:

i

∈

I₁

⊂ {

3

, . . . ,

ⁿ

}}

lie on

∂

^B2.

(ii) If verticesA⁰_iandB⁰_i

(

ⁱ

=

3

, . . . ,

ⁿ

)

^ofP1satisfy

{

A⁰₃

, . . . ,

^A0n

,

^B03

, . . . ,

^B0n

} ⊂

intB²

,

then we transformP₁intoP₂

⊂

B². The transformation lengthens the segmentA⁰₁A⁰₂intoA⁰⁰₁A⁰⁰₂and satisfies that some vertices

{

A⁰⁰_i

:

i

∈

I₁

⊂ {

3

, . . . ,

ⁿ

}}

lie on

∂

^B2.

The third step, transformP₂intoP⁰(seeFig. 3.3). IfA⁰⁰₁

,

^A002

,

^A00i

(

ⁱ

∈

I₁

)

are three contiguous vertices on

∂

^B2, then this theorem has already been proved; otherwise we rotateP₂about the origin, we can get a new polygonP₃⁰ such thatA⁰⁰₂A⁰⁰_i parallels theX-axis (see (4) inFig. 3.3). Then lengthen segmentsA⁰⁰₂B⁰⁰_i andA⁰⁰_iB⁰⁰₂ intoA⁽₂³⁾B⁽_i³⁾and A⁽_i³⁾B⁽₂³⁾, respectively, satisfying that some vertices

{

A⁽_j³⁾

:

j

∈

I₂

⊂ {

3

, . . . ,

ⁱ

−

1

}}

lie on

∂

^B2and

{

A⁽_j³⁾

:

j

∈ {

3

, . . . ,

ⁱ

−

1

} \

I₂

} ⊂

B². By Lemma 3.2, vertices

{

A⁽_i+³⁾₁

, . . . ,

^A(n³⁾

,

^B(1³⁾

}

are still in the internal ofB²(see (5) inFig. 3.3). LetP₃denote the new polygon. We getP₃

⊂

B².

There arei

−

3 vertices between the vertexA⁽₂³⁾and vertexA⁽_i³⁾. Ifi

−

3

=

1, thenP₃is the polygon satisfying the theorem.

Ifi

−

3

≥

2, consider the following two cases:

(i) Ifj

−

2

≥

i

−

j

(

^wherej

∈

I₂

)

^{, rotateP}3about the origin such thatA⁽₂³⁾A⁽_j³⁾parallels theX-axis.

(ii) Ifj

−

2

≤

i

−

j

(

^wherej

∈

I₂

)

^{, rotateP}3about the origin such thatA⁽_j³⁾A⁽_i³⁾parallels theX-axis.

Fig. 3.3. Transforms (4) into (5).

Fig. 3.4. PolygonP⁰and its polar body.

We denote the new polygon asP₄. It is clear that there are less thani

−

3 vertices betweenA⁽₂³⁾andA⁽_j³⁾(or between A⁽_j³⁾andA⁽_i³⁾). By induction, we can get a new polygonP⁰

⊂

B²with three contiguous vertices contained in

∂

^B2after finite transformations.

By the above theorem, we consider the volume-product of the polygon with three contiguous vertices contained in

∂

^B2. Theorem 3.3. Suppose that P⁰

⊂

B²is an origin-symmetric polygon and A

,

^C

,

B are three contiguous vertices of P⁰contained in

∂

^{B2, thenP}

(

^P00

) ≤

_P

(

^P0

)

^{, where P00}is a new polygon obtained from P⁰by removing vertices C and C⁰.

Proof. Suppose that the sideABparallels theX-axis (seeFig. 3.4.) and straight linesl

,

^l1andl₂are three tangent lines to the unit circleB²passing through pointsC

,

^AandB, respectively. LetA

= ( −

x₀

,

^y0

)

^{, thenB}

= (

^x0

,

^y0

)

^{, wherex}02

+

y₀²

=

1. Let

θ =

⁶ XOC. It is clear that

π/

²

≤ θ ≤ π −

arctan

(

^y0

/

^x0

)

when the pointCis in the second quadrant. We have the following equations of straight lines:

l₁

:

y

−

y₀

=

^x0

y₀

(

^x

+

x₀

),

l₂

:

y

−

y₀

= −

^x0

y₀

(

^x

−

x₀

),

Y. Lin, G. Leng / Discrete Mathematics 310 (2010) 3018–3025 3023

l

:

y

−

sin

θ = −

^cos

θ

sin

θ (

^x

−

cos

θ).

Let the pointNdenote the intersection ofland theY-axis and the pointMdenote the intersection ofl₁and theY-axis.

We can easily getN

(

⁰

,

¹

/

^sin

θ)

^andM

(

⁰

,

¹

/

^y0

)

^{. LetHandL}denote the intersections oflandl₁

,

^l2, respectively. Solve the following systems of equations:







y

−

sin

θ = −

^cos

θ

sin

θ (

^x

−

cos

θ)

y

−

y₀

=

^x0

y₀

(

^x

+

x₀

)

^(3.2)

and







y

−

sin

θ = −

^cos

θ

sin

θ (

^x

−

cos

θ)

y

−

y₀

= −

^x0

y₀

(

^x

−

x₀

).

^(3.3)

We get the abscissas of pointsHandL:

x₁

=

^y0

−

sin

θ

y₀cos

θ +

x₀sin

θ

and

x₂

=

^y0

−

sin

θ

y₀cos

θ −

x₀sin

θ .

LetS4MHLdenote the area of

4

MHL, it follows that S4MHL

=

^x0

y₀

·

^sin

θ −

y₀ sin

θ +

y₀

.

LetV

=

V

(

^P00

)

^andV0

=

V

(

^P00∗

)

^{, whereP00}denotes the new polygon obtained fromP⁰by removing verticesCandC⁰, thenP

(

^P0

)

is a functionf

(θ)

^{, where}

f

(θ) = (

^V

+

2x₀

(

^sin

θ −

y₀

))

V⁰

−

^2x0

y₀

·

^sin

θ −

y₀ sin

θ +

y₀

(3.4) and

π

2

≤ θ ≤ π −

arctan y₀

x₀

.

We have

f⁰

(θ) =

2x₀cos

θ · (

^V0y0

−

2x₀

)(

^sin

θ +

y₀

)

²

+

2y₀

(

^4x0y₀

−

V

)

y₀

(

^sin

θ +

y₀

)

²

.

^(3.5)

In(3.4), since cos

θ ≤

0 andy₀

(

^sin

θ +

y₀

)

²

≥

0, to provef⁰

(θ) ≤

0, lett

=

sin

θ

, we need only to show thatg

(

^t

) ≥

0, where

g

(

^t

) = (

^V0y0

−

2x₀

)(

^t

+

y₀

)

²

+

2y₀

(

^4x0y₀

−

V

),

^t

∈ [

y₀

,

¹

] .

^(3.6) LetS_ABA⁰_B⁰andS4ABMdenote the area of a rectangleABA⁰B⁰and triangleABM, respectively. Since

V

(

^P00∗

) ≥

V

(

^conv

{

A

,

^M

,

^B

,

^A0

,

^M0

,

^B0

} ),

we get

V⁰

≥

S_ABA⁰_B⁰

+

2S4ABM

=

4x₀y₀

+

2x₀ 1

y₀

−

y₀

,

therefore,

V⁰y₀

−

2x₀

≥

4x₀y₀

+

2x₀ 1

y₀

−

y₀

y₀

−

2x₀

=

2x₀y²₀

>

⁰

.

^(3.7)

Thus, the graph ofg

(

^t

)

is a parabola opening upward. Since the axis of symmetry of the parabola ist

= −

y₀

,

^g

(

^t

)

is increasing fort

∈ [

y₀

,

¹

]

. To proveg

(

^t

) ≥

0, it suffices to prove

g

(

^y0

) =

2y₀

(

^2V0y20

−

V

) ≥

0

.

^(3.8)

LetDdenote the area of a circular segment enclosed by the arcBA⁰and the chordBA⁰, then V⁰

≥

S_ABA⁰_B⁰

+

2S4ABM

+

2D

=

4x₀y₀

+

2x₀ 1

y₀

−

y₀

+

2D (3.9)

and

V

≤

S_ABA0B0

+

2D

=

4x₀y₀

+

2D

.

^(3.10)

To prove(3.8), it suffices to prove 2

4x₀y₀

+

2x₀ 1

y₀

−

y₀

+

2D

y²₀

≥

4x₀y₀

+

2D

,

^(3.11)

which is equivalent to

2x₀y³₀

≥

_D

(

¹

−

2y²₀

).

^(3.12)

And since

D

≤ (

¹

−

x₀

) ·

2y₀

,

^(3.13)

hence, it suffices to prove

x₀y³₀

≥

y₀

(

¹

−

x₀

)(

¹

−

2y²₀

),

^(3.14)

which is equivalent to

x³₀

−

2x²₀

+

1

≥

0

,

^(3.15)

which is clearly correct when 0

<

^x0

<

^1.

We get thatf

(θ)

is decreasing when

θ ∈ [ π/

²

, π −

arctan

(

^y0

/

^x0

) ]

, hence the functionf

(θ)

has a minimum value at

θ = π −

arctan

(

^y0

/

^x0

)

. The pointCcoincides with the pointAwhen

θ = π −

arctan

(

^y0

/

^x0

)

. ThereforeP

(

^P00

) ≤

_P

(

^P0

)

^.

ByTheorem 3.3, we can get the followingCorollaries 3.4and3.5.

Corollary 3.4. If P

⊂

_R²is an origin-symmetric polygon, thenP

(

^P

) ≥

_P

(

^S

)

, where S is a square.

Proof. ByTheorems 3.1and3.3and the linear invariance ofP

(

^P

)

, if the number of sides of the polygonPis 2n, there exists a polygonP₁with 2

(

ⁿ

−

1

)

sides satisfyingP

(

^P1

) ≤

_P

(

^P

)

. Repeating this processn

−

2 times, we can get a squareSsatisfying P

(

^P

) ≥

_P

(

^S

)

^.

Corollary 3.5. If K

⊂

_R²is an origin-symmetric convex body and S

⊂

_R²is an origin-symmetric square, thenP

(

^K

) ≥

_P

(

^S

)

^. Proof. For any origin-symmetric convex body K

⊂

_R², there exists a sequence of origin-symmetric polytopes

{

P_i

}

converging toKunder the Hausdorff metric (see p. 54 in [12]). ByCorollary 3.4andLemma 2.2., we have

P

(

^K

) =

lim

n→∞P

(

^Pi

) ≥

_P

(

^S

).

(3.16)

Acknowledgements

The authors are most grateful to the referee for his many excellent suggestions for improving the original manuscript.

The authors would like to acknowledge the support from the National Natural Science Foundation of China (10971128) and Shanghai Leading Academic Discipline Project (S30104).

References

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