A low-technology estimate in convex geometry

arXiv:math/9211216v1 [math.MG] 1 Nov 1992A low-technology estimate in convex geometryGreg Kuperberg∗Department of Mathematics, University of Chicago, Chicago, IL 60637†Let K be an n-dimensional symmetric convex body with n ≥4 and let K◦be its polar body. We present anelementary proof of the fact that(Vol K)(Vol K◦) ≥b2n(log2 n)n ,where bn is the volume of the Euclidean ball of radius 1.

The inequality is asymptotically weaker than the esti-mate of Bourgain and Milman, which replaces the log2 n by a constant. However, there is no known elementaryproof of the Bourgain-Milman theorem1.Let V be a finite-dimensional vector space over R with avolume element and let V ∗denote the dual vector space withthe dual volume element.

A convex body is a compact convexset with nonempty interior. A convex set is symmetric if it isinvariant under x 7→−x.

We define a ball to be a symmetricconvex body. We define K◦, the dual of a ball K ⊂V, byK◦= {y ∈V ∗y(K) ⊂[−1,1]}.A ball K is an ellipsoid if it is a set of the form {x⟨x,x⟩K ≤1}for some positive-definite inner product ⟨·,·⟩K on V.In this paper we will present a low-technology proof of thefollowing estimate:Theorem 1.

Let K be a symmetric convex body in an n-dimensional space V and suppose that there are two ellipsoidsE1 and E2 such that E1 ⊆K ⊆E2 and (Vol E2)/(Vol E1) = rnwith r ≥2. Then(Vol K)(Vol K◦)(Vol B)(Vol B◦) ≥(2log2 r)−n,where B is an ellipsoid.If K is an arbitrary convex body of dimension n, then thelargest-volume ellipsoid J ⊆K, which is called the John ellip-soid, satisfies K ⊆√nK.

(Proof: If x /∈√nJ but x ∈K, then Jis not the largest ellipsoid in the convex hull of J ∪{x,−x}. )It follows that a corollary.Corollary 2.

For symmetric convex body K of dimension n ≥4,(Vol K)(Vol K◦)(Vol B)(Vol B◦) ≥(log2 n)−n.It is not surprising that this estimate is asymptotically infe-rior to a high-technology estimate due to Bourgain and Mil-man [1] (see also Pisier [2]) which says that, for some fixedconstant C independent of n and K,(Vol K)(Vol K◦)(Vol B)(Vol B◦) ≥C−n.∗Supported by a Sloan Foundation Graduate Fellowship in Mathematics†Current email:greg@math.ucdavis.edu1The abstract is adapted from the Math Review by Keith Ball, MR 93h:52010.These estimates can be considered a partial inverse of San-tal´o’s inequality [4], which states that:(Vol K)(Vol K◦) ≤(Vol B)(Vol B◦).There is a nice proof of Santal´o’s inequality due to Saint-Raymond [3].We begin with some notation which will be used in theproof of the theorem. If X and Y are two vector spaces, letPX,Y denote the projection from X ×Y to Y and interpret Xand Y as also being the subsets X ×{0} and {0}×Y of X ×Y.If K is a symmetric convex body, we define the norm || · ||Kby setting ||x||K to be the least positive number t such thatx/t ∈K; in other words, K is the unit ball of || · ||K.

If A andB are two symmetric convex sets in the same vector space andp ≥1, letA+p B def= {sa +tba ∈A,b ∈B, and |s|p + |t|p ≤1},and if A and B are convex bodies, let A ∩p B be the convexbody C such that||x||pC = ||x||pA + ||x||pB. (These definitions are obviously related to the ℓp norms andhave the usual interpretation when p = ∞.) If A is a symmet-ric convex set in X and B is a symmetric convex body in Y,let A ×p B denote A +p B ⊂X ×Y.

Thus, +∞, ∩∞, and ×∞coincide with the usual operations of +, ∩, and × for sets,and A +1 B is the convex hull of A and B. Note that the re-sult of any of these operations is always a symmetric convexbody.

Finally, a standard computation shows that, if A is n-dimensional and B is k-dimensional,Vol A×p B = (Vol A)(Vol B)(n+k)/pn/p,where a fractional binomial coefficient is interpreted by thefactorial formula, i.e., x! = Γ(x+ 1).Proof of theorem.

The result is clearly true if 2 ≤r ≤4, be-cause in this case r ≤2log2 r, and the volume ratio is at leastr−n because E◦2 ⊆K◦. Otherwise, let F be the unique ellipsoidsuch that if we identify V with V ∗by the inner product ⟨·,·⟩F,then E1 = E◦2.

We will maintain this identification between V

2and V ∗for the rest of the proof, and we can assume to avoidconfusion that the volume elements on V and V ∗are equal.Consider the convex body S(K ×2 K◦) ⊂V ×V ∗, where S isthe linear operator given by S(x,y) = (x,x+ y). Observe thatV ∩S(K ×2 K◦) = K ∩2 K◦and thatPV,V∗(S(K ×2 K◦)) = K +2 K◦.Thus:(Vol K)(Vol K◦) = nn/2(Vol S(K ×2 K◦))> nn/22nn (Vol K ∩2 K◦)(Vol K +2 K◦)> 2−n(Vol K ∩2 K◦)(Vol (K ∩2 K◦)◦).The first inequality follows from an estimate of Rogers andShepard: If C is a symmetric convex body in X ×Y, where Xand Y are vector spaces, then C is at least as big as (C ∩X)×1PX,Y(C).

(Proof: For all x ∈PX,Y(C), C ∩(x + X) contains atranslate of (C ∩X)(1 −||x||PX,Y(C)). )Finally, observe thatVol FVol E1=sVol E2Vol E1and that1√2F ⊇K ∩2 K◦⊇E1 ∩2 E1 = 1√2E1.The first inclusion follows from the observation that||x||2F = ⟨x,x⟩F < ||x||K||x||K◦,which implies that||x||2K + ||x||2K◦≥2||x||2F.The theorem follows by induction.AcknowledgmentsI would like to thank the Institut des Hautes ´Etudes Scien-tifiques for their hospitality during my stay there.

I would alsolike to thank Sean Bates for his encouragement and interest inthis work. [1] Jean Bourgain and Vitaly D. Milman, New volume ratio proper-ties for convex symmetric bodies in Rn, Invent.

Math. 88 (1987),319–340.

[2] Gilles Pisier, The volume of convex bodies and Banach spacegeometry, Cambridge Tracts in Mathematics, vol. 94, CambridgeUniversity Press, 1989.

[3] Jean Saint-Raymond, Sur le volume des corps convexes sym´et-riques, Initiation seminar on analysis, 20th year: 1980/1981(G. Choquet, M. Rogalsky, and J. Saint-Raymond, eds. ), Exp.No.

11, 25, Univ. Paris VI, Paris, 1981.

[4] Luis A. Santal´o, Un invariante afin para los cuerpos convexosdel espacio de n dimensiones, Portugaliae Math. 8 (1949), 155–161.

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