Convex Body Isoperimetric Conjecture

Conjecture. The least perimeter to enclose given volume inside an open ball in Rn is greater than inside any other convex body of the same volume.

See  e.g. Hallard T. Croft, Kenneth J. Falconer, and Richard K. Guy, Unsolved problems in geometry, Problem Books in Mathematics, Unsolved Problems in Intuitive Mathematics, II, Springer-Verlag, New York, 1991, 1994; and [EFKNT] Introduction. Hutchings cites Wichiramala.


1. We thank Ken Brakke for pointing out the necessity of “convex,” to avoid the region between concentric spheres (joined by a thin tube if you want to keep the region simply connected).

2. Inside the ball, an isoperimetric surface is a spherical cap perpendicular to the boundary, as was first proved by Burago and Maz’ja [BM]. Simplest proof: Least-perimeter surface exists. By spherical symmetrization, may assume meets sphere in a circle. Must be spherical cap. Must meet sphere orthogonally.

3. In R2 it probably suffices to assume region simply connected rather than convex. In R2 there is a similar conjecture for the complement of a ball. Thanks to Mohammad Ghomi for pointing out the necessity of restricting to dimension 2; in R3 the conjecture fails in the complement of a long thin cylinder.

4. added 1 October 2010. Ernst Kuwert [K, Cor. 1.2] proved that the least perimeter to enclose given volume inside a halfspace in Rn is greater than or equal to inside any smooth convex body, which proves the conjecture for smooth convex bodies up to a constant  nan/2an-1 ~ c√n, where an is the volume of the unit ball in Rn and c = √(π/2). Ritoré and Stratos (preprint, 2012) describe how to generalize such results to nonsmooth convex bodies by approximation.

(For sharp lower bounds, see my post Sharp Isoperimetric Bounds for Convex Bodies.)

[BM] Ju. D. Burago and V. G. Maz’ja, Some questions of potential theory and function theory for domains with non-regular boundaries. Zap. Nauchn. Semin. Leningr. Otd. Mat. Inst. Steklova 3 (1967) 1-152 (Russian). English translation: Seminars in Math., V. A. Steklov Math. Inst., Leningrad 3 (1969) 1-68. Consultants Bureau, New York.

[K] Ernst Kuwert, Note on the isoperimetric profile of a convex body, Geometric Analysis and Nonlinear Partial Differential Equations, S. Hildebrandt and H. Karcher, eds., Springer, Berlin, 2003, 195-200.

Discussion with Emanuel Milman provides an improvement by √(log n). The Kuwert bound follows from concavity of the n/(n-1) power of the least-perimeter as a function of volume, together with behavior for small volume. To this one adds two more ingredients. First, in Rn it takes at least as much area to cut the ball in half as any other convex body of the same volume, for centrally symmetric convex bodies even by hyperplanes [L] (which are optimal for the ball but not in general) and within a constant (e/2) in general (see Prop. B below at end of post). This bound, which holds for all volume fractions by concavity, meets the Kuwert bound at volume fraction on the order of 1/√n. Second, the least perimeter to cut the ball (by a spherical cap) in any fraction is well approximated by the perimeter of balls in the sphere (because most of the volume of high dimensional balls is concentrated near the bounding sphere), which is in turn approximated by Gaussian density on R, with the perimeter of a halfline asymptotic to v √(log (1/v2 )) for small v (see below), yielding the desired bound for v ~ 1/√n and also below, by a second application of concavity or the fact that from 0 to 1/√n, Pn/(n-1)  lies above the secant line, i.e. that the perimeter of a smaller spherical cap in the unit ball is greater than the perimeter from scaling everything down to the same volume. To see that as the ball gets smaller the requisite perimeter P decreases note that if you shrink the ball just a little, one thing you could do is just move the cap down a little so that its boundary still fits on the boundary of the shrunken ball. The volume enclosed, however, would then increase, so that the least perimeter for the given volume would decrease as claimed. For the approximation by Gaussian density, see e.g. [C, Prop. 2.2]; perimeter and volume on the sphere and on the line with Gaussian density are relatively close for x2/n small, x2/n is small at the perimeter and most of the volume because spherical volume is concentrated near the equator in high dimensions.

[C] Joe Corneli, Ivan Corwin, Stephanie Hurder, Vojislav Sesum, Ya Xu, Elizabeth Adams, Diana Davis, Michelle Lee, Regina Visocchi, Double bubbles in Gauss space and spheres, Houston J. Math.  34  (2008) 181-204.

[KLS] Ravi Kannan, László Lovász, and Miklós Simonovits, Isoperimetric problems for convex bodies and a localization lemma, Discrete Comput. Geom. 13 (1995), 541–559.

See also earlier post.

Update 1 December 2010. For exactly half the volume the two-dimensional case has been proved by

[EFKNT] L. Esposito, V. Ferone, B. Kawohl, C. Nitsch, and C. Trombetti, “The longest shortest fence and sharp Poincaré-Sobolev inequalities,” Arch. Ration. Mech. Anal. 206 (2012), 821–851, (2010),

though it fails for area-bisecting chords, for which Auerbach triangles provide the extreme case. For exactly half the volume the n-dimensional case was previously known for centrally symmetric convex bodies, even among bisecting chords or hyperplanes, as follows from work on the Buseman-Petty problem. Busemann-Petty considers two convex bodies B and C and asks whether the hypothesis that all hyperplanes through the origin have larger intersection with C than with B implies the conclusion that vol C > vol B. Although it is false in general, Lutwak [L, Thm. 2] proved it if B is a so-called intersection body, e.g., a ball. Therefore if the least perimeter to enclose half the volume of C and hence the intersection of every central hyperplane with C, if C is centrally symmetric, is greater than the intersection of every central hyperplane with the ball B, the volume of C cannot equal the volume of B (see Prop. B below). In the case of equality, all central hyperplanes cut C isoperimetrically and hence meet ∂C normally (where ∂C is C1, almost everywhere) and C is a ball.

The still open Hyperplane Conjecture says that there is a dimension-independent positive constant c such that every unit-volume convex body has a central hypersection of area at least c, equivalent to an upper bound on the so-called isotropic constant.

[L]  Erwin Lutwak, Intersection bodies and dual mixed volumes, Adv. Math. 71 (1988), 232-261.

(Update: My 2013 Williams College SMALL undergraduate research Geometry Group (completed by my 2015 Geometry Group) proved the 2D result for arbitrarily small areas and for triangles, parallelograms, and regular n-gons:

John Berry, Eliot Bongiovanni, Wyatt Boyer, Bryan Brown, Matthew Dannenberg, Paul Gallagher, David Hu, Jason Liang, Alyssa Loving, Zane Martin, Maggie Miller, Byron Perpetua, Sarah Tammen, and Yingyi Zeng, The convex body isoperimetric conjecture in R2, Rose-Hulman Und. Math. J., 2018.)

Well-known trivial asymptotic estimate for volume V and perimeter P of halfline in line with Gaussian density (2π)-1/2exp(-x2/2),

                       P ~ V√(log V-2)   as V -> 0,

follows immediately from the following lemma.

Lemma. V(t) ~ P(t)/t.

Proof. On the one hand,

V = (2π)-1/2texp(-x2/2)dx = (2π)-1/20exp(-(y+t)2/2)dy

= P(t) ∫0exp(-(2yt+y2)/2)dy ≤ P(t) ∫0exp(-ytdy = P(t)/t.

On the other hand,

V =  P(t) ∫0exp(-(2yt+y2)/2)dy ≥ P(t) ∫02exp(-(t+1)ydy

    = P(t)[1-exp(-2(t+1)]/(t+1) ~ P(t)/t.

15 May 2013. The following proposition is more or less well known; compare [KLS, Thm. 5.4], for which our argument improves the constant from 10 to 2√2 (since their psi is at most 2A/V). The Cheeger constant is defined as the infimum of the boundary area to volume ratio of regions in the interior of K with at most half the volume. For K convex, for regions bounded by hyperplanes, the cross-sectional area is a concave function of volume, and the infimum ratio occurs at half the volume.

Proposition A. For a convex body K in  Rnthe Cheeger constant is less than √π times the Cheeger constant of the ball of the same volume, even dividing the body by a hyperplane; in other words, K has such a cross-section splitting its volume in half with less than √π times the area of such a cross-section of the ball.

Proof. For given volume, the expected value of ∑xi2 attains its minimum of n/(n+2) for the unit ball about the origin. We may assume that the expected value of xn2 is at least 1/(n+2) for K, and that half the integral of xn2 and at most half the volume lies in {xn≥0}, where the expected value of xn2 is hence at least 1/(n+2). By Schwarz symmetrization etc., for this given volume V and expected value of xn2, the portion K+ of K in {xn≥0} has smallest base area A if it is a right round spherical cone of height say h, in which case E(xn2) = 2h2/(n+1)(n+2), A/V = n/h, and hence E(xn2)A2/V2 = 2n2/(n+1)(n+2). By the concavity of areas of horizontal sections of a convex body, it follows that there is a horizontal hyperplane splitting K in half with A/V at most  √2 n/√(n+1). For the ball the corresponding ratio is the volume of the unit (n-1)-ball divided by half the volume of the unit n-ball (which approaches √(2n/π)). One checks that the ratio of these ratios approaches √π from below as n approaches infinity.

Remarks. With orthogonal coordinates chosen to minimize E(xn2), our proof shows that some horizontal hyperplane slice satisfies E(xn2)A2/V2 = 2n2/(n+1)(n+2) ≤ 2. Similarly, the portion K+ of K in {xn≥0} has smallest base area A if it is a cylinder of height say h, in which case E(xn2) = h2/3, A/V = 1/h, and hence E(xn2)A2/V2 ≥ 1/3.

The Kannan-Lovász-Simonovits conjecture says that the Cheeger constant of any logconcave density is
achieved to within a universal, dimension-independent constant factor by a hyperplane-induced subset.

22 May 2013. The following proposition, which we got from Emanuel Milman, improves the constant of Proposition A from √π ~ 1.7725 to e/2 ~1.3591. The result is more or less well known, the centrally symmetric case a special case of the result of Lutwak described above.

Proposition B. For a convex body K in  Rnthe Cheeger constant is less than e/2 times the Cheeger constant of the ball of the same volume, even dividing the body by a hyperplane; in other words, K has such a cross-section splitting its volume in half with less than e/2 times the area of such a cross-section of the ball. If K is centrally symmetric, e/2 may be improved to the sharp value of 1.

Proof. With center of mass at the origin, K is the graph in polar coordinates over {θ in Sn-1} of the distance r in that direction from the origin. Let Aθ denote the area of the cross-section perpendicular to θ. Then by Jensen’s inequality, averages over  Sn-1 satisfy

ave Aθ = aave rn-1 ≤ a(ave rn)(n-1)/n = c|K|(n-1)/n ,

with equality when K is the ball. It follows that some Aθ has no more area than the (area-minimizing) central slice of the ball. If K is centrally symmetric, slices through the center of mass divide the volume in half, and we have the sharp constant 1. In generally, the slice in a given direction splitting the volume in half may have more area than the slice through the center of mass. By Schwarz symmetrization etc., the extreme case is a spherical cone, more extreme as the dimension increases, for which the ratio is 21/n/2(1-1/(n+1))n-1, which converges up to e/2 as n approaches infinity.

Remark. As n approaches infinity, the cone slices approach the log-convex density exx ≥ 0. The log of the area of the slices convex is equivalent to area a convex function of volume, which implies that the ratio area/volume is decreasing.

9 February 2021. See “Longest minimal length partitions” by Beniamin Bogosel and Edouard Oudet (preprint February 4, 2021) for some numerical evidence and the consideration of multiple regions.

13 April 2021. Bo-Hshiung Wang and Ye-Kai Wang have proved the 2D conjecture for axis-symmetric curves with four vertices and for smooth perturbations of the circle: “A note on the convex body isoperimetric conjecture in the plane”


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