A STABILITY RESULT FOR A VOLUME RATIO

(1)

A STABILITY RESULT FOR A VOLUME RATIO

By

Daniel Hug and Rolf Schneider

Mathematisches Institut, Albert-Ludwigs-Universit¨at Eckerstr. 1, D-79104 Freiburg. i. Br., Germany

e-mail: daniel.hug@math.uni-freiburg.de, rolf.schneider@math.uni-freiburg.de

ABSTRACT

For a convex body K in Rⁿ, the volume quotient is the ratio of the smallest volume of the circumscribed ellipsoids to the largest volume of the inscribed ellipsoids, raised to power 1/n. It attains its maximum if and only if K is a simplex. We improve this result by estimating the Banach–Mazur distance of K from a simplex if the volume quotient of K is close to the maximum.

Introduction and Result

We work in Euclidean spaceRⁿ(n≥2) with scalar producth·,·iand normk·k. LetKⁿ denote the set of compact, convex sets in Rⁿ with nonempty interiors (convex bodies).

ForK ∈ Kⁿ, letE^J(K) denote the ellipsoid of maximal volume contained inK (the John ellipsoid). By a result of John [1], the concentric homothetic ellipsoid n(EJ(K)− c)+c, wherecdenotes the center ofE^J(K), contains the bodyK. IfK is a simplex, then the factorn cannot be decreased, but the simplex is not characterized by this extremal property. This changes if also shifts are allowed. The extended Banach–Mazur distance of not necessarily symmetric convex bodies K, L∈ Kⁿ is defined by

dBM(K, L) := inf{λ≥1 :∃α∈Aff(n)∃x∈Rⁿ:L⊆αK ⊆λL+x}

where Aff(n) denotes the set of bijective affine transformations of Rⁿ. If Bⁿ denotes the Euclidean unit ball, then John’s result implies that

dBM(K, Bⁿ)≤n (1)

for K ∈ Kⁿ. Here, equality holds if and only if K is a simplex. This was proved by Leichtweiß [2] and was rediscovered by Palmon [3].

This work was supported in part by the European Network PHD, FP6 Marie Curie Actions, RTN, Contract MCRN-511953.

(2)

As soon as one has uniqueness, the question for a stability improvement of the inequality can be raised. For the inequality (1), such a stability result seems to be un- known, but we will prove stability for a weaker version of (1), the inequality (2) below.

LetE^L(K) be the ellipsoid of minimal volume containingK(the L¨owner ellipsoid), and let

vq(K) :=

V(EL(K)) V(E^J(K))

¹/n

,

where V denotes the volume (and vq stands for ‘volume quotient’). Clearly, vq(K)≤dBM(K, Bⁿ)≤n.

For the inequality

vq(K)≤n, (2)

mentioned by Leichtweiß [2] (Korollar), in which equality holds precisely for simplices, we establish an improvement in the form of a stability estimate. By Tⁿ we denote an n-dimensional simplex in Rⁿ.

Theorem. There exist constants c0(n), 0(n)>0 depending only on the dimension n such that the following holds. If 0≤≤0(n) and

vq(K)≥(1−)n, then

d_BM(K, Tⁿ)≤1 +c0(n)¹^/⁴.

Rough estimates yield that the constant c0(n) can be chosen of order n⁷. We do not know whether the order of is optimal.

Proof of the Theorem

We write Sⁿ⁻¹ = ∂Bⁿ for the boundary of the unit ball Bⁿ with center 0. Suppose that K ∈ Kⁿ is such that vq(K) ≥(1−)n, where ∈ [0, 0] and 0 = 0(n) ≤1 will be specified in the course of the proof. We assume, without loss of generality, that EJ(K) =n⁻¹Bⁿ.

Lemma 1. Under the above assumptions, the support function h(K,·) of K satisfies h(K, u)h(K,−u)≥ 1

n −6√

(3)

for all u∈Sⁿ⁻¹.

Proof. Since E^J(K) =n⁻¹Bⁿ, it follows from vq(K)≥(1−)n that

V(EL(K))≥[(1−)n]ⁿn⁻ⁿκ_n = (1−)ⁿκ_n, (4) where κn :=V(Bⁿ), and it follows from John’s theorem that K ⊆Bⁿ.

(3)

Let u ∈ Sⁿ⁻¹. We set a :=h(K, u) and b := h(K,−u), hence a, b ∈ [1/n,1]. For x∈K, we have

kxk²−1≤ 0 (5)

since K ⊆Bⁿ, and

hx, ui −a≤0, hx, ui+b≥0. (6) Combining (5) and (6), we obtain for x∈K and allλ≥0 that

kxk²−1 +λ(hx, ui −a)(hx, ui+b)≤0. (7) The set of all x∈Rⁿ satisfying (7) is an ellipsoid Ea,b(λ) withV(Ea,b(λ)) =fa,b(λ)κn, where

fa,b(λ) :=

"

1 + (1 +ab)λ+

a+b 2

² λ²

#ⁿ₂

(1 +λ)⁻ⁿ⁺¹² , λ ≥0 (8) (so far, we essentially followed Leichtweiß [2]). From (4) and the fact thatK ⊆Ea,b(λ), we have

fa,b(λ)≥(1−)ⁿ, λ≥0. (9)

We derive a lower bound for ab. By (8) and (9), 1 + (1 +ab)λ+

a+b 2

²

λ² ≥(1 +λ)ⁿ⁺¹ⁿ (1−)². Put (1−)² =:β. Since a+b ≤2, we deduce that

1 + (1 +ab)λ+λ² ≥[1 + (1 +n⁻¹)λ]β, which is equivalent to

1−β+ [1 +ab−β(1 +n⁻¹)]λ+λ² ≥0, λ≥0. (10) We remark that for = 0 this yields ab ≥ 1/n (as also obtained in [2] and [3], in different ways). Now we assume ≤1 and assert that

ab≥ 1 n −6√

. (11)

The polynomial of degree two in (10) has discriminant

D= [1 +ab−β(1 +n⁻¹)]²−4(1−β).

If D≤0, then −1−ab+β(1 +n⁻¹)≤2√

1−β and hence ab≥(1−)²(1 +n⁻¹)−1−2p

1−(1−)². (12)

If D > 0, the condition (10) implies that the linear term of the polynomial has a nonnegative coefficient, and this also implies (12). From (12) we get

ab≥ 1

n −3−2√

2≥ 1 n −6√

,

(4)

which establishes (11) and thus proves Lemma 1.

The ball n⁻¹Bⁿ is a maximal ball contained in K. It is known that this implies 0 ∈ conv(∂K ∩n⁻¹Sⁿ⁻¹). (Otherwise, conv(∂K ∩ n⁻¹Sⁿ⁻¹) and 0 can be strictly separated by a hyperplane. Hence, there is closed half ball of n⁻¹Bⁿ with positive distance from ∂K. The union of a suitable neighborhood of this half ball and ofn⁻¹Bⁿ is contained in K and contains a ball larger than n⁻¹Bⁿ, a contradiction.) Hence, by Carath´eodory’s theorem, there are mutually distinct vectors u1, . . . , um ∈ Sⁿ⁻¹ and numbers α1, . . . , α_m >0 such that m≤n+ 1,

m

X

i=1

αiui = 0,

m

X

i=1

αi = 1 (13)

and

n⁻¹ui ∈∂K∩n⁻¹Sⁿ⁻¹, i= 1, . . . , m. (14) In the following, the unit vectors u1, . . . , um and numbers α1, . . . , αm are fixed. Our aim is to show that, if is sufficiently small, then m=n+ 1 and u1, . . . , un+1 are close to the vertex vectors of a regular simplex with centroid 0. For this, we first use Lemma 1 to estimate how close the values of hui, uji, i6=j, and αi are to those in the regular case.

Lemma 2. There are positive constants 3 ≤ 1 and c3 with the following properties.

If ∈[0, 3], then m =n+ 1,

hui, uji+ 1 n

≤c3¹^/⁴ for i, j = 1, . . . , n+ 1, i6=j, (15)

and

αi− 1 n+ 1

≤c3¹^/⁴ for i= 1, . . . , n+ 1. (16) Proof. We set β_i := h(K,−u_i) for i = 1, . . . , m. Since h(K, u_i) = 1/n by (14), the inequality (3) implies that

βi ≥1−6n√

, i= 1, . . . , m.

We choose zi ∈ K∩H(K,−ui), where H(K,−ui) is the supporting hyperplane of K with outer normal vector −ui. Since K ⊆ Bⁿ, we can write zi = −βiui+γiwi, where wi ∈Sⁿ⁻¹∩u^⊥_i and γi ∈

0, c1¹^/⁴

with c1 :=√ 12n.

The following figure illustrates the situation: For = 0, the simplex S would be regular and inscribed to the outer ball Bⁿ, and K would coincide with S. For small >0, we show that K is close to such a regular simplex.

(5)

u_i

n⁻¹u_i

z_i

−β_iu_i o

H(K,−u_i) S K

For anyx∈K andi∈ {1, . . . , m}, we havehui, xi ≤h(K, ui) = 1/n. Since zj ∈K, we deduce that, for i, j ∈ {1, . . . , m} and i6=j,

−βjhui, uji+γjhui, wji ≤1/n, (17) which implies

hui, uji ≥ − 1 nβj

+γj

βjhui, wji ≥ − 1

n(1−6n¹^/²) + γj

βjhui, wji. We set 1 := (12n)⁻² and assume ∈[0, 1]. Then we can estimate

1

1−6n¹^/² ≤1 + 12n¹^/²

and

γj

β_jhui, wji

≤ c1¹^/⁴

1−6n¹^/² ≤2c1¹^/⁴, to obtain

hui, uji ≥ −1

n −c2¹^/⁴, i6=j (18)

(6)

with c2 := 12 + 2c1.

Next, we introduce the auxiliary vectors:=u1+. . .+um. Then, fori∈ {1, . . . , m}, hs, uii = 1 +

m

X

j=1 j6=i

hui, uji

≥ 1 + (m−1)

−1

n −c2¹^/⁴

= n+ 1−m

n −(m−1)c2¹^/⁴.

We set 2 := (n²c2)⁻⁴ (which is < 1) and require that ∈[0, 2]. Then we can deduce that m=n+ 1. In fact, otherwise hs, u_ii>0 for i= 1, . . . , m, and by (13)

0 =

* s,

m

X

i=1

αiui

+

=

m

X

i=1

αihs, uii>0, a contradiction. As another consequence, we have

hs, uii ≥ −nc2¹^/⁴, i= 1, . . . , n+ 1. (19) Our next purpose is to estimatehui, ujialso from above fori6=j. For this, we first establish upper and lower bounds for the coefficients α1, . . . , αn+1 in (13):

0 =

* ui,

n+1

X

j=1

αjuj

+

=αi+

n+1

X

j=1 j6=i

αjhui, uji

≥ αi+

n+1

X

j=1 j6=i

αj

−1

n −c2¹^/⁴

= αi

1 + 1

n +c2¹^/⁴

− 1

n −c2¹^/⁴, hence

αi ≤ 1 +nc2¹^/⁴

n+ 1 +nc2¹^/⁴ ≤ 1

n+ 1 +c2¹^/⁴, i= 1, . . . , n+ 1. (20) On the other hand,

αi = 1−

n+1

X

j=1 j6=i

αj ≥1 +n

− 1

n+ 1 −c2¹^/⁴

= 1

n+ 1 −nc2¹^/⁴. (21) Define 3 := (2n(n+ 1)c2)⁻⁴ (which is < 2). Then the obtained bounds for αi imply

that 1

2(n+ 1) ≤αi ≤ 2

n+ 1, i= 1, . . . , n+ 1, (22)

(7)

if ∈[0, 3], which we assume in the following.

Now we are in a position to assert that hui, uji ≤ −1

n + 2n²c2¹^/⁴, i6=j. (23) To check this, we assume to the contrary that, for some pair i, j ∈ {1, . . . , n+ 1}with i6=j,

hui, uji>−1

n + 2n²c2¹^/⁴. Then, using (18), we get

hs, uii >1 +

−1

n + 2n²c2¹^/⁴

+ (n−1)

−1

n −c2¹^/⁴

= (2n² −n+ 1)c2¹^/⁴, and the same estimate is obtained for hs, u_ji. From this we infer, using (19) and (22), that

0 =

* s,

n+1

X

k=1

α_ku_k +

=

n+1

X

k=1

α_khs, u_ki

≥ α_ihs, u_ii+α_jhs, u_ji −(n−1) 2n

n+ 1c2¹^/⁴

> c2¹^/⁴ ≥0,

a contradiction. Setting c3 := 2n²c2, we obtain (15) from (18) and (23), and (16) from (20) and (21). This completes the proof of Lemma 2.

In the course of this proof, we have obtained points z1, . . . , zn+1 such that conv{z1, . . . , z_n+1} ⊆K ⊆

n+1

\

i=1

H⁻(u_i,1/n) (24)

with H⁻(u, t) :={x∈Rⁿ :hx, ui ≤t}, and

kzi+uik=k(1−βi)ui+γiwik ≤1−βi+γi ≤c4¹^/⁴ (25) with c4 :=√

12n+ 6n. These points will be needed later.

By Lemma 2, the unit vectors u1, . . . , un+1 have scalar products which are close to those of the vertex vectors of a regular simplex with centroid 0. From this, we must now deduce that u1, . . . , un+1 are, in fact, close to the vertices of a suitable regular simplex.

Lemma 3. There are vectors v1, . . . , vn+1 ∈Sⁿ⁻¹ satisfying hvi, vji=−1/n for i, j ∈ {1, . . . , n+ 1} with i 6= j and ku_i −v_ik ≤ c5¹^/⁴ for i = 1, . . . , n+ 1, where c5 = 6(n+ 1)³^/²c3.

(8)

Proof. In the Euclidean space Rⁿ×R(with the standard scalar product, also denoted by h·,·i) we define the vectors Ui :=p

n/(n+ 1)(ui,1/√

n) for i = 1, . . . , n+ 1. They satisfy kU_ik= 1 and

hUi, Uji= n n+ 1

hui, uji+ 1 n

for i6=j.

By (15),

|hUi, Uji| ≤c3¹^/⁴ =:η fori6=j. (26) We write the vectors of Rⁿ×R as columns and define M as the matrix with columns U1, . . . , Un+1. Let (E1, . . . , En+1) be an orthonormal basis of Rⁿ×R, and denote by k|Mk|the Hilbert-Schmidt norm of M, thus

k|Mk|² =

n+1

X

i=1

kMEik².

By polar decomposition and diagonalization, the nonsingular matrix M has a re- presentation M = S1DS2 with orthogonal matrices S1, S2 and a diagonal matrix D = diag(d1, . . . , dn+1) with di > 0. Let I denote the (n + 1, n + 1) unit matrix.

By (26) and the invariance properties of the Hilbert-Schmidt norm, we have

(n+ 1)η≥ k|M^>M −Ik|=k|D²−Ik|=

n+1

X

i=1

(d²_i −1)²

!¹/2

≥

n+1

X

i=1

(di−1)²

!¹/2

.

Setting D−I =: ˜D, we get

M =S1(I + ˜D)S2 =S+S1DS˜ 2

with the orthogonal matrix S = S1S2. The columns of S yield an orthonormal basis (X1, . . . , Xn+1) ofRⁿ×R. Fori∈ {1, . . . , n+ 1} we have, ifEi is theith vector of the standard basis of Rⁿ×R,

kUi−Xik=kMEi−SEik ≤ k|M −Sk|=k|S1DS˜ 2k|=k|D˜k|=

n+1

X

i=1

(di−1)²

!¹/2

, hence

kUi −Xik ≤(n+ 1)η.

The hyperplane H through the orthonormal vectors X1, . . . , X_n+1 has a normal vector

N_X := 1

n+ 1(X1+· · ·+X_n+1) of lengthkN_Xk= 1

√n+ 1. We compare this vector with the vectors

NU := 1

n+ 1(U1+· · ·+Un+1) and N0 :=

n+1

X

i=1

αiUi.

(9)

By (13), N0 = (0,1/√

n+ 1). Further, kNX −NUk ≤(n+ 1)η and, by (16), kNU −N0k=

n+1

X

i=1

1

n+ 1 −αi

Ui

≤(n+ 1)η.

We deduce that kNX −N0k ≤2(n+ 1)η.

By the latter estimate, there exists a rotationϑofRⁿ×Rfor which ϑN_X =N0 and kϑZ −Zk ≤2(n+ 1)η√

n+ 1kZk= 2(n+ 1)³^/²ηkZk

for each vector Z ∈Rⁿ×R. The hyperplane H passes throughNX and is orthogonal to it, so ϑH passes through N0 and is orthogonal to it, and it has the same distance from the origin as H. It follows that all points of ϑH have last coordinate 1/√

n+ 1.

Therefore, we can define vectors v1, . . . , vn+1 ∈Rⁿ by ϑX_i =

r n n+ 1

v_i, 1

√n

.

From kϑXik= 1 it follows that kvik= 1, and for i6=j it follows from hϑXi, ϑXji= 0 that hvi, vji=−1/n. Finally,

r n

n+ 1kvi−uik=kϑXi−Uik ≤ kϑXi−Xik+kXi−Uik ≤3(n+ 1)³^/²η, hence

kui−vik ≤c5¹^/⁴ (27) with c5 := 6(n+ 1)³^/²c3. This proves Lemma 3.

To complete the proof of the theorem, we now define the n-dimensional simplices T :=

n+1

\

i=1

H⁻(vi,1/n) and S :=

n+1

\

i=1

H⁻(ui,1/n),

whereT is regular and has vertices−v1, . . . ,−vn+1. Our aim is to show thatKhas small Banach-Mazur distance from T. The definition of the Banach-Mazur distance involves inclusions of convex bodies, and these can be approached via estimates of support functions, or Hausdorff distances. Therefore, we first transfer the available information (27) on the unit normal vectors of the simplices T, S into estimates for their support functions, via the polar simplices. For the following, observe that the regular simplex T has inradius 1/n and hence circumradius 1; its polar then has circumradiusn.

Let∈[0, 3]. The polar bodiesS^oandT^oare simplices with verticesnu1, . . . , nun+1

and nv1, . . . , nvn+1, respectively. If δ denotes the Hausdorff metric, we deduce from (27) that δ(S^o, T^o) ≤ nc5¹^/⁴. Assume that ∈ [0, 4] with 4 := (2nc5)⁻⁴ (which is

< 3). Then δ(Sô, Tô) ≤ 1/2, and since Bⁿ ⊂ Tô, this implies 2⁻¹Bⁿ ⊂ Sô. Clearly, n⁻¹Bⁿ ⊂ S, T and hence Sô, Tô ⊂ nBⁿ. For the radial function ρ and for u ∈ Sⁿ⁻¹ this gives

|ρ(Tô, u)−ρ(Sô, u)| ≤2nδ(Tô, Sô). (28)

(10)

To see this, assume that, say, ρ(Sô, u)< ρ(Tô, u), and letH⁻be a supporting halfspace of Sô at the boundary pointρ(Sô, u)u, with outer unit normal vectorν. The translated halfspace H⁻+δ(Tô, Sô)ν contains Tô, hence ρ(Tô, u)−ρ(Sô, u) ≤ δ(Tô, Sô)/hu, νi. On the other hand, hu, νi is bounded from below by the ratio of the radii of the inner and the outer ball. This proves (28).

From this we deduce

S ⊆T +c6¹^/⁴Bⁿ ⊆T +c6¹^/⁴nT = 1 +nc6¹^/⁴ T, and since K ⊆S,

K ⊆ 1 +nc6¹^/⁴

T. (29)

The points z1, . . . , z_n+1 appearing in (24) can be written in the form zi =−vi+pi with kpik ≤c7¹^/⁴,

for i = 1, . . . , n+ 1, where c7 := c4 +c5, by (25) and (27). Assume that ∈ [0, 5] with 5 := (2nc7)⁻⁴ (which is < 4). Then kpik ≤ (2n)⁻¹, hence the polytope P := conv{z1, . . . , zn+1} satisfies δ(P, T) ≤ (2n)⁻¹. Since n⁻¹Bⁿ ⊆ T, this implies (2n)⁻¹Bⁿ⊆P. We deduce that

1−2nc7¹^/⁴

(−vi) = 1−2nc7¹^/⁴

(zi−pi)∈ 1−2nc7¹^/⁴

P +c7¹^/⁴Bⁿ

⊆ 1−2nc7¹^/⁴

P + 2nc7¹^/⁴P =P ⊆K.

Since −v1, . . . ,−vn+1 are the vertices of T, it follows that 1−2nc7¹^/⁴

T ⊆K. (30)

From (29) and (30) we obtain

1−2nc7¹^/⁴

T ⊆K ⊆ 1 +nc6¹^/⁴ T.

This finally shows that

dBM(K, T)≤ 1 +nc6¹^/⁴

1−2nc7¹^/⁴ ≤1 +c8¹^/⁴ for ∈[0, 6], where 6 := (4nc7)⁻⁴ < 5 and c8 := 2n(c6+ 2c7).

Acknowledgement. We thank the referee for his valuable suggestions, which led to an improvement of the paper.

(11)

References

[1] F. John, Extremum problems with inequalities as subsidiary conditions, Courant Anniversary Volume, pp. 187–204, Interscience, New York, 1948.

[2] K. Leichtweiß, Uber die affine Exzentrizit¨¨ at konvexer K¨orper, Arch. Math. 10 (1959), 187–199.

[3] O. Palmon,The only convex body with extremal distance from the ball is the simplex, Israel J. Math. 80 (1992), 337–349.