A STABILITY RESULT FOR A VOLUME RATIO
By
Daniel Hug and Rolf Schneider
Mathematisches Institut, Albert-Ludwigs-Universität
Eckerstr. 1, D-79104 Freiburg. i. Br., Germany
e-mail: [email protected], [email protected]
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 space Rn (n ≥ 2) with scalar product h·, ·i and norm k·k. Let Kn
denote the set of compact, convex sets in Rn with nonempty interiors (convex bodies).
For K ∈ Kn , let EJ (K) denote the ellipsoid of maximal volume contained in K (the
John ellipsoid). By a result of John [1], the concentric homothetic ellipsoid n(EJ (K) −
c)+c, where c denotes the center of EJ (K), contains the body K. If K is a simplex, then
the factor n 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 B n denotes
the Euclidean unit ball, then John’s result implies that
dBM (K, B n ) ≤ 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.
1
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 unknown, but we will prove stability for a weaker version of (1), the inequality (2) below.
Let EL (K) be the ellipsoid of minimal volume containing K (the Löwner ellipsoid), and
let
1/n
V (EL (K))
,
vq(K) :=
V (EJ (K))
where V denotes the volume (and vq stands for ‘volume quotient’). Clearly,
vq(K) ≤ dBM (K, B n ) ≤ 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 T n 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
dBM (K, T n ) ≤ 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 S n−1 = ∂B n for the boundary of the unit ball B n 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−1 B n .
Lemma 1. Under the above assumptions, the support function h(K, ·) of K satisfies
h(K, u)h(K, −u) ≥
√
1
−6 n
(3)
for all u ∈ S n−1 .
Proof. Since EJ (K) = n−1 B n , it follows from vq(K) ≥ (1 − )n that
V (EL (K)) ≥ [(1 − )n]n n−n κn = (1 − )n κn ,
where κn := V (B n ), and it follows from John’s theorem that K ⊆ B n .
2
(4)
Let u ∈ S n−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 ⊆ B n , and
hx, ui − a ≤ 0,
hx, ui + b ≥ 0.
(6)
Combining (5) and (6), we obtain for x ∈ K and all λ ≥ 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) with V (Ea,b (λ)) = fa,b (λ)κn ,
where
"
2 # n2
n+1
a+b
λ≥0
(8)
fa,b (λ) := 1 + (1 + ab)λ +
λ2 (1 + λ)− 2 ,
2
(so far, we essentially followed Leichtweiß [2]). From (4) and the fact that K ⊆ Ea,b (λ),
we have
fa,b (λ) ≥ (1 − )n ,
λ ≥ 0.
(9)
We derive a lower bound for ab. By (8) and (9),
2
n+1
a+b
λ2 ≥ (1 + λ) n (1 − )2 .
1 + (1 + ab)λ +
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
− 6 .
n
(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
p
ab ≥ (1 − )2 (1 + n−1 ) − 1 − 2 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
1
− 3 − 2 2 ≥ − 6 ,
n
n
3
which establishes (11) and thus proves Lemma 1.
The ball n−1 B n is a maximal ball contained in K. It is known that this implies
0 ∈ conv(∂K ∩ n−1 S n−1). (Otherwise, conv(∂K ∩ n−1 S n−1 ) and 0 can be strictly
separated by a hyperplane. Hence, there is closed half ball of n−1 B n with positive
distance from ∂K. The union of a suitable neighborhood of this half ball and of n−1 B n
is contained in K and contains a ball larger than n−1 B n , a contradiction.) Hence, by
Carathéodory’s theorem, there are mutually distinct vectors u1 , . . . , um ∈ S n−1 and
numbers α1 , . . . , αm > 0 such that m ≤ n + 1,
m
X
m
X
αi ui = 0,
i=1
αi = 1
(13)
i = 1, . . . , m.
(14)
i=1
and
n−1 ui ∈ ∂K ∩ n−1 S n−1 ,
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, uj i, i 6= 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,
1
hui, uj i + ≤ c3 1/4
for i, j = 1, . . . , n + 1, i 6= j,
(15)
n
and
αi −
1 ≤ c3 1/4
n + 1
for i = 1, . . . , n + 1.
(16)
Proof. We set βi := h(K, −ui) for i = 1, . . . , m. Since h(K, ui) = 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 ⊆ B n ,√we can write zi = −βi ui + γi wi, where
1/4
wi ∈ S n−1 ∩ u⊥
with c1 := 12n.
i and γi ∈ 0, c1 The following figure illustrates the situation: For = 0, the simplex S would be
regular and inscribed to the outer ball B n , and K would coincide with S. For small
> 0, we show that K is close to such a regular simplex.
4
ui
n−1 ui
S
K
o
H(K, −ui )
−βi ui
zi
For any x ∈ K and i ∈ {1, . . . , m}, we have hui, xi ≤ h(K, ui) = 1/n. Since zj ∈ K,
we deduce that, for i, j ∈ {1, . . . , m} and i 6= j,
−βj hui, uj i + γj hui , wj i ≤ 1/n,
(17)
which implies
hui , uj i ≥ −
1
γj
1
γj
+ hui , wj i ≥ −
+ hui , wj i.
1/2
nβj
βj
n(1 − 6n ) βj
We set 1 := (12n)−2 and assume ∈ [0, 1 ]. Then we can estimate
and
to obtain
1
≤ 1 + 12n1/2
1 − 6n1/2
1/4
γj
1/4
hui, wj i ≤ c1 1 − 6n1/2 ≤ 2c1 ,
βj
hui , uj i ≥ −
1
− c2 1/4 ,
n
5
i 6= j
(18)
with c2 := 12 + 2c1 .
Next, we introduce the auxiliary vector s := u1 +. . .+um . Then, for i ∈ {1, . . . , m},
hs, uii = 1 +
m
X
j=1
j6=i
hui, uj i
1
1/4
≥ 1 + (m − 1) − − c2 n
=
n+1−m
− (m − 1)c2 1/4 .
n
We set 2 := (n2 c2 )−4 (which is < 1 ) and require that ∈ [0, 2 ]. Then we can deduce
that m = n + 1. In fact, otherwise hs, uii > 0 for i = 1, . . . , m, and by (13)
* m
+
m
X
X
0 = s,
αi ui =
αi hs, uii > 0,
i=1
i=1
a contradiction. As another consequence, we have
hs, ui i ≥ −nc2 1/4 ,
i = 1, . . . , n + 1.
(19)
Our next purpose is to estimate hui, uj i also from above for i 6= j. For this, we first
establish upper and lower bounds for the coefficients α1 , . . . , αn+1 in (13):
* n+1
+
n+1
X
X
0 =
ui ,
αj uj = αi +
αj hui , uj i
j=1
≥ αi +
n+1
X
j=1
j6=i
j=1
j6=i
1
1/4
αj − − c2 n
1
1
1/4
= αi 1 + + c2 − − c2 1/4 ,
n
n
hence
1 + nc2 1/4
1
≤
+ c2 1/4 ,
1/4
n + 1 + nc2 n+1
On the other hand,
αi ≤
αi = 1 −
n+1
X
j=1
j6=i
i = 1, . . . , n + 1.
1
1
1/4
αj ≥ 1 + n −
− c2 =
− nc2 1/4 .
n+1
n+1
(20)
(21)
Define 3 := (2n(n + 1)c2 )−4 (which is < 2 ). Then the obtained bounds for αi imply
that
1
2
≤ αi ≤
,
i = 1, . . . , n + 1,
(22)
2(n + 1)
n+1
6
if ∈ [0, 3 ], which we assume in the following.
Now we are in a position to assert that
hui , uj i ≤ −
1
+ 2n2 c2 1/4 ,
n
i 6= j.
(23)
To check this, we assume to the contrary that, for some pair i, j ∈ {1, . . . , n + 1} with
i 6= j,
1
hui, uj i > − + 2n2 c2 1/4 .
n
Then, using (18), we get
1
1
2
1/4
1/4
hs, uii > 1 + − + 2n c2 + (n − 1) − − c2 = (2n2 − n + 1)c2 1/4 ,
n
n
and the same estimate is obtained for hs, uj i. From this we infer, using (19) and (22),
that
* n+1
+ n+1
X
X
0 =
s,
αk uk =
αk hs, uk i
k=1
k=1
≥ αi hs, uii + αj hs, uj i − (n − 1)
2n
c2 1/4
n+1
> c2 1/4 ≥ 0,
a contradiction. Setting c3 := 2n2 c2 , 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 , . . . , zn+1 } ⊆ K ⊆
n+1
\
H − (ui, 1/n)
(24)
i=1
with H − (u, t) := {x ∈ Rn : hx, ui ≤ t}, and
with c4 :=
√
kzi + ui k = k(1 − βi )ui + γi wi k ≤ 1 − βi + γi ≤ c4 1/4
(25)
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 ∈ S n−1 satisfying hvi , vj i = −1/n for i, j ∈
{1, . . . , n + 1} with i 6= j and kui − vi k ≤ c5 1/4 for i = 1, . . . , n + 1, where c5 =
6(n + 1)3/2 c3 .
7
Proof. In the Euclidean space Rn × Rp(with the standard
√ scalar product, also denoted
by h·, ·i) we define the vectors Ui := n/(n + 1)(ui , 1/ n) for i = 1, . . . , n + 1. They
satisfy kUi k = 1 and
1
n
hUi , Uj i =
hui, uj i +
for i 6= j.
n+1
n
By (15),
|hUi , Uj i| ≤ c3 1/4 =: η
for i 6= 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
2
k|Mk| =
n+1
X
i=1
kMEi k2 .
By polar decomposition and diagonalization, the nonsingular matrix M has a representation M = S1 DS2 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|D 2 − Ik| =
n+1
X
i=1
(d2i − 1)2
!1/2
≥
n+1
X
i=1
(di − 1)2
!1/2
.
Setting D − I =: D̃, we get
M = S1 (I + D̃)S2 = S + S1 D̃S2
with the orthogonal matrix S = S1 S2 . The columns of S yield an orthonormal basis
(X1 , . . . , Xn+1 ) of Rn × R. For i ∈ {1, . . . , n + 1} we have, if Ei is the ith vector of the
standard basis of Rn × R,
kUi − Xi k = kMEi − SEi k ≤ k|M − Sk| = k|S1 D̃S2 k| = k|D̃k| =
n+1
X
i=1
(di − 1)2
!1/2
,
hence
kUi − Xi k ≤ (n + 1)η.
The hyperplane H through the orthonormal vectors X1 , . . . , Xn+1 has a normal
vector
1
1
NX :=
(X1 + · · · + Xn+1 )
of length kNX k = √
.
n+1
n+1
We compare this vector with the vectors
1
NU :=
(U1 + · · · + Un+1 ) and
n+1
8
N0 :=
n+1
X
i=1
αi Ui .
√
By (13), N0 = (0, 1/ n + 1). Further, kNX − NU k ≤ (n + 1)η and, by (16),
n+1 X
1
kNU − N0 k = − αi Ui ≤ (n + 1)η.
n+1
i=1
We deduce that kNX − N0 k ≤ 2(n + 1)η.
By the latter estimate, there exists a rotation ϑ of Rn × R for which ϑNX = N0 and
√
kϑZ − Zk ≤ 2(n + 1)η n + 1kZk = 2(n + 1)3/2 ηkZk
for each vector Z ∈ Rn × R. The hyperplane H passes through NX 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
r
n
1
ϑXi =
vi , √
.
n+1
n
From kϑXi k = 1 it follows that kvi k = 1, and for i 6= j it follows from hϑXi , ϑXj i = 0
that hvi , vj i = −1/n. Finally,
r
n
kvi − ui k = kϑXi − Ui k ≤ kϑXi − Xi k + kXi − Ui k ≤ 3(n + 1)3/2 η,
n+1
hence
kui − vi k ≤ c5 1/4
(27)
with c5 := 6(n + 1)3/2 c3 . This proves Lemma 3.
To complete the proof of the theorem, we now define the n-dimensional simplices
T :=
n+1
\
−
H (vi , 1/n)
and
i=1
S :=
n+1
\
H − (ui , 1/n) ,
i=1
where T is regular and has vertices −v1 , . . . , −vn+1 . Our aim is to show that K has 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 circumradius n.
Let ∈ [0, 3 ]. The polar bodies S o and T o are simplices with vertices nu1 , . . . , nun+1
and nv1 , . . . , nvn+1 , respectively. If δ denotes the Hausdorff metric, we deduce from
(27) that δ(S o , T o ) ≤ nc5 1/4 . Assume that ∈ [0, 4 ] with 4 := (2nc5 )−4 (which is
< 3 ). Then δ(S o , T o ) ≤ 1/2, and since B n ⊂ T o , this implies 2−1 B n ⊂ S o . Clearly,
n−1 B n ⊂ S, T and hence S o , T o ⊂ nB n . For the radial function ρ and for u ∈ S n−1
this gives
|ρ(T o , u) − ρ(S o , u)| ≤ 2nδ(T o , S o ).
(28)
9
To see this, assume that, say, ρ(S o , u) < ρ(T o , u), and let H − be a supporting halfspace
of S o at the boundary point ρ(S o , u)u, with outer unit normal vector ν. The translated
halfspace H − + δ(T o , S o )ν contains T o , hence ρ(T o , u) − ρ(S o , u) ≤ δ(T o , S o )/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
|h(T, u) − h(S, u)| = |ρ(T o , u)−1 − ρ(S o , u)−1 | ≤ 2|ρ(T o , u) − ρ(S o , u)| ≤ c6 1/4
with c6 := 4n2 c5 . But then
S ⊆ T + c6 1/4 B n ⊆ T + c6 1/4 nT = 1 + nc6 1/4 T,
and since K ⊆ S,
K ⊆ 1 + nc6 1/4 T.
(29)
The points z1 , . . . , zn+1 appearing in (24) can be written in the form
zi = −vi + pi
with kpik ≤ c7 1/4 ,
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 kpi k ≤ (2n)−1 , hence the polytope
P := conv{z1 , . . . , zn+1 } satisfies δ(P, T ) ≤ (2n)−1 . Since n−1 B n ⊆ T , this implies
(2n)−1 B n ⊆ P . We deduce that
1 − 2nc7 1/4 (−vi ) = 1 − 2nc7 1/4 (zi − pi ) ∈ 1 − 2nc7 1/4 P + c7 1/4 B n
⊆ 1 − 2nc7 1/4 P + 2nc7 1/4 P = P ⊆ K.
Since −v1 , . . . , −vn+1 are the vertices of T , it follows that
1 − 2nc7 1/4 T ⊆ K.
(30)
From (29) and (30) we obtain
This finally shows that
1 − 2nc7 1/4 T ⊆ K ⊆ 1 + nc6 1/4 T.
dBM (K, T ) ≤
1 + nc6 1/4
≤ 1 + c8 1/4
1 − 2nc7 1/4
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.
10
References
[1] F. John, Extremum problems with inequalities as subsidiary conditions, Courant
Anniversary Volume, pp. 187–204, Interscience, New York, 1948.
[2] K. Leichtweiß, Über die affine Exzentrizität konvexer Körper, 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.
11