Table Of ContentA convex body whose centroid and Santalo´
point are far apart
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0 Mathieu Meyer, Carsten Schu¨tt and Elisabeth M. Werner †
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Abstract
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We give an example of a convex body whose centroid and Santal´o point
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A are “far apart”.
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∗Keywords: Santal´o point, . 2000 Mathematics Subject Classification: 52A20, 53A15
†Partially supported by an NSF grant, a FRG-NSF grant and a BSF grant
1
1 Introduction. The main theorem.
Inhissurveypaper[7],Gru¨nbaumintroducedmeasuresofasymmetryforconvexbodies
in Rn, i.e. compact, convex sets in Rn with nonempty interior. Measures of asymmetry
determine the degree of non-symmetricity of a non symmetric convex body. Recall that
a convex set K in Rn is centrally symmetric with respect to the origin if x K implies
∈
x K. More generally, K is centrally symmetric with respect to the point z if x K
− ∈ ∈
implies 2z x K.
− ∈
Gru¨nbaum argues that a measure of asymmetry should be a function that is 0 for
centrallysymmetricconvexbodiesandmaximalforthesimplexandonlyforthesimplex.
However, the simplex has enough symmetries to ensure that all affine invariant points
are identical. This leads to the question: How far apart can specific affine invariant
points be? This distance can be used as a measure for the asymmetry of the convex
body.
Of particular interest are the centroid g(K) and the Santal´o point s(K) of a convex
bodyK inRn. Thesepointsareofmajorimportanceandhavebeenwidelystudied. For
the (less well known) Santal´o point see e.g. [4, 12, 14, 13, 20]. Recent developments in
this direction include e.g. the works [1, 2, 3, 5, 6, 10, 15]. We show here that the points
can be very far apart, namely
Theorem 1 There is an absolute constant c > 0 and n N such that for all n n
0 0
there is a convex body C =C in Rn with ∈ ≥
n
g(C) s(C) g(C) s(C)
c k − k2 = k − k2 . (1)
≤ vol (ℓ C) w (u)
1 C
∩
is the Euclidean norm, l is the line through g =g(C) and s=s(C) and w (u)=
2 C
hkC·k(u)+hC(−u) is the width of C in directon of the unit vector u= kgg−−ssk2.
The proof actually shows that we can asymptotically determine the constant c of the
theorem.
g(C) s(C)
2
k − k
vol (ℓ C)
1
∩
is asymptotically with respect to the dimension, greater than or equal to
1 √eπ 2
1 − =0.142673...
− e √eπ+ 2
(cid:18) (cid:19) e 1
−
and is asymptotically with respect to the dimension, less than or equal to
1 √e 1
1 − =0.18383...
− e √e+ 1
(cid:18) (cid:19) e 1
−
Throughout the paper we use the following notations. Bn(a,r) is the Euclidean ball in
2
Rn centered at a with radius r. We denote Bn = Bn(0,1). Sn 1 is the boundary ∂Bn
2 2 − 2
2
of the Euclidean unit ball. For x Rn, let x = ( n x p)p1, if 1 p < and
x = max x , if p = an∈d let Bn =k kxp Rn :i=1x| i| 1 be th≤e unit∞ball of
tkhek∞space ln.1≤i≤n| i| ∞ p { ∈ P k kp ≤ }
p
For a convex body K in Rn we denote the volume by vol (K) (if we want to emphasize
n
the dimension) or by K . co[A,B] = λa+(1 λ)b : a A,b B,0 λ 1 is the
| | { − ∈ ∈ ≤ ≤ }
convex hull of A and B.
h (u)=max u,x is the supportfunction of K in direction u.
K x K
For x K, Kx∈=h(Ki x) = y Rn : y,z x 1 for all z K is the polar
◦
∈ − { ∈ h − i ≤ ∈ }
body of K with respect to x.
Let ξ Rn with ξ = 1 and s R. Then the (n 1)-dimensional section of K
2
∈ k k ∈ −
orthogonal to ξ through sξ is
K(s,ξ)= x K ξ,x =s .
{ ∈ |h i }
We just write K(s) if it is clear which direction ξ is meant.
The centroid g(K) of a convex body K in Rn is the point
1
g(K)= xdx.
vol (K)
n ZK
The Santal´o point s(K) of a convex body K is the unique point s(K) for which
vol (K)vol (Kx)
n n
attains its minimum. It is more difficult to compute the Santal´o point as it is defined
implicitly.
Before we go into the construction of the convex bodies that fulfill the properties
of Theorem 1, we discuss an example that shows that a more elaborate construction is
necessary to get the statement of Theorem 1.
For instance, it is not enough to take one half of a Euclidean ball
B = x Rn x 1,x 0 .
2 1
{ ∈ |k k ≤ ≥ }
We compute the centroid of B. The coordinates of the centroid are g(B)(2) = =
···
g(B)(n)=0 and
g(B)(1) = |B22n|Z01tvoln−1(cid:16)p1−t2B2n−1(cid:17)dt= 2v(onln+−11(cid:0))|BB2n2n−|1(cid:1) .
NowwecomputetheSantal´opointofB. ThepolarbodyoftheEuclideanballBn(λe ,1)
2 1
with center λe , 0 λ<1, and radius 1 is
1
≤
λ 2
x Rn (1 λ2)2 x + +(1 λ2)x2+ +(1 λ2)x2 1 . (2)
( ∈ (cid:12)(cid:12) − (cid:18) 1 1−λ2(cid:19) − 2 ··· − n ≤ )
(cid:12)
Therefore, the(cid:12) polar body of the half ball B with respect to λe is the convex hull of
(cid:12) 1
the vector −eλ1 and the ellipsoid (2). We estimate the volume of (B −λe1)◦. Since
(B λe ) contains the ellipsoid (2)
1 ◦
−
(1−λ2)−n+21 voln(B2n)≤voln((B−λe1)◦).
3
On the other hand, as (B λe ) is contained in the union of the ellipsoid (2) and a
1 ◦
−
cone with height 1 and with a base that is a Euclidean ball with radius (1+λ2) 1,
λ −
voln((B−λe1)◦)≤ (1volnλ(2B)2nn+2)1 + nvλol(n1−+1(λB22n)−n−1)1
−
For λ= 1 we get
√n
voln((B−λe1)◦)≤ 1− n1 −n+21 voln(B2n)+ √vonln(1−1+(B12n)−n1)1 .
(cid:18) (cid:19) n −
Since
voln 1(B2n−1) πn
−
vol (Bn) ∼ 2
n 2 r
we get
1 π
vol ((B λe ) ). √e+ vol (Bn) .
n − 1 ◦ e 2 n 2
(cid:18) r (cid:19)
On the other hand, for λ= γ
√n
n+1
γ2 − 2
vol ((B λe ) ) 1 vol (Bn).
n − 1 ◦ ≥ − n n 2
(cid:18) (cid:19)
With 1 t e t
−
− ≤ voln((B−λe1)◦)≥eγ2n2+n1 voln(B2n) .
Therefore,thereisaγ suchthatforalln Nwehaves(B)(1) γ . Thus kg(B)−s(B)k2
is of order 1 . ∈ ≤ √n vol1(ℓ∩B)
√n
The next lemma is well known (see [20]).
Lemma 2 For any convex body K, an interior point x of K is the Santalo´ point if and
only if 0 is the centroid of (K x) .
◦
−
This lemma can be rephrased as follows:
Let K be a convex body. Then 0 is the Santalo´ point of Kg(K).
Indeed, Kg(K) =(K g(K))◦ and(K g(K))◦◦ =K g(K). Since0isthecentroidof
− − −
(K g(K)) ,itfollowsbyLemma2that0istheSantal´opointof(K g(K)) =Kg(K).
◦◦ ◦
− −
For convex bodies K and L in Rn and natural numbers k, 0 k n, the coefficients
≤ ≤
V (K,L)=V(K,...,K,L,...,L)
n k,k
−
n k k
−
| {z } | {z }
4
in the expansion
n
n
voln(λ1K+λ2L)= k λ1n−kλk2Vn−k,k(K,L) (3)
k=0(cid:18) (cid:19)
X
are the mixed volumes of K and L (see [20]).
Now we introduce the convex bodies which will serve as candidates for Theorem 1.
For convex bodies K and L in Rn and real numbers a > 0 and b > 0, we construct a
convex body M in Rn+1
n
M =co[(K, a),(L,b)]= t(x, a)+(1 t)(y,b)x K,y L,0 t 1 . (4)
n
− { − − | ∈ ∈ ≤ ≤ }
The bodies we are using in Theorem 1 will be the polar bodies to M . The polar of M
n n
can be described as follows: For 1 s 1, the sections of M orthogonal to e
−a ≤ ≤ b n◦ n+1
and containing se are
n+1
M (s)=M (s,e )=(1+s a)K (1 s b)L . (5)
n◦ n◦ n+1 ◦∩ − ◦
We show this.
M = (z,s) Rn R x K :<z,x> sa 1and y L:<z,y >+sb 1
n◦ { ∈ × |∀ ∈ − ≤ ∀ ∈ ≤ }
= (z,s) Rn Rz (1+sa)K (1 bs)L
◦ ◦
{ ∈ × | ∈ ∩ − }
ThebodyM istheconvexhulloftwon-dimensionalfacesK andL. Inthefollowing
n
proposition we choose specific bodies for those faces. We choose them in such a way
that their volume product differs greatly, which has as effect that the centroid and the
Santal´o point of Mg(Mn) are “far apart”. One face is the Euclidean ball and the other
n
the unit ball of ℓn . We normalize both balls so that their volume is 1.
∞
Proposition 3 Let
Bn 1
K = 2 L= Bn .
|B2n|n1 2 ∞
and a=1 and b= 1 . Let M be the convex body in Rn+1 defined in (4). Then
e 1 n
−
(i) lim g(M )=0.
n n
→∞
(ii) Let
√e 1 2 √eπ
s = − = 0.290815... s = − = 0.225705...
0 −√e+ 1 − 1 √eπ+ 2 −
e 1 e 1
− −
Then for every ε>0 there is n such that for all n n the n+1-st coordinate of the
0 0
≥
centroid g(M ) of M satisfies
n◦ n◦
s ε g(M )(n+1) s +ε.
0− ≤ n◦ ≤ 1
5
The Santalo´ point of Mg(Mn), s = s(Mg(Mn)) = 0 and therefore the centroid g and
n ∗n n n∗
the Santalo´ point s of Mg(Mn) satisfy
∗n n
√e 1 2 √eπ
√e+−e−11 ≥linm→s∞up|gn∗ −s∗n|≥linm→i∞nf|gn∗ −s∗n|=linm→i∞nf|gn∗|≥ √e−π+ e−21 .
As for the proof of Proposition 3, it follows from Lemma 2 that s(Mg(Mn))=0. For the
n
centroid g we actually estimate the coordinates of g(M ) and not g(Mg(Mn)). Since
n∗ n◦ n
limg(M ) = 0 it suffices to use a perturbation argument. The remaining part of the
n
proof of the proposition is in the next section.
Proof of Theorem 1. The proof follows immediately from Proposition 3: For every
n, let C =M .
n n◦
2 Proof of Proposition 3
This section is devoted to the proof of Proposition 3. We will need several lemmas and
then give the proof of the proposition at the end of the section.
Lemma 4 Let K and L be convex bodies in Rn such that their centroid are at 0. Let
c>0 and let M be the convex body in Rn+1
n
M =co[(K,0),(L,c)].
n
Then the (n+1)-coordinate of the centroid g(M )(n+1) of M satisfies
n n
c n (k+1)V (K,L)
g(Mn)(n+1)=hg(Mn),en+1i= n+2 k=0n V n−(kK,k,L) .
P k=0 n−k,k
P
Proof. By definition
cw vol (M (w))dw
g(M ),e = 0 n n .
h n n+1i cvol (M (w))dw
R 0 n n
Note that R
w w
co[(K,0),(L,c)]= (1 )K+ L,w 0 w c .
− c c ≤ ≤
n(cid:16) (cid:17)(cid:12) o
Therefore (cid:12)
(cid:12)
cw vol ((1 w)K+ wL)dw
g(M ),e = 0 n − c c
h n n+1i cvol ((1 w)K+ wL)dw
R 0 n − c c
= cR01t voln((1−t)K+tL)dt.
R 01voln((1−t)K+tL)dt
R
6
Now we use the mixed volume formula (3). Thus
1 n n tk+1 (1 t)n k V (K,L) dt
0 k=0 k − − n−k,k
g(M ),e = c (cid:18) (cid:19)
h n n+1i R 1P n (cid:0) (cid:1)n tk (1 t)n k V (K,L) dt
0 k=0 k − − n−k,k
(cid:18) (cid:19)
cR Pn (k(cid:0)+(cid:1)1)V (K,L)
= n+2 k=0n V n−(kK,k,L) ,
P k=0 n−k,k
where we have also used the BetafunctPion
1 Γ(k)Γ(l)
B(k,l)= tk 1(1 t)l 1dt= , k,l>0.
− −
− Γ(k+l)
Z0
(cid:3)
The next lemma is well known ([8], p. 216, formula 54).
Lemma 5 For all n N and t 0
∈ ≥
n
n
voln(B2n+tB∞n )= k 2kvoln−k B2n−k tk,
k=0(cid:18) (cid:19)
X (cid:0) (cid:1)
with the convention that vol (B0)=1. Therefore, for 0 k n,
0 2 ≤ ≤
Vn k,k(B2n,Bn )=2k voln k B2n−k .
− ∞ −
(cid:0) (cid:1)
Lemma 6 The following formula holds
n k+1
1 k=0 Γ(1+n)nkΓ(1+n−k) 1
lim 2 2 =1 .
n n+2 Pn 1 − e
→∞ k=0 Γ(1+n)nkΓ(1+n−k)
2 2
P
Proof. One has
n k+1 n n k+1
1 k=0 Γ(1+n)nkΓ(1+n−k) 1 k=0 Γ(1+n)−n−nkΓ(1+k)
2 2 = 2 2
n+2 Pn 1 n+2 Pn 1
k=0 Γ(1+n2)nkΓ(1+n−2k) k=0 Γ(1+n2)n−nkΓ(1+k2)
P P
1 nk=0 (n−k+Γ(11)+Γ(k1+) n2)nk n+1 1 nk=0kΓΓ(1(1++n2k))nk
= 2 = 2 .
n+2P n Γ(1+n)nk n+2 − n+2Pn Γ(1+n)nk
2 2
k=0 Γ(1+k) k=0 Γ(1+k)
2 2
It is thus needed to proPve that P
n kΓ(1+n2)nk
1 k=0 Γ(1+k) 1
lim 2 = . (6)
n→∞nPn Γ(1+n2)nk e
k=0 Γ(1+k)
2
P
7
For every x 0,
≥
n kxk n kxk n xk 1
= =2x −
Γ(1+ k) kΓ(k) Γ(k)
k=0 2 k=1 2 2 k=1 2
X X X
= 2x 1 + n xk−1 =2x 1 +xn−2 xk .
Γ(1) Γ(k) Γ(1) Γ(1+ k)
(cid:16) 2 kX=2 2 (cid:17) (cid:16) 2 kX=0 2 (cid:17)
Thus for x=Γ(1+ n)n1
2
n kΓ(1+ n2)nk =2Γ 1+ n n1 1 +Γ 1+ n n1 n−2Γ(1+ n2)nk . (7)
Γ(1+ k) 2 Γ(1) 2 Γ(1+ k) !
kX=0 2 (cid:16) (cid:17) 2 (cid:16) (cid:17) kX=0 2
Let
n Γ(1+ n)nk n Γ(1+ n)nk
A = k 2 and B = 2 .
n Γ(1+ k) n Γ(1+ k)
k=0 2 k=0 2
X X
(6) is equivalent to
A 1
lim n = .
n→∞nBn e
By Stirling’s formula
n n n n n
2 e−n2√πn Γ 1+ 2 e−n2√πne6n−112
2 ≤ 2 ≤ 2
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
and thus
n n 1 n
(πn)21n Γ 1+ n (πn)21nen(6n1−12)
2e ≤ 2 ≤ 2e
r r
(cid:16) (cid:17)
Moreover,
Γ(1+ n)n2 n
B 2 .
n ≥ Γ(2) ≥ 2e
(In fact, it can be proved that Bn 2ec2n 2e2ne, where we write a(n) b(n), to mean
∼ ∼ ∼
that there are absolute constants c ,c >0 such that c a(n) b(n) c a(n). But for
1 2 1 2
≤ ≤
our purposes the above estimate is enough.) Also, for n 3
≥
Γ(1+ n)nn−1 Γ(1+ n)
2 = 2
Γ(1+ n−21) Γ(1+ n2)n1Γ(1+ n−21)
(n2)n2e−n2√πn e6n−112
≤ (πn)21n 2ne(n−21)n−21e−n−21 π(n−1)
p(n)n2√n ep 21n
≤ √n(n 1)n−21 (n 1) πn
− − (cid:16) (cid:17)
n n
= e 21n np 2 e21n 1+ 1 2 exp n+1 e.
πn n 1 ≤ n 1 ≤ 2(n 1) ≤
(cid:16) (cid:17) (cid:18) − (cid:19) (cid:18) − (cid:19) (cid:18) − (cid:19)
8
1
Therefore, by (7) and with c = Γ(1+ n) n,
n 2
(cid:0) (cid:1) Γ(1+n2)n−n1 + Γ(1+n2)nn
lim An = lim 2cn 1 +c 1 Γ(1+n−21) Γ(1+n2)
n→∞nBn n→∞ n Γ(12)Bn n − Bn
2c2 2 n 2 1
= lim n = lim = .
n n n n 2e e
→∞ →∞ (cid:18)r (cid:19)
(cid:3)
Lemma 7 Let a,b R, a>0, b>0 and
∈
Bn Bn
K = voln(B22n)n1 L= 2∞.
Let M = co[(K, a),(L,b)] be defined as in (4). Then the center of gravity g(M )
n n
−
satisfies
1 1
lim g(M )= ( a)+ 1 b.
n n e − − e
→∞ (cid:18) (cid:19)
Proof. By symmetry, the centroid of M is an element of the (n+1)-axis. Therefore
n
we only need to compute its (n+1)-th coordinate.
Instead of M = co[(K, a),(L,b)], we consider M˜ = co[(K,0),(L,a+b)] with
n n
−
centroid g˜=g˜ . The centroid g of M is then given by g =g˜ a.
n n n n n
−
We use Lemma 4 with c=a+b to get
a+b n (k+1)V (K,L)
hg˜(M˜n),en+1i= n+2 k=0n V n−(kK,k,L) .
P k=0 n−k,k
By Lemma 5 and the linearity of the mixedPvolumes in each component, we get
hg˜(M˜n),en+1i = na++b2 Pnk=0(nkk=0+v1o)lnv(oBln2n()Bnk2nv)onklnvokln−Bk2n(cid:0)−Bk2n−k(cid:1)
−
n k+1
a+b k=P0 k (cid:0) (cid:1)
= (Γ(1+n/2)n Γ(1+(n−k)/2). (8)
n+2 Pn 1
k=0 k
(Γ(1+n/2)n Γ(1+(n k)/2)
−
P
Now we apply Lemma 6. (cid:3)
Eventually we will have to investigate expressions of the form
Bn Bn
vol 2 t 1 , t 0.
n voln(B2n)n1 ∩ voln(B1n)n1 ! ≥
9
Schechtman, Schmuckenschl¨ager and Zinn established asymptotic formulas for large n
forthevolumesofBn tBn [17,18,19]. Todoso, theyconsideredindependentrandom
variables hp,...,hp pwi∩th dqensity
1 n p
e tp (9)
2Γ(1) −| |
p
Here, we need uniform estimates instead of asymptotic ones.
Lemma 8 Let 0<p,q < and let hp be a random variable with density (9). Then
∞
Ehp q = p ∞ tqe tpdt= p ∞tqe tpdt= 1 Γ q+1 (10)
| | 2Γ(1) | | −| | Γ(1) − Γ(1) p
p Z−∞ p Z0 p (cid:18) (cid:19)
In particular, for p=1 and q =1 respectively q =2 we get
Eh1 =1 respectively Eh1 2 =2. (11)
| | | |
For p=2 and q =2 we get
1
Eh2 2 = .
| | 2
The next lemma treats the Gaußian case of a more general statement which we will
prove below.
Lemma 9 Let (Ω,P) be a probability space. Let g :Ω R, 1 i n, be independent
i
→ ≤ ≤
N(0,1)-random variables. Then, for all γ >0 there is n such that for all n n
0 0
≥
1 n g (ω) 2 1
P ω n i=1| i | γ (12)
(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n1 Pni=1|gi(ω)|2 21 −rπ(cid:12)(cid:12)(cid:12)≤ ≥ 2
(cid:12)(cid:12) (cid:12)
(cid:12)(cid:12)(cid:12)(cid:12)(cid:0) P (cid:1) (cid:12)(cid:12)
Proof. By Chebyshev’s inequality. (cid:3)
Lemma 10 Let (Ω,P) be a probability space. Let h1 : Ω Rn, 1 i n, be indepen-
i → ≤ ≤
dent random variables with density e t. Then for all γ >0 there is n such that for all
−| | 0
n n
0
≥ P ω √2 γ (n1 ni=1|h1i(ω)|2)12 √2+γ 1 (13)
( (cid:12)(cid:12) − ≤ n1P ni=1|h1i(ω)| ≤ )≥ 2
(cid:12)
(cid:12) P
(cid:12)
Proof. Again, by Chebyshev’s inequality. (cid:3)
Lemma 11 Let (Ω,P) be a probability space. Let g :Ω R, 1 i n, be independent
i
→ ≤ ≤
N(0,1)-random variables. Then
1 n g t
voln(B2n∩t B1n)=n voln(B2n)Z0 rn−1P ( Pni=i=11|g|i|i2|)21 ≤ r! dr (14)
P
10
