arXiv:0812.4284v1 [math.MG] 22 Dec 2008
A REMARK ABOUT MAHLER’S CONJECTURE AND
THE MAXIMUM VALUE OF BOX SPLINES
ZHIQIANG XU
Abstract. In this paper, we recast a special case of Mahler’c
conjecture by the maximum value of box splines. This is the case
of polytopes with at most 2n + 2 facets. An asymptotic formula
for univariate box splines is given. Based on the formula, Mahler’s
conjecture is proved in this case provided n is big enough.
1. introduction
Let K be a symmetric convex body in Rn, and let K∗ be its polar
{x :  hy, xi  ≤ 1 for all y ∈ K}. An old conjecture of Mahler is
(1) vol(K) · vol(K∗) ≥
4n
n!
.
We note that an ndimensional parallelepiped, which has 2n facets,
gives equality in (1). So, the first nontrivial case of Mahler’s conjecture
is the symmetric convex body K with 2n + 2 facets. Such K can be
realized, up to affine invariance, as a onecodimensional section of an
(n + 1)dimensional cube. This case has been raised as a separate
problem by Ball [1]. Ball also shows an interesting relation between the
special case of Mahler’s conjecture and solutions of a scaling equation.
For each r ∈ R, we set
(2) 'A(r) := vol ((H + rA) ∩ Qn)
where Qn := [−1
2 , 1
2 ]n, A := (a1, . . . , an) is an unit vector and H :=
hAi⊥. Throughout this paper, without loss of generality, we suppose
0 < a1 ≤ a2 ≤ · · · ≤ an. In particular, Ball shows that the special case
of Mahler’s conjecture is equivalent to
(3) 'A(0) · E
Xn
k=1
ak"k
!
≥ 1,
where "k is a sequence of mutually independent random variables with
distribution P{1} = P{−1} = 1/2. In [7], the authors prove (3) for
n ≤ 8 by reducing the problem to a search over a finite set of polytopes
for each fixed dimension.
1
2 ZHIQIANG XU
An interesting observation is that 'A(·) is a box spline, a popular
tool in approximation theory. Therefore, we can recast the special case
of Mahler’s conjecture by the maximum value of box splines. Using
the saddle point approximation in statistics, we give an asymptotic
formula of 'A(·). Based on the asymptotic formula, we obtain the
following result:
Theorem 1. Suppose that there is a constant c0 so that an/a1 < c0 for
any n. Then there exists a positive integer N0(c0) so that
'A(0) · E
Xn
k=1
ak"k
!
≥ 1
when n ≥ N0(c0).
This paper is organized as follows. In Section 2, we introduce box
splines and show the relation between box splines and the special case
of Mahler’s conjecture. In Section 3, we use the saddle point approximation
to give an asymptotic formula of univariate box splines. Section
4 presents the proof of Theorem 1.
2. Box splines
Suppose that M is a s×n matrix. The box spline B(·M) associated
with M is the distribution given by the rule [2, 3]
(4)
Z
Rs
B(xM)(x)dx =
Z
[0,1)n
(Mu)du, ∈ D(Rs).
By taking = exp(−i·) in (4), we obtain the Fourier transform of
B(·M) as
(5) b B(M) =
Yn
j=1
1 − exp(−iTmj)
iTmj
, ∈ Cs.
The following formula shows the relation between box splines and the
volume of the section of unit cube (see [4], page 2):
(6) B(xM) =
voln(P ∩ [0, 1)n) p
 det(MMT )
,
where P := {y : My = x, y ∈ Rn
P +}. Set A := (a1, . . . , an). Recall that
j a2j
= 1 and 0 < a1 ≤ a2 ≤ · · · ≤ an. Then combining (2) and (6)
we have that
'A(·) = B(· + (a1 + · · · + an)/2A).
MAHLER’S CONJECTURE AND BOX SPLINES 3
Also, noting B(·A) reaching the maximum value at (a1+a2+· · ·+an)/2,
(3) is equivalent to
max
x
B(xA) ≥
1
E (
Pn
k=1 ak"k)
.
3. An asymptotic formula of univariate box splines
In this section, we shall present an asymptotic formula of B(·A). In
[8], Unser et. al. proved that B(·A) tends to the Gaussian function
as n increase provided a1 = a2 = · · · = an. Here, using the saddle
point approximation in statistics, for the general matrix A, we can
show the box spline B(·A) also converges to the Gaussian function as
n increases:
Theorem 2.
lim
n→∞
B(xA) =
p
6/ exp(−6(x −
X
j
aj/2)2),
where the limit may be taken pointwise or in Lp(R), p ∈ [2,∞).
Proof. The saddle point approximation of B(·A) is (see Theorem 6.1
and 6.2 in [5])
(7)
1
(2)1/2K′′(s0)1/2 exp(K(s0) − s0x).
Here,
K(s) := ln
Y
i
exp(ais) − 1
ais
and s0 satisfies K′(s0) = x. We can consider s0 as a function of x. So,
(8) K′(s0) − x = 0
defines an implicit relationship between s0 and x. Noting K′(0) =
(a1+· · ·+an)/2, the equation (8) implies that s0((a1+· · ·+an)/2) = 0.
Also, by (8), we have
(9) K′′(s0)s′
0 − 1 = 0,
which implies that
s′
0((a1 + · · · + an)/2) = 1/K′′(0) = 12/
X
j
a2j
= 12.
Using the similar method, we have
s′′
0((a1 + · · · + an)/2) = 0,
4 ZHIQIANG XU
and
s′′′
0 ((a1 + · · · + an)/2) = 864
Xn
j=1
a4j
!
/5 = O(1/n).
Using Taylor expansion at (a1 + · · · + an)/2, one has
(10) s0(x) = 12 · (x − (a1 + · · · + an)/2) + O(1/n).
Also, by (9), we have
K′′(s0) = 1/s′
0 = 1/12 + O(1/n).
Combining Taylor expansion of K(·) at 0 and (10), one has
K(s0) = 6(
X
j
aj)(x−(a1+· · ·+an)/2)+6(x−(a1+· · ·+an)/2)2+O(1/n).
By (10), we have
s0 · x = 12(x − (a1 + · · · + an)/2)x + O(1/n)
= 12(x − (a1 + · · · + an)/2)2 + 6(
X
j
aj)(x − (a1 + · · · + an)/2) + O(1/n).
From (7), the saddle point approximation of B(xA) is
p
6/ exp(−6(x −
X
j
aj/2)2) + O(1/n).
The properties of the saddle point approximation imply this theorem.
4. proof of the main result
To prove the main result, we firstly introduce a lemma.
Lemma 3. Put
F(s) :=
2
Z ∞
0
(1 −
cos(t/√s)
s
)
t
−
2dt, s
>
0.
Then
E
Xn
k=1
ak"k
!
≥ F(a−2
n ).
Proof. By Lemma 1.3 in [6], we have
E

Xn
k=1
ak"k
!
≥
Xn
k=1
a2
kF(a−2
k ).
MAHLER’S CONJECTURE AND BOX SPLINES 5
Since F is an increasing function (Lemma 1.4 [6]), we have
E

Xn
k=1
ak"k
!
≥
Xn
k=1
a2
kF(a−2
k ) ≥ F(a−2
n ).
Proof of Theorem 1. To prove the theorem, we only need to prove that
there exists a positive integer N0 so that
max
x
B(xA) ≥
1
E (
Pn
k=1 ak"k)
when n ≥ N0. Theorem 2 implies that
(11) lim
n→∞
max
x
B(xA) =
p
6/.
Since
P
j a2j
= 1 and an/a1 < c0, we have limn→∞ 1/an = ∞. By [6],
we have lims→∞ F(s) =
p
2/. We choose "0 so that
0 < "0 <
p
6/ −
p
/2
2
.
By Lemma 3, there is a positive integer N1 so that
(12)
1
E (
Pn
k=1 ak"k) ≤
1
F(a−2
n ) ≤
p
/2 + "0
provided n ≥ N1. The equation (11) implies that there is a positive
integer N2(c0) so that
(13) max
x
B(xA) ≥
p
6/ − "0
provided n ≥ N2(c0). We set N0(c0) := max{N1,N2(c0)}. Noting that
p
6/ − "0 ≥
p
/2 + "0,
combining (12) and (13), we have
maxB(xA) ≥
1
E (
Pn
k=1 ak"k)
when n ≥ N0(c0).
6 ZHIQIANG XU
References
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42(1982/83)99115.
4. C. de Boor, K. H¨ollig and S. Riemenschneider, Box Splines, SpringerVerlag,
New York, 1993.
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Friday, December 26, 2008
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