14 I. Convex Sets at Large
Hence we have pc H2k,n. Let
K = conv pc : c
Sn−1
be the convex hull of all polynomials pc. Since the sphere
Sn−1
is compact and the
map c −→ pc is continuous, the set pc : c Sn−1 is a compact subset of H2k,n.
Therefore, by Corollary 2.4, we conclude that K is compact.
Let us prove that γ x 2k K for some γ 0. The idea is to average the
polynomials pc over all possible vectors c Sn−1. To this end, let dc be the
rotation invariant probability measure on Sn−1 and let
(3.3.1) p(x) =
Sn−1
pc(x) dc =
Sn−1
c, x
2k
dc
be the average of all polynomials pc. We observe that p H2k,n. Moreover, since dc
is a rotation invariant measure, we have U(p) = p for any orthogonal transformation
U of
Rn
and hence by Lemma 3.2, we must have
p(x) = γ x
2k
for some γ R.
We observe that γ 0. Indeed, for any x = 0, we have pc(x) 0 for all c
Sn−1
except from a set of measure 0 and hence p(x) 0.
The integral (3.3.1) can be approximated with arbitrary precision by a finite
Riemann sum:
p(x)
1
N
N
i=1
pci (x) for some ci
Sn−1.
Therefore, p lies in the closure of K. Since K is closed, p By 2.3, we
can write p(x) = γ x
2k
as a convex combination of some
(K.
n+2k−1
2k
)Theorem
+1 polynomials
pci (x) = ci,x
2k.
Dividing by γ, we complete the proof.
It is not always easy to come up with a particular choice of ci in the identity
of Proposition 3.3.
PROBLEMS.
1. Prove Liouville’s identity:
(ξ1
2
+ ξ2
2
+ ξ3
2
+ ξ4
2)2
=
1
6
1≤ij≤4
(ξi +
ξj)4
+
1
6
1≤ij≤4
(ξi
ξj)4.
2. Prove Fleck’s identity:
(ξ1
2
+ξ2
2
+ξ3
2
+ξ4
2)3
=
1
60
1≤ijk≤4
(
ξi ±ξj ±ξk
)6
+
1
30
1≤ij≤4
(
ξi ±ξj
)6
+
3
5
1≤i≤4
ξi
6,
where the sums containing ± signs are taken over all possible independent choices
of pluses and minuses.
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