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.