3. An Application: Positive Polynomials 13

PROBLEMS.

1◦.

Check that (U1U2)(p) = U1(U2(p)).

2◦. Let

p(x) = x

2k

=

(

ξ1

2

+ . . . + ξn

2

)k

.

Prove that U(p) = p for any orthogonal transformation U.

It turns out that the polynomial of Problem 2, Section 3.1, up to a scalar

multiple, is the only polynomial that stays invariant under any orthogonal trans-

formation.

(3.2) Lemma. Let p ∈ H2k,n be a polynomial such that U(p) = p for every or-

thogonal transformation U. Then

p(x) = γ x

2k

= γ

(

ξ1

2

+ . . . + ξn

2

)k

for some γ ∈ R.

Proof. Let us choose a point y ∈

Rd

such that y = 1 and let γ = p(y). Let us

consider

q(x) = p(x) − γ x

2k.

Thus q is a homogeneous polynomial of degree 2k and q(Ux) = q(x) for any or-

thogonal transformation U and any vector x. Moreover, q(y) = 0. Since for every

vector x such that x = there is an orthogonal transformation Ux such that

Uxy = x, we have q(x) = q

(1

Uxy

)

= q(y) = 0 and hence q(x) = 0 for all x such that

x = 1. Since q is a homogeneous polynomial, we have q(x) = 0 for all x ∈

Rn.

Therefore, p(x) = γ x

2k

as claimed.

We are going to use Theorem 2.3 to deduce the existence of an interesting

identity.

(3.3) Proposition. Let k and n be positive integers. Then there exist vectors

c1,... , cm ∈

Rn

such that

x

2k

=

m

i=1

ci,x

2k

for all x ∈

Rn.

In words: the k-th power of the sum of squares of n real variables is a sum of 2k-th

powers of linear forms in the variables.

Proof. We are going to apply Carath´ eodory’s Theorem in the space H2k,n.

Let

Sn−1

= c ∈

Rn

: c = 1

be the unit sphere in

Rn.

For a c ∈

Sn−1,

let

pc(x) = c, x

2k

where x = (ξ1, . . . , ξn).