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).
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