3. An Application: Positive Polynomials 15
3. Prove that one can choose m
(n+2k−1)
2k
in Proposition 3.3.
Remark: In his solution of Waring’s problem, for all positive integers k and n,
D. Hilbert constructed integer vectors ci and rational numbers γi such that
x
2k
=
m
i=1
γi ci,x
2k
for all x
Rn;
see, for example, Chapter 3 of [N96].
We apply Proposition 3.3 to study positive polynomials.
(3.4) Definition. Let p H2k,n be a polynomial. We say that p is positive
provided p(x) 0 for all x = 0. Equivalently, p H2k,n is positive provided
p(x) 0 for all x
Sn−1.
Similarly, a polynomial p H2k,n is non-negative if
p(x) 0 for all x.
PROBLEM.
1◦.
Prove that the set of all positive polynomials is a non-empty open convex
set in H2k,n and that the set of all non-negative polynomials is a non-empty closed
convex set in H2k,n.
We apply Proposition 3.3 to prove that a homogeneous polynomial is positive
if and only if it can be multiplied by a sufficiently high power of x 2 to produce a
sum of even powers of linear functions. The proof below is due to B. Reznick [R95]
and [R00].
(3.5) Proposition. Let p H2k,n be a positive polynomial. Then there exist a
positive integer s and vectors c1,... , cm Rn such that
x
2s−2kp(x)
=
m
i=1
ci,x
2s
for all x
Rn.
Sketch of Proof. For a polynomial f H2k,n,
f(x) =
a=(α1,... ,αn)
λaξ1
α1
. . . ξn
αn
,
let us formally define the differential operator
f(∂) =
a=(α1,... ,αn)
λa
∂α1
∂ξ1
α1
· · ·
∂αn
∂ξn
αn
.
Let us choose a positive integer s 2k and the corresponding identity of Proposition
3.3:
(3.5.1) x
2s
=
m
i=1
ci,x
2s.
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