16 I. Convex Sets at Large

Let us see what happens if we apply f(∂) to both sides of the identity.

It is not very hard to see that

(3.5.2) f(∂) c, x

2s

=

(2s)!

(2s − 2k)!

f(c) · c, x

2s−2k.

It suﬃces to check the identity when f is a monomial and then it is straightforward.

One can also see that

(3.5.3) f(∂) x

2s

=

22ks!

(s − 2k)!

g(x) · x

2s−2k

for some g ∈ H2k,n.

The correspondence f −→ g defines a linear transformation

Φs : H2k,n −→ H2k,n

and the crucial observation is that Φs converges to the identity operator I as s

grows. Again, it suﬃces to check this when f is a monomial, in which case Φs(f) =

f + O(1/s) by the repeated application of the chain rule.

Since I−1 = I, for all suﬃciently large s the operator Φs is invertible and Φs −1

converges to the identity operator I as s grows. Now we note that the set of positive

polynomials is open; see Problem 1 of Section 3.4. Therefore, for a suﬃciently large

s the polynomial q = Φs

−1(p)

lies in a suﬃciently small neighborhood of p = I(p)

and hence is positive. Applying q(∂) to both sides of (3.5.1), by (3.5.2) and (3.5.3)

we get

22ks!

(s − 2k)!

Φs(q) · x

2s−2k

=

(2s)!

(2s − 2k)!

m

i=1

q(ci) ci,x

2s−2k.

Now Φs(q) = p and q(ci) 0 for i = 1, . . . , m. Rescaling, we obtain a representa-

tion of p · x

2s−2k

as a sum of powers of linear forms.

PROBLEMS.

1◦. Check formulas (3.5.2) and (3.5.3).

2◦. Check that Φs indeed converges to the identity operator on H2k,n as s

grows.

3. For polynomials f, g ∈ H2k,n, let us define

f, g = f(∂)g.

Note that since deg f = deg g, we get a number. Prove that f, g is a scalar product

in H2k,n and that

U(f),U(g) = f, g

for every orthogonal transformation of

Rn.

4. Construct an example of a non-negative polynomial p ∈ H2k,n for which the

conclusion of Proposition 3.5 does not hold true.

5. Using Proposition 3.5, deduce Polya’s Theorem: