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:
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