12 I. Convex Sets at Large
(2.4) Corollary. If S
Rd
is a compact set, then conv(S) is a compact set.
Proof. Let Δ Rd+1 be the standard d-dimensional simplex; see Problem 1 of
Section 2.2:
Δ = (α1, . . . , αd+1) :
d+1
i=1
αi = 1 and αi 0 for i = 1, . . . , d + 1 .
Then Δ is compact and so is the direct product
Sd+1
×Δ =
(
u1,... , ud+1; α1,... , αd+1
)
: ui S and (α1, . . . , αd+1) Δ .
Let us consider the map Φ : Sd+1 × Δ −→ Rd,
Φ(u1, . . . , ud+1; α1,... , αd+1) = α1u1 + . . . + αd+1ud+1.
Theorem 2.3 implies that the image of Φ is conv(S). Since Φ is continuous, the
image of Φ is compact, which completes the proof.
PROBLEMS.
1. Give an example of a closed set in R2 whose convex hull is not closed.
2. Prove that the convex hull of an open set in Rd is open.
3. An Application: Positive Polynomials
In this section, we demonstrate a somewhat unexpected application of Cara-
th´ eodory’s Theorem (Theorem 2.3). We will use Carath´ eodory’s Theorem in the
space of (homogeneous) polynomials.
Let us fix positive integers k and n and let H2k,n be the real vector space of
all homogeneous polynomials p(x) of degree 2k in n real variables x = (ξ1, . . . , ξn).
We choose a basis of H2k,n consisting of the monomials
ea = ξ1
α1
· · · ξn
αn
for a = (α1, . . . , αn) where α1 + . . . + αn = 2k.
Hence dim H2k,n =
(
n+2k−1
2k
)
. At this point, we are not particularly concerned
with choosing the “correct” scalar product in H2k,n. Instead, we declare {ea} the
orthonormal basis of H2k,n, hence identifying H2k,n = Rd with d =
(
n+2k−1
2k
)
.
We can change variables in polynomials.
(3.1) Definition. Let U :
Rn
−→
Rn
be an orthogonal transformation and let
p H2k,n be a polynomial. We define q = U(p) by
q(x) = p
(
U
−1
x
)
for x = (ξ1, . . . , ξn).
Clearly, q is a homogeneous polynomial of degree 2k in ξ1,... , ξn.
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