26 I. Convex Sets at Large
Proof. Let A(τ) : τ T be the sets defined in the proof of Proposition 6.2. Let
A = A(σ1) . . . A(σn).
First, we prove that A is compact. Indeed, by Problem 1, Section 6.2, the set A is
closed. It remains to show that A is bounded. Let us define a function
N :
Rm
−→ R, N(x) = max |fx(σi)| : i = 1, . . . , n .
Then N(λx) = |λ|N(x) for λ R, N(x) 0 for x = 0 and N is continuous (in
fact, N is a norm in
Rm).
Therefore,
min N(x) : x = 1 = δ 0
and N(x) δ x .
Now, if |g(σi) fx(σi)| for i = 1, . . . , n, we have |fx(σi)| |g(σi)| + for
i = 1, . . . , n. Letting
R = + max |g(σi)| : i = 1, . . . , n ,
we conclude that N(x) R, and, therefore, x R/δ for any x A. Thus A is
compact.
For τ T let A(τ) = A(τ)∩A. Then each set A(τ) is compact. Applying Helly’s
Theorem as in the proof of Proposition 6.2, we conclude that every intersection of
a finite family of sets A(τ) is non-empty. Therefore, every intersection A(τ1)
. . . A(τm+1) is a non-empty compact convex set. Therefore, By Corollary 4.3, the
intersection of all the sets A(τ) is non-empty and so is the intersection of all the
sets A(τ). A point
x = (ξ1, . . . , ξm)
τ∈T
A(τ)
gives rise to a function
fx = ξ1f1 + . . . + ξmfm,
which approximates g uniformly within the error .
PROBLEMS.
In the problems below, T = [0, 1] and fi(τ) = τ
i,
i = 0, . . . , m (note that we
start with f0).
1◦.
Prove that for any m + 1 distinct points τ1,τ2,... , τm+1 from [0, 1] the
intersection A(τ1) . . . A(τm+1) is compact.
2◦. Let g(τ) = for τ [0, 1]. Let us choose = 0. Check that each
intersection A(τ1)∩...∩A(τm+1) is not empty for any choice of τ1,... , τm+1 [0, 1],
but
τ∈[0,1]
= ∅. In other words, for every m + 1 points τ1,... , τm+1 there is a
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