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
−→ 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
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
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) ∈
gives rise to a function
fx = ξ1f1 + . . . + ξmfm,
which approximates g uniformly within the error .
In the problems below, T = [0, 1] and fi(τ) = τ
i = 0, . . . , m (note that we
start with f0).
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) = eτ 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],
Aτ = ∅. In other words, for every m + 1 points τ1,... , τm+1 there is a