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We have shown that V ∩ W ⊂ {z ∈ X : θ(z) ∩ G = ∅}. By the foregoing, we have that {z ∈ X : θ(z) ∩ G = ∅} ⊂◦ X. 4 Lemma Let X be a fully normal space, Y a normed linear space and ϕ : X → P(Y ) an lsc carrier such that every ϕ(x) is convex. For every ǫ > 0, there exists a continuous mapping f : X → Y such that we have, for every x ∈ X, that d(f (x), ϕ(x)) < ǫ, where d denotes the norm distance in Y . Proof. For every y ∈ Y , let Uy = {x ∈ X : ϕ(x) ∩ Bd (y, ǫ) = ∅}, and note that we have Uy ⊂◦ X because ϕ is lsc.

Let ǫ > 0. Then there exists a finite set ∅ = B ⊂ A such that α∈B fα (x) > 1 − ǫ 2 and fα (x) > 0 for every α ∈ B. We denote by m the number |B| and by δ the minimum of the numbers ǫ 2m and fα (x), α ∈ B. As the functions fα , α ∈ B, are continuous, there exists k ∈ N such that fα (xn ) > fα (x) − δ 2 for all n ≥ k and α ∈ B. Now we have, for all n ≥ k and α ∈ B, that fα (xn )fα (x) > fα (x) − δ 2 fα (x) ≥ fα (x)2 − δ 2 ≥ fα (x) − δ ≥ fα (x) − ǫ 2m . By the foregoing, we have, for every n ≥ k, that α∈A fα (xn )fα (x) ≥ fα (xn )fα (x) > α∈B = i∈B Since we have, for every ℓ ∈ N, that α∈A α∈B fα (x) − α∈A ǫ 2 ≥1− ǫ 2 − ǫ 2 fα (x) − ǫ 2m = 1 − ǫ.

Xk+1. Assume that we have f (x0 , xk+1 ) > 2f (x0 , x1 ) and f (x0 , xk+1 ) > 2f (xk , xk+1 ). From the last inequality it follows, since f (x0 , xk+1 ) ≤ 2 max(f (x0 , xk ), f (xk , xk+1 )), that we have f (x0 , xk+1 ) ≤ 2f (x0 , xk ). Denote by ℓ the least i such that f (x0 , xk+1 ) ≤ 2f (x0 , xi ). Note that we have 1 < ℓ < k + 1 and f (x0 , xk+1 ) > 2f (x0 , xℓ−1 ). Since f (x0 , xk+1 ) ≤ 2 max(f (x0 , xℓ−1 ), f (xℓ−1 , xk+1 )), it fol- lows that f (x0 , xk+1 ) ≤ 2f (xℓ−1 , xk+1 )). By the inductive assumption, we have that ℓ−2 f (x0 , xℓ ) ≤ 2f (x0 , x1 ) + 4 f (xi , xi+1 ) + 2f (xℓ−1 , xℓ ) and i=1 k−1 f (xℓ−1 , xk+1 ) ≤ 2f (xℓ−1 , xℓ ) + 4 f (xi , xi+1 ) + 2f (xk , xk+1 ) .