502 CHAPTER 18. TOPOLOGICAL VECTOR SPACES

covers f (K). Letφ i (y)≡ (r−ρA (y− yi))

+.

Thus φ i (y)> 0 if y ∈ yi +U and φ i (y) = 0 if y /∈ yi +U . For x ∈ f (K), let

ψ i (x)≡ φ i (x)

(n

∑j=1

φ j (x)

)−1

.

Then 18.5.21 is satisfied. Now let fU be given by 18.5.22 for x ∈ K. For such x,

f (x)− fU (x) = ∑{i: f (x)−yi∈U}

( f (x)− yi)ψ i ( f (x))

+ ∑{i: f (x)−yi /∈U}

( f (x)− yi)ψ i ( f (x))

= ∑{i: f (x)−yi∈U}

( f (x)− yi)ψ i ( f (x)) =

∑{i: f (x)−yi∈U}

( f (x)− yi)ψ i ( f (x))+ ∑{i: f (x)−yi /∈U}

0ψ i ( f (x)) ∈U

because 0 ∈U , U is convex, and 18.5.21.We think of fU as an approximation to f .

Lemma 18.5.7 For each U ∈B0, there exists xU ∈ convex hull of f (K)⊆ K such that

fU (xU ) = xU .

Proof: If fU (xU ) = xU and

xU =n

∑i=1

aiyi

for ∑ni=1 ai = 1, we need

n

∑j=1

y jψ j

(f

(n

∑i=1

aiyi

))=

n

∑j=1

a jy j.

Also, if this is satisfied, then we have the desired fixed point. This will be satisfied if foreach j = 1, · · · ,n,

a j = ψ j

(f

(n

∑i=1

aiyi

)); (18.5.23)

so, let

Σn−1 ≡

{a ∈ Rn :

n

∑i=1

ai = 1, ai ≥ 0

}