498 CHAPTER 18. TOPOLOGICAL VECTOR SPACES

are exactly those which are infinite dimensional. The above reference claims this retractionproperty holds for any infinite dimensional normed linear space. I think it is fairly clearto see from the above example that this is not a surprising assertion. Recall that one ofthe proofs of the Brouwer fixed point theorem used the non existence of such a retraction,obtained using integration theory, to prove the theorem.

We let K be a closed convex subset of X a Banach space and let

f be continuous, f : K→ K, and f (K) is compact.

Lemma 18.5.2 For each r > 0 there exists a finite set of points

{y1, · · · ,yn} ⊆ f (K)

and continuous functions ψ i defined on f (K) such that for x ∈ f (K),

n

∑i=1

ψ i (x) = 1, (18.5.18)

ψ i (x) = 0 if x /∈ B(yi,r) , ψ i (x)> 0 if x ∈ B(yi,r) .

If

fr (x)≡n

∑i=1

yiψ i ( f (x)), (18.5.19)

then whenever x ∈ K,∥ f (x)− fr (x)∥ ≤ r.

Proof: Using the compactness of f (K), there exists

{y1, · · · ,yn} ⊆ f (K)⊆ K

such that{B(yi,r)}n

i=1

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

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

ψ i (x)≡ φ i (x)

(n

∑j=1

φ j (x)

)−1

.

Then 18.5.18 is satisfied. Indeed the denominator is not zero because x is in one of theB(yi,r). Thus it is obvious that the sum of these equals 1 on K. Now let fr be given by18.5.19 for x ∈ K. For such x,

f (x)− fr (x) =n

∑i=1

( f (x)− yi)ψ i ( f (x))