164 CHAPTER 6. FIXED POINT THEOREMS

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

n

∑i=1

ψ i (x) = 1, (6.6)

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)), (6.7)

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

Proof: Using the compactness of f (K) which implies this set is totally bounded, thereexists an r net

{y1, · · · ,yn} ⊆ f (K)⊆ K

such that {B(yi,r)}ni=1 covers f (K). Let

φ 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 6.6 is satisfied. Indeed the denominator is not zero because x is in one of the B(yi,r).Thus it is obvious that the sum of these ψ i ( f (x)) equals 1 for x ∈ K. Now let fr be givenby 6.7 for x ∈ K. For such x,

f (x)− fr (x) =n

∑i=1

( f (x)− yi)ψ i ( f (x))

Thusf (x)− fr (x) = ∑

{i: f (x)∈B(yi,r)}( f (x)− yi)ψ i ( f (x))

+ ∑{i: f (x)/∈B(yi,r)}

( f (x)− yi)ψ i ( f (x))

= ∑{i: f (x)−yi∈B(0,r)}

( f (x)− yi)ψ i ( f (x)) =

∑{i: f (x)−yi∈B(0,r)}

( f (x)− yi)ψ i ( f (x))+ ∑{i: f (x)/∈B(yi,r)}

0ψ i ( f (x)) ∈ B(0,r)

because 0 ∈ B(0,r), B(0,r) is convex, and 6.6. It is just a convex combination of things inB(0,r). ■

Note that we could have had the yi in f (K) in addition to being in f (K). This wouldmake it possible to eliminate the assumption that K is closed later on. All you really needis that K is convex.

We think of fr as an approximation to f . In fact it is uniformly within r of f on K. Thenext lemma shows that this fr has a fixed point. This is the main result and comes from theBrouwer fixed point theorem in Rn. This will be an approximate fixed point.