792 CHAPTER 23. DEGREE THEORY

Fε (−x) =mε

∑k=−mε

F (xk)φ k (F (−x))

∑i φ i (F (−x))=

mε

∑k=−mε

F (xk)φ−k (F (x))

∑i φ−i (F (x))

= −mε

∑k=−mε

F (−xk)φ−k (F (x))

∑i φ−i (F (x))=−

mε

∑k=−mε

F (x−k)φ−k (F (x))

∑i φ−i (F (x))

= −mε

∑k=−mε

F (xk)φ k (F (x))

∑i φ i (F (x))=−Fε (x)

The rest of the argument is the same.Now let F : Ω→ X be compact and consider I−F. Is (I−F)(∂Ω) closed? Suppose

(I−F)xk→ y. Then K ≡ y∪{(I−F)xk}∞

k=1 is a compact set because if you have any opencover, one of the open sets contains y and hence it contains all (I−F)xk except for finitelymany which can then be covered by finitely many open sets in the open cover. Hence, since(I−F) is proper, (I−F)−1 (K) is compact. It follows that there is a subsequence, stillcalled xk such that xk→ x ∈ (I−F)−1 (K). Then by continuity of F,

(I−F)(xk) → (I−F)(x)

(I−F)(xk) → y

It follows y = (I−F)x and so in fact (I−F)(∂Ω) is closed.

Lemma 23.9.4 If F : Ω→ X is compact and Ω is a bounded open set in X , then

(I−F)(∂Ω)

is closed.

Justification for definition of Leray Schauder Degree

Now let y /∈ (I−F)(∂Ω) , a closed set. Hence dist(y,(I−F)(∂Ω))> 4δ > 0. Now letFk be a sequence of approximations to F which have values in an increasing sequence offinite dimensional subsets Vk each of which contains y. Thus

limk→∞

supx∈Ω∥F (x)−Fk (x)∥= 0. Consider

d(I−Fk|Vk ,Ω∩Vk,y

)Each of these is a well defined integer according to Theorem 23.8.3. For all k large enough,

supx∈Ω

∥(I−F)(x)− (I−Fk)(x)∥< δ

Hence, for all such k,

B(y,3δ )∩ (I−Fk)(∂Ω) = /0, that is dist(y,(I−Fk)(∂Ω))> 3δ (23.9.18)