23.9. THE LERAY SCHAUDER DEGREE 795

Hence dist(y,(I−Fk)∂Ω)≥ 5δ while ∥y− z∥< δ . Hence

d((I−Fk) |Vk ,Ω∩Vk,y

)= d

((I−Fk) |Vk ,Ω∩Vk,z

)by Theorem 23.2.2.

From compactness of h, there is an ε net for h(Ω× [0,1]

),{h(xk, tk)} such that

h(Ω× [0,T ]

)⊆ ∪n

k=1B(h(xk, tk) ,ε) .

Say the tk are ordered. Then, as before,

φ k (x)≡ (ε−∥h(x, t)−h(xk, tk)∥)+

hε (x, t)≡n

∑k=1

h(xk, tk)φ k (h(x, t))

∑i φ i (h(x, t))

Then this is clearly continuous and has values in span({h(xk, tk)}nk=1) . How well does it

approximate? Say h(x, t) ∈ h(Ω× [0,T ]

). Then it is in some

B(h(xk, tk) ,ε) ,

maybe several. Thus letting K (x, t) be those indices k such that

h(x, t) ∈ B(h(xk, tk) ,ε)

∥hε (x, t)−h(x, t)∥ ≤ ∑k∈K (x,t)

∥h(xk, tk)−h(x, t)∥ φ k (h(x, t))∑i φ i (h(x, t))

≤ ε

n

∑k=1

φ k (h(x, t))∑i φ i (h(x, t))

= ε

Now here is a claim.Claim: There exists δ > 0 such that for all t ∈ [0,1] ,

dist(y(t) ,(I−h)(∂Ω, t))> 6δ

Proof of claim: If not, there is (xn, tn) ∈ ∂Ω× [0,1] such that

∥y(tn)− (I−h)(xn, tn)∥< 1/n

Then h(xn, tn) is in a compact set because of compactness of h. Also, the y(tn) are ina compact set because y is continuous and y([0,T ]) must therefore be compact. It fol-lows that (xn, tn) must be in a compact subset of ∂Ω× [0,1]. It follows there is a subse-quence, still denoted as (xn, tn) which converges to (x, t) in ∂Ω× [0,1]. then by continuity,∥y(t)− (I−h)(x, t)∥= 0 contrary to assumption. This proves the claim.

As with h there exists a sequence {yk (t)} such that yk (t)→ y(t) uniformly in t ∈ [0,1]but yk has values in a finite dimensional subspace of X ,Yk. Choose k0 large enough that forall t ∈ [0,1] ,

∥∥y(t)− yk0 (t)∥∥< δ . Thus by the first claim,

D(h(·, t) ,Ω,y(t)) = D(h(·, t) ,Ω,yk0 (t)

)

23.9. THE LERAY SCHAUDER DEGREE 795Hence dist (y, (J — Fy.) 0Q) > 56 while ||y — z|| < 6. Henced((I—F,) ly 2AVe.y) =d((I—-F) lvyj» 2AVe,z)by Theorem 23.2.2. _From compactness of h, there is an € net for h (Q x [0, 1]) , {A (xz, t,) } such thath(Qx [0,7]) C Ug, B (A (xe, te) .€)-Say the ¢, are ordered. Then, as before,, (x) = (€ = ||A (x,t) — A (xe, te) ||) >1») =¥ nse) feo)= Pent OW) Sp tice *))Then this is clearly continuous and has values in span ({/ (xx, tx) };-1) - How well does itapproximate? Say h(x,t) € h(Q x [0,7]) . Then it is in someB(h(xz, tk) ,€),maybe several. Thus letting .% (x,t) be those indices k such thath(x,t) € B(h(xe,th) ,€)0, (h(x,t))he (x,t) —h(x,t)|| < h(xp,t,) —h(xI ( t) ( t)|| rn! ( k tk) ( ,t)|| Yi; (h (x, t))a Suloy ete =Now here is a claim.Claim: There exists 6 > 0 such that for all t € [0,1],dist (y(t) ,(—h) (Q,t)) > 66Proof of claim: If not, there is (x;,t,) € AQ x [0,1] such thatlly tn) — (Zh) nstn)|| < 1/nThen /(xy,t,) is in a compact set because of compactness of h. Also, the y(t,) are ina compact set because y is continuous and y({0,7]) must therefore be compact. It fol-lows that (x,,t,) must be in a compact subset of dQ x [0,1]. It follows there is a subse-quence, still denoted as (x,,t,) which converges to (x,t) in dQ x [0,1]. then by continuity,||y (t) — (I-A) (x,t) || = 0 contrary to assumption. This proves the claim.As with A there exists a sequence {y,; (t)} such that y, (t) > y(t) uniformly in ¢ € [0, 1]but y; has values in a finite dimensional subspace of X , Y,. Choose kg large enough that forall ¢ € (0, 1], ||y (t) yx, (¢)|| < 6. Thus by the first claim,D(h(-,t),Q,y(t)) =D (h(.,t) ,Q,Vko (t))