Kenneth Kuttler

23.9. THE LERAY SCHAUDER DEGREE 797

Proof: By Theorem 16.2.5, K is a retract. Thus there is a continuous function R : X→Kwhich leaves points of K unchanged. Then you consider F ◦R. It is still a compact mappingobviously. Let B(0,r) be so large that it contains K. Then from the above theorem, it hasa fixed point in B(0,r) denoted as x. Then F (R(x)) = x. But F (R(x)) ∈ K and so x ∈ K.Hence Rx = x and so Fx = x.

There is an easy modification of the above which is often useful. If F : X→F (X) whereF (X) is bounded and in a compact set, and F is a compact map, then you could consider F :conv(F (X))→ F (X)⊆ conv(F (X)) where here conv(F (X)) is a closed bounded convexsubset of X . Then by the Schauder theorem, there is a fixed point for F .

Here is an easy application of this theorem to ordinary differential equations.

Theorem 23.9.9 Let g : [0,T ]×Rn→ Rn be continuous. Let

F : C ([0,T ] ;Rn)→C ([0,T ] ;Rn)

be given by

F (y)(t) = y0 +∫ t

0g(s,y(s))ds

Suppose that whenevery(s) = F (y)(s) , for s≤ t,

it follows that maxs∈[0,t] |y(s)| < M, |y0| < M. Then there exists a solution to the integralequation

y(t) = y0 +∫ t

0g(s,y(s))ds

for t ∈ [0,T ] .

Proof: Let rM be the radial projection in Rn onto B(0,M) . Then F ◦ rM is compactbecause |g(s,rMy)| is bounded. It also maps into a compact subset of C ([0,T ] ;Rn) thanksto the Arzela Ascoli theorem. Then by the Schauder fixed point theorem, there exists asolution y = F ◦ rM to

y(t) = y0 +∫ t

0g(s,rMy(s))ds

Then for s ∈[0, T̂]

where T̂ is the largest such that ∥y(s)∥ ≤ M for s ∈[0, T̂]. Thus on[

0, T̂],rM has no effect. If T̂ < T, then by the estimate,

∣∣y(T̂)∣∣< M. Hence T̂ is not reallythe last. Thus T̂ = T .

The Schauder alternative or Schaefer fixed point theorem is as follows [38].

Theorem 23.9.10 Let f : X → X be a compact map. Then either

1. There is a fixed point for t f for all t ∈ [0,1] or

2. For every r > 0, there exists a solution to x = t f (x) for t ∈ (0,1) such that ∥x∥> r.

23.9. THE LERAY SCHAUDER DEGREE 797Proof: By Theorem 16.2.5, K is aretract. Thus there is a continuous function R: X > Kwhich leaves points of K unchanged. Then you consider F oR. It is still a compact mappingobviously. Let B(0,r) be so large that it contains K. Then from the above theorem, it hasa fixed point in B(0,r) denoted as x. Then F (R(x)) =x. But F (R(x)) € K and sox eK.Hence Rx =xandsoFx=x. JThere is an easy modification of the above which is often useful. If F : X — F (X) whereF (X) is bounded and in a compact set, and F is a compact map, then you could consider F :conv (F (X)) > F (X) C conv (F (X)) where here conv (F (X)) is a closed bounded convexsubset of X. Then by the Schauder theorem, there is a fixed point for F.Here is an easy application of this theorem to ordinary differential equations.Theorem 23.9.9 Let g : [0,7] x R" > R" be continuous. LetF :C((0,7];R") > C((0,7];R")be given byFoy) =w+ [ glsy(o))dsSuppose that whenevery(s) =F (y)(s), fors <t,it follows that maxs<jo,) |\y(s)| <M, |yo| <M. Then there exists a solution to the integralequationy(t) = ['s(sy(o))asfort €[0,T].Proof: Let ry be the radial projection in R” onto B(0,M) . Then F ory is compactbecause |g(s,ry)| is bounded. It also maps into a compact subset of C ([0, 7]; IR”) thanksto the Arzela Ascoli theorem. Then by the Schauder fixed point theorem, there exists asolution y = F ory toy(t) = 0+ ['s(sruy(o))dsThen for s € [0,7] where 7 is the largest such that ||y(s)|| <M for s € [0,7]. Thus on[0,7] ,rw has no effect. If T < T, then by the estimate, |y (7)| <M. Hence T is not reallythe last. Thus7=T. JjThe Schauder alternative or Schaefer fixed point theorem is as follows [38].Theorem 23.9.10 Let f :X — X be a compact map. Then either1. There is a fixed point for tf for allt € [0,1] or2. For every r > 0, there exists a solution to x = tf (x) fort € (0,1) such that \|x|| > r.