Bai Tap Diem Bat Dong

1. Exercises-Fixed points of c.c. maps (1) Let K be a bounded, open, convex subset of E. Let F : K → E be completely continuous and be such that F (∂K) ⊂ K. Then F has a fixed point in K. Solution 1:(Berton Earnshaw) This solution reduces the exercise to an application of the Schauder Fixed Point Theorem. Choose c ∈ K and R > 0, such that the open ball of radius R centered at c, BR (c) ⊂ K. For each b ∈ ∂BR (c) define r(b) = {tb + (1 − t)c : t ≥ 0}, the ray starting at c and containing b Since K is bounded, it follows that r(b) ∩ ∂K 6= ∅, and equals exactly one point, since K is convex. Call this point p(b). For each x ∈ r(b), define  F (x), x∈K F˜ (x) = F (p(b)), x∈ / K. Notice that F˜ is continuous and that F˜ (E) ⊂ F (K), implying that F˜ is completely continuous. Put
A = co F (K) ∪ K . Then A is a bounded, closed, and convex set and F˜ : A → A, hence by the Schauder Fixed Point Theorem, F˜ has a fixed point in the set A, call it y. Notice that if y ∈ / K, then F˜ (y) = F (p(b)), for some b ∈ ∂BR (c), hence, by hypothesis (F (∂K) ⊂ K) y ∈ K. Thus the fixed point y must lie in K, but then F˜ (y) = F (y). Solution 2: Choose c ∈ K and make the change of variables v = u − c, then the fixed point equation u = F (u) is equivalent to the fixed point equation v = F (v + c) − c =: G(v), where G is a completely continuous mapping. Define the set ˜ := {v = u − c : u ∈ K}. K This set is a bounded open neighborhood of 0 ∈ E and ˜ u ∈ K ⇐⇒ v = u − c ∈ K and ˜ u ∈ ∂K ⇐⇒ v = u − c ∈ ∂ K. We furthermore conclude that ˜ ˜ ⊂ K. G(∂ K) 1

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If we have that F has a fixed point u ∈ ∂K, then the proof is complete and ˜ Thus, assume that G has no fixed points in ∂ K. ˜ G has a fixed point in ∂ K. Consider the family of mappings g(t, v) := v − tG(v), 0 ≤ t ≤ 1. This is a family of completely continuous perturbations of the identity and ˜ g(t, v) 6= 0, t ∈ [0, 1]. Thus by the homotopy invariance for v ∈ ∂ K, principle of the Leray-Schauder degree, we have that ˜ 0) = d(id, K, ˜ 0) = 1. d(g(t, ·), K, We therefore conclude that the equation v − G(v) = 0 ˜ has a solution in K. We remark that in the above proof the convexity of the set K may be replaced by the weaker requirement that K be starlike with respect to a point c ∈ K, i.e. that the ray emenating from c will intersect the boundary of K in exactly one point.

(2) Let Ω be a bounded open set in E with 0 ∈ Ω. Let F : Ω → E be completely continuous and satisfy kx − F (x)k2 ≥ kF (x)k2 − kxk2 , x ∈ ∂Ω. then F has a fixed point in Ω. Solution: Let us assume that F has no fixed points in ∂Ω. Consider the family of c.c. perturbations of the identity f (t, x) := x − tF (x), 0 ≤ t ≤ 1. This family has no zeros on ∂Ω, for t = 0, 1. If, on the other hand f (t, x) = 0, for some t ∈ (0, 1), and some x ∈ ∂Ω, then x = tF (x), and the inequality in the exercise becomes (1 − t)2 kF (x)k2 ≥ (1 − t2 )kF (x)k2 . But F (x) 6= 0, and thus (1 − t)2 ≥ 1 − t2 , i.e., t ≥ 1, contradicting that t ∈ (0, 1). We hence may conclude, by the homotopy invariance principle of Leray-Schauder degree that d(id, Ω, 0) = 1 = d(id − F, Ω, 0). Which implies that F has a fixed point in Ω. Note a particular case, where the above condition hold is the following: kF (x)k ≤ kxk, x ∈ ∂Ω. As an example, where this condition holds, consider the following:

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Let F :E→E be a completely continuous mapping such there exist nonnegative constants a and b, a < 1, such that kF (x)k ≤ akxk + b. Choose R >> 1, so that b ≤ (1 − a)R. Then for kxk ≥ R kF (x)k ≤ akxk + b ≤ akxk + (1 − a)kxk = kxk. Thus, for such R, we may choose Ω = BR (0) and apply the above result.