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**Additional info for Brouwer's Fixed Point Theorem. Methods of proof and generalizations**

**Example text**

Then f has a ﬁxed point in X. CHAPTER 3. GENERALIZATIONS Proof. Fix x ∈ X. Let K = ∞ n n=1 {f (x)}. 45 Since f is condensing, if α(K) = 0 we have α(K) = α(f (K) ∪ {f (x)}) ≤ α(f (K)) < α(K). Thus α(K) = 0, and so K is compact. Also f (K) ⊆ K. 7, there is a nonempty subset M ⊂ K such that M = f (M ). We use M to produce a compact convex set to which we can apply Schauder’s theorem. To this end, set C = {B ⊆ X : M ⊆ B, B closed, convex, and invariant under f }. Note that C is nonempty, since X ∈ C.

1 has a Lipschitz constant L on A. √ √ Fix w ∈ 1 + t2 Sn−1 , and let z ∈ Sn−1 be such that w = 1 + t2 z. Then for to all of A by setting v(x) = x v t < min { 13 , L−1 }, the function g : x → z − tv(x) maps A into A, and is a contraction mapping. Thus by the Banach Contraction Principle, g has a ﬁxed point, say x, in A. So x = z − tv(x). Since z ∈ Sn−1 , we have x + tv(x) 2 = 1. Expanding the inner product 1 2 1 x + tv(x), x + tv(x) gives x = (1 + t2 ) . Then y = (1 + t2 ) 2 x ∈ Sn−1 , and y + tv(y) = w.

Then there exists x ∈ X such that f (x), x − y ≥ 0 for all y ∈ X. Proof. Suppose the contrary, and let F (x) = {y ∈ X : f (x), x − y < 0}. Then by assumption, for any x ∈ X, F (x) = 0. Further, if x1 , x2 ∈ F (x), then for 0 ≤ λ ≤ 1, we have f (x), x − (λx1 + (1 − λ)x2 ) = λ f (x), x − x1 + (1 − λ) f (x), x − x2 < 0. Thus F (x) is convex. 3, for a ﬁxed y ∈ X, the function deﬁned by g(x) = f (x), x − y is continuous. Thus g−1 (−∞, 0) = F −1 (y) is open. 2, F has a ﬁxed point. But x ∈ F (x) means 0 > f (x), x − x = 0, a contradiction.