By Ewald C.-O.

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**Sample text**

E. a point xn which satisfies xn = hn (xn ) = f (xn , g n (xn )). g. we can as well assume that these two sequences already converge and x˜ := limn xn as well as y˜ := lim f n (xn ). By the continuity of f we have x˜ = f (˜ x, y˜. Furthermore since γ is closed Gr(γ) is closed and for all n we have (x , g n (xn )) ∈ B 1 (Gr(γ)). Therefore (˜ x, y˜) = limn (xn g n (xn )) must lie in n Gr(γ) and therefore y˜ ∈ γ(˜ x). By the assumption on the correspondence µ we have x˜ = f (˜ x, y˜ ∈ µ(˜ x) so that x˜ is a fixed point of µ.

We can assume that ||x|| > ||y||. Since 0 ∈ X ◦ it follows that there exists ǫ > 0 such that Dǫm := {z ∈ Rm : ||z|| ≤ ǫ} ⊂ X ◦ . Then the convex hull co(x, Dǫm ) contains an open neighborhood of y. Since X is convex and closed, we have co(x, Dǫm ) ⊂ X and hence y ∈ X ◦ which is a contradiction to y ∈ ∂X. Let us now consider the following function : f : ∂X → S m−1 := {z ∈ Rm : ||z|| = 1} x x → ||x|| Since X contains an open ball around the origin, it follows that the map f is well defined 0 ∈ ∂X and furthermore surjective.

Is lhc ⇔ xn → x and y ∈ γ(x) implies there exists a sequence yn ∈ γ(xn ) with yn → y. Proof. 1. “⇒” : Assume xn → x and yn ∈ γ(xn ). Since γ(x) is compact it has a bounded neighborhood V . t. γ(U ) ⊂ V . t. ∀n ≥ n0 we have xn ∈ U . Then since yn ∈ γ(xn ) ⊂ γ(U ) ⊂ V for all n ≥ n0 and V is bounded, (yn ) has a convergent subsequence ynk → y. Clearly y ∈ V . By making V smaller and smaller ( for example one can take V ǫ := x∈γ(x) Bǫ (x) and let ǫ go to zero ) we see that y lies in any neighborhood of γ(x).