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Then for any K problem ˜ w , η v, w − a(v, v, w) = K, w ∈ X. 43) ˜ < η 2 this solution is unique. Moreover, if 2αβ|K| Proof. It is no loss to set η = 1. 45) Here the second equality follows since the embedding X → Y is continuous. Abbreviate F (v) = B(v, v). Observe that F is locally Lipschitz continuous since if |u|, |v| ≤ ρ we have |F (u) − F (v)| = |B(u − v, u) − B(v, u − v)| ≤ 2α ρ |u − v|Y ≤ 2αβ 2 ρ|u − v|. 46) Moreover, let vn be a bounded sequence in X . After passing to a subsequence we can assume that vn is Cauchy in Y and hence F (vn ) is Cauchy in X by |F (u) − F (v)| ≤ 2α ρ|u − v|Y .

Let x ∈ (1l+ f˜)−1 (y), then x = y −f (x) ∈ Rm implies (1l+ f˜)−1 (y) = (1l+ f˜m )−1 (y). 51) shows deg(1l + f, U, y) = deg(1l + f˜, U, y) = deg(1l + f˜m , Um , y) = deg(1l + fm , Um , y) as desired. ✷ Let U ⊆ Rn and f ∈ C(U , Rn ) be as usual. 2 we know that deg(f, U, y) is the same for every y in a connected component of Rn \f (∂U ). We will denote these components by Kj and write deg(f, U, y) = deg(f, U, Kj ) if y ∈ Kj . 19 (Product formula) Let U ⊆ Rn be a bounded and open set and denote by Gj the connected components of Rn \f (∂U ).

Consider H(t, x) = x − x0 − t(F (x) − x0 ), then we have H(t, x) = 0 for x ∈ ∂U and t ∈ [0, 1] by assumption. If H(1, x) = 0 for some x ∈ ∂U , then x is a fixed point and we are done. Otherwise we have deg(1l − F, U, 0) = deg(1l − x0 , U, 0) = deg(1l, U, x0 ) = 1 and hence F has a fixed point. ✷ Now we come to the anticipated corollaries. 9 Let U ⊂ X be open and bounded and let F ∈ C(U , X). Then F has a fixed point if one of the following conditions holds. 1. U = Bρ (0) and F (∂U ) ⊆ U (Rothe).

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Nonlinear functional analysis (lecture notes) by Teschl G.

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