RENDICONTI DEL SEMINARIO MATEMATICO

136 D. Papini - F. Zanolinare satisfied.We consider now the continuous map g : [0, 1] ∋ s ↦→ ξ 1 (s) − φ 1 (ξ(s)), whereξ 1 (s) = p 1 (ξ(s)). Since ξ(s) ∈ ¯γ ⊆ ¯σ, we have that ξ 1 (0) ≥ −a and thereforeg(0) ≥ −a −(−a) = 0. Similarly, one can check that g(1) ≤ a −a = 0. By Bolzano’sTheorem, we conclude that there exists some ŝ = ŝ ε ∈ [0, 1] such that, settingwe find thatˆt = ˆt ε := ξ 1 (ŝ), ˆx = ˆx ε := ξ 2 (ŝ), ẑ = ẑ ε := (ˆt, ˆx),ẑ ∈ V ε ∩ W, ˆt = φ 1 (ẑ), φ 2 (ẑ) ∈ B[0, R].By the definition of ˜φ and ψ, it is clear thatẑ ∈ V ε ∩ W, ˆt = ψ 1 (ẑ), φ 2 (ẑ) ∈ B[0, R].Moreover, for each ẑ ∈ V ε ∩ W, there is ẑ i ∈ S such that ||ẑ − ẑ i || < 2ε.Then, letting ε = ε n ց 0 and passing to a subsequence on the corresponding ẑ n ’s(thanks to the compactness of S), we can find a point¯z = (¯t, ¯x) ∈ Ssuch that (by the continuity of φ and ψ and the closure of W)¯z ∈ S ∩ W, ¯t = ψ 1 (¯z), φ 2 (¯z) ∈ B[0, R],follows. The fact that S is contained in the solution set of (7) implies that¯x = ψ 2 (¯t, ¯x)and therefore ¯z ∈ W is a fixed point of ψ (since we have already proved that ¯t =ψ 1 (¯t, ¯x)). The fact that φ 2 (¯t, ¯x) ∈ B[0, R], implies (in view of (6) and the remarks atthe beginning of the proof) that ¯z ∈ W is a fixed point of φ.REMARK 1. The first part of the proof of Theorem 6 can be used to obtainthe following result where we use a slightly different expansive condition which isinspired from [33, 36, 37]. We recall that here by a continuum we mean a compact andconnected set.THEOREM 7. With the notation of Theorem 6, assume that φ is compact on D∩ B[a, R] and there is a closed subset W ⊆ D ∩ B[a, R] such that the assumption(H ′ ) for every continuum σ ⊆ B[a, R] with σ ∩B l ̸= ∅ and σ ∩B r ̸= ∅, there is acontinuum Ŵ ⊆ σ∩W with φ(Ŵ) ⊆ B[a, R] and φ(Ŵ)∩B l ̸= ∅, φ(Ŵ)∩B r ̸= ∅,holds. Then there exists ˜z = (˜t, ˜x) ∈ W ⊆ D, with φ(˜z) = ˜z.Proof. We follow the proof of Theorem 6 till to (8). Now, assumption (H ′ ) guaranteesthe existence of a continuum Ŵ ⊆ S ∩ W with φ(Ŵ) ⊆ B[a, R] and φ(Ŵ) ∩ B l ̸= ∅,