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RENDICONTI DEL SEMINARIO MATEMATICO

RENDICONTI DEL SEMINARIO MATEMATICO

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Periodic solutions of difference equations 21Suppose now that α (resp. β) is a strict lower (resp. upper) solution of (10).Define the open set(16) αβ = {(x 1 ,..., x n−1 ) ∈ R n−1 : α m < x m < β m (1 ≤ m ≤ n − 1)},and the continuous mapping H : R n−1 → R n−1 by(17)H m (x) = (Lx) m + f m (x m ) (1 ≤ m ≤ n − 1).COROLLARY 1. Assume that the conditions of Theorem 2 hold with strict lowerand upper solutions. Then(18)|d B [H, αβ , 0]| = 1,with αβ defined in (16).Proof. If ρ is large enough, then, using the additivity-excision property of Brouwerdegree [7], we have|d B [G, αβ , 0]| = |d B [G, B(ρ), 0]| = 1.On the other hand, H is equal to G on αβ , and then|d B [G, αβ , 0]| = |d B [H, αβ , 0]|.A simple but useful consequence of Theorem 2, goes as follows.COROLLARY 2. Assume that there exists numbers α ≤ β such thatf m (α) ≥ 0 ≥ f m (β) (1 ≤ m ≤ n − 1).Then problem (10) has at least one solution with α ≤ x m ≤ β (1 ≤ m ≤ n − 1).Proof. Just observe that (α,...,α) is a lower solution and (β,...,β) an upper solutionfor (10).COROLLARY 3. For each p > 0, a m > 0 and b m ∈ R (1 ≤ m ≤ n − 1) theproblemDx m − a m |x m | p−1 x m = b m (1 ≤ m ≤ n − 1), x 1 = x nhas at least one solution.() 1/p|bProof. If R ≥ max m |1≤m≤n−1 a m , then (−R,...,−R) is a lower solution and(R,..., R) an upper solution.

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