RENDICONTI DEL SEMINARIO MATEMATICO

22 C. Bereanu - J. MawhinREMARK 1. When β m ≤ α m (1 ≤ m ≤ n − 1), one can try to repeat theargument of Theorem 2 by definingand considering the modified problem⎧⎨ α m if x > α mδ m (x) = x if β m ≤ x ≤ α m⎩β m if x < β m ,(19) Dx m + x m + f m ◦ δ m (x m ) − δ m (x m ) = 0 (1 ≤ m ≤ n − 1), x 1 = x n .As L + I is invertible, the degree argument still gives the existence of at least onesolution for (19). If one tries to show that, say, β m ≤ x m (1 ≤ m ≤ n − 1), and assumeby contradiction that β i − x i > 0 for some 1 ≤ i ≤ n, one gets no contradiction withDβ m + f m (β m ) ≤ 0. This is in contrast with the ordinary differential equation case,for which the argument works independently of their order [13]. The reason of thisdifference comes from the fact that a local extremum is characterized by an equality(vanishing of the first derivative) in the differential case and by two inequalities (withonly one usable in the argument) in the difference case. This raises the question ofthe validity of the method of upper and lower solutions with reversed upper and lowersolutions in the difference case. This question is solved by the negative in the next twosections.5. Spectrum of the linear partThe construction of the counter-example proving the last assertion above is clarifiedby analyzing the spectral properties of the first order difference operator with periodicboundary conditions.DEFINITION 2. An eigenvalue of the first order difference operator with periodicboundary conditions is any λ ∈ C such that the problem(20)Dx m = λx m (1 ≤ m ≤ n − 1), x 1 = x nhas a nontrivial solution.Explicitly, system (20) can be written as(21)x 1 − x n = 0x 2 − (1 + λ)x 1 = 0... ... ... ... ...x n − (1 + λ)x n−1 = 0