RENDICONTI DEL SEMINARIO MATEMATICO

Periodic solutions of difference equations 29LEMMA 5. If the functions f m (1 ≤ m ≤ n − 1), are all bounded from belowor all bounded from above, say by c, and if for some R > 0(39)n−1∑m=1f m (x m ) ̸= 0 whenever min1≤ j≤n−1 x j ≥ R or max1≤ j≤n−1 x j ≤ −R,then, for each λ ∈]0, 1] each possible zero x of L + λF is such that(40)max |x j| < R + 2(n − 1)|c|.1≤ j≤n−1Proof. Let (λ, x) ∈]0, 1] × R n−1 be a possible zero of L + λN. It is a solution of theequivalent system(41)n−1∑m=1f m (x m ) = 0, Dx m + λf m (x m ) = 0, x 1 = x n , (1 ≤ m ≤ n − 1).On the other hand, if we assume, say, that each f m (1 ≤ m ≤ n − 1) is bounded frombelow, say by c, we have, for all 1 ≤ m ≤ n − 1, and all u ∈ R,and hence(42)| f m (u)| − |c| ≤ | f m (u) − c| = f m (u) − c,| f m (u)| ≤ f m (u) + 2|c|.Consequently, using (41) and (42), we obtain(43)We deduce(44)n−1∑ ∑ ∑|Dx m | = λ | f m (x m )| ≤ | f m (x m )|m=1≤n−1m=1n−1∑m=1n−1m=1f m (x m ) + 2(n − 1)|c| = 2(n − 1)|c|.max x m ≤ min x ∑n−1m + |Dx m |1≤m≤n−1 1≤m≤n−1Using (41) and assumption (39), we obtainCombined with (44), this gives−[R + 2(n − 1)|c|]