RENDICONTI DEL SEMINARIO MATEMATICO

30 C. Bereanu - J. MawhinDefine ϕ : R → R byso that, for u(1,...,1) ∈ N(L),ϕ(u) = 1n − 1( ∑n−1)f m (u) ,m=1P F(u(1,...,1)) = ϕ(u)(1,...,1).The following theorem slightly sharpens a result of [2].THEOREM 5. Suppose that the functions f m (1 ≤ m ≤ n − 1) are all boundedfrom below or all bounded from above, and that for some R > 0 and ǫ ∈ {−1, 1},(45)ǫǫn−1∑m=1n−1∑m=1f m (x m ) ≥ 0 whenever min1≤ j≤n−1 x j ≥ Rf m (x m ) ≤ 0 whenever max1≤ j≤n−1 x j ≤ −R.Then, problem (35) has at least one solution.Proof. For definiteness, assume that each f m is bounded from below by c. For eachk ≥ 1, let us definef m (k) (x ǫx mm) = f m (x m ) +k(1 + |x m |)(1 ≤ m ≤ n − 1),so that each f (k)m is bounded from below by c − 1 and, using assumption (45),(46)ǫǫn−1∑m=1n−1∑m=1f (k)m (x m) > 0 whenever min1≤ j≤n−1 x j ≥ Rf (k)m (x m) < 0 whenever max1≤ j≤n−1 x j ≤ −R.Define F (k) : R n−1 → R n−1 byF m(k) (x 1,..., x n−1 ) = f m (k) (x m) (1 ≤ m ≤ n − 1).Lemma 5 implies that each possible solution x (k) of each equation(47)Lx + λF (k) (x) = 0 (k = 1, 2,...), (λ ∈ ]0, 1])is such thatmax1≤m≤n−1 |x(k) m| < R + 2(n − 1)(|c| + 1) := ρ (k = 1, 2,...).