RENDICONTI DEL SEMINARIO MATEMATICO

26 C. Bereanu - J. MawhinTHEOREM 3. If the functions f m (1 ≤ m ≤ n − 1) satisfy (29), there existss 1 ∈ R such that (28) has zero, at least one or at least two solutions according tos < s 1 , s = s 1 , s > s 1 .Proof. LetS j = {s ∈ R : (28) has at least j solutions } ( j ≥ 1).(a) S 1 ̸= ∅.Take s ∗ > max 1≤m≤n−1 f m (0) and use (29) to find R− ∗ < 0 such thatmin f m(R− ∗ ) > 1≤m≤n−1 s∗ .Then α with α j = R ∗ − < 0 (1 ≤ j ≤ n) is a strict lower solution and β with β j = 0(1 ≤ j ≤ n) is a strict upper solution for (28) with s = s ∗ . Hence, using Theorem 2,s ∗ ∈ S 1 .(b) If˜s ∈ S 1 and s >˜s then s ∈ S 1 .Let ˜x = (˜x 1 ,...,˜x n ) be a solution of (28) with s =˜s, and let s >˜s. Then ˜x is a strictupper solution for (28). Take now R − < min 1≤m≤n ˜x m such that min 1≤m≤n−1 f m (R − )> s It follows that α with α j = R − (1 ≤ j ≤ n) is a strict lower solution for (28), andhence, using Theorem 2, s ∈ S 1 .(c) s 1 = inf S 1 is finite and S 1 ⊃ ]s 1 ,∞[.Let s ∈ R and suppose that (28) has a solution (x 1 ,..., x n ). Then (33) holds, fromwhere we deduce that s ≥ c, with c ∈ R given in (31). To obtain the second part ofclaim (c) S 1 ⊃]s 1 ,∞[ we apply (b).(d) S 2 ⊃ ]s 1 ,∞[.We reformulate (28) to apply Brouwer degree theory. Consider the continuous mappingG : R × R n−1 → R n−1 defined byG m (s, x) = (Lx) m + f m (x m ) − s (1 ≤ m ≤ n − 1).Then (x 1 ,..., x n−1 , x 1 ) is a solution of (28) if and only if (x 1 ,..., x n−1 ) ∈ R n−1 isa zero of G(s,·). Let s 3 < s 1 < s 2 . Using Lemma 3 we find ρ > 0 such that eachpossible zero of G(s,·) with s ∈ [s 3 , s 2 ] is such that max 1≤m≤n−1 |x m | < ρ. Consequently,d B [G(s,·), B(ρ), 0] is well defined and does not depend upon s ∈ [s 3 , s 2 ].However, using (c), we see that G(s 3 , x) ̸= 0 for all x ∈ R n−1 . This implies thatd B [G(s 3 ,·), B(ρ), 0] = 0, so that d B [G(s 2 ,·), B(ρ), 0] = 0 and, by excision property,d B [G(s 2 ,·), B(ρ ′ ), 0] = 0 if ρ ′ > ρ. Let s ∈ ]s 1 , s 2 [ and ̂x = (̂x 1 ,...,̂x n ) be a solutionof (28) (using (c)). Then ̂x is a strict upper solution of (28) with s = s 2 . LetR < min 1≤ j≤n ̂x j be such that min 1≤m≤n−1 f m (R) > s 2 . Then (R,..., R) ∈ R n is astrict lower solution of (28) with s = s 2 . Consequently, using Corollary 1, (28) withs = s 2 has a solution in R̂x and|d B [G(s 2 ,·), R̂x , 0]| = 1.Taking ρ ′ sufficiently large, we deduce from the additivity property of Brouwer degree