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For these reasons, we have given a definition of solution weaker than the distributional one and we have considered the subspace Wu as the space of the admissible test functions. Notice that if u ∈ H01 (Ω) is a generalized solution of problem (P1 ) (resp. (P2 )) and u ∈ L∞ (Ω), then u is a distributional solution of (P1 ) (resp. (P2 )). e. g1 and g2 ) just to present - in a simple case - the main difficulties we are going to tackle. 20). To prove these general results we will use an abstract critical point theory for lower semi-continuous functionals developed in [50, 58, 60].

113) Then the following conclusions hold: N (i) If c ∈ L 2 (Ω) and f ∈ Lr (Ω), with r ∈ (2N/(N + 2), N/2), then u belongs ∗∗ to Lr (Ω), where r∗∗ = N r/(N − 2r); (ii) if c ∈ Lt (Ω) with t > N/2 and f ∈ Lq (Ω), with q > N/2, then u belongs to L∞ (Ω). Proof. Let us first prove conclusion (i). 114) ηk (s) = (s − R) , if R < s < k,   bk s + ck , if s > k, where bk and ck are constant such that ηk is C 1 . 113), v = ηk (u) belongs to Wu . Then we can take it as test function, moreover, js (x, u, ∇u)ηk (u) ≥ 0.

46. 113 for suitable f (x) and c(x). 5 of [36], then we will give here a sketch of the proof of [36] just for clearness. We set g0 (x, s) = min{max{g(x, s), −a(x)}, a(x)}, g1 (x, s) = g(x, s) − g0 (x, s). It follows that g(x, s) = g0 (x, s) + g1 (x, s) and |g0 (x, s)| ≤ a(x) so that we can set f (x) = g0 (x, u(x)). Moreover, we define c(x) = , if u(x) = 0, − g1 (x,u(x)) u(x) 0, if u(x) = 0. 4 N −2 N 2 Then |c(x)| ≤ b|u(x)| , so that c(x) ∈ L (Ω). 46 implies that conclusion (a) holds. Now, if r > N/2 we have that f (x) ∈ Lr (Ω) with r > N/2.

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