“…) where (z m 0 ) k denotes, as usual, the k th component of the vector z m 0 0. Since 2 a i 1 k (ω 1 , x 1 ) (z m 0 ) k > 0, lemma 3.15 in [29] ensures that 0 is a strong sub-equilibrium for (5.11). Then, as we have mentioned before, there exists a t i 1 > 0 and a z 0i 1 > 0 such that, if h(t, ω, x, 0) is the solution of (5.11) with initial condition 0 ∈ R, then h(t, ω, x, 0) 0 for any t 0 and (ω, x) ∈ K, and besides h(t, ω, x, 0) > 2 z 0i 1 for any t t i 1 , (ω, x) ∈ K.…”