“…Taking any point (x, y, Ψ c ′ (x, y)) ∈ epi Ψ c ′ , we have (y, Ψ c ′ (x, y)) ∈ C and, for all λ ≥ 0, (y, Ψ c ′ (x, y)) + λ(0, 1) = (y, Ψ c ′ (x, y) + λ) ∈ C. According to Lemma 2.1, (0, 1) ∈ rec C, and since C is functionally representable by assumption, due to Remark 2.1 we obtain C = epi h with h(x) = inf {a ∈ R : (x, a) ∈ C} . If we show that h = e-conv p, we have, using also (8), the chain epi Ψ((0, ·), (0, ·), ·) c ′ = epi(e-conv p) = epi h = C = e-conv(Pr Y ×R (epi Ψ c ′ )).…”