... > 2(R −B + 1) + (B −1) = 2R + 1 −B,since parts (1) and (4) of Lemma 2.11 give R −B + 1 < 0 and B − 1 < 0.Because the right-hand inequality is a consequence of Theorem 3.6, part (2), ... Vuorinen was supported by the Academy of Finland, Project 2600066611. Thisproject also supported Slavko Simi´c’ visit to Finland.3Theorem 1.6. For a, b > 0 and 0 < x, y < 1 let DFbe as ... on this observation and a few computer experiments, we posed in [2] the followingquestion:Question 3.1. Fix c, d > 0 and let g(x) = xF (c, d; c + d; x) for x ∈ (0, 1) and setQg(x, y) =g(x)...