Proposition 2.1. Let f be one of the functions θ, ψ or ψk , k ≥ 2. There is no M > 0 such that
f (x + y) > f (x) + f (y) for all x, y > M or f (x + y) < f (x) + f (y) for all x, y > M .
Proof. For f = ψ or f = ψk , k ≥ 2, since between (2n)! and (2n)! + n there are no prime powers, we have f (x + y) < f (x) + f (y) for all x = (2n)! − 1 and 4 ≤ y < n + 2, so the first
statement is true.