Can a small Gaussian perturbation break subadditivity?

Given an integer a≥ 1, a function f: R→ R is said to be a-subadditive if f(ax+y) ≤ af(x)+f(y) \,\,\, for all x,y ∈ R. Of course, $1$-subadditive functions (which correspond to ordinary subadditive functions) are $2$-subadditive. % and $3$-subadditive. Answering a question of Matkowski, we show that there exists a continuous function f satisfying $f(0)=0$ which is $2$-subadditive but not $1$-subadditive. In addition, the same example is not $3$-subadditive, which shows that the sequence of families of continuous a-subadditive functions passing through the origin is not increasing with respect to a. The construction relies on a perturbation of a given subadditive function with an even Gaussian ring, which will destroy the original subadditivity while keeping the weaker property. Lastly, given a positive rational cone H⊆ (0,∞) which is not finitely generated, we prove that there exists a subadditive bijection f:H→ H such that x→ 0f(x)=0 and x→ 0f(x)=1. This is related an open question of Matkowski and {Ś}wi{ a}tkowski in [Proc. Amer. Math. Soc. 119 (1993), 187--197].