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 ( a x + y ) ≤ a f ( x ) + f ( y ) for all x , y ∈ R . Of course, 1-subadditive functions (which correspond to ordinary subadditive functions) are 2-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 lim inf x → 0 f ( x ) = 0 and lim sup x → 0 f ( x ) = 1 . This is related an open question of Matkowski and Świątkowski in [Proc. Amer. Math. Soc. 119 (1993), 187–197].