Abstract
It is well known that boundedness of a subadditive function need not imply its continuity. Here we prove that each subadditive function f:Xrightarrow {mathbb {R}} bounded above on a shift–compact (non–Haar–null, non–Haar–meagre) set is locally bounded at each point of the domain. Our results refer to results from Kuczma’s book (An Introduction to the theory of functional equations and inequalities. Cauchy’s equation and Jensen’s inequality, 2nd edn, Birkhäuser Verlag, Basel, 2009, Chapter 16) and papers by Bingham and Ostaszewski [Proc Am Math Soc 136(12):4257–4266, 2008, Aequationes Math 78(3):257–270, 2009, Dissert Math 472:138pp., 2010, Indag Math (N.S.) 29:687–713, 2018, Aequationes Math 93(2):351–369, 2019).
Highlights
Darji [23] shows that in each locally compact complete Abelian metric group the notion of a Haar–meagre set is equivalent to the notion of a meagre set
Let us recall some basic properties of subadditive functions
A function f : X → R defined on an Abelian metric group X is WNT if and only if for every sequencen∈N convergent to 0 in X there exist k ∈ N, an infinite set M ⊂ N and t ∈ X such that {t + um : m ∈ M} ⊂ Hk
Summary
A universally Baire set B ⊂ X is called Haar–meagre if there exists a continuous map f : K → X from a non-empty compact metric space K such that the set f −1(B + x) is meagre in K for every x ∈ X. Darji [23] shows that in each locally compact complete Abelian metric group the notion of a Haar–meagre set is equivalent to the notion of a meagre set. The converse of Theorem 1.6 does not hold: see [19, Theorem 12], [2, Example 7.1] and [38] It emerges that in non-locally compact complete Abelian metric groups there exist sets which are neither Haar–null nor Haar–meagre and with k-fold sums that are meagre for each k ∈ N. By the Steinhaus–Pettis–Piccard Theorem (see [40,41,43] or [31, Theorems 2.9.1, 3.7.1]), such a situation is not possible in the case of locally compact Abelian
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