Another approach to K-subadditivity

Abstract

In the paper the notion of weakly K-subadditive set-valued maps is introduced in such a way that F is weakly K-superadditive if and only if $-F$ is weakly K-subadditive. This new definition is a natural generalization of K-subadditive set-valued maps from Jabłońska and Nikodem (Aequ Math 95:1221–1231, 2021), for which opposite set-valued maps need not be K-subadditive. Among others, we prove that every weakly K-subadditive set-valued map which is K–upper bounded on a “large” set has to be locally weakly K-upper bounded and weakly K-lower bounded at every point of the domain. This theorem completes an analogous result for K-subadditive set-valued maps which are weakly K-upper bounded on “large” sets from Jabłońska and Nikodem (Aequ Math 95:1221–1231, 2021).

1 Introduction and preliminaries

It is rather well known that every subadditive function $f:\mathbb {R}^n\rightarrow \mathbb {R}$ bounded above on a “large” set (i.e. a set of positive Lebesgue measure or a non–meager set with the Baire property) has to be locally bounded at every point of the domain .¹ However, such a function need not be continuous; as an example it is enough to choose the characteristic function of the set of irrational numbers (see e.g. [11, Theorem 16.2.3, Example 16.2.1]).

In the paper [10] the notions of K–subadditive set–valued maps ² and K–superadditive s.v. maps were introduced.

Definition 1

Let X, Y be monoids ³ and K be a submonoid of Y. Denote by n(Y) the family of all nonempty subsets of Y. A set–valued map $F:X\rightarrow n(Y)$ is called K–subadditive, if

$$\begin{aligned} F(x)+F(y)\subset F(x+y)+K,\qquad x,y\in X, \end{aligned}$$

(1)

and K–superadditive, if

$$\begin{aligned} F(x+y)\subset F(x) + F(y)+K, \qquad x,y\in X. \end{aligned}$$

(2)

These notions look analogous to the notions of K–convexity and K–concavity introduced by Nikodem in [12] and generalize the notions of subadditivity and superadditivity of real functions (for $K=[0,\infty )$ and $Y=\mathbb {R}$ provided F is single–valued).

Unfortunately, if an s.v. map F is K–subadditive, the opposite s.v. map $-F$ need not be K–superadditive (or $-K$–superadditive) (see [10, Examples 2 and 3]) as opposed to the case of real functions where the subadditivity of f means the superadditivity of $-f$.

It makes some results concerning K–subadditive s.v. maps not analogous to respective results concerning K–superadditive s.v. maps. To show an example, let us recall the definitions of K–boundedness from the paper [12].

Definition 2

Let X, Y be metric groups ⁴ and K be a submonoid of Y. Denote by $\mathcal {B}(Y)$ the family of all nonempty bounded subsets of Y. An s.v. map $F:X\rightarrow n(Y)$ is called:

• K–upper bounded on a nonempty set $A\subset X$, if

  $$\begin{aligned} \exists \;B\in \mathcal {B}(Y)\;\;\forall \;x\in A\qquad F(x)\subset B-K, \end{aligned}$$

• weakly K–upper bounded on a nonmepty set $A\subset X$, if

  $$\begin{aligned} \exists \;B\in \mathcal {B}(Y)\;\;\forall \;x\in A\qquad F(x)\cap (B-K)\ne \emptyset , \end{aligned}$$

• K–lower bounded on a set $A\subset X$, if

  $$\begin{aligned} \exists \;B\in \mathcal {B}(Y)\;\;\forall \;x\in A\qquad F(x)\subset B+K, \end{aligned}$$

• weakly K–lower bounded on a set $A\subset X$, if

  $$\begin{aligned} \exists \;B\in \mathcal {B}(Y)\;\;\forall \;x\in A\qquad F(x)\cap (B+K)\ne \emptyset . \end{aligned}$$

Directly from the definition we derive a very useful observation that F is [weakly] K–upper bounded on A if and only if $-F$ is [weakly] K–lower bounded on A.

Clearly, in the case when $K=\{0\}$, [weak] K–upper boundedness and [weak] K–lower boundedness are equivalent properties. Moreover, for $K=[0,\infty )$, $Y=\mathbb {R}$ and a single–valued F, K–upper (K–lower) boundedness as well as weak K–upper (K–lower, resp.) boundedness means boundedness above (below, resp.) of a real function.

Now, let us come back to non–analogous properties of K–subadditive and K–superadditive s.v. maps.

In [10, Theorems 2 and 3] we proved that for metric groups X, Y every s.v. map $F:X\rightarrow \mathcal {B}(Y)$ which is

• K–subadditive and weakly K–upper bounded on a “large” set ,⁵ or

• K–superadditive and K–lower bounded on a “large” set,

has to be locally weakly K–upper bounded and locally K–lower bounded at every point of X. Moreover, in [10, Example 1 and Remark 1] it is showed that we can not obtain K–upper boundedness (in place of weak K–upper boundedness).

Here, we would like to complete the above result proving that every s.v. map $F:X\rightarrow \mathcal {B}(Y)$ which is

• weakly K–subadditive and K–upper bounded on a “large” set, or

• weakly K–superadditive and K–lower bounded on a “large” set,

has to be locally weakly K–upper bounded and locally weakly K–lower bounded at every point of X.

So, let us introduce new definitions of K–subadditivity and K–superadditivity in such a way that the K–superadditivity of F means the K–subadditivity of $-F$. We will call this kind of K–subadditivity weak K–subadditivity.

Definition 3

Let X, Y be monoids and K be a submonoid of Y. The s.v. map $F:X\rightarrow n(Y)$ is called:

• weakly K–subadditive, if

  $$\begin{aligned} \big (F(x+y)+K\big )\cap \big (F(x)+F(y)\big )\ne \emptyset , \qquad x,y\in X, \end{aligned}$$

  (3)

• weakly K–superadditive, if

  $$\begin{aligned} F(x+y)\cap \big (F(x)+F(y)+K\big )\ne \emptyset , \qquad x,y\in X. \end{aligned}$$

  (4)

In this approach, if Y is a group, F is weakly K–superadditive if and only if $-F$ is weakly K–subadditive because of the following obvious property of the Minkowski sum:

$$\begin{aligned} (A+B)\cap C=\emptyset \;\; \Longleftrightarrow \;\; A\cap (C-B)=\emptyset \end{aligned}$$

for $A,B,C \in n(Y).$

Clearly, every K–subadditive (K–superadditivite) s.v. map has to be weakly K–subadditive (weakly K–superadditive, resp.), but the converse implication does not hold, e.g.

• (i) $F(x)=[-|x|,0]$, $x\in \mathbb {R}$, is weakly K–subadditive, but not K–subadditive for $K=[0,\infty )$,

• (ii) $F(x)=[|\sin x|,|\sin x|+2]$, $x\in \mathbb {R}$, is weakly K–superadditive, but not K–superadditive for $K=[0,\infty )$.

Let us also notice that

• if $K=\{0\}$, weak K–subadditivity as well as weak K–superadditivity coincide and both generalize the old notion of additivity of s.v. maps [6]–[9] and subadditivity of s.v. maps [13]–[14];

• if $K=[0,\infty )$ and $Y=\mathbb {R}$, for a single–valued map weak K–subadditivity (weak K–superadditivity) as well as K–subadditivity (K–superadditivity, resp.) means the classical subadditivity (superadditivity, resp.)

2 Basic properties

We begin with some examples which allow us to compare the behaviour of K–subadditive s.v. maps and weakly K–subadditive s.v. maps.

Example 1

Let $K=[0,\infty )$ and $F(x)=[m(x),M(x)]$, where $m,M:\mathbb {R}\rightarrow \mathbb {R}$, $m(x)\le M(x)$ for $x\in \mathbb {R}$. Then:

• the K–subadditivity of F is equivalent to the subadditivity of m,

• the weak K–subadditivity of F is equivalent to the condition

  $$\begin{aligned} m(x+y)\le M(x)+M(y),\qquad x,y\in X. \end{aligned}$$

Remark 1

From the above example we can easy derive that if $K=[0,\infty )$ and $F(x)=[m(x),M(x)]$, where $m,M:\mathbb {R}\rightarrow \mathbb {R}$, and there is a subadditive function $f:\mathbb {R}\rightarrow \mathbb {R}$ such that $m(x)\le f(x)\le M(x)$ for $x\in \mathbb {R},$ then F is weakly K–subadditive.

The converse implication does not hold. Let $F:\mathbb {R}\rightarrow n(\mathbb {R})$ be given by

$$\begin{aligned} F(x) = \left\{ {\begin{array}{*{20}l} {[ - 1,2],} &{} {x < 1,} \\ {[ - 4, - 2],} &{} {x \ge 1.}\\ \end{array} } \right. . \end{aligned}$$

(5)

This s.v. map is weakly K–subadditive. On the other hand, if there existed a subadditive function f such that $m(x)\le f(x)\le M(x)$ for $x\in \mathbb {R}$, then $f(x+y),f(x),f(y)\in [-4,-2]$ for $x,y\ge 1,$ and hence

$$\begin{aligned} f(x+y)\le f(x)+f(y)\le -4\le f(x+y),\qquad x,y\ge 1. \end{aligned}$$

Thus $f(t)=-4$ for $t\ge 2$. But then $f(t_1+t_2)=-4>-8=f(t_1)+f(t_2)$ for $t_1,t_2\ge 2$, which contradicts the subadditivity of f.

Example 2

Let X, Y be monoids, K be a submonoid of Y, $A\in n(Y)$ and $G:X\rightarrow n(Y)$. Let $(G+A)(x):=G(x)+A$, $x\in X$.

• If G is weakly $\{0\}$–additive and $A\cap K\ne \emptyset$, then $G+A$ is weakly K–subadditive.

• If G is $\{0\}$–subadditive and $A\subset K$, then $G+A$ is K–subadditive.

Example 3

Let X be a submonoid of $\mathbb {R}$, Y be a real vector space, K be a submonoid of Y and $A\in n(Y)$. Let $F_A(x):=xA,$ $x\in X$. Then $F_A$ is K–superadditive and weakly K–subadditive, because (e.g. in view of [12, Lemma 1.1])

$$\begin{aligned} F(x+y)=(x+y)A\subset xA+yA=F(x)+F(y),\qquad x,y\in X. \end{aligned}$$

However, such a map need not be K–subadditive; e.g. for $X=Y=\mathbb {R}$, $K=[0,\infty )$ and $A=\{-1,2\}$,

$$\begin{aligned} F(2)+F(-1)=2A+(-A)=\{-4,-1,2,5\}\not \subset [-1,\infty )=F(1)+[0,\infty ). \end{aligned}$$

Example 4

Let X be a monoid, Y be a real vector space and K be a convex cone in Y (i.e. $K+K\subset K$ and $tK\subset K$ for $t\ge 0$). Let $t\ge 0$ and $(tF)(x):=tF(x)$, $x\in X$. Since $tA+tB=t(A+B)$ for $A,B\subset Y$ (see e.g. [12, Lemma 1.1]), if F is [weakly] K–subadditive, then tF is also [weakly] K–subadditive.

Lemma 1

Let X, Y be monoids, K be a submonoid of Y. If $F,G:X\rightarrow n(Y)$ are [weakly] K–subadditive, then

$$\begin{aligned} (F+G)(x):=F(x)+G(x),\qquad x\in X, \end{aligned}$$

and, in particular,

$$\begin{aligned} (F+K)(x):=F(x)+K, \qquad x\in X, \end{aligned}$$

are also [weakly] K–subadditive.

The proof of the above lemma is obvious.

Lemma 2

Let X be a monoid, Y be a real topological vector space and K be a submonoid of Y. If $F:X\rightarrow n(Y)$ is [weakly] K–subadditive and the sets F(x) are relatively compact for $x\in X$, then $(\mathrm { cl\,}F)(x):=\mathrm { cl\,}F(x)$, $x\in X$, is [weakly] $\mathrm { cl\,}K$–subadditive.

Proof

Assume that F is weakly K–subadditive (the proof for K–subadditive F runs in the same way). Since $\mathrm { cl\,}(A+B)=\mathrm { cl\,}A+\mathrm { cl\,}B$ for $A,B\subset Y$ such that $\mathrm { cl\,}A+\mathrm { cl\,}B$ is closed (see [12, Lemma 1.9]), for every $x,y\in X$ we have

$$\begin{aligned} \begin{array}{ll} \emptyset \!\!\!\!&{}{}\ne \big (F(x)+F(y)\big )\cap \big (F(x+y)+K\big )\subset \mathrm { cl\,}\big [ \big (F(x)+F(y)\big )\cap \big (F(x+y)+K\big )\big ]\\ {} &{}{}\subset \mathrm { cl\,}\big (F(x)+F(y)\big )\cap \mathrm { cl\,}\big (F(x+y)+K\big )\\ {} &{}{}=\big (\mathrm { cl\,}F(x)+\mathrm { cl\,}F(y)\big )\cap \big (\mathrm { cl\,}F(x+y)+\mathrm { cl\,}K\big ), \end{array} \end{aligned}$$

which proves the weak $\mathrm { cl\,}K$–subadditivity of $\mathrm { cl\,}F$. $\square$

Lemma 3

Let X be a monoid, Y be a real topological vector space and K be a convex cone in Y. If $F:X\rightarrow n(Y)$ is K–subadditive and F(x) are convex sets with non–empty interiors for $x\in X$, then $(\mathrm { int\,}F)(x):=\mathrm { int\,}F(x)$, $x\in X$, is also K–subadditive.

Proof

Assume that F is K–subadditive. Since $\mathrm { int\,}(A+B)=\mathrm { int\,}A+B$ for convex sets $A,B\subset Y$ such that $\mathrm { int\,}A\ne \emptyset$ (see [12, Lemma 1.11]), for every $x,y\in X$ we get

$$\begin{aligned} \mathrm { int\,}F(x)+\mathrm { int\,}F(y)&= {} \mathrm { int\,}\big (F(x)+F(y)\big )\subset \mathrm { int\,}\big (F(x+y)+K\big )\\ {}&= {} \mathrm { int\,}F(x+y)+K, \end{aligned}$$

which proves the K–subadditivity of $\mathrm { int\,}F$. $\square$

The next example shows that we are not able to obtain an analogous result to the above lemma for weak K–subadditivity.

Example 5

Let $K=[0,\infty )$. The s.v. map. $F:\mathbb {R}\rightarrow n(\mathbb {R})$ given by (5) is weakly K–subadditive. But $\mathrm { int\,}F$ is not weakly K–subadditive. Indeed, for $x,y\ge 1$ we get $x+y>1$,

$$\begin{aligned} \mathrm { int\,}F(x)=\mathrm { int\,}F(y)=\mathrm { int\,}F(x+y)=(-4,-2), \end{aligned}$$

and whence,

$$\begin{aligned} \big (\mathrm { int\,}F(x)+\mathrm { int\,}F(y)\big )\cap \big (\mathrm { int\,}F(x+y)+K\big )=(-8,-4)\cap (-4,\infty )=\emptyset . \end{aligned}$$

Lemma 4

Let X, Y, Z be monoids, K be a submonoid of Y and L be a submonoid of Z. If $F:X\rightarrow n(Y)$ is [weakly] K–subadditive and $G:X\rightarrow n(Z)$ is [weakly] L–subadditive then $(F\times G)(x):=F(x)\times G(x)$, $x\in X$, is [weakly] $K\times L$–subadditive.

Proof

If F is K–subadditive and G is L–subadditive, then

$$\begin{aligned} \begin{array}{ll} \big (F(x)\times G(x)\big )+\big (F(y)\times G(y)\big )\!\!\!\!&{}{} = \big (F(x)+F(y)\big )\times \big (G(x)+G(y)\big )\\ {} &{}{}\subset \big (F(x+y)+K\big )\times \big (G(x+y)+L\big )\\ {} &{}{}=\big (F(x+y)\times G(x+y)\big )+(K\times L) \end{array} \end{aligned}$$

for every $x,y\in X$.

If F is weakly K–subadditive and G is weakly L–subadditive, for $x,y\in X$ we obtain

$$\begin{aligned} \begin{array}{ll} \emptyset \!\!\!\!&{}{}\ne \big (F(x)+F(y)\big )\cap \big (F(x+y)+K\big )\times \big (G(x)+G(y)\big )\cap \big (G(x+y)+L\big ) \\ {} &{}{}\subset \big [\big (F(x)+F(y)\big )\times \big (G(x)+G(y)\big )\big ]\cap \big [\big (F(x+y)+K\big )\times \big (G(x+y)+L\big )\big ] \\ {} &{}{}=\big [\big (F(x)\times G(x)\big )+\big (F(y)\times G(y)\big )\big ]\cap \big [\big (F(x+y)\times G(x+y)\big )+(K\times L)\big ] \end{array} \end{aligned}$$

which ends the proof. $\square$

The reader can easily check the analogous properties of [weakly] K–superadditive s.v. maps (some of them can slightly differ from that for [weak] K–subadditivity).

3 Main results

In the Introduction we mentioned Theorems 2 and 3 from [10] which were in fact a starting point for the whole paper. Now, let us recall these results in their original form.

Theorem 5

([10, Theorems 2 and 3]) Let X, Y be abelian metric groups and K be a submonoid of Y. Assume that $A\subset X$ is a non–null–finite set. If an s.v. map $F:X\rightarrow \mathcal {B}(Y)$ satisfies one of the following conditions:

• (i) F is K–subadditive and weakly K–upper bounded on A,

• (ii) F is K–superadditive and K–lower bounded on A,

then F is locally weakly K–upper bounded and locally K–lower bounded at every point of X.

Null–finite sets were extensively studied in [1] and [2] (see also the notion of shift–compact sets in [3]).

Definition 4

The set A in a topological group X is called null–finite, if there is a sequence $(x_n)_{n\in \mathbb {N}}$ convergent to 0 in X such that the set $\{n\in \mathbb {N}:x+x_n\in A\}$ is finite for every $x\in X$.

The family of all null–finite sets in a metric group is very important, because its complement contains “large” sets in two various senses – topological and measured. More precisely, if X is a complete metric group, then every Borel null–finite set is:

• Haar–null, i.e. there exists a probability $\sigma$–additive Borel measure $\nu$ on X such that $\nu (B+x)=0$ for each $x\in X$ (see [4] and [2, Theorem 6.1]),

• Haar–meager, i.e. there exists a continuous function $f:2^\omega \rightarrow X$ such that $f^{-1}(A+x)$ is meager for each $x\in X$ (see [5, 2, Theorem 5.1], [1, Proposition 5.1]).

Consequently, such a set has to be meager and, if X is additionally locally compact, has Haar measure zero.

Now, we prove a same type of result as Theorem 5 for weakly K–subadditive s.v. maps, but we need a new idea for the proof.

Theorem 6

Let X, Y be abelian metric groups. Assume that $A\subset X$ is a set which is not null–finite and K is a submonoid of Y. If an s.v. map $F:X\rightarrow \mathcal {B}(Y)$ is weakly K–subadditive and K–upper bounded on A, then F is locally weakly K–upper bounded and weakly K–lower bounded at every point of X.

Proof

By the K–upper boundedness of F on A, there exists a set $B\in \mathcal {B}(Y)$ such that

$$\begin{aligned} F(x)\subset B-K, \qquad x\in A. \end{aligned}$$

(6)

First, for the proof by contradiction, suppose that F is not weakly K–upper bounded on any neighborhood of some $x_0$. Consequently, for every $n\in \mathbb {N},$ the open ball $U_n:=B\left( x_0,\frac{1}{2^n}\right)$ and $B_n:=B+B(0,n)\in \mathcal {B}(Y),$ there exists $x_n\in U_n$ such that

$$\begin{aligned} F(x_n)\cap (B_n-K)=\emptyset . \end{aligned}$$

(7)

Moreover, by the definition of $U_n$ the sequence $(x_n)_{n\in \mathbb {N}}$ is convergent to $x_0$ in X, and hence $x_n-x_0\rightarrow 0$. The set A is not null–finite, so there exists $b\in X$ such that the set $\mathbb {N}_0:=\{n\in \mathbb {N}:b+x_n-x_0\in A\}$ is not finite. Put $a=b-x_0$. By (6) we have

$$\begin{aligned} F(a+x_n)\subset B-K, \qquad n\in \mathbb {N}_0. \end{aligned}$$

(8)

Since $F(-a)\in \mathcal {B}(Y)$, we can find $n_0 \in \mathbb {N}$, such that

$$\begin{aligned} F(-a)\subset B(0,n_0)\subset B(0,n)\qquad \text{ for }\; n\in \mathbb {N},\;n\ge n_0. \end{aligned}$$

In view of the weak K–subadditivity of F,

$$\begin{aligned} \begin{array}{ll} \emptyset &{}\ne \big (F(a+x_n)+F(-a) -K\big )\cap F(x_n)\\ &{}\displaystyle \subset \big (B-K+B(0,n)-K\big )\cap F(x_n)\subset (B_n-K)\cap F(x_n) \end{array} \end{aligned}$$

(9)

for $n\in \mathbb {N}_0,$ $n\ge n_0,$ which contradicts (7).

In this way we have proved that F is locally weakly K–upper bounded at every point of X.

Next, for the proof by contradiction, suppose that F is not weakly K–lower bounded on any neighborhood of some $y_0$. We can find $k_0 \in \mathbb {N}$ such that $B\subset B(0,k_0)$ and, for every $n\in \mathbb {N}$, $U_n'=B(y_0,\frac{1}{2^n})$ and $B_n':=B(0,k_0)+B(0,n)\in \mathcal {B}(Y),$ there exists $y_n\in U_n'$ such that $F(y_n)\cap (B_n'+K)=\emptyset .$ Hence

$$\begin{aligned} (F(y_n)-K)\cap B_n'=\emptyset . \end{aligned}$$

(10)

Moreover, by the definition of $U_n'$, $y_0-y_n\rightarrow 0$. The set A is not null–finite, so there exists $c\in X$ such that the set $\mathbb {N}_1:=\{n\in \mathbb {N}:c+y_0-y_n\in A\}$ is not finite. Put $d=c+y_0$. By (6)

$$\begin{aligned} F(d-y_n)\subset B(0,k_0)-K, \qquad n\in \mathbb {N}_1. \end{aligned}$$

(11)

Since $F(d)\in \mathcal {B}(Y)$, we can find $n_1 \in \mathbb {N}$, such that

$$\begin{aligned} F(d)\subset B(0,n_1)\subset B(0,n)\qquad \text{ for }\; n\in \mathbb {N},\;n\ge n_1. \end{aligned}$$

Then, in view of the weak K–subadditivity of F,

$$\begin{aligned} \emptyset \ne \big (F(d-y_n)+F(y_n)-K\big )\cap F(d)\subset \big (B(0,k_0)+F(y_n)-K\big )\cap F(d) \end{aligned}$$

(12)

for $n\in \mathbb {N}_1,$ $n\ge n_1,$ and thus,

$$\begin{aligned} \emptyset \ne (F(y_n)-K)\cap (F(d)-B(0,k_0))\subset (F(y_n)-K)\cap (B(0,n)+B(0,k_0)) \end{aligned}$$

which contradicts (10). $\square$

This result seems to be optimal; we are not able to strengthen the thesis, or obtain the same thesis weakening the assumption, which we present in the next two examples.

The first of them shows that an s.v. map $F:X\rightarrow \mathcal {B}(Y)$ which is weakly K–subadditive and K–upper bounded on a non–null–finite set need not be either locally K–upper bounded at every point of X, or locally K–lower bounded at every point of X.

Example 6

Let $K=[0,\infty )$ and

$$\begin{aligned} F(x) = \left\{ {\begin{array}{*{20}l} {\left[ { - |a(x)|,\frac{1}{{|x|}}} \right] ,} &{} {x \in \mathbb {R}{ \setminus }\{ 0\} ,} \\ {[ - |a(x)|,0],} &{} {x = 0,} \\ \end{array} } \right. \end{aligned}$$

where $a:\mathbb {R}\rightarrow \mathbb {R}$ is a discontinuous additive function. Then F is weakly K–subadditive and K–upper bounded on [1, 2]. However, F is neither K–upper bounded, nor K–lower bounded at 0.

Indeed, if for some neighborhood $U_0$ of 0 and $B\in \mathcal {B}(\mathbb {R})$ $F(x)\subset B+K,\; x\in U_0,$ then $-|a(x)|\ge \inf B$ for $x\in U_0$, which contradicts the discontinuity of a (see e.g. [11, Lemma 9.3.1]).

On the other hand, if for some neighborhood $U_0$ of 0 and $B\in \mathcal {B}(\mathbb {R})$ $F(x)\subset B-K,\; x\in U_0,$ then $\frac{1}{|x|}\le \sup B$ for $x\in U_0$, which also gives a contradiction.

The second example shows that an s.v. map $F:X\rightarrow \mathcal {B}(Y)$ which is weakly K–subadditive and weakly K–upper bounded on a non–null–finite set, need not be locally weakly K–upper bounded at every point of X.

Example 7

In view of [10, Example 4], if $K=[0,\infty )$ and $a:\mathbb {R}\rightarrow \mathbb {R}$ is a discontinuous additive function, then

$$\begin{aligned} G(x)={\left\{ \begin{array}{ll} \big [-2|a(x)|, -|a(x)|\big ], &{}|x|<1,\\ \big [-2|a(x)|,0\big ], &{} |x|\ge 1, \end{array}\right. } \end{aligned}$$

is K–superadditive and weakly K–lower bounded on [1, 2], but it is not weakly K–lower bounded at 0.

Hence, $F(x)=-G(x)$, $x\in \mathbb {R}$, is weakly K–subadditive and weakly K–upper bounded on [1, 2]. However, F is not locally weakly K–upper bounded at 0.

Clearly, according to Theorem 6, we obtain the following corollary for weakly K–superadditive s.v. maps.

Corollary 7

Let X, Y be abelian metric groups and K be a submonoid of Y. If an s.v. map $F:X\rightarrow \mathcal {B}(Y)$ is weakly K–superadditive and K–lower bounded on a non–null–finite set, then F is locally weakly K–upper bounded and weakly K–lower bounded at every point of X.

In view of Examples 6–7 this result is optimal.