Properties of K-Additive Set-Valued Maps

Abstract

For monoids X, Y and a submonoid $K\subset Y$ we define a K-additive set-valued map $F:X\rightarrow 2^Y$ as a map which is additive “modulo K”. In the paper fundamental properties of K-additive set-valued maps are studied. Among others, we prove that in the class of K-additive set-valued maps K-lower (or weakly K-upper) boundedness on a “large” set implies K-continuity on the domain, as well as K-continuity implies K-homogeneity. We also study an algebraic structure of the K-homogeneity set for K-additive set-valued maps.

1 Introduction

In the paper [7] the notions of K-subadditive set-valued maps (shortly called s.v. maps) and K-superadditive s.v. maps have been introduced, which generalize the well known notions of subadditive and superadditive real functions.

Definition 1

Let X, Y be commutative monoids and $K\subset Y$ be a submonoid.¹ 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

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

(1)

and K-superadditve, if

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

(2)

Here, we would like to introduce the notion of K-additivity for s.v. maps in such a way to generalize the notion of additivity of real functions.

Since additive real functions can be characterized as functions which are simultaneously subadditive and superadditive, the natural definition of K-additivity is the following one.

Definition 2

Let X, Y be commutative monoids and $K\subset Y$ be a submonoid. A s.v. map $F:X\rightarrow n(Y)$ is called K-additive, if it is simultaneously K-subadditive and K-superadditive.

In the case $K=\{0\}$ the notion of K-additivity coincides with the definition of additivity of s.v. maps introduced by Nikodem in [10].

If $K=[0,\infty )$, $Y={\mathbb {R}}$, and F is additionally single-valued, K-additivity of F means classical additivity of the real function F.

The following properties of real additive functions defined on a real normed space X seem to be well known (see e.g. [9, Theorems 5.2.1, 5.4.1, 5.4.2, 9.3.1, 9.3.2, 13.2.1, Lemma 13.2.3.]):

• (i) each additive function satisfies Jensen’s equation, i.e.

  $$\begin{aligned} f\left( \frac{x+y}{2}\right) =\frac{f(x)+f(y)}{2},\quad x,y\in X,\end{aligned}$$

  (3)

• (ii) each function satisfying Jensen’s equation and condition $f(0)=0$ is additive,

• (iii) each additive function bounded above (or below) on a “large” set (i.e. non-meager with the Baire property or of the positive Lebesgue measure) has to be continuous,

• (iv) each continuous additive function is linear,

• (v) if X is additionally finite dimensional, each linear functional is continuous,

• (vi) the set $H_f:=\{t\in {\mathbb {R}}:\,f(tx)=tf(x)\;\text{ for } \text{ all }\;x\in X\}$ is a field (called the homogeneity field of f),

• (vii) for every field $L\subset {\mathbb {R}}$ there is an additive function $f:X\rightarrow {\mathbb {R}}$ such that $H_f=L$.

The aim of the paper is to show properties of K-additive s.v. maps which are in some sense analogous to those mentioned above, and even are far-reaching generalizations of them.

At the beginning of the paper (in the Sect. 2) we show some examples and basic properties of K-additive s.v. maps. Next, in the Sect. 3, we prove that every K-additive s.v. map is K-Jensen. Moreover, we check that the converse implication generally does not hold, however under some additional assumptions we can get K-additivity of a K-Jensen s.v. map. In the Sect. 4 we prove that in the class of K-additive s.v. maps weak K-upper boundedness as well as K-lower boundedness on a “large” set imply K-continuity on the whole domain and, moreover, K-continuity implies K-homogeneity. Finally, we show that under some additional assumptions K-homogeneity implies K-continuity of a K-additive s.v. map. At the end of the paper, in the Sect. 5, we study an algebraic structure of the K-homogeneity set of a K-additive s.v. map.

All necessary notions such as: K-Jensen s.v. map, K-upper/K-lower boundedness, K-continuity and K-homogeneity, we explain in relevant sections for the convenience of the reader.

2 Basic Properties of K-Additive s.v. Maps

Lets start with some examples and basic properties of K-additive s.v. maps.

Example 1

Let X be a submonoid of $\big ([0,\infty ),+\big )$, Y be a real vector space and A be a nonempty convex subset of Y. Then

$$\begin{aligned} F_A(x):=xA,\quad x\in X, \end{aligned}$$

is $\{0\}$-additive, because (e.g. in view of [10, Lemma 1.1])

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

Example 2

Let X be a commutative monoid, Y be a real vector space and $K\subset Y$ be a convex cone (i.e. $K+K\subset K$ and $tK\subset K$ for $t\ge 0$). Fix $t\ge 0$ and define

$$\begin{aligned} (tF)(x):=tF(x), \quad x\in X. \end{aligned}$$

Since $tA+tB=t(A+B)$ for $A,B\subset Y$ (see e.g. [10, Lemma 1.1]), if F is K-additive, then tF is also K-additive.

Lemma 1

Let X, Y be commutative monoids and $K\subset Y$ be a submonoid. If $F,G:X\rightarrow n(Y)$ are K-additive, then

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

is also K-additive. In particular, for every $A\in n(Y)$ satisfying $0\in A\subset K$,

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

is K-additive, too.

The proof of the above lemma is obvious.

Lemma 2

Let X, Y, Z be commutative monoids and $K\subset Y$, $L\subset Z$ be submonoids. If $F:X\rightarrow n(Y)$ is K-additive and $G:X\rightarrow n(Z)$ is L-additive, then

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

is $K\times L$-additive.

Proof

For every $x,y\in X$ we get

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

which ends the proof. $\square$

Lemma 3

Let X be a commutative monoid, Y be a real topological vector space and K be a submonoid of $(Y,+)$. If $F:X\rightarrow n(Y)$ is K-additive and sets F(x) are relatively compact for $x\in X$, then

$$\begin{aligned} (\mathrm { cl\,}F)(x):=\mathrm { cl\,}F(x),\quad x\in X, \end{aligned}$$

is $\mathrm { cl\,}K$-additive.

Proof

Assume that F is K-additive. Since $\mathrm { cl\,}(A+B)=\mathrm { cl\,}A+\mathrm { cl\,}B$ for $A,B\subset Y$ such that the set $\mathrm { cl\,}A+\mathrm { cl\,}B$ is closed (see [10, Lemma 1.9]), for every $x,y\in X$ we get

$$\begin{aligned} \begin{array}{ll} \mathrm { cl\,}F(x)+\mathrm { cl\,}F(y)=\mathrm { cl\,}\big (F(x)+F(y)\big )\subset \mathrm { cl\,}\big (F(x+y)+K\big )=\mathrm { cl\,}F(x+y)+\mathrm { cl\,}K, \end{array} \end{aligned}$$

and

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

which proves $\mathrm { cl\,}K$-additivity of $\mathrm { cl\,}F$. $\square$

Lemma 4

Let X be a commutative monoid, Y be a real topological vector space and K be a convex cone in Y. If $F:X\rightarrow n(Y)$ is K-additive and F(x) are convex sets with non-empty interiors for $x\in X$, then

$$\begin{aligned} (\mathrm { int\,}F)(x):=\mathrm { int\,}F(x), \quad x\in X, \end{aligned}$$

is also K-additive.

Proof

Assume that F is K-additive. Since $\mathrm { int\,}(A+B)=\mathrm { int\,}A+B$ and $\mathrm { int\,}(A+C)=\mathrm { int\,}A+\mathrm { int\,}C$ for convex sets $A,B,C\subset Y$ such that $\mathrm { int\,}A\ne \emptyset$ and $\mathrm { int\,}C\ne \emptyset$ (see [10, Lemma 1.11]), for every $x,y\in X$ we get

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

which proves K-additivity of $\mathrm { int\,}F$. $\square$

Lemma 5

Let X be a commutative monoid, Y be a real vector space and $K\subset Y$ be a convex cone satisfying $K\cap (-K)=\{0\}$. Assume that $z_0\in K\setminus \{0\}$ and $F:X\rightarrow n(Y)$ is the s.v. map given by

$$\begin{aligned} F(x)=[m(x),M(x)]z_0,\quad x\in X, \end{aligned}$$

with $m,M:X \rightarrow {\mathbb {R}}$ satisfying $m(x)\le M(x)$ for $x\in X$. Then F is K-additive if and only if m is additive.

Proof

First assume that m is additive. Then

$$\begin{aligned} \begin{array}{ll} F(x+y)&{}{}=[m(x+y),M(x+y)]z_0\subset [m(x)+m(y),\infty )z_0\\ {} &{}{}=[m(x),M(x)]z_0+ [m(y),M(y)]z_0+[0,\infty )z_0\subset F(x)+F(y)+K \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{ll} F(x)+F(y)&{}{}= [m(x)+m(y),M(x)+M(y)]z_0\subset [m(x+y),\infty )z_0\\ {} &{}{}\subset [m(x+y),M(x+y)]z_0+[0,\infty )z_0 \subset F(x+y)+K \end{array} \end{aligned}$$

for every $x,y\in X$, which means that F is K-additive.

Now, assume that F is K-additive. Then, for every $x,y\in X$,

$$\begin{aligned} \begin{array}{ll} m(x+y)z_0\in F(x+y)\subset F(x)+F(y)+K=[m(x)+m(y),M(x)+M(y)]z_0+K \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{ll} \big (m(x)+m(y)\big )z_0\in F(x)+F(y)\subset F(x+y)+K=[m(x+y),M(x+y)]z_0+K. \end{array} \end{aligned}$$

Hence, for $x,y\in X$,

$$\begin{aligned} \big (m(x+y)-\alpha \big ) z_0\in K,\qquad \big (m(x)+m(y)-\beta \big ) z_0\in K, \end{aligned}$$

with some $\alpha \in [m(x)+m(y),M(x)+M(y)]$ and $\beta \in [m(x+y),M(x+y)]$. Since $z_0\in K{\setminus }\{0\}$ and $K\cap (-K)=\{0\}$,

$$\begin{aligned} m(x+y)\ge \alpha \ge m(x)+m(y),\qquad m(x)+m(y)\ge \beta \ge m(x+y) \end{aligned}$$

for $x,y\in X$, which proves additivity of m. $\square$

From the above lemma we can easy derive the following useful corollary.

Corollary 6

Let X be a commutative monoid, $K=[0,\infty )$ and

$$\begin{aligned} F(x)=[m(x),M(x)],\qquad x\in X, \end{aligned}$$

where $m,M:X \rightarrow {\mathbb {R}}$ satisfy $m(x)\le M(x)$ for $x\in X$. Then K-additivity of F is equivalent to additivity of m.

Let us recall that a subset C of a uniquely 2-divisible commutative monoid Y² is called mid-convex, if $\frac{1}{2}C+\frac{1}{2}C\subset C$. It is well known (see e.g. [9, Lemma 5.1.1]) that mid-convexity is equivalent to ${\mathbb {D}}$-convexity, i.e. $dC+(1-d)C\subset C$ for any $d\in {\mathbb {D}}\cap [0,1]$, where ${\mathbb {D}}$ is the set of dyadic numbers,

$$\begin{aligned} {\mathbb {D}}=\left\{ \frac{k}{2^n}:\;k\in {\mathbb {Z}},\;n\in {\mathbb {N}}\cup \{0\}\right\} . \end{aligned}$$

Denote by ${\mathbb {D}}(A)$ ${\mathbb {D}}$-convex hull of a subset A of a uniquely 2-divisible commutative monoid Y (i.e. ${\mathbb {D}}(A)$ is the smallest ${\mathbb {D}}$-convex set containing A). Since ${\mathbb {D}}(A+B)={\mathbb {D}}(A)+{\mathbb {D}}(B)$ for any $A,B\subset Y$, hence we can obtain the following lemma.

Lemma 7

Let X be a commutative monoid, Y be a uniquely 2-divisible commutative monoid and $K\subset Y$ be a uniquely 2-divisible submonoid. If $F:X\rightarrow n(Y)$ is a K-additive s.v. map, then

$$\begin{aligned} {\mathbb {D}}F(x):={\mathbb {D}}(F(x)),\quad x\in X, \end{aligned}$$

is K-additive.

Proof

By K-additivity, for $x,y\in X$ we obtain

$$\begin{aligned} {\mathbb {D}}(F(x+y))\subset {\mathbb {D}}(F(x)+F(y)+K)={\mathbb {D}}(F(x))+{\mathbb {D}}(F(y))+{\mathbb {D}}(K) \end{aligned}$$

and

$$\begin{aligned} \begin{array}{ll} {\mathbb {D}}(F(x))+{\mathbb {D}}(F(y))={\mathbb {D}}(F(x)+F(y))\subset {\mathbb {D}}(F(x+y)+K) ={\mathbb {D}}(F(x+y))+{\mathbb {D}}(K). \end{array} \end{aligned}$$

But K is a uniquely 2-divisible submonoid, so it is mid-convex and hence ${\mathbb {D}}$-convex. Consequently, ${\mathbb {D}}(K)=K$, which ends the proof. $\square$

Since $\mathrm { conv\,}(A+B)=\mathrm { conv\,}A+\mathrm { conv\,}B$ for any subsets A, B of a real vector space Y, in the same way as Lemma 7 we can prove the next lemma.

Lemma 8

Let X be a commutative monoid, Y be a real vector space and $K\subset Y$ be a convex cone. If $F:X\rightarrow n(Y)$ is a K-additive s.v. map, then

$$\begin{aligned} \mathrm { conv\,}F(x):=\mathrm { conv\,}(F(x)),\quad x\in X, \end{aligned}$$

is K-additive.

At the end of the section, let us introduce a relation $=_K$ in the family n(Y) of all nonempty subsets of a monoid Y with a given submonoid $K\subset Y$:

$$\begin{aligned} A=_K B\;\;\Longleftrightarrow \;\;(A\subset B+K\;\;\wedge \;\;B\subset A+K) \end{aligned}$$

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

First let us observe that

$$\begin{aligned} A=_KB\;\;\Longleftrightarrow \;\;A+K=B+K \end{aligned}$$

for $A,B\in n(Y)$. Indeed, if $A=_KB$, then $A+K\subset B+K+K\subset B+K$ and, analogously, $B+K\subset A+K+K\subset A+K$, which means that $A+K=B+K$. On the other hand, if $A+K=B+K$, then $A\subset A+K=B+K$ and $B\subset B+K=A+K$, so $A=_KB$.

Lemma 9

If  Y is a commutative monoid and $K\subset Y$ is a submonoid, then $=_K$ is an equivalence relation in n(Y) and for every $A,B,C,D\in n(Y)$ the following properties hold:

• (i) if $A=_KB$ and $C=_KD$, then $A+C=_KB+D,$

• (ii) if $0\in C\subset K$ and $A=_KB+C$, then $A=_KB$,

• (iii) if $A=_KB$, then $\frac{1}{2}A=_K \frac{1}{2}B$, provided Y, K are uniquely 2-divisible,

• (iv) if $A=_KB$, then $tA=_K tB$ for every $t>0$, provided Y is a real vector space and K is a convex cone in Y.

Proof

Let $A,B,C,D\in n(Y)$. Reflexivity and symmetry of the relation $=_K$ is obvious. We show that this relation is transitive.

If $A=_KB$ and $B=_KC$, then

$$\begin{aligned} \begin{array}{l} A\subset B+K,\quad B\subset A+K,\quad B\subset C+K,\quad C\subset B+K, \end{array} \end{aligned}$$

and hence

$$\begin{aligned} \begin{array}{l} A\subset B+K\subset C+K+K\subset C+K,\\ C\subset B+K\subset A+K+K\subset A+K,\end{array} \end{aligned}$$

which means that $A=_KC$. Consequently, $=_K$ is an equivalence relation.

(i) If $A=_KB$ and $C=_KD$, then

$$\begin{aligned} \begin{array}{l} A\subset B+K,\quad B\subset A+K,\quad C\subset D+K,\quad D\subset C+K, \end{array} \end{aligned}$$

and hence

$$\begin{aligned} \begin{array}{l} A+C\subset B+D+K+K\subset B+D+K,\\ B+D\subset A+C+K+K\subset A+C+K,\end{array} \end{aligned}$$

which means that $A+C=_KB+D$.

(ii) Let $0\in C\subset K$ and $A=_KB+C$. Then

$$\begin{aligned} \begin{array}{c} A\subset B+C+K\subset B+K+K\subset B+K,\\ B\subset B+C\subset A+K, \end{array} \end{aligned}$$

which means that $A=_KB$.

(iii) Now, assume that Y and K are uniquely 2-divisible. If $A=_KB$, then

$$\begin{aligned} A\subset B+K,\quad B\subset A+K, \end{aligned}$$

(4)

and hence

$$\begin{aligned} \frac{1}{2}A\subset \frac{1}{2}B+\frac{1}{2}K\subset \frac{1}{2}B+K,\quad \frac{1}{2}B\subset \frac{1}{2}A+\frac{1}{2}K\subset \frac{1}{2}A+K, \end{aligned}$$

so $\frac{1}{2}A=_K\frac{1}{2}B$.

(iv) Finally, assume that Y is a real vector space and K is a convex cone in Y. If $A=_KB$ and $t>0$, then (4) holds, and hence

$$\begin{aligned} tA\subset tB+tK\subset tB+K,\quad tB\subset tA+tK\subset tA+K, \end{aligned}$$

so $tA=_KtB$. $\square$

Now, for commutative monoids X, Y and a submonoid $K\subset Y$, we can easy write that a s.v. map $F:X\rightarrow n(Y)$ is K-additive, if

$$\begin{aligned} F(x+y)=_K F(x)+F(y),\quad x,y\in X. \end{aligned}$$

We will use this clear notation during the whole paper.

3 Connection Between K-Additivity and K-Jensen s.v. Maps

The next definition generalizes the notion of a Jensen s.v. map which has been introduced in [10].

Definition 3

Let X, Y be uniquely 2-divisible commutative monoids and $K\subset Y$ be a submonoid. A s.v. map $F:X\rightarrow n(Y)$ is called K-Jensen if it is simultaneously K-midconvex and K-midconcave, i.e.

$$\begin{aligned} F\left( \frac{x+y}{2}\right) =_K \frac{1}{2}\big (F(x)+F(y)\big ),\quad x,y,\in X. \end{aligned}$$

Example 3

If X is a uniquely 2-divisible commutative monoid, $K=[0,\infty )$ and $F(x)=[m(x),M(x)]$, where $m,M:X\rightarrow {\mathbb {R}}$ satisfy $m(x)\le M(x)$ for $x\in X$, then F is K-Jensen if and only if m satisfies Jensen’s equation (3).

Lemma 10

Let X, Y be uniquely 2-divisible commutative monoids and $K\subset Y$ be a uniquely 2-divisible submonoid. If $F:X\rightarrow n(Y)$ is a K-additive mid-convex-valued map, then it is K-Jensen.

Proof

For every $x,y\in X$, by K-additivity of F, mid-convexity of $F\left( \frac{x+y}{2}\right)$ and transitivity of $=_K$, we get

$$\begin{aligned} 2F\left( \frac{x+y}{2}\right) = F\left( \frac{x+y}{2}\right) +F\left( \frac{x+y}{2}\right) =_K F(x+y)=_K F(x)+F(y). \end{aligned}$$

To end the proof it is enough to apply Lemma 9 (iii). $\square$

It is clear that there are s.v. maps which are K-Jensen but not K-additive; it is enough to choose $K=[0,\infty )$ and $F:{\mathbb {R}}\rightarrow n({\mathbb {R}})$ given by $F(x)=[x+1,x+2]$ for $x\in {\mathbb {R}}$.

However, under some additional assumptions, K-Jensen s.v. maps have to be K-additive.

Theorem 11

Let X, Y be uniquely 2-divisible commutative monoids and $K\subset Y$ be a submonoid. If $F:X \rightarrow n(Y)$ is a K-Jensen mid-convex-valued map such that $0\in F(0)\subset K$, then it is K-additive.

Proof

According to Lemma 9 (i),

$$\begin{aligned} \begin{array}{ll}F(x)+F(y)&{}{}\displaystyle =F\left( \frac{2x+0}{2}\right) +F\left( \frac{2y+0}{2}\right) \\[2ex] &{}{}\displaystyle =_K \frac{1}{2}\big (F(2x)+F(0)\big )+\frac{1}{2}\big (F(2y)+F(0)\big )=\frac{F(2x)+F(2y)}{2}+F(0)\\[2ex] &{}{} \displaystyle =_K F\left( \frac{2x+2y}{2}\right) +F(0)= F(x+y)+F(0) \end{array} \end{aligned}$$

for any $x,y\in X$. Hence, since $0\in F(0)\subset K$, by Lemma 9 (ii) we get $F(x+y)=_K F(x)+F(y)$ for every $x,y\in X$. $\square$

The converse theorem to Theorem 11 holds under some additional assumptions on Y, K and F.

Theorem 12

Let X be a uniquely 2-divisible commutative monoid, Y be a real vector metric space and K be a closed convex cone in Y such that $K\cap (-K)=\{0\}$. Let $F:X\rightarrow n(Y)$ be a s.v. map such that F(x) are compact convex sets for $x\in X$. Then F is K-additive if and only if F is K-Jensen and $0\in F(0)\subset K$.

Proof

By Theorem 11 and Lemma 10 it is enough to show that if F is K-additive, then $0\in F(0)\subset K$.

In view of K-additivity,

$$\begin{aligned} 2F(0)=F(0)+F(0)=_K F(0), \end{aligned}$$

and hence, according to Lemma 9 (iii),

$$\begin{aligned} \begin{array}{c} \displaystyle F(0)=_K \frac{1}{2}F(0),\\ \displaystyle F(0)=_K\frac{1}{2}F(0)=_K \frac{1}{4}F(0), \end{array} \end{aligned}$$

and, using induction, we get

$$\begin{aligned} F(0)=_K\frac{1}{2^n}F(0)\quad \text{ for } \text{ every }\;n\in {\mathbb {N}}. \end{aligned}$$

In the first step we show that $F(0)\subset K$. So, take any $y\in F(0)$. Since

$$\begin{aligned} F(0)\subset \frac{1}{2^n}F(0)+K\quad \text{ for } \; n\in {\mathbb {N}}, \end{aligned}$$

we can find sequences $(y_n)_{n\in {\mathbb {N}}}\subset F(0)$ and $(k_n)_{n\in {\mathbb {N}}}\subset K$ such that $y=\frac{y_n}{2^n}+k_n$. But F(0) is compact, so there is a convergent subsequence $(y_{s_n})_{n\in {\mathbb {N}}}$ of $(y_{n})_{n\in {\mathbb {N}}}$. Thus

$$\begin{aligned} k_{s_n}=y-\frac{y_{s_n}}{2^{s_n}}\rightarrow y \end{aligned}$$

and whence $y\in \mathrm { cl\,}K=K$.

Next, we prove that $0\in F(0)$. Since

$$\begin{aligned} \frac{1}{2^n}F(0)\subset F(0)+K\quad \text{ for } \; n\in {\mathbb {N}}, \end{aligned}$$

for fixed $y\in F(0)$ the sequence $\left\{ \frac{y}{2^n}\right\} _{n\in {\mathbb {N}}}$ is contained in $F(0)+K$ and converges to 0. Hence $0\in \mathrm { cl\,}(F(0)+K)\subset F(0)+K.$ It means that there is some $y_0\in F(0)\subset K$ such that $-y_0\in K$. Hence $y_0\in K\cap (-K)=\{0\}$, which ends the proof. $\square$

At the end of the section let us mention another well known property (see [9, Theorem 13.2.1]) that each real function satisfying Jensen’s equality is a translation of an additive function by a constant. A similar result holds also for $\{0\}$–Jensen s.v. maps (see [10, Theorem 5.6]). Unfortunately, we are not able to answer the following question.

Problem 1

Let X, Y, K be as in Theorem 12 and $F:X\rightarrow n(Y)$ be a convex-valued K-Jensen map. Are there a K-additive s.v. map $A:X\rightarrow n(Y)$ and a set $B\subset Y$ such that $F(x)=_K A(x)+B$ for $x\in X$?

4 K-Continuity of K-Additive s.v. Maps

From now on we will use the following notations for families of subsets of a real vector space Y:

• ${\mathcal {B}}(Y)$ – the family of all nonempty bounded subsets of Y,

• $\mathcal{B}\mathcal{C}(Y)$ – the family of all nonempty bounded convex subsets of Y,

• $\mathcal{C}\mathcal{C}(Y)$ – the family of all nonempty compact convex subsets of Y.

First, let us recall definitions of K-boundedness and K-continuity of s.v. maps from the paper [10].

Definition 4

Let X, Y be real topological vector spaces and K be a convex cone in Y. A s.v. map $F:X\rightarrow n(Y)$ is called:

• K-upper bounded on a set $A\subset X$, if there exists a set $B\in {\mathcal {B}}(Y)$ such that

  $$\begin{aligned} F(x)\subset B-K\;\;\;\text{ for } \text{ all }\;\; x\in A; \end{aligned}$$

• weakly K-upper bounded on a set $A\subset X$, if there exists a set $B\in {\mathcal {B}}(Y)$ such that

  $$\begin{aligned} F(x)\cap (B-K)\ne \emptyset \;\;\;\text{ for } \text{ all }\;\; x\in A; \end{aligned}$$

• [weakly] K-lower bounded on a set $A\subset X$, if it is [weakly] $(-K)$-upper bounded on this set,

• K-continuous at $x_0\in X$, if for every neighborhood $W\subset Y$ of 0 there is a neighborhood U of $x_0$ such that

  $$\begin{aligned} F(x)\subset F(x_0)+W+K\quad \text{ and } \quad F(x_0)\subset F(x)+W+K \end{aligned}$$

  for all $x\in U.$

Remark 1

Clearly, in the case when $K=\{0\}$, [weak] K-upper boundedness and [weak] K-lower boundedness are equivalent assumptions and K-continuity means continuity with respect to the Hausdorff topology on n(Y).

Remark 2

For a real topological vector space X and $K=[0,\infty )$ define

$$\begin{aligned} F(x)=[m(x),M(x)],\quad x\in X, \end{aligned}$$

where $m,M:X\rightarrow {\mathbb {R}}$ satisfy $m(x)\le M(x)$ for $x\in X$. It is very easy to show that:

• (a) F is K-lower (weakly K-upper) bounded on a set if and only if m is bounded below (above, resp.) on this set,

• (b) F is K-continuous at a point if and only if m is continuous at this point.

Let us also recall the notion of “smalness” introduced in [1, 2] (see also the notion of a shift-compact set in [3]).

Definition 5

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 $x\in X$.

If X is a complete metric group, then every Borel null-finite set $B\subset X$ is “small” in topological and measured senses, which means that it 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]); consequently, B has Haar measure zero provided X is additionally locally compact,

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

In [6, Theorems 3.1 and 3.2] the authors proved that for real vector metric spaces X, Y each s.v. map $F:X\rightarrow \mathcal{B}\mathcal{C}(Y)$ which is K-midconvex and weakly K-upper bounded on a non-null finite set, or K-midconcave and K-lower bounded on a non-null finite set, has to be K-continuous. Hence, directly from Lemma 10, we can obtain the following important result, which generalizes [6, Theorem 3.4] and also [10, Theorem 5.1].

Theorem 13

Let X, Y be real vector metric spaces. Assume that K is a convex cone in Y. If a s.v. map $F:X\rightarrow \mathcal{B}\mathcal{C}(Y)$ is K-additive and satisfies one of the following conditions:

• F is weakly K-upper bounded on a non-null finite set,

• F is K-lower bounded on a non-null-finite set,

then F is K-continuous on X.

In the above theorem the assumption that F is K-lower bounded on a non-null-finite set can not be replaced by a weaker one, i.e. by weak K-lower boundedness on such a set.

Example 4

Let $a:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be a discontinuous additive function, $K = [0,\infty )$ and $F :{\mathbb {R}}\rightarrow \mathcal{B}\mathcal{C}({\mathbb {R}})$ be given by

$$\begin{aligned} F(x)=[a(x),\max \{1,a(x)+1\}],\quad x\in {\mathbb {R}}. \end{aligned}$$

In view of Corollary 6 and Remark 2b) such a map is K-additive, but is not K-continuous at any point of ${\mathbb {R}}$. Moreover, $F(x)\cap (\{0\}+K)\ne \emptyset$ for $x\in {\mathbb {R}}$, so F is weakly K-lower bounded on the whole domain.

It is known that K-continuity at a point of a s.v. map implies weak K-upper boundedness, as well as K-lower boundedness, on a neighbourhood of this point (see e.g. [8, Section 3]). Hence, by Theorem 13, we can easy derive the following corollary.

Corollary 14

Let X, Y be real vector metric spaces and $K\subset Y$ be a convex cone. If a s.v. map $F:X\rightarrow \mathcal{B}\mathcal{C}(Y)$ is K-additive and K-continuous at a point, then F is K-continuous on X.

Knowing that real continuous additive functions have to be linear, we would like to show some kind of K-homogeneity of K-continuous K-additive s.v. maps.

Theorem 15

Let X, Y be real vector metric spaces and K be a closed convex cone in Y such that $K\cap (-K)=\{0\}$. If $F:X\rightarrow \mathcal{C}\mathcal{C}(Y)$ is a K-continuous and K-additive s.v. map, then it is K-homogeneous, i.e.

$$\begin{aligned} F(tx)=_K tF(x),\quad x\in X, \;t\ge 0. \end{aligned}$$

(5)

Moreover, if F is not single-valued,

$$\begin{aligned} \left\{ t\in {\mathbb {R}}:\;F(tx)=_K tF(x)\;\;\text{ for }\; x\in X\right\} =[0,\infty ). \end{aligned}$$

Proof

According to Theorem 12, $0\in F(0)\subset K$ and F is K-midconvex and K-midconcave. Thus, in view of [10, Theorems 3.1 and 4.1], F is K-convex and K-concave, i.e.

$$\begin{aligned} tF(x)+(1-t)F(y)=_K F(tx+(1-t)y),\quad x,y\in X,\;t\in [0,1]. \end{aligned}$$

Hence, for $y=0$,

$$\begin{aligned} tF(x)+(1-t)F(0)=_K F(tx),\quad x\in X,\;t\in [0,1]. \end{aligned}$$

Clearly, since $0\in F(0)\subset K$, we get $0\in (1-t)F(0)\subset (1-t)K\subset K,$ and thus, in view of Lemma 9 (ii),

$$\begin{aligned} tF(x)=_KF(tx),\quad x\in X,\;t\in [0,1]. \end{aligned}$$

Now, we can complete the proof of (5) by induction. Fix $n\in {\mathbb {N}}$ and assume that $F(tx)=_KtF(x)$ for every $t\in [n,n+1]$ and $x\in X$. Take any $t\in [n+1,n+2]$. Then $t-1\in [n,n+1]$, so, by K-additivity,

$$\begin{aligned} F(tx)=_K F((t-1)x)+F(x)=_K (t-1)F(x)+F(x)= tF(x) \end{aligned}$$

for every $x\in X.$

Finally, assume that F is not single-valued. For the proof by contradiction suppose that for some $t>0$

$$\begin{aligned} F(-tx)=_K-tF(x),\quad x\in X. \end{aligned}$$

Then, by (5), $F(tx)=_KtF(x)$ and, in view of Theorem 12 and K-additivity of F,

$$\begin{aligned} \begin{array}{ll} t\big (F(x)-F(x)\big )&{}{}=tF(x)+(-t)F(x)\subset \big (F(tx)+K\big )+\big (F(-tx)+K\big )\\ {} &{}{}\subset F(tx-tx)+K=F(0)+K\subset K \end{array} \end{aligned}$$

for $x\in X$. Hence

$$\begin{aligned} -\big (F(x)-F(x)\big )=F(x)-F(x)\subset \frac{1}{t}K\subset K, \end{aligned}$$

and, consequently, $F(x)-F(x)\subset K\cap (-K)=\{0\}$ for $x\in X$, which means that all sets F(x) are singletons. This contradiction ends the proof. $\square$

Theorem 15 with $K=\{0\}$ was proved by Nikodem (see [10, Theorem 5.3]).

Now, the question is what about the converse result? More precisely, is it true that every K-additive and K-homogeneous s.v map has to be K-continuous?

In the next example we show that generally it is not true.

Example 5

Let $K=[0,\infty )$, X be an infinite dimensional real normed space, $m,M:X\rightarrow {\mathbb {R}}$, $m(x)\le M(x)$ for $x\in X$. If m is a discontinuous linear functional, then $F(x)=[m(x),M(x)]$, $x\in X$, is a K-additive and K-homogeneous s.v. map, which is not K-continuous.

However, under some natural additional assumptions, we can obtain the converse result to Theorem 15.

Theorem 16

Let X be a finite dimensional real normed space, Y be a real normed space and K be a convex cone in Y. If $F:X\rightarrow \mathcal{C}\mathcal{C}(Y)$ is a K-additive and K-homogeneous s.v. map, it is K-continuous on X.

Proof

Let $K^*$ be the set of all continuous linear functionals on Y which are non-negative on K. For every $y^*\in K^*$ define

$$\begin{aligned} f_{y^*}(x)=\inf y^*(F(x)),\quad x\in X. \end{aligned}$$

First, observe that for every $y^*\in K^*$ and $A,B\in \mathcal{C}\mathcal{C}(Y)$,

$$\begin{aligned} \text{ if }\;A=_KB,\;\;\text{ then }\;\inf y^*(A)=\inf y^*(B).\end{aligned}$$

(6)

Indeed, let $A=_KB$. Then

$$\begin{aligned} y^*(A)\subset y^*(B+K)\subset y^*(B)+[0,\infty ),\quad \end{aligned}$$

and hence $\inf y^*(A)\ge \inf y^*(B)$. In view of symmetry of $=_K$, we get $\inf y^*(B)\ge \inf y^*(A)$. Thus condition (6) holds.

Now, fix $y^*\in K^*.$ We prove that $f_{y^*}$ is a linear functional. Since F is K-additive, $F(x_1+x_2)=_KF(x_1)+F(x_2)$ for $x_1,x_2\in X$, thus, by (6),

$$\begin{aligned} \begin{array}{rl} f_{y^*}(x_1)+f_{y^*}(x_2)&{}{}= \inf y^*(F(x_1))+\inf y^*(F(x_2))=\inf \big (y^*(F(x_1))+ y^*(F(x_2))\big )\\ {} &{}{} =\inf y^*(F(x_1)+F(x_2))=\inf y^*(F(x_1+x_2))=f_{y^*}(x_1+x_2) \end{array} \end{aligned}$$

for all $x_1,x_2\in X$, which proves additivity of $f_{y^*}$. Moreover, by K-homogeneity of F, $F(tx)=_KtF(x)$ for $t\ge 0$ and $x\in X$. Hence, in view of (6),

$$\begin{aligned} f_{y^*}(tx)=\inf y^*(F(tx))=\inf y^*(tF(x))=\inf ty^*(F(x))=tf_{y^*}(x) \end{aligned}$$

for $t\ge 0$ and $x\in X$. Since every additive real function is odd, $f_{y^*}$ is a linear functional.

In this way we obtained that every functional $f_{y^*}:X\rightarrow {\mathbb {R}}$ is linear, so it is continuous because X is finite dimensional. But in view of [8, Theorem 4] K-subadditive s.v. map for which $f_{y^*}$ is continuous for every $y^*\in K^*$ has to be weakly K-upper bounded on an open set. Thus, according to Theorem 13, F is K-continuous, which ends the proof. $\square$

5 Algebraic Structure of a K-Homogeneity Set

As we mentioned in the introduction, for every additive function $f:X\rightarrow {\mathbb {R}}$ defined on a real vector space X the set

$$\begin{aligned} H_f:=\left\{ t\in {\mathbb {R}}:\,f(tx)=tf(x)\;\text{ for } \text{ all }\;x\in X\right\} \end{aligned}$$

is a field, called the homogeneity field of f. Moreover, the following result holds.

Theorem 17

[9, Theorem 5.4.2]. Let  X be a real vector space. For every field $L\subset {\mathbb {R}}$ there is an additive function $f:X\rightarrow {\mathbb {R}}$ such that $H_f=L$.

Now, for real vector spaces X, Y, a s.v. map $F:X\rightarrow n(Y)$ and a closed convex cone $K\subset Y$ such that $K\cap (-K)=\{0\}$, we would like to define in the same way the K-homogeneity set of F,

$$\begin{aligned} H_{F,K}:=\left\{ t\in {\mathbb {R}}:F(tx)=_K tF(x)\;\text{ for } \text{ all }\;x\in X\right\} . \end{aligned}$$

First, let us prove basic properties of $H_{F,K}$.

Theorem 18

Let X be a real vector space, Y be a real vector metric space and $K\subset Y$ be a closed convex cone such that $K\cap (-K)=\{0\}$. If a s.v. map $F:X\rightarrow \mathcal{C}\mathcal{C}(Y)$ is K-additive and not single-valued, then the following conditions hold:

• (i) $\{0,1\}\subset H_{F,K}\subset [0,\infty )$,

• (ii) $s+t\in H_{F,K}$ for $s,t\in H_{F,K}$,

• (iii) $\frac{s}{t}\in H_{F,K}$ for $s,t\in H_{F,K}$ with $t\ne 0$;

i.e. $H_{F,K}$ is a submonoid of $\big ([0,\infty ),+\big )$ and $H_{F,K}{\setminus }\{0\}$ is a subgroup of $\big ((0,\infty ),\cdot \big )$.

Proof

(i) Clearly, $1\in H_{F,K}$ and, in view of Theorem 12, $0\in F(0)\subset K$, so $0\in H_{F,K}$. We have to prove yet that $H_{F,K}\subset [0,\infty )$.

For the proof by contradiction suppose that there is $t>0$ such that $-t\in H_{F,K}$. Fix arbitrary $x\in X$. Notice that in view of K-additivity of F

$$\begin{aligned} F(x)+F(-x)\subset F(0)+K\subset K, \end{aligned}$$

which easy implies

$$\begin{aligned} F(x)\subset -F(-x)+K. \end{aligned}$$

(7)

Moreover, since $-t\in H_{F,K}$,

$$\begin{aligned} -tF(-x)\subset F(-t(-x))+K=F(tx)+K \end{aligned}$$

(8)

and

$$\begin{aligned} F(tx)-tF(x)\subset F(tx)+F(-tx)+K\subset F(0)+K\subset K. \end{aligned}$$

(9)

Hence, according to (7)–(9),

$$\begin{aligned} \begin{array}{ll} tF(x)-tF(x)&{}{} \subset t\big (\!\!-F(-x)+K\big )-tF(x)\subset -tF(-x)-tF(x)+K\\ {} &{}{}\subset \big (F(tx)+K\big )-tF(x)+K\subset F(tx)-tF(x)+K\subset K. \end{array} \end{aligned}$$

Thus

$$\begin{aligned} -\big (F(x)-F(x)\big )=F(x)-F(x)\subset \frac{1}{t}K\subset K, \end{aligned}$$

which means that $F(x)-F(x)\subset K\cap (-K)=\{0\}$. But this implies that F(x) is a singleton for each $x\in X$, which contradicts the assumption.

(ii) Let $s,t\in H_{F,K}$ and $x\in X$. Then

$$\begin{aligned} F(tx)=_K tF(x),\quad F(sx)=_KsF(x), \end{aligned}$$

and hence, by K-additivity of F,

$$\begin{aligned} F((s+t)x)=F(sx+tx)=_K F(sx)+F(tx)=_K sF(x)+tF(x)= (s+t)F(x), \end{aligned}$$

which means that $s+t\in H_{F,K}$.

(iii) Now, assume that $s,t\in H_{F,K}$, $t\ne 0$ and $x\in X$. Then

$$\begin{aligned} F\left( \frac{s}{t}x\right) =_K sF\left( \frac{1}{t}x\right) . \end{aligned}$$

Hence, according to Lemma 9 (iv),

$$\begin{aligned} tF\left( \frac{s}{t}x\right) =_K tsF\left( \frac{1}{t}x\right) =_KsF\left( t\frac{1}{t}x\right) = sF(x), \end{aligned}$$

and, consequently,

$$\begin{aligned} F\left( \frac{s}{t}x\right) =_K\frac{s}{t}F(x), \end{aligned}$$

which means that $\frac{s}{t}\in H_{F,K}$. $\square$

For a s.v. map F in a special form we can obtain further important properties of $H_{F,K}$.

Corollary 19

Let X be a real vector space, Y be a real vector metric space and $K\subset Y$ be a closed convex cone such that $K\cap (-K)=\{0\}$. Assume that $z_0\in K\setminus \{0\}$ and

$$\begin{aligned} F(x)=[m(x),M(x)]z_0,\quad x\in X, \end{aligned}$$

where $m,M:X\rightarrow {\mathbb {R}}$ satisfy $m(x)< M(x)$ for $x\in X$. If F is K-additive, then m is additive and the following conditions hold:

• (i) $\{0,1\}\subset H_{F,K}\subset [0,\infty )$,

• (ii) $s+t\in H_{F,K}$ for $s,t\in H_{F,K}$,

• (iii) $\frac{s}{t}\in H_{F,K}$ for $s,t\in H_{F,K}$ with $t\ne 0$,

• (iv) $H_{F,K}=H_m\cap [0,\infty )$, where $H_m$ is the homogeneity field of m,

• (v) $s-t\in H_{F,K}$ for $s,t\in H_{F,K}$ such that $s-t\ge 0$.

Proof

Since F is K-additive, in view of Lemma 5, m is additive. According to Theorem 18, it is enough to show (iv) and (v).

(iv) Fix $t\in H_{F,K}$. By Theorem 18 (i), $t\ge 0$ and, for every $x\in X$, we get

$$\begin{aligned}{}[m(tx),M(tx)]z_0=F(tx)=_K tF(x)=[tm(x),tM(x)]z_0. \end{aligned}$$

Hence

$$\begin{aligned} \big (m(tx)-\alpha \big ) z_0\in K,\quad \big (tm(x)-\beta \big ) z_0\in K \end{aligned}$$

for some $\alpha \in [tm(x),tM(x)]$ and $\beta \in [m(tx),M(tx)]$. Since $z_0\in K\setminus \{0\}$ and $K\cap (-K)=\{0\}$,

$$\begin{aligned} m(tx)\ge \alpha \ge tm(x),\quad tm(x)\ge \beta \ge m(tx), \end{aligned}$$

which proves that $t\in H_m$.

On the other hand, if $t\in H_m\cap (0,\infty )$, then

$$\begin{aligned} \begin{array}{ll} F(tx)&{}{}=[m(tx),M(tx)]z_0\subset [m(tx),\infty )z_0= t[m(x),\infty )z_0\\ {} &{}{}=t[m(x),M(x)]z_0+[0,\infty )z_0\subset tF(x)+K \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{ll} tF(x)&{}{}=t[m(x),M(x)]z_0=[m(tx),tM(x)]z_0\subset [m(tx),\infty )z_0\\ {} &{}{}= [m(tx),M(tx)]z_0+[0,\infty )z_0\subset F(tx)+K \end{array} \end{aligned}$$

for every $x\in X$. It means that $t\in H_{F,K}$. Moreover, if $t=0$, in view of (i) $0\in H_{F,K}$.

(v) Take $s,t\in H_{F,K}$ such that $s-t\ge 0$. Then, according to (iv), s, t belongs to the field $H_m$ and hence $s-t\in H_m\cap [0,\infty )=H_{F,K},$ which ends the proof. $\square$

Problem 2

Is it true that under assumptions of Theorem 18 condition (v) of Corollary 19 holds?

We can give the positive answer to the above problem only in the case when F is additionally K-continuous and not single-valued, because then, in view of Theorem 15, $H_{F,K}=[0,\infty )$.

Finally, let us prove a result which seems to be (in some sense) analogous to Theorem 17.

Theorem 20

Let $S\subset {\mathbb {R}}$ be a set satisfying the following conditions:

• (i) $\{0,1\}\subset S\subset [0,\infty )$,

• (ii) $s+t\in S$ for $s,t\in S$,

• (iii) $\frac{s}{t}\in S$ for $s,t\in S$ with $t\ne 0$,

• (iv) $s-t\in S$ for $s,t\in S$ with $s-t\ge 0$.

Assume that X is a real vector space, Y is a real vector metric space and $K\subset Y$ is a closed convex cone such that $K\cap (-K)=\{0\}.$ Then there exists a s.v. map $F:X\rightarrow \mathcal{C}\mathcal{C}(Y)$ such that $S=H_{F,K}$.

Proof

First we prove that $L:=S\cup (-S)$ is a subfield of ${\mathbb {R}}$, i.e.

$$\begin{aligned} s-t\in L,\quad s,t\in L,\end{aligned}$$

(10)

$$\begin{aligned} \frac{s}{t}\in L,\quad s,t\in L,\; t\ne 0. \end{aligned}$$

(11)

Let $s,t\in L$. If $s\in S$ and $t\in -S$, then $-t\in S$ and $s-t\in S\subset L$ by (ii). If $s,t\in S$ or $s,t\in -S$, then either $s-t\ge 0$ or $s-t\le 0$, hence, according to (iv), $s-t\in S\cup (-S)=L.$ Thus (10) holds.

Now, let $s,t\in L$ with $t\ne 0$. If $s\in S$ and $t\in -S$, then $-t\in S$ and $\frac{s}{t}=-(\frac{s}{-t})\in -S\subset L$. If $s,t\in S$ or $s,t\in -S$, then, by (iii), $\frac{s}{t}=\frac{-s}{-t}\in S\subset L$. Hence (11) holds.

Knowing that L is a field, according to Theorem 17 we can find an additive function $f:X\rightarrow {\mathbb {R}}$ such that $H_f=L$.

Fix $z_0\in K\setminus \{0\}$ and define $F:X\rightarrow {\mathcal{C}\mathcal{C}}(Y)$ by

$$\begin{aligned} F(x)=[f(x),f(x)+1]z_0,\quad x\in X. \end{aligned}$$

Since f is additive, in view of Lemma 5, F is K-additive, and hence, according to Corollary 19,

$$\begin{aligned} H_{F,K}=H_f\cap [0,\infty )=L\cap [0,\infty )=S, \end{aligned}$$

which ends the proof $\square$