Further remarks on local K-boundedness of K-subadditive set-valued maps

Abstract

Let X be an abelian metric group with an invariant metric, Y be a real normed space and K be a convex cone in Y. We prove that a K-subadditive (K-superadditive) compact- and convex-valued map $F:X\rightarrow \mathcal{C}\mathcal{C}(Y)$, for which the functionals $f_{y^*}(x)=\inf y^*(F(x))$ are lower (upper, resp.) semicontinuous for any real continuous and non-negative on K functional $y^*$, has to be locally K-bounded on X. Our results refer to the papers Banakh and Jabłońska (Israel J Math 230:361–386, 2019), Jabłońska and Nikodem (Math Inequal Appl 22:1081–1089, 2019) and Nikodem (Aequationes Math 62:175–183, 2001).

1 Introduction

Let X be an abelian metric group with an invariant metric and Y be a real topological vector space. Assume that K is a subsemigroup of Y (i.e. $K+K\subset K$). Denote by n(Y) the family of all nonempty subsets of Y, and by ${\mathcal {B}}(Y)$ and $\mathcal{C}\mathcal{C}(Y)$ its subfamilies of all bounded subsets of Y and all convex compact subsets of Y, respectively.

Definition 1

A set-valued map (s.v. map for short) $F:X\rightarrow n(Y)$ is called K-subadditive if

$$\begin{aligned} F(x_1)+F(x_2) \subset F(x_1+x_2)+K \end{aligned}$$

(1)

for all $x_1,x_2 \in X$. If F satisfies

$$\begin{aligned} F(x_1+x_2)\subset F(x_1)+F(x_2)+K \end{aligned}$$

(2)

for all $x_1,x_2 \in X$ then it is called K-superadditive.

The concepts of K-subadditivity and K-superadditivity were introduced in [11], following the notions of K-midconvexity and K-midconcavity from [13]. Clearly, if F is a single-valued function, $Y={\mathbb {R}}$ and $K=[0,\infty )$, then K-subadditivity means the classical subadditivity, i.e. $f:X\rightarrow {\mathbb {R}}$ satisfies

$$\begin{aligned} f(x_1+x_2)\le f(x_1)+f(x_2)\;\;\;\text{ for } \text{ every }\;x_1,x_2\in X, \end{aligned}$$

as well as K-superadditivity means the classical superadditivity, i.e. $-f$ is subadditive.

Let us recall also the notion of K-lower (K-upper) semicontinuity from [13] which generalizes the classical notion of upper (lower, resp.) semicontinuity of a single-valued real function.

Definition 2

Let $x_0\in X$. The s.v. map $F:X\rightarrow n(Y)$ is called:

• K-lower semicontinuous at $x_0$, if for every neighborhood W of 0 in Y there exists a neighborhood U of 0 in X such that

  $$\begin{aligned} F(x_0)\subset F(x)+W+K\;\;\;\text{ for }\;x\in x_0+U, \end{aligned}$$

• K-upper semicontinuous at $x_0$, if for every neighborhood W of 0 in Y there exists a neighborhood U of 0 in X such that

  $$\begin{aligned} F(x)\subset F(x_0)+W+K\;\;\;\text{ for }\;x\in x_0+U. \end{aligned}$$

Moreover, F is K-lower (K-upper) semicontinuous on X, if it is K-lower (K-upper, resp.) semicontinuous at each point $x\in X$.

Definition 3

A s.v. map $F:X\rightarrow {\mathcal {B}}(Y)$ is called:

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

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

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

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

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

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

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

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

The above idea of (weak) K-upper/K-lower boundedness was introduced in [13] as a generalization of the notion of boundedness from above/below of single-valued real functions. More precisely, weak $[0,\infty )$-upper boundedness, as well as $[0,\infty )$-upper boundedness of a single-valued real function means its classical boundedness from above. The same holds for (weak) $[0,\infty )$-lower boundedness of a single-valued real function.

Definition 4

A s.v. map $F:X\rightarrow {\mathcal {B}}(Y)$ is called:

• locally (weakly) K-upper (K-lower) bounded at $x\in X$, if it is (weakly) K-upper (K-lower, resp.) bounded on some neighborhood of x,

• locally K-bounded at $x\in X$, if it is both locally weakly K-upper bounded and locally K-lower bounded at x,

• locally K-bounded on X, if it is locally K-bounded at each point of X.

It is well known that for every subadditive real function defined on an abelian metric group boundedness from above on a “large” (in the sense of category or measure) set implies its local boundedness on the whole domain (see e.g. [12, Theorem 16.2.3]). Some recent results on subadditive functions can be found e.g. in [3]–[7].

In [2] the following generalization of the mentioned classical result was proved.

Theorem 1

[2, Theorem 2.2] Let X be an abelian metric group with an invariant metric and $f:X\rightarrow {\mathbb {R}}$ be a subadditive function. If f is bounded from above on a non-null-finite set $A\subset X$ then f is locally bounded on X.

The concept of null-finite sets was introduced in [1].¹

Definition 5

A subset A of an abelian metric group X is called null-finite if there exists a sequence $(x_n)_{n\in {\mathbb {N}}}$ tending to zero in X such that the set $\{n\in {\mathbb {N}}:x+x_n\in A\}$ is finite for every $x\in X$.

In a complete abelian metric group X with an invariant metric the following sets are not null-finite: open sets, non-meager sets with the Baire property, sets of positive Haar measure provided X is locally compact, universally measurable sets which are not Haar-null, Borel sets which are not Haar-meager (see [1, Theorems 5.1 and 6.1]).

The notions of a Haar-null set and a Haar-meager set have been introduced by Christensen [8] and Darji [9], respectively. A subset B of an abelian Polish group X is called:

• Haar-meager if there exist a Borel set $A\supset B$, a compact metric space K and a continuous function $f:K\rightarrow X$ such that $f^{-1}(A+x)$ is meager in K for every $x\in X$;

• Haar-null if there exist a universally measurable set $A\supset B$ and a $\sigma$-additive probability Borel measure $\mu$ on X such that $\mu (A+x)=0$ for every $x\in X$.

It was proved in [8] and [9] that every Haar-meager set is meager and, moreover, in every locally compact abelian Polish group the notions of a Haar-meager set and a Haar-null set are equivalent to the notions of a meager set and a set of Haar measure zero, respectively.

In [11] we proved the following generalization of Theorem 1.

Theorem 2

[11, Theorems 2 and 3] Let X be an abelian metric group with an invariant metric and Y be a locally convex real topological vector space. Assume that $A\subset X$ is a non-null-finite set, K is a subsemigroup of Y, and $F:X\rightarrow {\mathcal {B}}(Y)$ is a s.v. map. If F 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 K-bounded on X.²

Here we use the above theorem to show some relationships between a K-subadditivie or K-superadditive s.v. map $F:X\rightarrow \mathcal{C}\mathcal{C}(Y)$ and the functionals $f_{y^*}:X\rightarrow {\mathbb {R}}$ defined by

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

(3)

where $y^*\in K^*$ and $K^*$ means the set of all real continuous functionals on a real topological vector space Y which are non-negative on K, i.e.

$$\begin{aligned} K^*=\{y^*\in Y^*:y^*(y)\ge 0 \ \text{ for } \text{ every } \ y\in K \}. \end{aligned}$$

The results obtained refer to [10, Theorem 5] and [14, Theorem 1], where the continuity of a K-midconvex (K-midconcave) s.v. map was proved under the assumption that the functionals (3) are lower (upper, resp.) semicontinuous. This paper relates also to [1, Theorem 9.1].

2 Main results

Theorem 3

Let X be an abelian metric group with an invariant metric, and $A\subset X$ be a non-null-finite set. Assume that Y is a locally convex real topological vector space, K is a subsemigroup of Y, and $F:X\rightarrow {\mathcal {B}}(Y)$ is a s.v. map. If one of the following conditions holds:

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

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

then for every $y^*\in K^*$ the functional $f_{y^*}:X\rightarrow {\mathbb {R}}$ defined by (3) is locally bounded on X.

Proof

Assume that F satisfies (i) (if F satisfies (ii), the proof runs in the same way). Then there exists a bounded set $B\subset Y$ such that

$$\begin{aligned} F(x)\cap (B-K)\ne \emptyset , \ \ x\in A. \end{aligned}$$

(4)

Fix any $y^*\in K^*$ and take the functional $f_{y^*}$ defined by (3). Since F is K-subadditive and $y^*\in K^*$, we have

$$\begin{aligned} y^*(F(x_1))+ y^*(F(x_2)){} & {} = y^*(F(x_1)+F(x_2))\subset y^*(F(x_1+x_2)+K ) \\{} & {} \subset y^*(F(x_1+x_2))+[0,\infty ) \end{aligned}$$

for all $x_1,x_2\in X$. Hence

$$\begin{aligned} f_{y^*}(x_1)+f_{y^*}(x_2){} & {} = \inf y^*(F(x_1))+\inf y^*(F(x_2))\\{} & {} =\inf \left( y^*(F(x_1))+ y^*(F(x_2))\right) \ge \inf y^*(F(x_1+x_2) \\{} & {} =f_{y^*}(x_1+x_2), \end{aligned}$$

which means that $f_{y^*}$ is subadditive. By (4), for every $x\in A$ we have

$$\begin{aligned} y^*(F(x))\cap y^*(B-K)\ne \emptyset . \end{aligned}$$

Hence

$$\begin{aligned} y^*(F(x))\cap \big ( y^*(B)+(-\infty , 0]\big ) \ne \emptyset . \end{aligned}$$

(5)

But the set $y^*(B)$ is bounded, i.e. $y^*(B)\subset [m, M]$ for some $m<M$. Then, by (5),

$$\begin{aligned} y^*(F(x))\cap (-\infty , M] \ne \emptyset , \end{aligned}$$

which means that

$$\begin{aligned} f_{y^*}(x)\le M, \ \ x\in A. \end{aligned}$$

Consequently, in view of Theorem 1, $f_{y^*}$ is locally bounded on X. $\square$

Now, which assumptions on the functionals $f_{y^*}$ defined by (3) for $y^*\in K^*$ imply the local K-boundedness on X of a K-subadditive s.v. map F? The next theorem gives an answer.

Let us recall that in a real vector space by a convex cone we mean the set K satisfying $K+K\subset K$ and $tK\subset K$ for every $t\in [0,\infty )$.

Theorem 4

Let X be a complete metric space with an invariant metric, Y be a real normed space, and K be a convex cone in Y. If a s.v. map $F:X\rightarrow \mathcal{C}\mathcal{C}(Y)$ is K-subadditive and for every $y^*\in K^*$ the functional $f_{y^*}:X\rightarrow {\mathbb {R}}$ defined by (3) is lower semicontinuous on X, then F is locally K-bounded on X.

Proof

Let B be the unit ball in Y and $B_n:=nB$, $n\in {\mathbb {N}}$. Denote

$$\begin{aligned} A_n:=\{x\in X:F(x)\cap \textrm{ cl}\,({B}_n - K)\ne \emptyset \},\;\; n\in {\mathbb {N}}. \end{aligned}$$

Clearly, $\bigcup _{n\in {\mathbb {N}}} A_n = X.$ We will prove that the sets $A_n$ are closed.

If $A_n=X$, it is obvious. So, fix $n\in {\mathbb {N}}$ such that $A_n\ne X$ and fix $x_0\in X\setminus A_n$. Then

$$\begin{aligned} F(x_0)\cap \textrm{ cl}\,(B_n-K)=\emptyset . \end{aligned}$$

Since $F(x_0)$ is compact convex and $B_n$ is convex, by the separation theorem (see [14, Lemma 1]) there exists a functional $y^* \in K^*$ such that

$$\begin{aligned} \inf y^*(F(x_0)) > \sup y^*(B_n). \end{aligned}$$

Let

$$\begin{aligned} \varepsilon := \inf y^*(F(x_0))-\sup y^*(B_n). \end{aligned}$$

Since $y^*$ is continuous at zero, we can find a neighborhood V of zero in Y with $y^*(V)\subset \left( -\frac{\varepsilon }{2},\frac{\varepsilon }{2}\right)$. By the lower semicontinuity of $f_{y^*}$ at $x_0$ there exists a neighborhood U of $x_0$ such that

$$\begin{aligned} f_{y^*}(x) > f_{y^*}(x_0)-\frac{\varepsilon }{2}\;\;\;\text{ for } \text{ every }\;x\in U. \end{aligned}$$

Now, we prove that

$$\begin{aligned} F(x)\cap \textrm{ cl}\,(B_n -K) =\emptyset \;\;\;\text{ for } \text{ every }\; x\in U. \end{aligned}$$

(6)

Indeed, if for some $x\in U$ it was not true, there would exist a point

$$\begin{aligned} z\in F(x)\cap \textrm{ cl}\,(B_n -K)\subset F(x)\cap (B_n-K-V). \end{aligned}$$

Let $z=b-k-v$ with some $b\in B_n$, $k\in K$ and $v\in V$. Then

$$\begin{aligned} y^*(b)= & {} y^*(z)+y^*(k)+y^*(v)> y^*(z)-\frac{\varepsilon }{2} \\\ge & {} f_{y^*}(x)-\frac{\varepsilon }{2}>f_{y^*}(x_0)-\varepsilon = \sup y^*(B_n), \end{aligned}$$

a contradiction.

Thus (6) holds and, consequently, $U\subset X{\setminus } A_n$ which shows that $X{\setminus } A_n$ is open and hence $A_n$ is closed.

Since X is a complete metric space and $X=\bigcup _{n\in {\mathbb {N}}} A_n$, by the Baire category theorem there exists $n_0 \in {\mathbb {N}}$ such that $\textrm{ int}\,A_{n_0} \ne \emptyset .$ By the definition of $A_{n_0}$ we have

$$\begin{aligned} F(x)\cap \big ((B_{n_0}+B)-K\big )\ne \emptyset \;\;\;\text{ for }\; x\in A_{n_0}. \end{aligned}$$

Since the set $B_{n_0}+B$ is bounded, this means that F is weakly K-upper bounded on the set $A_{n_0}$ with non-empty interior. Consequently, by Theorem 2, F is locally K-bounded on X. This finishes the proof. $\square$

The next example shows that it is not possible to get the same result in Theorem 4 if we weaken the assumption on the functionals $f_{y^*}$, i.e. replace lower semicontinuity on X of $f_{y^*}$ by local boundedness from below on X.

Example 1

Let $K=[0,\infty )$ and $F:{\mathbb {R}}\rightarrow \mathcal{C}\mathcal{C}({\mathbb {R}})$ be defined by

$$\begin{aligned} F(x)=\big [|a(x)|, |a(x)|+1 \big ]\;\;\;\text{ for } \text{ every } \;x\in {\mathbb {R}}, \end{aligned}$$

where $a:{\mathbb {R}}\rightarrow {\mathbb {R}}$ is a discontinuous additive function. Clearly, then F is K-subadditive, even locally K-lower bounded on X, and every $y^* \in K^*$ is given by $y^*(x)=cx$ with some $c\ge 0$. Hence

$$\begin{aligned} f_{y^*}(x)= \inf y^*\big (F(x)\big ) = \inf c\big [|a(x)|, |a(x)|+1\big ]=c|a(x)|\ge 0 \end{aligned}$$

which means that $f_{y^*}$ is bounded from below on the whole ${\mathbb {R}}$, but F is not locally weakly K-upper bounded at any point.

Indeed, if for some open set $U\subset {\mathbb {R}}$ and some $[m,M]\subset {\mathbb {R}}$ we had $F(x)\cap \big ([m,M]-K\big )\ne \emptyset$ for $x\in U$, then $a(x) \le |a(x)| \le M$ for $x\in U$, which would be impossible because of the discontinuity of a.

Problem 1

Let X, Y, K and F be as in Theorem 4. Can we obtain the same result replacing the lower semicontinuity of $f_{y^*}$ on X by upper semicontinuity on X for every $y^*\in K^*$, or by local boundedness on X for every $y^*\in K^*$?

We know that the answer to Problem 1 is positive in the case when $Y={\mathbb {R}}$ and $K=[0,\infty )$ (even under the weaker assumptions that for some nonzero $y^* \in K^*$ the functional $f_{y^*}$ is upper semicontinuous at a point of X or it is locally bounded at a point of X).

Indeed, let $F(x)=[m(x),M(x)]$ for $x\in X$. Then the K-subadditivity of F means the subadditivity of m. Moreover, $f_{y*}(x)=c\,m(x)$, $x\in X$, with $c>0$ and the upper semicontinuity (local boundedness) of $f_{y*}$ at some $x_0\in X$ implies the upper semicontinuity (local boundedness, resp.) of m at $x_0$, and hence m is locally bounded on X in view of Theorem 1. Thus, for every $x\in X$ we can find a neighborhood U of x such that $m(U)\subset [a,A]$ with some $a<A$. For every $t\in U$ we get

$$\begin{aligned} F(t)\cap ([a,A]-K){} & {} =[m(t),M(t)]\cap (-\infty , A]\ne \emptyset ;\\ F(t){} & {} = [m(t),M(t)]\subset [a,\infty )=[a,A]+K, \end{aligned}$$

which means that F is locally K-bounded on X.

Let us recall that the subadditivity of a single-valued real function $f:X\rightarrow {\mathbb {R}}$ means the superadditivity of the function $-f$. Unfortunately there is no analogous property for K-subadditive s.v. maps, i.e. the K-superadditivity of F does not mean the $\pm K$-subadditivity of $-F$ (see [11, Examples 2 and 3]). That is why we have to prove an analogous result for K-superadditive s.v. maps independently.

Theorem 5

Let X be a complete metric space with an invariant metric, Y be a real normed space and K be a convex cone in Y. If a s.v. map $F:X\rightarrow \mathcal{C}\mathcal{C}(Y)$ is K-superadditive and for every $y^*\in K^*$ the functional $f_{y^*}:X\rightarrow {\mathbb {R}}$ defined by (3) is upper semicontinuous on X, then F is locally K-bounded on X.

Proof

Let B be the unit ball in Y and $B_n:=nB$ for $n\in {\mathbb {N}}$. Define

$$\begin{aligned} A_n:=\{x\in X:F(x)\subset \textrm{ cl}\,({B}_n + K)\},\;\; n\in {\mathbb {N}}. \end{aligned}$$

(7)

Then $\bigcup _{n\in {\mathbb {N}}} A_n = X.$ We will show that $F(x)\subset \textrm{ cl}\,(B_n+K)$ for every $x\in \textrm{ cl}\,A_n$ and $n\in {\mathbb {N}}$.

For a proof by contradiction suppose that there are $n_0\in {\mathbb {N}}$, $x_0\in \textrm{ cl}\,A_{n_0}$ and $z\in F(x_0) {\setminus } \textrm{ cl}\,(B_{n_0}+K).$ Since the set $\textrm{ cl}\,(B_{n_0}+K)$ is convex and closed, by the separation theorem (see e.g [15, Theorem 3.4]) there exists a continuous linear functional $y^* \in Y^*$ such that

$$\begin{aligned} y^*(z) < \inf y^*\big (\textrm{ cl}\,(B_{n_0}+ K) \big ). \end{aligned}$$

(8)

Since

$$\begin{aligned} y^*(k)\ge y^*(z)-y^*(b_0)=:M\;\;\;\text{ for } \text{ all }\;k\in K \end{aligned}$$

with arbitrarily fixed $b_0 \in B_{n_0},$ and, moreover,

$$\begin{aligned} y^*(k) = \frac{1}{m}y^*(mk) \ge \frac{1}{m} M\;\;\;\text{ for } \text{ every }\; m\in {\mathbb {N}}, \end{aligned}$$

for $m\rightarrow \infty$ we obtain $y^*(k)\ge 0$ for all $k\in K$ which means that $y^* \in K^*.$ Now, put

$$\begin{aligned} \varepsilon :=\inf y^*\big (\textrm{ cl}\,(B_{n_0}+K)\big )-y^*(z). \end{aligned}$$

By the upper semicontinuity of $f_{y^*}$ at $x_0$ there exists a neighborhood U of $x_0$ such that

$$\begin{aligned} f_{y^*}(x) < f_{y^*}(x_0) + \varepsilon \;\;\;\text{ for } \text{ every }\; x\in U. \end{aligned}$$

(9)

Since $x_0\in \textrm{ cl}\,A_{n_0}$, there exists an $x_1 \in A_{n_0} \cap U.$ Then, according to (9) and the definition of $\varepsilon$, we obtain

$$\begin{aligned} \begin{array}{ll} f_{y^*}(x_1)\!\!\! &{}< f_{y^*}(x_0) + \varepsilon \le y^*(z) +\varepsilon =\inf y^* \big (\textrm{ cl}\,(B_{n_0}+K) \big ) \\ &{}\le \inf y^*\big (F(x_1) \big )= f_{y^*}(x_1). \end{array} \end{aligned}$$

This contradiction proves that F is K-lower bounded on $\textrm{ cl}\,A_n$ for every $n\in {\mathbb {N}}$, i.e.

$$\begin{aligned} F(x)\subset \textrm{ cl}\,(B_{n}+K)\;\;\;\text{ for } \text{ every }\; x\in \textrm{ cl}\,A_{n}. \end{aligned}$$

Since $X=\bigcup _{n\in {\mathbb {N}}} \textrm{ cl}\,A_n$ and X is complete, in view of the Baire category theorem there exists $n_0 \in {\mathbb {N}}$ such that $\textrm{ int}\,\textrm{ cl}\,A_{n_0} \ne \emptyset .$ Moreover,

$$\begin{aligned} F(x)\subset \textrm{ cl}\,(B_{n_0}+K)\subset B_{n_0}+B+K\;\;\;\text{ for } \text{ every }\; x\in \textrm{ cl}\,A_{n_0}, \end{aligned}$$

so F is K-lower bounded on an open set. Now, to complete the proof it is enough to apply Theorem 2. $\square$

Modifying Example 1, it is easy to observe that it is impossible to get the same result in Theorem 5 assuming the local boundedness from above on X (instead of upper semicontinuity on X) of functionals $f_{y^*}$.

Example 2

Let $K=(-\infty ,0]$ and $F:{\mathbb {R}}\rightarrow \mathcal{C}\mathcal{C}({\mathbb {R}})$ be defined by

$$\begin{aligned} F(x)=\big [|a(x)|, |a(x)|+1 \big ]\;\;\;\text{ for } \text{ every } \;x\in {\mathbb {R}}, \end{aligned}$$

where $a:{\mathbb {R}}\rightarrow {\mathbb {R}}$ is a discontinuous additive function. Clearly, then F is K-superadditive, even locally K-upper bounded on X, and every $y^* \in K^*$ is given by $y^*(x)=cx$ with some $c\le 0$. Hence

$$\begin{aligned} f_{y^*}(x)= \inf y^*\big (F(x)\big ) = \inf c\big [|a(x)|, |a(x)|+1\big ]=c\big (|a(x)|+1\big )\le 0 \end{aligned}$$

which means that $f_{y^*}$ is bounded from above on the whole ${\mathbb {R}}$, but F is not locally K-lower bounded at any point.

The following question arises which is analogous to Problem 1.

Problem 2

Let X, Y, K and F be as in Theorem 5. Can we obtain the same result replacing the upper semicontinuity of $f_{y^*}$ on X by lower semicontinuity on X for every $y^*\in K^*$, or by local boundedness on X for every $y^*\in K^*$?

For now we can give a positive answer only in the case $Y={\mathbb {R}}$ and $K=[0,\infty )$ (the solution runs in the same way as the solution of Problem 1).

Remark 1

Notice that in the conclusion of Theorems 4 and 5 we are not able to get the local K-upper boundedness (instead of the local weak K-upper boundedness) of F on X.

For example, let $K=[0,\infty )$ and $F:{\mathbb {R}}\rightarrow \mathcal{C}\mathcal{C}({\mathbb {R}})$ be given by

$$\begin{aligned} F(x)=\left\{ \begin{array}{ll} \left[ 0,\frac{1}{|x|}\right] , &{}\quad x\ne 0,\\[1.5ex] \{0\},&{}\quad x=0. \end{array} \right. \end{aligned}$$

Clearly, F is K-subadditive and K-superadditive. Moreover, for every $y^*\in K^*$ (i.e. $y^*(x)=cx$ with $c\ge 0$) the functional $f_{y^*}=0$, so it is continuous on ${\mathbb {R}}$. But F is not K-upper bounded at 0.

3 Applications and final remarks

One can easily observe that if $F:X\rightarrow {\mathcal {B}}(Y)$ is a s.v. map, where X is a metric space and Y is a real normed space, then

• (i) the K-upper semicontinuity of F at a point $x_0\in X$ implies local K-lower boundedness at this point,

• (ii) the K-lower semicontinuity of F at a point $x_0\in X$ implies local weak K-upper boundedness at this point.

Indeed, to obtain (i) it is enough to put the bounded set $B:=F(x_0)+W$ with a fixed neighbourhood $W\subset Y$ of 0. The proof of (ii) runs by contradiction. Fix a neighborhood $W\subset Y$ of 0. Then, by the K-lower semicontinuity of F, there is a neighbourhood U of $x_0$ such that

$$\begin{aligned} F(x_0)\subset F(x)+W+K\;\;\;\text{ for } \text{ every }\;\;x\in U. \end{aligned}$$

If for the bounded set $B:=F(x_0)-W$ we could find $x_1\in U$ such that $F(x_1)\cap (B-K)=\emptyset$, then

$$\begin{aligned} (F(x_1)+W+K)\cap F(x_0)=\emptyset . \end{aligned}$$

This contradiction proves that F is weakly K-upper bounded at $x_0$.

Moreover, in view of Examples 1 and 2, a K-subadditive s.v. map which is locally K-lower bounded on X, as well as K-superadditive s.v. map which is locally K-upper bounded on X, needn’t be locally K-bounded on X.

However, as an immediate consequence of Theorems 4 and 5, we obtain the following result.

Corollary 6

Let X be a complete metric space with an invariant metric, Y be a real normed space, and K be a convex cone in Y. If a s.v. map $F:X\rightarrow \mathcal{C}\mathcal{C}(Y)$ satisfies one of the following two conditions:

• (i) F is K-subadditive and K-upper semicontinuous on X,

• (ii) F is K-superadditive and K-lower semicontinuous on X,

then F is locally K-bounded on X.

Proof

Assume that F satisfies (i) (if F satisfies (ii) the proof runs in the same way). Let B be the unit ball in Y. Fix arbitrary $x_0\in X$, $y^* \in K^*$ and $\varepsilon >0$. Since F is K-upper semicontinuous at $x_0$, there exists a neighborhood U of 0 in X such that

$$\begin{aligned} F(x) \subset F(x_0) + \frac{\varepsilon }{\Vert y^*\Vert }B + K\;\;\;\text{ for } \text{ every }\;x\in x_0 +U. \end{aligned}$$

Hence

$$\begin{aligned} y^*\big (F(x)\big ) \subset y^*\big (F(x_0)\big ) + \frac{\varepsilon }{\Vert y^*\Vert }[-\Vert y^*\Vert , \Vert y^*\Vert ] + [0,\infty ), \ \ \ x\in x_0 +U. \end{aligned}$$

Consequently,

$$\begin{aligned} f_{y^*}(x)\ge f_{y^*}(x_0) -\varepsilon , \ \ \ x\in x_0 +U, \end{aligned}$$

which shows that $f_{y^*}$ is lower semicontinuous at $x_0$. To complete the proof it is enough to apply Theorem 4. $\square$

In Theorems 4 and 5 we assume that K is a convex cone in a real normed space, but considerations on K-subadditivity/K-superadditivity seem to be much more natural for a subsemigroup K.

It is easy to check that if K is a subsemigroup with 0, then $\textrm{ conv}\,K$ is a convex cone. Clearly then

• K-subadditivity implies $\textrm{ conv}\,K$-subadditivity,

• K-superadditivity implies $\textrm{ conv}\,K$-superadditivity,

• local K-boundedness at a point implies local $\textrm{ conv}\,K$-boundedness at the same point,

• $y^*\in K^*$ implies $y^*\in (\textrm{ conv}\,K)^*$.

Consequently, by Theorems 4 and 5, we get the following corollary.

Corollary 7

Let X be a complete metric space with an invariant metric, Y be a real normed space, and K be a subsemigroup of Y with 0. Assume that $F:X\rightarrow \mathcal{C}\mathcal{C}(Y)$ is a s.v. map and $f_{y^*}:X\rightarrow {\mathbb {R}}$ is the functional defined by (3) for every $y^*\in K^*$. If one of the following conditions holds:

• (i) F is K-subadditive and $f_{y^*}$ is lower semicontinuous on X for every $y^*\in K^*$,

• (ii) F is K-superadditive and $f_{y^*}$ is upper semicontinuous on X for every $y^*\in K^*$,

then F is locally $\textrm{ conv}\,K$-bounded on X.

Problem 3

Is it possible to get the stronger result that F is locally K-bounded on X in Corollary 7?

The next corollary gives an answer in the case $Y={\mathbb {R}}.$

Corollary 8

Let X be a complete metric space with an invariant metric, and K be a subsemigroup of ${\mathbb {R}}$ with 0. Assume that $F:X\rightarrow \mathcal{C}\mathcal{C}({\mathbb {R}})$ is a s.v. map and $f_{y^*}:X\rightarrow {\mathbb {R}}$ is the functional defined by (3) for every $y^*\in K^*$. If one of the following conditions holds:

• (i) F is K-subadditive and $f_{y^*}$ is lower semicontinuous on X for every $y^*\in K^*$,

• (ii) F is K-superadditive and $f_{y^*}$ is upper semicontinuous on X for every $y^*\in K^*$,

then F is locally K-bounded on X.

Proof

Assume that F satisfies (i) (the proof for (ii) is similar). Let $F(x)=[m(x),M(x)]$ for $x\in X$.

The case when $\textrm{ conv}\,K=\{0\}$ is trivial because then $K=\{0\}$, so it is enough to use Corollary 7.

Now, consider the case when $\textrm{ conv}\,K=[0,\infty )$. By Corollary 7F is locally $\textrm{ conv}\,K$-bounded on X; i.e., for every $x\in X$ there are a neighborhood U of x and intervals $[m_1,M_1],[m_2,M_2]$ with $m_i<M_i$ for $i=1,2$ such that for any $t\in U$

$$\begin{aligned}{} & {} {[}m(t),M(t)]\cap (-\infty ,M_1]=F(t)\cap ([m_1,M_1]-\textrm{ conv}\,K)\ne \emptyset ,\\{} & {} {[}m(t),M(t)]=F(t)\subset [m_2,M_2]+\textrm{ conv}\,K=[m_2,\infty ). \end{aligned}$$

Hence $m(U)\subset [m_2,M_1].$ Since $0\in K$, for $B=[m_2,M_1]$ and every $t\in U$

$$\begin{aligned} F(t)\cap (B-K)\supset [m(t),M(t)] \cap B=[m(t),M(t)]\cap [m_2,M_1]\ne \emptyset , \end{aligned}$$

which means that F is weakly K-upper bounded at $x\in X$. Hence, in view of Theorem 2, F is locally K-bounded on X.

If $\textrm{ conv}\,K=(-\infty ,0]$, then the proof runs in a similar way (then M is bounded at x).

Finally, let $\textrm{ conv}\,K={\mathbb {R}}$. Then there are $k_1,k_2\in K$ with $k_1<0$ and $k_2>0$. Moreover, for every $x\in X$

$$\begin{aligned}{} & {} F(x)\cap \big ([k_1,k_2]-K\big )=F(x)\cap {\mathbb {R}}\ne \emptyset ,\\{} & {} F(x)\subset {\mathbb {R}}=[k_1,k_2]+K, \end{aligned}$$

which means that F is locally K-bounded on X. $\square$