2F (s) + 2F (t) ⊂ F (s + t) + F (s − t) + K, s, t ∈ X,

Katarzyna Troczka–Pawelec

K-subquadratic set-valued functions

Abstract. Let X = (X, +) be an arbitrary topological group. The aim of the paper is to prove a regularity theorem for K-subquadratic set-valued functions, that is, solutions of the inclusion

2F (s) + 2F (t) ⊂ F (s + t) + F (s − t) + K, s, t ∈ X,

with values in a topological vector space and where K is a cone in this space.

2000 Mathematics Subject Classification: 54C05, 54C60, 39B72, 26E25, 39B62, 26A15.

Key words and phrases: Subquadratic set-valued functions; superquadratic set-valued functions; regularity theorem, set-valued functions.

1. Introduction. Let X = (X, +) be an arbitrary topological group. A real- valued function F , is called subquadratic if it fulfills the inequality

(1) F (x + y) + F (x − y) ¬ 2F (x) + 2F (y), x, y ∈ X.

The continuity problem of functions of this kind was considered in [1]. This problem can be also considered in the class of set-valued functions. In this paper we will consider a K−subquadratic set-valued function , that is, solution of the inclusion (2) 2F (x) + 2F (y) ⊂ F (x + y) + F (x − y) + K, x, y ∈ X

which is defined on 2−divisible locally bounded topological group X with non-

empty, compact and convex values in a locally convex topological vector space Y ,

where K denotes a cone in this space. We will present here some conditions under

which such multifunctions are K−continuous. The concept of K-subquadraticity is

related to real-valued subquadratic functions. Note that, in the case when F is a

single-valued real function and K = [0, ∞), we obtain the standard definition of subquadratic functionals (1). Assuming K = {0} in (2) we obtain the inclusion (3) 2F (x) + 2F (y) ⊂ F (x + y) + F (x − y), x, y ∈ X.

The continuity problem of set-valued functions of this kind was considered in [4].

Let us start with the notations used in this paper. Let Y be a topological vector space. Let n(Y ) denotes the family of all non-empty subsets of Y , cc(Y )−the family of all compact and convex members of n(Y ). The term set-valued function will be abbreviated to the form s.v.f.

Now we shall present some examples of K−subquadratic s.v. functions.

Examples 1.1 The s.v. function F : R → n(R) defined by

F (x) =

 R , x 6= 0 {0} , x = 0.

is K−subquadratic, where K = {0}.

Examples 1.2 The s.v. function F : R → n(R) defined by

F (x) =

 [0, 1] , x ∈ R \ Q {0} , x ∈ Q.

is K-subquadratic, where K = [0, ∞).

Now we present here some definitions for the sake of completeness. Recall that a set K ⊂ Y is called a cone iff K + K ⊂ K and sK ⊂ K for all s ∈ (0, ∞).

Definition 1.3 (cf. [2]) A cone K in a topological vector space Y is said to be a normal cone iff there exists a base W of zero in Y such that

W = (W + K) ∩ (W − K) for all W ∈ W.

Definition 1.4 (cf. [2]) An s.v.f. F : X → n(Y ) is said to be K-upper semi- continuous (abbreviated K−u.s.c.) at x ⁰ ∈ X iff for every neighbourhood V of zero in Y there exists a neighbourhood U of zero in X such that

F (x) ⊂ F (x ⁰ ) + V + K

for every x ∈ x ⁰ + U.

Definition 1.5 (cf. [2]) An s.v.f. F : X → n(Y ) is said to be K-lower semi- continuous (abbreviated K−l.s.c.) at x ⁰ ∈ X iff for every neighbourhood V of zero in Y there exists a neighbourhood U of zero in X such that

F (x 0 ) ⊂ F (x) + V + K for every x ∈ x ⁰ + U.

Definition 1.6 (cf. [2]) An s.v.f. F : X → n(Y ) is said to be K−continuous at x 0 ∈ X iff it is both K−u.s.c. and K−l.s.c. at x 0 . It is said to be K−continuous iff it is K−continuous at each point of X.

Note that K−continuity of F in the case where K = {0} means its continuity with respect to the Hausdorff topology on n(Y ).

We will frequently use the following lemma.

Lemma 1.7 Let Y be a topological vector space and K be a cone in Y . Let A, B, C be non-empty subsets of Y such that A + C ⊂ B + C + K. If B is convex and C is bounded then for every a ∈ A there exists a sequence {d ⁿ } n ∈N such that d n ∈ B + K for every n ∈ N and d ⁿ −→ a.

Proof Let a ∈ A. Suppose that c ¹ ∈ C. Since a + c ¹ ∈ B + C + K there exist b 1 ∈ B, c ² ∈ C, k ¹ ∈ K such that

a + c 1 = b 1 + c 2 + k 1 . Because K is a cone, the following inclusion holds

A + C + K ⊂ B + C + K.

For the same reason we can find b 2 ∈ B, c ³ ∈ C, k ² ∈ K for which a + c 2 + k 1 = b 2 + c 3 + k 2 .

Repeat this procedure infinitely. Adding the first n of the obtained equalities we get

na + X n

i=1

c i +

n X −1 i=1

k i = X n i=1

b i +

n+1 X

i=2

c i + X n i=1

k i . Hence

a = 1 n

X n i=1

b i − 1 n c 1 + 1

n c n+1 + 1 n k n . Define

d n := 1 n

X n i=1

b i + 1 n k n .

Because of convexity of the set B, we obtain d n ∈ B + K. On the other hand d n = a + 1

n c 1 − 1 n c n+1 .

Letting n −→ ∞ we have d ⁿ −→ a. This completes the proof. 

As an immediate consequence of the above lemma we obtain the following lemma.

Lemma 1.8 Let Y be a topological vector space and K be a cone in Y . Let A, B, C be non-empty subsets of Y such that A + C ⊂ B + C + K. If B is convex and C is bounded then A ⊂ B + K.

In our proofs we will often use four known lemmas ( see Lemma 1.1, Lemma 1.3 , Lemma 1.6 and Lemma 1.9 in [2]). The first lemma says that for a convex subset A of an arbitrary real vector space Y the equality (s + t)A = sA + tA holds for every s, t ­ 0 ( or s, t ¬ 0 ). The second lemma says that in a real vector space Y for two convex subsets A, B the set A + B is also convex. The next lemma says that if A ⊂ Y is a closed set and B ⊂ Y is a compact set, where Y denotes a real topological vector space, then the set A + B is closed. For any sets A, B ⊂ Y , where Y denotes the same space as above, the inclusion clA + clB ⊂ cl(A + B) holds.

2. The main result.

Remark 2.1 Let Y be a real topological vector space. If K is a closed cone, then it is a cone with zero.

Let us adopt the following two definitions which are natural extension of the concept of the lower boundedness for real-valued functions.

Definition 2.2 An s.v. function F : X → n(Y ) is said to be K−lower bounded on a set A ⊂ X iff there exists a bounded set B ⊂ Y such that F (x) ⊂ B + K for all x ∈ A.

Definition 2.3 An s.v. function F : X → n(Y ) is said to be K−upper bounded on a set A ⊂ X iff there exists a bounded set B ⊂ Y such that F (x) ⊂ B − K for all x ∈ A.

Definition 2.4 An s.v. function F : X → n(Y ) is said to be locally K−lower (upper) bounded in X iff for every x ∈ X there exists a neighbourhood U ^x of zero in X such that F is K−lower (upper) bounded on a set x + U ^x .

The idea of the first part of the proof below is due to W. Smajdor (Theorem 4.3 in [3]).

Lemma 2.5 Let X be a 2−divisible topological group, Y locally convex topological

vector space and K ⊂ Y a closed normal cone. If a K−subquadratic s.v.f. F : X →

cc(Y ) is K−continuous at zero, F (0) = {0} and locally K-lower bounded in X, then

it is K-u.s.c. in X.

Proof Suppose that F is not K-u.s.c. at a point z ∈ X, i.e. there exists a neigh- bourhood V of zero in Y such that for every neighbourhood U of zero in X we can find x u ∈ U for which

F (z + x u ) * F (z) + V + K.

Take a balanced convex neighbourhood W of zero in Y such that W ⊂ V and F (z) + W + K ⊂ F (z) + V + K. Then also

(4) F (z + x u ) * F (z) + W + K.

Let a neighbourhood U of zero in X be fixed arbitrary. Suppose that (5) F (z + 2 ^k x u ) * F (z) + 2 ^k (2 ^k − 1) F (x ^u ) + 2 ^k W + K

for some k ∈ N ⁰ := {0, 1, 2, ...}. The proof of (5) runs by induction. For k = 0 condition (5) holds with respect to (4). Putting y = x in (2) we obtain

2F (x) + 2F (x) ⊂ F (2x) + K,

for all x ∈ X. By convexity of the sets F (x), for all x ∈ X, the following inclusion holds

4F (x) ⊂ F (2x) + K.

An easy induction shows that

(6) 4 ⁿ F (x) ⊂ F (2 ⁿ x) + K,

for x ∈ X and for all positive integers n. By K-subquadraticity of F and (6), we have

(7) F (z + 2 ^k+1 x u ) + F (z) + K = F (z + 2 ^k x u + 2 ^k x u ) + F (z + 2 ^k x u − 2 ^k x u ) + K ⊃ 2F (z + 2 ^k x u ) + 2F (2 ^k x u ) + 2K ⊃ 2F (z + 2 ^k x u ) + 2 ^2k+1 F (x u ).

In view of the fact that for any sets A, B ⊂ Y , A + B ⊂ A + B, we get F (z) + 2 ^k (2 ^k − 1) F (x ^u ) + 2 ^k W + K + K ⊂

(8) ⊂ F (z) + 2 ^k (2 ^k − 1) F (x ^u ) + 2 ^k W + K.

By (5) and (8), we obtain

F (z + 2 ^k x u ) * F (z) + 2 ^k (2 ^k − 1) F (x ^u ) + 2 ^k W + K + K.

Notice, that for a cone K, the equality aK = K holds, for every a ∈ (0, ∞). Hence (9) 2F (z + 2 ^k x u ) * 2F (z) + 2 ^k+1 (2 ^k − 1) F (x ^u ) + 2 ^k+1 W + K + K.

Using (9) and Lemma 1.8, we obtain

(10) 2F (z + 2 ^k x u ) + 2 ^2k+1 F (x u ) *

* 2F (z) + 2 ^k+1 (2 ^k − 1) F (x ^u ) + 2 ^k+1 W + K + 2 ^2k+1 F (x u ) + K.

In view of the fact that the sum of closed and compact sets is closed (Lemma 1.6 in [2]), F has convex values and by (10), we get

2F (z + 2 ^k x u ) + 2 ^2k+1 F (x u ) *

(11) * 2F (z) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K + K.

Consequently, by (7) and (11), we obtain

F (z + 2 ^k+1 x u ) + F (z) + K *

(12) * 2F (z) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K + K.

In view of the fact that for any sets A, B ⊂ Y , A + B ⊂ A + B and by convexity of the set F (z), we obtain

F (z) + F (z) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K ⊂

(13) ⊂ 2F (z) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K.

Inclusions (12) and (13) imply

F (z + 2 ^k+1 x u ) + F (z) + K *

* F (z) + F (z) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K + K.

Consequently

F (z + 2 ^k+1 x u ) * F (z) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K.

We have proved that (5) holds for every neighbourhood U of zero and k = 0, 1, 2, 3, ...

Because F is K-lower bounded on a neighbourhood of z, then there exists a neighbourhood U of zero in X and bounded set B ₁ ⊂ Y , such that

F (z + t) ⊂ B ¹ + K, t ∈ U.

Since the sets B 1 and F (z) are bounded, there exist λ 1 > 0 and λ 2 > 0 such that F (z) ⊂ λ ¹

W , B 1 ⊂ λ ¹

W .

Let λ := min{λ ¹ , λ 2 }. Since W is balanced , we get

(14) F (z + t) ⊂ 1

λ W + K, t ∈ U and

(15) F (z) ⊂ 1

λ W.

Let k ∈ N be so large that

(16) 2 ^k > 3

λ .

Since K is a normal cone there exists a base W of neighbourhoods of zero in Y such that M = (M + K) ∩ (M − K), for all M ∈ W.

We can choose a set V ₁ ∈ W and a symmetric neighbourhood V ² of zero in Y such that

(17) V 2 ⊂ V ¹ ⊂ 1

λ2 ^k (2 ^k − 1) W.

Since F is K-l.s.c. at zero and F (0) = {0} there exists a neighbourhood U ¹ of zero in X such that

0 ∈ F (t) + V ² + K, t ∈ U ¹ .

In view of above there exists a ∈ F (t) such that a ∈ V ² − K ⊂ V ¹ − K and therefore

(18) F (t) ∩ (V ¹ − K) 6= ∅, t ∈ U ¹ .

Since F is K-u.s.c. at zero and F (0) = {0} there exists a neighbourhood U ² of zero in X such that

(19) F (t) ⊂ V ¹ + K, t ∈ U ² .

It follows by (18) and (19) that

∅ 6= F (t) ∩ (V ¹ − K) ⊂ (V ¹ + K) ∩ (V ¹ − K) = V ¹ , t ∈ U ¹ ∩ U ² . Now we consider a set e U := ₂ ¹

U

∩ U ¹ ∩ U ² . There exists x u ∈ e U such that (5) holds. Let a ∈ F (z + 2 ^k x u ), b ∈ F (z) and c ∈ F (x ^u ) ∩ (V ¹ − K). Notice, that

a = b + (a − b − 2 ^k (2 ^k − 1)c) + 2 ^k (2 ^k − 1)c.

By (14)-(19), we get

a − b − 2 ^k (2 ^k − 1)c ∈ 1

λ W + K + 1

λ W − 2 ^k (2 ^k − 1)V ¹ ⊂ 3

λ W + K ⊂ 2 ^k W + K.

Therefore

a ∈ F (z) + 2 ^k W + 2 ^k (2 ^k − 1)F (x ^u ) + K.

We have proved that

F (z + 2 ^k x u ) ⊂ F (z) + 2 ^k W + 2 ^k (2 ^k − 1)F (x ^u ) + K,

which contradicts (5). _

Definition 2.6 We say that 2-divisible topological group X has the property ( ¹ ₂ ) iff for every neighbourhood V of zero there exists a neighbourhood W of zero such that ¹ ₂ W ⊂ W ⊂ V .

Lemma 2.7 Let X be a 2−divisible topological group satisfying condition ¹ 2

, Y locally convex topological vector space and K ⊂ Y a closed normal cone. If a K −subquadratic s.v.f. F : X → cc(Y ) is K−continuous at zero, F (0) = {0} and locally K-lower and K-upper bounded in X, then it is K-l.s.c. in X.

Proof Proceeding likewise as in the proof of Lemma 2.5, suppose that F is not K-l.s.c. at a point z ∈ X, i.e. there exists a neighbourhood V of zero in Y such that for every neighbourhood U of zero in X we can find x u ∈ U for which

F (z) * F (z + x u ) + V + K.

Take a balanced convex neighbourhood W of zero in Y such that W ⊂ V and F (z + x u ) + W + K ⊂ F (z + x ^u ) + V + K.

Then also

(20) F (z) * F (z + x ^u ) + W + K.

Let a neighbourhood U of zero in X be fixed arbitrary. Suppose that (21) F (z + (1 − 2 ^k )x u ) * F (z + x ^u ) + 2 ^k (2 ^k − 1) F (x ^u ) + 2 ^k W + K

for some k ∈ N ⁰ := {0, 1, 2...} . The proof of (21) runs by induction. For k = 0 condition (21) holds with respect to (20). An easy induction shows that

(22) 4 ⁿ F (x) ⊂ F (2 ⁿ x) + K,

for x ∈ X and n ∈ N. By (2) and (22), we obtain

F (z + (1 − 2 ^k+1 )x u ) + F (z + x u ) + K ⊃ 2F (z + x ^u − 2 ^k x u ) + 2F (2 ^k x u ) + 2K ⊃

(23) ⊃ 2F (z + (1 − 2 ^k )x u ) + 2 ^2k+1 F (x u ).

In view of the fact that for any sets A, B ⊂ Y A + B ⊂ A + B, we obtain (24) F (z + x u ) + 2 ^k (2 ^k − 1) F (x ^u ) + 2 ^k W + K + K ⊂

⊂ F (z + x ^u ) + 2 ^k (2 ^k − 1) F (x ^u ) + 2 ^k W + K.

By (21) and (24), we have

F (z + (1 − 2 ^k )x u ) * F (z + x ^u ) + 2 ^k (2 ^k − 1) F (x ^u ) + 2 ^k W + K + K.

Since aK = K, for every a ∈ (0, ∞), the inclusion above is equivalent to the following

(25) 2F (z + (1 − 2 ^k )x u ) * 2F (z + x ^u ) + 2 ^k+1 (2 ^k − 1) F (x ^u ) + 2 ^k+1 W + K + K.

By (25) and Lemma 1.8, we get

2F (z + (1 − 2 ^k )x u ) + 2 ^2k+1 F (x u ) *

(26) 2F (z + x u ) + 2 ^k+1 (2 ^k − 1) F (x ^u ) + 2 ^k+1 W + K + 2 ^2k+1 F (x u ) + K.

In view of the fact that the sum of closed and compact sets is closed (Lemma 1.6 in [2]), F has convex values and by (26), we get

2F (z + (1 − 2 ^k )x u ) + 2 ^2k+1 F (x u ) *

(27) * 2F (z + x ^u ) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K + K.

By (23) and (27), we have

F (z + (1 − 2 ^k+1 )x u ) + F (z + x u ) + K *

(28) * 2F (z + x ^u ) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K + K.

In view of the fact that for any sets A, B ⊂ Y A + B ⊂ A + B and by convexity of the set F (z + x u ), we get

F (z + x u ) + F (z + x u ) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K ⊂

(29) ⊂ 2F (z + x ^u ) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K.

By (28) and (29), we obtain

F (z + (1 − 2 ^k+1 )x u ) + F (z + x u ) + K *

* F (z + x ^u ) + F (z + x u ) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K + K and consequence

F (z + (1 − 2 ^k+1 )x u ) * F (z + x ^u ) + 2 ^k+1 (2 ^k+1 − 1) F (x ^u ) + 2 ^k+1 W + K.

We have proved that (21) holds for every neighbourhood U of zero and k = 1, 2, 3, ...

Let W ₁ ∈ W, where W denotes the same as in the proof of Lemma 2.5 and let W 2 be a balanced neighbourhood of zero in Y such that W 2 ⊂ W ¹ ⊂ W . Because of K-lower and K-upper boundedness of F on a neighbourhood of z, progressing likewise as in the proof of Lemma 2.5, we obtain

F (z + t) ⊂ 1

λ 1 W 2 + K, t ∈ U ¹ ,

F (z + t) ⊂ 1

λ 2 W 2 − K, t ∈ U ² ,

where U 1 , U 2 denote neighbourhoods of zero in X and for some positive λ 1 , λ 2 . Let λ := min {λ ¹ , λ 2 }. Since W ² is balanced, we get

F (z + t) ⊂ 1

λ W 2 + K ⊂ 1

λ W 1 + K, t ∈ U ¹ and

F (z + t) ⊂ 1

λ W 2 − K ⊂ 1

λ W 1 − K, t ∈ U ² , which implies

(30) F (z + t) ⊂

 1

λ W 1 + K



∩

 1

λ W 1 − K



= 1

λ W 1 ⊂ 1

λ W, t ∈ U 1 ∩ U 2 . Let k ∈ N be so large that

(31) 2 ^k > 3

λ .

Analogously as in the proof of Lemma 2.5, we can choose a balanced neighbourhood V 2 of zero in Y and V 1 ∈ W such that

(32) V 2 ⊂ V ¹ ⊂ 1

λ2 ^k (2 ^k − 1) W.

Similarly as in the proof of Lemma 2.5, we have

(33) F (t) ∩ (V ¹ − K) 6= ∅, t ∈ U ³ ,

where U ₃ denotes a neighbourhood of zero in X (cf.17). By (33) and by K-u.s.c. of F at the point zero, we obtain

(34) ∅ 6= F (t) ∩ (V ¹ − K) ⊂ (V ¹ + K) ∩ (V ¹ − K) = V ¹ , t ∈ U ³ ∩ U ⁴ , where U 4 denotes a neighbourhood of zero in X.

Let U be a symmetric neighbourhood of zero in X, such that U + U ⊂ U ¹ ∩ U ² and ¹ ₂ U ⊂ U. Now we consider a set e U := ₂ ¹

U

∩ U ³ ∩ U ⁴ . There exists x _u ∈ e U such that (21) holds and

(35) (1 − 2 ^k )x u = x u − 2 ^k x u ∈ e U − 2 ^k U e ⊂ U + U ⊂ U ¹ ∩ U ² . By (34), we get

(36) ∅ 6= F (x ^u ) ∩ (V ¹ − K) ⊂ V ¹ .

Let a ∈ F (z + (1 − 2 ^k )x u ), b ∈ F (z + x ^u ) i c ∈ F (x ^u ) ∩ (V ¹ − K) and let a has the same representation as in the proof of Lemma 2.5, then by (35), (30), (36), (32) i (31), we obtain

a − b − 2 ^k (2 ^k − 1)c ∈ 1 λ W + 1

λ W − 2 ^k (2 ^k − 1)V ¹ ⊂ 3

λ W ⊂ 2 ^k W

and consequently

F (z + (1 − 2 ^k )x u ) ⊂ F (z + x ^u ) + 2 ^k W + 2 ^k (2 ^k − 1)F (x ^u ).

By Remark 2.1, K is a cone with zero, then

F (z + (1 − 2 ^k )x u ) ⊂ F (z + x ^u ) + 2 ^k W + 2 ^k (2 ^k − 1)F (x ^u ) + K

which contradicts (21). _

As an immediate consequence of Lemmas 2.5 and 2.7 we obtain the following the- orem.

Theorem 2.8 Let X be a 2−divisible topological group satisfying condition ¹ 2

, Y locally convex topological vector space and K ⊂ Y a closed normal cone. If a K −subquadratic s.v.f. F : X → cc(Y ) is K−continuous at zero, F (0) = {0} and locally K-lower and K-upper bounded in X, then it is K-continuous in X.

Remark 2.9 There exist K-subquadratic s.v. functions which satisfy conditions given in Theorem 2.8 , but need not be continous. For example the s.v. function F : R → n(R) given by

F (x) =

 [0, 1] , x ∈ R \ Q {0} , x ∈ Q.

is K-subquadratic (for K = [0, ∞)), K-continuous, but it is not continuous.

References

[1] Z. Kominek and K. Troczka-Pawelec, Continuity of real valued subquadratic functions, Com- mentationes Mathematicae, Vol. 51, No. 1 (2011), 71-75.

[2] K. Nikodem, K-convex and K-concave set-valued functions, Zeszyty Naukowe Politechniki Łódzkiej 559, Łódź 1989.

[3] W. Smajdor, Subadditive and subquadratic set-valued functions, Prace Naukowe Uniwersytetu Ślaskiego w Katowicach, 889, Katowice, 1987.

[4] K. Troczka-Pawelec, Continuity of superquadrqtic set-valued functions, Scientific Issues Jan Długosz University in Cz¸estochowa, Mathematics XVII, 2012.

Katarzyna Troczka–Pawelec

Institute of Mathematics and Computer Science, Jan Długosz University in Cze¸stochowa al. Armii Krajowej 13/15, 42-200 Cze¸stochowa,Poland

E-mail: k.troczka@ajd.czest.pl

(Received: 17.02.2013)