J. Fixed Point Theory Appl. 18 (2016) 133–145 DOI 10.1007/s11784-015-0265-9 Journal of Fixed Point Published online October 17, 2015 and Applications © 2015 The Author(s) This article is published with open access at Springerlink.com
Theory
Selections of set-valued functions satisfying the general linear inclusion Andrzej Smajdor and Joanna Szczawi´ nska Abstract. Applying the classical Banach fixed point theorem we prove that a set-valued function satisfying a general linear functional inclusion admits a unique selection fulfilling the corresponding functional equation. We also adopt the method of the proof for investigating the Rassias stability of general linear equation. Mathematics Subject Classification. 54C65, 54C60, 39B82. Keywords. Set-valued function, selection, stability.
1. Introduction Let (Y, ∥ · ∥) be a real normed space. We denote by n(Y ) the family of all nonempty subsets of Y and by cl(Y ), ccl(Y ), c(Y ) and cp(Y ) we denote collections of all closed, convex closed, compact and complete members of n(Y ), respectively. The number diam A := sup{∥a − b∥ : a, b ∈ A}
is said to be the diameter of A ∈ n(Y ). We say that a set-valued function F : K → n(Y ) (an “s.v. function” for abbreviation) is with bounded diameter if the function K ∋ x �→ diam F (x) ∈ R is bounded. In [5], the authors proved that if S is a commutative semigroup with zero and (Y, ∥ · ∥) is a real Banach space, then F : S → ccl(Y ) is a subadditive s.v. function; i.e., F (x + y) ⊂ F (x) + F (y),
x, y ∈ S,
with bounded diameter admits a unique additive selection. Popa in [11] proved that if ∅ = ̸ K is a convex cone in a real vector space X (i.e., t1 K + t2 K ⊂ K for all t1 , t2 ≥ 0) and F : K → ccl(Y ) (where (Y, ∥ · ∥) is a real Banach space) is an s.v. function with bounded diameter fulfilling the inclusion F (αx + βy) ⊂ αF (x) + βF (y),
x, y ∈ K,
for α, β > 0, α + β ̸= 1, then there exists exactly one additive selection of F .
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Nikodem and Popa in [9] and Piszczek in [10] proved the following theorem. Theorem 1.1. Let K be a convex cone in a real vector space X, (Y, ∥· ∥) a real Banach space and α, β, p, q > 0. Consider an s.v. function F : K → ccl(Y ) with bounded diameter fulfilling the inclusion αF (x) + βF (y) ⊂ F (px + qy),
x, y ∈ K.
If α+β < 1, then there exists a unique selection f : K → Y of F satisfying the equation αf (x) + βf (y) = f (px + qy), x, y ∈ K. If α + β > 1, then F is single valued.
The case of p + q = 1 was investigated by Popa in [13], Inoan and Popa in [7] and recently, by Smajdor and Szczawi´ nska in [14]. In this paper, we determine the conditions for which an s.v. function F : K → n(Y ) satisfying the inclusion αF (x) + βF (y) ⊂ F (px + qy),
x, y ∈ K,
(1.1)
where |α| < p and diam F (x) ≤ M ∥x∥, x ∈ K (for some M > 0) admits a selection satisfying the corresponding functional equation. It is easy to check that if F satisfies the opposite inclusion, then it is single valued. Theorem 2.1 and also the method of its proof are used for the investigation of the Rassias stability of general linear functional equation.
2. The main theorem Let (X, ∥ · ∥) and (Y, ∥ · ∥) be real normed spaces and let K be a nonempty subset of X. Consider an s.v. function F : K → n(Y ). A function f : K → Y is a selection of the s.v. function F if and only if f (x) ∈ F (x), x ∈ K. Let Sel(F ) := {f : K → Y : f (x) ∈ F (x), x ∈ K}. It is easy to check that if there exists a constant M > 0 such that diam F (x) ≤ M ∥x∥ for all x ∈ K, then the function } { ∥f (x) − g(x)∥ , x ∈ K \ {0} , f, g ∈ Sel(F ), d(f, g) := sup ∥x∥
is a metric in Sel(F ). Moreover, if F (x) is complete for every x ∈ K, the metric space (Sel(F ), d) is complete. Obviously, the convergence in the space (Sel(F ), d) implies the pointwise convergence on the set K.
Theorem 2.1. Let (X, ∥ · ∥) and (Y, ∥ · ∥) be real normed spaces and let K ⊂ X be such that 0 ∈ K. Assume that p, q > 0 and α, β ∈ R are fixed and one of the following conditions holds: (i) |α| < p and K ⊂ pK, (ii) |β| < q and K ⊂ qK.
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Consider an s.v. function F : K → cp(Y ) such that 0 ∈ F (0) and diam F (x) ≤ M ∥x∥,
for some positive constant M . If
x ∈ K,
αF (x) + βF (y) ⊂ F (px + qy),
x, y ∈ K, px + qy ∈ K,
αf (x) + βf (y) = f (px + qy),
x, y ∈ K, px + qy ∈ K.
(2.1)
then there exists a unique selection f : K → Y of F such that
Proof. Assume that |α| < p and K ⊂ pK. Since diam F (0) = 0 and 0 ∈ F (0), F (0) = {0}. Putting y = 0 in (2.1), we obtain ( ) x αF ⊂ F (x), x ∈ K. (2.2) p Let
( ) x T (g)(x) := αg , x ∈ K, g ∈ Sel(G). p By (2.2), T (g) ∈ Sel(F ), g ∈ Sel(F ). Moreover, for every g1 , g2 ∈ Sel(F ), } { ( ) ( ) ∥g1 xp − g2 xp ∥ ) ( , x ∈ K \ {0} d T (g1 ), T (g2 ) = |α| sup ∥x∥ { } ( ) ( ) ∥g1 xp − g2 xp ∥ |α| sup , x ∈ K \ {0} = p ∥ xp ∥ ≤
|α| d(g1 , g2 ). p
Since |α| < p, the map T : Sel(F ) → Sel(F ) is contractive in the complete metric space (Sel(F ), d), so by the Banach theorem, it has a unique fixed point f and limn→∞ T n (g) = f for each g ∈ Sel(F ). Hence f : K → Y is the unique selection of the s.v. function F such that ( ) x f (x) = αf , x ∈ K. p
Fix g ∈ Sel(F ) and x, y ∈ K such that px + qy ∈ K. Then x y px + qy , , ∈ K. p p p By (2.1),
Hence
( ) ( ) ( ) ( ) y px + qy px + qy x + βg ,g ∈F . αg p p p p
� ( ) ( ( ) ( )� ) � � �αg x + βg y − g px + qy � ≤ diam F px + qy � � p p p p � � � px + qy � �. ≤M� � � p
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Thus � � �αT (g)(x) + βT (g)(y) − T (g)(px + qy)� ≤ |α| M ∥px + qy∥ p
for every x, y ∈ K such that px + qy ∈ K. Proceeding by induction, we get ( )n � � �αT n (g)(x) + βT n (g)(y) − T n (g)(px + qy)� ≤ |α| M ∥px + qy∥ p
for every n ∈ N and all x, y ∈ K with px+qy ∈ K. Letting n → ∞, we obtain αf (x) + βf (y) = f (px + qy),
x, y ∈ K, px + qy ∈ K.
Let X and Y be real vector spaces and K a convex cone in X. It is easy to check that if f : K → Y satisfies the equation x, y ∈ K,
αf (x) + βf (y) = f (px + qy), and f (0) = 0, then f is additive; i.e.,
x, y ∈ K.
f (x) + f (y) = f (x + y),
Corollary 2.2. Let p, q > 0 and |α| < p (or |β| < q). Assume that (X, ∥ · ∥) and (Y, ∥ · ∥) are normed spaces, K a convex cone in X and F : K → cp(Y ) an s.v. function such that 0 ∈ F (0) and diam F (x) ≤ M ∥x∥,
x ∈ K,
for some constant M > 0. If αF (x) + βF (y) ⊂ F (px + qy),
x, y ∈ K,
then the unique selection f : K → Y of F fulfilling the equation αf (x) + βf (y) = f (px + qy),
x, y ∈ K,
is additive. Remark 2.3. If |α| = p and |β| = q, the above corollary is not true. Proof. Let K = [0, +∞) and let F : K → c(R) be an s.v. function defined by F (x) = [0, x],
x ∈ K,
and diam F (x) = |x|, x ∈ K. Then F (x + y) = F (x) + F (y),
x, y ∈ K,
and for every a ∈ [0, 1], the function f (x) = ax, x ∈ K, is an additive selection of F .
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3. Stability results In the first part of this section we present an application of Theorem 2.1 to the investigation of the Rassias stability of the general linear functional equation αf (x) + βf (y) = f (px + qy). (3.1) For definition and more results in the Rassias stability theory see, for instance, [6, 8]. The general linear equation was considered by several authors (for example, [1, 2, 12]). A set-valued approach can be found in [3]. Gajda showed in [4] (see also [8, Theorem 2.6]) that for ε > 0 there is a function f : R → R satisfying the inequality f (x) + f (y) − f (x + y) ≤ ε(|x| + |y|), x, y ∈ R,
while there is no constant M ≥ 0 and no additive function f0 : R → R such that f (x) − f0 (x) ≤ M |x|, x ∈ R.
We apply the method used in the proof of Theorem 2.1 to obtain the stability result for equation (3.1). We will denote by R+ and Q+ the set of all nonnegative real and rational numbers, respectively.
Theorem 3.1. Let K be a convex cone in a real normed space (X, ∥ · ∥), (Y, ∥ · ∥) a real Banach space and ε : R+ × R+ → R+ an R+ -homogenous function. Assume that p, q > 0, α, β ∈ R are fixed and |α| < p. If a function f : K → Y satisfies ∥αf (x) + βf (y) − f (px + qy)∥ ≤ ε(∥x∥, ∥y∥),
x, y ∈ K,
then there exists a unique function f0 : K → Y fulfilling the functional equation αf0 (x) + βf0 (y) = f0 (px + qy), x, y ∈ K, and such that
∥f (x) − f0 (x) − f (0)∥ ≤ M ∥x∥,
x ∈ K,
for some M > 0. Moreover, ∥f (x) − f0 (x) − f (0)∥ ≤
ε(1, 0) ∥x∥, p−α
x ∈ K,
and f0 is an additive function. In particular, if ε(1, 0) = 0, then f (x) = f0 (x) − f (0),
x ∈ K.
Proof. Let g : K → Y be given by g(x) = f (x) − f (0), x ∈ K. It is easy to check that g(0) = 0 and ∥αg(x) + βg(y) − g(px + qy)∥ ≤ ε(∥x∥, ∥y∥),
x, y ∈ K.
(3.2)
Hence, for all x, y ∈ K, αg(x) + βg(y) ∈ g(px + qy) + ε (∥x∥, ∥y∥)B,
(3.3)
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where B denotes the unit closed ball in the space Y . Setting y = 0 and replacing x by x/p in (3.3), we obtain ( ) ( ) ε(1, 0) ∥x∥ x ∈ g(x) + ε , 0 B = g(x) + ∥x∥B, x ∈ K. αg p p p Consider an s.v. function G : K → cl(Y ) given by G(x) = g(x) + Then diam G(x) = and
ε(1, 0) ∥x∥B, p−α
2ε (1, 0) ∥x∥, p−α
x ∈ K. x ∈ K,
� � ( ) ( ) � ε (1, 0) � x x �x� B = αg +α αG � p p p − α p� ( ) ε (1, 0) α ⊂ g(x) + 1+ ∥x∥B = G(x) p p−α
for all x ∈ K. The idea of the proof is the same as before so we only give a sketch. The function T : Sel(G) → Sel(G), given by ( ) x , x ∈ K, h ∈ Sel(G), T (h)(x) := αh p is contraction with the constant |α|/p. By the Banach theorem, there exists a unique function f0 ∈ Sel(G) such that f0 (px) = αf0 (x),
x ∈ K,
and limn→∞ T n (g) = f0 . By the definition of G, ∥f0 (x) − g(x)∥ ≤
ε (1, 0) ∥x∥, p−α
x ∈ K.
Since g satisfies (3.2) and ε is positively homogeneous, � � �αT (g)(x) + βT (g)(y) − T (g)(px + qy)� ≤ |α| ε (∥x∥, ∥y∥) p
for all x, y ∈ K. Proceeding by induction, we get
� n � �αT (g)(x) + βT n (g)(y) − T n (g)(px + qy)� ≤
(
|α| p
)n
ε(∥x∥, ∥y∥)
for every x, y ∈ K and n ≥ 1. It follows that
αf0 (x) + βf0 (y) = f0 (px + qy),
x, y ∈ K.
Since f0 (0) = 0, f0 is additive. For the end assume that f1 : K → Y is a solution of equation (3.1) such that ∥f1 (x) − g(x)∥ ≤ M ∥x∥, x ∈ K,
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for some positive M . Then
( ) x = f1 (x), αf1 p
f1 (0) = 0, and
∥f1 (x) − f0 (x)∥ ≤ It is easy to check that ∥f1 (x) − f0 (x)∥ ≤
(
|α| p
(
M+
)n (
M+
Hence f1 = f0 , which ends the proof.
ε(1, 0) p−α
)
ε(1, 0) p−α
x∈K
∥x∥,
)
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x ∈ K.
∥x∥,
x ∈ K, n ∈ N.
In the second part of this section, we apply Theorem 2.1 to obtain a stability result for equation (3.1). Theorem 3.2. Let K be a convex cone in a real normed space (X, ∥·∥), (Y, ∥·∥) a real Banach space and p, q > 0, α, β ∈ R fixed constants such that |α| < p. Consider η ≥ 0 and a function f : K → Y satisfying the inequality ∥αf (x) + βf (y) − f (px + qy)∥ ≤ η (∥x∥ + ∥y∥),
x, y ∈ K.
If there exists a constant λ > 0 such that (1 + λα)∥x∥ + (1 + λβ)∥y∥ ≤ λ∥px + qy∥,
x, y ∈ K,
(3.4)
then there exists a unique function f0 : K → Y fulfilling the functional equation αf0 (x) + βf0 (y) = f0 (px + qy), x, y ∈ K, and such that
∥f (x) − f0 (x) − f (0)∥ ≤ M ∥x∥,
x ∈ K,
for some M > 0. Moreover, ∥f (x) − f0 (x) − f (0)∥ ≤ λη ∥x∥,
x ∈ K,
and f0 is an additive function. Proof. Let g : K → Y be given by g(x) = f (x) − f (0), x ∈ K. As in the proof of Theorem 3.1, g(0) = 0 and ∥αg(x) + βg(y) − g(px + qy)∥ ≤ η (∥x∥ + ∥y∥),
In particular, for all x, y ∈ K,
x, y ∈ K.
αg(x) + βg(y) ∈ g(px + qy) + η(∥x∥ + ∥y∥)B,
where B is the unit closed ball in Y . Define an s.v. function G : K → cl(Y ) as follows: G(x) = g(x) + λη ∥x∥B, x ∈ K.
Then 0 ∈ G(0), diam G(x) = 2ηλ ∥x∥, x ∈ K, and αG(x) + βG(y) ⊂ G(px + qy),
x, y ∈ K.
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Indeed, let x, y ∈ K be fixed. By condition (3.4), ( ) αG(x) + βG(y) = αg(x) + βg(y) + η λα ∥x∥ + λβ ∥y∥ B ( ) ⊂ g(px + qy) + η (∥x∥ + ∥y∥)B + η λα ∥x∥ + λβ ∥y∥ B ⊂ g(px + qy) + ηλ ∥px + qy∥B = G(px + qy).
By Theorem 2.1, there exists exactly one selection f0 : K → Y of the s.v. function G satisfying equation (3.1). Hence ∥f (x) − f0 (x) − f (0)∥ ≤ λη ∥x∥,
x ∈ K.
∥f (x) − f0 (x) − f (0)∥ ≤ M ∥x∥,
x ∈ K,
Since f0 (0) = 0, f0 must be additive. The proof that f0 is the only solution of equation (3.1) such that for some M > 0 runs as before.
Example. Let |α| < p and |β| < q. Every convex cone { } K ⊂ (x1 , . . . , xn ) ∈ Rn : x1 ≥ 0 ∧ · · · ∧ xn ≥ 0 ∨ x1 ≤ 0 ∧ · · · ∧ xn ≤ 0 satisfies condition (3.4) with
λ ≥ max and with the norm in Rn given by
{
1 1 , p−α q−β
∥(x1 , . . . , xn )∥ = |x1 | + · · · + |xn |,
}
(x1 , . . . , xn ) ∈ Rn .
4. The selection theorem In this section we present a certain consequence of Theorem 2.1 for the case of a compact-valued s.v. function F satisfying the inclusion (1.1) and such that the function x �→ diam F (x) maps bounded sets onto bounded ones.
Lemma 4.1. Let p, q > 0 and α, β ∈ R be fixed. Assume that X and Y are real vector spaces, K a convex cone in X and K a collection of nonempty subsets of Y closed under intersections of chains. Let F : K → K be an s.v. function fulfilling the inclusion (1.1) and 0 ∈ F (0). There exists a minimal s.v. function F0 : K → K such that (i) F0 (x) ⊂ F (x), x ∈ K, (ii) 0 ∈ F0 (0), (iii) αF0 (x) + βF0 (y) ⊂ F0 (px + qy), x, y ∈ K. Proof. Define the set F of all s.v. functions H : K → K such that 0 ∈ H(0) and
H(x) ⊂ F (x),
x ∈ K,
αH(x) + βH(y) ⊂ H(px + qy),
x, y ∈ K.
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The set F is nonempty, because F ∈ F , and partially ordered by the relation H1 ≼ H2 ⇐⇒ H1 (x) ⊂ H2 (x),
x ∈ K, H1 , H2 ∈ F.
It is enough to prove that there is a minimal element in F. Let C be a chain in F. Put ∩ H0 (x) = {H(x) : H ∈ C}, x ∈ K.
Since K is closed under intersections, H0 (x) ∈ K, x ∈ K. Obviously, 0 ∈ H0 (0) and H0 (x) ⊂ H(x), x ∈ K. Fix now x, y ∈ K. For every H ∈ C, hence
αH0 (x) + βH0 (y) ⊂ αH(x) + βH(y) ⊂ H(px + qy),
αH0 (x) + βH0 (y) ⊂ H0 (px + qy). Therefore, the s.v. function H0 : K → K belongs to F and H0 ≼ H, H ∈ C. By the Kuratowski–Zorn lemma , there exists a minimal element F0 in the set F. Lemma 4.2. Let p, q > 0 and α, β ∈ R be fixed. Assume that X and Y are real vector spaces, K a convex cone in X and K a family of nonempty subsets of Y such that αK ⊂ K and λK ⊂ K, λ ∈ Q+ . Consider an s.v. function F : K → K. If F0 : K → K is a minimal s.v. function such that (i) F0 (x) ⊂ F (x), x ∈ K, (ii) 0 ∈ F0 (0), (iii) αF0 (x) + βF0 (y) ⊂ F0 (px + qy), x, y ∈ K, then F0 is superadditive and Q+ -homogeneous; i.e., F0 (x) + F0 (y) ⊂ F0 (x + y),
and F0 (λx) = λF0 (x), x ∈ K, λ ∈ Q+ .
Proof. Putting y = 0 in (iii) and by (ii), ( ) x ⊂ F0 (x), αF0 p
x, y ∈ K,
x ∈ K.
Define
( ) x , x ∈ K. H(x) = αF0 p Then H(x) ∈ K, H(x) ⊂ F0 (x) ⊂ F (x), x ∈ K and 0 ∈ αF0 (0) = H(0). Observe that the s.v. function H : K → K satisfies the inclusion (iii). Indeed, let x, y ∈ K be fixed. Then ( ( ) ( )) x y αH(x) + βH(y) = α αF0 + βF0 p p ) ( px + qy = H(px + qy). ⊂ αF0 p Thus, by the minimality of F0 , F0 (x) = H(x), x ∈ K; i.e., ( ) x , x ∈ K. F0 (x) = αF0 p
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Similarly we get F0 (y) = βF0 Hence, by (iii), F0 (x) + F0 (y) = αF0
( ) y , q
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y ∈ K.
( ) ( ) x y + βF0 ⊂ F0 (x + y) p q
for every x, y ∈ K; i.e., F0 is superadditive. In particular, (x) ⊂ F0 (x), x ∈ K, n ∈ N. nF0 n Fix now n ∈ N and consider the s.v. function (x) ∈ K. K ∋ x �→ nF0 n We have (x) nF0 ⊂ F0 (x) ⊂ F (x), x ∈ K, n 0 ∈ nF0 (0) and ) ( (x) (y) px + qy + βnF0 ⊂ nF0 , x, y ∈ K. αnF0 n n n Therefore,
(x)
= F0 (x), x ∈ K, n by the minimality of F0 . Consequently, ( ) F0 λx = λF0 (x), x ∈ K, λ ∈ Q+ . nF0
Remark 4.3. Let (X, ∥·∥) be a real normed space and K ⊂ X a cone. Assume that f : K → R is Q+ -homogeneous. The function f maps bounded subsets of K onto bounded sets if and only if there exists a positive constant M such that |f (x)| ≤ M ∥x∥, x ∈ K.
Proof. Assume that f : K → R is Q+ -homogeneous (in particular, f (0) = 0) and maps bounded subsets of K onto bounded sets. Let M = sup{|f (x)| : x ∈ K ∧ ∥x∥ ≤ 1}.
Fix x ∈ K \ {0}. There exists a decreasing sequence (λn )n such that lim λn = 1
n→∞
and
λn ∥x∥ ∈ Q+ ,
n ∈ N.
Since
� � � x � 1 � � � λn ∥x∥ � = λn < 1, n ∈ N, we have � ( )� � � x � ≤ M, n ∈ N. �f � λn ∥x∥ � Consequently |f (x)| ≤ M ∥x∥. The opposite implication is obvious.
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Theorem 4.4. Let p, q > 0, α, β ∈ R and |α| < p. Assume that (X, ∥ · ∥) and (Y, ∥ · ∥) are real normed spaces, K is a convex cone in X and F : K → c(Y ) an s.v. function such that 0 ∈ F (0) and αF (x) + βF (y) ⊂ F (px + qy),
x, y ∈ K.
If K ∋ x �→ diam F (x) ∈ R maps bounded sets onto bounded sets, then there exists a unique selection f : K → Y of F fulfilling the equation αf (x) + βf (y) = f (px + qy),
x, y ∈ K.
The selection f is additive. Proof. The family c(Y ) is closed under intersections of chains and λ ∈ R.
λ c(Y ) ⊂ c(Y ),
By Lemma 4.1, there exists a minimal s.v. function F0 : K → c(Y ) such that F0 (x) ⊂ F (x), x ∈ K, 0 ∈ F0 (0) and αF0 (x) + βF0 (y) ⊂ F0 (px + qy),
x, y ∈ K.
Lemma 4.2 shows now that F0 is superadditive and Q+ -homogeneous. Since diam F0 (x) ≤ diam F (x), x ∈ K, the function K ∋ x �→ diam F0 (x) ∈ R maps bounded sets onto bounded sets and it is Q+ -homogeneous. Hence, by Remark 4.3, there exists a positive constant M such that diam F0 (x) ≤ M ∥x∥,
x ∈ K.
Consequently, by Theorem 2.1, the s.v. function F0 admits a unique selection f0 : K → Y such that αf0 (x) + βf0 (y) = f0 (px + qy),
x, y ∈ K.
Since f0 (0) = 0 and f0 satisfies the above equation, f0 must be additive. Assume that f1 , f2 : K → Y are selections of F such that αfi (x) + βfi (y) = fi (px + qy),
x, y ∈ K, i = 1, 2.
They are additive and fi (x) = αfi Hence, for every n ∈ N, n
fi (x) = α fi
( ) x , p (
x pn
)
x ∈ K, i = 1, 2.
,
x ∈ K, i = 1, 2.
Fix x ∈ K arbitrary. Since K ∋ z �→ diam F (z) ∈ R maps bounded sets onto bounded sets, there exists a constant M such that diam F (y) ≤ M,
∥y∥ ≤ 2∥x∥.
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A. A. Smajdor and and J. Szczawin ´skanska Smajdor J. Szczawi´
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For every n ∈ N there exists a number qn ∈ (1, 2) such that qn pn ∈ Q+ . Obviously, ∥qn x∥ ≤ 2∥x∥, n ∈ N. Hence, for every n ∈ N, � ( ) ( )� � x x � n� � − f2 ∥f1 (x) − f2 (x)∥ = |α| �f1 n p pn � � ( ) ( )� � qn x qn x � n� � − f2 = |α| �f1 qn pn qn pn � ( )n � � 1 |α| �f1 (qn x) − f2 (qn x)� = qn p ( )n ( )n |α| |α| ≤ diam F (qn x) ≤ M. p p Letting n → +∞ ends the proof.
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[14] A. Smajdor and J. Szczawi´ nska, Selections of generalized convex set-valued functions with bounded diameter. Fixed Point Theory, to appear. Andrzej Smajdor Institute of Mathematics Pedagogical University Podchor¸az˙ ych 2 PL-30-084 Krak´ ow Poland e-mail:
[email protected] Joanna Szczawi´ nska Institute of Mathematics Pedagogical University Podchor¸az˙ ych 2 PL-30-084 Krak´ ow Poland e-mail:
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