- Research Article
- Open access
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On Some Properties of Hyperconvex Spaces
Fixed Point Theory and Applications volume 2010, Article number: 213812 (2010)
Abstract
We are going to answer some open questions in the theory of hyperconvex metric spaces. We prove that in complete -trees hyperconvex hulls are uniquely determined. Next we show that hyperconvexity of subsets of normed spaces implies their convexity if and only if the space under consideration is strictly convex. Moreover, we prove a Krein-Milman type theorem for
-trees. Finally, we discuss a general construction of certain complete metric spaces. We analyse its particular cases to investigate hyperconvexity via measures of noncompactness.
1. Introduction
It is hard to believe that although hyperconvex metric spaces have been investigated for more that fifty years, some basic questions in their theory still remain open (let us recall that hyperconvex metric spaces were introduced in [1] (see also [2]), but from formal point of view it has to be emphasized that the notion of hyperconvexity was investigated earlier by Aronszajn in his Ph.D. thesis [3] which was never published). The main purpose of this paper is to answer some of these questions.
Let us begin with the notion of hyperconvex hull which was introduced by Isbell in [4] (see Definition 2.7). This notion is more difficult to investigate than the classical notion of convex hull, since the former one is not uniquely determined (see Proposition 2.8). In Section 3 we are going to prove that in hyperconvex metric spaces with the unique metric segments property, hyperconvex hulls are uniquely determined. Let us recall that such hyperconvex spaces were characterized by Kirk (see [5]) as complete -trees (see Theorem 2.15). This led to a surprising application of the theory of hyperconvex spaces to graph theory (see [6]).
Another interesting question is about the relation between the notion of convexity and hyperconvexity (cf. Remark 4.1). In particular, it is inspired by the following Sine's remark [7, page 863], stated without a proof: "The term hyperconvex does have some unfortunate aspects. First, a hyperconvex subset of even (with the
norm) need not be convex. Also convex sets can fail to be hyperconvex (but for this one must go to at least
)." It turns out that all hyperconvex subsets of a given normed space are convex if and only if the space in question is strictly convex; this fact is proved in Section 4.
In Section 5 we turn our attention to the classical Krein-Milman theorem (see [8]). We prove that a bounded complete -tree is a convex hull of its extremal points (note that a similar result, but with the assumption of compactness, is proved in [9]). Hence, in particular, such a property holds for bounded hyperconvex metric spaces with unique metric segments.
Let us denote by and
the Kuratowski and Hausdorff measures of noncompactness, respectively, (see [10, 11] for the definition and basic properties). It was noticed by EspĂnola (see [12]) that if a metric space is hyperconvex, then
for all its bounded subsets
. The question is about the inverse implication. More precisely, assume that
for every bounded subset of a given metric space
. Does this equality imply that
is hyperconvex? (Obviously, we mean nontrivial cases, i.e., we exclude spaces in which every bounded set is relatively compact.) In Sections 6 and 7 we introduce a few metric spaces which are not hyperconvex, but
for all their bounded subsets. Hence the answer to the above question is negative. Let us emphasize that the metrics considered in Sections 6 and 7 are extensions and generalizations of commonly known radial metric and river metric, which were proved in [13] to be hyperconvex.
Let us notice that in general it is not easy to provide explicit formulae which would allow to evaluate the measures of noncompactness in particular spaces. We are going to state such formulae for the metric spaces considered in Sections 6 and 7.
Let us emphasize that another motivation to consider those metrics comes from the real world. Let us consider an example of the transmission of phone signals, when one person (say, ) calls another (say,
), assuming there are two base transceiver stations (say,
and
). We may have two cases. If
and
are in the range of one of the BTS's, say
, then the signal is first transmitted from
to
and then from
to
âeven if
and
are "close" to each other. If
and
are located in the ranges of
and
respectively, then the signal is transmitted from
to
, then from
to
and finally from
to
. Hence we have the metric considered in Definition 7.4.
In Section 8 we provide a general scheme to construct metrics similar to these of Sections 6 and 7. This scheme is a generalization of a construction from [14].
For completeness, in Section 2 we collect some basic definitions and facts used in the sequel.
2. Preliminaries
In what follows we will denote the Euclidean metric on by
and a "maximum" norm on any suitable space by
.
Let us begin with some classical definitions and facts.
Definition 2.1.
Let be a metric space. We call a set
a metric segment(joining the points
) if there exists an isometric embedding
such that
and
.
Definition 2.2 (see [1, page 410, Definition â1]).
We call a metric space (X, d) hyperconvex, if any family of closed balls with centers at
's and radii of
's, respectively, such that
for any
has a nonempty intersection.
Hyperconvex spaces possessâamong othersâthe following properties.
Proposition 2.3 (see [1, page 417, Theorem â1']).
A hyperconvex space is complete.
Proposition 2.4 (see [1, page 423, Theorem â9]).
A nonexpansive retract (i.e., a retract by a nonexpansive retraction) of a hyperconvex space is hyperconvex.
Proposition (see [1, page 422, Corollary â4]).
Each hyperconvex metric space is an absolute nonexpansive retract, that is, it is a nonexpansive retract of any metric space it is isometrically embedded in. In particular, hyperconvex spaces are absolute retracts.
The following theorem gives a characterization of hyperconvex real Banach spaces.
Theorem 2.6 (Nachbin-Kelley, see [15, 16]).
A real Banach space is hyperconvex if and only if it is isometrically isomorphic to some space of all real continuous functions on a Hausdorff, compact and extremally disconnected topological space
with the
norm.
Now let us state the definition of a hyperconvex hull. We will not need the general version of this notion, investigated by Isbell in [4]; instead, the notion of a hyperconvex hull of a subset of a hyperconvex space will suffice for our considerations.
Definition 2.7 (see, e.g., [17, page 408]).
Let be a nonempty subset of a hyperconvex space
. We call
a hyperconvex hull of
(in
) if
, the set
is hyperconvex (as a metric subspace) and there exists no hyperconvex
such that
.
A hyperconvex hull always exists, but needs not to be unique. It is, however, unique up to an isometry. To be more precise, the following holds.
Proposition 2.8 (cf. [17, page 408, Proposition â5.6]).
Each nonempty subset of a hyperconvex metric space possesses a hyperconvex hull. If and
are hyperconvex spaces,
,
are isometric and
is an isometry, then for any hyperconvex hulls
,
of
and
respectively, the isometry
extends to an isometry
.
In what follows, we will also need the definitions of total and strict convexity.
Definition 2.9 (see, e.g. [1, page 407] and [18, page 6, Definition â2.1]).
A metric space is called totally convex if for any two points
and for all
such that
there exists a point
satisfying the equalities
and
. If this point is unique for all possible combinations of
, we call the space
convex and denote this point by
.
Remark 2.10 (see [1, page 410]).
A hyperconvex space is totally convex.
Remark 2.11 (see, e.g., [18, page 7]).
For normed spaces, the above definition of strict convexity (Definition 2.9) coincides with the usual one.
Proposition 2.12 (see, e.g., [18, page 7]).
In a strictly convex metric space, intersection of any family of totally convex subsets is itself totally convex.
The above proposition lets us define the notion of a convex hull in any strictly convex metric space in a natural way.
Definition 2.13.
Let be a nonempty subset of a strictly convex metric space
. The convex hull of A (in
is the set
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ1_HTML.gif)
When the underlying space is obvious from the context, we will usually write
instead of
.
Now, let us recall the definition of an -tree.
Definition 2.14 (see, e.g., [5, page 68, Definition â1.2]).
A metric space is called an
-tree, if the following conditions are satisfied:
(1)any two points are joined by a unique metric segment (denoted by
);
(2)if and
then
;
(3)for any there exists
such that
.
(Let us note that (3) follows from (1); it is, however, useful to have it among the basic properties of -trees.) We will also use the notation
for an open metric segment joining
and
and
for a left-open one.
Theorem 2.15 (see [5, Theorem â3.2]).
For a metric space the following conditions are equivalent:
(1) is a complete
-tree;
(2) is hyperconvex and any two points in
are joined by a unique metric segment.
In what follows, we will also use the classical notions of Chebyshev subset of a metric space, a metric projection onto such a set (which we will denote by
), Kuratowski and Hausdorff measures of noncompactness (which we will denote by
and
, resp.), and the radial and river metrics (which we will denote by
and
, resp.). The reader may find the relevant definitions, for instance, in the papers [11, 19, 20].
3.
-Trees
Let us begin this section with the following three simple propositions, which will enable us to characterize -trees as exactly these hyperconvex spaces in which hyperconvex hulls are unique.
Proposition 3.1.
A hyperconvex hull of a two-point subset of a hyperconvex metric space is a metric segment joining
and
.
Proof.
It is enough to consider as a subset of
and apply the uniqueness (up to isometry) of hyperconvex hulls (Proposition 2.8).
Proposition 3.2.
-trees are strictly convex.
Proof.
Let be an
-tree. Assume that
,
,
,
,
and
for
. Then
But we have
for
and therefore
, which is a contradiction.
Proposition 3.3.
For a subset of an
-tree, the following conditions are equivalent:
(1) is hyperconvex;
(2) is closed and totally convex.
Proof.
For , it is enough to use Proposition 2.3 and Remark 2.10. On the other hand, if a subset
of an
-tree
is closed and totally convex, it is a complete sub-
-tree of
. Indeed, it is enough to show that for each
, the metric segment
. But in view of the strict convexity of
, we have
. Now, in view of Theorem 2.15,
is hyperconvex.
A natural question to ask is: in which hyperconvex metric spaces the hyperconvex hulls are unique? The following theorem answers this question.
Theorem 3.4.
Let be a hyperconvex metric space. The following conditions are equivalent:
(1)for each there exists exactly one hyperconvex hull of
in
;
(2) is an
-tree.
Proof.
Necessity follows easily from Proposition 3.1 and Theorem 2.15. Sufficiency. Let be a subset of an
-tree
. Notice that
. Using Propositions 3.2, 2.12 and 3.3, we arrive at the conclusion that
is the hyperconvex hull of
in
.
4. Normed Spaces
In the first part of this section we will give an answer to the following question: In which spaces closed and convex subsets are hyperconvex?
Remark 4.1.
Note that the question whether all closed and convex subsets of some normed space are hyperconvex makes sense only in spaces which are themselves hyperconvex, so we will now restrict our attention to such spaces.
Theorem 4.2 (see [21, page 474, Theorem â1]).
If is a two-dimensional real normed space, then each nonempty, closed, and convex subset of
is a nonexpansive retract of
.
Corollary 4.3.
Each nonempty, closed and convex subset of endowed with any hyperconvex norm is hyperconvex.
Remark 4.4.
Notice that "any hyperconvex norm on " means essentially (i.e., up to an isometric isomorphism) the maximum norm; this follows from Theorem 2.6 and can also be proved using a geometric argument (see [19, Theorem â4.1]).
Theorem 4.5.
Let be a hyperconvex normed space. If
is not isometrically isomorphic to
or
, then there exists a two-dimensional linear subspace of
which is not hyperconvex.
Proof.
Since is not isometrically isomorphic to
, its dimension must be at least 2. Further, since the only (up to an isometric isomorphism) two-dimensional hyperconvex space is
, we may assume
. By Theorem 2.6 we may assume that
is the space
for some Hausdorff, compact and extremally disconnected topological space
. Since
, the space
has at least three points, so
includes a copy of
. This means that it is enough to prove the theorem in case of
with the "maximum" norm.
For simplicity, we will construct an affine non-hyperconvex subspace of ; by an appropriate translation one can obtain a linear one. Let
. Consider the following three balls in
:
,
,
. Since the corresponding balls in
intersect only at
the space
is not hyperconvex.
Corollary 4.3 and Theorem 4.5 yield the following characterization.
Corollary 4.6.
Let be a real normed space. The following conditions are equivalent:
(1)each nonempty, closed, and convex subset of is hyperconvex;
(2) is isometrically isomorphic to
or
.
We will now turn our attention to the problem of describing the spaces in which hyperconvexity implies convexity. We will start with an observation suggested to us by Grzybowski [22].
Proposition 4.7.
If a real normed space is strictly convex, then all its hyperconvex subsets are one-dimensional.
Proof.
Let be at least two-dimensional. Therefore there exist three noncollinear points
. Put
and let
,
. It is clear that
and similarly for other distances. But
is strictly convex, so we have
and
, so
. It must be therefore
, which finishes the proof.
Corollary 4.8.
If a real normed space is strictly convex, then all its hyperconvex subsets are convex.
Proof.
From Proposition 4.7 we know that hyperconvex subsets of are one dimensional; but from Proposition 2.5 we infer that hyperconvex sets are connected, which for one-dimensional sets is equivalent to their convexity.
To prove the inverse implication, we will need a simple lemma.
Lemma 4.9 (see [23, page 44, Lemma â15.1]).
Let be a metric space and
be such that
. If there exist metric segments:
joining the points
and
and
joining the points
and
, then
is a metric segment joining the points
and
.
Now we are ready to prove the following theorem.
Theorem 4.10.
If all hyperconvex subsets of a real normed space are convex, then
is strictly convex.
Proof.
Assume that is not strictly convex; we will construct a nonconvex, hyperconvex subset of
. There exist points
and positive numbers
such that
,
and the equalities
and
hold. From Lemma 4.9, both sets
and
, where
means an affine segment with endpoints
,
, are metric segments joining
and
(and hence hyperconvex sets). They cannot be, however, both convex, so at least one of them is the desired counterexample.
Again, combining Corollary 4.8 and Theorem 4.10, we obtain the following characterization of strictly convex normed spaces.
Theorem 4.11.
A normed space is strictly convex if and only if each its hyperconvex subset is convex.
5. Krein-Milman Type Theorem
In this short section, we will show that a Krein-Milman type theorem holds for -trees. It turns out that instead of compactness we only need a weaker boundedness condition.
For completeness, let us state the definition of an extremal point in the setting of -trees.
Definition 5.1.
Let be a subset of an
-tree
. We call a point
an extremal point of
if no open metric segment included in
contains
.
Theorem 5.2.
A complete and bounded -tree is a convex hull of the set of its extremal points.
Proof.
It is enough to show that each point of lies on a metric segment joining some two extremal points of
. Let
. We may assume that
is not extremal; let
. The family of all metric segments having
as one of its endpoints satisfies the assumptions of the Kuratowski-Zorn lemma. Let
and
be maximal metric segments containing the respective given metric segments. We will first show that
and
are extremal points.
If, say, were not extremal, we would have
for some
,
. Let
and
If
, we would have
and
, so
but
, so
âcontradiction. This means that
or
; assume
. Now
, so
, which contradicts the maximality of
.
Now let us show that . We will prove that
. Assume
and
. Let
. Choose
such that
. We have
and hence
; analogously,
. This means that
and
; but
âcontradiction.
Since closed and convex subsets of an -tree are hyperconvex (Proposition 3.3), Corollary 4.6 might give the impression that
-trees are somehow similar to 1- or 2-dimensional vector spaces and that completeness and boundedness of an
-tree imply its compactness. As the following example shows, this analogy is misleading.
Example 5.3.
Let be
with the radial metric. It is easy to see that
is an
-tree and so is
, which is both complete and bounded, but not compact.
6. Hyperconvexity and Measures of Noncompactness
Let us begin this section with the following definition.
Definition 6.1.
Let be some point in the Euclidean plane. Let us define a function
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ2_HTML.gif)
for all . If
, we will write
instead of
.
It is easy to prove the following lemma.
Lemma 6.2.
is a complete metric space.
We will call the function (resp.,
) introduced in Definition 6.1, the modified radial metric (resp., centered at
).
Remark 6.3.
The topology of with the metric
is strictly stronger than the topology of the same space induced by the radial metric.
Lemma 6.4.
The space with the metric
is not hyperconvex.
Proof.
Let us consider two closed balls and
Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ3_HTML.gif)
but
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ4_HTML.gif)
This shows that the metric fails to be hyperconvex.
Now we are going to examine the measures of noncompactness in the space . For this purpose we are going to use a similar approach as in the case of the measures of noncompactness in
with the radial metric (cf. [20, Theorem â4]). First let us introduce the following definition.
Definition.
Let be a bounded subset of
. We say that
satisfies
(1) condition, if for every
, there exist infinitely many pairwise distinct points
such that
;
(2) condition, if for every
, there exist infinitely many pairwise distinct points
such that
.
Let us put .
Using above conditions we can prove the following theorem
Theorem 6.6.
For any bounded subset of
with the metric
we have
and
.
Proof.
If there exists no nonnegative number satisfying either
or
, then clearly
consists of a finite number of points. Hence
in this case.
Now consider a bounded set such that there exists a
satisfying
or
condition. To prove that
, let us first show that
. For this, consider a covering
of
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ5_HTML.gif)
for some . Consider the sets
, where
. Then for every
there exists a
and
such that
. Since
for every
. Hence
.
Next we prove that . Obviously, if
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ6_HTML.gif)
then is contained in the closed ball of center
and radius
. So in this case
.
Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ7_HTML.gif)
then according to Definition 6.5, for every there exist at most finitely many points
with the property
. Hence
. Moreover,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ8_HTML.gif)
Since is arbitrary, we get
in this case. Finally, we get
. This implies
and
Example 6.7.
Using the previous formulae, we can calculate that in we have
; in particular, the closed unit ball is noncompact.
Remark.
It is known (see [12, page 135] for the details) that if a space is hyperconvex, then for any of its bounded subset , the following equality holds
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ9_HTML.gif)
The above theorem shows that even in the nontrivial cases (i.e., in cases, when bounded sets are not necessarily relatively compact), the above equality does not have to imply that the space in question is hyperconvex.
Definition 6.1 can be slightly modified. Namely, let us introduce the following definition.
Definition 6.9.
Let us define a function as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ10_HTML.gif)
for all .
Remark 6.10.
It can be easily checked that is a complete metric space. Its topology is also stronger than the topology of
with the radial metric. On the other hand this topology is obviously equivalent to the topology induced by the metric
.
Lemma 6.11.
The space with the metric
is not hyperconvex.
Proof.
Let us consider two closed balls and
Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ11_HTML.gif)
It shows that the metric fails to be hyperconvex.
For the measures of noncompactness in the space of bounded subsets in the space we have similar formulas to those given in Theorem 6.6.
Definition 6.12.
Let D be a bounded subset of with the metric
. We say that
satisfies
(1) condition, if for every
, there exist infinitely many pairwise distinct points
such that
;
(2) condition, if for every
, there exist infinitely many pairwise distinct points
such that
.
Let us put .
Theorem 6.13.
For any bounded subset of
with the metric
one has
and
The proof of Theorem 6.13 is similar to the proof of Theorem 6.6 and therefore we omit it.
The metric we are going to consider to the end of this section is, roughly speaking, like between the radial metric and the river metric. We will call it a modified river metric.
Definition 6.14.
Let . Define a function
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ12_HTML.gif)
for all If
we will write
instead of
.
The following fact can be easily checked.
Lemma 6.15.
is a complete metric space.
Remark 6.16.
The topology of is strictly stronger than the topology of
induced by the river metric.
It is interesting to consider a closed ball where
and
Such a ball consists of two disjoint closed sets (a square and a segment) which, in particular, means that it is not connected.
Lemma 6.17.
The space with the metric
is not hyperconvex.
Proof.
Let us consider two closed balls and
Then
but
. This shows that
is not hyperconvex.
To evaluate the measures of noncompactness of any bounded subset of one can use a similar approach as in the case of
(cf. Definition 6.12 and Theorem 6.13).
In connection with Remark 6.8 let us notice that as well as
are also examples of metric spaces such that
for any bounded subset
or
, but those spaces are not hyperconvex.
7. Generalized Modified Radial and River Metrics
The metric spaces as well as
are special cases of a general construction provided in [19]. More precisely, let
be a normed space and
its Chebyshev subset.
Definition 7.1.
Let be a Chebyshev set in a normed space
and let
be any metric defined on
. Let us define
by the formula
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ13_HTML.gif)
The above defined function is a metric (see [19, Lemmaâ 3.1]). Now, the following question can be risen. Is it possible to consider two disjoint Chebyshev sets, instead of one Chebyshev set
, in such a way to get a variant of the metric defined above? The following two examples show that in the case of classical hyperconvex metrics: the radial metric as well as the river metric, this problem seems not to be easy.
Example 7.2.
Let be a fixed segment in
and
the perpendicular bisector of
dividing the whole plane
into two open half-planes
and
. Let us define a function
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ14_HTML.gif)
for all , where
,
are the radial metrics on the plane centered at
and
, respectively. Then this
is not a metric. Indeed it does not satisfy the triangle inequality in the following case.
Let us consider three points such that
;
and
are collinear;
and
are collinear;
and
Then
Example 7.3.
Let and
, where
, be two points in
. Let
be the perpendicular bisector of
; it divides the whole plane
into two open half-planes
and
. Let us define a function
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ15_HTML.gif)
for all , where
denotes the river metric. Then this
is not a metric. Indeed, it does not satisfy the triangle inequality in the following case. Let
and let us take three points
Then, by the definition
and
but
, which shows
However, it appears that all the metrics introduced in Section 6 (Definitions 6.1, 6.9 and 6.14) are appropriate to define new metrics using the idea described at the beginning of this section.
Let us begin with the following definition
Definition 7.4.
Let be a segment in
and
be the perpendicular bisector of
dividing the whole plane
into two open half-planes
and
. Let us define a function
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ16_HTML.gif)
for all , where
and
denote modified radial metrics centered at
and
respectively.
Let us note that if and
both are in
, then
, so
is well-defined.
Lemma 7.5.
is a complete metric space.
Proof.
It is easy to check that is a metric. Now to verify that it is complete, let us consider a Cauchy sequence
in the space
. Then there exists
such that for all
, the points
belong to the same closed half-plane
or
. Hence, by Lemma 6.2,
is convergent, which completes the proof.
Remark 7.6.
It is clear that the topologies of induced by the metric
and the modified radial metric are not comparable.
Lemma 7.7.
The space with the metric
is not hyperconvex.
Proof.
For convenience consider and consider two closed balls
and
. Then
but
.
Remark 7.8.
It is easy to evaluate the Kuratowski and Hausdorff measures of noncompactness of bounded sets in with the metric
.
Indeed, let us consider a bounded set D in with this metric. Then we can write
as the union of two sets
and
, where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ17_HTML.gif)
Then, by the maximum property of the measures of noncompactness, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ18_HTML.gif)
To evaluate and
it is enough to apply formulas similar to the one given in Theorem 6.6.
Remark 7.9.
It is clear that in Definition 7.4 one can replace by
respectively, (cf. Definition 6.9) getting again a complete metric space which is not hyperconvex.
Now, using the metric from Definition 6.14, let us introduce the following metric.
Definition 7.10.
Let be a fixed segment in
parallel to the
-axis and
perpendicular bisector of
dividing the whole plane into two open half-planes
and
Let us define a function
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ19_HTML.gif)
for all where
and
denote the metrics from Definition 6.14.
One can prove the following lemma.
Lemma 7.11.
is a complete metric space.
The proof of this Lemma is similar to the proof of Lemma 7.5 and therefore we omit it.
Remark 7.12.
The metric is a variant of the metric
defined in Definition 6.14. The topologies induced by these metrics are not comparable. The space
is not hyperconvex, either. Finally, to find the Kuratowski and the Hausdorff measures of noncompactness of bounded sets in
with the metric
, it is enough to use the same approach as in Remark 7.8.
In Definitions 7.4 and 7.10, we considered two Chebyshev sets. Now one can think of the following question. Is it possible to increase the number of suitably chosen Chebyshev sets? The answer is "yes." Let us introduce the following definition.
Definition 7.13.
Let us consider the square in
with vertices:
,
where
Denote
,
,
,
,
. Let
be the "maximum" metric on
. By
,
, and so forth, we will mean a metric defined as in Definition 7.4, but using
,
, and so forth, instead of
,
and so forth. Denote the four open quadrants by
,
,
and
. Let us define a function
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ20_HTML.gif)
and eight more similar expressions involving ,
, and
for all
, where
and
denote the closed quadrants and
and
denote the modified radial metrics defined in Definition 6.1.
Lemma 7.14.
is a complete metric space.
Proof.
To prove that is a metric on
is straightforward, although quite long, so we omit this proof. To prove that
is complete, let us consider a Cauchy sequence
in the space
Then for every
there exists
such that
for every
It means there exists
such that for every
belongs to the same closed quadrant, because if
and
were in different quadrants (without loss of generality suppose
and
), then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ21_HTML.gif)
So, if we choose then
which contradicts that
is a Cauchy sequence. Hence almost all the terms of any Cauchy sequence must be in the same closed quadrant. Thus by Lemma 6.2,
is convergent, which shows that the space
is complete.
For the convenience of the reader, let us present a figure of a closed ball in
where
and
.
Obviously, the following lemma holds.
Lemma 7.15.
The space with the metric
is not hyperconvex.
Proof.
For convenience let us consider and
and two closed balls
and
Then
but
Remark 7.16.
It is easy to evaluate the Kuratowski and Hausdorff measures of noncompactness of bounded sets in Indeed, one can use a similar approach as in Remark 7.8.
8. Linking Construction
In this section we will give a slight generalization of the so-called linking construction described by Aksoy and Maurizi in [14] and show how this generalization includes the metrics of Section 7. Notice that a similar concept appears in [24], where it is used to study existence of certain mappings between Banach spaces.
Definition 8.1 (cf. [14, page 221, Theorem â2.1]).
Let be a metric space and
a collection of pairwise disjoint metric spaces, each disjoint with
. Let
be an arbitrary function and let
be a function satisfying
for each
. Define
for
. Let
. Define the function
by the formula
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ22_HTML.gif)
Theorem 8.2 (cf. [14, page 221, Theorem â2.1]).
The function defined above is a metric on
. If all the metric spaces
,
for
are hyperconvex, then so is
.
Remark 8.3.
The paper [14] contains the above theorem only for hyperconvex spaces. It is obvious that is a metric also in the general case.
Remark 8.4.
The authors of the paper [14] applied their version of Theorem 8.2 to obtain the hyperconvexity of the metric of Definition 7.1 (see [14, Theorem â2.2]). Let us notice that an identical result was given in an earlier work [19].
Proposition 8.5.
The metric from Definition 8.1 is complete if all the spaces
and
are complete.
Proof.
Let be a Cauchy sequence in
. We will show that
has a convergent subsequence. If
has infinitely many terms in
, we are done. If
has infinitely many terms in some
, it must be convergent in
to some
; if
, the proof is complete, and if
, it is easily seen that
in
as
. Therefore we may assume that
includes only a finite number (possibly zero) of points from
and each
. Define
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ23_HTML.gif)
Observe that ; for if that were not the case, there would exist a subsequence
and an
such that each
would lie in different
and
; this would mean that
for all
âcontradiction with
being Cauchy.
Now notice that for
, so the sequence
is also Cauchy and hence convergent to some
We have
as
and the proof is complete.
Remark 8.6.
To evaluate the Kuratowski and Hausdorff measures of noncompactness of bounded sets in with the metric
, when the set
is finite, we use following procedure.
Let us consider a bounded set in
with the metric
. Then we can write
as the following union:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ24_HTML.gif)
Then, by the maximum property of the measures of noncompactness, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ25_HTML.gif)
Example 8.7.
Notice that the metric from Definition 7.4 can be obtained as a special case of Definition 8.1. Indeed, put and
. For each
, define
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ26_HTML.gif)
for
and
for
.
In a similar way, other metrics from Sections 6 and 7 are special cases of Definition 8.1. As an example, let us provide a way to construct the metric from Definition 6.14.
Example 8.8.
Let and
for
. Define the metric
by the formula
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ27_HTML.gif)
For each , let
be the metric defined by
. Further, let
be an identity mapping and
. It is easily seen that applying Definition 8.1 we obtain the metric space
.
At the beginning of Section 7 we posed a question whether it is possible to construct a metric analogous to that from Definition 7.1, but with more than one Chebyshev subset. In all our examples, however, these subsets were singletons. Let us now show an example of two similar metrics constructed using two disjoint Chebyshev subsets consisting of more than one point.
Example 8.9.
Define the following two Chebyshev subsets of the Euclidean plane: and
. Put
. Let
and
. Let
and
be metric projections and define
by the formula
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F213812/MediaObjects/13663_2009_Article_1227_Equ28_HTML.gif)
For each , let
. Let
the identity map and
be defined by
for
. The metrics on
and
's are inherited from
. Applying Definition 8.1 we obtain a certain metric on
. Let us notice that it is not complete; taking
and
for
,
and
as before we obtain another metric, this time complete. Let us finish by observing that since
, and hence
, is disconnected, in both cases
cannot be hyperconvex.
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Borkowski, M., Bugajewski, D. & Phulara, D. On Some Properties of Hyperconvex Spaces. Fixed Point Theory Appl 2010, 213812 (2010). https://doi.org/10.1155/2010/213812
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DOI: https://doi.org/10.1155/2010/213812