1. Introduction
In the present paper, we intend to consider the Banach space consisting of real functions that are defined on a metric space being locally compact and countable at infinity. We will discuss, further on, in detail some classes of metric spaces of such a type. Moreover, we will assume that the considered real functions have increments tempered by a given modulus of continuity. It turns out that the described function space can be normed by a suitable defined norm which is complete; i.e., the mentioned function space creates the Banach space under that norm.
Next, we will formulate a theorem containing a condition sufficient for relative compactness in the above-described Banach space.
We will discuss a few particular cases of the mentioned Banach space. Namely, we will consider the space consisting of functions satisfying the Lipschitz or the Hölder condition on a given locally compact and countable at infinity metric space. For example, as that metric space, we can consider the half-axis , the set of real numbers , or the Euclidean space .
It is worthwhile mentioning that in earlier papers, we considered the case when as the above-mentioned metric space we took the metric space .
Finally, let us pay attention to the fact that the Banach function space described above with the metric space taken as
finds a lot of applications in the theory of functional integral equations (cf. [
1,
2,
3,
4,
5], among others). We expect that also the Banach space studied in this paper finds some applications in the theory of functional, differential, and integral equations.
Moreover, let us also pay attention to the fact that the results of the paper have some connections with recently published papers concerning fractional differential and integral equations (cf. [
6,
7], for example).
2. Locally Compact and Countable at Infinity Metric Spaces
In the theory of topological spaces, there are considered spaces being locally compact and which are called countable at infinite or
compact. Let us recall [
8] that a topological space
X is said to be countable at infinity if
X can be represented as the union of a sequence of compact subsets of
X.
On the other hand, we can encounter a topological space being countable at infinity but not locally compact (cf. [
8]).
In this paper, we restrict ourselves to the case of metric spaces. In such a case, we can start with the following definition [
9].
Definition 1. Let X be a given metric space (with a metric d). We say that X satisfies the condition if there exists a sequence of open sets in X such that and for (the symbol denotes the closure of the set A). Moreover, is a compact set for .
It can be shown [
9] that a metric space
X satisfies condition
if and only if it is locally compact and separable.
Let us pay attention to the fact that the concept of a metric space satisfying condition can be defined equivalently in the following way suggested by the definition of a space being locally compact and countable at infinity.
Indeed, we have the following well-known theorem [
8].
Theorem 1. Let X be a metric space that is locally compact and countable at infinity. Then, there exists a sequence of relatively compact and open subsets of X such that and for .
As an immediate consequence of Theorem 1, we infer that any metric space that is locally compact and countable at infinity satisfies condition . Obviously, the converse implications is also true.
Thus, we have the following useful conclusion.
Theorem 2. A metric space satisfies condition if and only if it is locally compact and countable at infinity.
Proof. First, let us assume that the metric space
X satisfies condition
. Then, in view of the above-mentioned result from [
9], the space
X is locally compact. Furthermore, keeping in mind that
X satisfies condition
, we infer that there exists a sequence of open subsets
of the space
X such that
is compact and
. Obviously, this implies that
; thus, the space
X is countable at infinity.
Conversely, let us assume that X is locally compact and countable at infinity. Then, according to Theorem 1, we have that there exists a sequence of open and relatively compact subsets of X such that . Hence, we deduce that is compact. This allows us to conclude that the space X satisfies condition .
The proof is complete. □
For our further purposes, the following corollary will be very crucial.
Corollary 1. Let be a metric space that is locally compact and countable at infinity. Then, there exists an increasing sequence (i.e., for ) of compact subsets of X such that .
In what follows, we will often base our considerations on Corollary 1.
3. The Space of Functions Defined on a Locally Compact and Countable at Infinity Metric Space with Increments Tempered by a Modulus of Continuity
This section is devoted to introducing and studying the Banach space consisting of real functions defined on a locally compact and countable at infinity metric space and having increments tempered by a given modulus of continuity.
To this end, consider a function such that for . Moreover, we will assume that is nondecreasing on . Any such function will be called a modulus of continuity.
In what follows, we will also assume that the modulus of continuity is continuous at ; i.e., as .
Let us observe (cf. [
1,
3]) that the functions
and
(
is fixed) can serve as examples of moduli of continuity.
Now, let us take a metric space
which is locally compact and countable at infinity (cf.
Section 2). For a given modulus of continuity
, consider the linear space
consisting of functions
such that there exists a constant
(depending on the function
x) such that
for all
. In other words, we have that
if and only if the quantity
is finite.
Obviously, the set forms a linear space over the field of real numbers .
Let us notice that functions belonging to the linear space
are uniformly continuous on the metric space
X. On the other hand, if we take a real function
which is defined and uniformly continuous on the metric space
X, then for any number
, we can define the quantity
by the following formula
The function is well defined in view of assumption on uniform continuity of the function x and is said to be the modulus of continuity of the function x.
Let us notice that the function
is an element of the space
if and only if the modulus of continuity of
x is majorized by the modulus of continuity
; i.e., there exists a constant
such that
for any
.
Further on, let us fix an arbitrary element
Next, for an arbitrary function
, we define the quantity
by the following formula
Observe that
for any
. We can also show that
is a norm in the space
; i.e.,
forms a real normed space with the norm defined by (
1).
Remark 1. Notice that up to now, we have not utilized the assumption on the local compactness and the countability at infinity of the metric space . Indeed, in the definition of the normed space , we can dispense with the mentioned assumption. Nevertheless, this assumption plays an essential role in the forthcoming theorem.
Theorem 3. Let be a metric space that is locally compact and countable at infinity. Then, the space is the Banach space with the norm defined by (1). Proof. Let us take a Cauchy sequence
in the space
. This means that the following condition is satisfied:
Hence, in particular, we obtain
or, equivalently
The above established fact implies that for an arbitrarily fixed
, there exists a natural number
such that for any number
and for
, the following inequality is satisfied
Hence, we obtain
for
, and
.
Now, keeping in mind Corollary 1, we can find an increasing sequence of compact subsets of the metric space X such that .
Thus, let us fix a number
; i.e., let us fix a compact set
of the above established sequence
. On the base of inequality (
3), we deduce that the functions from the set
are equicontinuous on the set
. Obviously, this implies that the functions of the sequence
are equicontinuous on the set
.
In the similar way, putting in (
3)
(or utilizing inequality (
2)), we infer that
Hence, in virtue of (
2) we get
From the above inequality, we deduce that the functions of the set are equibounded on the set . Obviously, this allows us to infer that the functions from the sequence are equibounded on the set .
The above established properties of the functions of the sequence and the Ascoli–Arzelá theorem allows us to conclude that the sequence is relatively compact on the set for any . Thus, applying the diagonal procedure, we can select from the sequence a subsequence , which converges nearly uniformly on the metric space X. This means that the subsequence is uniformly convergent on each set to a function defined on X.
Next, let us take into account inequality (
3) being valid for
and for arbitrary
. Fixing
u and
v and passing with
, from that inequality, we get
On the other hand, keeping in mind that
, we deduce that there exists a constant
such that
for
.
Now, joining (
4) and (
5), we derive the following estimate
for
. This shows that
.
In what follows, taking into account inequality (
2), for an arbitrarily fixed
and for a natural number
chosen according to (
2), we have that
for
.
Letting in the above inequality with
, we obtain the following estimate
This shows that .
Thus, the function
is the limit of the function sequence
with respect to the norm
of the space
defined by (
1). The proof is complete. □
In what follows, let us mention that as the modulus of continuity
indicated in our earlier considerations conducted in this paper, we can take the function
. In this case, the function space
represents the space consisting of functions
satisfying the Lipschitz condition on the metric space
X (locally compact and countable at infinity). Thus,
if and only if there exists a constant
such that
for arbitrary
.
If we take as the modulus of continuity, the modulus generated by the Hölder condition—i.e., if we take the function
(where
is a fixed number such that
)—then the suitable space
consists of functions
such that there exists a constant
(depending on the function
x) such that
for arbitrary
.
Now, let us pay attention to some particular cases of the metric space
being locally compact and countable at infinity. The simple example of such a metric space is the set
with the natural metric
. Such a case was considered in paper [
3].
Further, let us observe that if we take a metric space with the metric d generated by a norm i.e., if X is a normed space, then the assumption on the local compactness of X implies that X is finite dimensional. In the real case, we get that X is isometric to the Euclidean space . Hence, we conclude that the most natural metric space X being locally compact and countable at infinity seems to be the Euclidean space .
Thus, it is natural to consider as the most representable Banach space the space , where the metric in can be considered as the Euclidean metric or a metric equivalent to that metric.
In our further consideration, it is reasonable to consider as the Banach space the space , where or for .
4. A Sufficient Condition for Relative Compactness in the Space
In this section, we are going to describe a criterion being a sufficient condition for relative compactness in the Banach space
, where
X (with a metric
d) is a metric space being locally compact and countable at infinity. The space
of such a type was described in detail in
Section 3.
In view of Corollary 1, we will assume in this section that is a sequence of nonempty and compact subsets of X such that the sequence is increasing (i.e., for ) and .
Taking into account the practical utility of the space
, we restrict ourselves to the case when
, since only in such a case we are in a position to apply such a criterion for relative compactness in a concrete situation (cf.
Section 3). Thus, in this section, we will consider the Banach space
described in the previous section. Obviously, in this case, instead of the sequence
of compact subset of
such that
for
and
, it will be convenient to take the family of balls
centered at the zero point
and with radius
. To fix our attention, we will consider in the Euclidean space
the classical maximum metric; i.e., if
, then we take
.
Now, we are prepared to formulate the announced sufficient condition.
Theorem 4. Let A be a bounded subset of the space satisfying the following to conditions:
Then, the set A is relatively compact in the space .
Proof. Let us fix arbitrarily a number
. Further, choose a number
according to condition
. Next, keeping in mind condition
, we can find a number
. Consider the set
, where the symbol
denotes the restriction of the function
x to the set
. Taking into account the fact that the set
satisfies condition
, in view of Theorem 4 in [
1], we infer that the set
is relatively compact in the space
. This means that there exists a finite
net of this set in the space
which consists of functions
being restrictions of functions
belonging to the space
i.e.,
for
and for
. This implies (cf. Lemma 2.6 in [
3]) that there exists a finite
net of the set
A that consists of functions
; i.e., there exist functions
being restrictions of functions
belonging to the space
, which forms an
net of the set
A.
Now, let us take an arbitrary function
. Then, taking into account the above reasonings, we infer that there exists a number
such that
This means that the following inequality is satisfied
Next, let us choose a number
to the number
according to condition
. Then, for arbitrary
, we have the following inequality
which is satisfied for any function
.
Now, taking the function
, which was chosen previously to the function
, for
, we get:
Hence, taking into account (
7) and the fact that
we obtain
Further, let us assume that
are such that
(the symbol
denotes the interior of the ball
),
, and
. Let us consider the segment
joining the points
u and
v. Denote by
the intersection of the segment
with the sphere
. Then,
Then, keeping in mind (
9) and (
10), for a fixed function
, we obtain:
Consequently, in view of condition
and inequality (
7), we deduce the following estimate
Next, taking into account the fact that
for
, we arrive at the following inequality
for
and
Further, in virtue of (
11) and (
12), we get
Finally, combining (
6), (
8), and (
13), we conclude that for any
there exists a function
such that the following inequality holds
This means that the functions form a finite net of the set A in the space . Thus, the set A is relatively compact in the space .
The proof is complete. □