On the Space of Functions with Growths Tempered by a Modulus of Continuity and Its Applications

We are going to study the space of real functions defined on a bounded metric space and having growths tempered by a modulus of continuity. We prove also a sufficient condition for the relative compactness in the mentioned function space. Using that condition and the classical Schauder fixed point theorem, we show the existence theorem for some quadratic integral equations of Fredholm type in the space of functions satisfying the Hölder condition. An example illustrating the mentioned existence result is also included.


Introduction
The principal aim of the paper is to discuss the function space consisting of functions with growths tempered by a given modulus of continuity.Functions belonging to such a space have their moduli of continuity in some sense proportional to a given modulus of continuity.We assume here that the mentioned function space consists of real functions defined on a bounded metric space and, as we pointed out above, having growths tempered by a given modulus of continuity.Typical examples of function spaces of such a type are the space of functions satisfying the Lipschitz condition and the space of Hölder functions, that is, the space of functions satisfying the Hölder condition with a given exponent belonging to the interval (0, 1]. It is a surprising fact that those spaces, although they are frequently used in mathematical investigations, were not intensively studied and described in mathematical literature (cf.[1][2][3][4]).It is probably caused by the fact that the norm in the mentioned function space is not convenient for use in comparison with the norm in the classical space of real functions being continuous on a compact set.On the other hand, there are no known convenient sufficient conditions for relative compactness of bounded subsets of the function space in question.
It is also worthwhile mentioning that it is very difficult to encounter some applications of the space of functions with growths tempered by a modulus of continuity in the theory of operator equations (functional, differential, integral, etc.), (cf.[5,6]).
In this paper we are going to discuss in detail the above mentioned function space consisting of real functions having growths tempered by a given modulus of continuity.We describe a general form of such a space and we point out some important particular cases of spaces of the described type.
Apart from this, we indicate some essential properties of the considered function space.Namely, we show that such a space is not a closed subspace of the space of all real continuous functions with the supremum norm, with respect to this norm.The consequence of this property is the fact that the norm introduced in the space of functions with tempered growths is not equivalent to the classical supremum norm.This causes that it is not possible to use the supremum norm in all considerations conducted in the mentioned space of functions with tempered growths.
The main result proved in the paper is a sufficient condition for relative compactness of bounded subset of the discussed function space.We also point out several particular cases of that result, especially in the space of functions satisfying the Hölder condition.

The Space of Functions with Tempered Moduli of Continuity
In this section we discuss the space of real functions defined on a given bounded metric space and having the growths tempered by a given modulus of continuity.Moreover, we will also study some important particular cases of that space.
To this end denote by R the set of all real numbers and put R + = [0, ∞).A function  : R + → R + is said to be a modulus of continuity if (0) = 0, () > 0 for  > 0, and  is nondecreasing on R + .
We will also assume (but not always) that the modulus of continuity  = () is continuous at  = 0, that is, () → 0 as  → 0.
In order to clarify the concept of the modulus of continuity, let us assume that (, ) is a given bounded metric space.Denote by () the space of all real functions defined and continuous on the metric space .For  ∈ () and for an arbitrary fixed number  > 0 let us define the quantity ](, ) by the formula ] (, ) = sup {| () −  (V)| : , V ∈ ,  (, V) ⩽ } .(1) The function  → ](, ) is called the modulus of continuity of the function .Observe that in order to define this concept it is sufficient to consider the space () consisting of real functions defined and bounded on .
Observe that if  ∈ (), then the modulus of continuity of , that is, the function  → ](, ), is the modulus of continuity in the above defined sense.Obviously, this modulus is continuous at  = 0 if and only if  is uniformly continuous on .
In what follows let us fix a modulus of continuity  = () and assume, as before, that (, ) is a given bounded metric space.Denote by   () the set of all real functions defined on  such that their growths are tempered by the modulus of continuity .More precisely, a function  = () belongs to the set   () provided  :  → R and there exists a constant   > 0 such that for all , V ∈ .In other words, we have that  ∈   () if and only if the quantity is finite.
It is easy to check that the set   () forms a linear space over the field of real numbers R. Obviously,   () is a linear subspace of the space ().
In the sequel, let us notice that if  ∈   (), then in view of (2) there exists a constant   > 0 such that for any  > 0. Obviously the converse implication is also true.Thus, we can say equivalently that the set   () consists of all real functions defined on X such that the moduli of continuity of those functions are tempered by the given modulus of continuity  = ().Now, fix arbitrarily an element  0 ∈ .For an arbitrary function  ∈   () we define the quantity ‖‖ by the formula Notice that ‖‖ < ∞ for any  ∈   ().Moreover, it can be shown that ‖ ⋅ ‖ is a norm in the space   (); that is,   () is a normed space with the norm defined by (5).We show that the norm defined by ( 5) is complete.To this end, take a Cauchy sequence (  ) in the space   ().Fix arbitrarily a number  > 0 and denote  = max{1, (diam )}, where diam  denotes the diameter of the metric space .Then, we can find a natural number  0 such that for ,  ⩾  0 , we have that ‖  −   ‖ ⩽ /2 or, by (5), Further, let us notice that from (6) we infer that This means that the real sequence {  ( 0 )} satisfies the Cauchy condition in R with natural metric.Hence we obtain that this sequence is convergent to a real number, say Putting in the above inequality V =  0 , we get that for each  ∈ ,  ̸ =  0 , the following estimate holds This yields the inequality which is valid for  ∈  and ,  ∈ N, ,  ⩾  0 (N denotes the set of natural numbers).
From that last inequality we get Hence we obtain for an arbitrary  ∈  and ,  ⩾  0 .Linking the above inequality and ( 7), we deduce that for  ∈  and for ,  ∈ N, ,  ⩾  0 .But this fact means that (  ()) is a real Cauchy sequence for arbitrarily fixed  ∈ .Thus this sequence is convergent in R. Denote its limit by (), that is, Passing in (13) with  → ∞, we obtain for any  ∈  and for  ∈ N,  ⩾  0 .This means that the sequence (  ) converges uniformly to the function  on the metric space .Now, we show that the sequence (  ) converges to the function  in the sense of norm (5).To this end we prove first that  ∈   ().Indeed, taking into account the fact that (  ) is a Cauchy sequence in the space   (), we infer that (  ) is bounded in   (); that is, there exists a constant  > 0 such that ‖  ‖ ⩽  for  = 1, 2, . . . .Hence, in view of (5) we have This implies that for arbitrarily fixed , V ∈ ,  ̸ = V and for  ∈ N, we get and, consequently, for all , V ∈  and for  ∈ N. Now, fix arbitrarily , V ∈ .Then, passing in the above inequality with  → ∞, we obtain for all , V ∈ .But this means that  ∈   ().
In what follows, using the fact that the sequence (  ) satisfies inequality (6), we derive that for arbitrary , V ∈ ,  ̸ = V and for arbitrary ,  ∈ N, ,  ⩾  0 , the following inequality holds Fixing , V and passing with  → ∞, we get Hence, in virtue of the arbitrariness of the choice of points , V ∈ ,  ̸ = V, we have the estimate being valid for  ⩾  0 .
In a similar way, keeping in mind inequality (6) for all , V ∈ .In other words, the function  satisfies the Hölder condition with a constant   and with an exponent .Notice that the norm in the space    () has the from Example 3. Now, let us take into account the modulus of continuity  having the form Note that  ∈   () if and only if there exists a constant Obviously, this condition is equivalent to the boundedness of the function .Indeed, fix  0 ∈ .Then, for any  ∈  we have This yields for  ∈ .The converse implication is also obvious.Thus we can write that   () = (); that is,   () contains all functions defined and bounded on .Further, observe that the norm in the space   () has now the form It is easy to show that this norm is equivalent to the classical supremum norm ‖ ⋅ ‖ ∞ in the space ().Indeed, we have for any function  ∈ (), where The examples of the functions () =  − 1 and () =  in the space ([0, 2]) with  0 = 0 show that the equality signs in (34) are attained.

Some Properties of the Space of Hölder Functions
In order to simplify the considerations in this section, we restrict ourselves to spaces of real functions defined on a fixed interval In what follows, for  ∈   [, ], the symbol    will denote the least possible constant for which inequality (35) is satisfied.In other words, we can write Notice that in the case  = 1, the linear space  1 [, ] coincides with the space of real functions defined on [, ] and satisfying the Lipschitz condition, that is, for some constant   > 0 and for all ,  ∈ [, ].This space will be denoted by the symbol Lip[, ].
Our principal aim is to show that the spaces   [, ] (for  ∈ (0, 1]) do not form closed subspaces of the space [, ] with respect to the norm ‖ ⋅ ‖ ∞ .In order to prove this fact, observe that the space   [, ] can be normed in a natural way (cf.Example 2) if for  ∈   [, ] we put Observe that in view of (36), the above definition can be written in the from Obviously, the space   [, ] with the norm (38) is the Banach space (cf.Section 2).
Further on let us notice that for an arbitrarily fixed  ∈   [, ] and for an arbitrary  ∈ [, ] we obtain Hence we get The above inequality means that the norm ‖⋅‖ ∞ is dominated by the norm ‖ ⋅ ‖  .
If the space   [, ] would be a closed subspace of [, ] with respect to the norm ‖ ⋅ ‖ ∞ , then this would mean that   [, ] (as the linear space) is a Banach space with respect to the norm ‖ ⋅ ‖ ∞ .In other words, the space   [, ] would be a Banach space both with respect to the norm ‖ ⋅ ‖ ∞ and with respect to the norm ‖ ⋅ ‖  .Hence, keeping in mind (41) and taking into account the theorem on the equivalence of dominated norms [4], we would infer that the norms ‖ ⋅ ‖ ∞ and ‖ ⋅ ‖  are equivalent in the space   [, ].We show that such a conclusion is not true.
To this end fix a number  ∈ (0, 1] and consider the sequence (  ) of real functions defined on the interval [0, 1] by the formula In what follows, let us observe that for 0 <  <  ⩽ 1 the following inclusions hold: Particularly, taking into account that for  = 1, we have which yields for 0 <  <  < 1.
This shows that  ∈   [, ] and hence we infer that inclusions (47) hold.(49) Hence we get We show that the inverse assertion is not true.To this end, take the function sequence (  ) ⊂   [0, 1], where   is defined by the formula

A Sufficient Condition for Relative Compactness in the Space 𝐶 𝜔 (𝑋)
This section is devoted to present a criterion for relative compactness in the space   () of functions with growths tempered by a given modulus of continuity  = ().More precisely, we will discuss a sufficient condition for relative compactness in   ().
It is worthwhile mentioning that as far as we know, there are no known conditions of such a type (e.g., cf.[1][2][3][4]).
Let us recall that the complete description of the space   () was given in Section 2.
Theorem 4. Assume that  = () is a given modulus of continuity such that () → 0 as  → 0, and assume that  is a compact metric space.Let  be a bounded subset of the space   () such that functions belonging to  are equicontinuous with respect to the modulus of continuity ; that is, the following condition is satisfied: Then the set  is relatively compact in the space   ().
Proof.At first, let us observe that since  is bounded, then there exists a constant  > 0 such that for an arbitrary  ∈  the following inequality holds: where  0 is a fixed element of the metric space  (cf.( 5)).
Particularly, we conclude that for an arbitrary  ∈ .Moreover, on the basis of (56) we infer that for any  ∈  and for arbitrary , V ∈ ,  ̸ = V, the following inequality holds: The above inequality yields for an arbitrary function  ∈  and for all , V ∈ .
Since lim  → 0 () = 0, from (59) we deduce that all functions belonging to the set  are equicontinuous on the set . Apart from that, from (59) we have that for an arbitrary  ∈  the inequality holds for all , V ∈ .Putting in the above inequality V =  0 , we infer that for arbitrary  ∈  and  ∈  the following estimate is satisfied: This implies that for  ∈  and  ∈ .This means that functions from the set  are equibounded.Now, let us consider an arbitrary sequence (  ) in the set . Taking into account the above established facts, we infer that functions of the sequence (  ) are equibounded and equicontinuous on the set . Hence, in view of Ascoli-Arzéla criterion we derive that there exists a subsequence of the sequence (  ) which is uniformly convergent on the set  to a function  = ().To avoid complicated notation, we will denote the mentioned subsequence of the sequence (  ) by the same symbol (  ).Observe that the function  = () is continuous on the set .
In what follows, we show that  ∈   ().To this end observe that from the fact that functions belonging to the sequence (  ) satisfy inequality (59), we have that for an arbitrary  ∈ N and for , V ∈ .Since lim  → ∞   () = () for an arbitrary element  ∈ , thus keeping in mind the continuity of the absolute value and other standard facts from mathematical analysis, we deduce from the last inequality that for arbitrary , V ∈ .
Further on we show that the sequence (  ) is convergent to the function  in the sense of the norm of the space   ().
To this end for convenience, let us denote where  > 0 is a fixed number.Obviously we have that  2 0 = for all  ⩾  0 and for any  ∈ .
Further, let us observe that (, V) >  for (, V) ∈ X2  .Hence we infer that ((, V)) ⩾ ().Thus, for an arbitrary  ⩾  0 , in view of ( 71) and ( 72 (73) Next, taking into account the fact that the functions of the sequence (  ) belong to the set , in virtue of the choice of the number  to the number /4, we infer that for an arbitrary pair (, V) ∈  2  and for an arbitrary natural number  the following inequality is satisfied: Hence we obtain for (, V) ∈  such that (, V) ⩽ .
Letting in the above inequality with  → ∞, we get for (, V) ∈  such that (, V) ⩽ .Consequently, we obtain for arbitrary (, V) ∈  2  .Now, in view of ( 74) and (77), we derive the following estimates: Combining (70) with inequalities (72), (73), and (78), we obtain that ||  − || ⩽  for  ∈ N,  ⩾  0 .This means that the sequence (  ) is convergent to the function  with respect to the norm of the space   ().Finally, we conclude that the set  is relatively compact in the space   () and the proof is complete.Now, based on Theorem 4, we prove a manageable and handy sufficient condition for relative compactness in the space   ().Theorem 5. Assume that  1 ,  2 are moduli of continuity being continuous at zero and such that  2 () = ( 1 ()) as  → 0, that is, Further, assume that (, ) is a compact metric space.Then, if  is a bounded subset of the space   2 () then  is relatively compact in the space   1 ().
Next, taking , V ∈  such that  ̸ = V, (, V) ⩽  and keeping in mind (81), we derive the following estimate: Hence we have for , V ∈  such that  ̸ = V and (, V) ⩽ .Finally, in view of Theorem 4 we conclude that the set  is relatively compact in the space   1 ().This completes the proof.
Example 6.In order to illustrate the applicability of Theorem 5, let us consider two moduli of continuity of Hölder type having the form  1 () =   ,  2 () =   , where 0 <  <  ⩽ 1 (cf.Section 3).Then we have lim This shows that the moduli of continuity  1 () and  2 () satisfy the conditions of Theorem 5. Thus, if we assume that a set  is bounded in the space In other words, if there exist numbers ,  with 0 <  < 1 and  <  ⩽ 1 such that  is a bounded set in the Hölder space with the exponent , that is, there exists a constant  > 0 such that for any  ∈  and for all , V ∈ , then the set  is relatively compact in the Hölder space with the exponent .

Application to a Quadratic Integral Equation
In this final section we are going to present an application of the results obtained in previous sections to derive an existence result for a quadratic integral equation of Fredholm type.Equations of such a type occur naturally in connection with some problems investigated in the theories of radiative transfer, neutron transport, and the kinetic theory of gases [7][8][9][10][11].Those integral equations are known as Chandrasekhar quadratic integral equations (cf.[10]).
Let us pay attention to some important facts connected with our considerations.
First of all, notice that the study of the existence of solutions of functional, differential, and integral equations is very complicated in the Lipschitz and Hölder function spaces which were considered in Section 3. To justify such an opinion, let us recall a result due to Matkowski [6] asserting that the so-called superposition operator generated by the function  = (, ) maps the space Lip[, ] into itself and is Lipschitzian (with respect to the norm in the space Lip[, ] described in Example 1) if and only if the function  is linear; that is, there exist functions ,  ∈ Lip[, ] such that  (, ) =  () +  () for all  ∈ [, ] and  ∈ R. Other results in this direction can be found in [5], for instance.This shows that the investigations on the existence of solutions of operator equations in the space   () consisting of functions with growths tempered by a modulus of continuity are rather a delicate matter.
As we indicate above, the object of the study conducted in this section is the quadratic integral equation of Fredholm type having the form where  ∈ [, ].
In our further considerations we will assume that  is a fixed number in the interval (0, 1].Now, we formulate the assumptions under which we will study (94).In what follows, based on the above assumptions, we can define the finite constant  by putting Now, we are prepared to present our last assumption.
(iii) The following inequality holds: where ||||  denotes the norm of the function  in the space   [, ].Now, we can formulate our existence result concerning (94).Theorem 7.Under assumptions ()-() (94) has at least one solution belonging to the space   [, ], where  is arbitrarily fixed number such that 0 <  < .
Proof.At the beginning, let us notice that in view of inclusions (47) we have that if a function  satisfies the Hölder condition with the exponent , then it satisfies this condition with the exponent .Thus, from assumptions (i) and (ii) it follows that there exists a constant   such that for all , ,  ∈ [, ], where the constant    associated with the function  is defined by (36).Now, let us consider the operator  defined on the space   [, ] by the formula Hence, in view of the inequalities |||| ∞ ⩽ max{1, ( − )  }‖‖  ,    ⩽ ‖‖  (cf.(39) and ( 41)), we derive the following estimate: In what follows, we will treat the ball   as a subset of the space   [, ].
We now show that the operator  is continuous on the space   [, ].(112) Hence we deduce that the inequality from assumption (iii) is satisfied provided It is easy to check that the above inequality is satisfied if we put, for example,  =  = 1/216,  = 1/128.Finally, applying Theorem 7, we conclude that (107) has at least one solution in the space   [0, 1] with 0 <  < 1/2, provided inequality (113) is satisfied.
Let   denote the modulus of continuity which corresponds to the Hölder condition.This means that   () =   for  ⩾ 0, where  is a fixed number from the interval (0, 1].Observe that  ∈    () if and only if there exists a constant   > 0 such that Further, if  (0 <  ⩽ 1) is a fixed number, then the symbol   [, ] will denote the space    ([, ]) (see Example 2) with   =   ; that is,   [, ] is the collection of all real functions  defined on [, ] and satisfying the Hölder condition.More precisely,  ∈   [, ] if there exists a constant    such that Observe that   [, ] forms a linear subspace of the linear space [, ].
Let  > 0 denote a number corresponding to /4 according to the assumption about equicontinuity of functions from the set  with respect to the modulus of continuity  = ().Now, denote by () the number defined as follows: () −  () −   (V) +  (V)      ( (, V)) This shows that the operator  transforms the space   [, ] into itself.Moreover, in view of assumption (iii), we infer that  transforms into itself the ball   (in the space   [, ]) centered at the zero function  and with radius  0 , where  0 is an arbitrary number from the interval [ 1 ,  2 ], while  }) 2 (2 +   ( − )).Now, observe that in view of (47) the ball   is contained in the space   [, ].Further, keeping in mind the fact established in Example 6, we conclude that the set   is relatively compact in the space   [, ].Moreover, taking into account estimate (50), we deduce that   is also closed in the space   [, ].Thus, gathering the above obtained facts, we infer that the set   is a compact and (obviously) convex subset of the space   [, ].