Abstract

A scale of spaces exists connecting the class of functions of bounded th -variation in the sense of Riesz-Merentes with the Sobolev space of functions with -integrable th derivative. This scale is generated by the generalized functionals of Merentes type. We prove some limiting relations for these functionals as well as sharp estimates in terms of the fractional modulus of smoothness of order .

1. Introduction

Let be a 1-periodic function on the real line. A set of points such that , where , will be called a partition. For a partition we denote by its diameter. Let , , and . For any , we setWe denote by the class of all 1-periodic functions such that , where the supremum is taken over all partitions of ; see, for example, [1].

If we get the class of functions of bounded -variation introduced by Wiener [2]. A celebrated result of Riesz (see, e.g., [3, 4]) states that the class of absolutely continuous 1-periodic functions such that the first derivative belongs to the Lebesgue space coincides with the class and .

Another characterization of the class was given by Terehin in [5] in terms of the fractional continuity modulus of order , introduced by Wiener in [2]. If , then , and conversely if , then .

The concept of functions of bounded second variation was introduced by de la Vallée Poussin in [6]. By a refined partition we mean a partition of the formMotivated by the result of Riesz, Merentes in [7] introduced, for , the class of functions of bounded second variation as the class of those functions for which is finite. He proved a similar result to that of Riesz (see, e.g., [7, Theorem, page 122]) and gave a variational characterization of the Sobolev space in terms of functions of bounded second variation, namely, a function belongs to if and only if and . Observe that from the characterization of Riesz and Merentes it follows that . The people interested in this topic can see [8].

The above results motivated Barza and Silvestre to study the scale of spaces between the class of functions of bounded second -variation and the Sobolev space (see [9]) as an extension of [1].

Following [9, 10], let and . Let be a partition of the formLet for any where the th divided differences are defined by We define the space as the space of 1-periodic functions such that where the supremum is taken over all partitions of type (5). The number is called the th variation of . For , we denote it by and call it the th -variation of . We have that is the space in [10]. The spaces form a scale between the space of functions of bounded th -variation and the Sobolev space . In [10] the authors extend the results of Riesz and Merentes by showing that if and only if .

Terehin studied in [5] properties of the function where , . The function is called the fractional modulus of smoothness of order . A characterization of the space in terms of was also given in [5]: if and only if .

We extend the results proved in [1, 9] to the case of functions of bounded th -variation. In Section 2 we prove some results which we will use in the proofs of our main theorems. In Section 3 we study sharp estimates of in terms of the fractional modulus of smoothness of order , and we study limits in the scales of spaces for .

2. Preliminary Results

Recall that a function satisfying is called -continuous and the class of -continuous functions is denoted by .

Lemma 1 (see [11, Lemma  1]). Let and such that for . Then

Remark 2. Let be a partition as in (5). Since   , then and since , it follows thatThen, . The space is the space in [10]. Then, .

Proposition 3. Let , , and . If , then exists everywhere and is continuous on .

Proof. Suppose that . Then, there exists a constant such thatfor any partition of type (5) and .
Let . It follows by (12) that for any partition of type (5) such that and . So that, Cauchy’s convergence criterion establishes the existence of the right-hand derivative when . Similarly, the left-hand derivative exists when . Let us consider the following partition It follows that which it tends to zero as and . Then, we obtain that and so that, exists.
Let us consider and the following type of partition where . Then, by (12), Letting , we obtain that from which the continuity of in follows. Also, is right and left continuous in and , respectively.

Remark 4. Let , , and . If , then it follows thatfor some . Indeed, by Proposition 3 we have that exists everywhere on and it is continuous. Then, . The equality (19) follows by [10, (2.8)].

Theorem 5. Let and .(i)If , then
(ii)If , then

Proof. (i) If , then and exists everywhere on by Proposition 3 and it is continuous. To prove the inequality, let be a partition of and take . Define the partition of the subinterval as Then, we have thatThen, by Remark 4, there exist and such thatBy letting , since is continuous, it follows that By taking then supremum over all ’s, the inequality follows.
(ii) Given , there exists a partition of type (5) such that Since , it follows that . Moreover, as we can see in [9, Theorem  2.3], we have that Moreover, by [9, Proposition  2.4], it follows that . Hence, . Therefore, is Riemann integrable in . Indeed, it can be written as a difference of two monotone functions and then, is Riemann integrable.
By using [10, Proposition  2.5], since is continuous, for and , we have that Letting tend to zero, the inequality follows.

Proposition 6. Let and assume that exists everywhere. Then(For simplicity, from now on we omit the domain in the functionals .)

Proof. The equality follows as in Theorem 5.

Let and be the Steklov average of the function (see, e.g., [1]) defined by We define the th Steklov average of as the Steklov average of the th Steklov average of ,

Lemma 7. Let and . Then(i)(ii)(iii)

Proof (the argument is inductive). (i) The inequality follows by the first inequality of [1, Lemma  2.2] and the definition of . Indeed, since and it follows that (ii) Let . We have that , In the computations in [9, Lemma  2.5], we find thatWe have that since Then, it follows that Therefore, by (38) (see [9, Lemma  2.5])and the inequality holds for 3. Let us assume thatand let us see that the inequality holds for .
We have that Therefore, by (43), it follows that (iii) The inequality holds for 2 by [9]. We have that
By Theorem 5 and [9, Proposition  2.4] it follows thatbecause since by [9, Lemma  2.5(iii)] . Therefore, since in [9, Lemma  2.5(iii)] we see that , then we have thatTherefore,for . By [1, Lemma  2.2 (2.15)] it follows thatand then since , it follows that Then we see that the inequality holds for 3 as well.
In particular we have by [9, Lemma  2.5(iii)] thatand we have proven thatLet us assume that the inequality holds for and let us take . We can see that the inequality holds also for . Indeed, by the same kind of argument than for 3 we see that (recall )since .
Recall thatThen . Hence, . Let us assume that this type of identity holds for , that is, . By (54), we see that and the same identity holds for any .
Therefore, by (53) and the previous identity, we obtain that

In the next proposition we give an exact description of Peetre’s -functional for the pair for . By [12, page 172]with for any .

Proposition 8. Let . For any

Proof. For let and . Since , then . Hence, by (ii) and (iii) in Lemma 7 it follows thatTherefore,and by taking infimum over all , it follows that We prove now the left inequality. Let . By [11, page 19], it follows thatsince . Taking infimum over all , we obtain that

3. Main Results

Proposition 9. Let and . Then

Proof. By Lemma 1 and [1, Proposition  3.1] it follows that Now we prove the reverse inequality. Let , , and . ThenWe set , which is the th -modulus of continuity; see [1, 9]. By Hölder’s inequality and Fubini’s theorem Moreover, it follows by Hölder’s inequality and Fubini’s theorem that We repeat this type of argument to the other terms in the sum in the right hand side of inequality (66), except for the last one. For instance, in the last case we see that Since , belong to , then by all the previous inequalities it follows that all the intermediate terms except the last one in the sum of the right hand side in (66) tend to zero as . It was proved in [13] that . Then

Theorem 10. Let and be a 1-periodic function.(i)If , then (ii)If and then .

Proof. (i) Let and . By Proposition 9 we have that, for any , there exists such thatMultiplying (73) by and then integrating from to and subtracting from each side, it follows thatOn the other hand, since , then and by Lemma 1 and Theorem 5 we have that Hence, it follows thatBy adding up (74) and (76) we get As , the right hand side tends to . Since was arbitrary, by letting it tend to zero we get the first part of the theorem.
(ii) Applying the first part of the theorem to and Lemma 7(i), we have that there exists a constant such that and then Since , then by a deep result of Hardy and Littlewood (see [14, Theorem  24]) it follows that .

In the following theorem we prove the continuity at the end points of the scale determined by in the same manner as in [1, 9].

Theorem 11. Let . If is 1-periodic function, then(i), and(ii)if , for some , then

Proof. The proof of (i) and (ii) is similar to the proof of [1, Theorem  3.4] (see also [9, Theorem  3.3]). It is enough to define for a partition of type (5), , , andBy summation by parts, the result follows as in [9, Theorem  3.3].

The condition for some is necessary for (80) to hold. If has a discontinuous derivative, then for any . Therefore, for all , while .

Terehin in [13] showed that, for any function , the modulus is equivalent to the nondecreasing function on given by Clearly, is a decreasing function and (see, e.g., [1])

Similarly, for any function such that , it is possible to prove that the following function defined byis increasing and is decreasing. Moreover, since for any (see [15]), we have that

In the next theorem we prove a sharp estimate of in terms of , which in particular implies the result of Merentes [7].

Theorem 12. Let , , and be a 1-periodic function in . Assume thatThen, andwhere is constant dependent of and .

Proof. Condition (86) implies that . Without loss of generality (see [1, Theorem  4.1]), we may suppose thatLet and , . As in [1, page 5], we define the sequence of natural numbers . Set andBy the monotonicity properties of we have that eitherorAssume first that (90) holds. Let us define For a fixed partition , we setWe define For any , we have that and, by [10, Proposition  2.3], Then, by Hölder’s inequality it follows that where and .
Thus, by Lemma 7(ii) and by (85), it follows that Thenand moreoverBy (98) and (99), it follows that We estimate the second term of the above sum. Obviously, . By Lemma 7(iii), (85), and (90), it follows that Hence,Assume now that (91) holds. Let us define Thus by Lemma 7(ii) and (85) and by the definition of and (91) Thenand moreoverBy (105) and (106), it follows that We estimate now the second term of the above sum. Obviously, we have . By Lemma 7(iii) and (85) we get that Hence,By (102) and (109), Applying [1, Lemma  2.1] with , and , it follows that since and .
On the other hand, by (85), and also Therefore,

Remarks. If , then, by Theorem 10(i), we have that Moreover, by the celebrated result of Riesz and Theorem 5, The upper limit as of the right hand side of (87) is less than

Also, by Theorem 11(i), the left hand side of (87) tends to . Thus, taking limit as in it, it follows thatThis implies the result of Merentes [7] that states that , and it shows that the order of the constant is optimal.

Let us assume that for some . Since for , it follows thatMoreover, since , by Theorem 12, . Hence, tends to as by Theorem 11(ii). Thus, the left hand side of (87) agree in behavior with the right hand side when which means that the order of the constant is also optimal.

4. Further Results and Applications

Lemma 13. Let , , and be a 1-periodic function. Assume that Then whereif and if .

Proof. From the pointwise estimate of Lemma 1, we have that We prove now the second inequality. Since , it follows that . Then, . We use the following Marchaud-Timan type inequality in [15, Theorem  4],where is an absolute constant independent on and .
Assume first that , . Let . By raising (124) to , then multiplying both sides of it by , and then integrating it from to we obtain We estimate now the first term of the above sum by the use of the sharp weighted Hardy inequality (see, e.g, [16, Lemma  3.9]) and we getFor the second term, by Fubini’s theorem we haveTherefore, (126) and (127) imply the following estimate Let now , that is, . By raising (124) to , then multiplying both sides of it by , and then integrating it from to we get The first term of the right-hand side of the inequality can be estimated as before (see (126)). The second term instead can be estimated by the fact that for any the following inequality holdsThen, Therefore,

Theorem 12 can be derived by [1, Theorem  4.1] and the previous lemma, but the constant will not have right asymptotic behaviour. Indeed, since , then, as in Proposition 14,

Proposition 14. Let , , and be a 1-periodic function such that belongs to . Assume thatThen, andwhere is an absolute constant and is in Lemma 13.

Proof. The result follows by [1, Theorem  4.1], Lemma 13, and Theorem 5. Indeed, since and , then by [1, Theorem  4.1] it follows that . Then, by Theorem 5. By [1, Theorem  4.1] and Lemma 13, it follows thatwhere is the absolute constant in [1, Theorem 4.1]. Finally, applying again Theorem 5, it follows that The inequality is sharp in the sense that (134) cannot be removed.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors are very grateful to the professors N. Merentes and J. L. Sánchez for useful discussions. They would like to thank Professor V. Kolyada for sharing [15] and the referees by their appropriate comments. This research has been partially supported by the Central Bank of Venezuela. They want to thank the library staff of B.C.V for compiling the references.