Functions of Bounded th -Variation and Continuity Modulus
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 .
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, .
If we get the class of functions of bounded -variation introduced by Wiener . 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 .
The concept of functions of bounded second variation was introduced by de la Vallée Poussin in . By a refined partition we mean a partition of the formMotivated by the result of Riesz, Merentes in  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 .
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 . The spaces form a scale between the space of functions of bounded th -variation and the Sobolev space . In  the authors extend the results of Riesz and Merentes by showing that if and only if .
Terehin studied in  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 : 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
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.
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., ) 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 . 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  that . Then
Theorem 10. Let and be a 1-periodic function.(i)If , then (ii)If and