Hindawi Publishing Corporation Journal of Function Spaces and Applications Volume 2013, Article ID 718507, 12 pages http://dx.doi.org/10.1155/2013/718507
Research Article Functions of Bounded 𝜅𝜑-Variation in the Sense of Riesz-Korenblum Mariela Castillo,1 Sergio Rivas,2 María Sanoja,1 and Iván Zea1 1 2
Escuela de Matem´atica, Universidad Central de Venezuela, Los Chaguaramos, Caracas 1050, Venezuela ´ Area de Matem´atica, Universidad Nacional Abierta, San Bernandino, Caracas 1010, Venezuela
Correspondence should be addressed to Mar´ıa Sanoja; sanoja
[email protected] Received 4 December 2012; Revised 6 February 2013; Accepted 7 February 2013 Academic Editor: J´ozef Bana´s Copyright © 2013 Mariela Castillo et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We present the space of functions of bounded 𝜅𝜑-variation in the sense of Riesz-Korenblum, denoted by 𝜅BV 𝜑 [a,b], which is a combination of the notions of bounded 𝜑-variation in the sense of Riesz and bounded 𝜅-variation in the sense of Korenblum. Moreover, we prove that the space generated by this class of functions is a Banach space with a given norm and we prove that the uniformly bounded composition operator satisfies Matkowski’s weak condition.
1. Introduction The concept of functions of bounded variation has been well known since Jordan [1] gave the complete characterization of functions of a bounded variation as the difference of two increasing functions in 1881. This class of functions immediately proved to be important in connection with the rectification of curves and with Dirichlet’s theorem on the convergence of Fourier series. Functions of a bounded variation exhibit many interesting properties that make them a suitable class of functions in a variety of contexts with wide applications in pure and applied mathematics (see [2–4]). Riesz [5] in 1910 generalized the notion of Jordan and introduced the concept of bounded 𝑝-variation (1 < 𝑝 < ∞) and showed that, for 1 < 𝑝 < ∞, this class coincides with the class of functions absolutely continuous with the derivative in the space 𝐿 𝑝 . On the other hand, this notion of bounded 𝑝-variation was generalized by Medvedev [6] in 1953 who introduced the concept of bounded 𝜑-variation in the sense of Riesz and also showed a Riesz’s lemma for this class of functions. Korenblum [7] in 1975 introduced the notion of bounded 𝜅-variation. This concept differs from others due to the fact that it introduces a distortion function 𝜅 that measures intervals in the domain of the function and not in the range. In 1985, Cyphert and Kelingos [8] showed that a function 𝑢 is of bounded 𝜅-variation if it can be written as the difference
of two 𝜅-decreasing functions. In 1986, S. K. Kim and J. Kim [9] and Park [10], in 2010, introduced the notion of functions of 𝜅𝜙-bounded variation on compact interval [𝑎, 𝑏] ⊂ R which is a combination of concepts of bounded 𝜅-variation and bounded 𝜙-variation in the sense of Schramm [11], and in 2011 Aziz et al. [12] showed that the space of bounded 𝜅variation satisfies Matkowski’s weak condition. Recently in [13] Castillo et al. introduce the notion of bounded 𝜅-variation in the sense of Riesz-Korenblum, which is a combination of the notions of bounded 𝑝-variation in the sense of Riesz and bounded 𝜅-variation in the sense of Korenblum. The purpose of this paper is twofold. First, to introduce the concept of bounded 𝜅𝜑-variation in the sense of Riesz-Korenblum, which is a combination of the notions of bounded 𝜑-variation in the sense of Riesz and bounded 𝜅variation in the sense of Korenblum. We prove some properties of this class of functions and its relation with the functions of bounded 𝜅-variation and bounded 𝜑-variation in the sense of Riesz. Second we prove that the space generated by this class of functions is a Banach space with a given norm and that the uniformly bounded composition operator satisfies Matkowski’s weak condition in this space. The Matkowski property has been studied by several authors (see [14–16]), and for Matkowski’s weak property, see also [3, 17–21]. In [22–24] Matkowski, Merentes, and others authors have been studying a weaker condition on the composition operator
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such as uniformly bounded and uniformly continuous composition operators.
The notion of bounded variation due to Jordan (Definition 1) was generalized by Medvedev (see [6]) as follows.
2. Preliminaries
Definition 4. Let 𝜑 be a 𝜑-function and 𝑢 : [𝑎, 𝑏] → R be a function. For each partition 𝜋 : 𝑎 = 𝑡0 < 𝑡1 < ⋅ ⋅ ⋅ < 𝑡𝑛 = 𝑏 of the interval [𝑎, 𝑏], we define
In this section we present some definitions and preliminary results related to the notion of functions of bounded 𝜅𝜑variation in the sense of Riesz-Korenblum. Definition 1. Let 𝑢 : [𝑎, 𝑏] → R be a function. For each partition 𝜋 : 𝑎 = 𝑡0 < 𝑡1 < ⋅ ⋅ ⋅ < 𝑡𝑛 = 𝑏 of the interval [𝑎, 𝑏], we define 𝑛
𝑉 (𝑢; [𝑎, 𝑏]) := sup∑ 𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) , 𝜋 𝑖=1
(1)
where the supremum is taken over all partitions 𝜋 of the interval [𝑎, 𝑏]. If 𝑉(𝑢; [𝑎, 𝑏]) < ∞, we say that 𝑢 has bounded variation. We denote by 𝐵𝑉[𝑎, 𝑏] the collection of all functions of bounded variation on [𝑎, 𝑏]. Now, we will give some well-known properties of the space of functions 𝐵𝑉[𝑎, 𝑏]. (1) If the function 𝑢 is monotone, then 𝑉(𝑢; [𝑎, 𝑏]) = |𝑢(𝑏) − 𝑢(𝑎)|. (2) If 𝑢 ∈ 𝐵𝑉[𝑎, 𝑏], then 𝑢 is bounded on [𝑎, 𝑏]. (3) A function 𝑢 has bounded variation in an interval [𝑎, 𝑏] if and only if it can be decomposed as a difference of increasing functions. (4) Every function of bounded variation has left- and right-hand limits at each point of its domain. (5) If 𝑢 ∈ Lip [𝑎, 𝑏], then 𝑢 ∈ 𝐵𝑉[𝑎, 𝑏] that is, Lip [𝑎, 𝑏] ⊂ 𝐵𝑉[𝑎, 𝑏]. (6) 𝐵𝑉[𝑎, 𝑏] is a Banach space endowed with the norm ‖𝑢‖𝐵𝑉 = |𝑢 (𝑎)| + 𝑉 (𝑢; [𝑎, 𝑏]) ,
𝑢 ∈ 𝐵𝑉 [𝑎, 𝑏] .
(2)
In 1937, Young (see [25]) introduced the definition of 𝜑function as follows. Definition 2. A function 𝜑 : [0, ∞) → [0, ∞) is said to be a 𝜑-function if it satisfies the following properties.
𝑛 𝑢 (𝑡 ) − 𝑢 (𝑡𝑖−1 ) ) 𝑡𝑖 − 𝑡𝑖−1 , := sup∑𝜑 ( 𝑖 𝜋 𝑖=1 𝑡𝑖 − 𝑡𝑖−1
(4)
where the supremum is taken over all partitions 𝜋 of the interval [𝑎, 𝑏]. If 𝑉𝜑𝑅 (𝑢; [𝑎, 𝑏]) < ∞, we say that 𝑢 has bounded 𝜑-variation in the sense of Riesz. We denote by 𝑉𝜑𝑅 [𝑎, 𝑏] the class of all functions of bounded 𝜑-variation in the sense of Riesz on [𝑎, 𝑏]. Now we give some properties of the class of functions 𝑉𝜑𝑅 [𝑎, 𝑏]. (1) If 𝑢 ∈ Lip [𝑎, 𝑏], then 𝑢 ∈ 𝑉𝜑𝑅 [𝑎, 𝑏]; that is, Lip [𝑎, 𝑏] ⊂ 𝑉𝜑𝑅 [𝑎, 𝑏]. (2) If 𝜑 is convex then 𝑉𝜑𝑅 [𝑎, 𝑏] ⊂ 𝐵𝑉[𝑎, 𝑏] and if lim𝑡 → ∞ (𝜑(𝑡)/𝑡) = 𝛾 < ∞, then 𝑉𝜑𝑅 [𝑎, 𝑏] = 𝐵𝑉[𝑎, 𝑏]. (3) If 𝜑 satisfies the condition ∞1 and 𝑉𝜑𝑅 (𝑢; [𝑎, 𝑏]) < ∞, then 𝑢 is bounded on [𝑎, 𝑏]. (4) 𝑉𝜑𝑅 [𝑎, 𝑏] is a symmetric convex set. (5) If 𝜑 is convex, then 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] is a vector space if and only if 𝜑 satisfies the condition Δ 2 (∞). (6) If 𝜑 is convex, then the set 𝑅𝑉𝜑 [𝑎, 𝑏] = {𝑢 : [𝑎, 𝑏] → R : ∃𝜆 > 0, 𝑉𝜑𝑅 (𝜆𝑢) < ∞} is a Banach space endowed with the norm 𝑢 ||𝑢||𝜑 = |𝑢 (𝑎)| + inf {𝜀 > 0 : 𝑉𝜑 ( ) ≤ 1} . 𝜀
(5)
(7) Let 𝜑 be a convex 𝜑-function such that it satisfies the condition ∞1 . A function 𝑢 has bounded 𝜑-variation in an interval [𝑎, 𝑏] if and only if 𝑢 is absolutely
(a) 𝜑 is continuous on [0, ∞). (b) 𝜑(𝑡) = 0 if and only if 𝑡 = 0.
𝑏
(c) 𝜑 is strictly increasing.
continuous in [𝑎, 𝑏] and ∫𝑎 𝜑(|𝑢 (𝑡)|)𝑑𝑡 < ∞. Also,
(d) lim𝑡 → ∞ 𝜑(𝑡) = ∞. Definition 3 (conditions ∞1 and Δ 2 ). Let 𝜑 be a convex 𝜑function, then (a) 𝜑 satisfies the condition ∞1 if lim𝑡 → ∞ (𝜑(𝑡)/𝑡) = ∞, (b) 𝜑 satisfies the condition Δ 2 (∞) if there is 𝐶 > 0, 𝑥0 > 0 such that 𝜑 (2𝑡) ≤ 𝐶𝜑 (𝑡) ,
𝑉𝜑𝑅 (𝑢) = 𝑉𝜑𝑅 (𝑢; [𝑎, 𝑏])
𝑡 ≥ 𝑥0 .
(3)
𝑏
𝑉𝜑𝑅 (𝑢; [𝑎, 𝑏]) = ∫ 𝜑 (𝑢 (𝑡)) 𝑑𝑡. 𝑎
(6)
Other generalization of the notion of bounded variation was introduced by Korenblum. Korenblum employed a function 𝜅 : [0, 1] → [0, 1] called 𝜅-function. This function 𝜅 can be viewed as a rescaling of lengths of subintervals of [𝑎, 𝑏] such that the length of [𝑎, 𝑏] is 1 if 𝜅(1) = 1.
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Definition 5. A function 𝜅 : [0, 1] → [0, 1] is said to be a 𝜅-function if it satisfies the following properties: (a) 𝜅 is continuous with 𝜅(0) = 0 and 𝜅(1) = 1,
𝜅𝑉𝜑𝑅 (𝑢) = 𝜅𝑉𝜑𝑅 (𝑢; [𝑎, 𝑏])
(b) 𝜅 is concave (down), increasing, and (c) lim𝑡 → 0+ (𝜅(𝑡)/𝑡) = ∞. The set of all 𝜅-functions will be denoted by K. Note that, every 𝜅-function 𝜅 is subadditive; that is, 𝜅 (𝑡1 + 𝑡2 ) ≤ 𝜅 (𝑡1 ) + 𝜅 (𝑡2 ) ,
𝑡1 , 𝑡2 ∈ [0, 1] .
(7)
Then, for all partition 𝜋 : 𝑎 = 𝑡0 < 𝑡1 < ⋅ ⋅ ⋅ < 𝑡𝑛 = 𝑏 of [𝑎, 𝑏], we have 𝑛
𝑛 𝑡 −𝑡 𝑡𝑖 − 𝑡𝑖−1 ) ≤ ∑𝜅 ( 𝑖 𝑖−1 ) . 𝑏−𝑎 𝑖=1 𝑏 − 𝑎 𝑖=1
1 = 𝜅 (1) = 𝜅 (∑
(8)
Korenblum (see [7]) introduces the definition of bounded 𝜅-variation as follows. Definition 6. A real function 𝑢 on [𝑎, 𝑏] is said to be of bounded 𝜅-variation, if 𝜅𝑉 (𝑢) = 𝜅𝑉 (𝑢; [𝑎, 𝑏]) := sup 𝜋
∑𝑛𝑖=1 𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) < ∞, ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎))
Definition 7. Let 𝜑 be a 𝜑-function, 𝜅 ∈ K, and 𝑢 : [𝑎, 𝑏] → R be a function. For each partition 𝜋 : 𝑎 = 𝑡0 < 𝑡1 < ⋅ ⋅ ⋅ < 𝑡𝑛 = 𝑏 of the interval [𝑎, 𝑏], we define
(9)
where the supremum is taken over all partitions 𝜋 of the interval [𝑎, 𝑏]. We denote by 𝜅𝐵𝑉[𝑎, 𝑏] the collection of all functions of bounded 𝜅-variation on [𝑎, 𝑏]. Next, some properties of the space 𝜅𝐵𝑉[𝑎, 𝑏] are exposed (see [8]). (1) If the function 𝑢 is monotone, then 𝜅𝑉(𝑢; [𝑎, 𝑏]) = |𝑢(𝑏) − 𝑢(𝑎)|. (2) If 𝑢 ∈ 𝜅𝐵𝑉[𝑎, 𝑏], then 𝑢 is bounded on [𝑎, 𝑏]. (3) If 𝑢 ∈ 𝐵𝑉[𝑎, 𝑏], then 𝑢 ∈ 𝜅𝐵𝑉[𝑎, 𝑏]; that is, 𝐵𝑉[𝑎, 𝑏] ⊂ 𝜅𝐵𝑉[𝑎, 𝑏].
:= sup 𝜋
∑𝑛𝑖=1 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 , ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) (11)
where the supremum is taken over all partitions 𝜋 of the interval [𝑎, 𝑏]. If 𝜅𝑉𝜑𝑅 (𝑢; [𝑎, 𝑏]) < ∞, we say that 𝑢 has bounded 𝜅𝜑-variation in the sense of Riesz-Korenblum. We will denote by 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] the class of all functions of bounded 𝜅𝜑-variation in the sense of Riesz-Korenblum on [𝑎, 𝑏]. Remark 8. Note that the class 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] is not empty since for an affine function 𝑢 : [𝑎, 𝑏] → R is defined by 𝑢(𝑡) := 𝑑𝑡 + 𝑒, 𝑡 ∈ [𝑎, 𝑏], where 𝑑, 𝑒 are fixed real numbers. For a given partition 𝜋 : 𝑎 = 𝑡0 < 𝑡1 < ⋅ ⋅ ⋅ < 𝑡𝑛 = 𝑏 of [𝑎, 𝑏] we have ∑𝑛𝑖=1 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) 𝜑 (|𝑑|) (𝑏 − 𝑎) = 𝑛 . ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎))
(12)
Taking the supremum over all partitions 𝜋 of the interval [𝑎, 𝑏], the greater value of the right side of the above expression is obtain for the partition 𝜋 : 𝑎 = 𝑡0 < 𝑡1 = 𝑏 and in this case we get 𝜑 (|𝑑|) (𝑏 − 𝑎) 𝜑 (|𝑑|) (𝑏 − 𝑎) = = 𝜑 (|𝑑|) (𝑏 − 𝑎) . 𝜅 (1) 𝜅 ((𝑡1 − 𝑡0 ) / (𝑏 − 𝑎)) (13) Therefore, 𝜅𝑉𝜑𝑅 (𝑢; [𝑎, 𝑏]) = 𝜑 (|𝑑|) (𝑏 − 𝑎) .
(14)
(4) A function 𝑢 has bounded 𝜅-variation in an interval [𝑎, 𝑏] if and only if it can be decomposed as a difference of 𝜅-decreasing functions.
In the following proposition, we prove two important properties of the space 𝜅𝑅𝑉𝜑 [𝑎, 𝑏].
(5) Every function of bounded 𝜅-variation has left- and right-hand limits at each point of its domain.
Proposition 9. Let 𝜑 be a convex 𝜑-function, then
(6) 𝜅𝐵𝑉[𝑎, 𝑏] is a Banach space endowed with the norm
(a) 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] ⊂ 𝜅𝐵𝑉[𝑎, 𝑏].
||𝑢||𝜅 = |𝑢 (𝑎)| + 𝜅𝑉 (𝑢; [𝑎, 𝑏]) ,
(b) If lim𝑡 → ∞ (𝜑(𝑡)/𝑡) = 𝛾 < ∞, then 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] = 𝜅𝐵𝑉[𝑎, 𝑏].
𝑢 ∈ 𝜅𝐵𝑉 [𝑎, 𝑏] .
(10)
3. Main Results In this section we present the principal results of this paper. Next, we introduce the definition of function of bounded 𝜅𝜑variation in the sense of Riesz-Korenblum for the function 𝑢 : [𝑎, 𝑏] → R.
Proof. (a) Let 𝑢 ∈ 𝜅𝑉𝜑𝑅 [𝑎, 𝑏], 𝜋 : 𝑎 = 𝑡0 < 𝑡1 < ⋅ ⋅ ⋅ < 𝑡𝑛 = 𝑏, be a partition of the interval [𝑎, 𝑏] and 𝑢 (𝑡 ) − 𝑢 (𝑡𝑖−1 ) ≤ 1} . 𝜎 := {𝑖 = 1, . . . , 𝑛 : 𝑖 𝑡𝑖 − 𝑡𝑖−1
(15)
4
Journal of Function Spaces and Applications Since 𝜑 is a convex 𝜑-function and 𝜑(0) = 0, we have 𝑡 1 𝜑 (1) = 𝜑 ( ) ≤ 𝜑 (𝑡) , 𝑡 𝑡
𝑡 ≥ 1,
Now, we will show part (b). If
𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) ≤ 𝜑 (1) 𝑡𝑖 − 𝑡𝑖−1
𝜑 (𝑡) ≤ 𝑐𝑡, (17)
multiplied by |𝑡𝑖 − 𝑡𝑖−1 | and applying the sum on both sides of the above inequality, we have 𝑛 𝑢 (𝑡 ) − 𝑢 (𝑡𝑖−1 ) ∑ 𝑖 𝑡𝑖 − 𝑡𝑖−1 𝑡 − 𝑡 𝑖 𝑖−1 𝑖∉𝜎 𝑛 𝑢 (𝑡 ) − 𝑢 (𝑡𝑖−1 ) 1 ) 𝑡𝑖 − 𝑡𝑖−1 , ≤ ∑𝜑 ( 𝑖 𝜑 (1) 𝑖∉𝜎 𝑡𝑖 − 𝑡𝑖−1
(18)
∑𝑛𝑖=1 𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑛 (𝑢 (𝑡 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 = 𝑖=1 𝑛𝑖 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑ (𝑢 (𝑡 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 = 𝑖∈𝜎 𝑛𝑖 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑖∉𝜎 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 + ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑖∈𝜎 𝑡𝑖 − 𝑡𝑖−1 ≤ 𝑛 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) 1 ∑𝑖∉𝜎 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 + . 𝜑 (1) ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) 1 (𝑏 − 𝑎) + 𝜅𝑉𝑅 (𝑢) 𝜑 (1) 𝜑 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎))
≤ (𝑏 − 𝑎) +
1 𝜅𝑉𝑅 (𝑢) . 𝜑 (1) 𝜑 (19)
≤ (𝑏 − 𝑎) +
1 𝜅𝑉𝑅 (𝑢) . 𝜑 (1) 𝜑 (20)
Considering the supremum over all partitions 𝜋 of the interval [𝑎, 𝑏] in the above expression, we get 𝜅𝑉 (𝑢) ≤ 𝑏 − 𝑎 + therefore
𝜅𝑉𝜑𝑅 [𝑎, 𝑏]
(23)
Let us consider the partition 𝜋 : 𝑎 = 𝑡0 < 𝑡1 < ⋅ ⋅ ⋅ < 𝑡𝑛 = 𝑏 of the interval [𝑎, 𝑏], 𝑢 ∈ 𝜅𝐵𝑉[𝑎, 𝑏], and 𝑢 (𝑡 ) − 𝑢 (𝑡𝑖−1 ) (24) > 𝑥0 } , 𝛼𝑥0 := {𝑖 = 1, . . . , 𝑛 : 𝑖 𝑡𝑖 − 𝑡𝑖−1 ∑𝑛𝑖=1 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑖∈𝛼𝑥 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡i−1 0 = ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑖∉𝛼𝑥 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 0 + ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑖∈𝛼𝑥 𝑐 𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) ≤ 𝑛 0 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑖∉𝛼𝑥 𝜑 (𝑥0 ) 𝑡𝑖 − 𝑡𝑖−1 0 + 𝑛 , ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎))
(25)
thus, ∑𝑛𝑖=1 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) 𝑐 ∑𝑖∈𝛼𝑥 𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) 0 ≤ 𝑛 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) 𝜑 (𝑥0 ) ∑𝑖∉𝛼𝑥 𝑡𝑖 − 𝑡𝑖−1 0 + 𝑛 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ≤ 𝑐 ⋅ 𝜅𝑉 (𝑢) + 𝜑 (𝑥0 )
∑𝑛𝑖=1
(26)
(𝑏 − 𝑎) 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎))
≤ 𝑐 ⋅ 𝜅𝑉 (𝑢) + 𝜑 (𝑥0 ) (𝑏 − 𝑎) .
Then ∑𝑛𝑖=1 𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎))
𝑡 ≥ 𝑥0 .
then
then
∑𝑛𝑖=1
(22)
then there exist 𝑥0 > 0 and 𝑐 > 0, such that
so 𝑡 ≤ 𝜑(𝑡)/𝜑(1). Hence, for 𝑖 ∉ 𝜎 we get
≤
𝜑 (𝑡) = 𝛾 < ∞, 𝑡→∞ 𝑡
0 < lim
(16)
1 𝜅𝑉𝑅 (𝑢) , 𝜑 (1) 𝜑
⊂ 𝜅𝐵𝑉[𝑎, 𝑏].
(21)
Then by considering the supremum over all partitions 𝜋 of the interval [𝑎, 𝑏] of the left side, we get 𝜅𝑉𝜑𝑅 (𝑢) ≤ 𝑐 ⋅ 𝜅𝑉 (𝑢) + 𝜑 (𝑥0 ) (𝑏 − 𝑎) ,
(27)
𝜅𝐵𝑉 [𝑎, 𝑏] ⊂ 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] .
(28)
that is,
Therefore, from part (a) and (28) we have 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] = 𝜅𝐵𝑉[𝑎, 𝑏].
Journal of Function Spaces and Applications
5
Proposition 10. Let Lip [𝑎, 𝑏] be the Banach space of all Lipschitz functions 𝑢 : [𝑎, 𝑏] → R. Then Lip [𝑎, 𝑏] ⊂ 𝜅𝑉𝜑𝑅 [𝑎, 𝑏]. Proof. Let 𝜋 : 𝑎 = 𝑡0 < 𝑡1 < ⋅ ⋅ ⋅ < 𝑡𝑛 = 𝑏 be a partition of the interval [𝑎, 𝑏] and 𝑢 ∈ Lip [𝑎, 𝑏], then there exists 𝑐 ∈ R such that for any 𝑥, 𝑦 ∈ [𝑎, 𝑏] we have |𝑢(𝑥) − 𝑢(𝑦)| ≤ 𝑐|𝑥 − 𝑦|. Hence, ∑𝑛𝑖=1 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑛 𝜑 (𝑐 𝑡 − 𝑡 / 𝑡 − 𝑡 ) 𝑡 − 𝑡 ≤ 𝑖=1 𝑛 𝑖 𝑖−1 𝑖 𝑖−1 𝑖 𝑖−1 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) (29) ∑𝑛 𝜑 (𝑐) 𝑡𝑖 − 𝑡𝑖−1 = 𝑛 𝑖=1 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) 𝜑 (𝑐) (𝑏 − 𝑎) = 𝑛 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ≤ 𝜑 (𝑐) (𝑏 − 𝑎) , considering the supremum over all partitions 𝜋 of the interval [𝑎, 𝑏], we get 𝜅𝑉𝜑𝑅 (𝑢) ≤ 𝜑 (𝑐) (𝑏 − 𝑎) < ∞.
(30)
Therefore, Lip [𝑎, 𝑏] ⊂ 𝜅𝑉𝜑𝑅 [𝑎, 𝑏]. The class of functions of a bounded 𝜅𝜑-variation has many interesting properties as the following proposition showes. Proposition 11. Let 𝜑 be a 𝜑-function, 𝜅 ∈ K and 𝑢 : [𝑎, 𝑏] → R be a function, then (a) The function 𝜅𝑉𝜑𝑅 (⋅) : 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] → R is an even function, that is, 𝜅𝑉𝜑𝑅 (𝑢) = 𝜅𝑉𝜑𝑅 (−𝑢). (b) 𝜅𝑉𝜑𝑅 (𝑢) = 0 if and only if 𝑢 is constant. (c) If 𝜅𝑉𝜑𝑅 (𝑢) < ∞ then 𝑢 is bounded on [𝑎, 𝑏]. (d) 𝜑 is convex if and only if the function 𝜅𝑉𝜑𝑅 (⋅) : 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] → [0, ∞) defined by 𝜅𝑉𝜑𝑅 (𝑢)
:=
𝜅𝑉𝜑𝑅 (𝑢; [𝑎, 𝑏]) ,
𝑢∈
𝜅𝑉𝜑𝑅 [𝑎, 𝑏]
(b) If 𝑢 is constant, then 𝜅𝑉𝜑𝑅 (𝑢) = 0. Now if 𝜅𝑉𝜑𝑅 (𝑢) = 0, we get that 0 = 𝜅𝑉𝜑𝑅 (𝑢) ≥
𝜑 (|𝑢 (𝑡) − 𝑢 (𝑎)| / |𝑡 − 𝑎|) |𝑡 − 𝑎| , 𝜅 ((𝑡 − 𝑎) / (𝑏 − 𝑎)) + 𝜅 ((𝑏 − 𝑡) / (𝑏 − 𝑎)) (33)
for some 𝑡 ∈ (𝑎, 𝑏]. From Definitions 2 and 5, we have 0≥
𝜑 (|𝑢 (𝑡) − 𝑢 (𝑎)| / |𝑡 − 𝑎|) |𝑡 − 𝑎| ≥ 0, 𝜅 ((𝑡 − 𝑎) / (𝑏 − 𝑎)) + 𝜅 ((𝑏 − 𝑡) / (𝑏 − 𝑎))
hence 𝜑(
|𝑢 (𝑡) − 𝑢 (𝑎)| ) |𝑡 − 𝑎| = 0, |𝑡 − 𝑎|
𝑡 ∈ (𝑎, 𝑏] .
Proof. (a) From Definition 4, we have 𝜅𝑉𝜑𝑅 (𝑢) ∑𝑛𝑖=1 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 𝑛 𝜋 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑛 𝜑 (−𝑢 (𝑡𝑖 ) − (−𝑢 (𝑡𝑖−1 )) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 = sup 𝑖=1 𝜋 ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎))
= sup
= 𝜅𝑉𝜑𝑅 (−𝑢) . (32)
(35)
Since 𝜑(0) = 0 if and only if 𝑡 = 0, we get |𝑢 (𝑡) − 𝑢 (𝑎)| = 0, |𝑡 − 𝑎|
(36)
therefore, 𝑢(𝑡) = 𝑢(𝑎) 𝑡 ∈ [𝑎, 𝑏]. So, 𝑢 is constant. (c) Suppose that 𝑢 ∈ 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] is unbounded on [𝑎, 𝑏]. Then, there exists a sequence {𝑡𝑛 }𝑛≥1 , 𝑡𝑛 ∈ [𝑎, 𝑏], for all 𝑛 ≥ 1, such that lim𝑛 → ∞ |𝑢(𝑡𝑛 )| = ∞. Let {𝑡𝑛𝑘 }𝑛𝑘 ≥1 be a subsequence of {𝑡𝑛 }𝑛≥ 1 such that {𝑡𝑛𝑘 }𝑛𝑘 ≥1 converges to point 𝑥 ∈ [𝑎, 𝑏]. Then, {𝑢(𝑡𝑛𝑘 )}𝑛𝑘 ≥1 is a subsequence of {𝑢(𝑡𝑛 )}𝑛≥1 . So, lim𝑘 → ∞ |𝑢(𝑡𝑛𝑘 )| = ∞. Case 1. Suppose that 𝑥 ≠ 𝑎. Since 𝜑 (𝑢 (𝑡𝑛𝑘 ) − 𝑢 (𝑎) / 𝑡𝑛𝑘 − 𝑎) 𝑡𝑛𝑘 − 𝑎 𝜅 ((𝑡𝑛𝑘 − 𝑎) / (𝑏 − 𝑎)) + 𝜅 ((𝑏 − 𝑡𝑛𝑘 ) / (𝑏 − 𝑎))
(37)
≤ 𝜅𝑉𝜑𝑅 (𝑢) < ∞, for all 𝑛𝑘 ≥ 1 and since 𝜑 is continuous, we have 𝜑 (lim𝑘 → ∞ (𝑢 (𝑡𝑛𝑘 ) − 𝑢 (𝑎) / |𝑥 − 𝑎|)) |𝑥 − 𝑎| 𝜅 ((𝑥 − 𝑎) / (𝑏 − 𝑎)) + 𝜅 ((𝑏 − 𝑥) / (𝑏 − 𝑎)) 𝜑 (𝑢 (𝑡𝑛𝑘 ) − 𝑢 (𝑎) / 𝑡𝑛𝑘 − 𝑎) 𝑡𝑛𝑘 − 𝑎 = lim 𝑘 → ∞ 𝜅 ((𝑡 − 𝑎) / (𝑏 − 𝑎)) + 𝜅 ((𝑏 − 𝑡 ) / (𝑏 − 𝑎)) 𝑛𝑘 𝑛𝑘 ≤ 𝜅𝑉𝜑𝑅 (𝑢) . (38)
(31)
is convex.
(34)
On the other hand, |𝑢(𝑡𝑛𝑘 )−𝑢(𝑎)| tends to infinity as 𝑘 → ∞. Then, since 𝜑(𝑡) → ∞ as 𝑡 → ∞, we get 𝜑 (𝑢 (𝑡𝑛𝑘 ) − 𝑢 (𝑎) / |𝑥 − 𝑎|) |𝑥 − 𝑎| = ∞, (39) lim 𝑘 → ∞ 𝜅 ((𝑥 − 𝑎) / (𝑏 − 𝑎)) + 𝜅 ((𝑏 − 𝑥) / (𝑏 − 𝑎)) then 𝜅𝑉𝜑𝑅 (𝑢) = ∞, which is a contradiction. Case 2. Suppose that 𝑥 = 𝑎. Then, since 𝜑 (𝑢 (𝑏) − 𝑢 (𝑡𝑛𝑘 ) / 𝑏 − 𝑡𝑛𝑘 ) 𝑏 − 𝑡𝑛𝑘 𝜅 ((𝑡𝑛𝑘 − 𝑎) / (𝑏 − 𝑎)) + 𝜅 ((𝑏 − 𝑡𝑛𝑘 ) / (𝑏 − 𝑎)) ≤ 𝜅𝑉𝜑𝑅 (𝑢) < ∞,
(40)
6
Journal of Function Spaces and Applications 𝑛 (𝛼𝑢 + 𝛽V) (𝑡𝑖 ) − (𝛼𝑢 + 𝛽V) (𝑡𝑖−1 ) ) = sup (∑ 𝜑 ( 𝜋 𝑡𝑖 − 𝑡𝑖−1 𝑖=1
for all 𝑛𝑘 ≥ 1, and since 𝜑 is continuous, we have 𝜑 (lim𝑘 → ∞ 𝑢 (𝑏) − 𝑢 (𝑡𝑛𝑘 ) / |𝑏 − 𝑥|) |𝑏 − 𝑥|
𝑛
𝜅 ((𝑥 − 𝑎) / (𝑏 − 𝑎)) + 𝜅 ((𝑏 − 𝑥) / (𝑏 − 𝑎)) 𝜑 (𝑢 (𝑏) − 𝑢 (𝑡𝑛𝑘 ) / 𝑏 − 𝑡𝑛𝑘 ) 𝑏 − 𝑡𝑛𝑘 = lim 𝑘 → ∞ 𝜅 ((𝑡 − 𝑎) / (𝑏 − 𝑎)) + 𝜅 ((𝑏 − 𝑡 ) / (𝑏 − 𝑎)) 𝑛𝑘 𝑛𝑘 ≤
Since 𝜑(𝑡) → ∞ as 𝑡 → ∞ and lim𝑘 → ∞ |𝑢(𝑏)−𝑢(𝑡𝑛𝑘 )| = ∞, we get
lim
𝑖=1
= 𝜅𝑉𝜑𝑅 (𝛼𝑢 + 𝛽V) . (44)
𝜅𝑉𝜑𝑅 (𝑢) . (41)
𝜑 (𝑢 (𝑏) − 𝑢 (𝑡𝑛𝑘 ) / |𝑏 − 𝑥|) |𝑏 − 𝑥|
𝑘→∞
−1
𝑡 −𝑡 × 𝑡𝑖 − 𝑡𝑖−1 ) (∑ 𝜅 ( 𝑖 𝑖−1 )) 𝑏−𝑎
𝜅 ((𝑥 − 𝑎) / (𝑏 − 𝑎)) + 𝜅 ((𝑏 − 𝑥) / (𝑏 − 𝑎))
= ∞, (42)
so 𝜅𝑉𝜑𝑅 = ∞ which is a contradiction. In both cases a contradiction is reached. Hence 𝑢 is bounded. (d) Let 𝑢, V : [𝑎, 𝑏] → R and 𝛼, 𝛽 ∈ [0, 1] such that 𝛼 + 𝛽 = 1. Suppose that 𝜑 is convex, then 𝛼𝜅𝑉𝜑𝑅 (𝑢) + 𝛽𝜅𝑉𝜑𝑅 (V)
So 𝜅𝑉𝜑𝑅 (𝛼𝑢 + 𝛽V) ≤ 𝛼𝜅𝑉𝜑𝑅 (𝑢) + 𝛽𝜅𝑉𝜑𝑅 (V) for all 𝛼, 𝛽 ∈ [0, 1] such that 𝛼 + 𝛽 = 1, and therefore 𝜅𝑉𝜑𝑅 (⋅) is convex. In order to prove the order direction, let us consider the following functions. 𝑢 : [𝑎, 𝑏] → R defined by 𝑢(𝑡) := 𝑥𝑡 and V : [𝑎, 𝑏] → R defined by V(𝑡) := 𝑦𝑡, 𝑡 ∈ [𝑎, 𝑏]. Therefore, we have 𝜅𝑉𝜑𝑅 (𝑢) = 𝜑 (𝑥) (𝑏 − 𝑎) ,
𝜅𝑉𝜑𝑅 (V) = 𝜑 (𝑦) (𝑏 − 𝑎) ,
𝜅𝑉𝜑𝑅 (𝛼𝑢 + 𝛽V) = 𝜑 (𝛼𝑥 + 𝛽𝑦) (𝑏 − 𝑎) .
(45)
Since 𝜅𝑉𝜑𝑅 (⋅) is convex, we get 𝜑 (𝛼𝑥 + 𝛽𝑦) (𝑏 − 𝑎) = 𝜅𝑉𝜑𝑅 (𝛼𝑢 + 𝛽V)
∑𝑛𝑖=1 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 𝜋 ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑛 𝜑 (V (𝑡 ) − V (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 + 𝛽sup 𝑖=1 𝑛 𝑖 𝜋 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎))
≤ 𝛼𝜅𝑉𝜑𝑅 (𝑢) + 𝛽𝜅𝑉𝜑𝑅 (V)
= 𝛼sup
𝑢 (𝑡 ) − 𝑢 (𝑡𝑖−1 ) ) 𝑡𝑖 − 𝑡𝑖−1 ≥ sup (∑ [𝛼𝜑 ( 𝑖 𝜋 𝑡𝑖 − 𝑡𝑖−1 𝑖=1 V (𝑡 ) − V (𝑡𝑖−1 ) ) 𝑡 − 𝑡 ]) +𝛽𝜑 ( 𝑖 𝑖 𝑖−1 𝑡𝑖 − 𝑡𝑖−1
(46)
= 𝛼𝜑 (𝑥) (𝑏 − 𝑎) + 𝛽𝜑 (𝑦) (𝑏 − 𝑎) , which implies that
𝑛
𝑛
× (∑ 𝜅 ( 𝑖=1
𝜑 (𝛼𝑥 + 𝛽𝑦) ≤ 𝛼𝜑 (𝑥) + 𝛽𝜑 (𝑦) .
(47)
Therefore, 𝜑 is convex, and the proof of the theorem is completed.
−1
𝑡𝑖 − 𝑡𝑖−1 )) , 𝑏−𝑎
Remark 12. The part (c) of Proposition 11 is a consequence of the part (a) of Proposition 9 if the 𝜑-function is convex. (43)
where the supremum is taken over all partitions 𝜋 of the interval [𝑎, 𝑏]. Since 𝜑 is convex, we have
Proposition 13. Let 𝜑 be a 𝜑-function, 𝜅 ∈ K, and 𝑢, V : [𝑎, 𝑏] → R be functions, then 𝜅𝑉𝜑𝑅 (𝛼𝑢 + 𝛽V) ≤ 𝜅𝑉𝜑𝑅 (𝑢) + 𝜅𝑉𝜑𝑅 (V) 𝛼, 𝛽 ∈ [0, 1] , 𝛼 + 𝛽 = 1.
𝛼𝜅𝑉𝜑𝑅 (𝑢) + 𝛽𝜅𝑉𝜑𝑅 (V) 𝑛 𝛼 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 )) + 𝛽 (V (𝑡𝑖 ) − V (𝑡𝑖−1 )) ) ≥ sup (∑ 𝜑 ( 𝜋 𝑡𝑖 − 𝑡𝑖−1 𝑖=1 𝑛 𝑡 −𝑡 × 𝑡𝑖 − 𝑡𝑖−1 ) (∑ 𝜅 ( 𝑖 𝑖−1 )) 𝑏−𝑎 𝑖=1
(48)
Proof. Let 𝛼, 𝛽 ∈ [0, 1] such that 𝛼 + 𝛽 = 1 and 𝑥, 𝑦 ∈ [0, ∞). Since 𝜑 is nondecreasing and nonnegative and 𝛼𝑥 + 𝛽𝑦 is one of the segment joining point 𝑥 with 𝑦, then we have
−1
𝜑 (𝛼𝑥 + 𝛽𝑦) ≤ max {𝜑 (𝑥) , 𝜑 (𝑦)} ≤ 𝜑 (𝑥) + 𝜑 (𝑦) .
(49)
Journal of Function Spaces and Applications
7 Also, if 𝑥, 𝑦 ∈ 𝐴 and 𝛼 > 0, 𝛽 > 0, then
From the inequality above, we deduce that 𝜅𝑉𝜑𝑅 (𝑢)
+
𝜅𝑉𝜑𝑅 (V)
𝛼𝑥 + 𝛽𝑦 = (𝛼 + 𝛽) (
∑𝑛𝑖=1 𝜑 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 𝑛 𝜋 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) ∑𝑛 𝜑 (V (𝑡 ) − V (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 + sup 𝑖=1 𝑛 𝑖 𝜋 ∑𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎))
= sup
𝑛 𝛼 (𝑢 (𝑡𝑖 ) − 𝑢 (𝑡𝑖−1 )) + 𝛽 (V (𝑡𝑖 ) − V (𝑡𝑖−1 )) ) ≥ sup (∑ 𝜑 ( 𝜋 𝑡𝑖 − 𝑡𝑖−1 𝑖=1 −1
𝑛 𝑡 −𝑡 × 𝑡𝑖 − 𝑡𝑖−1 ) (∑ 𝜅 ( 𝑖 𝑖−1 )) , 𝑏−𝑎 𝑖=1
(50) +
Therefore, ⋃𝜆>0 𝜆𝐴 is a vector space. Let us now prove that ⋃𝜆>0 𝜆𝐴 = {𝑥 ∈ 𝑋 : ∃𝜆 > 0, 𝜆𝑥 ∈ 𝐴}. Indeed, if 𝑧 = 𝜆𝑥, 𝑥 ∈ 𝐴, 𝜆 > 0, then 1/𝑧 = 𝑥 ∈ 𝐴. Conversely, if 𝜆𝑥 ∈ 𝐴, for some 𝜆 > 0, then 𝑥 = (1/𝜆)(𝜆𝑥) ∈ (1/𝜆)𝐴. As a consequence of Lemma 15 and since 𝜅𝑉𝜑𝑅 is a convex and symmetric set, we have the following corollary. Corollary 16. Let 𝜑 be a 𝜑-function and 𝜅 ∈ K. Then, the vector space generated by the class 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] is equal to the following 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] = {𝑢 : [𝑎, 𝑏] → R : ∃𝜆 > 0, 𝜅𝑉𝜑𝑅 (𝜆𝑢) < ∞} . (55)
thus 𝜅𝑉𝜑𝑅 (𝑢)
𝛽 𝛼 𝑥+ 𝑦) ∈ (𝛼 + 𝛽) 𝐴. (54) 𝛼+𝛽 𝛼+𝛽
𝜅𝑉𝜑𝑅 (V)
𝑛 (𝛼𝑢 + 𝛽V) (𝑡𝑖 ) − (𝛼𝑢 + 𝛽V) (𝑡𝑖−1 ) ) = sup (∑ 𝜑 ( 𝜋 𝑡𝑖 − 𝑡𝑖−1 𝑖=1 𝑛
𝑡 −𝑡 × 𝑡𝑖 − 𝑡𝑖−1 ) (∑ 𝜅 ( 𝑖 𝑖−1 )) 𝑏−𝑎 𝑖=1
−1
Theorem 17. Let 𝜑 be a 𝜑-function and 𝜅 ∈ K, then (51)
= 𝜅𝑉𝜑𝑅 (𝛼𝑢 + 𝛽V) .
𝑅𝑉𝜑 [𝑎, 𝑏] ⊂ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] ⊂ 𝜅𝐵𝑉 [𝑎, 𝑏] .
Proof. First we prove that 𝑅𝑉𝜑 [𝑎, 𝑏] ⊂ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏]. Let 𝑢 ∈ 𝑅𝑉𝜑 [𝑎, 𝑏], then there exists 𝜆 > 0 such that 𝑉𝜑𝑅 (𝜆𝑢) < ∞ and 𝜋 : 𝑎 = 𝑡0 < 𝑡1 < ⋅ ⋅ ⋅ < 𝑡𝑛 = 𝑏 is a partition of the interval [𝑎, 𝑏], then from inequality (8) we have 𝑛 𝜆𝑢 (𝑡𝑖 ) − 𝜆𝑢 (𝑡𝑖−1 ) ) 𝑡𝑖 − 𝑡𝑖−1 ∑𝜑 ( 𝑡 − 𝑡 𝑖 𝑖−1 𝑖=1
Therefore, 𝜅𝑉𝜑𝑅 (𝛼𝑢 + 𝛽V) ≤ 𝜅𝑉𝜑𝑅 (𝑢) + 𝜅𝑉𝜑𝑅 (V) for all 𝛼, 𝛽 ∈ [0, 1] such that 𝛼 + 𝛽 = 1. Remark 14. From Propositions 11 and 13, we have 𝜅𝑉𝜑𝑅 [𝑎, 𝑏] to be convex and symmetric.
≤ 𝑉𝜑𝑅 (𝜆𝑢) 𝑛
𝑡𝑖 − 𝑡𝑖−1 ) 𝑖=1 𝑏 − 𝑎
= 𝑉𝜑𝑅 (𝜆𝑢) 𝜅 (∑
The following lemma allows us to give a characterization of the space 𝜅𝑅𝑉𝜑 [𝑎, 𝑏].
𝑛
Lemma 15. Let 𝑋 be a vector space and 𝐴 ⊂ 𝑋 a nonempty convex and symmetric set. Then one has the following. (a) 0 ∈ 𝐴. (b) The vector space associated by 𝐴 is iqual to: ⟨𝐴⟩ = {𝑥 ∈ 𝑋 : ∃𝜆 > 0, 𝜆𝑥 ∈ 𝐴} = ⋃ 𝜆𝐴. 𝜆>0
(52)
𝛼 > 0,
𝜆>0
𝛼 (𝜆𝑥) = 0 ∈ ⋃ 𝜆𝐴,
𝛼 = 0,
𝜆>0
𝛼 (𝜆𝑥) = (−𝛼𝜆) (−𝑥) ∈ ⋃ 𝜆𝐴, 𝜆>0
𝛼 < 0.
≤ 𝑉𝜑𝑅 (𝜆𝑢) ∑𝜅 ( 𝑖=1
(57)
𝑡𝑖 − 𝑡𝑖−1 ). 𝑏−𝑎
Thus, 𝜑 (𝜆𝑢 (𝑡𝑖 ) − 𝜆𝑢 (𝑡𝑖−1 ) / 𝑡𝑖 − 𝑡𝑖−1 ) 𝑡𝑖 − 𝑡𝑖−1 ≤ 𝑉𝜑𝑅 (𝜆𝑢) . ∑𝑛𝑖=1 𝜅 ((𝑡𝑖 − 𝑡𝑖−1 ) / (𝑏 − 𝑎)) (58)
∑𝑛𝑖=1
Proof. (a) For all 𝑥 ∈ 𝐴, we have that −𝑥 ∈ 𝐴 0 = (1/2)𝑥 + (1/2)(−𝑥) ∈ 𝐴. (b) By definition, 𝐴 ⊂ ⋃𝜆>0 𝜆𝐴 ⊂ ⟨𝐴⟩. In order to show the other inclusion, we have to prove that ⋃𝜆>0 𝜆𝐴 is a vector space. Indeed, if 𝛼 ∈ R, we get 𝛼 (𝜆𝑥) ∈ ⋃ 𝜆𝐴,
(56)
(53)
Then, considering the supremum of the left side, we get 𝜅𝑉𝜑𝑅 (𝜆𝑢) ≤ 𝑉𝜑𝑅 (𝜆𝑢) ,
(59)
therefore, 𝑢 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] and 𝑅𝑉𝜑 [𝑎, 𝑏] ⊂ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏]. On the other hand, by part (a) of Proposition 9 we get that 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] ⊂ 𝜅𝐵𝑉[𝑎, 𝑏]. Theorem 18. Let 𝜑 be a convex 𝜑-function, then Λ = {𝑢 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] : 𝜅𝑉𝜑𝑅 (𝑢) ≤ 1} is a symmetric convex absorbent subset of 𝜅𝑅𝑉𝜑 [𝑎, 𝑏].
(60)
8
Journal of Function Spaces and Applications
Proof. First we show the convexity. Let 𝛼, 𝛽 ∈ [0, 1] such that 𝛼 + 𝛽 = 1 and 𝑢, V ∈ Λ. Then, by Proposition 11 we get 𝜅𝑉𝜑𝑅 (𝛼𝑢 + 𝛽V) ≤ 𝛼𝜅𝑉𝜑𝑅 (𝑢) + 𝛽𝜅𝑉𝜑𝑅 (V) ≤ 𝛼 ⋅ 1 + 𝛽 ⋅ 1 = 1, (61)
(c) ||𝑢 + V||𝜅𝜑 = |(𝑢 + V) (𝑎)| + 𝜇Λ (𝑢 + V) = |(𝑢) (𝑎) + V (𝑎)| + 𝜇Λ (𝑢 + V) ≤ |𝑢 (𝑎)| + |V (𝑎)| + 𝜇Λ (𝑢) + 𝜇Λ (V)
thus 𝛼𝑢 + 𝛽V ∈ Λ. Now let 𝑢 ∈ Λ and 0 < |𝜆| ≤ 1. If 0 < 𝜆 ≤ 1 by Proposition 11, we have 𝜅𝑉𝜑𝑅 (𝜆𝑢) ≤ 𝜆𝜅𝑉𝜑𝑅 (𝑢) ≤ 𝜆 ⋅ 1 ≤ 1.
(62)
For the case −1 ≤ 𝜆 < 0, by the symmetric and convexity of the functional 𝜅𝑉𝜑𝑅 (⋅) given in Proposition 11 we get that 𝜅𝑉𝜑𝑅 (𝜆𝑢) = 𝜅𝑉𝜑𝑅 (−𝜆 (−𝑢)) ≤ −𝜆𝜅𝑉𝜑𝑅 (−𝑢) = |𝜆| 𝜅𝑉𝜑𝑅 (𝑢) . (63) Hence, we have shown that Λ is balance. Now we will show that Λ is absorbent. Let 𝑢 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] then there exist 𝛼 > 0 such that 𝜅𝑉𝜑𝑅 (𝛼𝑢) < ∞. If 𝜅𝑉𝜑𝑅 (𝛼𝑢) ≤ 1 then 𝛼𝑢 ∈ Λ. On the other hand, if 𝜅𝑉𝜑𝑅 (𝛼𝑢) > 1 we have 𝜅𝑉𝜑𝑅 (
𝛼𝑢 𝜅𝑉𝜑𝑅 (𝛼𝑢)
)≤
1 𝜅𝑉𝜑𝑅 (𝛼𝑢)
𝜅𝑉𝜑𝑅 (𝛼𝑢) = 1,
(64)
in this case, (𝛼/𝜅𝑉𝜑𝑅 (𝛼𝑢))𝑢 ∈ Λ. Hence, Λ is absorbent. Remark 19. As a consequence of Theorem 18, the Minkowski functional associated to the set Λ defines a seminorm on 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] and is defined by 𝑢 𝜇Λ (𝑢) = inf {𝜆 > 0 : 𝜅𝑉𝜑𝑅 ( ) ≤ 1} . 𝜆
(65)
Theorem 20. Let 𝜑 be a convex 𝜑-function. Then, (𝜅𝑅𝑉𝜑 [𝑎, 𝑏], || ⋅ ||𝜅𝜑 ), where the functional || ⋅ ||𝜅𝜑 : 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] → R, defined by ||𝑢||𝜅𝜑 = |𝑢 (𝑎)| + 𝜇Λ (𝑢) ,
𝑢 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] ,
(66)
is a normed space. Proof. Let 𝑢, V ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏], 𝛼 ∈ R. Then, we have the following. (a) ||𝑢||𝜅𝜑 ≥ 0 since |𝑢(𝑎)| ≥ 0 and 𝜇Λ (𝑢) ≥ 0. ||𝛼𝑢||𝜅𝜑 = |𝛼𝑢 (𝑎)| + |𝛼| 𝜇Λ (𝑢)
= |𝛼| (|𝑢 (𝑎)| + 𝜇Λ (𝑢)) = |𝛼| ||𝑢||𝜅𝜑 . Therefore, ||𝛼𝑢||𝜅𝜑 = |𝛼|||𝑢||𝜅𝜑 .
= ||𝑢||𝜅𝜑 + ||V||𝜅𝜑 . Thus, ||𝑢 + V||𝜅𝜑 ≤ ||𝑢||𝜅𝜑 + ||V||𝜅𝜑 . (d) Let us now prove that ||𝑢||𝜅𝜑 = 0 if and only if 𝑢 = 0. Indeed, suppose that ||𝑢||𝜅𝜑 = 0, that is, |𝑢 (𝑎)| + 𝜇Λ (𝑢) = 0,
(69)
then |𝑢(𝑎)| = 0 and 0 = 𝜇Λ (𝑢) = inf{𝜆 > 0 : 𝜅𝑉𝜑𝑅 (𝑢/𝜆) ≤ 1}. This implies that for each positive integer 𝑛, there exists 𝜆 𝑛 such that 1/𝜆 𝑛 > 𝜆 𝑛 > 0 and 𝜅𝑉𝜑𝑅 (𝑢/𝜆 𝑛 ) ≤ 1. As the function 𝜅𝑉𝜑𝑅 (⋅) is convex, we have 𝜅𝑉𝜑𝑅 (𝑢) ≤ 𝜆 𝑛 . Taking the limit as 𝑛 → ∞, we get 𝜅𝑉𝜑𝑅 (𝑢) = 0. Moreover, by part (b) of Proposition 11 we have that 𝑢 is constant; that is, 𝑢(𝑡) = 𝑢(𝑎), 𝑡 ∈ [𝑎, 𝑏], therefore 𝑢 = 0. Now, suppose that 𝑢 = 0. Then |𝑢(𝑎)| = 0 and 𝜇Λ (𝑢) = 0. Hence ||𝑢||𝜅𝜑 = 0. Therefore (𝜅𝑅𝑉𝜑 [𝑎, 𝑏], || ⋅ ||𝜅𝜑 ) is a normed space.
Lemma 21. Let 𝑢 ∈ 𝜅𝑅𝑉𝜑𝑅 [𝑎, 𝑏]. If 𝜆 > 0, then 𝜇Λ (𝑢) ≤ 𝜆 if and only if 𝜅𝑉𝜑𝑅 (𝑢/𝜆; [𝑎, 𝑏]) ≤ 1. Proof Case 1. Let 𝑢 ∈ 𝜅𝑅𝑉𝜑𝑅 [𝑎, 𝑏] and 𝜇Λ (𝑢) < 𝜆. Then, by the infimum property, there is 0 < 𝑘 < 𝜆 such that 𝜅𝑉𝜑𝑅 (𝑢/𝑘) ≤ 1. Hence, by the convexity of 𝜅𝑉𝜑𝑅 (⋅), 𝑢 𝑢𝑘 𝑢 𝑘 𝑘 𝜅𝑉𝜑𝑅 ( ) = 𝜅𝑉𝜑𝑅 ( ) ≤ 𝜅𝑉𝜑𝑅 ( ) ≤ ≤ 1. 𝜆 𝑘𝜆 𝜆 𝑘 𝜆
(70)
Case 2. Let 𝑢 ∈ 𝜅𝑅𝑉𝜑𝑅 [𝑎, 𝑏] and 𝜇Λ (𝑢) = 𝜆. Then, by the infimum property, there exists a sequence {𝜆 𝑛 }𝑛∈N such that 𝜆 𝑛 > 𝜆,
𝑢 𝜅𝑉𝜑𝑅 ( ) ≤ 1, 𝑘
𝑛 ∈ N,
lim 𝜆 𝑛→∞ 𝑛
= 𝜆.
(71)
Since 𝑢/𝜆 𝑛 pointwise converges to 𝑢/𝜆 on [𝑎, 𝑏] as 𝑛 → ∞, by the lower semicontinuity of 𝜅𝑉𝜑𝑅 (⋅), we obtain that
(b)
= |𝛼| |𝑢 (𝑎)| + |𝛼| 𝜇Λ (𝑢)
(68)
(67)
𝑢 𝑢 𝜅𝑉𝜑𝑅 ( ) ≤ lim 𝜅𝑉𝜑𝑅 ( ) ≤ 1. 𝑛→∞ 𝜆 𝜆𝑛
(72)
As, by the definition of the infimum, the converse is obvious, this completes the proof. In the next theorem, we prove that (𝜅𝑅𝑉𝜑 [𝑎, 𝑏], || ⋅ ||𝜅𝜑 ) is a Banach space.
Journal of Function Spaces and Applications
9
Theorem 22. Let 𝜑 be a convex 𝜑-function such that 𝜑 satisfies the condition ∞1 , then the space (𝜅𝑅𝑉𝜑 [𝑎, 𝑏], || ⋅ ||𝜅𝜑 ) is a Banach space. Proof. Let {𝑢𝑛 }𝑛∈N be a Cauchy sequence in (𝜅𝑅𝑉𝜑 [𝑎, 𝑏], || ⋅ ||𝜅𝜑 ), then given that 0 < 𝜀 < 1, there is 𝑁 ∈ N, such that for 𝑛, 𝑚 ≥ 𝑁 we have 𝑢𝑛 − 𝑢𝑚 𝜅𝜑 < 𝜀, 𝑛, 𝑚 ≥ 𝑁, (73) that is, (𝑢𝑛 − 𝑢𝑚 ) (𝑎) + 𝜇Λ (𝑢𝑛 − 𝑢𝑚 ) < 𝜀,
−1
𝑛, 𝑚 ≥ 𝑁.
𝜇Λ (𝑢𝑛 − 𝑢𝑚 ) < 𝜀
𝑛 𝑡 −𝑡 × 𝑡𝑖 − 𝑡𝑖−1 ) (∑ 𝜅 ( 𝑖 𝑖−1 )) 𝑏−𝑎
(74)
𝑖=1
Then, 𝑢𝑛 (𝑎) − 𝑢𝑚 (𝑎) < 𝜀,
We defined on [𝑎, 𝑏] the function 𝑢(𝑡) := lim𝑛 → ∞ 𝑢𝑛 (𝑡). We claim that 𝑢 ∈ 𝜅𝑅𝑉𝑝 [𝑎, 𝑏]. In fact, let 𝜋 : 𝑎 = 𝑡1 < 𝑡2 < ⋅ ⋅ ⋅ < 𝑡𝑛 = 𝑏 be a partition of the interval [𝑎, 𝑏]. Then, for 𝑛 ≥ 𝑁 and 𝜀 > 0 one has 𝑢 −𝑢 𝜅𝑉𝜑𝑅 ( 𝑛 ) 𝜀 𝑛 (𝑢 − 𝑢) (𝑡𝑖 ) − (𝑢𝑛 − 𝑢) (𝑡𝑖−1 ) ) = sup (∑ 𝜑 ( 𝑛 𝜋 𝜀 𝑡𝑖 − 𝑡𝑖−1 𝑖=1
𝑛, 𝑚 ≥ 𝑁, (75)
𝑛 = sup (∑ 𝜑 ((𝑢𝑛 − lim 𝑢𝑚 ) (𝑡𝑖 ) 𝑚→∞ 𝜋 𝑖=1
−1 − (𝑢𝑛 − lim 𝑢𝑚 ) (𝑡𝑖−1 ) (𝜀 𝑡𝑖 − 𝑡𝑖−1 ) ) 𝑚→∞
by Lemma 21 and the last inequality, we have 𝜅𝑉𝜑𝑅 (
𝑢𝑛 − 𝑢𝑚 ; [𝑎, 𝑏]) < 1. 𝜀
(76)
𝑛
𝑡 −𝑡 × 𝑡𝑖 − 𝑡𝑖−1 ) (∑ 𝜅 ( 𝑖 𝑖−1 )) 𝑏−𝑎 𝑖=1
Then, for a partition 𝜋 : 𝑎 ≤ 𝑥 < 𝑦 ≤ 𝑏 we get that
𝑛 (𝑢 − 𝑢𝑚 ) (𝑡𝑖 ) − (𝑢𝑛 − 𝑢𝑚 ) (𝑡𝑖−1 ) ) = sup lim (∑ 𝜑 ( 𝑛 𝜋 𝑚 → ∞ 𝑖=1 𝜀 𝑡𝑖 − 𝑡𝑖−1
(𝑢 − 𝑢𝑚 ) (𝑦) − (𝑢𝑛 − 𝑢𝑚 ) (𝑥) ) 𝑦 − 𝑥 𝜑 ( 𝑛 𝜀 𝑦 − 𝑥 × (𝜅 (
𝑦−𝑥 𝑏 − 𝑦 −1 𝑥−𝑎 ) + 𝜅( ) + 𝜅( )) 𝑏−𝑎 𝑏−𝑎 𝑏−𝑎 ≤ 𝜅𝑉𝜑𝑅 (
𝑢𝑛 − 𝑢𝑚 ; [𝑎, 𝑏]) < 1. 𝜀
𝑦−𝑥 𝑏−𝑦 𝑥−𝑎 ) + 𝜅( ) + 𝜅( ) 𝑏−𝑎 𝑏−𝑎 𝑏−𝑎
(78)
−1
𝑛
𝑡 −𝑡 × 𝑡𝑖 − 𝑡𝑖−1 ) (∑ 𝜅 ( 𝑖 𝑖−1 )) . 𝑏−𝑎 𝑖=1 (82)
(77)
Hence, we have (𝑢 − 𝑢𝑚 ) (𝑦) − (𝑢𝑛 − 𝑢𝑚 ) (𝑥) ) 𝑦 − 𝑥 𝜑 ( 𝑛 𝜀 𝑦 − 𝑥 < 𝜅(
−1
Since for 𝑛, 𝑚 ≥ 𝑁, 𝑢 − 𝑢𝑚 1 ≥ 𝜅𝑉𝜑𝑅 ( 𝑛 ) 𝜀 𝑛 (𝑢 − 𝑢𝑚 ) (𝑡𝑖 ) − (𝑢𝑛 − 𝑢𝑚 ) (𝑡𝑖−1 ) ) 𝑡𝑖 − 𝑡𝑖−1 ) ≥ (∑𝜑 ( 𝑛 𝑡 𝜀 − 𝑡 𝑖 𝑖−1 𝑖=1 𝑛
−1
× (∑𝜅 (
≤ 3 ⋅ 𝜅 (1) = 3,
𝑖=1
𝑡𝑖 − 𝑡𝑖−1 )) , 𝑏−𝑎 (83)
therefore 3 −1 (𝑢𝑛 − 𝑢𝑚 ) (𝑦) − (𝑢𝑛 − 𝑢𝑚 ) (𝑥) < 𝜑 ( ) 𝑦 − 𝑥 𝜀. 𝑦 − 𝑥 (79)
then 𝑛 (𝑢 − 𝑢𝑚 ) (𝑡𝑖 ) − (𝑢𝑛 − 𝑢𝑚 ) (𝑡𝑖−1 ) ) lim (∑ 𝜑 ( 𝑛 𝑚→∞ 𝜀 𝑡𝑖 − 𝑡𝑖−1 𝑖=1
Since 𝜑 satisfies the condition ∞1 and 𝜑−1 is continuous, we obtain that 3 𝜑−1 ( ) 𝑦 − 𝑥 , 𝑦 − 𝑥
−1
𝑛 𝑡 −𝑡 × 𝑡𝑖 − 𝑡𝑖−1 ) (∑ 𝜅 ( 𝑖 𝑖−1 )) 𝑏−𝑎
(84)
𝑖=1
𝑥, 𝑦 ∈ [𝑎, 𝑏] , 𝑥 ≠ 𝑦,
(80)
𝑛 ≥ 𝑁.
Thus,
is bounded. Let 𝑀 > 0 be a upper bound of (80), then (𝑢𝑛 − 𝑢𝑚 ) (𝑦) − (𝑢𝑛 − 𝑢𝑚 ) (𝑥) < 𝑀𝜀.
≤ 1,
(81)
As a consequence, the sequence {𝑢𝑛 }𝑛∈N is a uniformly Cauchy sequence, on the interval [𝑎, 𝑏], and by the completeness of R lim𝑛 → ∞ 𝑢𝑛 (𝑡) exist for all 𝑡 ∈ [𝑎, 𝑏].
𝜅𝑉𝜑𝑅 (
𝑢𝑛 − 𝑢 ) ≤ 1. 𝜀
(85)
Therefore, (𝑢𝑛 − 𝑢)/𝜀 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] for all 𝑛 ≥ 𝑁. As 𝑢𝑛 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] for all 𝑛 ≥ 𝑁, and 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] is a vector space, then 𝑢 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏].
10
Journal of Function Spaces and Applications
Finally, let us prove that {𝑢𝑛 }𝑛≥1 converge in norm to 𝑢. Let 𝜀 > 0 be arbitrary, then 𝑢 𝑢𝑛 − 𝑢𝜅𝜑 = 𝑢𝑛 − 𝑚lim → ∞ 𝑚 𝜅𝜑 = lim 𝑢𝑛 − 𝑢𝑚 𝜅𝜑 < 𝜀, 𝑚→∞
(86) ∀𝑛 ≥ 𝑁.
Therefore, the sequence {𝑢𝑛 }𝑛≥1 converges to 𝑢 in the norm || ⋅ ||𝜅𝜑 and thus 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] is a Banach space. 3.1. Uniformly Bounded Composition Operator. In this section, we present the other main result of this paper; namely, we show that any uniformly bounded composition operator that maps the space the 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] into itself necessarily satisfies the so-called Matkowski’s weak condition. Definition 23. For a given function ℎ : [𝑎, 𝑏] × R → R, the composition operator 𝐻 : R[𝑎,𝑏] → R[𝑎,𝑏] generated by ℎ is defined by 𝐻𝑢 (𝑡) = ℎ (𝑡, 𝑢 (𝑡)) ,
𝑢 ∈ R[𝑎,𝑏] , 𝑡 ∈ [𝑎, 𝑏] .
(87)
Here R[𝑎,𝑏] denotes the family of all functions 𝑢 : [𝑎, 𝑏] → R. Remark 24. Since 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] ⊂ 𝜅𝐵𝑉[𝑎, 𝑏], then every function of bounded 𝜅𝜑-variation in the sense of RieszKorenblum has left- and right-hand limits at each point of its domain (see [8]).
Theorem 28. Let 𝜑 be a 𝜑-function, 𝜅 ∈ K, and ℎ : [𝑎, 𝑏] × R → R be a function continuous with respect to the second variable. Suppose that the composition operator 𝐻 generated by ℎ maps 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] into itself and satisfies the following inequality: ||𝐻 (𝑢) − 𝐻 (V)||𝜅𝜑 ≤ 𝛾 (||𝑢 − V||𝜅𝜑 ) ,
for some function 𝛾 : [0, ∞) → [0, ∞). Then, there exist functions 𝛼, 𝛽 ∈ 𝜅𝑅𝑉𝜑− [𝑎, 𝑏] such that ℎ− (𝑡, 𝑥) = 𝛼 (𝑡) 𝑥 + 𝛽 (𝑡) ,
lim 𝑢 (𝑠) , 𝑡 ∈ (𝑎, 𝑏] , 𝑢− (𝑡) := {𝑠 → 𝑡− 𝑢 (𝑎) , 𝑡 = 𝑎.
(88)
We will denote by 𝜅𝑅𝑉𝜑− [𝑎, 𝑏] the subset in 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] which consists of those functions that are left continuous on (𝑎, 𝑏].
𝜇Λ (𝐻 (𝑢1 ) − 𝐻 (𝑢2 )) ≤ 𝐻 (𝑢1 ) − 𝐻 (𝑢2 )𝜅𝜑 ≤ 𝛾 (𝑢1 − 𝑢2 𝜅𝜑 ) .
diam 𝐵 ≤ 𝑡 ⇒ diam 𝐻 (𝐵) ≤ 𝛾 (𝑡) .
(89)
(92)
From the inequality (92) and Lemma 21, if 𝛾(||𝑢1 − 𝑢2 ||𝜅𝜑 ) > 0, then 𝜅𝑉𝜑𝑅 (
𝐻 (𝑢1 ) − 𝐻 (𝑢2 ) ) ≤ 1. 𝛾 (𝑢1 − 𝑢2 𝜅𝜑 )
(93)
On the other hand, if 𝑎 ≤ 𝑟 < 𝑠 ≤ 𝑏, then from the definitions of the operator 𝐻, the functional 𝜅𝑉𝜑𝑅 , and inequality (93), we have 𝐻 (𝑢1 ) (𝑠) − 𝐻 (𝑢2 ) (𝑠) − 𝐻 (𝑢1 ) (𝑟) + 𝐻 (𝑢2 ) (𝑟) 𝜑 ( ) 𝛾 (𝑢1 − 𝑢2 𝜅𝜑 ) |𝑠 − 𝑟| × |𝑠 − 𝑟| (𝜅 (
𝑠−𝑟 𝑏 − 𝑠 −1 𝑟−𝑎 ) + 𝜅( ) + 𝜅( )) 𝑏−𝑎 𝑏−𝑎 𝑏−𝑎 ≤ 𝜅𝑉𝜑𝑅 (
Definition 27 (see [24, Definition 1]). Let X and Y be two metric (or normed) spaces. One says that a mapping 𝐻 : X → Y is uniformly bounded if, for any 𝑡 > 0, there exists a nonnegative real number 𝛾(𝑡) such that for any nonempty set 𝐵 ⊂ X we have
(91)
Proof. By hypothesis, for 𝑥 ∈ R fixed the constant function 𝑢(𝑡) = 𝑥, 𝑡 ∈ [𝑎, 𝑏] belongs to 𝜅𝑅𝑉𝜑 [𝑎, 𝑏]. Since 𝐻 maps 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] into itself, we have that the function (𝐻𝑢)(𝑡) = ℎ(𝑡, 𝑢(𝑡)) = ℎ(𝑡, 𝑥) ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏]. By Lemma 26, the left regularization ℎ− (⋅, 𝑥) ∈ 𝜅𝑅𝑉𝜑− for every 𝑥 ∈ R. From inequality (90) and the definition of the norm ||⋅||𝜅𝜑 we obtain for 𝑢1 , 𝑢2 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] that
Lemma 26. If 𝑢 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏], then 𝑢− ∈ 𝜅𝑅𝑉𝜑− [𝑎, 𝑏]. Thus, if a function 𝑢 has bounded 𝜅𝜑-variation in the sense of Riesz-Korenblum, then its left regularization is a left continuous function.
𝑡 ∈ [𝑎, 𝑏] , 𝑥 ∈ R,
where ℎ− (⋅, 𝑥) : (𝑎, 𝑏] → R is the left regularization of ℎ(⋅, 𝑥) for all 𝑥 ∈ R.
Now, we will give the definition of left regularization of a function. Definition 25. Let 𝑢 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏], one defined its left regularization 𝑢− : (𝑎, 𝑏] → R of mapping 𝑢 by the following:
𝑢, V ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] (90)
𝐻 (𝑢1 ) − 𝐻 (𝑢2 ) ) ≤ 1, 𝛾 (𝑢1 − 𝑢2 𝜅𝜑 ) (94)
whence ℎ (𝑠, 𝑢1 (𝑠)) − ℎ (𝑠, 𝑢2 (𝑠)) − ℎ (𝑟, 𝑢1 (𝑟)) + ℎ (𝑟, 𝑢2 (𝑟)) 𝜑 ( ) 𝛾 (𝑢1 − 𝑢2 𝜅𝜑 ) |𝑠 − 𝑟| × |𝑠 − 𝑟| (𝜅 (
𝑠−𝑟 𝑏 − 𝑠 −1 𝑟−𝑎 ) + 𝜅( ) + 𝜅( )) ≤ 1. 𝑏−𝑎 𝑏−𝑎 𝑏−𝑎 (95)
Journal of Function Spaces and Applications
11
Let 𝑥1 , 𝑥2 ∈ R, 𝑥1 < 𝑥2 and consider the functions 𝑢1 , 𝑢2 : [𝑎, 𝑏] → R defined by 𝑢𝑘 (𝑡) :=
𝑥 + 𝑥2 𝑥1 − 𝑥2 , (𝑡 − 𝑟) + 𝑘 2 (𝑠 − 𝑟) 2
(96)
(97)
𝑥 − 𝑥2 . 𝑢1 − 𝑢2 𝜅𝜑 = 1 2
(98)
𝑥1 + 𝑥2 = 𝑢2 (𝑠) , 2
𝑢2 (𝑟) = 𝑥2 .
therefore,
Moreover,
(99)
From (95) we get 𝑥 + 𝑥2 𝑥 + 𝑥2 𝜑 (ℎ (𝑠, 𝑥1 ) − ℎ (𝑠, 1 ) − ℎ (𝑟, 1 ) + ℎ (𝑟, 𝑥2 ) 2 2 −1 × (𝛾 ((𝑥1 − 𝑥2 ) /2) |𝑠 − 𝑟|) ) ≤
3 , |𝑠 − 𝑟|
𝑥 − 𝑥2 3 ) |𝑠 − 𝑟| . ) 𝛾 ( 1 |𝑠 − 𝑟| 2
(102)
then from the definition of ℎ− and letting 𝑟 tend to 𝑠 in (101), we obtain 𝑥 + 𝑥2 − ℎ (𝑠, 𝑥1 ) − 2ℎ− (𝑠, 1 (103) ) + ℎ− (𝑠, 𝑥2 ) ≤ 0, 2 therefore ℎ− (𝑠, 𝑥1 ) + ℎ− (𝑠, 𝑥2 ) 𝑥 + 𝑥2 ℎ− (𝑠, 1 )= , 2 2
(104)
𝑠 ∈ (𝑎, 𝑏] , 𝑥1 , 𝑥2 ∈ R which proves that for every fixed 𝑠 ∈ (𝑎, 𝑏] the function ℎ− (𝑠, ⋅) satisfies the Jensen functional equation in R (see [26, page 315]). The continuity of ℎ with respect to the second variable implies that for every 𝑡 ∈ [𝑎, 𝑏] there exist 𝛼(𝑡), 𝛽(𝑡) ∈ R such that ℎ− (𝑡, 𝑥) = 𝛼 (𝑡) 𝑥 + 𝛽 (𝑡) ,
𝑡 ∈ [𝑎, 𝑏] , 𝑥 ∈ R.
𝑡 ∈ [𝑎, 𝑏] ,
(107)
as 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] is a vector space which implies that 𝛼 ∈ 𝜅𝑅𝑉𝜑− [𝑎, 𝑏]. Theorem 29. Let 𝜑 be a function and 𝜅 ∈ K. Suppose that a function ℎ : [𝑎, 𝑏] × R → R is continuous with respect to the second variable. If the composition operator 𝐻 generated by ℎ maps the space 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] into itself and is uniformly bounded, then there exist the functions 𝛼, 𝛽 ∈ 𝜅𝑅𝑉𝜑− [𝑎, 𝑏], such that ℎ− (𝑡, 𝑥) = 𝛼 (𝑡) 𝑥 + 𝛽 (𝑡) ,
𝑡 ∈ [𝑎, 𝑏] , 𝑥 ∈ R.
(108)
Proof. Take any 𝑡 ≥ 0 and 𝑢, V ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏] such that ||𝑢 − V||𝜅𝜑 ≤ 𝑡.
(109)
Since diam{𝑢, V} ≤ 𝑡, by the uniform boundedness of 𝐻, we have diam 𝐻 ({𝑢, V}) ≤ 𝛾 (𝑡) ,
(110)
that is, ||𝐻 (𝑢) − 𝐻 (V)||𝜅𝜑 = diam 𝐻 ({𝑢, V}) ≤ 𝛾 (||𝑢 − V||𝜅𝜑 ) , (111) therefore, by the Theorem 28 we have
(101)
Since 𝜑 satisfies the condition ∞1 , we have 1 lim 𝜑−1 ( ) 𝑡 = 0, 𝑡→0 𝑡
(106)
therefore 𝛽 ∈ 𝜅𝑅𝑉𝜑− [𝑎, 𝑏]. On the other hand, considering 𝑥 = 1 we obtain that
(100)
then ℎ (𝑠, 𝑥 ) − ℎ (𝑠, 𝑥1 + 𝑥2 ) − ℎ (𝑟, 𝑥1 + 𝑥2 ) + ℎ (𝑟, 𝑥 ) 1 2 2 2 ≤ 𝜑−1 (
𝑡 ∈ [𝑎, 𝑏] ,
𝛼 (𝑡) = ℎ− (𝑡, 𝑥) − 𝛽 (𝑡) ,
𝑥 − 𝑥2 𝑢1 − 𝑢2 = 1 , 2
𝑢1 (𝑟) =
𝛽 (𝑡) := ℎ− (𝑡, 0)
𝑡 ∈ [𝑎, 𝑏] , 𝑘 = 1, 2
then 𝑢𝑘 ∈ 𝜅𝑅𝑉𝜑 [𝑎, 𝑏], 𝑘 = 1, 2. From (96), we have
𝑢1 (𝑠) = 𝑥1 ,
Since ℎ− (⋅, 𝑥) ∈ 𝜅𝑅𝑉𝜑− [𝑎, 𝑏] for all 𝑥 ∈ R in particular for 𝑥 = 0,
(105)
ℎ− (𝑡, 𝑥) = 𝛼 (𝑡) 𝑥 + 𝛽 (𝑡) ,
𝑡 ∈ [𝑎, 𝑏] , 𝑥 ∈ R.
(112)
Remark 30. Observe that similar results hold for the right regularization ℎ+ of ℎ defined by ℎ+ (𝑡, 𝑥) := lim+ ℎ (𝑠, 𝑥) 𝑠→𝑡
𝑡 ∈ [𝑎, 𝑏) .
(113)
Acknowledgments Thanks are due to the referee for the useful suggestion and comments to improve this paper. This research has been partly supported by the Central Bank of Venezuela. The authors want to thanks the library staff of B.C.V for compiling the references.
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