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Regulated Functions with Values in Banach Space 144 (2019) MATHEMATICA BOHEMICA No. 4, 437–456 REGULATED FUNCTIONS WITH VALUES IN BANACH SPACE Dana Fraňková, Zájezd Received September 2, 2019. Published online November 22, 2019. Communicated by Jiří Šremr Cordially dedicated to the memory of Štefan Schwabik Abstract. This paper deals with regulated functions having values in a Banach space. In particular, families of equiregulated functions are considered and criteria for relative compactness in the space of regulated functions are given. Keywords: regulated function; bounded variation; function with values in a Banach space; ϕ-variation; relative compactness; equiregulated function MSC 2010 : 26A45, 46E40 Introduction This paper is an extension of the previous one (see [2]), where regulated functions with values in Euclidean spaces were considered. Here, we deal with regulated func- tions having values in a Banach space. We discuss some of the properties of the space of such regulated functions, including compactness theorems. Classic results of mathematical analysis are being used (see [4]) and some ideas from previous works on the topic of regulated functions appear here (see [3], [5]). 1. Notation and definitions N (i) The symbol N will denote the set of all positive integers, N0 = N ∪{0}; R (where N ∈ N) is the N-dimensional Euclidean space with the usual norm |·|N . 1 We write R and |·| instead of R and |·|1. (ii) Throughout the paper, the symbol X will denote a Banach space with a norm k·kX and C([a,b]; X) is the set of all continuous functions f :[a,b] → X. DOI: 10.21136/MB.2019.0124-19 437 c The author(s) 2019. This is an open access article under the CC BY-NC-ND licence cbnd (iii) We say that a function h:[a,b] → R is increasing if a 6 s<t 6 b implies h(s) <h(t); the function h is non-decreasing if a 6 s<t 6 b implies h(s) 6 h(t). (iv) We say that g :[a,b] → X is a finite step function, or shortly step function, if it is piecewise constant; i.e., there is a division a = a0 <a1 <...<ak = b such that the function g is constant on each of the intervals (ai−1,ai), i =1, 2,...,k. (v) We denote by Da,b the set of divisions {a0,...,ak} such that a = a0 < a1 <...<ak = b. (vi) For any function f :[a,b] → X, we write kfk∞ = sup{kf(t)kX : t ∈ [a,b]}. If kfk∞ < ∞, we say that the function f is bounded; k·k∞ is called the sup-norm. (vii) We say that a sequence of functions fn :[a,b] → X, n ∈ N, is uniformly conver- gent to a function f0 :[a,b] → X (or that f0 is the uniform limit of {fn}n∈N) if kfn − f0k∞ → 0 with n → ∞; we denote fn ⇒ f0. 2. Basic properties of a regulated function Definition 2.1. We say that a function f :[a,b] → X is regulated if the limit f(t−) = lim f(τ) exists for every t ∈ (a,b], and the limit f(t+) = lim f(τ) exists τ→t− τ→t+ for every t ∈ [a,b). We denote by G([a,b]; X) the set of all regulated functions f :[a,b] → X. Obviously, any finite step function on [a,b] and any continuous function on [a,b] are regulated on [a,b]. Moreover, any function with bounded variation on [a,b] and any monotone real valued function are regulated on [a,b]. Proposition 2.2. Assume that fn :[a,b] → X, n ∈ N, are regulated functions and f0 :[a,b] → X is a function such that fn ⇒ f0. Then the function f0 is regulated and fn(t−) → f0(t−) for each t ∈ (a,b], fn(t+) → f0(t+) for each t ∈ [a,b). Proof. The proof follows easily from the classical Moore-Osgood theorem on exchanging the order of limits, cf. e.g. [4]. Theorem 2.3. The following properties of a function f :[a,b] → X are equiva- lent: (i) The function f is regulated. (ii) The function f is the uniform limit of a sequence of step functions. (iii) For every ε> 0 there is a step function g :[a,b] → X such that kf − gk∞ <ε. (iv) For every ε > 0 there is a division a = a0 < a1 <...<ak = b such that if ′ ′′ ′′ ′ ai−1 <t <t <ai for some i ∈{1, 2,...,k} then kf(t ) − f(t )kX <ε. 438 P r o o f. (i) ⇒ (iv): Let ε> 0 be given. For every x ∈ (a,b], define ε s = inf s ∈ (a, x): if τ ′, τ ′′ ∈ (s, x) then kf(τ ′) − f(τ ′′)k < . x n X 2o For every x ∈ [a,b), define ε (2.1) t = sup t ∈ (x, b): if τ ′, τ ′′ ∈ (x, t) then kf(τ ′) − f(τ ′′)k < . x n X 2o It follows from the existence of the limits f(x−), f(x+) that sx < x and tx > x. Obviously, [a,t ) ∪ (s ,t ) ∪ (s ,b] = [a,b] a [ x x b x∈(a,b) and, since [a,b] is compact, there are k ∈ N and a finite set {a1,...,ak−1} of points in (a,b) such that a1 <a2 <...<ak−1, k−1 (2.2) [a,t ) ∪ (s ,t ) ∪ (s ,b] = [a,b]. a [ ai ai b i=1 We shall verify that sai <tai−1 for i ∈{1, 2,...,k}. On the contrary, assume that there is σ such that tai−1 6 σ 6 sai . Thanks to (2.2), there is j∈ / {i − 1,i} such ′ ′′ 1 that σ ∈ (saj ,taj ). If j<i − 1 then by (2.1) we have kf(τ ) − f(τ )kX < 2 ε for ′ ′′ ′ ′′ all τ , τ ∈ (aj ,taj ), which specifically holds also for all τ , τ ∈ (ai−1,taj ). Hence taj 6 tai−1 6 σ<taj which is a contradiction. Similarly, if j>i we find that this leads to a contradiction as well. Consequently, for any i ∈ {1, 2,...,k}, the intersection (sai ,tai−1 ) ∩ (ai−1,ai) is nonempty and we choose bi ∈ (sai ,tai−1 ) ∩ (ai−1,ai). ′ ′′ Now, if ai−1 < t < t < ai for some i ∈ {1,...,k}, there are three possibilities: ′ ′′ ′ ′′ ′ ′′ either ai−1 < t < t 6 bi or bi 6 t < t < ai or ai−1 < t 6 bi 6 t < ai. In the ′ ′′ first case, both t , t are in (ai−1,tai−1 ), and thanks to (2.1) ε kf(t′′) − f(t′)k < . X 2 ′ ′′ ′ ′′ Similarly, if bi 6 t <t <ai for some i then t ,t ∈ (sai ,ai) and ε kf(t′′) − f(t′)k < ; X 2 ′ ′′ ′ ′′ and, if ai−1 < t 6 bi 6 t < ai for some i then t ,bi ∈ (ai−1,tai−1 ), and bi,t ∈ (sai ,ai) and hence ′′ ′ ′′ ′ kf(t ) − f(t )kX 6 kf(t ) − f(bi)kX + kf(bi) − f(t )kX < ε. To summarize, (iv) is true. 439 (iv) ⇒ (iii): Given ε > 0 we can find the described division a = a0 <a1 <...< ak = b; choose points τi ∈ (ai−1,ai) and define g(τ) = f(τi) for τ ∈ (ai−1,ai), i =1, 2,...,k; g(ai)= f(ai), i = 0, 1,...,k. Then g is a step function and kg(τ) − f(τ)kX <ε for every τ ∈ [a,b]. (iii) ⇒ (ii): For ε =1/n, we can find a step function gn such that kf −gnk∞ < 1/n. Hence, gn ⇒ f. (ii) ⇒ (i): This implication follows from Proposition 2.2. Let us notice that the equivalences contained in Theorem 2.3 have been already proved in [3] in a slightly different way. The following result also can be found in [3], but no detailed proof is provided therein. Proposition 2.4. If a function f :[a,b] → X is regulated, then (i) for any c > 0, the sets {t ∈ [a,b): kf(t+) − f(t)kX > c} and {t ∈ (a,b]: kf(t−) − f(t)kX > c} are finite; (ii) the sets J + = {t ∈ [a,b): f(t+) 6= f(t)} and J − = {t ∈ (a,b]: f(t−) 6= f(t)} are at most countable. P r o o f. (i) By Theorem 2.3 (iv), there is a division a = a0 <...<ak = b such that c kf(u) − f(t)k < whenever u,t ∈ (a − ,a ) for some i. X 2 i 1 i Passing to the limit u → t+ we get c kf(t+) − f(t)k 6 <c for all t ∈ [a,b] \{a ,...,a }. X 2 0 k + (ii) It is evident that J = {t ∈ [a,b): kf(t+) − f(t)kX > 1/n}; this is a nS∈N countable union of finite sets, therefore at most countable. Similarly for the left- sided limits. In the following theorem we are going to use the notion of total ϕ-variation which appears in [1]. Definition 2.5. Let us denote by Φ the set of all increasing functions ϕ: [0, ∞) → [0, ∞) such that ϕ(0) = ϕ(0+) = 0, ϕ(∞) = ∞. For f :[a,b] → X, given ϕ ∈ Φ and a division d = {t0,t1,...,tm}; d ∈ Da,b, we define m ϕ V (f)= ϕ(kf(t ) − f(t − )k ), d X j j 1 X j=1 and the total ϕ-variation of f by ϕ ϕ- Var(f) = sup Vd (f): d ∈ Da,b . [a,b] 440 Theorem 2.6. The following properties of a function f :[a,b] → X are equiva- lent: (i) The function f is regulated. (ii) There is a continuous function g :[c, d] → X and a non-decreasing function h:[a,b] → [c, d] such that f(t)= g(h(t)) for every t ∈ [a,b].
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