The Jordan decomposition of bounded variation functions valued in vector spaces

: In this paper we show the Jordan decomposition for bounded variation functions with values in Riesz spaces. Through an equivalence relation, we prove that this decomposition is satisﬁed for functions valued in Hilbert spaces. This result is a generalization of the real case. Moreover, we prove that, in general, the Jordan decomposition is not satisﬁed for vector-valued functions.


Introduction
The concept of bounded variation function was introduced in 1881 by Camille Jordan [1] for real functions defined in a closed interval I = [a, b] ⊂ R. He proves that a function is of bounded variation if and only if it can be represented as the difference of two increasing functions. This representation is known as the Jordan decomposition.
Because of the existence of several kinds of functions, mainly due to variations of domain and codomain, it has been necessary to define different types of bounded variation. We can mention Vitali, Hardy, Arzela, Pierpont, Frechet, and Tonelli who give different definitions of bounded variation for real functions of two variables. C. R. Adams [2,3] studied the relation between the concepts defined by the previous authors.
For bounded variation functions f : [a, b] → X, where X is a metric space, V. V. Chistyakov studies many aspects around those functions [4,5,6,7]. In the first reference, he proves an alternative result to the Jordan decomposition, affirming that for bounded variation functions valued in metric spaces the decomposition as the difference of two monotone functions is inapplicable. On the other hand, Bianchini and Tonon [8] assert that there is no hope for a further generalization of this decomposition to vector valued BV functions, apart from the case of a function f : R → R m where the analysis is straightforward.
Defining the bounded variation with respect to the order in the first part of this paper we show that the Jordan decomposition is possible for functions valued in Riesz spaces. Additionally, as an alternative to affirmations of Chistyakov and Bianchini-Tonon, we prove that for functions valued in Hilbert spaces, proposition 2.10, the Jordan decomposition is satisfied in a generalized sense from an equivalence relation, being the decomposition for real-valued functions a particular case. This result allows us to give a negative answer to the Jordan decomposition problem of a bounded variation function f : I → (H, H + ), where the Hilbert space H is ordered by a given extensible cone H + .

Preliminaries
There are vector spaces in which is possible to define a natural order relation, for instance for continuous real functions defined on a compact interval [a, b], denoted by C([a, b], R). In this case: Nevertheless, there are some vector spaces where a natural order relation cannot be defined. This has led to creating mechanisms that permit comparison vectors associated with the order.
We listed some concepts that will be useful in our exposition and that are linked to order in vector spaces. The notation [a, b] always will be reserved for compact intervals in R.
• Let X be a partially ordered set. We say that X is a lattice if every subset consisting of two points has a supremum and an infimum. • A vector space X is called ordered if it is partially ordered in such a manner that the structure of vector space and the order structure are compatible, that is to say: i) x ≤ y implies x + z ≤ y + z, for every z ∈ X, ii) x ≥ 0 implies αx ≥ 0, for every α ≥ 0 in R.
If, in addition, X is a lattice with respect to the partial order, then X is called a Riesz space.
• Let X be a Riesz space We say that f is bounded with respect to the order if it is at the same time bounded above and bounded below.
• Let X be a normed space. X + a closed subset of X is called a cone if X + + X + ⊆ X + , X + (−X + ) = {0} and cX + ⊆ X + , for all c ≥ 0. The order relation ≤ defined by x ≤ y if and only if y − x ∈ X + is an order partial in X. The pair (X, X + ) is called ordered normed space. • Let X be an ordered normed space with a cone X + . We say that f : • Assume that X + is a cone in X. If there exists a cone X 1 in X and b > 0 such that for any x ∈ X + : B(x, b x ) ⊂ X 1 , then X + is called an extensible cone.
The following characterization of extensible cones is useful for our purposes.
Theorem 1.1. [9] Assume that X + is a cone in X. Then X + is extensible if and only if there exists g ∈ X * and a constant α > 0 such that g(x) ≥ α x , for all x ∈ X + .
A partition of [a, b] is a finite ordered set of points in [a, b]: Definition 1.2. Let X be a normed space. We say that f : The expression (1) is This can be seen examining, for instance, the case m .
By the inequality where α ji ≥ 0 for i = 1, 2, ..., n; j = 1, 2; and taking Evaluating the supremum over all the partitions in [a, b] , we obtain the left-hand inequality of (2). This inequality may be strict. For example, let The affirmation is proved observing that for the partition P 0 = {−1, 0, 1/2, 1} : The above example shows that this fact may not be possible. We know that if X is a normed space and ϕ ∈ X * , then ϕ(x) ≤ ϕ x X . Therefore, we can observe that the following result is satisfied. Theorem 1.6. Let X be an ordered normed space with an extensible cone X + . Then every monotone (increasing or decreasing) function is of bounded variation.
Proof. We prove the case when f is an increasing function, the proof for decreasing functions is similar.
Since X + is an extensible cone, then, by theorem 1.1, there exists g ∈ X * and a constant α > 0 such that Therefore we conclude the proof with the following inequalities.
As a consequence of the previous theorem we have that the difference of two increasing functions is of bounded variation. In subsection 2.2 we will see that the reciprocal of this result may not be true.
Supposing that "≤" is a natural order in the normed space X, then X + = {x ∈ X : 0 ≤ x} will be a cone. In this case we have that a monotone function in the ordered normed space (X, X + ) may not be of bounded variation. This fact contrasts with theorem 1.6. In the following example we use the function defined in [[10], Example 7.1.8].
L ∞ [0, 1] has a natural order in the following sense.
We conclude that F(t) is an increasing function in the natural sense but is not of bounded variation.
2. The Jordan decomposition of functions with values...
The previous notation makes sense because a Riesz space is a lattice with respect to the partial order. We define the concept of bounded variation with respect to the order as follows.
for all partition P ∈ P([a, b]).
By analogy with the case X = R, it is not difficult to be convinced of the validity of the following results.  ii , it follows that This means that D(t 2 ) − D(t 1 ) ≥ 0, and ii) holds. Example 2.8. Let f : [− π 2 , π 2 ] → R 2 be given by f (t) = (cos t, t), where R 2 is considered as a Riesz space with the order (x 1 , x 2 ) ≤ (y 1 , y 2 ), whenever x 1 ≤ y 1 and x 2 ≤ y 2 . Let P ∈ P [a, b].
If P contains to 0, then Suppose that exist t i−1 , t i ∈ P such that 0 ∈ (t i−1 , t i ). Then Therefore, f is of bounded variation with respect to the order and (cos t, R), π .

...in Hilbert spaces
We will analyze the Jordan decomposition for functions with values in a Hilbert space.
Definition 2.9. Let H be a Hilbert space on R and let x 0 be fixed in H. We say that x, y ∈ H are related with respect to x 0 , and we use the notation The previous relation is of equivalence, so that, for each x 0 ∈ H, we can divide H in disjoint classes. Since any two elements x, y ∈ H are related with respect to 0, then the only equivalence class will be H.
Using this relation, in the following proposition we prove a generalization of the Jordan decomposition.
Proof. By the Riesz lemma, for x 0 ∈ H there is only one h 0 ∈ H * such that h 0 (x) = x, x 0 . By lemma Because of x 0 0, the functional h 0 is not identically zero. Let α ∈ (0, x 0 ) and This set is a cone because if x and −x ∈ H x 0 + , then x 0 belongs to H x 0 + .
Let f x 0 1 (t) = g 1 (t) x 0 and f x 0 2 (t) = g 2 (t) x 0 . Because 0 ≤ g 1 (t 2 ) − g 1 (t 1 ) for t 1 < t 2 , then we have Making a similar observation for f x 0 2 , we have that f x 0 1 (t) and f x 0 2 (t) are increasing functions. On the other hand, considering that h 0 • f = g 1 − g 2 : Since H x 0 + is a cone and 1 x 0 2 > 0, then the functions 1 x 0 2 f x 0 i (t) , i = 1, 2, are increasing in H, H x 0 + . Redefining f x 0 1 (t) and f x 0 2 (t), respectively by the previous multiple functions, then, by (4) and (5), we get Due to Riesz's lemma and taking x 0 ∈ H associated with the functional h 0 , we have the following corollary.
Remark 2.12. Because g 1 and g 2 may be V t and This opens the possibility of defining the right side of (6) as the variation function of f with respect to x 0 . Remark 2.13. We can make a variant of the proof of proposition 2.10, if we consider the cone H x 0 + should be extend to H x 00 , and this last one is non-extensible.
Remark 2.14. We can observe that the proposition 2.10 generalizes the case H =R. The cone H x 0 + associated to each x 0 ∈ R, x 0 0, has the form We have: x, y ∈ R are x 0 −related if and only if x = y. Thus f ∼ x 0 f x 0 1 − f x 0 2 if and only if f = f x 0 1 − f x 0 2 . We observe that for any other x 1 0, we have by (7) that H x 0 + = H x 1 + ; although x 0 x 1 .
Proof. Let x 1 ∈ H with x 1 0 such that x 1 , x 0 = 0. Let λ : [a, b] → R an increasing function and f (t) = λ(t)x 1 . Then f is a bounded variation function in H. Suppose that there exist f 1 , f 2 increasing functions in H, H x 0 + such that f ∼ x 0 [ f 1 − f 2 ]. By lemma 2.16, we have which is equivalent to Thus: Therefore: h 0 • f 1 (t) = h 0 • f 2 (t), for all t ∈ [a, b]. By (8), we conclude that As consequence of the previous proposition, we can find bounded variation functions with values in a Hilbert space (therefore normed) for which the Jordan decomposition is satisfied only if they are related with the zero function. The difference of the previous proposition with proposition 2.10 is that, at this last, the extensible cone is predetermined, while in proposition 2.10 the cone depends on x 0 .