ABOUT AN EXISTENCE THEOREM OF THE HENSTOCK-FOURIER TRANSFORM

We show that if f is lying on the intersection of the space of Henstock-Kurzweil integrable functions and the space of the bounded variation functions in the neighborhood of ±∞, then its Fourier Transform exists in all R. This result is more general than the classical result which enunciates that if f is Lebesgue integrable, then the Fourier Transform of f exists in all R, because we also have proved that there are functions which belong to the intersection of the space of the Henstock-Kurzweil integrable functions and the space of the bounded variation functions which are not Lebesgue integrable.


Introduction
In [6] (which explains Henstock-Kurzweil Fourier transforms), Erik Talvila shows several results about Fourier Transform in HK(R), the space of the Henstock-Kurzweil integrable functions over R. A particular result which has captured our attention is the proposition 2 (b) which talks about the existence of Fourier Transform in the HK loc (R) space, which is enunciated as follows If f is locally HK-integrable and of bounded variation in a neighborhood of infinity with limit zero at infinity, then, b f exists in R. It is obvious that under these conditions, it is not necessarily true that the Fourier Transform exists in zero (for instance).The hypothesis proposed by Talvila leads to the idea that if we impose a condition where f ∈ HK(R) ∩ BV ( [±∞] ), here BV ( [±∞] ) is the space of the functions which are of bounded variation in some neighborhood of ±∞, then b f (s) exists for all s ∈ R. A problem which might arise is that HK(R) ∩ BV ( [±∞] ) ⊆ L(R), where L(R) is the Lebesgue integrablefunctions space, for which the existence of the Fourier Transform, for all s ∈ R, in HK(R) ∩ BV ( [±∞] ) could not represent a progress.Just this situation, we have developed two points in this paper: 1 Contention relations of the space HK(I) ∩ BV (I) for unbounded intervals.
With respect to the point 1, we know that if I is a compact interval, every bounded variation function is Lebesgue integrable.Because every Lebesgue integrable function is HK-integrable, we will have that, for bounded intervals: for which HK(I) ∩ BV (I) ⊂ L(I), and this contention is proper.For instance, the function f : [0, 1] → R defined as then it is a natural question to ask ourselves if the contention relations of the spaces BV (I), L(I), and HK(I) are the same as when I is compact.To answer to this, we show a family of functions which is useful to us for showing that the former contentions behave in a different way.Among these relations, the more significative is: With respect to the point 2, we will prove that: If f ∈ HK(R) ∩ BV ( [±∞]) , then b f (s) exists for all s ∈ R.
1.The Spaces HK(I) and BV (I) The function γ ε is commonly called gauge, and if the partition P complies with the condition (1.1), we will say that it is γ ε −fine.The number A is called as the integral of f over [a, b] and it is denoted as Thus, the function will be integrable if it satisfies definition 1.1, and also the condition of that the partition P be γ ε −fine acording to the previous incises.In addition, these conditions cause that the last term of ) is zero and thus this sum is finite.For functions defined over intervals [−∞, a] or [−∞, +∞] we do similar considerations.
Through the theory of this integral we have that f We denote as

I}
Some features of HK(I) are the following: 1.It is a vector space, i.e.: the sum of functions and the product by scalars of Henstock-Kurzweil integrable functions are integrable.The integral is a linear functional over this space.
2. It contains the Riemann and Lebesgue integrable functions.Also, the functions whose Riemann or Lebesgue improper integrals exist, and its values coincide.
3. It generalizes the Fundamental Theorem of Calculus, in the sense that every derivative function is integrable.This does not happen with Riemann and Lebesgue integrals.In this case we have: 4. Since we know, in Riemann's integral, if two functions are integrable, then their product is also integrable.In the case of the integral of HK, this is not true.Nevertheless, the product of a HK-integrable function by a bounded variation function, is in fact integrable.
5. The HK integral is not an absolute integral.This asseveration is in the sense that if f is HK-integrable, it does not imply that |f | is so.When |f | and f are integrable, we say that f is absolutely HK-integrable.
6.The space of the absolutely HK-integrable functions is L(I), the space of the Lebesgue-integrable functions.
The following theorems plays an important roll into this theory of integration, and also in the demonstration of our results.The intervals we have considered could be compact or unbounded.

Bounded Variation Functions in Unbounded Intervals
We suppose that the reader knows the concept of bounded variation for functions defined in bounded intervals.We will denote the total Variation of f in I as V (f, I).For unbounded intervals, the definition of the concept about bounded variation is: Definition 1.2.Let f : [a, ∞] → R be a function.We say that f is of bounded variation in I = [a, ∞] if there exists M > 0 in such a way that, for all b ≥ a, it is fulfilled: The total variation of f in I will be: , the total variation is defined as See that in (1.3) and in (1.4), we have, respectively, that: Let BV (I) = {f : I → R | f is of bounded variation in I} be, where I is an unbounded interval.Some features belonging to BV (I) are the following: a) BV (I) is a vector space.b) f ∈ BV (I), if and only if, there exist f 1 and f 2 which are increasing bounded-functions such that we have similar results.

Contention relations of the space HK(I) ∩ BV (I), for unbounded intervals
In the same way as commented in the introductory section, if I is a compact interval, then It is easy to demonstrate that the function f (x) = 1/x defined in [1, ∞], is of bounded variation, and and This implies that (2.1) is true.For (2.2), we consider the function Next, we will prove that: ) is done in a similar way.Suppose that n 0 is the natural minor number for which a < (1+4n 0 )π/4.For x ∈ [a, ∞] we have ¸.
Given n ∈ N, as (3 + 4n)π/4 < (1 + n)π, we have that: On the other hand, we have that: Considering that 1 − α > 0, then using Theorem 1.1.and (2.9) we have that Let: f (s) exists for all s ∈ R. As mentioned in the introductory section, with these conditions, it is not necessarily true the existence of b f (0).If s 6 = 0, then the result is true.
For finding existence conditions of b f (s) for all s ∈ R, the main idea is considering a proper subspace of HK loc (R) ∩ BV 0 ( [±∞] ).This subspace should be in such a way that it is not contained or it equals L(R), because in this last way we will not have any advance.On the other hand, in order to have the existence of b f (0) we need that f ∈ HK(R.The above sentence leads to us to think that the subspace might be Taking into account (3.1) and the result from section 2, then the following theorem is true for a larger space.If s 6 = 0, we follow the proof from [5] and [6].The result is true for s = 0 because f ∈ HK(R).

Theorem 1 . 1 .Theorem 1 . 2 .
Let f ∈ HK(I) be a function.Then |f | is integrable, if and only if, F (x) = [ R x a f ] ∈ BV (I).In this case we have that the total variation of F in I is: If f, g ∈ HK(I) and |f (x)| ≤ g(x) for all x ∈ I, then f ∈ L(I).In addition, Theorem 1.3 (Hake's).Let I = [a, ∞] and f : I → R be a function, then f ∈ HK(I), if and only if, f ∈ HK( [a, c] ) for any compact interval [a, c] ⊂ [a, ∞), and there exists A ∈ R such that lim c→∞ Z c a f = A, In this case we have that: A = R ∞ a f .This last theorem is also valid for intervals [−∞, b] or [−∞, ∞], with its corresponding adaptation.Theorem 1.4 (Chartier-Dirichlet). Let f, ϕ : [a, ∞] → R be functions, and suppose that: a) f ∈ HK ³ [a, c] ´for all c ≥ a, and