# Henstock-Kurzweil Integral Transforms

Authors: Salvador Sánchez-Perales; Francisco J. Mendoza Torres; Juan A. Escamilla Reyna
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We show conditions for the existence, continuity, and differentiability of functions defined byΓ(s)=∫-∞∞f(t)g(t,s)dt, where f is a function of bounded variation on ℝ with lim|t|→∞f(t)=0.


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## Body

## 1. Introduction

Letg be a complex function defined on a certain subset of ℝ2. Many functions on functional analysis are integrals of the following form:(1.1)Γ(s)=∫-∞∞‍f(t)g(t,s)dt.We discuss the above functionΓ, where the integral that we use is that of Henstock-Kurzweil. This integral introduced independently by Kurzweil and Henstock in 1957-58 encompasses the Riemann and Lebesgue integrals, as well as the Riemann and Lebesgue improper integrals.In Lebesgue theory, there are well-known results about the existence, continuity, and differentiability ofΓ. For Henstock-Kurzweil integrals also there are results about this, for example, Theorems 12.12 and 12.13 of [1]. However, they all need the stronger condition: f(t)g(t,s) is bounded by a Henstock-Kurzweil integrable function r(t). We provide other conditions for the existence, continuity, and differentiability of Γ.


## 2. Preliminaries

Let us begin by recalling the definition of Henstock-Kurzweil integral. For finite intervals inℝ it is defined in the following way.Definition 2.1.
Letf:[a,b]→ℝ be a function. One can say that f is Henstock-Kurzweil (shortly, HK-) integrable, if there exists A∈ℝ such that, for each ϵ>0, there is a function γϵ:[a,b]→(0,∞) (named a gauge) with the property that for any δϵ-fine partition P={([xi-1,xi],ti)}i=1n of [a,b] (i.e., for each i, [xi-1,xi]⊂[ti-γϵ(ti),ti+γϵ(ti)]), one has
(2.1)|∑i=1nf(ti)(xi-xi-1)-A|<ϵ.
The numberA is the integral of f over [a,b] and it is denoted as A=∫ab‍f.
In the unbounded case, the Henstock-Kurzweil integral is defined as follows.Definition 2.2.
Given a gauge functionγ:[a,∞]→(0,∞), one can say that a tagged partition P={([xi-1,xi],ti)}i=1n+1 of [a,∞] is γ-fine, if (a)
a = x0,xn+1=tn+1=∞,(b)
[xi-1,xi]⊂[ti-γ(ti),ti+γ(ti)]    for alli=1,2,…,n,(c)
[xn,∞]⊆[1/γ(tn+1),∞].Definition 2.3.
A functionf:[a,∞]→ℝ is Henstock-Kurzweil integrable on [a,∞], if there exists A∈ℝ such that, for each ϵ>0, there is a gauge γϵ:[a,∞]→(0,∞) for which (2.1) is satisfied for all tagged partition P which is δϵ-fine according to Definition 2.2.
Letf be a function defined on an infinite interval [a,∞), One can suppose that f is defined on [a,∞] assuming that f(∞)=0. Thus, f is Henstock-Kurzweil integrable on [a,∞) if f extended on [a,∞] is HK-integrable. For functions defined over intervals (-∞,a] and (-∞,∞)One can makes similar considerations.
LetI be a finite or infinite interval. The space of all Henstock-Kurzweil integrable functions over I is denoted by ℋ𝒦(I). This space will be considered with the Alexiewicz seminorm, which it is defined as follows:
(2.2)∥f∥I=supJ⊆I|∫J‍f|,
where the supremum is being taken over all intervals J contained in I.Definition 2.4.
Letφ:I→ℝ be a function, where I⊆ℝ is a finite interval. The variation of φ over the interval I is defined as follows:
(2.3)VIφ=sup{∑‍ni=1|φ(xi)-φ(xi-1)|:PispartitionofI}.
We say that the functionφ is of bounded variation on I if VIφ<∞. Now if φ is a function defined on an infinite interval I, then φ is of bounded variation on I, if φ is of bounded variation on each finite subinterval of I and there is M>0 such that V[a,b]φ≤M for all [a,b]⊆I. The variation of φ on I is VIφ=sup{V[a,b]φ|[a,b]⊆I}.
Given an intervalI, the space of all bounded variation functions on I is denoted by ℬ𝒱(I). We set ℬ𝒱0(ℝ)={f∈ℬ𝒱(ℝ)∣lim|t|→∞f(t)=0}. The following are some classical theorems that are used throughout this paper. The first is given in [2, Lemma 24] and is an immediate consequence of [1, Theorem 10.12,and Corollary H.4].Theorem 2.5.
Ifg is a HK-integrable function on [a,b]⊆ℝ and f is a function of bounded variation on [a,b], then fg is HK-integrable on [a,b] and
(2.4)|∫ab‍fg|≤inft∈[a,b]|f(t)||∫ab‍g(t)dt|+∥g∥[a,b]V[a,b]f.Theorem 2.6 ([1] Chartier-Dirichlet's test).
Letf and g be functions defined on [a,∞). Suppose that(i)
g∈ℋ𝒦([a,c]) for every c≥a, and G defined by G(x)=∫ax‍g is bounded on [a,∞);(ii)
f is of bounded variation on [a,∞) and limx→∞f(x)=0.Then fg∈ℋ𝒦([a,∞)).Definition 2.7 (see [3]).
LetE⊆[a,b]. A function f:[a,b]→ℝ is ACδ on E, if for every ϵ>0, there exist ηϵ>0 and a gauge δϵ on E such that
(2.5)∑‍si=1|f(vi)-f(ui)|<ϵ,
whenever P={([ui,vi],ti)}i=1s is a (δϵ,E)-fine subpartition of [a,b] (i.e., P is δϵ-fine and the tags ti belong to E) and ∑i=1s|vi-ui|<ηϵ.
We say thatf is ACGδ on [a,b], if [a,b] can be written as a countable union of sets on each of which the function f is ACδ.
Ifh(t,s) is a function on ℝ×ℝ, then we use the notation D2h for the partial derivative of h with respect to the second component s.Theorem 2.8 (4, Theorem4).
Leta,b∈ℝ. If h:ℝ×[a,b]→ℂ is such that (i)
h(t,·) is ACGδ on [a,b] for almost all t∈ℝ;(ii)
h(·,s) is HK-integrable on ℝ for all s∈[a,b].
ThenH:=∫-∞∞‍h(t,·)dt is ACGδ on [a,b] and H'(s)=∫-∞∞‍D2h(t,s)dt for almost all s∈(a,b), if and only if,
(2.6)∫st‍∫-∞∞‍D2h(t,s)dtds=∫-∞∞‍∫st‍D2h(t,s)dsdt,
for all [s,t]⊆[a,b]. In particular,
(2.7)H′(s0)=∫-∞∞‍D2h(t,s0)dt,
when H2:=∫-∞∞‍D2h(t,·)dt is continuous at s0.


## 3. Main Results

All results in this paper are based on functions in the vector spaceℬ𝒱0(ℝ). Note that ℬ𝒱0(ℝ)⊈L(ℝ), where L(ℝ) is the space of Lebesgue integrable functions. Indeed, the function f:ℝ→ℝ defined by
(3.1)f(x)={0ifx∈(-∞,1),1xifx∈[1,∞),
is in ℬ𝒱0(ℝ)∖L(ℝ). However, for bounded intervals I, functions in ℬ𝒱(I) are Lebesgue integrables on I.To facilitate the statement of these results, it seems appropriate to introduce some additional terminology. Ifg:ℝ×ℝ→ℂ is a function and s0∈ℝ, we say that s0 satisfies Hypothesis (H) relative to g if (H)
there existδ=δ(s0)>0 and M=M(s0)>0, such that, if |s-s0|<δ then
(3.2)|∫uv‍g(t,s)dt|≤M,for all [u,v]⊆ℝ.This type of condition plays a major role in the results of the present work.Theorem 3.1.
Letf:ℝ→ℝ and g:ℝ×ℝ→ℂ   be functions. If f∈ℬ𝒱0(ℝ), and s0∈ℝ satisfies Hypothesis  (H) relative to g, then
(3.3)Γ(s)=∫-∞∞‍f(t)g(t,s)dt
exists for all s in a neighborhood of s0.Proof.
It follows by Theorem2.6.Theorem 3.2.
Letf:ℝ→ℝ and g:ℝ×ℝ→ℂ be functions such that (i)
f∈ℬ𝒱0(ℝ), g is bounded, and(ii)
g(t,·) is continuous for all t∈ℝ. If s0∈ℝ satisfies Hypothesis  (H) relative to g, then the function Γ defined in Theorem 3.1 is continuous at s0.Proof.
There existδ1>0 and M>0, such that, if |s-s0|<δ1 then
(3.4)|∫uv‍g(t,s)dt|≤M,
for all [u,v]⊆ℝ. From Theorem 3.1, Γ(s) exists for all s∈Bδ1(s0).
Letϵ>0 be given. By Hake's Theorem, there exists K1>0 such that
(3.5)|∫|t|≥u‍f(t)g(t,s0)dt|<ϵ3,
for all u≥K1. On the other hand, as
(3.6)limt→-∞V(-∞,t]f=0,limt→∞V[t,∞)f=0,
there is K2>0 such that for each t>K2,
(3.7)V(-∞,-t]f+V[t,∞)f<ϵ3M.
LetK=max{K1,K2}. From Theorem 2.5, it follows that for every v≥K and every s∈Bδ1(s0),
(3.8)|∫Kv‍f(t)g(t,s)dt|≤∥g(·,s)∥[K,v][inft∈[K,v]|f(t)|+V[K,v]f]≤M[|f(v)|+V[K,∞)f],
where the second inequality is true due to (3.4). This implies, since limt→∞|f(t)|=0, that
(3.9)|∫K∞‍f(t)g(t,s)dt|≤M·V[K,∞)f.
Analogously we have that(3.10)|∫-∞-K‍f(t)g(t,s)dt|≤M·V(-∞,-K]f.
Therefore, for eachs∈Bδ1(s0),
(3.11)|∫|t|≥K‍f(t)g(t,s)dt|≤M[V(-∞,-K]f+V[K,∞)f]<Mϵ3M=ϵ3.
By hypothesis,f is Lebesgue integrable on [-K,K], g is bounded, and g(t,·) is continuous for all t∈ℝ. From this it is easy to see, for example using [1, Theorem 12.12], that ΓK:ℝ→ℝ defined as
(3.12)ΓK(s)=∫-KK‍f(t)g(t,s)dt,s∈ℝ,
is continuous at s0. This implies that there is δ2>0 such that for every s∈Bδ2(s0),
(3.13)|∫-KK‍f(t)[g(t,s)-g(t,s0)]dt|<ϵ3.
Letδ=min{δ1,δ2}. Then for all s∈Bδ(s0),
(3.14)|Γ(s)-Γ(s0)|≤|∫-KK‍f(t)[g(t,s)-g(t,s0)]dt|+|∫|t|≥K‍f(t)g(t,s)dt|+|∫|t|≥K‍f(t)g(t,s0)dt|.
Thus, from (3.5), (3.11), and (3.13), |Γ(s)-Γ(s0)|<ϵ/3+ϵ/3+ϵ/3=ϵ, for all s∈Bδ(s0).Theorem 3.3.
Leta,b∈ℝ. If f:ℝ→ℝ and g:ℝ×[a,b]→ℂ are functions such that (i)
f∈ℬ𝒱0(ℝ), g is measurable, bounded, and(ii)
for alls∈[a,b], s satisfies hypothesis (H) relative to g, then
(3.15)∫ab‍∫-∞∞‍f(t)g(t,s)dtds=∫-∞∞‍∫ab‍f(t)g(t,s)dsdt.Proof.
From condition  (ii) and by the compactness of[a,b], we claim that there exists M>0 such that, for each s∈[a,b], |∫uv‍g(t,s)dt|≤M, for all [u,v]⊆ℝ.
For eachr>0 and s∈[a,b], let Γr(s)=∫-rr‍f(t)g(t,s)dt. Observe, by Theorem 2.5,
(3.16)|Γr(s)|=|∫-rr‍f(t)g(t,s)dt|≤∥g(·,s)∥[-r,r][inft∈[-r,r]|f(t)|+V[-r,r]f]≤M[|f(0)|+Vf],
for all s∈[a,b].
So, for eachr>0, Γr is HK-integrable on [a,b] and is bounded for a fixed constant. Moreover, by Theorem 3.1 and Hake's Theorem,
(3.17)limr→∞Γr(s)=Γ(s),
for all s∈[a,b].
Therefore, by dominated convergence theorem,Γ is HK-integrable on [a,b] and
(3.18)∫ab‍Γ(s)ds=limr→∞∫ab‍Γr(s)ds.
Now, sincef is Lebesgue integrable on [-r,r], g is measurable and bounded; it follows by Fubini's Theorem that
(3.19)∫ab‍∫-rr‍f(t)g(t,s)dtds=∫-rr‍∫ab‍f(t)g(t,s)dsdt.
Consequently,(3.20)limr→∞∫-rr‍∫ab‍f(t)g(t,s)dsdt=limr→∞∫ab‍Γr(s)ds=∫ab‍Γ(s)ds.
So by Hake's Theorem,(3.21)∫-∞∞‍∫ab‍f(t)g(t,s)dsdt=∫ab‍Γ(s)ds=∫ab‍∫-∞∞‍f(t)g(t,s)dtds.Theorem 3.4.
Considerf:ℝ→ℝ and g:ℝ×ℝ→ℂ functions, where f∈ℬ𝒱0(ℝ) and the partial derivative D2g exists on ℝ×ℝ and is bounded and continuous. If s0∈ℝ such that(i)
there isK>0 for which ∥g(·,s0)∥[u,v]≤K for all [u,v]⊆ℝ, and(ii)
s0 satisfies Hypothesis (H) relative to D2g;then Γ is differentiable at s0, and
(3.22)Γ′(s0)=∫-∞∞‍f(t)D2g(t,s0)dt.Proof.
It is not difficult to prove, using conditions  (i), (ii), and the Mean Value Theorem, that there existδ>0 and M>0 such that, for each s∈(s0-δ,s0+δ),
(3.23)|∫uv‍D2g(t,s)dt|<M,|∫uv‍g(t,s)dt|<M,
for all [u,v]⊆ℝ.
Considera,b∈ℝ with s0-δ<a<s0<b<s0+δ. In order to show (3.22), we use Theorem 2.8. The function f(t)g(t,·) is differentiable on [a,b] for each t∈ℝ, so f(t)g(t,·) is ACGδ on [a,b] for all t∈ℝ. Also, by (3.23) and Theorem 2.6, f(·)g(·,s) is HK-integrable on ℝ for all s∈[a,b]. Then
(3.24)Γ′(s0)=∫-∞∞‍f(t)D2g(t,s0)dt,
if
(3.25)Γ2:=∫-∞∞‍f(t)D2g(t,·)dt
is continuous at s0, and
(3.26)∫st‍∫-∞∞‍f(t)D2g(t,s)dtds=∫-∞∞‍∫st‍f(t)D2g(t,s)dsdt,
for all [s,t]⊆[a,b]. The first affirmation is true by (3.23) and Theorem 3.2. The second affirmation is true due to (3.23) and Theorem 3.3.Remark 3.5.
In the previous theorems the kernelg(t,s) satisfies |∫uv‍g(t,s)dt|≤M, for all [u,v]⊆ℝ. Moreover, if g will satisfy
(3.27)|∫uv‍g(t,s)dt|≤M0|s|,
for all [u,v]⊆ℝ, thenlim|s|→∞Γ(s)=0, when f∈ℬ𝒱0(ℝ) (a version of Riemann-Lebesgue Lemma).


## 4. Applications

Iff:ℝ→ℝ, then its Fourier transform at s∈ℝ is defined as follows:
(4.1)f^(s)=∫-∞∞‍f(t)e-itsdt.Talvila in [2] has done an extensive work about the Fourier transform using the Henstock-Kurzweil integral: existence, continuity, inversion theorems and so forth. Nevertheless, there are some omissions in those results that use [2, Lemma 25(a)]. Also Mendoza Torres et al. in [5] have studied existence, continuity, and Riemann-Lebesgue Lemma about the Fourier transform of functions belonging to ℋ𝒦(R)∩ℬ𝒱(ℝ). Following the line of [5], in Theorem 4.2, we include some results from them as consequences of theorems in the above section.Letf and g be real-valued functions on ℝ. The convolution of f and g is the function f*g defined by
(4.2)f*g(x)=∫-∞∞‍f(x-y)g(y)dy,
for all x such that the integral exists. Various conditions can be imposed on f and g to guarantee that f*g is defined on ℝ, for example, if f is HK-integrable and g is of bounded variation.Lemma 4.1.
Iff∈ℋ𝒦(ℝ)∩ℬ𝒱(ℝ), then lim|x|→∞f(x)=0.Proof.
Sincef is of bounded variation on ℝ, then limx→-∞f(x) and limx→∞f(x) exist. Suppose that limx→∞f(x)=α≠0. If α>0, there exists A>0 such that α/2<f(x), for all x>A. If α<0, there is B>0 such that -α/2<-f(x), for all x>B. This shows that f∉ℋ𝒦([A,∞)) or -f∉ℋ𝒦([B,∞)), which contradicts f∈ℋ𝒦(ℝ), so limx→∞f(x)=0. Using a similar argument, we show that limx→-∞f(x)=0.As consequence of Lemma4.1, the vector space ℋ𝒦(ℝ)∩ℬ𝒱(ℝ) is contained in ℬ𝒱0(ℝ). So the next theorem is an immediate consequence of  the above section.Theorem 4.2.
Iff∈ℋ𝒦(ℝ)∩ℬ𝒱(ℝ), then (a)
f^exists on ℝ.(b)
f^ is continuous on ℝ∖{0}.(c)
lim|s|→∞f^(s)=0.(d)
Defineg(t)=tf(t) and suppose that g∈ℋ𝒦(ℝ)∩ℬ𝒱(ℝ), then f^ is differentiable on ℝ∖{0}, and
(4.3)f^′(s)=-ig^(s),∀s∈ℝ∖{0}.(e)
Ifh∈L(ℝ)∩ℬ𝒱(ℝ), then f*h^(s)=f^(s)h^(s) for all s∈ℝ.Proof.
First observe that(4.4)|∫uv‍e-itsdt|≤2|s|,
for all [u,v]⊆ℝ. Then, each s0≠0 satisfies Hypothesis (H) relative to e-its.
(
a
) Theorem 3.1 implies that f^(s0) exists for all s0≠0 and, since f∈ℋ𝒦(ℝ), f^(0) exists. Thus, f^ exists on ℝ.
(
b
) From Theorem 3.2, f^ is continuous at s0, for all s0≠0.
(
c
) It follows by Remark 3.5 and (4.4).
(
d
) It follows by Theorem 2.8 in a similar way to the proof of Theorem 3.4.
(
e
) Take s∈ℝ and let k(x,y)=f(y-x)e-iys. Then, for each y∈ℝ and all [u,v]⊆ℝ,
(4.5)|∫uv‍k(x,y)dx|=|∫uv‍f(y-x)dx|=|∫y-uy-v‍f(z)dz|≤∥f∥.
Thus, for everyy∈ℝ, y satisfies Hypothesis (H) relative to k. Now, observe that h∈ℬ𝒱0(ℝ) and k is measurable and bounded. So by Theorem 3.3,
(4.6)∫-aa‍∫-∞∞‍h(x)k(x,y)dxdy=∫-∞∞‍∫-aa‍h(x)k(x,y)dydx,
for all a>0.
On the other hand,(4.7)|h(x)∫-aa‍f(y-x)e-iysdy|≤|h(x)||∫-a-xa-x‍f(z)e-izsdz|≤|h(x)|∥f(·)e-i(·)s∥.
Thus, sinceh∈L(ℝ), dominated convergence theorem implies that
(4.8)f^(s)h^(s)=∫-∞∞‍h(x)∫-∞∞‍f(y-x)e-iysdydx=lima→∞∫-∞∞‍h(x)∫-aa‍f(y-x)e-iysdydx,
but from (4.6), we have that
(4.9)f^(s)h^(s)=lima→∞∫-aa‍∫-∞∞‍h(x)f(y-x)e-iysdxdy=lima→∞∫-aa‍(f*h)(y)e-iysdy.
Therefore, by Hake's Theorem,
(4.10)f*h^(s)=f^(s)h^(s).Iff:[0,∞)→ℝ, then its Laplace transform at z∈ℂ is defined as follows:
(4.11)L(f)(z)=∫0∞‍f(t)e-ztdt.
Here, also the Laplace transform is considered as Henstock-Kurzweil integral.Theorem 4.3.
Iff∈ℋ𝒦([0,∞))∩ℬ𝒱([0,∞)), then (a)
L(f)(z) exists for all z∈ℂ with Rez≥0.(b)
IfF(x,y)=L(f)(x+iy), then F(·,y) is continuous on ℝ+∪{0} for all y≠0, and F(x,·) is continuous on ℝ for all x>0.Proof.
It is an easy consequence from Theorems3.1 and 3.2, since |∫uv‍e-(x+iy)tdt|≤2/|x+iy| for all u,v,x∈ℝ+∪{0}, y∈ℝ with x+iy≠0.Moreover, the Riemann-Lebesgue Lemma holds the following.Theorem 4.4.
Iff∈ℋ𝒦([0,∞))∩ℬ𝒱([0,∞)) and z=x+iy, with x≥0, then limy→∞L(f)(z)=0.Proof.
It results by Remark3.5 and (4.4), because f(·)e-x(·) is in ℋ𝒦([0,∞))∩ℬ𝒱([0,∞)).



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