Published online: June 2, 2014
LAPLACE TRANSFORM USING THE HENSTOCK-KURZWEIL INTEGRAL
SALVADOR S ´ANCHEZ-PERALES AND JES ´US F. TENORIO
Abstract. We consider the Laplace transform as a Henstock-Kurzweil in-tegral. We give conditions for the existence, continuity and differentiability of the Laplace transform. A Riemann-Lebesgue Lemma is given, and it is proved that the Laplace transform of a convolution is the pointwise product of Laplace transforms.
1. Introduction
Given a function f : [0,∞) → R, its Laplace transform at x is defined by L{f}(x) = R∞
0 f(t)e
−txdt. In this paper we consider this transform using the
Henstock-Kurzweil integral. This integral generalizes the Riemann and Lebesgue integrals, as well as the Riemann and Lebesgue improper integrals. We prove some properties of the Laplace transform (existence, continuity and differentiability). A Riemann-Lebesgue Lemma is given, and various conditions are imposed on the functions so that the Laplace transform of a convolution is the pointwise product of Laplace transforms.
Throughout this paper we use the variablesto indicate thatL{f}(s) is defined on a real numbers, and we usezto indicate thatL{f}(z) is defined on a complex numberz.
2. Preliminaries
Let us begin by recalling the definition of the Henstock-Kurzweil integral. For finite intervals inRit is defined in the following way:
Definition 2.1. Let f : [a, b] → R be a function. We say that f is
Henstock-Kurzweil (shortly, HK-) integrable, if there existsA∈Rsuch that, for each >0,
there is a functionγ: [a, b]→(0,∞) (named a gauge) with the property that for
anyγ-fine partitionP ={( [xi−1, xi], ti)}ni=1 of [a, b] (i.e.{[xi−1, xi] :i= 1, . . . , n}
is a non-overlapping partition of [a, b] and for each i, [xi−1, xi]⊂[ti−γ(ti), ti+
γ(ti) ]), one has
|Σni=1f(ti)(xi−xi−1)−A|< . (2.1)
2010Mathematics Subject Classification. 44A10, 26A39.
The numberAis the integral of f over [a, b] and it is denoted asA=Rb
af.
In the unbounded case, we require the following definition.
Definition 2.2. Given a gauge functionγ: [a,∞]→(0,∞) we say that a tagged partitionP ={( [xi−1, xi], ti)}ni=1+1 of [a,∞] isγ-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 function f : [a,∞] → R is Henstock-Kurzweil integrable on
[a,∞], if there existsA∈Rsuch that, for each >0, there is a gaugeγ: [a,∞]→
(0,∞) for which (2.1) is satisfied for every tagged partition P which is γ-fine
according to Definition 2.2.
Letf be a real function defined on an infinite interval [a,∞); we can suppose thatf is defined on [a,∞] assuming thatf(∞) = 0. Thusf is Henstock-Kurzweil integrable on [a,∞) iff extended to [a,∞] is HK-integrable. For functions defined over intervals (−∞, a] and (−∞,∞) we make similar considerations.
Let I be a finite or infinite interval. The space of all Henstock-Kurzweil inte-grable functions onIis denoted byHK(I). This space will be considered with the Alexiewicz semi-norm, which is defined as
kfkI = sup J⊆I
Z
J
f
, f ∈ HK(I),
where the supremum is being taken over all intervalsJ contained inI.
Definition 2.4. Letϕ:I→Cbe a function, whereI⊆Ris a finite interval. The variation ofϕon the intervalI is defined as
VIϕ= sup
( n
X
i=1
|ϕ(xi)−ϕ(xi−1)|
P is partition ofI )
.
We say that the function ϕis 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 ϕis of bounded variation on each finite subinterval ofI and there exists some M > 0 such that V[a,b]ϕ≤M for all [a, b] ⊆I. The variation of ϕ onI is
VIϕ= sup{V[a,b]ϕ|[a, b]⊆I}. The space of all bounded variation functions onI is denoted byBV(I).
The following theorems are classical and will be used throughout this paper.
Theorem 2.5. [11, Lemma 24] Ifg is a HK-integrable function on[a, b]⊆Rand
f is a function of bounded variation on [a, b], then f g is HK-integrable on [a, b] and
Z b
a
f g
≤ inf
t∈[a,b]|f(t)|
Z b
a
g(t)dt
+kgk[a,b]V[a,b]f.
(i) g ∈ HK([a, c])for every c≥a, and Gdefined by G(x) =Rx
a g is bounded on
[a,∞);
(ii) f is of bounded variation on[a,∞)and lim
x→∞f(x) = 0. Thenf g∈ HK([a,∞)).
Theorem 2.7 (Du Bois-Reymond’s Test, [2]). Let f and ϕ be functions defined on[a,∞)and suppose that:
(1) f ∈ HK([a, c])for all c≥aandF(x) =Rx
a f is bounded on [a,∞);
(2) ϕis differentiable on[a,∞)andϕ0 is Lebesgue integrable on [a,∞); (3) lim
x→∞F(x)ϕ(x)exists. Thenf ϕ∈ HK([a,∞)).
Definition 2.8 ([4]). LetE ⊆[a, b]. A functionf : [a, b]→RisACδ onE, if for
every >0, there existη>0 and a gauge δ onE such that s
X
i=1
|f(vi)−f(ui)|< ,
whenever P = {( [ui, vi], ti)}si=1 is a (δ, E)-fine subpartition of [a, b] (i.e., P is
δ-fine and the tagsti belong toE) and s
P
i=1
|vi−ui|< η.
We say thatf isACGδ on [a, b], if [a, b] can be written as a countable union of
sets on each of which the functionf isACδ.
If h is a function in the variables (t, s), then we use the notation D2h for the partial derivative ofhwith respect to the variables.
Theorem 2.9. [10, Theorem 4]Let a, b∈R. Ifh:R×[a, b]→Cis such that
(i) h(t,·)isACGδ on[a, b] for almost allt∈R,
(ii) h(·, s)is HK-integrable onRfor alls∈[a, b],
thenH:=R∞
−∞h(t,·)dtisACGδ on[a, b]andH
0(s) =R∞
−∞D2h(t, s)dtfor almost alls∈(a, b), if and only if,
Z t
s
Z ∞
−∞
D2h(t, s)dtds=
Z ∞
−∞
Z t
s
D2h(t, s)dsdt for all[s, t]⊆[a, b]. In particular,
H0(s0) =
Z ∞
−∞
D2h(t, s0)dt, whenH2 :=
R∞
−∞D2h(t,·)dt is continuous ats0. 3. Main results
LetI be a finite or infinite interval. We use the following notation:
• L(I) ={f | f is Lebesgue integrable onI},
• Lloc={f |f is Lebesgue integrable on each [a, b]⊆[0,∞)},
• HKloc={f |f is HK-integrable on each [a, b]⊆[0,∞)},
• B(I) ={f |f is bounded onI},
• BV(I) ={f |f is of bounded variation onI},
• BV0([0,∞)) ={f ∈ BV([0,∞))| lim
t→∞f(t) = 0},
• BV0[∞] ={f |f ∈ BV([b,∞)) for someb >0 and lim
b≤t→∞f(t) = 0}. Talvila in [11] introduced the spaceHKloc∩ BV0[∞] to study the Fourier trans-form. Also Mendoza, Escamilla and S´anchez (see [5, 6, 7]) have studied this trans-form on the spaces HK([0,∞))∩ BV([0,∞)) and BV0([0,∞)). All these spaces satisfied the following inclusion relations:
(1) HK([0,∞))∩ BV([0,∞))⊆ BV0([0,∞))⊆ HKloc∩ BV0[∞];
(2) HK([0,∞))∩ BV([0,∞))6⊆L([0,∞)), (see [5, Example 2.1 (i)]);
(3) HKloc∩ BV0[∞]6⊆ HK([0,∞)) andHK([0,∞))6⊆ HKloc∩ BV0[∞] (see [5,
Example 2.1 (ii)]).
In this paper, we analyse the Laplace transform on the spaces above.
It is well known that if f ∈ Lloc and there exist constants c ∈ R, M >0 and T >0 such that|f(t)| ≤M ectfor all t > T, thenL{f}(s) exists for alls > c. We
give other existential conditions.
Observe that ifs∈R+, thene−st, as a function of a real variablet, is monotone
and lim
t→∞e
−st = 0. Therefore, from Chartier-Dirichlet’s Test (Theorem 2.6), it
follows that iff ∈ HKlocandF(x) =
Rx
0 f(t)dt, 0≤x <∞, is bounded on [0,∞), thenL{f}(s) exists for alls >0. So the next theorem follows.
Theorem 3.1. If f ∈ HK([0,∞)), thenL{f}(s)exists for alls∈[0,∞).
Proof. Its clear that if f ∈ HK([0,∞)), then L{f}(0) exists. Moreover, since
Rx
0 f(t)dt
≤ kfk[0,∞) for all x ≥ 0, it follows that L{f}(s) exists for all s ∈
[0,∞).
Example 3.2. For anya >0, sintt ∈ HK([a,∞))\L([a,∞)). Thus the function
fa : [0,∞)→Rdefined by
fa(t) =
(sint
t , ift≥a,
0, if 0≤t < a
belongs toHK([0,∞)). ThereforeL{fa}(s) exists for alls∈[0,∞).
The functiont7→e−zt, now whenz∈C, is not of bounded variation on [0,∞),
so in order to prove the existence of the Laplace transform we can’t use Chartier-Dirichlet’s Test. In [3] the next result is proved using Du Bois-Reymond’s Test. For the sake of completeness, here we will give its proof.
Theorem 3.3. If f ∈ HKloc and F(x) =
Rx
0 f(t)dt, 0 ≤ x <∞, is bounded on [0,∞), thenL{f}(z) exists for allz∈CwithRez >0.
Proof. The condition Rez > 0 implies that lim
t→∞e
−zt = 0 and so d dt(e
−zt) is
Lebesgue integrable on [0,∞). Also lim
x→∞F(x)e
−zx= 0. Therefore by Theorem 2.7,
L{f}(z) =R∞
0 f(t)e
−ztdtexists.
Corollary 3.4. If f ∈ HK([0,∞)), thenL{f}(z) exists for all z ∈ Csuch that Rez >0.
Another condition for the existence of L{f}(z), that does not requireF(x) =
Rx
0 f(t)dt to be bounded on [0,∞), is shown in the following theorem.
Theorem 3.5. If f ∈ HKloc∩ BV0[∞] then L{f}(z) exists for all z ∈ C with
Rez >0.
Proof. There exists b > 0 such that f ∈ BV([b,∞)) and lim
t→∞f(t) = 0. Since Rez > 0, it follows that
Rx
b e
−tzdt
< |2z| for all x ∈ [b,∞). This implies, from
Theorem 2.6, that R∞
b f(t)e
−tzdt exists. Now since f ∈ HK
loc and e−tz is of
bounded variation on [0, b], the integralR0bf(t)e−tzdtalso exists.
Example 3.6. Let f : [1,∞) → R be a function defined by f(t) = t−p, where
0 < p < 1. Observe that lim
x→∞
Rx
0 t
−pdt = ∞ but f ∈ HK
loc∩ BV0[∞]. Thus,
Corollary 3.4 does not apply; however from Theorem 3.5, L{f}(z) exists for all
z∈Cwith Rez >0.
Remark 3.7. Iff ∈ HK([0,∞))∩ BV([0,∞)), thenL{f}(z) exists for allz∈C
with Rez ≥ 0. It is clear that this condition is satisfied when Rez > 0, be-cause HK([0,∞))∩ BV([0,∞)) ⊆ HKloc ∩ BV0[∞]. Now, if Rez = 0, then
L{f}(z) = f χ\[0,∞)(Imz), where the latter expression is the Fourier transform of f χ[0,∞) evaluated at Imz. From [6, Theorem 3.1], f χ\[0,∞)(Imz) exists, since
f ∈ HK([0,∞))∩BV([0,∞)). ThusL{f}(z) also exists whenz∈Cwith Rez= 0.
Now, we prove that the Laplace transform is continuous on [0,∞) when f ∈ HK([0,∞)), and is continuous on (0,∞) whenf ∈ HKloc∩ BV0[∞].
Proposition 3.8.Letf ∈ HK([0, b]),b >0. IfFbis defined byFb(s) =
Rb
0 f(t)e −ts,
thenFb is continuous on[0,∞).
Proof. Takes0∈[0,∞), and note that fors∈[0,∞) with|s−s0|< δ,
|Fb(s)−Fb(s0)|=
Z b
0
f(t)e−ts0[e−t(s−s0)−1]dt
≤ kf(·)e−(·)s0k
[0,b]
inf
t∈[0,b]|e
−t(s−s0)−1|+V
[0,b][e−t(s−s0)−1]
≤ kf(·)e−(·)s0k
[0,b]
h
V[0,b][e−t(s−s0)−1]
i .
Since |d dt(e
−t(s−s0) −1)| = |s −s
0|e−t(s−s0) ≤ |s −s0|eδb, it follows that
V[0,b][e−t(s−s0)−1]≤2|s−s0|eδbb. Thus
|Fb(s)−Fb(s0)| ≤2b|s−s0|eδbkf(·)e−(·)s0k[0,b], and hence lim
s→s0
Proposition 3.9. If f ∈ HK([a, b]), then Z v u
f(t)e−tsdt
≤e−askfk[a,b], for all0≤a≤u≤v≤b ands≥0. Proof. Takea≤u≤v ≤b ands≥0. Since e−ts, as a function of the variablet,
is decreasing, it follows thatV[u,v]e−ts ≤e−us−e−vs; thus by Theorem 2.5,
Z v u
f(t)e−tsdt
≤ kfk[u,v]
inf
t∈[u,v]
|e−ts|+V
[u,v]e−ts
≤ kfk[u,v]
e−vs+ (e−us−e−vs)
≤ kfk[u,v]e−us ≤ e−askfk[a,b].
Theorem 3.10. If eitherf ∈ HK([0,∞))or f ∈ HKloc∩ BV0[∞], thenL{f} is
continuous on (0,∞).
Proof. Take s0 ∈(0,∞) and let >0 be given. Consider 0 < δ1 < s0/2. First observe that for eachK >0 ands∈(s0−δ1, s0+δ1),
|L{f}(s)−L{f}(s0)| ≤ |FK(s)−FK(s0)|
+ Z ∞ K
f(t)(e−ts−e−ts0)dt
. (3.1)
We claim that there exists K0 > 0 such that
R∞
K0f(t)(e
−ts−e−ts0)dt <
2 independently ofs∈(s0−δ1, s0+δ1).
Assumption 1: f ∈ HK([0,∞)). By Hake’s Theorem, there exists K1 >0 such that
kfk[K1,∞)<
2. (3.2)
Note that, for allv≥K1ands0−δ1< s≤s0,
Z v K1
f(t)(e−ts−e−ts0)dt = Z v K1
f(t)e−ts[e−t(s0−s)−1]dt
≤ kf(·)e−(·)sk[K1,v]
inf
t∈[K1,v]
|e−t(s0−s)−1|
+V[K1,v][e
−t(s0−s)−1]
.
(3.3)
From Proposition 3.9,kf(·)e−(·)sk
[K1,v] ≤ kfk[K1,v]. Also, sinces≤s0, it follows
that the functione−t(s0−s)−1 is decreasing and not positive. Thus, the right-hand
side of the inequality (3.3) is bounded by
kfk[K1,v] h
1−e−K1(s0−s)+ (e−K1(s0−s)−e−v(s0−s))i
and this equalskfk[K1,v]
1−e−v(s0−s). So, taking the limit (asv→ ∞) produces Z ∞ K1
f(t)(e−ts−e−ts0)dt
for alls0−δ1< s≤s0. Of course this inequality is also true whens0≤s < s0+δ1.
Assumption 2: f ∈ HKloc∩ BV0[∞]. First observe that for all u, v ≥ 0 and
s∈(s0−δ1, s0+δ1),
Z v
u
(e−ts−e−ts0)dt
≤ 6 s0
.
We setM to the right side of this inequality. Since lim
t→∞V[t,∞)f = 0,it follows that there existsK2>0 such thatV[K2,∞)f <
2M. Then, from Theorem 2.5, it follows
that for everyv≥K2ands∈(s0−δ1, s0+δ1),
Z v
K2
f(t)(e−ts−e−ts0)dt
≤M
inf
t∈[K2, v]
|f(t)|+V[K2, v]f
≤M|f(v)|+V[K2,∞)f
.
This implies, since limt→∞|f(t)|= 0, that
Z ∞
K2
f(t)(e−ts−e−ts0)dt
≤M·V[K2,∞)f < M
2M =
2.
Therefore, with any of the two hypotheses, we see that our initial assertion is true. Finally, from Proposition 3.8, FK0 is continuous, so there exists δ2 > 0
such that for every s ∈ (0,∞) with |s−s0| < δ2, |FK0(s)−FK0(s0)| <
2. Let
δ= min{δ1, δ2}; then by (3.1) we have that for alls∈(s0−δ, s0+δ),
|L{f}(s)−L{f}(s0)|<
2 +
2 =.
It can be shown that iff ∈ HKlocandF(x) =
Rx
0 f(t)dt, 0≤x <∞, is bounded on [0,∞), then
lim
s→s+0L
{f}(s) =L{f}(s0) for alls0∈[0,∞).
We have already seen the continuity of the Laplace transform of a function, now we give another feature of this transform. This property indicates that not every arbitrary continuous function is a Laplace transform of some function.
Given 0≤a≤bwe set
Γ([a, b]) =
(
f ∈ HK([a, b])| lim
s→∞
Z b
a
f(t)e−tsdt= 0
)
.
It is clear thatL([a, b])⊆Γ([a, b]); however Γ([a, b])6⊆L([a, b]), see Example 3.12 below. Moreover, if 0 < a ≤ b, then, from Proposition 3.9, it follows that
HK([a, b]) = Γ([a, b]). Whena= 0 the veracity of the equalityHK([a, b]) = Γ([a, b]) is still an open question. A class of functions belonging to Γ([0, b]) is given in the next theorem:
then
lim
s→∞
Z b
0
g(t)e−tsdt= 0.
Proof. LetG(t) =tf(t), then G0(t) =g(t) for allt∈[0, b]\A. From [2, Theorem 4.7], g ∈ HK([0, b]) and by [2, Theorem 10.12], g(t)e−ts∈ HK([0, b]). Integrating
by parts we obtain
Z b
0
g(t)e−tsdt=e−tsG(t)
b
0 +
Z b
0
G(t)se−tsdt
=e−bsbf(b) +
Z b
0
f(t)tse−tsdt.
Let h: [0,∞) → R defined by h(x) = xe−x, then his a measurable function
satisfying
lim
r→∞ 1
r Z r
0
h(x)dx= lim
r→∞ 1
r
1− 1 er −
r er
= 0;
so, sincef ∈ L[0, b], by the generalized Riemman-Lebesgue Lemma ([1, Theorem 4.4.1]),
lim
s→∞
Z b
0
h(ts)f(t)dt= 0.
Thuse−bsbf(b) +Rb
0 f(t)tse
−tsdt→0 whens→ ∞.
Example 3.12. Defineg: [0,1]→Rby
g(t) =
(
cosπ
t
+π
t sin π
t
, ift∈(0,1],
0, ift= 0.
Theng∈ HK([0,1]), butg6∈L([0,1]). Note thatg(t) = cos πt+tdtd cos πt. Thus by Theorem 3.11,
lim
s→∞
Z 1
0
g(t)e−tsdt= 0,
and henceg∈Γ([0,1]).
Theorem 3.13 (Riemann-Lebesgue Lemma). Let f : [0,∞) → R be a function
such that f ∈Γ([0, b])for someb >0. If either
(1) f ∈ BV0([b,∞))or (2) f ∈ HK([b,∞)),
then lim
s→∞L{f}(s) = 0.
Proof. Lets >0. Iff ∈ BV0([b,∞)), thenf(·)e−(·)s∈ HK([b,∞)) and
Z ∞
b
f(t)e−tsdt
≤ 2
Therefore the conclusion of the theorem follows sincef ∈Γ([a, b]). On the other hand, iff ∈ HK([b,∞)), then by Proposition 3.9,
Z ∞
b
f(t)e−tsdt
≤ kfk[b,∞)e−bs,
and again the conclusion of the theorem is satisfied.
The following result is shown in [8, Theorem 3.3].
Lemma 3.14. Let a, b ∈ R. If g : [0,∞) → R and h : [0,∞)×[a, b] → C are
functions such that
(i) g∈ BV0([0,∞)),his measurable, bounded and (ii) there existsM >0 such that, for each y∈[a, b],
Z v
0
h(x, y)dx
≤M,
for all v≥0,
then
Z b
a
Z ∞
0
g(x)h(x, y)dxdy =
Z ∞
0
Z b
a
g(x)h(x, y)dydx.
Theorem 3.15. If f ∈ BV0([0,∞))and s0∈(0,∞), then L{f} is differentiable ats0, and
(L{f})0(s0) =− Z ∞
0
tf(t)e−ts0dt. (3.4)
Proof. There exista, b, M >0 witha < s0< bsuch that, for eachs∈[a, b],
Z v
0
e−tsdt
< M (3.5)
and
Z v
0
te−tsdt
< M (3.6)
for allv≥0.
In order to show (3.4) we use Theorem 2.9. The function f(t)e−t(·)is differen-tiable on [a, b] for all t∈ [0,∞), so f(t)e−t(·) isACGδ on [a, b] for allt ∈[0,∞).
By (3.5) and Theorem 2.6,f(·)e−(·)s is HK-integrable on [0,∞) for all s ∈[a, b].
Then
(L{f})0(s0) =
Z ∞
0
−tf(t)e−ts0dt,
if
Γ :=
Z ∞
0
−tf(t)e−t(·)dt
is continuous ats0, and
Z t
s
Z ∞
0
−tf(t)e−tsdtds=
Z ∞
0
Z t
s
for all [s, t]⊆[a, b]. The first assertion follows using (3.6) and a similar argument as in the proof of Theorem 3.10 (Assumption 2). The second claim is true due to
(3.6) and Lemma 3.14.
Iff andg are functions defined on the interval [0,∞), then their convolution is the functionf ∗g defined by
f∗g(y) =
Z y
0
f(y−x)g(x)dx.
It is clear that iff ∈ HK([0,∞)) andg∈ BV([0,∞)), thenf∗gexists on [0,∞) by the Multiplier Theorem (see [2]), and f ∗g(y) = g∗f(y) for all y ∈ [0,∞). In [9], the equalityL{f∗g}(s) =L{f}(s)L{g}(s) is established, under certain conditions. However the authors use [11, Lemma 25(a)] and it has an omission unlessf has compact support. We provide other conditions and a different proof.
Theorem 3.16. If f ∈ HK([0,∞))∩ B([0,∞)) and g ∈L([0,∞))∩ BV([0,∞)), then
L{f∗g}(z) =L{f}(z)L{g}(z)
for allz∈Cwith Rez >0.
Proof. Takez∈Cwith Rez >0. From Corollary 3.4, it follows thatL{f}(z) and L{g}(z) exist, andL{f}(z)L{g}(z) =R∞
0 g(x)
R∞
x f(y−x)e
−yzdydx.
LetD={(x, y)|0≤x, x≤y}and consider
h(x, y) =
(
f(y−x)e−yz, if (x, y)∈D,
0, if (x, y)∈([0,∞)×[0,∞))\D.
For eacha, x≥0,
g(x)
Z a
0
h(x, y)dy
≤ |g(x)|kf(·)e−(·)zk.
Thus, since g ∈ L([0,∞)), the Dominated Convergence Theorem and Hake’s Theorem imply the second equality of the following:
L{f}(z)L{g}(z) =
Z ∞
0
Z ∞
0
g(x)h(x, y)dydx
= lim
a→∞
Z ∞
0
Z a
0
g(x)h(x, y)dydx.
(3.7)
On the other hand, since Rv
0 h(x, y)dx
≤ kfk for all v ≥ 0, it follows, from
Lemma 3.14, that the right side of (3.7) is equal to
lim
a→∞
Z a
0
Z ∞
0
Therefore
lim
a→∞
Z a
0
g∗f(y)e−yzdy= lim
a→∞
Z a
0
Z y
0
g(x)f(y−x)e−yzdxdy
= lim
a→∞
Z a
0
Z ∞
0
g(x)h(x, y)dxdy
=L{f}(z)L{g}(z).
Again using Hake’s Theorem, we obtain that g∗f(·)e−(·)z ∈ HK([0,∞)) and
L{f∗g}(z) =L{f}(z)L{g}(z).
References
[1] G. Bachman, L. Narici, E. Beckenstein,Fourier and Wavelet Analysis, Springer-Verlag, 1991. MR 1729490
[2] R. G. Bartle,A Modern Theory of Integration, Grad. Studies in Math., Vol. 32, Amer. Math. Soc., Providence, 2001. MR 1817647
[3] M.S. Chaudhary and Sanket A. Tikare,On Gauge Laplace Transform, Int. J. Math. Anal. 5, No. 35, 2011, 1733–1740. MR 2853594
[4] R. A. Gordon,The Integrals of Lebesgue, Denjoy, Perron, and Henstock, Grad. Studies in Math., Vol. 4, Amer. Math. Soc., Providence, 1994. MR 1288751
[5] F. J. Mendoza Torres, J. A. Escamilla Reyna and S. S´anchez Perales,Inclusion Relations for the SpacesL(R),HK(R)∩ BV(R)andL2(R), Russ. J. Math. Phys. 16, No. 2, 2009, 287–289.
MR 2525413
[6] F. J. Mendoza Torres, J. A. Escamilla Reyna and S. S´anchez Perales,Some results about the Henstock-Kurzweil Fourier transform, Math. Bohem. 134 (2009), 379–386. MR 2597233 [7] F. J. Mendoza Torres,On pointwise inversion of the Fourier transform of BV0 functions,
Ann. Funct. Anal. 1, No. 2, 2010, 112–120. MR 2772044
[8] S. S´anchez-Perales, F. J. Mendoza Torres, J. A. Escamilla Reyna,Henstock-Kurzweil Integral Transforms, Int. J. Math. Math. Sci. vol. 2012, 11 pages. MR 2983789
[9] Sanket A. Tikare and M.S. Chaudhary, On some results of HK-convolution, International Electronic Journal of Pure and Applied Mathematics 2, No. 2, 2010, 93–102.
[10] E. Talvila, Necessary and sufficient conditions for differentiating under the integral sign, Amer. Math. Monthly, 108 (2001) 544–548. MR 1840661
[11] E. Talvila,Henstock-Kurzweil Fourier transforms, Illinois J. Math. 46 (2002), 1207–1226. MR 1988259
Salvador S´anchez-Perales
Instituto de F´ısica y Matem´aticas, Universidad Tecnol´ogica de la Mixteca, Km. 2.5, Carretera a Acatlima, 69000 Huajuapan de Le´on, Oaxaca, Mexico
Jes´us F. Tenorio
Instituto de F´ısica y Matem´aticas, Universidad Tecnol´ogica de la Mixteca, Km. 2.5, Carretera a Acatlima, 69000 Huajuapan de Le´on, Oaxaca, Mexico
