chapter 5 henstock-kurzweil laplace...
TRANSCRIPT
![Page 1: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/1.jpg)
Chapter 5
Henstock-Kurzweil Laplace
Transform
5.1 Introduction
The Laplace transform of a function f : [0,∞) → R, at s ∈ C, is defined as
the integral [15] ∫ ∞0
e−stf(t) dt. (5.1)
Many authors, in some previous decades, showed their interest in studying
the Laplace transform in a classical sense; by classical meaning Lebesgue
integral and/or Riemann integral on the real line.
In this chapter, we focus on an integral transform that generalizes the clas-
sical Laplace transform. Here the generalization is given by the replacement,
in the definition of Laplace transform, i.e., in (5.1), of Lebesgue integral with
Henstock-Kurzweil (HK) integral. The Henstock-Kurzweil integral is defined
in terms of Riemann sums but with slight modification, yet it includes the
Riemann, improper Riemann, and Lebesgue integrals as special cases. This
integral is equivalent to the Denjoy and Perron integrals.
117
![Page 2: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/2.jpg)
The first question arises is that whether (or in which conditions) the
Laplace transform when treated as HK integral make sense. The sufficient
condition for the existence of Laplace transform is that the function f is
locally integrable on [0,∞), because the product of locally integrable func-
tion and bounded measurable function is locally integrable. Since Henstock-
Kurzweil integral allows nonabsolute convergence, it makes an ideal setting
for the Laplace transform. It was proved [9] that, on a compact interval, the
product of nonabsolute integrable function and a bounded variation function
is still nonabsolute integrable and this remains valid in case of an unbounded
interval [1]. We observed in (5.1) that the kernel function e−st, s ∈ C, is
not bounded variation function on [0,∞) hence we cannot claim (as in the
classical sense) that the HK Laplace transform exists for any HK integrable
function.
The convolution plays an important role in pure and applied mathemat-
ics, approximation theory, differential and integral equations and many other
areas. Laplace transform has the property that it can interact with con-
volutions so we obtain various results on convolution. We give necessary
and sufficient conditions so that the convolution operator as HK integral is
continuous. Finally we solve the problem of inversion by using generalized
differentiation.
The results in this chapter are appeared in [2, 14].
5.2 Existential Conditions
In this section, we tackle the problem of existence of Laplace transform as
HK integral.
According to Zayed [16], the classical sufficient condition for the existence
118
![Page 3: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/3.jpg)
of Laplace transform is that the function f is locally integrable on [0,∞), i.e.,
f ∈ Lloc[0,∞). This is because the multiplier for the Lebesgue integrable
functions is bounded measurable function and the function e−st, s ∈ C, is
bounded measurable function. We know that the multipliers for HK inte-
grable functions are precisely the functions of bounded variation [1]. Note
that in (5.1), the function e−st, s ∈ C, is not of bounded variation on [0,∞)
but it is of bounded variation on any compact interval J ⊂ R. Hence we
can not consider (5.1) as HK integral directly. So we cannot get any specific
condition for the existence of Laplace transform as HK integral. However we
do have the following different conditions.
Theorem 5.2.1. Let f : [0,∞) → R be any continuous function such that
F (x) =∫ x
0f , 0 ≤ x <∞, is bounded on [0,∞). Then the Laplace transform
L{f}(s), i.e., L{f}(s) =∫∞
0e−stf(t) dt, Re.s > 0 exists.
Proof. Let 0 < x <∞, s = σ + iω, σ, ω ∈ R.
We have (F (t) e−st)′= −s F (t) e−st + f(t) e−st.
Since e−st is of bounded variation on any compact interval, say [a, b] and F (t)
is bounded on [0,∞). Then F (t) e−st is HK integrable on [a, b].
Therefore −s F (t) e−st is integrable over I = [a, b].
Now ∫ b
a
∣∣∣(e−st)′∣∣∣ dt =
∫ b
a
∣∣−s e−st∣∣ dt=
∫ b
a
√σ2 + ω2 e−σt dt
=
√σ2 + ω2
σ[e−σa − e−σb].
Without loss of generality let us take a = 0. Therefore∫ b
0
∣∣∣(e−st)′∣∣∣ dt =
√σ2 + ω2
σ[1− e−σb] <∞.
119
![Page 4: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/4.jpg)
Hence (e−st)′
is absolutely integrable over [a, b].
Now by Hake’s theorem [1], we can write
limb→∞
∫ b
0
∣∣∣(e−st)′∣∣∣ dt = limb→∞
√σ2 + ω2
σ[1− e−σb]
=
√σ2 + ω2
σ, σ > 0
=
√σ2 + ω2
σ, Re.s > 0.
This shows that (e−st)′
is absolutely integrable on [0,∞).
Again limt→∞
e−st = 0, Re.s > 0.
Hence the function e−st : [0,∞)→ R is continuous on [0,∞) with limt→∞
e−st =
0 and (e−st)′
is absolutely integrable over [0,∞) and we have assumed that
f : [0,∞)→ R is continuous function such that F (x) =∫ x
0f , 0 ≤ x <∞, is
bounded on [0,∞).
Therefore by Dedekind’s test [10], we can say that the integral∫ ∞0
e−stf(t) dt exists forRe.s > 0.
Remark 5.2.2. The above result can also be proved by using Du-Bois Rey-
mond’s test [1] as follows:
Take φ(t) = e−st, s = σ + iω, J = [0,∞).
Then clearly φ(t) is differentiable and φ′ ∈ L1(J).
Since F is bounded, we have limt→∞
F (t) e−st = 0, Re.S > 0.
Hence F (x) =∫ x
0f , 0 ≤ x <∞ is bounded on J , e−st is differentiable on J ,
(e−st)′ ∈ L1(J) and F (t) e−st → 0 as t→∞, Re.s > 0.
Therefore by Du-Bois Reymond’s test, f(t)e−st ∈ HK(J).
That is, the Laplace transform of f(t),∫∞
0e−stf(t) dt exists, Re.s > 0.
120
![Page 5: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/5.jpg)
Theorem 5.2.3. If the Laplace transform of f(t),∫∞
0e−stf(t) dt exists for
Re.s > 0, then the function f ∈ HKloc.
Proof. Suppose the integral∫∞
0e−stf(t) dt exists for Re.s > 0.
Note that the function e−st, s ∈ C, as a function of real variable t, is not
of bounded variation on [0,∞) but is of bounded variation on any compact
interval I = [a, b].
By Hake’s theorem [1], we can write∫ ∞0
e−stf(t) dt = lima→0b→∞
∫ b
a
e−stf(t) dt.
Since the limits on R.H.S. exist, we can say that the HK integral∫ bae−stf(t)dt
exists, Re.s > 0 for some compact interval [a, b].
Hence the function f(t) is HK integrable necessarily on the compact interval
I = [a, b], that is, f ∈ HKloc.
Theorem 5.2.4. Let f ∈ HK([0,∞)). Define F (x) =∫∞xf(t) dt. Then
L{f}(s) exists if and only if∫∞
0e−stF (t) dt exists at s ∈ C.
Proof. Let T > 0. The function e−st, s ∈ C, is of bounded variation on a
compact interval [0, T ] and the function f is HK integrable on [0,∞).
Hence e−stf(t), s ∈ C, is HK integrable on [0, T ] and by integrating by parts,
we get ∫ T
0
e−stf(t) dt = e−sTF (T )− F (0) + s
∫ T
0
e−stF (t) dt.
Since the function f is HK integrable on [0,∞), F (0) is finite, say A and by
[9.12(a), [5]], F is continuous function such that limT→∞
F (T ) = 0.
Therefore by Hake’s theorem [1],
limT→∞
∫ T
0
e−stf(t) dt = limT→∞
{e−sTF (T )− F (0) + s
∫ T
0
e−stF (t) dt
}.
121
![Page 6: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/6.jpg)
Hence ∫ ∞0
e−stf(t) dt = s
∫ ∞0
e−stF (t) dt− A.
This shows that L{f}(s) exists if and only if∫∞
0e−stF (t) dt exists at s ∈ C,
Re.s > 0.
Theorem 5.2.5. (Uniqueness theorem) Let L{f}(s) =∫∞
0e−stf(t) dt
and L{g}(s) =∫∞
0e−stg(t) dt. If f(t) = g(t) a.e. on [0,∞), then L{f}(s) =
L{g}(s).
5.3 Basic Properties
The usual elementary properties such as linearity, dilation, modulation, trans-
lation are remains same for the HK Laplace transform, the proofs of which
are quite similar to that of classical ones. Also there are some other re-
sults that are analogous to the classical one with some modification in the
hypothesis which we discuss below.
Theorem 5.3.1. (HK Laplace transform of derivative) Let f : [0,∞)→
R be a function which is in ACGδ(R) such that limt→∞
f(t)est
= 0. Then for Re.s >
0 both L{f}(s) and L{f ′}(s) fail to exist or L{f ′}(s) = s L{f}(s)− f(0).
Proof. Let T > 0. Consider the integral J =∫ T
0e−stf
′(t) dt.
This integral is valid since the function e−st, s ∈ C, is bounded variation on
any compact interval [0, T ] and the function f′
is HK integrable on R.
By integrating by parts, we get
J = e−sTf(T )− f(0) + s
∫ T
0
e−stf(t) dt.
By Hake’s theorem [1],
limT→∞
∫ T
0
e−stf′(t) dt = lim
T→∞e−sTf(T )− f(0) + lim
T→∞s
∫ T
0
e−stf(t) dt
122
![Page 7: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/7.jpg)
= s
∫ ∞0
e−stf(t) dt− f(0).
Therefore ∫ ∞0
e−stf′(t) dt = s
∫ ∞0
e−stf(t) dt− f(0).
That is, L{f ′}(s) = sL{f}(s)− f(0).
Remark 5.3.2. The above result can be generalize by imposing the condi-
tions on f as:
For k = 0, 1, 2, . . . , (n− 1), f (k) are in ACGδ(R) and limt→∞
f (k)
est= 0, Re.s > 0.
Then L{f (n)}(s) and L{f}(s) fails to exist or
L{f (n)}(s) = sn L{f}(s)− sn−1 f(0)− sn−2 f′(0)− . . .− f (n−1)(0).
Theorem 5.3.3. (Differentiation of HK Laplace transform) Let f :
[0,∞) → R be a function such that L{f}(s) exists on a compact interval,
say J = [α, β], α, β ∈ R. Define g(t) = t f(t) and assume that g ∈ HK(R+).
Then L{g}(s) exists and L{g}(s) = −(L{f}(s))′ a.e. Here s ∈ R.
Proof. Suppose
L{f}(s) =
∫ ∞0
e−stf(t) dt exists on [α, β], α, β ∈ R.
Define g(t) = t f(t).
We know the necessary and sufficient condition for differentiating under the
HK integral is that∫ ∞t=0
∫ b
s=a
e−st tf(t) ds dt =
∫ b
s=a
∫ ∞t=0
e−st tf(t) dt ds (5.2)
for all [a, b] ⊂ [α, β] [12].
Now ∫ ∞t=0
∫ b
s=a
e−st tf(t) ds dt =
∫ ∞t=0
(e−at − e−bt)f(t) dt
123
![Page 8: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/8.jpg)
= L{f}(a)− L{f}(b)
which exists.
Therefore the double HK integral∫∞t=0
∫ bs=a
e−st tf(t) ds dt exists.
Hence by [lemma 25(a), [13]], the equality in (5.2) holds and we can differ-
entiate under the HK integral. By (5.2), we have∫ b
s=a
∫ ∞t=0
e−st tf(t) dt ds = L{f}(a)− L{f}(b)
= −∫ b
a
(L{f}(s))′ds.
Therefore ∫ b
s=a
∫ ∞t=0
e−st tf(t) dt ds = −∫ b
a
(L{f}(s))′ds
for all [a, b] ⊂ [α, β]. Which implies that∫ ∞0
e−st tf(t) dt = −(L{f}(s))′ a.e. on [α, β].
That is, L{t f(t)}(s) = −(L{f}(s))′ a.e. on [α, β].
Hence the derivative of the HK Laplace transform of f(t) exists a.e. on
[α, β].
Remark 5.3.4. Let G(s) = L{t f(t)}(s) and H(s) = −L{f(t)}(s).
Then (H(s))′
= G(s) a.e. on [α, β]. And, we know that [5], A function
f : [a, b] → R is HK integrable on [a, b] if and only if there exists an ACGδfunction F on [a, b] such that F
′= f a.e. on [a, b].
Again, by fundamental theorem of calculus [10], (H(s))′
is HK integrable,
i.e., every derivative is HK integrable. Hence G(s) is also HK integrable.
Consequently, the function H(s) is ACGδ on [α, β].
Continuing in this way n-times, we can say that if tnf(t) is HK integrable,
then any order of derivative of L{f}(s) exists a.e. So L{f}(s) is analytic a.e.
124
![Page 9: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/9.jpg)
Theorem 5.3.5. (Multiplication by 1t) Let f : [0,∞)→ R be a function
such that L{f}(s) exists and assume that f(t)t∈ HK(R+). Then L
{f(t)t
}(s)
exists and
L
{f(t)
t
}(s) =
∫ ∞s
L{f}(u) du.
Here s ∈ R.
Proof. Assume that f(t)t∈ HK(R+).
Since the function e−st is of bounded variation on any compact interval I =
[a, b], we have e−st f(t)t∈ HK([a, b]). That is,
∫ bae−st f(t)
tdt exists.
Without loss of generality, let us take a = 0 so that∫ b
0e−st f(t)
tdt exists.
By Hake’s theorem [1], limb→∞
∫ b0e−st f(t)
tdt exists.
Hence L{f(t)t
}(s) exists.
Now consider the integrals
I1 =
∫ ∞0
∫ ∞s
e−ut f(t) du dt, I2 =
∫ ∞s
∫ ∞0
e−ut f(t) dt du.
I1 =
∫ ∞0
[e−st − 0]f(t)
tdt
=
∫ ∞0
e−stf(t)
tdt which exists.
Therefore I1 exists on R× R.
Also, note that the function e−st is of bounded variation on any compact
interval J ⊂ R. Therefore∫R VJ(e−st) dt < MJ , for some constant MJ > 0.
Hence by [lemma 25(a), [13]], I2 exists on R× R and I1 = I2 on R× R. So∫ ∞0
e−stf(t)
tdt =
∫ ∞s
∫ ∞0
e−ut f(t) dt du
=
∫ ∞s
L{f}(u) du.
That is, L{f(t)t
}(s) =
∫∞s
L{f}(u) du.
125
![Page 10: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/10.jpg)
Theorem 5.3.6. (HK Laplace transform of integral) Let f : [0,∞)→ R
be HK integrable function such that L{f}(s) exists. Then L{∫ t
0f(u) du
}(s)
exists and L{∫ t
0f(u) du
}(s) = 1
sL{f}(s).
Proof. Define F (t) =∫ t
0f(u) du, t ∈ R+. Then F
′(t) = f(t) a.e. on R+
[9.12(b), [5]].
Since f ∈ HK(R+), the integral∫ ∞0
f(u) du exists on R+.
That is,
limt→∞
∫ t
0
f(u) du exists on R+.
Hence limt→∞
F (t) is bounded on R+.
Now for T > 0, consider∫ T
0
e−st f(t) dt =
∫ T
0
e−st F′(t) dt.
This integral exists since e−st is of bounded variation on [0, T ] and the func-
tion f(t) is HK integrable on R+.
By integrating by parts, we get∫ T
0
e−st f(t) dt = e−sT F (T ) + s
∫ T
0
e−st F (t) dt.
Since limT→∞
F (T ) is bounded on R, set limT→∞
F (T ) = A.
By Hake’s theorem [1], we can write∫ ∞0
e−st f(t) dt = limT→∞
∫ T
0
e−st f(t) dt
= limT→∞
e−sT F (T ) + s limT→∞
∫ T
0
e−st F (t) dt
= s
∫ ∞0
e−st F (t) dt Re.s > 0.
126
![Page 11: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/11.jpg)
Therefore ∫ ∞0
e−st F (t) dt =1
s
∫ ∞0
e−st f(t) dt, Re.s > 0.
That is, ∫ ∞0
e−st∫ t
0
f(u) du dt =1
sL{f}(s), Re.s > 0.
Hence L{∫ t
0f(u) du
}= 1
sL{f}(s), Re.s > 0.
5.4 Convolution
In this section we shall discuss some classical results of Laplace convolution
in the HK integral setting, called HK-convolution. First we give two results
about the existence of HK-convolution.
Theorem 5.4.1. Let f ∈ HK(R+) and g ∈ BV(R+). Then we have f ∗ g
exists on R+.
Proof. By definition [4],
f ∗ g(t) =
∫ t
0
f(t− τ) g(τ) dτ
which exists as HK-integral by multiplier theorem [1].
By integrating by parts, we get
|f ∗ g(t)| ≤ A‖f‖{
infR+|g|+ V[0,t] g(τ)
}.
Hence f ∗ g exists on R+.
Theorem 5.4.2. Let f ∈ HKloc and g ∈ BV such that the support of g is
in the compact interval I = [a, b]. Then
|f ∗ g(t)| ≤ A‖f‖[t−b,t−a]
{inf[a,b]|g|+ V[a,b] g(τ)
}.
127
![Page 12: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/12.jpg)
Proof. Suppose supp(g) ⊂ [a, b]. Then we have
f ∗ g(t) =
∫ b
a
f(t− τ)g(τ) dτ.
Integrate by parts, we get
|f ∗ g(t)| ≤∣∣∣∣g(τ)
∫ b
a
f(t− τ) dτ
∣∣∣∣+
∣∣∣∣∫ b
a
∫ b
a
f(t− τ) dg(τ)
∣∣∣∣≤ inf
τ∈[a,b]|g(τ)| ·
∣∣∣∣∫ b
a
f(t− τ) dτ
∣∣∣∣+
∣∣∣∣∫ b
a
f(t− τ) dτ
∣∣∣∣ · V[a,b] g(τ)
≤{
infτ∈[a,b]
|g(τ)|+ V[a,b] g(τ)
}sup
[t−b,t−a]
∣∣∣∣∫ t−a
t−bf(u) du
∣∣∣∣ .Therefore
|f ∗ g(t)| ≤ A‖f‖[t−b,t−a]
{inf[a,b]|g|+ V[a,b] g(τ)
}.
The following result shows that the HK-convolution is bounded.
Theorem 5.4.3. Let f ∈ HK(R+) and g ∈ L1(R+) ∩ BV(R+). Then we
have A‖f ∗ g‖ ≤ A‖f‖ ‖g‖1.
Proof. For 0 ≤ a < b ≤ ∞, we have∫ b
a
f ∗ g(t) dt =
∫ b
a
∫ t
0
f(t− τ)g(τ) dτ dt.
Let
I1 =
∫ b
a
∫ t
0
f(t− τ)g(τ) dτ dt and I2 =
∫ t
0
∫ b
a
f(t− τ)g(τ) dt dτ.
Consider
I2 =
∫ t
0
∫ b
a
f(t− τ)g(τ) dt dτ.
Since f ∈ HK(R+), we have∫ baf(t− τ) dt exists and finite.
Let ∫ t−b
t−af(τ) dτ = M.
128
![Page 13: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/13.jpg)
Therefore
|I2| ≤M
∫ t
0
|g(τ)| dτ.
Since g ∈ L1(R+), we have for some finite K,∫ t
0
|g(τ)| dτ = K.
So |I2| ≤MK and hence I2 exists on R+ × R+.
Now as g ∈ L1(R+), we have ‖g‖1 ≤ K1 and also∫R+ V[0,t] g ≤M1.
Therefore by [lemma 25, [13]], I1 = I2 on R+ × R+.
Hence ∣∣∣∣∫ b
a
f ∗ g(t) dt
∣∣∣∣ ≤ sup[t−a,t−b]
∣∣∣∣∫ t−b
t−af(u) du
∣∣∣∣ · ∫ t
0
|g(τ)| dτ
≤ A‖f‖[t−a,t−b] · ‖g‖1.
Therefore
sup[a,b]
∣∣∣∣∫ b
a
f ∗ g(t) dt
∣∣∣∣ ≤ A‖f‖[t−a,t−b] · ‖g‖1.
That is, A‖f ∗ g‖ ≤ A‖f‖ ‖g‖1.
Now we give some basic elementary properties of HK-convolution.
Theorem 5.4.4. Suppose f ∈ HK(R+) and g ∈ BV(R+). Then the follow-
ing holds:
I) f ∗ g(t) = g ∗ f(t) for all t ∈ R+.
II) (λf) ∗ g = f ∗ (λg) = λ(f ∗ g) for some λ ∈ C.
III) (f ∗ g)x(t) = fx ∗ g(t) = f ∗ gx(t).
IV) h ∗ (f + g)(t) = h ∗ f(t) + h ∗ g(t) for some h ∈ HK(R+).
Proof. The proof is similar to that of the classical case.
129
![Page 14: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/14.jpg)
The associative property of HK-convolution is given in the next theorem.
Theorem 5.4.5. If f ∈ HK(R+), g ∈ BV(R+) and h ∈ L1(R+), then
(f ∗ g) ∗ h(t) = f ∗ (g ∗ h)(t), ∀ t ∈ R+.
Proof. Consider
(f ∗ g) ∗ h(t) =
∫ t
0
∫ τ
0
f(ξ) g(τ − ξ) h(t− τ) dξ dτ.
By changing the order of integration, we get
(f ∗ g) ∗ h(t) =
∫ t
ξ=0
f(ξ)
∫ t−ξ
0
g(u)h(t− u− ξ) du dξ
=
∫ t
ξ=0
f(ξ) g ∗ h(t− ξ) dξ
= f ∗ (g ∗ h)(t).
Now Consider
I1 =
∫ t
τ=ξ
∫ t
ξ=0
f(ξ) g(τ − ξ)h(t− τ) dξ dτ
and
I2 =
∫ t
ξ=0
f(ξ)
∫ t
τ=ξ
g(τ − ξ)h(t− τ) dτ dξ.
Let
J =
∫ t
τ=ξ
g(τ − ξ)h(t− τ) dτ.
Integrate by parts, we get
|J | ≤∣∣∣∣g(t− ξ)
∫ t
τ=ξ
h(t− τ) dτ
∣∣∣∣+
∣∣∣∣∫ t
τ=ξ
∫ u
τ=ξ
h(t− τ) dτ dg(τ − ξ)∣∣∣∣
≤ inf[ξ,t]|g| ·
∣∣∣∣∫ t
τ=ξ
h(t− τ) dτ
∣∣∣∣+
∣∣∣∣∫ t
τ=ξ
h(t− τ) dτ
∣∣∣∣ · V[ξ,t] g(τ − ξ)
≤ inf[ξ,t]|g| ·
∫ t
τ=ξ
|h(t− τ)| dτ +
∫ t
τ=ξ
|h(t− τ)| dτ · V[ξ,t] g(τ − ξ)
≤ inf[ξ,t]|g| · ‖h‖1 + ‖h‖1 · V[ξ,t] g(τ − ξ)
130
![Page 15: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/15.jpg)
≤ ‖h‖1
{inf[ξ,t]|g|+ V[ξ,t] g(τ − ξ)
}.
Therefore
I2 ≤ supR+
∣∣∣∣∫ t
0
f(ξ) dξ
∣∣∣∣ ·{‖h‖1
{inf[ξ,t]|g|+ V[ξ,t] g(τ − ξ)
}}≤ A‖f‖ · ‖h‖1
{inf[ξ,t]|g|+ V[ξ,t] g(τ − ξ)
}.
Hence I2 exists on R+ × R+. And, by [lemma 25, [13]] we are through.
Now we show that the HK-Laplace transform of convolution of two func-
tions is the product of their HK-Laplace transforms.
Theorem 5.4.6. Let f1 ∈ HK(R+), f2 ∈ L1(R+) ∩ BV(R+). Then f ∗ g
exists and if L {f1(t)} and L {f2(t)} exist at s ∈ C, then L {f1 ∗ f2(t)} exists
at s ∈ C and L {f1 ∗ f2(t)} (s) = L {f1(t)} (s) L {f2(t)} (s).
Proof. Consider
L {f1 ∗ f2(t)} (s) =
∫ ∞0
e−stf1 ∗ f2(t) dt
=
∫ ∞0
∫ t
0
e−sτf1(τ)e−s(t−τ)f2(t− τ) dτ dt.
By interchanging the iterated integrals, we have
L {f1 ∗ f2(t)} (s) =
∫ ∞τ=0
e−sτf1(τ)
∫ ∞t=τ
e−s(t−τ)f2(t− τ) dt dτ
=
∫ ∞0
e−sτf1(τ) dτ
∫ ∞0
e−suf2(u) du.
But by hypothesis both the integrals∫ ∞0
e−sτ f1(τ) dτ and
∫ ∞0
e−sτ f2(τ) dτ
exist at s ∈ C.
Therefore L {f1 ∗ f2(t)} exists at s ∈ C and
131
![Page 16: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/16.jpg)
L {f1 ∗ f2(t)} (s) = L {f1(t)} (s) L {f2(t)} (s).
Now it remains to show the justification for interchanging the iterated inte-
grals.
Since the function f2 ∈ L1(R+), ∃ KI such that ‖f2‖1 < KI , where I =
[a, b] ⊂ R+ is any compact interval.
Clearly, V[a,b] e−s(t−τ) f2(t− τ) ≤ e−σu V[t−b,t−a] f2(τ) + |f2(τ)| 2 e−σr for some
reals u and r and so∫ ∞0
V[t−b,t−a]e−sτ f2(τ) dτ ≤ e−σu
∫ ∞0
V[t−b,t−a]f2(τ) dτ+2e−σr∫ ∞
0
|f2(τ)| dτ.
Therefore∫ ∞0
V[t−b,t−a] e−sτ f2(τ) dτ ≤ e−σu
∫ ∞0
V[t−b,t−a] f2(τ) dτ + 2KI e−σr.
Since f2 ∈ BV(R+), there exists a constant MI such that∫ ∞0
V[t−b,t−a]f2(τ) dτ < MI .
Then ∫ ∞0
V[t−b,t−a]e−sτf2(τ) dτ ≤ e−σuMI + 2KI e
−σr.
Thus for each compact interval I = [a, b] ⊂ R+, the integral∫ ∞0
V[t−b,t−a]e−sτf2(τ) dτ
is finite and ‖f2‖1 ≤ KI .
Hence the integral ∫ ∞τ=0
∫ ∞t=τ
e−stf1(τ) f2(t− τ) dt dτ
exists on R+×R+. And, by [lemma 25, [13]], we can interchange the iterated
integrals.
132
![Page 17: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/17.jpg)
Theorem 5.4.7. In addition to the hypothesis of the above theorem if f1 ∗
f2(t) is in HK(R+) and f2 is continuous, then f1 ∗ f2(t) is continuous with
respect to the Alexiewicz norm.
Proof. Choose any δ > 0 such that 0 < δ ≤ 1 (The case for negative δ is
analogous).
Define D(t, δ) = f1 ∗ f2(t+ δ)− f1 ∗ f2(t) = I1 + I2.
Where
I1 =
∫ t
0
f1(τ) [f2(t+ δ − τ)− f2(t− τ)] dτ, I2 =
∫ t+δ
t
f1(τ)f2(t+δ−τ) dτ.
Since f2 ∈ BV(R+), f2 is bounded on R+.
So there exists M2 such that |f2(x)| ≤M2, for all x ∈ R+. Then
|I2| ≤M2
∣∣∣∣∫ t+δ
t
f1(τ) dτ
∣∣∣∣ .But f1 ∈ HK(R+) implies that
limδ→0
∣∣∣∣∫ t+δ
t
f1(τ) dτ
∣∣∣∣ −→ 0.
Therefore |I2| → 0 as δ → 0.
Now consider
I1 =
∫ t
0
f1(τ) [f2(t+ δ − τ)− f2(t− τ)] dτ.
By integrating by parts, we get
|I1| ≤∣∣∣∣[f2(t+ δ − τ)− f2(t− τ)]
∫ t
0
f1(τ) dτ
∣∣∣∣+
∣∣∣∣∫ t
0
∫ u
0
f1(τ) dτ d(f2(u+ δ − τ)− f2(u− τ))
∣∣∣∣≤ |f2(t+ δ − τ)− f2(t− τ)| · sup
t∈R+
∣∣∣∣∫ t
0
f1(τ) dτ
∣∣∣∣+ sup
u∈R+
∣∣∣∣∫ u
0
f1(τ) dτ
∣∣∣∣ · ∣∣∣∣∫ t
0
d(f2(u+ δ − τ)− f2(u− τ))
∣∣∣∣133
![Page 18: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/18.jpg)
= |f2(t+ δ − τ)− f2(t− τ)| · supt∈R+
∣∣∣∣∫ t
0
f1(τ) dτ
∣∣∣∣+ sup
u∈R+
∣∣∣∣∫ u
0
f1(τ) dτ
∣∣∣∣ · |f2(u+ δ − τ)− f2(u− τ)|.
Therefore
|I1| ≤ |f2(t+ δ − τ)− f2(t− τ)| · supt∈R+
∣∣∣∣∫ t
0
f1(τ) dτ
∣∣∣∣+ sup
u∈R+
∣∣∣∣∫ u
0
f1(τ) dτ
∣∣∣∣ · |f2(u+ δ − τ)− f2(u− τ)| .
Since f2 is continuous on R+, for given ε > 0 there exists η > 0 such that
|I1| ≤ ε, ∀ δ < η.
Therefore |D(t, δ)| ≤ |I1|+ |I2| < ε, ∀ δ < η.
That is, for given ε > 0 there exists η > 0 such that |D(t, δ)| < ε, ∀ δ < η.
Hence f1 ∗ f2 is continuous at t ∈ R+.
Now, let α, β ∈ R+.
Consider∫ β
α
[f1 ∗ f2(t+ δ)− f1 ∗ f2(t)] dt =
∫ α+δ
α
f1 ∗ f2(t) dt−∫ β+δ
β
f1 ∗ f2(t) dt.
Write F1,2(x) =∫ x
0f1 ∗ f2(t) dt. Then
supα, β∈R+
∣∣∣∣∫ β
α
[f1 ∗ f2(t+ δ)− f1 ∗ f2(t)] dt
∣∣∣∣ ≤ supα∈R+
|F1,2(α + δ)− F1,2(α)|
+ supβ∈R+
|F1,2(β + δ)− F1,2(β)| .
Since F1,2 is an indefinite HK integral of f1 ∗ f2 on R+, it is continuous on
R+ [5]. Therefore for given ε > 0 there exists η > 0 such that
|F1,2(ξ + δ)− F1,2(ξ)| < ε
2, ∀ δ < η and for all ξ ∈ R+.
Hence we have
supα, β∈R+
∣∣∣∣∫ β
α
[f1 ∗ f2(t+ δ)− f1 ∗ f2(t)] dt
∣∣∣∣ < ε, ∀ δ < η.
134
![Page 19: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/19.jpg)
Thus for given ε > 0 there exists η > 0 such that A‖f1∗f2(t+δ)−f1∗f2(t)‖ <
ε, ∀ δ < η.
This shows that f1 ∗ f2(t) is continuous at t ∈ R+ with respect to the Alex-
iewicz norm.
The following result concerns with necessary and sufficient conditions
on fn and gn : [0, t] → R so that the HK-convolution as an operator is
continuous. It involves either uniform boundedness or uniform convergence
of the indefinite HK integral of fn.
Theorem 5.4.8. Let {fn} be a sequence of HK-integrable functions, fn :
[0, t] → R, ∀ n such that∫ t
0fn =
∫ t0f as n → ∞ for some HK-integrable
function f : [0, t]→ R. Define Fn (x) =∫ x
0fn, F (x) =
∫ x0f .
Let {gn} be a sequence of uniform bounded variation functions, gn : [0, t]→
R, ∀ n, such that gn → g pointwise on [0, t], where g : [0, t] → R. Then
fn ∗ gn(t)→ f ∗ g(t) as n→∞ if and only if
i) Fn → F uniformly on [0, t] as n→∞.
ii) Fn → F pointwise on [0, t], {Fn} is uniformly bounded and V (gn − g)→
0.
Proof. i) Suppose Fn → F uniformly on [0, t] as n→∞.
By assumption and multiplier theorem [1], fn ∗ gn(t) exists at t ∈ R+ and
fn ∗ gn(t) =
∫ t
0
fn(u) gn(t− u) du and f ∗ g(t) =
∫ t
0
f(u) g(t− u) du.
Consider fn ∗ gn(t)− f ∗ g(t) = I1 + I2,
where
I1 =
∫ t
0
[fn(u)− f(u)] gn(t− u) du, I2 =
∫ t
0
[gn(t− u)− g(t− u)] f(u) du.
135
![Page 20: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/20.jpg)
By integrating by parts, we get
I1 = gn(t− u)
∫ t
0
[fn(u)− f(u)] du−∫ t
0
∫ y
0
[fn(u)− f(u)] dgn
= gn(t− u)[Fn(t)− F (t)]−∫ t
0
[Fn(u)− F (u)] dgn.
Then
|I1| ≤Max0≤u≤t
|Fn(u)− F (u)|∫ t
0
dgn +M |Fn(t)− F (t)|
−→ 0 as n→∞ (Since Fn → F uniformly).
Hence I1 → 0 as n→∞.
And by integrating by parts, we get
I2 = {gn(t− u)− g(t− u)}∫ t
0
f −∫ t
0
F dgn +
∫ t
0
F dg.
By assumption the first term tends to 0 and since F is continuous and each
gn is of bounded variation with gn → g as n → ∞, so we have∫ t
0F dgn →∫ t
0F dg. Hence I2 → 0 as n→∞. Therefore fn ∗ gn(t)→ f ∗ g(t) as n→∞.
Now for the converse part, suppose that either Fn 9 F on [0, t] or Fn−F → 0
not uniformly on [0, t].
Then there is a sequence {xi}i∈Λ in [0, t] on which Fn 9 F uniformly. So,
by Bolzano-Weierstrass theorem [7], we can find a subsequence {yi}i∈Γ of
{xi}i∈Λ, Γ ⊂ Λ, such that Fn(yi) 9 F (yi) for all i and yi → y as i→∞, i.e.,
Fn(yi)− F (yi) 9 0.
Let us assume without loss of generality that 0 < yn ≤ y ≤ t. Let H be
Heaviside step function.
Define
gn(t− u) =
H(u− yn), n ∈ Γ
H(u− y), otherwise
136
![Page 21: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/21.jpg)
and g(t− u) = H(u− y).
Then we have, for n ∈ Γ, fn∗gn(t) = Fn(t)−Fn(yn) and f∗g(t) = F (t)−F (y).
Since Fn(yn) 9 F (yn) as n → ∞, yn → y as n → ∞ and F is continuous,
we have Fn(yn) 9 F (y).
But Fn(t) → F (t) as n → ∞. Therefore fn ∗ gn(t) 9 f ∗ g(t) as n → ∞
which is contradiction. Hence we must have Fn → F uniformly on [0, t].
ii) Suppose Fn → F pointwise on [0, t], Fn is uniformly bounded and V (gn−
g)→ 0.
Consider fn ∗ gn(t)− f ∗ g(t) = I1 + I2,
where
I1 =
∫ t
0
fn(u) [gn(t− u)− g(t− u)] du, I2 =
∫ t
0
[fn(u)− f(u)] g(t− u) du.
By integrating by parts, we get
I1 = [gn(t− u)− g(t− u)]
∫ t
0
fn −∫ t
0
∫ y
0
fn d(gn − g)
= [gn(t− u)− g(t− u)] Fn(t)−∫ t
0
Fn(y) d(gn − g).
Therefore
|I1| ≤ |gn(t− u)− g(t− u)| |Fn(t)|+ |Fn(y)| V (gn − g).
Since {Fn} is uniformly bounded, we have |Fn| ≤ M , ∀ n and for some
constant M .
Therefore
|I1| ≤ |gn(t− u)− g(t− u)|M +M V (gn − g).
By assumption gn → g as n→∞ and also V (gn − g)→ 0.
Hence I1 → 0 as n→∞.
Again by integrating by parts, we write
I2 = g(t− u) {Fn(t)− F (t)} −∫ t
0
(Fn − F ) dg. (5.3)
137
![Page 22: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/22.jpg)
By assumption Fn(t)→ F (t), so the first term tends to 0. And, since |Fn| ≤
M , Fn → F and g is of bounded variation, by the dominated convergence
theorem for Riemann-Stieltjes integral [6], we have∫ t
0
Fn dg →∫ t
0
F dg as n→∞.
Therefore by (5.3), I2 → 0 as n→∞.
Hence fn ∗ gn(t)→ f ∗ g(t) as n→∞.
Now for the converse part suppose there is c ∈ (a, b) such that Fn(c)−F (c) 9
0 as n→∞.
Let gn(t− x) = g(t− x) = H(x− c).
Then∫ t
0
fn(u) gn(t− u) du = Fn(t)− Fn(c),
∫ t
0
f(u) g(t− u) du = F (t)− F (c).
Since Fn(t)→ F (t) and Fn(c) 9 F (c), we have Fn(t)−Fn(c) 9 F (t)−F (c)
as n→∞. That is,
fn ∗ gn(t) 9 f ∗ g(t) as n→∞
which is contradiction. Hence we must have Fn → F pointwise on [0, t].
Now if {Fn} is not uniformly bounded, then there is a sequence {xi}i∈Λ in
[0, t] on which |Fn| → ∞.
Therefore by Bolzano-Weierstrass theorem [7], we can find a subsequence
{yi}i∈Γ of {xi}i∈Λ, Γ ⊂ Λ, such that Fn(yi) ≥ 1 for all n and Fn(yi) → ∞
and yi → y.
Without loss of generality let us take 0 ≤ yi ≤ y ≤ t.
Define gn(t− u) = H(u−yi)√Fn(yi)
.
Then V (gn) ≤ 1, g = 0 and V (gn − g) = 0 and∫ t
0
fn(u) gn(t− u) du =
∫ t
0
fn(u)H(u− yi)√
Fn(yi)du
138
![Page 23: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/23.jpg)
=
∫ t
yi
fn(u)√Fn(yi)
du
=Fn(t)√Fn(yi)
−√Fn(yi).
Therefore ∫ t
0
fn(u)gn(t− u) du→ −∞ as n→∞
and ∫ t
0
f(u)g(t− u) du = 0.
Hence fn ∗ gn 9 f ∗ g as n→∞ which is contradiction and so we must have
{Fn} is uniformly bounded sequence.
5.5 Inversion
Since in case of Henstock-Kurzweil integral, the integration process and dif-
ferentiation process are inverse of each other. Here we shall establish the
inverse of Henstock-Kurzweil Laplace transform by using Post’s generalized
differentiation method [8]. For this we shall need following lemmas:
Lemma 5.5.1. Let η be such that 0 < η < b − a, and h(x) ∈ C2(a ≤ x ≤
a+ η), h′(a) = 0, h
′′(a) < 0, h(x) is nonincreasing on (a, b]. Then∫ b
a
ek h(x) dx −→ ek h(a)
(−π
2k h′′(a)
) 12
k →∞.
Proof. Since h(x) ∈ C2(a ≤ x ≤ a+ η), h′′(x) is continuous.
Let ε > 0 be such that 0 < ε < −h′′(a).
Choose δ > 0, δ < η such that |h′′(x)− h′′(a)| < ε, a ≤ x ≤ a+ δ.
That is,
h′′(a)− ε < h
′′(x) < h
′′(a) + ε, a ≤ x ≤ a+ δ. (5.4)
139
![Page 24: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/24.jpg)
Consider the integral
Ik =
∫ b
a
ek [h(x)−h(a)] dx.
We write Ik = I′
k + I′′
k ,
where
I′
k =
∫ a+δ
a
ek [h(x)−h(a)] dx, I′′
k =
∫ b
a+δ
ek [h(x)−h(a)] dx.
Since h is nonincreasing on (a, b], we have h(x) ≤ h(a+ δ) for all x > a+ δ.
Therefore
I′′
k ≤∫ b
a+δ
ek [h(a+δ)−h(a)] dx
and
|I ′′k | ≤ ek [h(a+δ)−h(a)] (b− a− δ).
Hence I′′
k → 0 as k →∞.
Now consider
I′
k =
∫ a+δ
a
ek [h(x)−h(a)] dx.
By Taylor’s series with remainder, we have
h(x)− h(a) ≈ h′′(ξ)
(x− a)2
2, a ≤ ξ ≤ a+ δ.
Therefore
I′
k =
∫ a+δ
a
ek h′′
(ξ)(x−a)2
2 dx.
So by (5.4), we can write∫ a+δ
a
ek [h′′
(a)−ε] (x−a)22 dx ≤
∫ a+δ
a
ek h′′
(ξ)(x−a)2
2 dx ≤∫ a+δ
a
ek [h′′
(a)+ε](x−a)2
2 dx.
That is, ∫ a+δ
a
ek [h′′
(a)−ε] (x−a)22 dx ≤ I
′
k ≤∫ a+δ
a
ek [h′′
(a)+ε](x−a)2
2 dx. (5.5)
Consider
J =
∫ a+δ
a
ek [h′′
(a)−ε] (x−a)22 dx.
140
![Page 25: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/25.jpg)
After substituting x− a = u, we get
J =
∫ δ
0
ek2
[h′′
(a)−ε] u2 du
=1√
−k2
(h′′(a)− ε)
∫ δ√−k2
(h′′ (a)−ε)
0
e−u2
du
−→ 1√−k2
(h′′(a)− ε)
∫ ∞0
e−u2
du as k →∞
=1√
−k2
(h′′(a)− ε)
√π
2as k →∞.
Therefore∫ a+δ
a
ek [h′′
(a)−ε] (x−a)22 dx =
(π
−2k (h′′(a)− ε)
) 12
as k →∞.
Similarly,∫ a+δ
a
ek [h′′
(a)+ε](x−a)2
2 dx =
(π
−2k (h′′(a) + ε)
) 12
as k →∞.
Therefore (5.5) becomes(π
−2k (h′′(a)− ε)
) 12
≤ I′
k ≤(
π
−2k (h′′(a) + ε)
) 12
.
Since ε > 0 is arbitrary, we have(π
−2k h′′(a)
) 12
≤ I′
k ≤(
π
−2k h′′(a)
) 12
.
Hence
I′
k −→(
π
−2k h′′(a)
) 12
as k →∞.
So ∫ b
a
ek [h(x)−h(a)] dx −→(
π
−2k h′′(a)
) 12
as k →∞.
That is, ∫ b
a
ek h(x) dx −→ ek h(a)
(π
−2k h′′(a)
) 12
as k →∞.
141
![Page 26: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/26.jpg)
Lemma 5.5.2. Let η be such that 0 < η < b − a, and h(x) ∈ C2(b − η ≤
x ≤ b), h′(b) = 0, h
′′(b) < 0, h(x) is nonincreasing on [a, b). Then∫ b
a
ek h(x) dx −→ ek h(b)
(π
−2k h′′(b)
) 12
as k →∞.
Proof. The proof is similar to that of previous lemma.
Lemma 5.5.3. Let η be such that 0 < η < b − a, and h(x) ∈ C2(a ≤
x ≤ a + η), h′(a) = 0, h
′′(a) < 0, h(x) is nonincreasing on (a, b]. Suppose
φ(x) ∈ HK([a, b]). Then∫ b
a
φ(x) ek h(x) dx −→ φ(a) ek h(a)
(π
−2k h′′(a)
) 12
as k →∞.
Proof. Since h(x) ∈ C2(a ≤ x ≤ a+ η), we have h′′(x) is continuous.
Let ε > 0 be such that 0 < ε < −h′′(a).
Choose δ > 0, δ < η such that |h′′(x)− h′′(a)| < ε, a ≤ x ≤ a+ δ.
That is,
h′′(a)− ε < h
′′(x) < h
′′(a) + ε, a ≤ x ≤ a+ δ.
Consider the integral
Ik =
∫ b
a
[φ(x)− φ(a)] ek [h(x)−h(a)] dx.
We show that Ik → 0 as k →∞.
Let
gk(x) =
ek [h(x)−h(a)], x ∈ [a, b]
0, x = a.
We claim that gk → 0 as k →∞.
Since h is nonincreasing on (a, b], we have h(x)− h(a) < 0 for all x ∈ (a, b].
Therefore gk(x) = ek [h(x)−h(a)] −→ 0 as k →∞.
Set g(x) = 0 for all x ∈ [a, b]. Then
gk(x) = ek [h(x)−h(a)] −→ 0 = g(x) for all x ∈ [a, b], as k →∞.
142
![Page 27: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/27.jpg)
Since φ(x) ∈ HK([a, b]), the integral∫ b
a
[φ(x)− φ(a)] dx
exists.
Observe that the sequence {gk(x)} is of uniform bounded variation on [a, b]
with gk(x)→ 0 as k →∞.
Then by [Cor.3.2, [11]], we have∫ b
a
[φ(x)− φ(a)] ek [h(x)−h(a)] dx −→∫ b
a
[φ(x)− φ(a)] · 0 dx as k →∞
−→ 0 as k →∞.
Therefore ∫ b
a
φ(x) ek [h(x)−h(a)] dx = φ(a)
∫ b
a
ek [h(x)−h(a)] dx.
That is, ∫ b
a
φ(x) ek h(x) dx = φ(a)
∫ b
a
ek h(x) dx.
But by lemma (5.5.1), we have∫ b
a
ek h(x) dx −→ ek h(a)
(−π
2k h′′(a)
) 12
as k →∞.
Therefore∫ b
a
φ(x) ek h(x) dx −→ φ(a) ek h(a)
(−π
2k h′′(a)
) 12
as k →∞.
Lemma 5.5.4. Let η be such that 0 < η < b − a, and h(x) ∈ C2(b − η ≤
x ≤ b), h′(b) = 0, h
′′(b) < 0, h(x) is nonincreasing on [a, b) and suppose
φ(x) ∈ HK([a, b]). Then∫ b
a
φ(x) ek h(x) dx −→ φ(b) ek h(b)
(−π
2k h′′(b)
) 12
as k →∞.
143
![Page 28: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/28.jpg)
Proof. The proof is similar to that of the previous lemma.
Now we are ready to obtain the inversion theorem for Henstock-Kurzweil
Laplace transform.
Theorem 5.5.5. (Inversion Theorem) If f(t) ∈ HK(R+) and if the HK-
Laplace transform of f(t), L{f}(s) exists, then
limk→∞
1
k!
(k
t
)k+1 ∫ ∞0
e−kut uk f(u) du = f(t).
Proof. The inversion theorem follows immediately, if we can prove the fol-
lowing:
I)
limk→∞
1
k!
(k
t
)k+1 ∫ ∞t
e−kut uk f(u) du =
f(t)
2(5.6)
II)
limk→∞
1
k!
(k
t
)k+1 ∫ t
0
e−kut uk f(u) du =
f(t)
2. (5.7)
I) We have the relation∫ b
a
f(x) ek φ(x) dx −→ f(a) ek φ(a)
(−π
2k φ′′(a)
) 12
as k →∞.
Take a = t and φ(x) = ln(x)− xt
for all x ∈ [t,∞). Then we have∫ b
t
f(x) ek [ln(x)−xt
] dx −→ f(t) ek [ln(t)−1]
(−π
2k (−1t2
)
) 12
as k →∞.
Therefore∫ b
t
f(x) xk e−kxt dx −→ f(t) tk e−k
(π t2
2k
) 12
as k →∞.
That is, ∫ b
t
f(x) xk e−kxt dx −→ f(t) tk+1 e−k
( π2k
) 12
as k →∞.
144
![Page 29: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/29.jpg)
Therefore as k →∞, we have
1
k!
(k
t
)k+1 ∫ b
t
f(x) xk e−kxt dx =
1
k!
(k
t
)k+1
f(t) tk+1 e−k( π
2k
) 12
=
(k
e
)kk
k!f(t)
( π2k
) 12.
The Stirling’s formula is given by(ne
)n √2πn→ n!, using this we can write
1
k!
(k
t
)k+1 ∫ b
t
f(x) xk e−kxt dx =
k√2πk
f(t)( π
2k
) 12
as k →∞
=f(t)
2as k →∞.
This is true for every b ∈ R, b > t. Therefore by Hake’s theorem [1], we have
limk→∞
1
k!
(k
t
)k+1 ∫ ∞t
f(x) xk e−kxt dx =
f(t)
2.
II) Again we have the relation∫ b
a
f(x) ek φ(x) dx −→ f(b) ek φ(b)
(−π
2 k φ′′(b)
) 12
as k →∞.
Take b = t and φ(x) = ln(x)− xt, for all x ∈ (0, t]. Then we get
limk→∞
1
k!
(k
t
)k+1 ∫ t
a
f(x) xk e−kxt dx =
f(t)
2.
By Hake’s theorem [1], we have
limk→∞
1
k!
(k
t
)k+1 ∫ t
0
f(x) xk e−kxt dx =
f(t)
2.
Therefore from (5.6) and (5.7), we can write
limk→∞
1
k!
(k
t
)k+1 ∫ t
0
f(x) xk e−kxt dx = f(t).
145
![Page 30: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/30.jpg)
5.6 Examples
Here we shall give some examples of HK-Laplace transformable functions
whose classical Laplace transform do not exist. However we could not able
to find the exact form of the HK-Laplace transform of these functions.
Example 5.6.1. Consider the function g(t) = e−st sin tln(t+2)
, s ∈ C.
Let f(t) = e−st sin t and φ(t) = 1ln(t+2)
.
Observe that the function φ(t) is differentiable on R+ and φ′ ∈ L1(R+) and
if F (t) =∫ t
0e−su sinu du, Re.s > 0, then |F (t)| ≤ M for all t ∈ (0,∞),
Re.s > 0 for some constant M > 0.
Now
F (t) φ(t) =
{e−st(s sin t− cos t)
s2 + 1+
1
s2 + 1
}· 1
ln(t+ 2).
Then
limt→∞
F (t) φ(t) = limt→∞
{e−st(s sin t− cos t)
s2 + 1+
1
s2 + 1
}· 1
ln(t+ 2)
−→ 0.
Therefore limt→∞
F (t) φ(t) exists.
Hence by Du-Bois-Reymond’s test [1], we have e−st sin tln(t+2)
∈ HK(R+), Re.s > 0.
That is, ∫ ∞0
e−st sin t
ln(t+ 2)dt exists, Re.s > 0.
Hence the HK-Laplace transform of the function g(t) = sin tln(t+2)
exists, Re.s >
0.
Example 5.6.2. Consider the function f(t) = cos t√t
.
We have
F (t) =
∫ t
0
cosu√udu = 2
∫ √t0
cosu2 du.
146
![Page 31: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/31.jpg)
Let φ(t) = e−st, s ∈ C. Then φ′(t) = −s e−st ∈ L1(R+), Re.s > 0.
Now
F (t)φ(t) = 2 e−st∫ √t
0
cosu2 du.
Then
limt→∞
F (t)φ(t) = 2 limt→∞
e−st∫ √t
0
cosu2 du
= 2 limt→∞
e−st · limt→∞
∫ √t0
cosu2 du
= 0, Re.s > 0.
Therefore by Du-Bois-Reymond’s test [1], we have f(t)φ(t) ∈ HK(R+),
Re.s > 0.
Hence the HK-Laplace transform of the function f(t) = cos t√t
exists for Re.s >
0.
Example 5.6.3. Let φ(t) = e−σt
t, t ∈ R+, σ > 0.
This function is continuous on R+ and
φ(t+ 1) =1
(t+ 1) eσ(t+1)<
1
t eσt= φ(t) for all t ∈ R+
and also limt→∞
φ(t) = limt→∞
1t eσt
= 0, σ > 0.
Therefore φ(t) is decreasing to zero as t→∞.
Observe that the integral ∫ ∞0
e−σt
tdt, σ > 0
diverges.
Now let n0 ∈ N be such that 0 < (1 + 4n0) π4ω
.
For t ∈ R, we have | sinωt| ≥ 1√2
if and only if ωt ∈ ∪∞n=n0[(1 + 4n)π
4, (3 +
4n)π4].
147
![Page 32: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/32.jpg)
If n ∈ N, then (3 + 4n) π4< (1 + n) π.∫ (1+n)π
0
e−σt
t| sin t| dt =
n∑j=n0
∫ (3+4j)π4
(1+4j)π4
e−σt
t| sinωt| dt
≥n∑
j=n0
∫ (3+4j)π4
(1+4j)π4
e−σt
t
1√2dt
≥ 1√2
n∑j=n0
∫ (3+4j)π4
(1+4j)π4
1
(3 + 4j) π4eσ(3+4j)π
4
dt
=1√2
n∑j=n0
4
(3 + 4j) π es(3+4j)π4
· π2
≥ π
2√
2
n∑j=n0
1
(1 + j) π eσ(1+j)π.
On the other hand, we have∫ (1+n)π
0
e−σt
tdt =
∫ n0π
0
e−σt
tdt+
∫ (1+n)π
n0π
e−σt
tdt
=
∫ n0π
0
e−σt
tdt+
n∑j=n0
∫ (1+j)π
jπ
e−σt
tdt
≤∫ n0π
0
e−σt
tdt+
n∑j=n0
∫ (1+j)π
jπ
e−σjπ
jπdt
=
∫ n0π
0
e−σt
tdt+
n∑j=n0
e−σjπ
jπ· π.
But since e−σt
t∈ HK(R+), we have
∫∞0
e−σt
tdt =∞.
Therefore∑∞
j=n0
e−σjπ
jπ=∞.
So e−σt
tsinωt /∈ L1(R+), σ > 0.
Similarly, e−σt
tcosωt /∈ L1(R+), σ > 0.
Now let f(t) = sinωt, φ(t) = e−σt
t, t ∈ R.
Observe that φ(t) is monotone function and φ(t)→ 0 as t→∞.
And∣∣∣∫ t0 sinωu du
∣∣∣ ≤ 2|ω| , ω 6= 0 for all t ∈ R+. That is, F is bounded.
Therefore by Chartier-Dirichlet’s test [1], we have e−σt
tsinωt ∈ HK(R+).
148
![Page 33: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/33.jpg)
Similarly, e−σt
tcosωt ∈ HK(R+).
Hence e−σt
t{cosωt− i sinωt} ∈ HK(R+).
That is,∫∞
0e−σt
te−iωt dt exists, σ > 0.
Thus∫∞
0e−st
tdt exists, Re.s > 0 as a HK integral, where s = σ + iω.
149
![Page 34: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/34.jpg)
References
[1] Bartle R. G., A Modern Theory of Integration, Grad. Stud. Math. 32,
Amer. Math. Soc. Providence, 2001.
[2] Chaudhary M. S., Tikare S. A., On Gauge Laplace Transform, Int. Jour-
nal of Math. Analysis, Vol. 5, No. 35 (2011), 1733-1740.
[3] Erdelyi A., Tables of Integral transforms, Vol. 1, McGraw Hill Book
Company, Inc. 1954.
[4] Gasquet C., P. Witomski P., Fourier analysis and applications, Springer-
Verlag, New York, Inc. 1999.
[5] Gordon R. A., The Integrals of Lebesgue, Denjoy, Perron, and Henstock,
Grad. Stud. Math. 4, Amer. Math. Soc. Providence, 1994.
[6] McLeod R. M., The Generalized Riemann Integral, Carus Math. Mono-
graphs, No. 20, The Mathematical Association of America, Washington
DC, 1980.
[7] Natanson I. P., Theory of functions of real variable, Vol. 1, Frederick
Unger Publishing Co. New York, 1964.
[8] Post E. L., Generalized differentiation, Trans. Amer. Math. Soc. Vol. 32
(1930), 723-781.
150
![Page 35: Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)](https://reader031.vdocuments.site/reader031/viewer/2022022716/5c201a2309d3f2f2538c5d7c/html5/thumbnails/35.jpg)
[9] Saks S., Theory of the Integral, Dover Publications, New York, 1964.
[10] Swartz C., Introduction to Gauge Integrals, World Scientific Publishing
Co. Singapore, 2001.
[11] Talvila E., Limits and Henstock integrals of products, Real Anal. Ex-
change 25 (1999-2000), 907-918.
[12] Talvila E., Necessary and sufficient conditions for differentiating under
the integral sign, Amer. Math. Monthly, 108 (2001), 544-548.
[13] Talvila E., Henstock-Kurzweil Fourier transforms, Ill. Jour. Math. Vol.
46, No. 4(2002), 1207-1226.
[14] Tikare S. A., Chaudhary M. S., On some results of HK-convolution, Int.
Electron. J. Pure Appl. Math. Vol. 2, No. 2 (2010), 93-102.
[15] Widder D. V., The Laplace transform, Princeton University Press,
Princeton, 1946.
[16] Zayed A. I., The Handbook of Function and Generalized Function trans-
formations, CRC Press, Inc. 1996.
151