Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 44, pp. 1–7.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
INTEGRATION BY PARTS FOR THE Lr
HENSTOCK-KURZWEIL INTEGRAL
PAUL MUSIAL, FRANCESCO TULONE
Abstract. Musial and Sagher [4] described a Henstock-Kurzweil type integral
that integrates Lr -derivatives. In this article, we develop a product rule for
the Lr -derivative and then an integration by parts formula.
1. Introduction
Definition 1.1 ([4]). A real-valued function f defined on [a, b] is said to be Lr
Henstock-Kurzweil integrable (f ∈ HKr [a, b]) if there exists a function F ∈ Lr [a, b]
so that for any ε > 0 there exists a gauge function δ(x) > 0 so that whenever
{(xi , [ci , di ])} is a δ-fine tagged partition of [a, b] we have
Z di
n 1/r
X
1
(L)
|F (y) − F (xi ) − f (xi )(y − xi )|r dy
< ε.
di − ci
ci
i=1
In the sequel, if an integral is not specified, it is a Lebesgue integral. It is shown
in [4] that if f is HKr -integrable on [a, b], the following function is well-defined for
all x ∈ [a, b]:
Z x
F (x) = (HKr )
f (t) dt
(1.1)
a
Here the function F is called the indefinite HKr integral of f . Our aim is to
establish an integration by parts formula for the HKr integral. In a manner similar
to L. Gordon [2] we state the following
Theorem 1.2. Suppose that f is HKr -integrable on [a, b], and G is absolutely
0
continuous on [a, b] with G0 ∈ Lr ([a, b]), where 1 ≤ r < ∞, r0 = r/(r − 1) if r > 1,
0
and r = ∞ if r = 1. Then f G is HKr -integrable on [a, b] and if F is the indefinite
HKr integral of f , then
Z b
Z b
(HKr )
f (t)G(t) dt = F (b)G(b) −
F (t)G0 (t) dt.
a
a
We note that if r = 1 so that r0 = ∞, the condition on G is that it is a Lipschitz
function of order 1 on [a, b].
2000 Mathematics Subject Classification. 26A39.
Key words and phrases. Henstock-Kurzweil; integration by parts.
c 2015 Texas State University - San Marcos.
Submitted December 8, 2014. Published February 16, 2015.
1
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P. MUSIAL, F. TULONE
EJDE-2015/44
In the classical case where f is Henstock-Kurzweil integrable (r = ∞, r0 = 1),
Theorem 1.2 holds, but it is enough to assume that G is of bounded variation on
Rb
[a, b]. In that case the integral on the right is the Riemann-Stieltjes integral a F dG.
See [3] for a proof of this statement.
To prove Theorem 1.2 we will need a product rule for the Lr -derivative. We
will also utilize a characterization of the space of HKr -integrable functions that
involves generalized absolute continuity in Lr sense (ACGr ([a, b])).
2. Product rule for the Lr -derivative
Definition 2.1 ([1]). For 1 ≤ r < ∞, a function F ∈ Lr ([a, b]) is said to be
Lr -differentiable at x ∈ [a, b] if there exists a ∈ R such that
Z h
|F (x + t) − F (x) − at|r dt = o(hr+1 ).
−h
It is clear that if such a number a exists, then it is unique. We say that a is the
Lr -derivative of F at x, and denote the value a by Fr0 (x).
Theorem 2.2. For 1 ≤ r < ∞, let x ∈ R and suppose F ∈ Lr (I) where I
is an interval having x in its interior, and suppose F is Lr -differentiable at x.
Suppose also that G ∈ L∞ (I) and that G is Lr -differentiable at x. Then F G is
Lr -differentiable at x and (F G)0r (x) = Fr0 (x)G(x) + F (x)G0r (x).
Proof. Let ε > 0. We need to choose γ so that for 0 < h < γ
Z h
|F (x + t)G(x + t) − F (x)G(x) − H(x)t|r dt < εhr+1
(2.1)
−h
Fr0 (x)G(x)+F (x)G0r (x).
where H(x) =
We add and subtract the terms F (x)G(x+t)
and Fr0 (x)G(x + t)t to the part of the integrand inside the absolute value signs. We
also note that if a, b and c are non-negative numbers then
(a + b + c)r ≤ C(ar + br + cr )
where C is a positive constant that depends on r.
Choose γ0 > 0 and N > 0 so that F ∈ Lr ([x − γ0 , x + γ0 ]) and that
esssup[x−γ0 ,x+γ0 ] G < N.
We then have that if 0 < h < γ0 then the integral in (2.1) is less than or equal to
Z h
C
|G(x + t)|r |F (x + t) − F (x) − Fr0 (x)t|r dt
(2.2)
−h
Z
h
+C
−h
Z h
+C
|F (x)|r |G(x + t) − G(x) − G0r (x)t|r dt
(2.3)
|Fr0 (x)|r |(G(x + t) − G(x))t|r dt.
(2.4)
−h
For (2.2), choose γ1 < γ0 so that if 0 < h < γ1 we have
Z h
εhr+1
|F (x + t) − F (x) − Fr0 (x)t|r dt <
4CN r
−h
so that
Z
h
C
−h
|G(x + t)|r |F (x + t) − F (x) − Fr0 (x)t|r dt <
εhr+1
.
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INTEGRATION BY PARTS
3
For (2.3), choose γ2 < γ1 so that if 0 < h < γ2 we have
Z h
εhr+1
|G(x + t) − G(x) − G0r (x)t|r dt <
4C(|F (x)|r + 1)
−h
so that
Z
h
|F (x)|r |G(x + t) − G(x) − G0r (x)t|r dt <
C
−h
εhr+1
4
For (2.4), we note that
Z h
C
|Fr0 (x)|r |(G(x + t) − G(x))t|r dt
−h
= C|Fr0 (x)|r
Z
h
|(G(x + t) − G(x) − G0r (x)t + G0r (x)t)t|r dt
−h
≤C
2
|Fr0 (x)|r hr
Z
h
|(G(x + t) − G(x) − G0r (x)t)|r dt
−h
Z
h
+
|G0r (x)t|r dt
−h
≤C
2
|Fr0 (x)|r hr
Z
h
|(G(x + t) − G(x) − G0r (x)t)|r dt
−h
+ 2C 2 |Fr0 (x)|r h2r+1 |G0r (x)|r .
Now we note that we can choose
1/r 0 < γ < min 1, γ2 , ε/ 8C 2 (|G0r (x)| + 1)(|Fr0 (x)| + 1)
so that if 0 < h < γ we have
Z h
|(G(x + t) − G(x) − G0r (x)t)|r dt <
−h
εhr+1
4C 2 (|Fr0 (x)|r + 1)
We then have that if 0 < h < γ, then
Z h
2
0
r r
C |Fr (x)| h
|(G(x + t) − G(x) − G0r (x)t)|r dt
−h
< (C 2 |Fr0 (x)|r hr )
≤
εhr+1
4C 2 (|Fr0 (x)|r + 1)
εh2r+1
εhr+1
<
4
4
and that
2C 2 |Fr0 (x)|r h2r+1 |G0r (x)|r
≤ 2C 2 |Fr0 (x)|r hr+1 |G0r (x)|r
ε
8C 2 (|Fr0 (x)| + 1)(|G0r (x)| + 1)
εhr+1
.
4
We can then conclude that (2.1) holds and the theorem is therefore proved.
≤
In [4] we find sufficient conditions for HKr -integrability. We will need the following definitions.
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P. MUSIAL, F. TULONE
EJDE-2015/44
Definition 2.3 ([4]). We say that F ∈ ACr (E) if for all ε > 0 there exist η > 0
and a gauge function δ(x) defined on E so that if P = {(xi , [ci , di ])} is a finite
collection of non-overlapping δ-fine tagged intervals having tags in E and satisfying
q
X
(di − ci ) < η
i=1
then
q X
i=1
di
Z
1
di − ci
|F (y) − F (xi )|r dy
1/r
< ε.
ci
Definition 2.4 ([4]). We say that F ∈ ACGr (E) if E can be written
E = ∪∞
i=1 Ei
and F ∈ ACr (Ei ) for all i.
Lemma 2.5. Suppose that F and G are in ACGr ([a, b]), and that G ∈ L∞ ([a, b]).
Then F G ∈ ACGr ([a, b]).
Proof. The function F ∈ ACGr ([a, b]) and so we can find a sequence of sets
∞
{An }∞
n=1 so that [a, b] = ∪n=1 An and F ∈ ACr (An ) for all n. Since G belongs to
∞
ACGr ([a, b]), we can also find a sequence of sets {Bm }∞
m=1 so that [a, b] = ∪m=1 Bm
and G ∈ ACr (Bm ) for all m. We can then write
∞
[a, b] = ∪∞
n=1 ∪m=1 (An ∩ Bm ).
We will rewrite the sequence {An ∩ Bm }n,m≥1 as {Ek }k≥1 . We then have that
both F and G are in ACr (Ek ) for all k ≥ 1. We will show that F G ∈ ACGr (Ek )
for all k.
Let N = 1 + kGk∞ and fix k. For j ≥ 1 let
Uj = {x ∈ Ek : j − 1 ≤ |F (x)| < j}
We then have
Ek = ∪∞
j=1 Uj .
We will show that F G ∈ ACr (Uj ) for all j.
Let ε > 0. There exist η > 0 and a gauge function δ(x) defined on Uj so that
if P = {xi , [ci , di ]} is a finite collection of non-overlapping δ-fine tagged intervals
having tags in Uj and satisfying
q
X
(di − ci ) < η
i=1
then
q X
Z
1
di − ci
Z
i=1
q X
i=1
di
1
di − ci
|F (y) − F (xi )|r dy
1/r
<
ε
,
2N
<
ε
.
2j
ci
di
|G(y) − G(xi )|r dy
1/r
ci
Then for such P,
q X
i=1
1
di − ci
Z
di
ci
|F (y)G(y) − F (xi )G(xi )|r dy
1/r
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INTEGRATION BY PARTS
≤
q X
i=1
1
di − ci
q X
+
≤N
i=1
di
Z
|F (y)G(y) − F (xi )G(y)|r dy
1/r
ci
di
Z
1
di − ci
i=1
q X
5
|F (xi )G(y) − F (xi )G(xi )|r dy
1/r
.
ci
1
di − ci
q
X
+ |F (xi )|
(
i=1
Z
di
|F (y) − F (xi )|r dy
1/r ci
Z
1
di − ci
di
|G(y) − G(xi )|r dy)1/r
ci
ε ε
≤N
+ j( ) = ε.
2N
2j
Now we can conclude that for P,
Z di
q 1/r
X
1
|F (y)G(y) − F (xi )G(xi )|r dy
<ε
di − ci ci
i=1
and so that F G ∈ ACGr ([a, b]).
3. Linearity of ACGr (E)
We now show that ACGr (E) is a linear space.
Theorem 3.1. Suppose F and G are in ACGr (E). Then for any constants a and
b we have that aF + bG ∈ ACGr (E).
Proof. Write E as ∪∞
n=1 En . We will show that aF + bG ∈ ACr (En ) for every n.
First we show that aF ∈ ACr (En ). Let ε > 0 and choose η > 0 and a gauge
function δ(x) defined on En so that if P = {xi , [ci , di ]} is a finite collection of
non-overlapping δ-fine tagged intervals having tags in E and satisfying
q
X
(di − ci ) < η
i=1
then
q X
i=1
1
di − ci
Z
di
|F (y) − F (xi )|r dy
1/r
ci
<
ε
.
|a| + 1
Then
q X
i=1
1
di − ci
q X
= |a|
Z
di
|aF (y) − aF (xi )|r dy
1/r
ci
1
di − ci
i=1
ε < |a|
< ε.
|a| + 1
Z
di
|F (y) − F (xi )|r dy
1/r ci
Now we show that F + G ∈ ACGr (E). Let ε > 0 and choose η > 0 and a
gauge function δ(x) defined on En so that if P = {xi , [ci , di ]} is a finite collection
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P. MUSIAL, F. TULONE
EJDE-2015/44
of non-overlapping δ-fine tagged intervals having tags in E and satisfying
q
X
(di − ci ) < η ,
i=1
then
q X
i=1
q X
i=1
1
di − ci
Z
1
di − ci
Z
di
|F (y) − F (xi )|r dy
1/r
|G(y) − G(xi )|r dy
1/r
<
ε
,
2
<
ε
.
2
ci
di
ci
Then we have for this P, using Minkowski’s inequality,
Z di
q 1/r
X
1
|F (y) + G(y) − (F (xi ) + G(xi ))|r dy
di − ci ci
i=1
Z di
q 1/r
X
1
≤
|F (y) + F (xi )|r dy
di − ci ci
i=1
q
1/r
X 1 Z di
+
|G(y) − G(xi )|r dy
d i − ci c i
i=1
ε ε
< + = ε.
2 2
We will use the following characterization of HKr -integrable functions.
Theorem 3.2 ([4]). Let 1 ≤ r < ∞. A function f is HKr -integrable on [a, b] if
0
and only if there exists a function F ∈ ACGr ([a, b]) so that Fr = f a.e.
4. Integration by Parts
We are now ready to give the proof of Theorem 1.2.
Proof. Define
V (x) = f (x)G(x),
Z x
J(x) = F (x)G(x) −
F (t)G0 (t) dt.
a
We note that F G0 is integrable by Hölder’s inequality [5]. Our task is to show
that J is the HKr -integral of V . By Theorem 3.2, we see that it is sufficient to
0
demonstrate that J ∈ ACGr ([a, b]) and that Jr = V a.e.
We note that the function
Z x
F (t)G0 (t) dt
a
is absolutely continuous on [a, b] and therefore is in ACGr ([a, b]) [4]. Its derivative,
and therefore its Lr -derivative, is equal to F (x)G0 (x) a.e. in [a, b].
Using Theorem 2.2 we can see that F G has an Lr -derivative equal to Fr0 G + F G0
0
a.e. in [a, b]. Using the linearity of the Lr -derivative, we have that Jr = V a.e.
Thus all that remains is to show that J ∈ ACGr ([a, b]). By Theorem 3.1 it is
sufficient to show that F G ∈ ACGr ([a, b]).
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INTEGRATION BY PARTS
7
The function F ∈ ACGr ([a, b]). Since G ∈ AC([a, b]), it is also in ACGr ([a, b])
and G is also in L∞ so by Lemma 2.5, F G ∈ ACGr ([a, b]) and Theorem 1.2 is
proved.
References
[1] Calderon, A. P.; Zygmund, A.; Local properties of solutions of elliptic partial differential
equations, Studia. Math. 20 (1961), pp. 171-225.
[2] Gordon, L.; Perron’s integral for derivatives in Lr , Studia Math. 28 (1966/1967), pp. 295-316.
[3] Gordon, R. A.; The Integrals of Lebesgue, Denjoy, Perron, and Henstock, Grad. Stud. Math.
4, Amer. Math. Soc., 1994.
[4] Musial, P.; Sagher, Y.; The Lr Henstock-Kurzweil integral, Stud. Math. 160 (1) (2004), pp.
53-81.
[5] Wheedan, R.; Zygmund, A.; Measure and Integral, Marcel Dekker, Inc., New York, 1977.
Paul Musial
Department of Mathematics and Computer Science, Chicago State University
9501 South King Drive, Chicago, Illinois 60628, USA
E-mail address: pmusial@csu.edu
Francesco Tulone
Department of Mathematics and Computer Science
University of Palermo, Via Archirafi, 34, 90132 Palermo, Italy
E-mail address: francesco.tulone@unipa.it