Regulated functions and the Perron-Stieltjes
integral
Milan Tvrd
Dedicated to Professor Otakar Boruvka on the occasion of his ninetieth birthday
Mathematical Institute
Academy of Sciences of the Czech Republic
115 67 PRAHA 1, itn 25, Czech Republic
(e-mail: [email protected])
Abstract.
Properties of the Perron-Stieltjes integral with respect to regulated functions are
investigated. It is shown that linear continuous functionals on the space G L ( ) of functions
regulated on ] and left-continuous on ( ) may be represented in the form
a b
a b
a b
( )= ( )+
F x
qx a
Z
a
b
( )d ( )]
p t
x t
where 2 R and ( ) is a function of bounded variation on ] Some basic theorems (e.g.
integration-by-parts formula, substitution theorem) known for the Perron-Stieltjes integral with
respect to functions of bounded variation are established.
AMS Subject Classication. 26 A 42, 26 A 45, 28 A 25, 46 E 99.
Keywords. regulated function, function of bounded variation, Perron-Stieltjes integral, Kurzweil
integral, left-continuous function, linear continuous functional.
p
q t
a b :
0 . Introduction
This paper deals with the space G (a b) of regulated functions on a compact interval
a b]: It is known that when equipped with the supremal norm G (a b) becomes a
Banach space, and linear bounded functionals on its subspace G L (a b) of functions
regulated on a b] and left-continuous on (a b) can be represented by means of
the Dushnik-Stieltjes (interior) integral. This result is due to H. S. Kaltenborn 7],
cf. also Ch. S. Hnig 5, Theorem 5.1]. Together with the known relationship between
the Dushnik-Stieltjes integral, the -Young-Stieltjes integral and the Perron-Stieltjes
integral (cf. Ch. S. Hnig 6] and . Schwabik 11],12]) this enables us to see that F
1
2
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
is a linear bounded functional on G L (a b) if and only if there exists a real number
q and a function p(t) of bounded variation on a b] such that
F (x) = qx(a) +
Zb
a
p(t)dx(t)] for any x 2 G L (a b)
where the integral is the Perron-Stieltjes integral. We will give here the proof of this
fact based only on the properties of the Perron-Stieltjes integral. To this aim, the
proof of the existence of the integral
Zb
a
f (t)dg(t)]
for any function f of bounded variation on a b] and any function g regulated on a b]
is crucial. Furthermore, we will prove extensions of some theorems (e.g. integrationby-parts and substitution theorems) needed for dealing with generalized dierential
equations and Volterra-Stieltjes integral equations in the space G (a b):
1 . Preliminaries
Throughout the paper R n denotes the space of real n-vectors, R 1 = R : Given x 2 R n
its arguments are denoted by x1 x2 : : : xn (x = (x1 x2 : : : xn)). N stands for
the set of all natural numbers (N = f1 2 : : : g). Given M R M denotes its
characteristic function (M (t) = 1 if t 2 M and M (t) = 0 if t 62 M:)
Let ;1 < a < b < 1: The sets d = ft0 t1 : : : tm g of points in the closed
interval a b] such that a = t0 < t1 < < tm = b are called divisions of a b]: Given
a division d of a b] its elements are usually denoted by t0 t1 : : : tm : The couples
D = (d ) where d = ft0 t1 : : : tm g is a division of a b] and = (1 2 : : : m) 2
R m is such that
tj;1 j tj for all j = 1 2 : : : m
are called partitions of a b]:
A function f : a b] 7! R which possesses nite limits
f ( )
f (t+) = lim
f ( ) and f (s;) = lim
!s;
!t+
for all t 2 a b) and all s 2 (a b] is said to be regulated on a b]: The set of all
regulated functions on a b] is denoted by G (a b): Given f 2 G (a b) we dene
f (a;) = f (a) f (b+) = f (b)
+f (t) = f (t+) ; f (t) ; f (t) = f (t) ; f (t;)
and
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
3
f (t) = f (t+) ; f (t;):
for any t 2 a b]: (In particular, we have ;f (a) = +f (b) = 0 f (a) = + f (a)
and f (b) = ; f (b).)
It is known (cf. 5, Corollary 3.2a]) that if f 2 G (a b) then for any " > 0 the
set of points t 2 a b] such that
j+ f (t)j > " or j; f (t)j > "
is nite. Consequently, for any f 2 G (a b) the set of its discontinuities in a b] is
countable. The subset of G (a b) consisting of all functions regulated on a b] and
left-continuous on (a b) will be denoted by G L (a b):
A function f : a b] 7! R is called a nite step function on a b] if there exists
a division ft0 t1 : : : tmg of a b] such that f is constant on every open interval
(tj;1 tj ) j = 1 2 : : : m: The set of all nite step functions on a b] is denoted by
S (a b): A function f : a b] 7! R is called a break function on a b] if there exist
sequences
ftk g1
k=1 a b]
+ 1
fc;k g1
k=1 R and fck gk=1 R
such that tk 6= tj for k 6= j c;k = 0 if tk = a c+k = 0 if tk = b
1 X
c;k + c+k < 1
and
(1.1)
or equivalently
k=1
f (t) =
X ; X +
ck + ck
tk t
tk <t
for t 2 a b]
1
X
f (t) = c;k tk b](t) + c+k (tk b](t) for
k=1
t 2 a b]:
Clearly, if f is given by (1.1), then + f (tk ) = c+k and ; f (tk ) = c;k for any
k 2 N and f (t;) = f (t) = f (t+) if t 2 a b] n ftk g1
k=1 : Furthermore, we have
f (a) = 0 and
var ba f
1 X
c;k + c+k =
k=1
for any such function. The set of all break functions on a b] is denoted by B (a b):
B V (a b) denotes the set of all functions with bounded variation on a b] and
kf kBV = jf (a)j + var ba f for f 2 B V (a b):
4
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
It is known that for any f 2 B V (a b) there exist uniquely determined functions
f C 2 B V (a b) and f B 2 B V (a b) such that f C is continuous on a b] f B is a break
function on a b] and f (t) = f C(t) + f B(t) on a b] (the Jordan decomposition of
f 2 B V (a b)). In particular, if W = fwk gk2N is the set of discontinuities of f in
a b] then
(1.2)
f B (t) =
1
X
k=1
;f (wk ) wkb](t) + +f (wk ) (wkb](t) 2 a b]:
Moreover, if we put
(1.3)
fn (t) =
B
n
X
k=1
;f (wk ) wkb](t) + +f (wk ) (wk b](t) on a b]
for any n 2 N then
lim kfnB ; f B kBV = 0
(1.4)
n!1
(cf. e.g. 14, the proof of Lemma I.4.23]). Obviously,
S (a
b) B (a b) B V (a b) G (a b):
Given f 2 G (a b) we dene
kF k = sup jf (t)j:
t2ab]
Clearly, kf k < 1 for any f 2 G (a b) and when endowed with this norm, G (a b)
becomes a Banach space (cf. 5, Theorem 3.6]). It is known that S(a b) is dense in
G (a b) (cf. 5, Theorem 3.1]). It means that f : a b] 7! R is regulated if and only
if it is a uniform limit on a b] of a sequence of nite step functions. Obviously,
G L (a b) is closed in G (a b) and hence it is also a Banach space. (Neither S (a b)
nor B (a b) are closed in G (a b) of course.)
For some more details concerning regulated functions see the monographs by
Ch. S. Hnig 5] and by G. Aumann 1] and the papers by D. Frakov 2] and 3].
The integrals which occur in this paper are the Perron-Stieltjes integrals. We
will work with the following denition which is a special case of the denition due
to J. Kurzweil 8].
Let ;1 < a < b < 1: An arbitrary positive valued function : a b] 7! (0 1)
is called a gauge on a b]: Given a gauge on a b] the partition (d ) of a b] is
said to be -ne if
tj;1 tj ] (j ; (j ) j + (j )) for any j = 1 2 : : : m:
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
5
Given functions f g : a b] 7! R and a partition D = (d ) of a b] let us dene
SD (f g) =
m
X
j =1
f (j )g(tj ) ; g(tj;1)]:
We say that I 2 R is the Kurzweil integral of f with respect to g from a to b and
denote
I=
Zb
a
f (t)dg(t)] or I =
Zb
a
f dg
if for any " > 0 there exists a gauge on a b] such that
I ; SD (f g ) < "
for all -ne partitions D of a b]:
The Perron-Stieltjes integral with respect to a function not necessarily of bounded variation was dened by A. J. Ward 15] (cf. also S. Saks 10, Chapter VI]). It
can be shown that the Kurzweil integral is equivalent to the Perron-Stieltjes integral
(cf. 11, Theorem 2.1], where the assumption g 2 B V (a b) is not used in the proof
and may be omitted). Consequently, the Riemann-Stieltjes integral (both of the
norm type and of the -type, cf. 4]) is its special case. The relationship between
the Kurzweil integral, the -Young-Stieltjes integral and the Perron-Stieltjes integral
was described by . Schwabik (cf. 11] and 12]).
Since we will make use of some of the properties of the -Riemann-Stieltjes integral, let us indicate here the proof that this integral is included in the Kurzweil integral. (For the denition of the -Riemann-Stieltjes integral, see e.g. 4, Sec. II.9].)
1.1. Proposition. RLet f g : a b] 7! R and I 2 R be such that the -Riemann-
Stieltjes
integral ab f dg exists and equals I . Then the Perron-Stieltjes integral
Rb
a f dg exists and equals I as well.
Proof. Let
Zb
a
f dg = I 2 R
i.e. for any " > 0 there is a division d0 = fs0 s1 : : : sm0 g of a b] such that for any
division d = ft0 t1 : : : tm g which is its renement (d0 d) and any 2 R m such
that D = (d ) is a partition of a b] the inequality
S (f g ) ; I < "
D
6
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
is satised. Let us dene
"(t) =
(
1
2
minfjt ; sj j j = 0 1 : : : m0 g if t 62 d0
"
if t 2 d0:
Then a partition D = (d ) of a b] is "-ne only if for any j = 1 2 : : : m0 there
is an index ij such that sj = ij : Furthermore,
SD (f g) =
m h
X
j =1
i
f (j )g(tj ) ; g(j )] + f (j )g(j ) ; g(tj;1)]
for any partition D = (d ) of a b]: Consequently, for any "-ne partition D = (d )
of a b] the corresponding integral sum SD (f g) equals the integral sum SD (f g)
corresponding to a partition D0 = (d0 0) where d0 is a division of a b] such that
d0 d0 and hence
0
S (f g ) ; I < ":
D
0
This means that the Kurzweil integral
Zb
a
Rb
a
f dg = f dg exists and
Zb
a
f dg = I
holds.
R
To prove the existence of the Perron-Stieltjes integral ab f dg for any f 2
B V (a b) and any g 2 G (a b) in Theorem 2.8 the following assertion is helpful.
1.2. Proposition. Let f 2 B V (a b) be continuous on a b] and let g 2 G (a b)
then both the -Riemann-Stieltjes integrals
Zb
a
f dg and Zb
a
g df
exist.
Proof. Let f 2 B V (a b) which is continuous on a b] and g 2 G (a b) be given
.According to the integration-by-parts formula 4, II.11.7] for -Riemann-Stieltjes
R
integrals to prove the lemma it is sucient to show that the integral ab g df exists.
First, let us assume that an arbitrary 2 a b] is given and g = a ]: Let us put
d0 =
fa bg if = a or = b
fa bg if 2 (a b):
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
7
It is easy to see that then for any partition D = (d ) such that
d0 d = ft0 t1 : : : tm g
we have = tk for some k 2 f0 1 : : : mg and
(
f ( ) ; f (a) if k+1 > SD (g f ) =
f (tk+1) ; f (a) if k+1 = :
Since f is assumed to be continuous, it is easy to show that for a given " > 0 there
exists a division d of a b] such that d0 d and
SD (g f ) ; f ( ) ; f (a)] < "
holds for any partition D = (d ) of a b] with d d i.e.
Zb
a
a ] df = f ( ) ; f (a) for all 2 a b]:
By a similar argument we could show the following relations:
and
Zb
a
Zb
a
Zb
a
a ) df = f ( ) ; f (a) for all 2 (a b]
b] df = f (b) ; f ( ) for all 2 a b]
( b] df = f (b) ; f ( ) for all 2 a b)
as well.
It follows that the integral
Zb
a
g df
exists for any f 2 B V (a b) continuous on a b] and any g 2 S(a b) (cf. Remark 2.2).
Now, if g 2 G (a b) is arbitrary, then there exists a sequence fgng1
k=1 S (a b)
such that
lim kgn ; gk = 0:
n!1
R
Since by the preceding part of the proof of the lemma all the integrals ab gn df
have a nite value, by means of the convergence theorem 4, Theorem II.15.1] valid
8
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
for -Riemann-Stieltjes integrals we obtain that the integral the relation
lim Zb
n!1
a
gn df = Zb
a
Rb
a
g df exists and
g df 2 R
holds. This completes the proof.
A direct corollary of Proposition 1.2 and of 4, Theorem II.13.17] is the following
assertion which will be helpful for the proof of the integration-by-parts formula
Theorem 2.15. (Of course, we could prove it by an argument similar to that used in
the proof of Proposition 1.2, as well.)
1.3. Corollary. Let f 2 B V (a b) and g 2 G (a b): Let
+ f (t)+g(t) = ; f (t);g(t) = 0 for all t 2 (a b):
Then both the -Riemann-Stieltjes integrals
Zb
a
f dg and Zb
a
g df
exist.
It is well known (cf. e.g. 14, Theorems I.4.17, RI.4.19 and Corollary I.4.27] that
if f 2 G (a b) and g 2 B V (a b) then the integral ab f dg exists and the inequality
Z b
f dg kf k;var b g a
a
holds. The Kurzweil integral is an additive function of intervals and possesses the
usual linearity properties. For the proofs of these assertions and some more details
concerning the Kurzweil integral with respect to functions of bounded variation see
e.g. 8], 9], 13] and 14].
2 . Perron-Stieltjes integral with respect to regulated functions
In this section we deal with the integrals
Zb
a
f (t)dg(t)] and
Zb
a
g(t)df (t)]
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
9
where f 2 B V (a b) and g 2 G (a b): we prove some basic theorems (integration by - parts formula, convergence theorems, substitution theorem and unsymmetric
Fubini theorem) needed in the theory of Stieltjes integral equations inR the space
b
G (a b): However, our rst task is the proof of existence of the integral a f dg for
any f 2 B V (a b) and any g 2 G (a b): First, we will consider some simple special
cases.
2.1. Proposition. Let g 2 G (a b) be arbitrary. Then for any 2 a b] we have
Zb
(2.1)
a
Zb
(2.2)
a
Zb
(2.3)
a
Zb
(2.4)
and
(2.5)
a
Zb
a
a ] dg = g( +) ; g(a)
a ) dg = g( ;) ; g(a)
b] dg = g(b) ; g( ;)
( b] dg = g(b) ; g( +)
] dg = g( +) ; g( ;)
where a) (t) (b](t) 0 and the convention g(a;) = g (a) g (b+) = g (b) is used.
Proof. Let g 2 G (a b) and 2 a b] be given.
a) Let f = a ]: It follows immediately from the denition that
Z
a
f dg = g( ) ; g(a):
In particular, 2.1 holds in the case = b: Let 2 a b) let " > 0 be given and let
( 1
j ; tj if < t b
"(t) = 2
"
if t = :
It is easy to see that any "-ne partition D = (d ) of b] must satisfy
1 = t0 = t1 < + " and SD (f g) = g(t1) ; g( ):
Consequently,
Zb
and
f dg = g( +) ; g( )
10
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
Zb
a
Z
Zb
f dg = f dg + f dg
a
= g( ) ; g(a) + g( +) ; g( ) = g( +) ; g(a)
i.e. the relation (2.1) is true for every 2 a b]:
b) Let f = a ): If = a then f 0 g( ;) ; g(a) = 0 and (2.2) is trivial. Let
2 (a b]: For a given " > 0 let us dene a gauge " on a ] by
"(t) =
(
1
2
j ; tj if a t < "
if t = :
Then for any "-ne partition D = (d ) of a ] we have
tm = m = tm;1 < ; " and SD (f g) = g(tm;1) ; g(a):
It follows immediately that
Z
a
f dg = g( ;) ; g(a)
and in view of the obvious identity
Zb
f dg = 0
this implies (2.2).
c) The remaining relations follow from 2.1, 2.2 and the equalities
and
b] = ab] ; a )
( b] = ab] ; a ]
] = a ] ; a ):
2.2. Remark. Since any nite step function is a linear combination of functions
b] (a b) and
(a < b), it follows immediately from Proposition 2.1
R b ( b]
that the integral a f dg exists for any f 2 S(a b) and any g 2 G (a b):
Other simple cases are covered by
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
11
2.3. Proposition. Let 2 a b]: Then for any function f : a b] 7! R the following
relations are true
(2.6)
(2.7)
(2.8)
(2.9)
and
(2.10)
Zb
a
Zb
a
Zb
a
Zb
a
Zb
a
f da ] =
f da ) =
f d b] =
f d( b] =
(
(
(
(
;f ( ) if < b
0
if = b
;f ( ) if > a
0
if = a
f ( ) if > a
0
if = a
f ( ) if < b
0
if = b
8
>
< ;f (a) if
f d ] = >
:
0
f (b)
=a
if a < < b
if = b
where a) (t) (b](t) 0 and the convention g (a;) = g (a) g (b+) = g (b) is used.
For the proof see 14, I.4.21 and I.4.22].
2.4. Corollary. Let W = fw1 w2 : : : wng a b] c 2 R and h : a b] 7! R be
such that
(2.11)
Then
(2.12)
h(t) = c for all t 2 a b] n W:
Zb
a
f dh = f (b)h(b) ; c] ; f (a)h(a) ; c]
holds for any function f : a b] 7! R :
Proof. A function h : a b] 7! R fulls (2.11) if and only if
h(t) = c +
n
X
k=1
h(wj ) ; c]wj ](t) on a b]:
Thus the formula (2.12) follows from (2.6) (with = b) and from (2.10) in Proposition 2.3.
12
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
2.5. Remark. It is well known (cf. 14, I.4.17] or 13, Theorem 1.22])
that if
R
g 2 B V (a b) h : a b] 7! R and hn : a b] 7! R R n 2 N are such that
for any n 2 N and limn!1 khn ; hk = 0 then ab h dg exists and
(2.13)
lim
Zb
n!1 a
Zb
hn dg =
a
b
a hn dg
exist
h dg
holds. To prove an analogous assertion for the case g 2 G (a b) we need the following
auxiliary assertion.
2.6. Lemma. Let f 2 B V (a b) and g 2 G (a b): The the inequality
;
jSD (f g )j jf (a)j + jf (b)j + var ba f kg k
(2.14)
holds for an arbitrary partition D of a b]:
Proof. For an arbitrary partition D = (d ) of a b] we have (putting 0 = a and
m+1 = b)
jSD (f g )j = jf (b)g (b) ; f (a)g (a) ;
jf (b)j + jf (a)j +
;
m
+1
X
j =1
m
+1
X
j =1
f (j ) ; f (j;1)]g(tj;1)j
jf (j ) ; f (j ;1)j kg k
jf (a)j + jf (b)j + var ba f kg k:
2.7. Theorem. Let g 2 G (a b) and let hn h : a b] 7! R be such that
Zb
Rb
a
hn dg exists for any n 2 N and nlim
kh ; hkBV = 0:
!1 n
h dg exists and (2.13) holds.
Proof. Since
Then
a
jf (b)j jf (a)j + jf (b) ; f (a)j jf (a)j + var ba f
we have by (2.14)
jSD ((hm ; hk )g )j 2khm ; hk kBV kg k
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
13
for all m k 2 N and all partitions D of a b]: Consequently,
Z b
(h ; h )dg 2kh ; h k kg k
m
k m
k BV
a
holds for all m k 2 N : This immediately implies that there is a q 2 R such that
lim
Zb
n!1 a
It remains to show that
q=
(2.15)
hn dg = q:
Zb
a
h dg:
For a given " > 0 let n0 2 N be such that
(2.16)
Z b
h dg ; q < " and kh ; hk
n0
n0
BV
a
<"
and let " be such a gauge on a b] that
Z
S (h g ) ; b h dg < "
D n0
n0 (2.17)
a
for all "-ne partitions D of a b]: Given an arbitrary "-ne partition D of a b]
we have by (2.16), (2.17) and Lemma 2.6
jq ; SD (h g )j
Zb
Z b
q ; hn0 dg + hn0 dg ; SD (hn0 g )
a
a
+ SD (hn0 g) ; SD (h g)
2" + jSD (hn0 ; h] g )j 2" + 2khn0 ; hkBV kg k 2" (1 + kg k)
wherefrom the relation (2.15) immediately follows. This completes the proof of the
theorem.
Now we can prove the following
2.8. Theorem. Let f 2 B V (a b) and g 2 G (a b): Then the integral
Zb
a
f dg
14
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
exists and the inequality
Z b
f dg ;jf (a)j + jf (b)j + var b f kg k
a
(2.18)
a
holds.
Proof. Let f 2 B V (a b) and g 2 G (a b) be given. Let W = fwk gk2N be the set of
discontinuities of f in a b] and let f = f C + f B be the Jordan decomposition of f
(i.e. f C is continuous on a b] and f B is given by (1.2)). We have
B
B
lim
k
f
;
f
kBV = 0
n
n!1
for fnB n 2 N given by (1.3). By (2.3) and (2.4),
Zb
(2.19)
a
fn dg =
B
n X
k=1
+f (wk )(g(b) ; g(wk +))
+ ;f (wk )(g(b) ; g(wk ;))
holds for any n 2 N : Thus according to Theorem 2.7 the integral
and
Zb
(2.20)
a
R
f dg = nlim
!1
B
Zb
a
Rb
a
f B dg exists
fnB dg:
The integral ab f CRdg exists as the -Riemann-Stieltjes integral by Proposition 1.2.
This means that ab f dg exists and
Zb
a
f dg =
Zb
a
f dg +
C
Zb
a
f dg =
B
Zb
a
f dg + nlim
!1
C
Zb
a
fnB dg:
The inequality (2.18) follows immediately from Lemma 2.6.
2.9. Remark. Since
1 X
+ f (wk )(g (b) ; g (wk +)) + ; f (wk )(g (b) ; g (wk ;))
k=1
2kg k
1
X
k=1
j+ f (wk )j + j; f (wk )j 2kg k(var ba f ) < 1
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
15
we have in virtue of (2.19) and (2.20)
(2.21)
Zb
a
f dg =
B
1 X
+f (wk )(g(b) ; g(wk+))
k=1
+ ; f (wk )(g(b) ; g(wk;)) :
As a direct consequence of Theorem 2.8 we obtain
2.10. Corollary. Let hn 2 G (a b) n 2 N and h 2 G (a b) be such that
lim khn ; hk = 0:
n!1
Then for any f 2 B V (a b) the integrals
Zb
and
a
Zb
f dh and
lim
Zb
n!1 a
a
f dhn =
Zb
a
f dhn n 2 N
exist
f dh:
2.11. Lemma. Let h : a b] ! R c 2 R and W = fwk gk2N a b] be such that
(2.11) and
1
X
(2.22)
k=1
jh(wk ) ; cj < 1
hold. Given n 2 N let us put Wn = fw1 w2 : : : wng and
(2.23)
hn(t) =
(
c if t 2 a b] n Wn
h(t) if t 2 Wn:
Then hn 2 B V (a b) for any n 2 N h 2 B V (a b) and
lim khn ; hkBV = 0
(2.24)
n!1
Proof. The functions hn n 2 N and h evidently have a bounded variation on a b].
For a given n 2 N we have
hn(t) ; h(t) =
(
0
if t 2 Wn or t 2 a b] n W
c ; hn(t) if t = wk for some k > n:
16
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
Thus,
(2.25)
lim h (t) = h(t)
n!1 n
and, moreover,
on a b]
m ;
1
X
X X
hn (tj ) ; h(tj ) ; ;hn (tj ;1 ) ; h(tj ;1 ) 2
jh(wk ) ; cj
j =1
k=n+1
holds for any n 2 N and any division ft0 t1 : : : tmg of a b]: Consequently,
var ba (hn ; h) 2
(2.26)
1
X
k=n+1
jh(wk ) ; cj
holds for any n 2 N : In virtue of the assumption (2.22) the right-hand side of (2.26)
tends to 0 as n ! 1: Hence (2.24) follows from (2.25) and (2.26).
2.12. Proposition. Let h : a b] 7! R c 2 R and W = fwk gk2N be such that
(2.11) and (2.22) hold. Then
Zb
a
1
X
h dg =
k=1
h(wk ) ; c]g(wk ) + c g(b) ; g(a)]
holds for any g 2 G (a b):
Proof. Let g 2 G (a b) be given. Let Wn = fw1 w2 : : : wng for n 2 N and let the
functions hn n 2 N be given by (2.23). Given an arbitrary n 2 N then (2.1) (with
= b) and (2.5) from Proposition 2.1 imply
Zb
a
hn dg =
Since (2.22) yields
n
X
k=1
h(wk ) ; c]g(wk ) + cg(b) ; g(a)]:
1
n X
X
h(wk ) ; c]g (wk ) 2 kg k
jh(wk ) ; cj < 1
k=1
k=1
and Lemma 2.11 implies
lim khn ; hkBV = 0
n!1
we can use Theorem 2.7 to prove that
Zb
a
h dg = nlim
!1
Zb
a
hn dg =
1
X
k=1
h(wk ) ; c]g(wk ) + c g(b) ; g(a)]:
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
17
2.13. Proposition. Let h : a b] 7! R c 2 R and W = fwk gk2N ful l (2.11). Then
Zb
(2.27)
a
f dh = f (b)h(b) ; c] ; f (a)h(a) ; c]
holds for any f 2 B V (a b):
Proof. Let f 2 B V (a b): For a given n 2 N let Wn = fw1 w2 : : : wng and let hn
be given by (2.23). Then
lim khn ; hk = 0:
(2.28)
n!1
Indeed, let " > 0 be given and let n0 2 N be such that k n0 implies
(2.29)
jh(wk ) ; cj < ":
(Such an n0 exists since jh(wk ) ; cj = j;h(wk )j = j+h(wk )j for any k 2 N and
the set of those k 2 N for which the inequality (2.29) does not hold may be only
nite.) Now, for any n n0 and any t 2 a b] such that t = wk for some k > n
(t 2 W n Wn) we have
jhn (t) ; h(t)j = jhn (wk ) ; h(wk )j = jc ; h(wk )j < ":
;
Since hn(t) = h(t) for all the other t 2 a b] (t 2 a b] n W Wn), it follows that
jhn (t) ; h(t)j < " on a b] i.e.
khn ; hk < ":
This proves the relation (2.28).
By Corollary 2.4 we have for any n 2 N
Zb
a
f dhn = f (b)h(b) ; c] ; f (a)h(a) ; c]:
Making use of (2.28) and Corollary 2.10 we obtain we obtain
Zb
a
f dh = nlim
!1
Zb
a
f dhn = f (b)h(b) ; c] ; f (a)h(a) ; c]:
2.14. Corollary. Let h 2 B V (a b) c 2 R and W = fwk gk2N ful l (2.11). Then
(2.27) holds for any f 2 G (a b):
Proof. By Proposition 2.12, (2.27) holds for any f 2 B V (a b): Making use of the
density of S(a b) B V (a b) in G (a b) and of the convergence theorem mentioned
in Remark 2.5 we complete the proof of our assumption.
18
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
2.15. Theorem. R(Integration-by-parts)
If f 2 B V (a b) and g 2 G (a b) then
R
both the integrals
Zb
(2.30)
a
b
a f dg
b
a g df
and
f dg +
Zb
a
+
exist and
g df = f (b)g(b) ; f (a)g(a)
X ;
f (t);g(t) ; +f (t)+g(t) :
t2ab]
R
b
Proof.
The
existence
of
the
integral
g df is well known, while the existence of
a
Rb
a g df is guaranteed by Theorem 2.8. Furthermore,
Zb
a
f dg +
=
Zb
a
;
Zb
a
g df
f (t)dg(t) + g(t)] +
+
Zb
a
f (t)d g(t)] +
+
Zb
a
Zb
a
g(t)df (t) ; ;f (t)]
g(t)d;f (t)]:
It is easy to verify that the function h(t) = +g(t) fulls the relation (2.11) with
c = 0 and h(b) = 0: Consequently, Proposition 2.13 yields
Zb
a
f (t)d+g(t)] = ;f (a)+g(a):
Similarly, by Corollary 2.14 we have
Zb
a
Hence
(2.31)
Zb
a
f dg +
Zb
a
g(t)d;f (t)] = ; f (b)g(b):
g(t)df =
Zb
a
f (t)dg(t+)] +
Zb
a
g(t)df (t;)]
+ f (a)+g(a) + ; f (b)g(b):
The rst integral on the right-hand side may be modied in the following way:
(2.32)
Zb
a
f (t)dg(t+)] =
Zb
a
f (t;)dg(t+)] +
Zb
a
; f (t)dg(t+)]:
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
19
Making use of Proposition 2.12 and taking into account that g1(t) = g(t) on
a b] for the function g1 dened by g1 (t) = g(t+) on a b] we further obtain
Zb
(2.33)
Similarly,
(2.34)
a
Zb
a
; f (t)dg(t+)] =
g(t)df (t;)] =
=
Zb
a
Zb
a
X
t2ab]
;f (t)g(t):
g(t+)df (t;) ;
Zb
g(t+)df (t;)] ;
a
+g(t)df (t;)]
X
t2ab]
+ g(t)f (t):
The function f (t;) is left-continuous on a b] while g(t+) is right-continuous on
a b): It means that both the integrals
Zb
a
f (t;)dg(t+)] and
Zb
a
g(t+)df (t;)]
exist as the -Riemann-Stieltjes integrals (cf. Corollary 1.3), and making use of the
integration-by-parts theorem for these integrals (cf. 4, Theorem II.11.7]) we get
(2.35)
Zb
a
f (t;)dg(t+)] +
Zb
a
g(t+)df (t;)] = f (b;)g(b) ; f (a)g(a+):
Inserting (2.32) - (2.35) into (2.31) we get
Zb
a
f dg +
Zb
a
+
;
g df = f (b;)g(b) ; f (a)g(a+)
X
t2ab]
X
t2ab]
; f (t);g(t) + + g(t)]
; f (t) + +f (t)]+g(t)
+ f (a)+g(a) + ; f (b)g(b)
= f (b)g(b) ; f (a)g(a)
+
X ;
f (t);g(t) ; + f (t)+g(t)
t2ab]
and this completes the proof
20
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
The following proposition describes some properties of indenite Perron-Stieltjes
integrals.
R
2.16. Proposition. Let f : a b] 7! R and g : a b] 7! R be such that ab f dg
exists. Then the function
h(t) =
Zt
a
f dg
is de ned on a b] and
(i) if g 2 G (a b) then h 2 G (a b) and
(2.36)
+h(t) = f (t)+g(t)
;h(t) = f (t);g(t) on a b]
(ii) if g 2 B V (a b) and f is bounded on a b] then h 2 B V (a b):
Proof. The former assertion follows from 8, Theorem 1.3.5]. The latter follows
immediately from the inequality
m Z tj
X
tj
j =1
1
m X
f dg kf k (var ttjj
j =1
;
1
;
g) = kf k(var ba g)
which is valid for any division ft0 t1 : : : tmg of a b]:
In the theory of generalized dierential equations the substitution formula
(2.37)
Zb
a
h(t)d
hZ t
a
i
f (s)dg(s)] =
Zb
a
h(t)f (t)dg(t)]
is often needed. In 4, II.19.3.7] this formula is proved for the -Young-Stieltjes
integral under
R b the assumption that g 2 G (a b) h is bounded on a b] and the
integral a f dg as well as one of the integrals in (2.37) exists. In 14, Theorem
I.4.25] this assertion was proved for the Kurzweil integral. Though it was assumed
there that g 2 B V (a b) this assumption was not used in the proof. We will give here
a slightly dierent proof based on the Saks-Henstock lemma (cf. e.g. 13, Lemma
1.11]).
2.17.
Lemma. (Saks-Henstock) Let f g : a b] 7! R be such that the integral
Rb
a f dg exists. Let " > 0 be given and let be a gauge on a b] such that
Z
S (f g ) ; b f dg < "
D
a
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
21
holds for any - ne partition D of a b]: Then for an arbitrary system f(
i i ] i)
i = 1 2 : : : kg of intervals and points such that
(2.38)
a 1 1 1 2 k k k b
and
i i] i ; (i) i + (i)] i = 1 2 : : : k
the inequality
(2.39)
holds.
Z
k
X
hf ( )g ( ) ; g (
)] ; i f dg i < "
i
i
i
i
i=1
Making use of Lemma 2.17 we can prove the following useful assertion
R
2.18. Lemma. If f : a b] 7! R and g : a b] 7! R are such that ab f dg exists,
then for any " > 0 there exists a gauge on a b] such that
(2.40)
Z
m X
f ( )g (t ) ; g (t )] ; tj
j
j
j ;1
tj
j =1
1
f dg < "
;
holds for any - ne partition (d ) of a b]:
Proof. Let : a b] 7! (0 1) be such that
Zb
Z tj
m
X
f (j )g(tj ) ; g(tj;1)] ;
SD (f g ) ; f dg = a
tj
j =1
1
f dg < 2"
;
for all -ne partitions D = (d ) of a b]: Let us choose an arbitrary -ne partition
D = (d ) of a b]: Let i = tpi and i = tpi;1 i = 1 2 : : : k be all the points of
the division d such that
f (pi )g(i) ; g(
i)] ;
Z i
i
f dg 0:
Then the system f(
i i] i) i = 1 2 : : : kg where i = pi fulls (2.38) and
(2.39) and hence we can use Lemma 2.17 to prove that the inequality
Z
k X
f ( )g ( ) ; g (
)] ; i f dg < "
pi
2
i
i
i
i=1
is true. Similarly, if !i = tqi and i = tqi;1 i = 1 2 : : : r are all points of the
division d such that
f (qi )g(!i) ; g(i)] ;
Z !i
i
f dg 0
22
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
the the inequality
Z
r X
f ( )g (! ) ; g ( )] ; !i f dg < "
qi
2
i
i
i
i=1
follows from Lemma 2.17, as well. Summarizing, we conclude that
Z
m X
f ( )g (t ) ; g (t )] ; tj
j
j
j ;1
tj
j =1
;
1
f dg Z i
k X
= f (pi )g(i) ; g(
i)] ;
f dg
i
i=1
Z !i
r X
+ f (qi )g(!i) ; g(i)] ;
f dg
"
<2
i=1
+"
2
i
= ":
This completes the proof.
2.19. Theorem. Let Rh : a b] 7! R be bounded on a b] and let f g : a b] 7! R be
such that the integral
b
a f dg
exists. Then the integral
Zb
a
h(t)f (t)dg(t)]
exists if and only if the integral
Zb
a
h(t)d
hZ t
a
i
f (s)dg(s)]
exists, and in this case the relation (2.37) holds.
Proof. Let jh(t)j C < 1 on a b]: Let us assume that the integral
Zb
a
h(t)f (t)dg(t)]
exists and let " > 0 be given. There exists a gauge 1 on a b] such that
Zb
m
X
h(k )f (j )g(tj ) ; g(tj;1)] ; h(t)f (t)dg(t)] < 2"
j =1
a
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
23
is satised for any 1-ne partition (d ) of a b]: By Lemma 2.18 there exists a
gauge on a b] such that (t) 1(t) on a b] and
Z
m X
f ( )g (t ) ; g (t )] ; tj
j
j
j ;1
tj
j =1
1
f dg <
;
"
2C
holds for any -ne partition (d ) of a b]: Let us denote
k(t) =
Zt
a
f dg for t 2 a b]:
Then for any -ne partition D = (d ) of a b] we have
Z
S (h k) ; b h(t)f (t)dg (t)]
D
a
Z tj
m
m
X
X
= h(j )
f dg ; h(j )f (j )g(tj ) ; g(tj;1)]
j =1
m
X
+
j =1
m
X
tj
j =1
1
h(j )f (j )g(tj ) ; g(tj;1)] ;
h Z tj
h(j )
j =1
m
X
+ ;
j =1
tj
1
Zb
a
h(t)f (t)dg(t)]
i
f dg ; f (j )g(tj ) ; g(tj;1)] ;
h(j )f (j )g(tj ) ; g(tj;1)] ;
Zb
a
hf dg < "
This implies the existence of the integral
Zb
a
h dk
and the relation (2.37). The second implication can be proved in an analogous
way.
The convergence result 2.10 enables us to extend the known theorems on the
change of integration order in iterated integrals
(2.41)
Zd
c
g(t)d
hZ b
a
i
h(t s)df (s)]
Zb Zd
a
c
g(t)dt h(t s)] df (s)]
where ;1 < c < d < 1 and h is of strongly bounded variation on c d] a b]
(cf. 14, Theorem I.6.20]). In what follows v(h) denotes the Vitali variation of the
24
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
function h on c d] a b] (cf. 4, Denition III.4.1] or 14, I.6.1]). For a given
t 2 c d] var bah(t :) denotes the variation of the function s 2 a b] 7! h(t s) 2 R on
a b]: Similarly, for s 2 a b] xed, var dch(: s) stands for the variation of the function
t 2 c d] 7! h(t s) 2 R on c d]:
2.20. Theorem. Let h : c d] a b] 7! R be such that
v(h) + var bah(c :) + var dch(: a) < 1:
Then for any f 2 B V (a b) and any g 2 G (c d) both the integrals (2.41) exist and
Zd
(2.42)
c
hZ b
g(t)d
a
i
h(t s)df (s)] =
Zb Zd
a
c
g(t)dth(t s)] df (s)]:
Proof. Let us notice that by 14, Theorem I.6.20] our assertion is true if g is also
supposed to be of bounded variation. in the general case of g 2 G (a b) there exists a
sequence fgng1
n=1 S (a b) such that limn!1 kg ; gn k = 0: Then, since the function
v(t) =
Zb
a
h(t s)df (s)]
is of bounded variation on c d] (cf. the rst part of the proof of 14, Theorem
I.6.20]), the integral on left-hand side of (2.42) exists and by Corollary 2.10 and 14,
Theorem I.6.20] we have
(2.43)
Zd
c
hZ b
g(t)d
a
i
h(t s)df (s)] = nlim
!1
= nlim
!1
Let us denote
wn(t) =
Zd
c
Zd
c
gn(t)d
Zb Zd
a
c
hZ b
a
h(t s)df (s)]
i
gn(t)dth(t s)] df (s)]:
gn(t)dth(t s)] for s 2 a b] and n 2 N :
Then wn 2 B V (a b) for any n 2 N (cf. 14, Theorem I.6.18]) and by 14, Theorem
I.4.17] mentioned here in Remark 2.5 we obtain
Zd
lim w (s) =
n!1 n
c
As
gn(t)dt h(t s)] := w(s) on a b]:
;
;
jwn (s) ; w(s)j kgn ; g k var dc h(: s) kgn ; g k v(h) + var dc h(: a)
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
25
for any s 2 a b] (cf. 14, Lemma I.6.6]), we have
lim kwn ; wk = 0:
n!1
It means that w 2 G (a b) and by Theorem 2.8 the integral
Zb
a
w(s)df (s)] =
Zb Zd
a
c
g(t)dt h(t s)] df (s)]
exists as well. Since obviously
lim
Zb Zd
n!1 a
c
gn(t)dt h(t s)] df (s)] = nlim
!1
=
Zb
a
Zb
a
w(s)df (s)] =
wn(s)df (s)]
Zb Zd
a
c
g(t)dth(t s)] df (s)]
the relation (2.42) follows from (2.43).
3 . Linear bounded functionals on GL(a,b)
By Theorem 2.8 the expression
(3.1)
F (x) = qx(a) +
Zb
a
p dq
is dened for any x 2 G (a b) and any = (p q) 2 B V (a b) R : Moreover, for any
2 B V (a b) R the relation (3.1) denes a linear bounded functional on G L (a b):
Proposition 2.3 immediately implies
3.1. Lemma. Let = (p q) 2 B V (a b) R be given. Then
(3.2)
F (ab]) = q
F (( b]) = p( ) for any 2 a b)
F (b]) = p(b):
3.2. Corollary. If = (p q) 2 B V (a b) R and F (x) = 0 for all x 2 S(a b)
which are left-continuous on (a b) then p(t) 0 on a b] and q = 0:
26
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
3.3. Lemma. Let x 2 G (a b) be given. Then for a given = (p q) 2 B V (a b) R
(3.3)
F (x) = x(a) if
F (x) = x(b) if
F (x) = x( ;) if
F (x) = x( +) if
p 0 on
p 1 on
p = a )
p = a ]
a b] and q = 1
a b] and q = 1
on a b] 2 (a b] and q = 1
on a b] 2 a b) and q = 1:
Proof follows from Proposition 2.1.
3.4. Corollary. If x 2 G (a b) and F (x) = 0 for all = (p q) 2 B V (a b) R
then
(3.4)
x(a) = x(a+) = x( ;) = x( +) = x(b;) = x(b)
holds for any 2 (a b): In particular, if x 2 G L (a b) (x is left-continuous on (a b))
and F (x) = 0 for all = (p q ) 2 B V (a b) R then x(t) 0 on a b]:
3.5. Remark. The space B V (a b) R is supposed to be equipped with the usual
norm (kkBV R = jqj + kpkBV for = (p q) 2
Banach space with respect to this norm.
B V (a
b) R ). Obviously, it is a
3.6. Proposition. The spaces G L (a b) and B V (a b) R form a dual pair with
respect to the bilinear form
(3.5)
x 2 G L (a b) 2 B V (a b) R 7! F (x):
Proof follows from Corollaries 3.2 and 3.4.
On the other hand, we have
3.7. Lemma. If F is a linear bounded functional on G L (a b) and
(3.6)
p(t) =
(
F ((tb]) if t 2 a b)
F (b]) if t = b
then p 2 B V (a b) and
(3.7)
where
jp(a)j + jp(b)j + var ba p 2kF k
kF k =
sup
x2G L (ab)kxk1
jF (x)j:
M.Tvrd: Regulated functions and the Perron-Stieltjes integral
27
Proof is analogous to that of part c (i) of 5, Theorem 5.1]. Indeed, for an arbitrary
division ft0 t1 : : : tmg of a b] we have
X
sup p(a)c0 + p(b)cm+1 + p(tj ) ; p(tj;1)]cj m
jcj j1cj 2R
=
j =1
sup
jcj j1cj 2R
sup
m;1
F (c + c ; X c 0 (ab] m+1 b]
j (tj 1 tj ] + cm (tm 1 b) )
khk2h2G L (ab)
j =1
jF (h)j = 2kF k:
;
;
In particular, for c0 = sgn p(a) cm+1 = sgn p(b) and cj = sgn(p(tj ) ; p(tj;1))
j = 1 2 : : : m we get
jp(a)j + jp(b)j +
m
X
j =1
jp(tj ) ; p(tj ;1 )j 2kF k
and the inequality (3.7) immediately follows.
Using the ideas from the proof of 5, Theorem 5.1] we may now prove the following
representation theorem.
3.8. Theorem. F is a linear bounded functional on G L (a b) (F 2 G L (a b)) if and
only if there is an = (p q) 2 B V (a b) R such that
(3.8)
F (x) = F (x) := qx(a) +
Zb
a
p dx
for any x 2 G L (a b):
The mapping
: 2 B V (a b) R 7! F 2 G L (a b)
is an isomorphism.
Proof. Let a linear bounded functional F on G L (a b) be given and let us put
(
F ((tb]) if t 2 a b)
F (b]) if t = b:
Then Lemma 2.6 implies = (p q) 2 B V (a b) R and by Lemma 3.1 we have
F (ab]) = F (ab])
F ((tb]) = F ((tb] ) for any t 2 a b)
and
(3.9)
q = F (ab]) and p(t) =
28
asopis pst. mat. 114 (1989), No. 2, pp. 187-209
F (b]) = F (b]):
Since all functions from S(a b) \ G L (a b) obviously are nite linear combinations of
the functions
ab] ( b] 2 a b) b]
it follows that F (x) = F (x) holds for any x 2 S(a b) \ G L (a b): Now, the density
of S(a b) \ G L (a b) in G L (a b) implies that
F (x) = F (x) for all x 2 G L (a b):
This completes the proof of the rst assertion of the theorem.
Given an x 2 G L (a b) then Lemma 2.6 yields
;
jF (x)j jp(a)j + jp(b)j + var ba p + jq j kxk
and consequently,
;
kF k jp(a)j + jp(b)j + var ba p + jq j 2 kpkBV + jq j = 2k kBV R :
On the other hand, according to Lemma 3.7 we have
;
kpkBV jp(a)j + jp(b)j + var ba p 2kF k:
Furthermore, in virtue of (3.9) we have jqj kF k and hence
k kBV R = kpkBV + jq j 2kF k:
It means that
1
2
kF k k kBV R 3kF k
and this completes the proof of the theorem.
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29
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