ARCHIVUM MATHEMATICUM (BRNO)
Tomus 51 (2015), 233–254
OSTROWSKI’S TYPE INEQUALITIES
FOR COMPLEX FUNCTIONS DEFINED ON UNIT CIRCLE
WITH APPLICATIONS FOR UNITARY OPERATORS
IN HILBERT SPACES
S.S. Dragomir
Abstract. Some Ostrowski’s type inequalities for the Riemann-Stieltjes inte
Rb
f eit du (t) of continuous complex valued integrands f : C (0, 1) → C
gral
a
defined on the complex unit circle C (0, 1) and various subclasses of integrators
u : [a, b] ⊆ [0, 2π] → C of bounded variation are given. Natural applications
for functions of unitary operators in Hilbert spaces are provided as well.
1. Introduction
Rb
The problem of approximating the Stieltjes integral a f (t) du (t) by the quantity
f (x) [u (b) − u (a)], which is a natural generalization of the Ostrowski problem
analyzed in 1937 (see [6]), was apparently first considered in the literature by the
author in 2000 (see [1]) where we obtained the following result:
Z b
x
b
h
i
_
_
r
r
[u
(b)
−
u
(a)]
f
(x)
−
f
(t)
du
(t)
≤
H
(x
−
a)
(f
)
+
(b
−
x)
(f )
a
a
x
W
i

h W
Wb
x
r
r
b

[(x−a) + (b − x) ] 12 a (f )+ 21 a (f )− x (f ) ;



hW
W
p i p1


p
x
b
qr
qr q1
[(x−a)
(
(f
))
+
(f
)
+
(b−x)
]
a
x
(1.1)
≤H×

1
1

if
p
>
1,
+
=
1;

p
q


Wb
 1

a+b r
a (f ) .
2 (b − a) + x − 2
Wb
for each x ∈ [a, b], provided f is of bounded variation on [a, b], a (f ) is its total
variation on [a, b], while u : [a, b] → R is r − H-Hölder continuous, i.e., we recall
that:
(1.2)
r
|u (x) − u (y)| ≤ H |x − y|
for each
x, y ∈ [a, b] ,
2010 Mathematics Subject Classification: primary 41A51; secondary 26D15, 47A63.
Key words and phrases: Ostrowski’s type inequalities, Riemann-Stieltjes integral inequalities,
unitary operators in Hilbert spaces, spectral theory, quadrature rules.
Received February 18, 2015, revised October 2015. Editor V. Müller.
DOI: 10.5817/AM2015-4-233
234
S.S. DRAGOMIR
where H > 0 and r ∈ (0, 1].
The dual case, i.e., when the integrand f is q − K-Hölder continuous and the
integrator u is of bounded variation was obtained by the author in 2001 and can
be stated as [2]
Z b
b
h1
a + b iq _
(1.3) [u (b) − u (a)] f (x) −
f (t) du (t) ≤ K (b − a) + x −
(u)
2
2
a
a
for each x ∈ [a, b].
The above inequalities provide, as important consequences, the following midpoint inequalities:
(
Wb
r
a + b Z b
1
(f )
r (b − a) H
(1.4)
−
f (t) du (t) ≤ 21
Wab
[u (b) − u (a)] f
q
2
a
a (u)
2q (b − a) K
which can be numerically implemented and provide a quadrature rule for approxiRb
mating the Stieltjes integral a f (t) du (t).
Let U be a selfadjoint operator on the complex Hilbert space (H, h·, ·i) with the
spectrum Sp (U ) included in the interval [m, M ] for some real numbers m < M and
let {Eλ }λ be its spectral family. Then for any continuous function f : [m, M ] → R,
it is well known that we have the following spectral representation in terms of the
Riemann-Stieltjes integral:
Z M
(1.5)
hf (U ) x, yi =
f (λ) d (hEλ x, yi) ,
m−0
for any x, y ∈ H. The function gx,y (λ) := hEλ x, yi is of bounded variation on the
interval [m, M ] and
gx,y (m − 0) = 0 and
gx,y (M ) = hx, yi
for any x, y ∈ H. It is also well known that gx (λ) := hEλ x, xi is monotonic
nondecreasing and right continuous on [m, M ].
On utilizing the spectral representation (1.5) and the Ostrowski’s type inequality (1.3) we obtained the following result for continuous functions of selfadjoint
operators:
Theorem 1 ([4]). Let A be a selfadjoint operator in the Hilbert space H with the
spectrum Sp (A) ⊆ [m, M ] for some real numbers m < M and let {Eλ }λ be its
spectral family. If f : [m, M ] → R is r − H-Hölder continuous on [m, M ], then we
have the inequality
M
_
h 1
m + M ir
|f (s) hx, yi − hf (A) x, yi| ≤ H
E(·) x, y
(M − m) + s −
2
2
m
h1
m + M ir
(1.6)
≤ H kxk kyk
(M − m) + s −
2
2
for any x, y ∈ H and s ∈ [m, M ].
The following dual result also holds:
OSTROWSKI’S TYPE INEQUALITIES FOR COMPLEX FUNCTIONS
235
Theorem 2 ([3]). Let A be a selfadjoint operator in the Hilbert space H with the
spectrum Sp (A) ⊆ [m, M ] for some real numbers m < M and let {Eλ }λ be its
spectral family. If f : [m, M ] → R is a continuous function of bounded variation
on [m, M ], then we have the inequality
1/2
|f (s) hx, yi − hf (A) x, yi | ≤ hEs x, xi
1/2
hEs y, yi
s
_
(f )
m
+ h(1H − Es ) x, xi
1/2
1/2
h(1H − Es ) y, yi
M
_
(f )
s
(1.7)
≤ kxk kyk
M
1 _
2
m
s
M
M
_
_
1 _
(f ) + (f ) −
(f ) ≤ kxk kyk (f )
2 m
s
m
for any x, y ∈ H and for any s ∈ [m, M ], where 1H is the identity operator on H.
Motivated by the above results, we investigate in the current paper the magnitude
of the difference
Z b
f eis [u (b) − u (a)] −
f eit du (t) with s ∈ [a, b] ⊆ [0, 2π]
a
for continuous complex valued function f : C (0, 1) → C defined on the complex unit
circle C (0, 1) and various subclasses of functions u : [a, b] ⊆ [0, 2π] → C of bounded
variation. Natural applications for functions of unitary operators in Hilbert spaces
are provided as well.
2. Scalar Ostrowski’s type inequalities
Theorem 3. Assume that f : C (0, 1) → C satisfies the following Hölder’s type
condition
r
|f (z) − f (w)| ≤ H |z − w|
(2.1)
for any w, z ∈ C (0, 1), where H > 0 and r ∈ (0, 1] are given.
If [a, b] ⊆ [0, 2π] and the function u : [a, b] → C is of bounded variation on [a, b],
then
Z b
is
f (eit ) du(t)
f (e )[u(b) − u(a)] −
a
b
s − t r _
≤ 2r H max sin
(u)
2
t∈[a,b]
a
(2.2)
for any s ∈ [a, b], where
b
W
(u) denotes the total variation of u on the interval [a, b].
a
Proof. Observe that
Z
(2.3)
f eis [u (b) − u (a)] −
a
for any s ∈ [a, b].
b
f eit du (t) =
Z
a
b
f eis − f eit du (t)
236
S.S. DRAGOMIR
It is known that if p : [c, d] → C is a continuous function and v : [c, d] → C is
Rd
of bounded variation, then the Riemann-Stieltjes integral c p (t) dv (t) exists and
the following inequality holds
d
Z d
_
p (t) dv (t) ≤ max |p (t)| (v)
(2.4)
t∈[c,d]
c
where
d
W
c
(v) denotes the total variation of v on [c, d].
c
Applying the property (2.4) to the identity (2.3) and utilizing the Hölder’s type
condition (2.1) we have successively
Z
is
f (e )[u(b) − u(a)] −
b
a
b
_
f (eit ) du(t) = max f eis − f eit (u)
t∈[a,b]
a
r
≤ H max eis − eit (2.5)
t∈[a,b]
b
_
(u) .
a
Since
is
e − eit 2 = eis 2 − 2 Re ei(s−t) + eit 2 = 2 − 2 cos (s − t) = 4 sin2 s − t
2
for any t, s ∈ R, then
r
is
e − eit r = 2r sin s − t 2
(2.6)
for any t, s ∈ R.
Now, by (2.5) and (2.6) we deduce the desired result (2.2).
Remark 1. If a = 0 and b = 2π, then for any s ∈ [0, 2π] there exists
a unique
= 1 for all
t ∈ [0, 2π] such that 12 |t − s| = π2 , therefore maxt∈[0,2π] sin s−t
2
s ∈ [0, 2π] and we deduce from (2.2) the following inequality of interest
(2.7)
Z
f eis [u (2π) − u (0)] −
0
2π
2π
_
f eit du (t) ≤ 2r H
(u)
0
that holds for each s ∈ [0, 2π].
Remark 2. If [a, b] ⊂ [0, 2π] and 0 < b − a ≤ π then for all t, s ∈
[a,π b]
we have
1
1
π
|t
−
s|
≤
(b
−
a)
≤
.
Since
the
function
sin
is
increasing
on
0,
2
2
2
2 , then we
have successively that
s − t 1
1
max sin
max {b − s, s − a}
= sin max |t − s| = sin
2
2
t∈[a,b]
t∈[a,b] 2
1
1
a + b (2.8)
= sin
(b − a) + s −
4
2
2
for any s ∈ [a, b].
OSTROWSKI’S TYPE INEQUALITIES FOR COMPLEX FUNCTIONS
237
Therefore, under the assumptions of Theorem 3 and if [a, b] ⊂ [0, 2π] with
0 < b − a ≤ π, then
Z b
is
(e
)[u(b)
−
u(a)]
−
f eit du (t) f
a
(2.9)
≤ 2r H sinr
b
b
h1
i_
a + b i _
1 (u) ≤ 2r H sinr
(b−a)+ s−
(b−a)
(u)
4
2
2
2
a
a
h1
for all s ∈ [a, b].
In particular, the best inequality we can get from (2.9) is incorporated in
(2.10)
Z b
b
a+b h1
i_
f eit du (t) ≤ 2r H sinr
(b − a)
(u) .
f e 2 i [u (b) − u (a)] −
4
a
a
The case when f : C (0, 1) → C satisfies the Lipschitz condition |f (z) − f (w)| ≤
L |z − w| for any w, z ∈ C (0, 1), where L > 0 is given, is of interest due to various
examples one can consider. Also in this case we can show that the corresponding
version of the inequality (2.11) is sharp.
Corollary 1. Assume that f : C (0, 1) → C is Lipschitzian with the constant L > 0
on the circle C (0, 1). If [a, b] ⊂ [0, 2π] with 0 < b − a ≤ π and the function
u : [a, b] → C is of bounded variation on [a, b], then we have
Z b
b
a+b i_
h1
(u) .
(2.11) f e 2 i [u (b) − u (a)] −
(b − a)
f eit du (t) ≤ 2L sin
4
a
a
The constant 2 cannot be replaced by a smaller quantity.
Proof. We need to prove only the sharpness of the constant.
If we consider the function f : C → C, f (z) = z, then obviously f is Lipschitzian
with the constant L = 1. Also, consider in (2.11) a = 0 and b = π to get
Z π
π
√ _
it
(2.12)
e du (t) ≤ 2 (u) .
i [u (π) − u (0)] −
0
0
Utilising the integration by parts formula for the Riemann-Stieltjes integral we
have
Z π
Z π
Z π
π
eit du (t) = eit u (t) 0 − i
eit u (t) dt = −u (π) − u (0) − i
eit u (t) dt
0
0
0
and replacing into the inequality (2.12) we deduce
Z π
π
√ _
eit u (t) dt ≤ 2 (u)
i [u (π) − u (0)] + u (π) + u (0) + i
0
which is equivalent with
Z
(2.13)
(i
−
1)
u
(π)
+
(i
+
1)
u
(0)
−
0
π
0
π
√ _
eit u (t) dt ≤ 2 (u)
0
238
S.S. DRAGOMIR
that holds for any functions of bounded variation u : [0, π] → C and is of interest
in itself.
Now, assume that there exists a constant C > 0 such that
Z π
π
_
(2.14)
eit u (t) dt ≤ C
(u)
(i − 1) u (π) + (i + 1) u (0) −
0
0
for any functions of bounded variation u : [0, π] → C.
Consider the function u : [0, π] → R with
(
0 if 0 ≤ t < π
u (t) =
1 if t = π .
π
Rπ
W
Then u is of bounded variation, 0 eit u (t) dt = 0,
(u) = 1 and from (2.14) we
0
√
get C ≥ 2 showing that (2.14) is sharp and therefore (2.11) is sharp.
Remark 3. The case of Riemann integral, namely when u (t) = t, t ∈ [a, b] ⊆
[0, 2π], is as follows
Z b
s − t r
1
is
(2.15)
f eit du (t) ≤ 2r H max sin
f e −
b−a a
2
t∈[a,b]
for any s ∈ [a, b] provided that f : C (0, 1) → C satisfies the Hölder’s type condition
(2.1).
When u is an integral, then the following weighted integral inequality also holds.
Remark 4. If w : [a, b] ⊆ [0, 2π] → C is Lebesgue integrable on [a, b] and
f : C (0, 1) → C satisfies the Hölder’s type condition (2.1), then
Z
Z b
b
is
e
w
(t)
dt
−
f eit w (t) dt
f
a
(2.16)
a
s − t r Z b
≤ 2 H max sin
|w (t)| dt
2
t∈[a,b]
a
r
for any s ∈ [a, b].
Rb
In particular, if w (t) ≥ 0 for t ∈ [a, b] and a w (t) dt > 0 then
Z b
s − t r
1
is
f eit w (t) dt ≤ 2r H max sin
(2.17)
f e − R b
2
t∈[a,b]
a
w
(t)
dt
a
for any s ∈ [a, b].
Theorem 4. Assume that f : C (0, 1) → C is Lipschitzian with the constant L > 0
on the circle C (0, 1). If [a, b] ⊆ [0, 2π] and the function u : [a, b] → C is Lipschitzian
with the constant K > 0 on [a, b], then
Z b
h
s − a
b − s i
f eit du (t) ≤ 4LK sin2
+ sin2
f eis [u (b) − u (a)] −
4
4
a
b − a
(2.18)
≤ 8LK sin2
4
OSTROWSKI’S TYPE INEQUALITIES FOR COMPLEX FUNCTIONS
239
for any s ∈ [a, b].
Proof. It is well known that if p : [a, b] → C is a Riemann integrable function and
v : [a, b] → C is Lipschitzian with the constant M > 0, i.e.,
|f (s) − f (t)| ≤ M |s − t| for any t, s ∈ [a, b] ,
Rb
then the Riemann-Stieltjes integral a p (t) dv (t) exists and the following inequality
holds
Z b
Z b
(2.19)
p (t) dv (t) ≤ M
|p (t)| dt .
a
a
Utilising the property (2.19), we have from (2.3) that
Z b
Z b f eit du (t) = f eis − f eit du (t) f eis [u(b) − u(a)] −
a
Z
(2.20)
≤K
a
b
f eis − f eit dt ≤ KL
Z
b
is
e − eit dt
a
a
for any s ∈ [a, b]. for any t, s ∈ R, then
Since, by (2.6), eis − eit = 2 sin s−t
2
Z b
Z b
s − t is
e − eit dt = 2
sin
dt
2
a
a
Z b
hZ s
s − t
t − s i
=2
sin
dt +
dt
sin
2
2
a
s
h
s − a i
h
b − s i
= 2 1 − cos
+ 2 1 − cos
2
2
h
s − a
b − s i
b − a
(2.21)
= 4 sin2
+ sin2
≤ 8 sin2
4
4
4
for any s ∈ [a, b] ⊆ [0, 2π], and the inequality (2.18) is proved.
The best inequality we can get from (2.18) is incorporated in
Corollary 2. With the assumptions in Theorem 4 we have the inequality
Z b
a+b b − a
i
2
[u
(b)
−
u
(a)]
−
f eit du (t) ≤ 8LK sin2
.
(2.22)
e
f
8
a
The multiplicative constant 8 cannot be replaced by a smaller quantity.
Proof. We need to prove only the sharpness of the constant.
If we consider the function f : C → C, f (z) = z, then obviously f is Lipschitzian
with the constant L = 1. Also, consider in (2.22) a = 0 and b = 2π to get
Z 2π
(2.23)
eit du (t) ≤ 4K .
− [u (2π) − u (0)] −
0
Utilising the integration by parts formula for the Riemann-Stieltjes integral, we
have
Z 2π
Z 2π
Z 2π
2π
eit du (t) = eit u (t) − i
eit u (t) dt = u (2π) − u (0) − i
eit u (t) dt ,
0
0
0
0
240
S.S. DRAGOMIR
which inserted in (2.23) produces the inequality
Z 2π
eit u (t) dt ≤ 4K
− 2 [u (2π) − u (0)] + i
0
which is equivalent with
Z 2π
2
eit u (t) dt − [u (2π) − u (0)] ≤ 4K
(2.24)
i
0
that holds for any K-Lipschitzian function u : [0, 2π] → C and is of interest in
itself.
Now, assume that the inequality (2.24) holds with a constant D > 0, namely
Z 2π
2
(2.25)
eit u (t) dt − [u (2π) − u (0)] ≤ DK
i
0
for any K-Lipschitzian function u : [0, 2π] → C.
Consider u : [0, 2π] → R, u (t) = |t − π|. Then, by the continuity property of the
modulus we have that u is Lipschitzian with the constant K = 1.
We also have that
Z 2π
Z 2π
Z 2π
eit u (t) dt =
eit |t − π| dt =
|t − π| (cos t + i sin t) dt
0
0
Z
0
2π
Z
|t − π| cos t dt + i
=
0
2π
|t − π| sin t dt .
0
R 2π
Observe that by symmetry reasons 0 |t − π| sin t dt = 0 and
Z 2π
Z π
Z
h
π
|t − π| cos t dt = 2
(π − t) cos t dt = 2 (π − t) sin t0 +
0
0
π
i
sin t dt = 4
0
and by (2.25) we get D ≥ 4 which proves the desired sharpness of the constant 8
in (2.22).
Remark 5. If u (t) = t, t ∈ [a, b], then we get from (2.18) and (2.22) the following
inequalities for the Riemann integral
Z b
h
s − a
b − s i
is
f eit dt ≤ 4L sin2
+ sin2
f e (b − a) −
4
4
a
2 b−a
(2.26)
≤ 8L sin
4
for any s ∈ [a, b] and
(2.27)
Z
a+b i
2
(b − a) −
f e
a
b
b − a
f eit dt ≤ 8L sin2
8
provided that f : C (0, 1) → C is Lipschitzian with the constant L > 0 on the circle
C (0, 1).
OSTROWSKI’S TYPE INEQUALITIES FOR COMPLEX FUNCTIONS
241
Remark 6. If w : [a, b] ⊆ [0, 2π] → C is essentially bounded on [a, b] and f : C (0, 1)
→ C is Lipschitzian with the constant L > 0 on the circle C (0, 1), then we have
the following weighted integral inequality
Z
Z b
h
s − a
b − s i
b
is
w (t) dt − f eit w (t) dt ≤ 4L kwk∞ sin2
+ sin2
f e
4
4
a
a
b − a
(2.28)
≤ 8L kwk∞ sin2
4
for any s ∈ [a, b] where kwk∞ := ess supt∈[a,b] |w (t)| .
In particular, we have
Z b
Z b
b − a
a+b (2.29)
w (t) dt −
f eit w (t) dt ≤ 8L kwk∞ sin2
.
f e 2 i
8
a
a
The case of monotonic integrators is as follows:
Theorem 5. Assume that f : C (0, 1) → C is Lipschitzian with the constant L > 0
on the circle C (0, 1). If [a, b] ⊆ [0, 2π] and the function u : [a, b] → R is monotonic
nondecreasing on [a, b], then
Z b
h
b − s
s − a
i
is
u (b) − sin
u (a)
f e [u (b)−u (a)]− f eit du (t) ≤ 2L sin
2
2
a
Z b
s − t
+L
(2.30)
sgn (s − t) cos
u (t) dt
2
a
for any s ∈ [a, b].
In particular, we have
Z
a+b f e 2 i [u (b) − u (a)] −
b
b − a
[u (b) − u (a)]
f eit du (t) ≤ 2L sin
4
a
Z b
a + b
a+b − t 2
+L
sgn
− t cos
u (t) dt .
2
2
a
(2.31)
Proof. It is well known that if p : [a, b] → C is a continuous function and
v : [a, b] → R is monotonic nondecreasing on [a, b], then the Riemann-Stieltjes
Rb
integral a p (t) dv (t) exists and the following inequality holds
Z b
Z b
(2.32)
p (t) dv (t) ≤
|p (t)| dv (t) .
a
a
Utilising the property (2.32), we have from (2.3) that
Z b
Z b
f eit du (t) = f eis − f eit du (t) f eis [u (b) − u (a)] −
Z
(2.33)
≤
f e
a
for any s ∈ [a, b].
a
b
a
is
− f eit du (t) ≤ L
Z
a
b
is
e − eit du (t)
242
S.S. DRAGOMIR
Since, by (2.6), eis − eit = 2 sin
b
Z
s−t
2
for any t, s ∈ R, then
b
s − t sin
du (t)
2
a
Z b
hZ s
s − t
t − s
i
=2
sin
du (t) +
sin
du (t)
2
2
a
s
is
e − eit du (t) = 2
a
(2.34)
Z
for any s ∈ [a, b] ⊆ [0, 2π].
Utilising the integration by parts formula for the Riemann-Stieltjes integral, we
have
Z s
s 1 Z s
s − t
s − t
s − t
sin
du (t) = sin
u (t) +
cos
u (t) dt
2
2
2 a
2
a
a
Z
s − a
s − t
1 s
= − sin
u (a) +
cos
u (t) dt
2
2 a
2
and
Z
b
sin
s
t − s
2
b 1 Z b
t − s
cos
u (t) −
u (t) dt
2
2 s
2
s
Z
b − s
t − s
1 b
= sin
u (b) −
cos
u (t) dt ,
2
2 s
2
du (t) = sin
t − s
which, by (2.34), produce the equality
Z
a
(2.35)
b
h
i
is
e − eit du (t) = 2 sin b − s u (b) − sin s − a u (a)
2
2
Z s
Z b
s − t
t − s
+
cos
u (t) dt −
cos
u (t) dt
2
2
a
s
h
b − s
s − a
i
u (b) − sin
u (a)
= 2 sin
2
2
Z b
s − t
+
sgn (s − t) cos
u (t) dt .
2
a
Utilising (2.33) we deduce the desired result (2.30).
Remark 7. We remark that if a = 0 and b = 2π, then we get from (2.30) and
(2.31) that
Z
is
u (2π) − u (0) −
f e
(2.36)
for any s ∈ [a, b].
2π
s
f eit du (t) ≤ 2L sin
[u (2π) − u (0)]
2
0
Z 2π
s − t
+L
sgn (s − t) cos
u (t) dt
2
0
OSTROWSKI’S TYPE INEQUALITIES FOR COMPLEX FUNCTIONS
In particular, we have
Z
f (−1) u (2π) − u (0) −
2π
0
√
f eit du (t) ≤ 2L [u (2π) − u (0)]
2π
Z
sgn (π − t) sin
+L
(2.37)
243
t
0
2
u (t) dt .
Corollary 3. Assume that f and u are as in Theorem 5 then for any [a, b] ⊂ [0, 2π]
with 0 < b − a ≤ π we have the sequence of inequalities
Z b
h
b − s
s − a
i
is
f (eit ) du(t) ≤ 2L sin
u (b) − sin
u (a)
f (e )[u(b) − u(a)] −
2
2
a
Z b
s − t
+L
u (t) dt
sgn (s − t) cos
2
a
h
b − s
s − a
i
(2.38)
≤ 2L sin
[u (b) − u (s)] + sin
[u (s) − u (a)] =: B (s)
2
2
where
( 1
[u (b) − u (a)]
sin 4 (b − a) + 12 s − a+b
2h
a+b
B (s) ≤ 2L ×
s−
u(b)−u(a)
2
cos
+
2 sin b−a
u (s) −
4
2
2
for any s ∈ [a, b].
In particular, we have
Z
a+b i
2
[u (b) − u (a)] −
f e
i
u(b)+u(a) 2
b
b − a
[u (b) − u (a)]
f eit du (t) ≤ 2L sin
4
a
Z b
a+b − t a + b
2
− t cos
u (t) dt =: M
+L
sgn
2
2
a
(2.39)
where
M ≤ 2L sin
Proof. Since 0 < b − a ≤ π, then
b − a
[u (b) − u (a)] .
4
|s−t|
2
≤
π
2
Z
s
fors, t ∈[a, b]. Utilising the fact that
u is monotonic nondecreasing on [a, b] and cos |s−t|
≥ 0 for s, t ∈ [a, b], then
2
Z
(2.40)
s
cos
s − t
2
a
u (t) dt ≤ u (s)
cos
s − t
2
a
dt = 2u (s) sin
s − a
2
and
Z
b
cos
s
s − t
2
Z
u (t) dt ≥ u (s)
b
cos
s
s − t
2
dt = 2u (s) sin
i.e.,
Z
(2.41)
−
b
cos
s
s − t
b − s
u (t) dt ≤ −2u (s) sin
.
2
2
b − s
2
244
S.S. DRAGOMIR
Summing (2.40) with (2.41) we deduce that
Z b
s − a
b − s
s − t
u (t) dt ≤ 2u (s) sin
− 2u (s) sin
sgn (s − t) cos
2
2
2
a
giving that
Z b
h
b − s
s − a
i
s − t
2L sin
u (b) − sin
u (a) + L
sgn (s − t) cos
u (t) dt
2
2
2
a
h
b − s
s − a
i
≤ 2L sin
[u (b) − u (s)] + sin
[u (s) − u (a)] ,
2
2
which proves the second inequality in (2.38).
The bounds for B (s) follows from the elementary property stating that
αx + βy ≤ max {α, β} (x + y)
where α, β, x, y ≥ 0.
3. A quadrature rule
We consider the following partition of the interval [a, b]
∆n : a = x0 < x1 < · · · < xn−1 < xn = b
and the intermediate points ξk ∈ [xk , xk+1 ] where 0 ≤ k ≤ n − 1. Define hk :=
xk+1 − xk , 0 ≤ k ≤ n − 1 and ν (∆n ) = max {hk : 0 ≤ k ≤ n − 1} the norm of the
partition ∆n .
For the continuous function f : C (0, 1) → C and the function u : [a, b] ⊆ [0, 2π] →
C of bounded variation on [a, b], define the quadrature rule
(3.1)
On (f, u, ∆n , ξ) :=
n−1
X
f eiξk [u (xk+1 ) − u (xk )]
k=0
and the remainder Rn (f, u, ∆n , ξ) in approximating the Riemann-Stieltjes integral
Rb
f eit du (t) by On (f, u, ∆n , ξ). Then we have
a
Z b
(3.2)
f eit du (t) = On (f, u, ∆n , ξ) + Rn (f, u, ∆n , ξ) .
a
The following result provides a priory bounds for Rn (f, u, ∆n , ξ) in several instances
of f and u as above.
Proposition 1. Assume that f : C (0, 1) → C satisfies the following Hölder’s type
condition
r
|f (z) − f (w)| ≤ H |z − w|
for any w, z ∈ C (0, 1), where H > 0 and r ∈ (0, 1] are given.
If [a, b] ⊆ [0, 2π] and the function u : [a, b] → C is of bounded variation on [a, b],
then for any partition ∆n : a = x0 < x1 < · · · < xn−1 < xn = b with the norm
OSTROWSKI’S TYPE INEQUALITIES FOR COMPLEX FUNCTIONS
ν (∆n ) ≤ π we have the error bound
n−1
h1
X
|Rn (f, u, ∆n , ξ)| ≤ 2r H
sinr (xk+1 − xk ) +
4
k=0
≤ 2r H
n−1
X
≤H
(3.3)
xk+1
xk + xk+1 i _
1 (u)
ξk −
2
2
x
k
sinr
h1
k=0
n−1
X
245
2
(xk+1 − xk )
k+1
i x_
(u)
xk
xk+1
r
(xk+1 − xk )
_
(u) ≤ Hν r (∆n )
xk
k=0
b
_
(u)
a
for any intermediate points ξk ∈ [xk , xk+1 ] where 0 ≤ k ≤ n − 1.
Proof. Since ν (∆n ) ≤ π, then on writing inequality (2.9) on each interval
[xk , xk+1 ] and for any intermediate points ξk ∈ [xk , xk+1 ] where 0 ≤ k ≤ n − 1, we
have
Z xk+1
iξk
f eit du (t) f (e )[u(xk+1 ) − u(xk )] −
xk
r
≤ 2 H sin
r
xk+1
1 xk + xk+1 i _
(u)
(xk+1 − xk ) + ξk −
4
2
2
x
h1
k
(3.4)
≤ 2r H sinr
h1
(xk+1 − xk )
2
k+1
i x_
xk+1
(u) ≤ H (xk+1 − xk )
xk
r
_
(u)
xk
where for the last inequality we have used the fact that sin x ≤ x for x ∈ 0, π2 .
Summing over k from 0 to n − 1 in (3.4) and utilizing the generalized triangle
inequality, we deduce the first part of (3.3). The second part is obvious.
Corollary 4. Assume that f, u and ∆n are as in Proposition 1. Define the midpoint
trapezoid type quadrature rule by
(3.5)
Tn (f, u, ∆n ) :=
n−1
X
f e
xk+1 +xk
2
i
[u (xk+1 ) − u (xk )]
k=0
and the error En (f, u, ∆n ) by
Z b
(3.6)
f eit du (t) = Tn (f, u, ∆n ) + En (f, u, ∆n ) .
a
Then we have the error bounds
x_
n−1
k+1
X
r
r 1
|En (f, u, ∆n )| ≤ 2 H
sin
(xk+1 − xk )
(u)
4
x
k=0
(3.7)
≤
1
H
2r
n−1
X
k=0
k
xk+1
(xk+1 − xk )
r
_
xk
b
(u) ≤
_
1
r
Hν
(∆
)
(u) .
n
2r
a
The case of both integrator and integrand being Lipschitzian is incorporated in
the following result:
246
S.S. DRAGOMIR
Proposition 2. Assume that f : C (0, 1) → C is Lipschitzian with the constant
L > 0 on the circle C (0, 1). If [a, b] ⊆ [0, 2π] and the function u : [a, b] → C is
Lipschitzian with the constant K > 0 on [a, b], then for any partition ∆n : a =
x0 < x1 < · · · < xn−1 < xn = b we have the error bound
|Rn (f, u, ∆n , ξ)| ≤ 4LK
n−1
Xh
sin2
k=0
≤ 8LK
n−1
X
sin2
x
ξ − x i
k+1 − ξk
k
k
+ sin2
4
4
x
n−1
k=0
(3.8)
X
− xk 1
2
≤ LK
(xk+1 − xk )
4
2
k+1
k=0
1
≤ LK (b − a) ν (∆n )
2
for any intermediate points ξk ∈ [xk , xk+1 ] where 0 ≤ k ≤ n − 1.
In particular, we have
|En (f, u, ∆n )| ≤ 8LK
n−1
X
sin2
k=0
x
− xk 8
k+1
n−1
≤
(3.9)
X
1
1
2
(xk+1 − xk ) ≤ LK (b − a) ν (∆n ) .
LK
8
8
k=0
The proof follows by Theorem 4 and the details are omitted.
Proposition 3. Assume that f : C (0, 1) → C is Lipschitzian with the constant
L > 0 on the circle C (0, 1). If [a, b] ⊆ [0, 2π] and the function u : [a, b] → R is
monotonic nondecreasing on [a, b], then for any partition ∆n : a = x0 < x1 < · · · <
xn−1 < xn = b with the norm ν (∆n ) ≤ π we have the error bound
|Rn (f, u, ∆n , ξ)| ≤ 2L
n−1
Xh
sin
x
k=0
+L
n−1
X Z xk+1
k=0
n−1
Xh
ξ − x i
− ξk k
k
u (xk+1 ) − sin
u (xk )
2
2
k+1
sgn (ξk − t) cos
xk
ξ − t
k
u (t) dt
2
− ξk [u (xk+1 ) − u (ξk )]
2
k=0
ξ − x i
k
k
+ sin
[u (ξk ) − u (xk )]
2
n−1
h1
X
1 xk +xk+1 i
≤ 2L
sin
(xk+1 −xk )+ ξk −
[u (xk+1 )−u (xk )]
4
2
2
≤ 2L
sin
x
k+1
k=0
≤ 2L
n−1
X
k=0
sin
h1
2
i
(xk+1 − xk ) [u (xk+1 ) − u (xk )]
OSTROWSKI’S TYPE INEQUALITIES FOR COMPLEX FUNCTIONS
(3.10)
≤L
n−1
X
247
(xk+1 − xk ) [u (xk+1 ) − u (xk )] ≤ ν (∆n ) L [u (b) − u (a)]
k=0
for any intermediate points ξk ∈ [xk , xk+1 ] where 0 ≤ k ≤ n − 1.
In particular, we have
n−1
x
X
k+1 − xk
[u (xk+1 ) − u (xk )]
|En (f, u, ∆n )| ≤ 2L
sin
4
k=0
n−1
x + x
xk +xk+1 − t X Z xk+1
k
k+1
2
+L
sgn
− t cos
u (t) dt
2
2
xk
k=0
≤ 2L
n−1
X
sin
k=0
x
− xk [u (xk+1 ) − u (xk )]
4
k+1
n−1
(3.11)
≤
1
1 X
(xk+1 − xk ) [u (xk+1 ) − u (xk )] ≤ Lν (∆n ) [u (b) − u (a)] .
L
2
2
k=0
The proof follows by Corollary 3 and the details are omitted.
4. Applications for functions of unitary operators
We recall that the bounded linear operator U on the Hilbert space H is unitary
iff U ∗ = U −1 .
It is well known that (see for instance [5, p. 275-p. 276]), if U is a unitary
operator, then there exists a family of projections {Eλ }λ∈[0,2π] , called the spectral
family of U with the following properties
a) Eλ ≤ Eµ for 0 ≤ λ ≤ µ ≤ 2π;
b) E0 = 0 and E2π = 1H (the identity operator on H);
c) Eλ+0 = Eλ for 0 ≤ λ < 2π;
R 2π
d) U = 0 eiλ dEλ where the integral is of Riemann-Stieltjes type.
Moreover, if {Fλ }λ∈[0,2π] is a family of projections satisfying the requirements
a)-d) above for the operator U, then Fλ = Eλ for all λ ∈ [0, 2π].
Also, for every continuous complex valued function f : C (0, 1) → C on the
complex unit circle, we have
Z 2π
(4.1)
f (U ) =
f eiλ dEλ
0
where the integral is taken in the Riemann-Stieltjes sense.
In particular, we have the equalities
Z 2π
(4.2)
f eiλ dEλ x ,
f (U ) x =
0
Z
(4.3)
hf (U ) x, yi =
0
2π
f eiλ d hEλ x, yi
248
S.S. DRAGOMIR
and
2
Z
2π
kf (U ) xk =
(4.4)
f eiλ 2 d kEλ xk2 ,
0
for any x, y ∈ H.
We consider the following partition of the interval [a, b]
∆n : 0 = λ0 < λ1 < · · · < λn−1 < λn = 2π
and the intermediate points ξk ∈ [λk , λk+1 ] where 0 ≤ k ≤ n − 1. Define hk :=
λk+1 − λk , 0 ≤ k ≤ n − 1 and ν (∆n ) = max {hk : 0 ≤ k ≤ n − 1} the norm of the
partition ∆n .
If U is a unitary operator on the Hilbert space H and {Eλ }λ∈[0,2π] , the spectral
family of U, then we can introduce the following sums
(4.5)
On (f, U, ∆n , ξ; x, y) :=
n−1
X
f eiξk
Eλk+1 − Eλk x, y
k=0
and
Tn (f, U, ∆n ; x, y) :=
(4.6)
n−1
X
λk+1 +λk
Eλk+1 − Eλk x, y
f e 2 i
k=0
where x, y ∈ H.
Theorem 6. With the above assumptions for U, {Eλ }λ∈[0,2π] , ∆n with ν (∆n ) ≤
π and if f : C (0, 1) → C satisfies the Hölder’s type condition |f (z) − f (w)| ≤
r
H |z − w| for any w, z ∈ C (0, 1), where H > 0 and r ∈ (0, 1] are given, then we
have the representation
(4.7)
hf (U ) x, yi = On (f, U, ∆n , ξ; x, y) + Rn (f, U, ∆n , ξ; x, y)
with the error Rn (f, U, ∆n , ξ; x, y) satisfying the bounds
Rn (f, U, ∆n , ξ; x, y)
≤ 2r H
n−1
X
sinr
k=0
≤ 2r H
n−1
X
k=0
≤H
n−1
X
λk+1
1 λk + λk+1 i _ (λk+1 − λk ) + ξk −
E(·) x, y
4
2
2
h1
λk
sinr
h1
2
(λk+1 − λk )
_ E(·) x, y
λk
λk+1
r
(λk+1 − λk )
k=0
(4.8)
i λk+1
2π
_ _
E(·) x, y ≤ Hν r (∆n )
E(·) x, y
λk
0
≤ Hν r (∆n ) kxk kyk
for any x, y ∈ H and the intermediate points ξk ∈ [λk , λk+1 ] where 0 ≤ k ≤ n − 1.
In particular we have
(4.9)
hf (U ) x, yi = Tn (f, U, ∆n ; x, y) + En (f, U, ∆n ; x, y)
OSTROWSKI’S TYPE INEQUALITIES FOR COMPLEX FUNCTIONS
249
with the error
En (f, U, ∆n ; x, y) ≤ 2r H
n−1
X
sinr
k=0
h1
4
(λk+1 − λk )
k+1
i λ_
E(·) x, y
λk
λk+1
≤
n−1
_ 1 X
r
H
(λ
−
λ
)
E(·) x, y
k+1
k
2r
k=0
≤
(4.10)
λk
1
Hν r (∆n )
2r
2π
_
E(·) x, y
≤
0
1
Hν r (∆n ) kxk kyk
2r
for any x, y ∈ H.
Proof. For given x, y ∈ H, define the function u (λ) := hEλ x, yi, λ ∈ [0, 2π]. We
will show that u is of bounded variation and
2π
2π
_
_
(4.11)
(u) =:
E(·) x, y ≤ kxk kyk .
0
0
It is well known that, if P is a nonnegative selfadjoint operator on H, i.e., hP x, xi ≥ 0
for any x ∈ H, then the following inequality is a generalization of the Schwarz
inequality in H
2
|hP x, yi| ≤ hP x, xi hP y, yi ,
(4.12)
for any x, y ∈ H.
Now, if d : 0 = t0 < t1 < · · · < tn−1 < tn = 2π is an arbitrary partition of the
interval [0, 2π], then we have by Schwarz’s inequality for nonnegative operators
(4.12) that
2π
n n−1
_
X o
Eti+1 − Eti x, y E(·) x, y = sup
d
0
(4.13)
≤ sup
n n−1
X
d
i=0
1/2 1/2 o
Eti+1 − Eti x, x
Eti+1 − Eti y, y
:= I .
i=0
By the Cauchy-Buniakovski-Schwarz inequality for sequences of real numbers we
also have that
h n−1
X
X
i1/2 h n−1
i1/2
I ≤ sup
Eti+1 − Eti x, x
Eti+1 − Eti y, y
d
i=0
≤ sup
h n−1
X
d
(4.14)
=
i=0
Eti+1
h n−1
X
i1/2
i1/2
− Eti x, x
sup
Eti+1 − Eti y, y
d
i=0
2π
h_
0
for any x, y ∈ H.
E(·) x, x
2π
i1/2 h _
0
E(·) y, y
i=0
i1/2
= kxk kyk
250
S.S. DRAGOMIR
On making use of (4.13) and (4.14) we deduce the desired result (4.11).
Now, applying Proposition 1 to the spectral representation (4.3) we deduce the
desired result (4.7) with the error bound (4.8). The details are omitted.
Remark 8. In the case when the partition reduces to the whole interval [0, 2π] ,
then utilizing the inequality (2.7) we deduce the bound
(4.15)
2π
_
f eis hx, yi − hf (U ) x, yi ≤ 2r H
E(·) x, y ≤ 2r H kxk kyk
0
for any s ∈ [0, 2π] and any vectors x, y ∈ H.
In the case when the division is
∆2 : 0 = λ0 < λ1 = π < λ2 = 2π
and we take the intermediate points u ∈ [0, π] and v ∈ [π, 2π] , then we get from
Theorem 6 that
iu
f (e ) hEπ x, yi + f (eiv ) h(I − Eπ ) x, yi − hf (U ) x, yi h
h1
≤ 2r H sinr π +
4
+ sinr
(4.16)
π
1 π i _
(hE(·) x, yi)
u − 2
2
0
2π
i
1 3π i _ E(·) x, y
π + v −
4
2
2
π
h1
for any vectors x, y ∈ H.
The best inequality we can get from (4.17) is obtained for u = π2 and v =
namely
f (i) hEπ x, yi + f (−i) h(1H − Eπ ) x, yi − hf (U ) x, yi r
≤ 22 H
(4.17)
3π
2 ,
2π
_
r
E(·) x, y ≤ 2 2 H kxk kyk
0
for any vectors x, y ∈ H.
If U is a unitary operator on the Hilbert space H and {Eλ }λ∈[0,2π] , the spectral
family of U, then we can introduce the following sums depending only of one vector
x∈H
n−1
X
(4.18)
Õn (f, U, ∆n , ξ; x) :=
f eiξk
Eλk+1 − Eλk x, x
k=0
and
(4.19)
T̃n (f, U, ∆n ; x, y) :=
n−1
X
f e
λk+1 +λk
2
i
Eλk+1 − Eλk x, x .
k=0
Theorem 7. With the above assumptions for U, {Eλ }λ∈[0,2π] , ∆n with ν (∆n ) ≤ π
and, if f : C (0, 1) → C is Lipschitzian with the constant L > 0 on the circle C (0, 1),
then we have the representation
OSTROWSKI’S TYPE INEQUALITIES FOR COMPLEX FUNCTIONS
251
hf (U ) x, xi = Õn (f, U, ∆n , ξ; x) + R̃n (f, U, ∆n , ξ; x)
(4.20)
with the error R̃n (f, U, ∆n , ξ; x) satisfying the bounds
R̃n (f, U, ∆n , ξ; x) ≤ 2L
n−1
Xh
i
ξ − λ − ξk k
k
hEλk x, xi
Eλk+1 x, x − sin
2
2
λ
k+1
sin
k=0
+L
n−1
X Z λk+1
sgn (ξk − t) cos
λk
k=0
n−1
Xh
ξ − t
k
hEt x, xi dt
2
− ξk Eλk+1 − Eξk x, x
2
k=0
i
ξ − λ k
k
h(Eξk − Eλk ) x, xi
+ sin
2
n−1
h1
X
1 λk + λk+1 i Eλk+1 − Eλk x, x
≤ 2L
sin
(λk+1 − λk ) + ξk −
4
2
2
≤ 2L
λ
k+1
sin
k=0
≤ 2L
n−1
X
sin
h1
2
k=0
(4.21)
≤L
n−1
X
(λk+1 − λk )
(λk+1 − λk )
i
Eλk+1 − Eλk x, x
2
Eλk+1 − Eλk x, x ≤ ν (∆n ) L kxk
k=0
for any x ∈ H and the intermediate points ξk ∈ [λk , λk+1 ] where 0 ≤ k ≤ n − 1.
In particular we have
hf (U ) x, xi = T̃n (f, U, ∆n ; x) + Ẽn (f, U, ∆n ; x)
(4.22)
with the error
|Ẽn (f, U,∆n ; x)| ≤ 2L
n−1
X
sin
k=0
+L
n−1
X Z xk+1
sgn
k=0
≤ 2L
n−1
X
sin
k=0
(4.23)
≤
λ
k+1 −λk
4
Eλk+1 − Eλk x, x
λ + λ
λk +λk+1 − t k
k+1
2
−t cos
hEt x, xi dt
2
2
− λk Eλk+1 − Eλk x, x
4
k+1
n−1
1
1 X
2
L
(λk+1 −λk ) Eλk+1 −Eλk x, x ≤ Lν (∆n ) kxk
2
2
k=0
for any x ∈ H.
xk
λ
252
S.S. DRAGOMIR
The proof follows by Proposition 3 applied for the monotonic nondecreasing
function u (t) := hEt x, xi, t ∈ [0, 2π].
Remark 9. We remark that if the partition reduces to the whole interval [0, 2π]
then we get from (2.36) that
is
f (e )kxk2 − hf (U )x, xi ≤ 2L sin s kxk2
2
Z 2π
s − t
(4.24)
+L
sgn (s − t) cos
hEt x, xi dt
2
0
for any s ∈ [a, b] and x ∈ H.
In particular, we have
√
f (−1)kxk2 − hf (U )x, xi ≤ 2Lkxk2
Z 2π
t
+L
sgn (π − t) sin
(4.25)
hEt x, xi dt
2
0
for any x ∈ H.
Example 1. In order to provide some simple examples for the inequalities above
we choose two complex functions as follows.
a) Consider the power function f : C\ {0} → C, f (z) = z m where m is a
nonzero integer. Then, obviously, for any z, w belonging to the unit circle C (0, 1)
we have the inequality
|f (z) − f (w)| ≤ |m| |z − w|
which shows that f is Lipschitzian with the constant L = |m| on the circle C (0, 1).
Then from (4.15), we get for any unitary operator U that
(4.26)
2π
_
ims
e
hx, yi − hU m x, yi ≤ 2 |m|
E(·) x, y ≤ 2 |m| kxk kyk
0
for any s ∈ [0, 2π] and x, y ∈ H.
Also, from (4.16) and the intermediate points u ∈ [0, π] and v ∈ [π, 2π], we have
for any unitary operator U
imu
e
hEπ x, yi + eimv h(1H − Eπ ) x, yi − hU m x, yi
π
h
h1
1 π i _ ≤ 2|m| sin π + u − E(·) x, y
4
2
2
0
(4.27)
+ sin
2π
i
1 3π i _ π + v −
E(·) x, y
4
2
2
π
h1
for any vectors x, y ∈ H, where {Eλ }λ∈[0,2π] is the spectral family of U .
OSTROWSKI’S TYPE INEQUALITIES FOR COMPLEX FUNCTIONS
The best inequality we can get from (4.27) is obtained for u =
namely
m
i hEπ x, yi + (−i)m h(1H − Eπ ) x, yi − hU m x, yi ≤
(4.28)
√
2 |m|
π
2
253
and v =
3π
2 ,
2π
_
√
E(·) x, y ≤ 2 |m| kxk kyk ,
0
for any vectors x, y ∈ H.
b) For a 6= ±1, 0 consider the function f : C (0, 1) → C, fa (z) =
that
|a| |z − w|
(4.29)
|fa (z) − fa (w)| =
|1 − az| |1 − aw|
1
1−az .
Observe
for any z, w ∈ C (0, 1).
If z = eit with t ∈ [0, 2π], then we have
2
2
|1 − az| = 1 − 2a Re (z̄) + a2 |z| = 1 − 2a cos t + a2
≥ 1 − 2 |a| + a2 = (1 − |a|)
2
therefore
(4.30)
1
1
≤
|1 − az|
|1 − |a||
and
1
1
≤
|1 − aw|
|1 − |a||
for any z, w ∈ C (0, 1).
Utilising (4.29) and (4.30) we deduce
(4.31)
|fa (z) − fa (w)| ≤
|a|
(1 − |a|)
2
|z − w|
for any z, w ∈ C (0, 1), showing that the function fa is Lipschitzian with the constant
|a|
La = (1−|a|)
2 on the circle C (0, 1).
Applying the inequality (4.15), we get for any unitary operator U that
(1 − aeis )−1 hx, yi − (1H − aU )−1 x, y (4.32)
≤
2π
_
2 |a|
2
(1 − |a|)
0
E(·) x, y
≤
2 |a|
2
(1 − |a|)
kxk kyk
for any s ∈ [0, 2π] and x, y ∈ H.
Also, from (4.16) and the intermediate points u ∈ [0, π] and v ∈ [π, 2π], we have
for any unitary operator U
(1 − aeiu )−1 hEπ x, yi + (1 − aeiv )−1 h(1H − Eπ )x, yi − (1H − aU )−1 x, y π
h1
2|a| h
1 π i _ ≤
sin
π
+
−
E(·) x, y
u
2
(1 − |a|)
4
2
2
0
(4.33)
2π
i
1 3π i _
+ sin π + v −
(hE(·) x, yi)
4
2
2
π
h1
254
S.S. DRAGOMIR
for any vectors x, y ∈ H, where {Eλ }λ∈[0,2π] is the spectral family of U .
The best inequality we can get from (4.27) is obtained for u = π2 and v =
namely
(1 − ai)−1 hEπ x, yi + (1 + ai)−1 h(1H − Eπ )x, yi − (1H − aU )−1 x, y √
(4.34)
≤
2π
_
2 |a|
(1 − |a|)
2
√
E(·) x, y
≤
0
2 |a|
2
(1 − |a|)
3π
2 ,
kxk kyk
for any vectors x, y ∈ H.
References
[1] Dragomir, S. S., On the Ostrowski’s inequality for Riemann-Stieltjes integral, Korean J. Appl.
Math. 7 (2000), 477–485.
Rb
[2] Dragomir, S. S., On the Ostrowski inequality for Riemann-Stieltjes integral
f (t) du (t)
a
where f is of Hölder type and u is of bounded variation and applications, J. KSIAM 5 (1)
(2001), 35–45.
[3] Dragomir, S. S., Ostrowski’s type inequalities for continuous functions of selfadjoint operators
on Hilbert spaces: a survey of recent results, Ann. Funct. Anal. 2 (1) (2011), 139–205.
[4] Dragomir, S. S., Ostrowski’s type inequalities for some classes of continuous functions of
selfadjoint operators in Hilbert spaces, Comput. Math. Appl. 62 (12) (2011), 4439–4448.
[5] Helmberg, G., Introduction to Spectral Theory in Hilbert Space, John Wiley and Sons, 1969.
[6] Ostrowski, A., Über die Absolutabweichung einer differentiierbaren Funktion von ihrem
Integralmittelwert (German), Comment. Math. Helv. 10 (1) (1937), 226–227.
Mathematics, College of Engineering and Science,
Victoria University, PO Box 14428,
Melbourne City, MC 8001, Australia
School of Computer Science and Applied Mathematics,
University of the Witwatersrand,
Private Bag 3, Johannesburg 2050, South Africa
E-mail: sever.dragomir@vu.edu.au