MATEMATIQKI VESNIK
originalni nauqni rad
research paper
64, 4 (2012), 336–346
December 2012
OSTROWSKI INEQUALITIES FOR COSINE AND SINE
OPERATOR FUNCTIONS
George A. Anastassiou
Abstract. Here we present Ostrowski type inequalities on Cosine and Sine Operator Functions for various norms. At the end we give some applications.
1. Introduction
The main motivation here is the famous Ostrowski inequality from 1938, see
[1, 2, 10], which follows:
Theorem 1. Let f : [a, b] → R be continuous on [a, b] and differentiable
on (a, b) whose derivative f 0 : (a, b) → R is bounded on (a, b), i.e. kf 0 k∞ :=
supt∈(a,b) |f 0 (t)| < ∞. Then
¯
¯ "
¢2 #
¡
¯ 1 Z b
¯
x − a+b
¯
¯
2
(b − a) kf 0 k∞ ,
(1)
f (t) dt − f (x)¯ ≤ C +
¯
2
¯b − a a
¯
(b − a)
for any x ∈ [a, b], where C is a positive constant. The constant C =
possible.
1
4
is the best
We generalize here (1) to Cosine and Sine operator functions and we expand
it in various directions. At the end we give some applications.
2. Background
For notions and results of this section see [4, 6, 8, 9, 11].
Let (X, k·k) be a real or complex Banach space. By definition, a cosine operator
function is a family {C (t) : t ∈ R} of bounded linear operators from X into itself,
satisfying
2010 AMS Subject Classification: 26D99, 47D09, 47D99.
Keywords and phrases: Ostrowski inequality; Cosine operator function; Sine operator
function.
336
Ostrowski inequalities for operator functions
337
(i) C (0) = I, I the identity operator;
(ii) C (t + s) + C (t − s) = 2C (t) C (s), for all t, s ∈ R (the last product is composition);
(iii) C (·) f is continuous on R, for all f ∈ X.
Notice that C (t) = C (−t), for all t ∈ R.
The associated sine operator function S (·) is defined by
Z t
S (t) f :=
C (s) f ds, for all t ∈ R and for all f ∈ X.
0
Notice that S (t) f ∈ X and it is continuous in t ∈ R.
The cosine operator function C (·) is such that kC (t)k ≤ M eωt , for some
M ≥ 1, ω ≥ 0, for all t ∈ R+ ; here k·k is the norm of the operator.
The infinitesimal generator A of C (·) is the operator from X into itself defined
as
2
Af := lim 2 (C (t) − I) f
t→0+ t
with domain D (A). The operator A is closed and D (A) is dense in X, i.e., D (A) =
X, and satisfies
Z t
Z t
S (s) f ds ∈ D (A) and A
S (s) f ds = C (t) f − f , for all f ∈ X.
0
0
00
Also, A = C (0) holds, and D (A) is the set of f ∈ X such that C (t) f is twice
differentiable at t = 0; equivalently,
©
ª
D (A) = f ∈ X : C (·) f ∈ C 2 (R, X) .
If f ∈ D (A), then C (t) f ∈ D (A), and C 00 (t) f = C (t) Af = AC (t) f , for all
t ∈ R; C 0 (0) f = 0, see [5, 12].
We define A0 = I, A2 = A ◦ A, . . . , An = A ◦ An−1 , n ∈ N. Let f ∈ D (An );
then C (t) f ∈ C 2n (R, X), and C (2n) (t) f = C (t) An f = An C (t) f , for all t ∈ R,
and C (2k−1) (0) f = 0, 1 ≤ k ≤ n, see [8].
For f ∈ D (An ), t ∈ R, we have the cosine operator function’s Taylor formula
[8, 9] saying that
Z t
n−1
2n−1
X t2k
(t − s)
Tn (t) f := C (t) f −
Ak f =
C (s) An f ds.
(2)
(2k)!
(2n − 1)!
0
k=0
By integrating (2) we obtain the sine operator function’s Taylor formula
Mn (t) f : = S (t) f − f t −
Z
=
0
and all f ∈ D (An ).
t
t3
t2n−1
Af − · · · −
An−1 f
3!
(2n − 1)!
2n
(t − s)
C (s) An f ds, for all t ∈ R,
(2n)!
(3)
338
G. A. Anastassiou
Integrals in (2) and (3) are vector valued Riemann integrals, see [3, 7]. Here
f ∈ D (An ), n ∈ N.
Let a > 0R and F ∈ C ([0, a] , X); then F is vector-Riemann integrable, see [11].
a
Clearly here 0 F (t) dt ∈ X.
3. Ostrowski type inequalities
We first present results on the natural interval; here [0, a], a > 0.
Theorem 2. Denote
kkC (·) An f kk∞,[0,a] := sup kC (t) An f k .
t∈[0,a]
Here t0 ∈ [0, a]. Then
°
° Z a
°
°1
°
Tn (t) f dt − Tn (t0 ) f °
(i) °
°
a 0
·µ
¶
¸
n
kkC (·) A f kk∞,[0,a]
4nt2n+1
+ a2n+1
0
≤
·
− at2n
,
0
a (2n)!
2n + 1
(ii)
(4)
° Z a
°
°1
°
°
Mn (t) f dt − Mn (t0 ) f °
°a
°
0
!
"Ã
#
2(n+1)
kkC (·) An f kk∞,[0,a]
2 (2n + 1) t0
+ a2(n+1)
·
− at2n+1
. (5)
≤
0
a (2n + 1)!
2 (n + 1)
Proof. By (2) we have (m := 2n − 1)
Z t
m
(t − s)
Tn (t) f =
C (s) An f ds,
m!
0
and
Z
t0
m
(t0 − s)
C (s) An f ds,
m!
0
for any t, t0 ∈ [0, a]. We estimate En (t) f := Tn (t) f − Tn (t0 ) f .
Case of t ≥ t0 : we have
° Z t0
m
°
(t − s)
kEn (t) f k = °
C (s) An f ds+
°
m!
0
°
Z t
Z t0
m
m
°
(t − s)
(t0 − s)
C (s) An f ds −
C (s) An f ds°
°
m!
m!
t0
0
°
°
Z t0
Z t
°
° 1
1
m
m
m
n
n
°
((t − s) − (t0 − s) ) C (s) A f ds +
(t − s) C (s) A f ds°
=°
°
m! 0
m! t0
·Z t0
¸
Z t
1
m
m
m
≤
((t − s) − (t0 − s) ) kC (s) An f k ds +
(t − s) kC (s) An f k ds
m! 0
t0
Tn (t0 ) f =
339
Ostrowski inequalities for operator functions
≤
kkC (·) An f kk∞,[0,a]
·Z
m!
Z
t0
m
m
t
((t − s) − (t0 − s) ) ds +
0
m
(t − s)
¸
ds
t0
kkC (·) An f kk∞,[0,a] £ m+1
¤
=
t
− tm+1
.
0
(m + 1)!
That is
kEn (t) f k ≤
kkC (·) An f kk∞,[0,a] £ m+1
¤
t
− tm+1
,
0
(m + 1)!
for t ≥ t0 , t, t0 ∈ [0, a].
Case of t < t0 : we similarly find that
°Z t
°
Z t0
m
m
°
°
(t − s)
(t0 − s)
n
n
°
kEn (t) f k = °
C (s) A f ds −
C (s) A f ds°
°
m!
m!
0
0
n
kkC (·) A f kk∞,[0,a] £ m+1
¤
≤
t0
− tm+1 ,
(m + 1)!
for t < t0 , t, t0 ∈ [0, a]. Therefore we get
kEn (t) f k ≤
kkC (·) An f kk∞,[0,a] ¯ m+1
¯
¯t
¯,
− tm+1
0
(m + 1)!
for all t, t0 ∈ [0, a].
Next we observe
° Z a
°
°Z
°
°1
° 1° a
°
°
°
°
Tn (t) f dt − Tn (t0 ) f ° = °
(Tn (t) f − Tn (t0 ) f ) dt°
°a
°
a
0
0
Z a
Z a
1
1
kTn (t) f − Tn (t0 ) f k dt =
kEn (t) f k dt
≤
a 0
a 0
kkC (·) An f kk∞,[0,a] Z a ¯ m+1
¯
¯t
¯ dt
≤
− tm+1
0
(m + 1)!a
0
·µ
¶
¸
kkC (·) An f kk∞,[0,a]
2 (m + 1) tm+2
+ am+2
0
− atm+1
=
.
0
(m + 1)!a
(m + 2)
So that
° Z a
°
°1
°
°
°
T
(t)
f
dt
−
T
(t
)
f
n
n 0
°a
°
0
·µ
¶
¸
kkC (·) An f kk∞,[0,a]
2 (m + 1) tm+2
+ am+2
m+1
0
≤
− at0
(m + 1)!a
(m + 2)
·µ
¶
¸
kkC (·) An f kk∞,[0,a]
4nt2n+1
+ a2n+1
2n
0
=
− at0 ,
a (2n)!
2n + 1
proving (4).
Letting m := 2n, we similarly obtain
°
° Z a
°
°1
°
°
M
(t)
f
dt
−
M
(t
)
f
n
n 0
°
°a
0
340
G. A. Anastassiou
≤
kkC (·) An f kk∞,[0,a]
·µ
(m + 1)!a
=
kkC (·) An f kk∞,[0,a]
(2n + 1)!a
¶
¸
2 (m + 1) tm+2
+ am+2
0
− atm+1
0
(m + 2)
"Ã
!
#
2(n+1)
2 (2n + 1) t0
+ a2(n+1)
2n+1
− at0
,
2 (n + 1)
proving (5).
When n = 1 we get
Proposition 3. Let f ∈ D (A), t0 ∈ [0, a], a > 0. We have that
(i)
° Z a
°
·µ 3
¶
¸
°1
° kkC (·) Af kk∞,[0,a]
4t0 + a3
2
°
°
C
(t)
f
dt
−
C
(t
)
f
≤
−
at
0
0 ,
°a
°
2a
3
0
and
(ii)
° Z a
°
³
°1
a ´°
°
°
S (t) f dt − S (t0 ) f + f t0 −
°a
2 °
0
·µ 4
¶
¸
kkC (·) Af kk∞,[0,a]
6t0 + a4
3
·
≤
− at0 .
6a
4
Next we call
Tn∗ (t) f := C (t) (f ) −
n−1
X
k=1
t ∈ R. Notice that
Tn∗
t2k k
A f,
(2k)!
(0) f = f . Furthermore we have
Tn∗ (t) f − Tn∗ (0) f = Tn∗ (t) f − f = Tn (t) f,
for all t ∈ R. Now we prove
Theorem 4. Let p, q > 1 be such that
(i)
1
p
+
1
q
= 1. Then
1
° Z a
°
kkC (·) An f kkq,[0,a] a2n− q
°1
°
∗
°
°
³
´,
Tn (t) f dt − f ° ≤
1
°a
0
(2n − 1)! (p (2n − 1) + 1) p 2n + p1
(ii)
°Z
°
°
°
0
a
1
°
kkC (·) An f kkq,[0,a] a2n+ p +1
°
°
³
´.
Mn (t) f dt° ≤
1
(2n)! (2pn + 1) p 2n + p1 + 1
Proof. (i) Set m = 2n − 1, then
Z
Tn (t) f =
0
t
m
(t − s)
C (s) An f ds.
m!
341
Ostrowski inequalities for operator functions
By Hölder’s inequality we obtain
°Z t
°
m
°
°
(t − s)
n
°
kTn (t) f k = °
C (s) A f ds°
°
m!
0
Z t
1
m
(t − s) kC (s) An f k ds
≤
m! 0
µZ t
¶ p1 µZ t
¶ q1
1
pm
q
n
≤
(t − s)
ds
kC (s) A f k ds
m!
0
0
Ã
! µZ
¶ q1
1
a
1
tm+ p
q
n
≤
kC (s) A f k ds
m! (pm + 1) p1
0
!
Ã
1
1
tm+ p
=
kkC (s) An f kkq,[0,a] ,
m! (pm + 1) p1
for all t ∈ [0, a]. That is we have
Ã
!
1
1
tm+ p
kTn (t) f k ≤
kkC (s) An f kkq,[0,a] ,
m! (pm + 1) p1
for all t ∈ [0, a]. Therefore
° Z a
° ° Z a
°
°1
° °1
°
∗
∗
∗
∗
°
°
°
Tn (t) f dt − Tn (0) f ° = °
(Tn (t) f − Tn (0) f ) dt°
°a
°
a
0
0
°
° Z a
Z a
° 1
°1
Tn (t) f dt°
kTn (t) f k dt
=°
°≤ a
°a
0
0
Ã
!
1
kkC (s) An f kkq,[0,a] am+ p
kkC (s) An f kkq,[0,a] Z a m+ 1
p
³
´.
dt =
≤
t
1
1
0
am! (pm + 1) p
m! (pm + 1) p m + p1 + 1
We have proved that
1
° Z a
°
kkC (s) An f kkq,[0,a] am+ p
°1
°
∗
∗
°
°
³
´
Tn (t) f dt − Tn (0) f ° ≤
1
°a
0
m! (pm + 1) p m + p1 + 1
1
kkC (s) An f kkq,[0,a] a2n− q
³
´,
=
1
(2n − 1)! (p (2n − 1) + 1) p 2n + p1
establishing the claim.
(ii) Set m = 2n, then similarly we get
1
° Z a
°
kkC (s) An f kkq,[0,a] am+ p
°1
°
°
°
³
´
Mn (t) f dt° ≤
1
°a
0
m! (pm + 1) p m + p1 + 1
1
kkC (s) An f kkq,[0,a] a2n+ p
³
´,
=
1
(2n)! (2pn + 1) p 2n + p1 + 1
proving the claim.
342
G. A. Anastassiou
Theorem 5.
° Z a
°
°1
°
a2n−1
∗
n
°
(i) °
Tn (t) f dt − f °
≤
° (2n − 1)! kkC (·) A f kk1,[0,a] .
a 0
° Z a
°
°1
° a2n+1
°≤
(ii) °
M
(t)
f
dt
kkC (·) An f kk1,[0,a] .
n
°a
°
(2n)!
0
Proof. Set m = 2n − 1. We notice that
°Z t
°
m
°
°
(t − s)
n
°
kTn (t) f k = °
C (s) A f ds°
°
m!
0
Z t
Z
1
am t
m
≤
(t − s) kC (s) An f k ds ≤
kC (s) An f k ds
m! 0
m! 0
Z
am
am a
kC (s) An f k ds ≤
kkC (s) An f kk1,[0,a] ,
=
m! 0
m!
for all t ∈ [0, a], i.e.
kTn (t) f k ≤
am
kkC (s) An f kk1,[0,a] ,
m!
for all t ∈ [0, a]. Therefore
° Z a
°
Z
°1
° 1 a
am
°
°
kkC (s) An f kk1,[0,a]
T
(t)
f
dt
kTn (t) f k dt ≤
≤
n
°a
° a
m!
0
0
a2n−1
=
kkC (s) An f kk1,[0,a] ,
(2n − 1)!
proving the claim.
(ii) Set m = 2n. Then similarly we get
°
° Z a
°
°1
a2n
n
°
Mn (t) f dt°
° ≤ (2n)! kkC (s) A f kk1,[0,a] ,
°a
0
proving the claim.
In the case that n = p = q = 2, we get
Corollary 6 (to Theorem 4).
°
°°
° Z aµ
° °
¶
2
°1
° °°C (·) A2 f °°2,[0,a] a3.5
t
√
(i) °
C (t) f − Af dt − f °
.
°a
°≤
2
27 7
0
°
°°
°Z a µ
¶ ° °
3
°
° °°C (·) A2 f °°2,[0,a] a5.5
t
S (t) f − f t − Af dt°
(ii) °
.
°
°≤
6
396
0
Next, we prove Ostrowski type inequality on the general interval [a, b], 0 ≤
a < b.
Ostrowski inequalities for operator functions
343
Theorem 7. Let t0 ∈ [a, b].
(i)
(ii)
°
°
° 1 Z b
° kkC (·) An f kk
∞,[0,b]
°
°
T (t) f dt − Tn (t0 ) f ° ≤
×
°
°b − a a n
°
(2n)! (b − a)
·
µ 2n+1
¶¸
4nt2n+1
a
+ b2n+1
0
− t2n
(a
+
b)
+
.
0
2n + 1
2n + 1
°
°
° kkC (·) An f kk
° 1 Z b
∞,[0,b]
°
°
Mn (t) f dt − Mn (t0 ) f ° ≤
×
°
°
°b − a a
(2n + 1)! (b − a)
"Ã
!
#
2(n+1)
2 (2n + 1) t0
+ a2(n+1) + b2(n+1)
2n+1
− (a + b) t0
.
2 (n + 1)
Proof. i) Set m := 2n−1, and En (t) f := Tn (t) f −Tn (t0 ) f , where t, t0 ∈ [a, b],
with a ≥ 0. As in the proof of Theorem 2 we obtain
kEn (t) f k ≤
kkC (·) An f kk∞,[0,b] ¯ m+1
¯
¯t
¯,
− tm+1
0
(m + 1)!
for all t, t0 ∈ [a, b] .
Next we observe
°Z
°
°
°
° 1 Z b
°
°
b
1 °
°
°
°
°
Tn (t) f dt − Tn (t0 ) f ° =
(Tn (t) f − Tn (t0 ) f ) dt°
°
°
°b − a a
° b−a° a
°
Z b
Z b
1
1
kTn (t) f − Tn (t0 ) f k dt =
kEn (t) f k dt
≤
b−a a
b−a a
kkC (·) An f kk∞,[0,b] Z b ¯ m+1
¯
¯t
¯ dt
≤
− tm+1
0
(m + 1)! (b − a)
a
"
#
Z b
kkC (·) An f kk∞,[0,b] Z t0 ¡ m+1
¢
¡
¢
t0
− tm+1 dt +
tm+1 − tm+1
=
dt
0
(m + 1)! (b − a)
a
t0
·
¶¸
µ m+2
kkC (·) An f kk∞,[0,b] m+1
a
+ bm+2 − 2tm+2
0
=
t0
[2t0 − a − b] +
(m + 1)! (b − a)
m+2
·
µ 2n+1
¶¸
2n+1
kkC (·) An f kk∞,[0,b] 2n
a
+b
− 2t2n+1
0
=
t0 [2t0 − a − b] +
(2n)! (b − a)
2n + 1
·
µ 2n+1
¶¸
n
2n+1
kkC (·) A f kk∞,[0,b] 4nt0
a
+ b2n+1
2n
− t0 (a + b) +
,
=
(2n)! (b − a)
2n + 1
2n + 1
proving the claim.
344
G. A. Anastassiou
(ii) Set m = 2n. Then similarly we find
°
°
°
° 1 Z b
°
°
Mn (t) f dt − Mn (t0 ) f °
°
°b − a a
°
·
µ m+2
¶¸
n
kkC (·) A f kk∞,[0,b] m+1
a
+ bm+2 − 2tm+2
0
[2t0 − a − b] +
≤
t0
(m + 1)! (b − a)
m+2
"Ã
!
#
n
2(n+1)
kkC (·) A f kk∞,[0,b]
2 (2n + 1) t0
+ a2(n+1) + b2(n+1)
2n+1
=
− (a + b) t0
,
(2n + 1)! (b − a)
2 (n + 1)
proving the claim.
When n = 1 we get
Proposition 8. Let f ∈ D (A), 0 ≤ a < b, t0 ∈ [0, a]. Then
°
°
° 1 Z b
°
°
°
(i) °
C (t) f dt − C (t0 ) f °
°b − a a
°
µ 3
· 3
¶¸
kkC (·) Af kk∞,[0,b]
a + b3
4t0
2
≤
·
− t0 (a + b) +
,
2 (b − a)
3
3
°
µ
¶°
° 1 Z b
a+b °
°
°
(ii) °
S (t) f dt − S (t0 ) f + f t0 −
°
°b − a a
°
2
·µ 4
¶
¸
kkC (·) Af kk∞,[0,b]
6t0 + a4 + b4
·
− (a + b) t30 .
≤
6 (b − a)
4
4. Applications
Let X be the Banach space of odd, 2π-periodic real functions in the space of
d2
bounded uniformly continuous functions from R into itself: BU C (R). Let A := dx
2
with D (An ) = {f ∈ X : f (2k) ∈ X, k = 1, . . . , n}, n ∈ N. A generates a Cosine
function C ∗ given by (see [6], p. 121)
1
[f (x + t) + f (x − t)] , for all x, t ∈ R.
2
The corresponding Sine function S ∗ is given by
·Z t
¸
Z t
1
[S ∗ (t) f ] (x) =
f (x + s) ds +
f (x − s) ds , for all x, t ∈ R.
2 0
0
[C ∗ (t) f ] (x) =
Here we consider f ∈ D (An ), n ∈ N, as above. By (2) we obtain
n−1
Tn (t) f :=
X t2k
1
[f (· + t) + f (· − t)] −
f (2k)
2
(2k)!
k=0
Z t
2n−1 h
i
(t − s)
=
f (2n) (· + s) + f (2n) (· − s) ds, for all t ∈ R.
0 2 (2n − 1)!
Ostrowski inequalities for operator functions
345
By (3) we get
Mn (t) f :=
1
2
·Z
Z
t
t
f (· + s) ds +
0
0
Z
=
0
t
2n
(t − s)
2 (2n)!
¸ X
n
f (· − s) ds −
k=1
t2k−1
f (2(k−1))
(2k − 1)!
h
i
f (2n) (· + s) + f (2n) (· − s) ds, for all t ∈ R.
Let g ∈ BU C (R), we define kgk = kgk∞ := supx∈R |g (x)| < ∞.
Notice also that
kkC ∗ (s)An f k∞ k∞ = kkC ∗ (s)f (2n) k∞ k∞
1
= kk(f (2n) (· + s) + f (2n) (· − s))k∞ k∞
2
1
(2n)
(· + s)k∞ k∞ + kkf (2n) (· − s)k∞ k∞ ] ≤ kf (2n) k∞ < ∞.
≤ [kkf
2
We have the following applications
Corollary 9 (to Theorem 7). Let t0 ∈ [a, b]. Then
°
°
° 1 Z b
°
°
°
(i) °
Tn (t) f dt − Tn (t0 ) f °
°b − a a
°
∞
° (2n) °
·
µ 2n+1
¶¸
2n+1
°f
°
4nt
a
+ b2n+1
2n
0
∞
·
− t0 (a + b) +
,
≤
(2n)! (b − a)
2n + 1
2n + 1
°
°
° 1 Z b
°
°
°
(ii) °
Mn (t) f dt − Mn (t0 ) f °
°b − a a
°
∞
"Ã
!
#
° (2n) °
2(n+1)
2(n+1)
2(n+1)
°f
°
2
(2n
+
1)
t
+
a
+
b
0
∞
·
− (a + b) t2n+1
.
≤
0
(2n + 1)! (b − a)
2 (n + 1)
Corollary 10 (to Proposition 8). Let f ∈ X : f (2) ∈ X, 0 ≤ a < b,
t0 ∈ [a, b]. Then
°
°
° 1 Z b
°
°
°
(i) °
[f (· + t) + f (· − t)] dt − [f (· + t0 ) + f (· − t0 )]°
°b − a a
°
∞
° (2) ° ·
µ 3
¶¸
°f °
3
a + b3
2
∞ 4t0
≤
− t0 (a + b) +
,
(b − a)
3
3
(ii)
°
µ
¶°
° 1 Z b
°
a
+
b
°
°
S ∗ (t) f dt − S ∗ (t0 ) f + f (·) · t0 −
°
°
°b − a a
°
2
∞
° (2) ° ·µ
¶
¸
4
4
4
°f °
6t
+
a
+
b
0
∞
− (a + b) t30 .
≤
6 (b − a)
4
346
G. A. Anastassiou
We finish with
Corollary 11 (to Corollary 6).
° Z ah
°
i
°1
°
2 (2)
°
(i) °
f (· + t) + f (· − t) − t f (·) dt − f (·)°
°
2a 0
∞
°° (4)
° °
°°f (· + t) + f (4) (· − t)° °
a3.5
∞ 2,[0,a]
√
≤
,
54 7
(ii)
°Z
°
°
°
a
0
µ
¶ °
°
t3
[S ∗ (t) f ] (x) − f (x) t − f (2) (x) dt°
°
6
∞,x
°
°°£
°
¤
°
°° ∗
a5.5
° C (t) f (4) (x)°∞,x °
2,[0,a],t
≤
.
396
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3781.
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(received 10.06.2011; in revised form 28.09.2011; available online 01.11.2011)
Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, U.S.A.
E-mail: ganastss@memphis.edu