# Algebra Structure of Operator-Valued Riesz Means

Authors: Pedro J. Miana
Journal: Journal of Operators
 (2014)

---

## Abstract

We characterize operator-valued Riesz means via an algebraic law of composition and establish their functional calculus accordingly. With this aim, we give a new integral expression of the Leibniz derivation rule for smooth functions.


---

## Body

## 1. Introduction

Every self-adjoint operator on a Hilbert space admits a spectral decomposition (on Borel subsets of its spectrum) which allows to defineL
∞-functional calculus. Spectral operators of scalar type (introduced in [1]) on arbitrary Banach spaces also have this functional calculus. If spectral projections (associated to a given operator) are required to be uniformly bounded only to closed intervals, these operators are said to be well-bounded. Roughly speaking, in reflexive Banach spaces X, a well-bounded operator (
A
,
D
(
A
)
) with σ
(
A
)
⊂
[
0
,
∞
) admits a projection-valued function E
:
[
0
,
∞
)
→
B
(
X
) such that E
(
t
)
~
χ
(
0
,
t
)
(
A
) [1–3].Well-bounded operators on[
0
,
∞
) have a functional calculus for absolutely continuous functions [2], and Riesz means, G
t
n
∶
=
(
1
/
n
!
)
(
t
-
A
)
+
n (for t
>
0 and n
∈
N
∪
{
0
}), are properly defined by this functional calculus; in some sense, G
t
n is the n-times integral of the decomposition of the identity associated to A; see [4, page 332].The aim of this paper is to introduce the operator-valued Riesz means(
G
t
n
)
t
>
0
⊂
B
(
X
) (for n
∈
N) in an axiomatic way taking into account the algebraic law of the composition G
t
n
G
s
n, the uniform boundedness, and the summability property; see Definition 5. The starting point is a multiplication identity for scalar Riesz means (Proposition 1) and led us to a new expression to the Leibniz formula, Proposition 3. To conclude, we show that certain holomorphic C
0-semigroups, A
C
(
n
)-functional calculus, and operator-valued Riesz means are equivalent concepts, essentially up some regularity; see Section 4.There exist alternative approaches to operator-valued Riesz means(
G
t
ν
)
t
>
0, (for ν
>
0) mainly closer to approximation theory and Fourier multipliers; see, for example, [5, page 193], [6, Section 2], [7, Section 3], and the references therein. Note that our point of view may be applied in some of these settings.Notation. In this paper, C
+
∶
=
{
z
∈
C
∣
R
z
>
0
}; χ
E is the characteristic function in the set E; we write (
s
-
|
a
|
)
+
=
(
s
-
|
a
|
)
χ
(
0
,
a
). X is a Banach space, and B
(
X
) is the set of linear and bounded operators on X; (
A
,
D
(
A
)
) is a closed operator on X, and σ
(
A
) is the spectrum of A.


## 2. Riesz Functions

Forn
∈
N, we consider Riesz functions (
R
t
n
)
t
≥
0 given by
(1)
R
t
n
(
u
)
∶
=
(
t
-
|
u
|
)
+
n
n
!
,
u
∈
R
,
for t
>
0 and R
0
n
=
0. Note that R
t
0
=
χ
(
-
t
,
t
) for t
>
0 and
(2)
R
t
n
(
ξ
)
=
1
(
n
-
j
)
!
∫
0
t
(
t
-
u
)
n
-
j
R
u
j
-
1
(
ξ
)
d
u
,
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
ξ
∈
R
,
for 1
≤
j
≤
n.Proposition 1.
Letn
∈
N and t
≥
s
≥
0. Then,
(3)
R
t
n
R
s
n
=
∑
j
=
1
n
(
t
-
s
)
n
-
j
(
n
+
j
)
!
n
!
2
(
n
j
)
×
∫
0
s
(
s
-
u
)
j
-
1
(
j
-
1
)
!
R
u
n
d
u
‍
+
(
t
-
s
)
n
n
!
R
s
n
.Proof.
Takeξ
∈
R such that |
ξ
|
≤
s; note that
(4)
∑
j
=
1
n
(
t
-
s
)
n
-
j
(
n
+
j
)
!
n
!
2
(
n
j
)
‍
∫
0
s
(
s
-
u
)
j
-
1
(
j
-
1
)
!
R
u
n
(
ξ
)
d
u
‍
+
(
t
-
s
)
n
n
!
R
s
n
(
ξ
)
=
∑
j
=
1
n
(
t
-
s
)
n
-
j
(
n
+
j
)
!
n
!
2
(
n
j
)
‍
×
∫
|
ξ
|
s
(
s
-
u
)
j
-
1
(
j
-
1
)
!
(
u
-
|
ξ
|
)
n
n
!
d
u
‍
+
(
t
-
s
)
n
n
!
(
s
-
|
ξ
|
)
n
n
!
=
(
s
-
|
ξ
|
)
n
n
!
×
(
1
n
!
∑
j
=
1
n
(
n
j
)
(
t
-
s
)
n
-
j
(
s
-
|
ξ
|
)
j
+
(
t
-
s
)
n
n
!
‍
)
=
R
s
n
(
ξ
)
R
t
n
(
ξ
)
.
For |
ξ
|
≥
s, the equality holds trivially.We consider functions(
f
0
,
s
)
s
≥
0 defined by
(5)
f
0
,
s
(
r
)
∶
=
sin
(
s
r
)
π
r
,
r
∈
R
,
and now define functions (
f
n
,
s
)
s
≥
0 by integration:
(6)
f
n
,
s
(
r
)
∶
=
1
(
n
-
1
)
!
∫
0
s
‍
(
s
-
u
)
n
-
1
f
0
,
u
(
r
)
d
u
,
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
r
∈
R
,
for n
∈
N and s
≥
0. It is straightforward to prove that
(7)
f
2
m
,
s
(
r
)
=
1
π
r
2
m
+
1
(
(
-
1
)
m
sin
⁡
(
r
s
)
∑
j
=
0
m
-
1
(
-
1
)
j
(
r
s
)
2
(
m
-
j
)
-
1
(
2
(
m
-
j
)
-
1
)
!
‍
iiiiiiiiiiiiiiiiiiiiiii
+
∑
j
=
0
m
-
1
(
-
1
)
j
(
r
s
)
2
(
m
-
j
)
-
1
(
2
(
m
-
j
)
-
1
)
!
‍
)
,
f
2
m
-
1
,
s
(
r
)
=
1
π
r
2
m
(
(
-
1
)
m
cos
⁡
(
r
s
)
∑
j
=
0
m
-
1
(
-
1
)
j
(
r
s
)
2
(
m
-
j
)
-
1
(
2
(
m
-
j
)
-
1
)
!
‍
iiiiiiiiiiiiiiiiiiii
+
∑
j
=
0
m
-
1
(
-
1
)
j
(
r
s
)
2
(
m
-
1
-
j
)
(
2
(
m
-
1
-
j
)
)
!
‍
)
,
for r
∈
R, m
≥
1, and s
≥
0. Note that these functions, f
2
m
,
s and f
2
m
-
1
,
s, up some factors, are the remainders of the Taylor series of sin
(
·
s
) and cos
⁡
(
·
s
), respectively.Note thatf
n
,
s
∈
L
1
(
R
) for n
∈
N and s
≥
0. We remind the reader that the Fourier transform of a function f in L
1
(
R
) is defined by
(8)
F
f
(
t
)
∶
=
∫
-
∞
∞
e
-
i
x
t
f
(
x
)
d
x
‍
,
t
∈
R
.
It is well known that F
f is continuous on R and F
f
(
t
)
→
0 when |
t
|
→
∞ (the Riemann-Lebesgue lemma). In the case that f
∈
L
p
(
R
) for some 1
<
p
≤
2, the Fourier transform of f is defined in terms of a limit in the norm of L
p
′
(
R
) of truncated integrals:
(9)
F
f
∶
=
lim
⁡
R
→
∞
F
(
f
χ
(
-
R
,
R
)
)
,
F
(
f
χ
(
-
R
,
R
)
)
(
t
)
=
∫
-
R
R
e
-
i
x
t
f
(
x
)
d
x
,
‍
t
∈
R
;
that is, F
f
∈
L
p
′
(
R
) and lim
⁡
R
→
∞
∥
F
f
-
F
(
f
χ
(
-
R
,
R
)
)
∥
p
′
=
0, where 1
/
p
+
1
/
p
′
=
1 and χ
(
-
R
,
R
) is the characteristic function of the interval (
-
R
,
R
); see, for example, [8, Volume 2, page 254]. Then, the existence of F
f
(
t
) is guaranteed only at almost every t, and F
f may be noncontinuous and the Riemann-Lebesgue lemma could not hold (unlike the case when f
∈
L
1
(
R
)).Remark 2.
Forn
∈
N and s
≥
0, we have F
(
f
n
,
s
)
=
R
s
n and
(10)
f
n
,
t
*
f
n
,
s
=
∑
j
=
1
n
(
t
-
s
)
n
-
j
(
n
+
j
)
!
n
!
2
(
n
j
)
‍
×
∫
0
s
(
s
-
u
)
j
-
1
(
j
-
1
)
!
f
n
,
u
d
u
‍
+
(
t
-
s
)
n
n
!
f
n
,
s
,
for t
,
s
≥
0. To show this, note that F
(
f
0
,
s
)
=
χ
(
-
s
,
s
) a.e., and then we have that
(11)
F
(
f
n
,
s
)
(
ξ
)
=
1
(
n
-
1
)
!
∫
0
s
(
s
-
u
)
n
-
1
χ
(
-
u
,
u
)
(
ξ
)
d
u
‍
=
(
s
-
|
ξ
|
)
+
n
n
!
=
R
s
n
(
ξ
)
for ξ
∈
R, n
∈
N, and s
≥
0. We apply the Fourier transform and Proposition 1 to get equality (10).


## 3. Leibniz Formula via Integrals

In this section, we give a new expression of the Leibniz formula(
f
g
)
(
n
) for functions f
,
g
∈
S, where the set S is the Schwartz class of functions on [
0
,
∞
).Proposition 3.
Forn
≥
2 and f
,
g
∈
S, one has that
(12)
(
f
g
)
(
n
)
(
t
)
=
f
(
n
)
(
t
)
g
(
t
)
+
f
(
t
)
g
(
n
)
(
t
)
+
∑
j
=
1
n
-
1
(
n
+
j
-
1
j
)
‍
×
∫
t
∞
(
-
1
)
j
(
u
-
t
)
j
-
1
(
j
-
1
)
!
‍
iiiii
×
(
f
(
n
)
(
u
)
g
(
j
)
(
u
)
+
f
(
j
)
(
u
)
g
(
n
)
(
u
)
)
d
u
,
for t
≥
0.Proof.
We write byH
n
(
f
,
g
)
(
t
) the right part of equality (12). We prove the formula by the induction method. For n
=
2, we have that
(13)
f
(
2
)
(
t
)
g
(
t
)
+
f
(
t
)
g
(
2
)
(
t
)
-
2
∫
t
∞
(
f
(
2
)
(
u
)
g
(
1
)
(
u
)
+
f
(
1
)
(
u
)
g
(
2
)
(
u
)
)
d
u
‍
=
f
(
2
)
(
t
)
g
(
t
)
+
f
(
t
)
g
(
2
)
(
t
)
+
2
f
(
1
)
(
t
)
g
(
1
)
(
t
)
=
(
f
g
)
(
2
)
(
t
)
,
for t
≥
0. Now, we suppose that (
f
g
)
(
n
)
=
H
n
(
f
,
g
), and we will show that (
H
n
(
f
,
g
)
)
′
=
H
n
+
1
(
f
,
g
), where we conclude that (
f
g
)
(
n
+
1
)
=
H
n
+
1
(
f
,
g
). By the induction method, we conclude equality (12).
Fort
≥
0, we have that
(14)
(
H
n
(
f
,
g
)
)
(
1
)
(
t
)
=
f
(
n
+
1
)
(
t
)
g
(
t
)
+
f
(
t
)
g
(
n
+
1
)
(
t
)
+
(
n
+
1
)
(
f
(
n
)
g
(
1
)
+
f
(
1
)
g
(
n
)
)
(
t
)
+
∑
j
=
2
n
-
1
(
n
+
j
-
1
j
)
‍
×
∫
t
∞
(
-
1
)
j
+
1
(
u
-
t
)
j
-
2
(
j
-
2
)
!
iiiiiiiiiiiii
×
(
f
(
n
)
(
u
)
g
(
j
)
(
u
)
+
f
(
j
)
(
u
)
g
(
n
)
(
u
)
)
d
u
.
Note that, for 1
≤
j
≤
n, we obtain that
(15)
f
(
n
)
(
u
)
g
(
j
)
(
u
)
=
-
∫
u
∞
(
f
(
n
)
g
(
j
)
)
(
1
)
(
s
)
d
s
‍
=
-
∫
u
∞
(
f
(
n
+
1
)
g
(
j
)
+
f
(
n
)
g
(
j
+
1
)
)
(
s
)
d
s
‍
for u
≥
0, and then
(16)
(
n
+
j
-
1
j
)
∫
t
∞
(
-
1
)
j
+
1
(
u
-
t
)
j
-
2
(
j
-
2
)
!
iiiiiiiiiiiiiiiiiiiii
×
f
(
n
)
(
u
)
g
(
j
)
(
u
)
d
u
=
(
-
1
)
j
(
n
+
j
-
1
j
)
×
∫
t
∞
(
s
-
t
)
j
-
1
(
j
-
1
)
!
‍
iiiiiiiiiiii
×
(
f
(
n
+
1
)
g
(
j
)
+
f
(
n
)
g
(
j
+
1
)
)
(
s
)
d
s
,
for 2
≤
j
≤
n and t
≥
0. Using these equalities and the combinatorial identity
(17)
(
n
+
j
j
)
+
(
n
+
j
j
+
1
)
=
(
n
+
1
+
j
j
+
1
)
,
we obtain that
(18)
(
n
+
1
)
(
f
(
n
)
g
(
1
)
)
(
t
)
+
∑
j
=
2
n
-
1
(
n
+
j
-
1
j
)
‍
×
∫
t
∞
(
-
1
)
j
+
1
(
u
-
t
)
j
-
2
(
j
-
2
)
!
f
(
n
)
(
u
)
g
(
j
)
(
u
)
d
u
‍
=
∑
j
=
1
n
-
1
(
n
+
j
j
)
(
-
1
)
j
‍
×
∫
t
∞
(
s
-
t
)
j
-
1
(
j
-
1
)
!
f
(
n
+
1
)
(
s
)
g
(
j
)
(
s
)
d
s
‍
+
(
2
n
-
1
n
-
1
)
(
-
1
)
n
+
1
×
∫
t
∞
(
s
-
t
)
n
-
2
(
n
-
2
)
!
f
(
n
)
(
s
)
g
(
n
)
(
s
)
d
s
‍
.
Similarly, we get
(19)
(
n
+
1
)
(
g
(
n
)
f
(
1
)
)
(
t
)
+
∑
j
=
2
n
-
1
(
n
+
j
-
1
j
)
‍
×
∫
t
∞
(
-
1
)
j
+
1
(
u
-
t
)
j
-
2
(
j
-
2
)
!
g
(
n
)
(
u
)
f
(
j
)
(
u
)
d
u
‍
=
∑
j
=
1
n
-
1
(
n
+
1
+
j
-
1
j
)
(
-
1
)
j
‍
×
∫
t
∞
(
s
-
t
)
j
-
1
(
j
-
1
)
!
g
(
n
+
1
)
(
s
)
f
(
j
)
(
s
)
d
s
‍
+
(
2
n
-
1
n
-
1
)
(
-
1
)
n
+
1
×
∫
t
∞
(
s
-
t
)
n
-
2
(
n
-
2
)
!
f
(
n
)
(
s
)
g
(
n
)
(
s
)
d
s
.
Taking into account that
(20)
2
(
2
n
-
1
n
-
1
)
(
-
1
)
n
+
1
×
∫
t
∞
(
s
-
t
)
n
-
2
(
n
-
2
)
!
f
(
n
)
(
s
)
g
(
n
)
(
s
)
d
s
‍
=
(
2
n
n
)
(
-
1
)
n
×
∫
t
∞
(
s
-
t
)
n
-
1
(
n
-
1
)
!
‍
iiiiiiiiiiiii
×
(
f
(
n
+
1
)
g
(
n
)
+
g
(
n
+
1
)
f
(
n
)
)
(
s
)
d
s
,
we conclude the proof.Remark 4.
Note that the Leibniz formula holds for functions which belong to larger sets thanS.Givenn
∈
N, we consider the norm ∥
·
∥
(
n
) defined by
(21)
∥
f
∥
(
n
)
∶
=
1
(
n
-
1
)
!
∫
0
∞
t
n
-
1
|
f
(
n
)
(
t
)
|
d
t
,
‍
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
f
∈
S
,
and the Banach space A
C
(
n
) obtained as the completion of S in the norm ∥
·
∥
(
n
). A function f
∈
A
C
(
n
) if and only if there exists f has all derivatives up to order n
-
1 satisfying lim
⁡
x
→
∞
x
k
f
(
k
)
(
x
)
=
0, for 0
≤
k
≤
n
-
1, and the (
n
-
1
)th derivative is absolutely continuous with ∫
0
∞
t
n
-
1
|
f
(
n
)
(
t
)
|
d
t
‍
<
∞; see [4, Corollary 3.2]. In particular, (
R
t
n
)
t
≥
0
⊂
A
C
(
n
) and ∥
R
t
n
∥
(
n
)
=
t
n
/
n
!, for t
≥
0; see more details in [4, page 319]. Consider functions (
e
z
)
z
∈
C
+
⊂
A
C
(
n
) and ∥
e
z
∥
(
n
)
=
(
|
z
|
/
R
z
)
n, where e
z
(
t
)
∶
=
e
-
z
t for z
∈
C
+ and t
≥
0.In fact, the spaceA
C
(
n
) is a Banach algebra under pointwise multiplication for n
∈
N (see [4, Proposition 3.4]). Applying Proposition 3, a second proof may be given.


## 4. Functional Calculus for Operator-Valued Riesz Means

We introduce the operator-valued Riesz means in an axiomatic way and use them to define a functional calculus which domain is the Banach algebraA
C
(
n
). Then, we may associate a closed operator (
A
,
D
(
A
)
) to operator-valued Riesz means.Definition 5.
Forn
∈
N, one callsoperator-valued Riesz mean of degree
n any strongly continuous family (
G
t
n
)
t
≥
0
⊂
B
(
X
) such that it satisfies the following conditions:(i)
su
p
t
≥
0
(
n
!
/
t
n
)
∥
G
t
n
∥
<
∞;
(ii)
lim
⁡
t
→
∞
(
n
!
/
t
n
)
G
t
n
(
x
)
=
x for x
∈
X;
(iii)
the equality(22)
G
t
n
G
s
n
(
x
)
=
∑
j
=
1
n
(
t
-
s
)
n
-
j
(
n
+
j
)
!
n
!
2
(
n
j
)
‍
×
∫
0
s
(
s
-
u
)
j
-
1
(
j
-
1
)
!
G
u
n
(
x
)
d
u
+
(
t
-
s
)
n
n
!
G
s
n
(
x
)
holds for t
≥
s
≥
0 and x
∈
X.Forn
=
1, we get the Fejer family considered in [9, Section 4]. If (
G
t
n
)
t
≥
0
⊂
B
(
X
) is an operator-valued Riesz mean of degree n, then (
G
t
m
)
t
≥
0
⊂
B
(
X
) is also an operator-valued Riesz mean of degree m, where
(23)
G
t
m
(
x
)
=
1
(
m
-
n
-
1
)
!
×
∫
0
t
(
t
-
s
)
m
-
n
-
1
G
s
n
(
x
)
d
s
,
x
∈
X
,
for m
>
n. Note that the Riesz function (
R
t
n
)
t
≥
0 is an operator-valued Riesz mean of degree n in the space A
C
(
n
) (Proposition 1) and (
f
n
,
s
)
s
≥
0 is an operator-valued Riesz mean of degree n in the space L
1
(
R
) (Remark 2).Theorem 6.
Letn
∈
N, and let X be a Banach space.(i)
Given an operator-valued Riesz mean of degreen, (
G
t
n
)
t
≥
0
⊂
B
(
X
), the map Φ
:
A
C
(
n
+
1
)
→
B
(
X
) defined by
(24)
Φ
(
f
)
x
∶
=
(
-
1
)
n
+
1
∫
0
∞
f
(
n
+
1
)
(
t
)
G
t
n
(
x
)
d
t
,
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
x
∈
X
,
f
∈
A
C
(
n
+
1
)
,
is a bounded functional calculus such that Φ
(
R
t
n
+
1
)
x
=
∫
0
t
G
s
n
(
x
)
d
s
‍ and lim
⁡
λ
→
0
+
Φ
(
e
λ
)
x
=
x for x
∈
X.
(ii)
Conversely, given a bounded functional calculusΦ
:
A
C
(
n
)
→
B
(
X
) such that lim
⁡
λ
→
0
+
Φ
(
e
λ
)
x
=
x for any x
∈
X, there exists an operator-valued Riesz mean of degree n
(
G
t
n
)
t
≥
0
⊂
B
(
X
) such that Φ is given by formula (24).Proof.
(i) It is clear that expression (24) defines a bounded homomorphism and ∥
Φ
∥
≤
sup
⁡
t
≥
0
(
n
!
/
t
n
)
∥
G
t
n
∥. Now, we check that Φ
(
f
g
)
=
Φ
(
f
)
Φ
(
g
) for f
,
g
∈
A
C
(
n
+
1
). Due to condition (i) of Definition 5, we apply the Fubini theorem to obtain that
(25)
Φ
(
f
)
Φ
(
g
)
x
=
∫
0
∞
∫
0
t
f
(
n
+
1
)
(
t
)
g
(
n
+
1
)
(
s
)
G
t
n
G
s
n
(
x
)
d
s
d
t
‍
‍
+
∫
0
∞
∫
0
s
f
(
n
+
1
)
(
t
)
g
(
n
+
1
)
(
s
)
G
t
n
G
s
n
(
x
)
d
t
d
s
‍
‍
for x
∈
X. Now, we apply condition (iii) of Definition 5 and the equality
(26)
∫
0
∞
f
(
n
+
1
)
(
t
)
∫
0
t
(
t
-
s
)
n
-
j
g
(
n
+
1
)
(
s
)
‍
iiiiiiiiiiiiiiiiiiiiii
×
∫
0
s
(
s
-
u
)
j
-
1
(
j
-
1
)
!
G
u
n
(
x
)
d
u
d
s
d
t
‍
=
(
n
-
j
)
!
(
-
1
)
n
+
1
+
j
×
∫
0
∞
G
u
n
(
x
)
‍
iiiiiiiiiiii
×
∫
u
∞
(
s
-
u
)
j
-
1
(
j
-
1
)
!
f
(
j
)
(
s
)
g
(
n
+
1
)
(
s
)
d
s
d
u
‍
for 1
≤
j
≤
n to get
(27)
∫
0
∞
∫
0
t
f
(
n
+
1
)
(
t
)
g
(
n
+
1
)
(
s
)
G
t
n
G
s
n
(
x
)
d
s
d
t
‍
‍
=
∑
j
=
1
n
(
n
+
j
j
)
(
-
1
)
n
+
1
+
j
‍
×
∫
0
∞
G
u
n
(
x
)
∫
u
∞
(
s
-
u
)
j
-
1
(
j
-
1
)
!
f
(
j
)
(
s
)
g
(
n
+
1
)
(
s
)
d
s
d
u
‍
+
(
-
1
)
n
+
1
∫
0
∞
f
(
s
)
g
(
n
+
1
)
(
s
)
G
s
n
(
x
)
d
s
.
It follows that(28)
Φ
(
f
)
Φ
(
g
)
x
=
(
-
1
)
n
+
1
×
∫
0
∞
(
f
(
s
)
g
(
n
+
1
)
(
s
)
+
g
(
s
)
f
(
n
+
1
)
(
s
)
)
G
s
n
(
x
)
d
s
‍
+
∑
j
=
1
n
(
n
+
j
j
)
(
-
1
)
n
+
1
+
j
‍
×
∫
0
∞
G
u
n
(
x
)
∫
u
∞
(
s
-
u
)
j
-
1
(
j
-
1
)
!
f
(
j
)
(
s
)
g
(
n
+
1
)
(
s
)
d
s
d
u
‍
‍
+
∑
j
=
1
n
(
n
+
j
j
)
(
-
1
)
n
+
1
+
j
‍
×
∫
0
∞
G
u
n
(
x
)
∫
u
∞
(
s
-
u
)
j
-
1
(
j
-
1
)
!
g
(
j
)
(
s
)
f
(
n
+
1
)
(
s
)
d
s
d
u
‍
‍
=
(
-
1
)
n
+
1
∫
0
∞
(
f
g
)
(
n
+
1
)
(
s
)
G
s
n
(
x
)
d
s
‍
=
Φ
(
f
g
)
x
,
where we have applied formula (12) in the last equality.
We get directly thatΦ
(
R
t
n
+
1
)
x
=
∫
0
t
G
s
n
(
x
)
d
s
, for x
∈
X. Now, take ε
>
0; then, there exists t
0
>
0 such that ∥
(
n
!
/
t
n
)
G
t
n
(
x
)
-
x
∥
≤
ε for t
>
t
0. Note that
(29)
∥
Φ
(
e
λ
)
x
-
x
∥
≤
λ
n
+
1
n
!
∫
0
t
0
t
n
e
-
λ
t
∥
n
!
t
n
G
t
n
x
-
x
∥
d
t
‍
+
ε
λ
n
+
1
n
!
∫
t
0
∞
t
n
e
-
λ
t
d
t
‍
≤
(
sup
⁡
t
>
0
n
!
t
n
∥
G
t
n
∥
+
1
)
∥
x
∥
n
!
∫
0
t
0
λ
u
n
+
1
e
-
u
d
u
+
ε
‍
,
and we conclude that lim
⁡
λ
→
0
+
Φ
(
e
λ
)
x
=
x for x
∈
X.
Now, we check part (ii). We defineG
t
n
∶
=
Φ
(
R
t
n
). It is clear that (
G
t
n
)
t
>
0 is a strongly continuous family of operators and G
0
n
=
0. Note that
(30)
n
!
t
n
∥
G
t
n
∥
≤
n
!
t
n
∥
Φ
∥
∥
R
t
n
∥
(
n
)
=
∥
Φ
∥
,
t
>
0
,
and the first condition of Definition 5 holds. To check the third condition, note that
(31)
G
t
n
G
s
n
(
x
)
=
∑
j
=
1
n
(
t
-
s
)
n
-
j
(
n
+
j
)
!
n
!
2
(
n
j
)
×
∫
0
s
(
s
-
u
)
j
-
1
(
j
-
1
)
!
G
u
n
(
x
)
d
u
‍
+
(
t
-
s
)
n
n
!
G
s
n
(
x
)
,
(for t
≥
s
≥
0 and x
∈
X) where we have applied Proposition 1. The proof of part (ii) of Definition 5 is similar to the proof of part (i) above.Recall that a holomorphicC
0-semigroup (
T
(
z
)
)
z
∈
C
+
⊂
B
(
X
) is a family of operators such that the map C
+
∋
z
↦
T
(
z
)
x
∈
X is analytic, lim
⁡
z
>
0
,
z
→
0
T
(
z
)
x
=
x, and T
(
z
+
z
′
)
x
=
T
(
z
)
T
(
z
′
)
x for x
∈
X and z
,
z
′
∈
C
+.Now, we consider(
T
(
z
)
)
z
∈
C
+
⊂
B
(
X
) a holomorphic C
0-semigroup such that
(32)
∥
T
(
z
)
∥
≤
C
(
|
z
|
R
z
)
α
,
z
∈
C
+
,
for some α
≥
0. There exists a big amount of these semigroups; see, for example, [10], [4, Section 7], [3, Lemma 2.3], [11], and the references therein. Then, the family of operators (
G
t
n
)
t
≥
0
⊂
B
(
X
) defined by
(33)
G
t
n
(
x
)
∶
=
1
2
π
i
∫
R
z
=
1
e
t
z
z
n
+
1
T
(
z
)
x
d
z
,
x
∈
X
is an operator-valued Riesz mean of degree n for n
>
α. Conversely, given an operator-valued Riesz mean of degree n, (
G
t
n
)
t
≥
0
⊂
B
(
X
), then the family of operators (
T
(
z
)
)
z
∈
C
+
⊂
B
(
X
) defined by
(34)
T
(
z
)
x
∶
=
z
n
+
1
∫
0
∞
e
-
z
t
G
t
n
(
x
)
d
t
,
x
∈
X
,
z
∈
C
+
,
is a holomorphic C
0-semigroup such that
(35)
∥
T
(
z
)
∥
≤
sup
⁡
t
>
0
n
!
t
n
∥
G
t
n
∥
(
|
z
|
R
z
)
n
+
1
,
z
∈
C
+
.
In this case, σ
(
A
)
⊂
[
0
,
∞
) and
(36)
(
λ
+
A
)
-
1
x
=
(
n
+
1
)
!
∫
0
∞
G
t
n
(
x
)
(
λ
+
t
)
n
+
2
d
t
,
λ
>
0
,
x
∈
X
,
where the operator (
-
A
,
D
(
A
)
) is the infinitesimal generator of a holomorphic C
0-semigroup (
T
(
z
)
)
z
∈
C
+
⊂
B
(
X
). These results are essentially given in [12, Theorem 3.1] for n
=
1,2 and extended [4, Theorem 4.1] under more general conditions than those considered here.Recall the family of functions(
f
n
,
s
)
n
∈
N
,
s
≥
0 considered in Section 2. The following result appears in [9, Corollary 4.5] for n
=
1. The proof for n
>
1 may be done directly using formula (10) and is also a consequence of the case n
=
1. Similar result for uniformly bounded cosine functions holds.Corollary 7.
Let(
T
(
t
)
)
t
∈
R be a uniformly bounded C
0-group generated by i
A on a Banach space X. Then, the family of operators (
G
s
n
)
s
>
0 defined by
(37)
G
s
n
(
x
)
∶
=
∫
-
∞
∞
f
n
,
s
(
t
)
T
(
t
)
x
d
s
,
x
∈
X
,
is an operator-valued Riesz mean of degree n associated to (
A
2
)
1
/
2.



---

*Source: 923616-2014-05-18.xml*