Algebra Structure of Operator-Valued Riesz Means

Every self-adjoint operator on a Hilbert space admits a spectral decomposition (on Borel subsets of its spectrum) which allows to define L-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, Gn t := (1/n!)(t − A) n


Introduction
Every self-adjoint operator on a Hilbert space admits a spectral decomposition (on Borel subsets of its spectrum) which allows to define  ∞ -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 , a well-bounded operator (, ()) with () ⊂ [0, ∞) admits a projection-valued function  : [0, ∞) → B() such that () ∼  (0,) () [1][2][3].
The aim of this paper is to introduce the operator-valued Riesz means (   ) >0 ⊂ B() (for  ∈ N) in an axiomatic way taking into account the algebraic law of the composition       , 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  0 -semigroups,  () -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 ]
) >0 , (for ] > 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 + := { ∈ C | R > 0};   is the characteristic function in the set ; we write ( − ||) + = ( − ||) (0,) . is a Banach space, and B() is the set of linear and bounded operators on ; (, ()) is a closed operator on , and () is the spectrum of .
We consider functions ( 0, ) ≥0 defined by and now define functions ( , ) ≥0 by integration: for  ∈ N and  ≥ 0. It is straightforward to prove that for  ∈ R,  ≥ 1, and  ≥ 0. Note that these functions,  2, and  2−1, , up some factors, are the remainders of the Taylor series of sin(⋅) and cos(⋅), respectively.Note that  , ∈  1 (R) for  ∈ N and  ≥ 0. We remind the reader that the Fourier transform of a function  in  1 (R) is defined by It is well known that F is continuous on R and F() → 0 when || → ∞ (the Riemann-Lebesgue lemma).In the case that  ∈   (R) for some 1 <  ≤ 2, the Fourier transform of  is defined in terms of a limit in the norm of    (R) of truncated integrals: Remark 2. For  ∈ N and  ≥ 0, we have F( , ) =    and for ,  ≥ 0. To show this, note that F( 0, ) =  (−,) a.e., and then we have that for  ∈ R,  ∈ N, and  ≥ 0. We apply the Fourier transform and Proposition 1 to get equality (10).

Leibniz Formula via Integrals
In this section, we give a new expression of the Leibniz formula () () for functions ,  ∈ S, where the set S is the Schwartz class of functions on [0, ∞).
Remark 4. Note that the Leibniz formula holds for functions which belong to larger sets than S.
In fact, the space  () is a Banach algebra under pointwise multiplication for  ∈ N (see [4,Proposition 3.4]).Applying Proposition 3, a second proof may be given.

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 algebra  () .Then, we may associate a closed operator (, ()) to operator-valued Riesz means.
Definition 5.For  ∈ N, one calls operator-valued Riesz mean of degree  any strongly continuous family (   ) ≥0 ⊂ B() such that it satisfies the following conditions: holds for  ≥  ≥ 0 and  ∈ .
For  = 1, we get the Fejer family considered in [9, Section 4].If (   ) ≥0 ⊂ B() is an operator-valued Riesz mean of degree , then (   ) ≥0 ⊂ B() is also an operator-valued Riesz mean of degree , where for  > .Note that the Riesz function (   ) ≥0 is an operator-valued Riesz mean of degree  in the space  () (Proposition 1) and ( , ) ≥0 is an operator-valued Riesz mean of degree  in the space  1 (R) (Remark 2).Theorem 6.Let  ∈ N, and let  be a Banach space.
Proof.(i) It is clear that expression (24) defines a bounded homomorphism and ‖Φ‖ ≤ sup ≥0 (!/  )‖   ‖.Now, we check that Φ() = Φ()Φ() for ,  ∈  (+1) .Due to condition (i) of Definition 5, we apply the Fubini theorem to obtain that for  ∈ .Now, we apply condition (iii) of Definition 5 and the equality It follows that where we have applied formula (12) in the last equality.We get directly that Φ( and the first condition of Definition 5 holds.To check the third condition, note that (for  ≥  ≥ 0 and  ∈ ) where we have applied Proposition 1.The proof of part (ii) of Definition 5 is similar to the proof of part (i) above.

where 1 /
+ 1/  = 1 and  (−,) is the characteristic function of the interval (−, ); see, for example, [8, Volume 2, page 254].Then, the existence of F() is guaranteed only at almost every , and F may be noncontinuous and the Riemann-Lebesgue lemma could not hold (unlike the case when  ∈  1 (R)).