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Polyanalytic functions are important not only from the theoretical point of view, but also in the theory of signals since they allow to encode n independent analytic functions into a single polyanalytic one using a special decomposition. This idea is similar to the problem of multiplexing signals. This is related to the construction of the polyanalytic Segal-Bargmann transform mapping L2(R) onto the poly-Fock space, see [5]. In quantum mechanics polyanalytic functions are relevant for the study of the Landau levels associated to Schr¨odinger operators, see [5, 13]. They were used also in [4] to study sampling and interpolation problems on Fock spaces using time frequency analysis techniques such as short-time Fourier transform (STFT) or Gabor transforms. This allows to extend Bargmann theory to the polyanalytic setting using Gabor analysis. The theory of slice hyperholomorphic functions started its full development from the beginning of this century [27, 38]. It has had a quite fast developments due to several authors and the main results, regarding the quaternionic setting, are contained in the books [10, 24, 36, 37], while for the Clifford algebra setting we refer the interested reader to the book [30] and the references therein. Nowadays the function theory has expanded in several directions but it is in operators theory where it has found its most profound applications and several monographs have been published in the last decade [9, 10, 22, 23, 30]. The slice monogenic functions were introduced with D. C. Struppa in [25, 26, 27, 28, 29] and in this paper we generalize this class of functions to the poly analytic setting. In order to state our results we need to explain the context in which we work and to highlight the importance of this branch of operator theory which is called quaternionic and Clifford operator theory. First of all we point out that the appropriate definition of the quaternionic spectrum for a quaternionic linear operator has been open problem at least since the paper [19] of G. Birkhoff, J. von Neumann, in 1936 on the logic of quantum mechanics, where the authors showed that quantum mechanics can be formulated also on quaternions. Moreover, consider a generalization of the gradient operators such as
T = a(x1,x2,x3)∂x1 i + b(x1,x2,x3)∂x2 j + c(x1,x2,x3)∂x3 k 3 where a, b and c are given real valued functions of the variables (x1,x2,x3) R , and i,j,k are the imaginary units of the quaternions. It is very interesting to observe that the∈ spectral theory for vector operators like the gradient operator or its generalizations such as the operator T defined above, has been unclear since long time, even∇ before 1936. Regarding the quaternionic spectral theorem we observe that some attempts have been done after the paper of G. Birkhoff, J. von Neumann, but all the approaches suffered of the lack of an appropriate notion of quaternionic spectrum. The turning point came in 2006 when it was introduced the S-spectrum and the S-functional calculus which are a cornerstone of quaternionic and Clifford operator theory. The S-spectrum was identified by purely hypercomplex analysis techniques and not on physical considerations as it is widely explained in the introduction of [23]. The spectral theorem for quaternionic normal operators based on the S-spectrum was finally proved 2015 and published in 2016, see [7], with D. P. Kimsey. This theorem is the most important tool for the formulation of quaternionic quantum mechanics and more recently there has been several efforts to study the perturbations of quaternionic normal operators in [20], moreover, the theory of quaternionic spectral operators has been developed in [33]. The theory of characteristic operator function has started its development in this setting not too long ago and the main advances can be found in the book [9]. 2 There are several applications of the spectral theory on the S-spectrum to fractional diffusion and fractional evolution problems because it is possible to define the fractional powers of vector operators so that we can generate fractional Fourier laws, see [22]. With this strategy we are able to write the fractional heat equation modifying just the Fourier’s law and preserving the conservation of energy law. Among the most developed areas in the slice hyperholomorphic setting there is the theory of slice hyperholomorphic reproducing kernel Hilbert spaces and more in general quaternionic Schur analysis has been largely investigated in the last decade. The material is spread over several papers, but the interested reader can find several results in the book [10] and in the references therein. The importance of hypercomplex analysis in operators theory is not just limited to the slice hypercomplex setting. In fact, using the classical theory of monogenic functions, see [34, 35, 40], A. Mc Intosh and his collaborators [42, 44] developed the monogenic functional calculus, which is based on the notion of monogenic spectrum. This monogenic calculus contains as particular cases the Weyl functional calculus and the Taylor functional calculus for commuting operators, see the book [45]. In harmonic analysis, the monogenic functional calculus plays a crucial role as shown in the book [48] and the references therein. Moreover, the monogenic functions theory has applications in boundary value problems [41], and in Clifford wavelets, singular integrals, and Hardy spaces as one can see in the book [46]. To complete this short introduction the hypercomplex spectral theories induced by hypercomplex function theories we recall that there is a link between the slice hyperholomorphic functions and the monogenic functions via the Fueter-Sce-Qian extension theorem and there is a link between the two spectral theories. This link is the F - functional calculus that generates a version of the monogenic functional calculus using the notion of S-spectrum, see [32]. We recall that also the Radon transform is a bridge between the monogenic and generalized slice monogenic functions, see [21]. It is important to point out that the results of this paper have to be seen in the framework of hyperholomorphic function theories and the associated spectral theories explained above. In fact, here we extend the slice monogenic function theory and its S-functional calculus to the poly slice monogenic setting. The quaternionic counterpart of the function theory started with the recent works [11, 12], while the functional calculus in this setting is introduced in this paper for the first time. The poly S-functional calculus, that in the following will be called the PS-functional calculus, is the polyanalytic version of the S-functional calculus. This calculus is based on the S-spectrum and it applies to (n + 1)-tuples of noncommuting operators (T0, T1, ..., Tn) written as the paravector operator T = T0 + T1e1 + ... + Tnen, where e1, ..., en are the units of the Clifford Rn. The quaternionic case is just a particular case. The contents of the paper are organized as follows. In Section 2 we recall some preliminary results on the theory of slice monogenic functions. In Section 3 we develop the theory of poly slice monogenic functions and we show some properties. In particular the slice monogenic integral representation of poly slice monogenic functions will be used for a representation of the PS-functional calculus. In Section 4 we prove the Cauchy formulas and product of poly slice monogenic functions. In Section 5 we give the formulations of the PS-functional calculus via the PS-resolvent operators and the poly slice monogenic Cauchy formula. In Section 6 we define and study the formulations of the PS-functional calculus via the modified S-resolvent operators. In Section 7 we show the equivalence of the two definitions of the PS-functional calculus and we prove the product rules of the PS-functional calculus. 3 2. Preliminary material In this section we recall some preliminary material useful to extend the theory of polyanalytic functions to the slice monogenic setting. The classical polyanalytic functions are those functions f :Ω C C of class M (Ω), for M N, such that ⊆ → C ∈ M ∂i f(z) = 0, for all z = u + iv Ω, i = √ 1 ∈ − where M 1 M ∂i = (∂u + i∂v) . (1) 2M A Cauchy-type formula for polyanalytic functions appeared for the first time in Th´eodoresco’s doctoral thesis [49] and recalled in the paper [15]: Theorem 2.1 ([15], Theorem 1.3). If a function f is polyanalytic of order M in a closed domain G bounded by a rectifiable closed contour Γ, then the value of f at any point z of the domain G is expressed, using values of the function itself and its formal derivatives at points t of the boundary Γ, by the formula M−1 ℓ 1 1 ¯ ℓ ∂ f f(z)= (¯z t) ℓ dt. (2) 2πi Γ ℓ!(t z) − ∂t¯ Xℓ=0 Z − The formula contains a finite sum in which appear the M kernels
1 ℓ πℓ(z,t)= (¯z t¯) , ℓ = 0, ..., M 1. ℓ!(t z) − − − An other the polyanalytic Cauchy formula is given by M−1 ℓ ℓ f(z)= ( 2) Lℓ(w z) dw ∂i f(w), ∂Ω − − Z Xℓ=0 where ∂Ω is the boundary of the smooth bounded domain Ω in C, the infinitesimal arc length is denoted by dw, and have set 1 (Re(z))ℓ Lℓ(z) := , ℓ = 0, ..., M 1. 2πi z ℓ! − M M We observe that when we define ∂i in (1) without the prefactor 1/2 then in the Cauchy formula the coefficients ( 2)ℓ have to be replaced by ( 1)ℓ. − − In the quaternionic setting or, more in general, in the Clifford algebra setting, one can extend the notion of holomorphic functions by considering functions in the kernel of a generalized Cauchy- Riemann operator thus obtaining the so-called regular or monogenic function or of its n-power thus obtaining poly-regular functions or poly-monogenic functions, see [17, 18]. Now, we recall the main definitions on Clifford algebras and the main facts on slice monogenic functions that are necessary to introduce and develop the theory of poly slice monogenic functions. Let Rn be the real Clifford algebra over n imaginary units e1,...,en satisfying the relations eℓem + 2 emeℓ = 0, ℓ = m, e = 1. An element in the Clifford algebra will be denoted by eAxA where 6 ℓ − A A = ℓ1 ...ℓr 1, 2,...,n , ℓ1 <...<ℓr is a multi-index and eA = eℓ1 eℓ2 ...eℓr , e∅ = 1. An { }∈P{ Rn+1} P n R element (x0,x1,...,xn) will be identified with the element x = x0+x = x0+ ℓ=1 xℓeℓ n ∈ n∈+1 called paravector and the real part x0 of x will also be denoted by Re(x). The norm of x R is 2 2 2 2 P ∈n defined as x = x0 + x1 + . . . + xn. The conjugate of x is defined byx ¯ = x0 x = x0 ℓ=1 xℓeℓ. We denote| by| S the sphere − − P S 2 2 = x = e1x1 + . . . + enxn x1 + . . . + xn = 1 ; { 4 | } 2 n+1 for j S we obviously have j = 1. Given an element x = x0 + x R let us set jx = x/ x if x = 0∈, and given an element x R−n+1, the set ∈ | | 6 ∈ [x] := y Rn+1 : y = x +j x , j S { ∈ 0 | | ∈ } is an (n 1)-dimensional sphere in Rn+1. The vector space R +jR passing through 1 and j S will be denoted− by C and an element belonging to C will be indicated by u +jv, for u, v ∈R. j j ∈ With an abuse of notation we will write x Rn+1. Thus, if U Rn+1 is an open set, a function n+1 ∈ ⊆ f : U R Rn can be interpreted as a function of the paravector x. ⊆ → In this paper we use the definition of poly slice monogenic functions that is the generalization of slice monogenic functions in the spirit of the Fueter-Sce-Qian mapping theorem, see [32]. This definition is the most appropriate for operator theory and the reason is widely explained in several papers and in the books [22, 23]. The corresponding function theory in real alternative algebras is developed in [39]. The same definition will be used for poly slice monogenic functions in the next section. Definition 2.2. Let U Rn+1. We say that U is axially symmetric if [x] U for every x U. ⊆ ∈ ∈ Definition 2.3 (Slice monogenic functions). Let U Rn+1 be an axially symmetric open set and 2 ⊆ let = (u, v) R : u + Sv U . A function f : U Rn is called a left slice function, if it is of theU form{ ∈ ⊂ } → f(x)= f (u, v)+jf (u, v) for x = u +jv U 0 1 ∈ with the two functions f ,f : Rn that satisfy the compatibility conditions 0 1 U → f (u, v)= f (u, v), f (u, v)= f (u, v). (3) 0 − 0 1 − − 1 If in addition f and f are 1 and satisfy the Cauchy-Riemann equations 0 1 C ∂uf0(u, v) ∂vf1(u, v)) = 0 − (4) ∂vf0(u, v)+ ∂uf1(u, v)) = 0 then f is called left slice monogenic. A function f : U Rn is called a right slice function if it is of the form → f(x)= f (u, v)+ f (u, v)j for x = u +jv U 0 1 ∈ 1 with the two functions f0,f1 : Rn that satisfy (3). If f0 and f1 are and satisfy the Cauchy-Riemann equations (4) thenU →f is called right slice monogenic. C If f is a left (or right) slice function such that f0 and f1 are real-valued, then f is called intrinsic. We denote the sets of left, right and intrinsic slice monogenic functions on U by L(U), SM R(U) and (U), respectively. SM N Remark 2.4. The set (U) is contained in both L(U) and R(U). N SM SM Definition 2.5. We define the notion of j-derivative by means of the operator: 1 ∂ := (∂u j∂v) . j 2 − For consistency, we will denote by 1 ∂ = (∂u +j∂v) j 2 the Cauchy- Riemann operator associated with the complex plane C , for j S. j ∈ Using the notations we have just introduced, the condition of left slice monogenicity will be expressed, in short, by ∂jf = 0. Right slice monogenicity will be expressed, with an abuse of notation, by f∂j = 0. 5 n+1 Definition 2.6. Let U be an open set in R and let f : U Rn be a slice monogenic function. → Its slice derivative ∂S is defined as
∂j(f)(x) x = u +jv, v = 0 ∂S(f)= 6 (5) ∂uf(u) u R. ∈ Lemma 2.7 (Splitting Lemma). Let U Rn+1 be an axially symmetric open set and let f ⊆ ∈ L(U). For every j=j1 S let j2,..., jn be a completion to a basis of Rn satisfying the defining SM ∈ n−1 relations jrjs +jsjr = 2δrs. Then there exist 2 holomorphic functions A : U Cj Cj such that for every z = u +j−v we have F ∩ → n−1 f (z)= A(z)jA, jA =ji . . . ji , j F 1 s |AX|=0 where A = i ...is is a subset of 2,...,n , with i <... −1 1 1 1 1 S (s,x)= 1 jxj + 1+jxj L 2 − s z 2 s z h i − h i − where we set x = u +jxv, z = u +jv, s = s0 +js1, and where j is the imaginary unit of the complex plane Cj. For slice monogenic functions have two equivalent ways to write the Cauchy kernels. Proposition 2.10. If x,s Rn+1 with x [s], then ∈ 6∈ (x2 2xRe(s)+ s 2)−1(x s) = (s q¯)(s2 2Re(x)s + x 2)−1 (8) − − | | − − − | | and (s2 2Re(x)s + x 2)−1(s x¯)= (x s¯)(x2 2Re(s)x + s 2)−1. (9) − | | − − − − | | So we can give the following definition to distinguish the two representations of the Cauchy kernels. Definition 2.11. Let x,s Rn+1 with x [s]. ∈ 6∈ We say that S−1(s,x) is written in the form I if • L −1 2 2 −1 SL (s,x) := (x 2Re(s)x + s ) (x s). − − 6 | | − We say that S−1(s,x) is written in the form II if • L S−1(s,x) := (s x¯)(s2 2Re(x)s + x 2)−1. L − − | | We say that S−1(s,x) is written in the form I if • R S−1(s,x) := (x s¯)(x2 2Re(s)x + s 2)−1. R − − − | | We say that S−1(s,x) is written in the form II if • R S−1(s,x) := (s2 2Re(x)s + x 2)−1(s x¯). R − | | − Rn+1 −1 Lemma 2.12. Let x,s with s / [x]. The left slice monogenic Cauchy kernel SL (s,x) is left slice monogenic in x∈and right slice∈ monogenic in s. The right slice monogenic Cauchy kernel −1 SR (s,x) is left slice monogenic is s and right slice monogenic in x. Definition 2.13 (Slice Cauchy domain). An axially symmetric open set U Rn+1 is called a slice ⊂ Cauchy domain, if U Cj is a Cauchy domain in Cj for any j S. More precisely, U is a slice Cauchy domain if, for∩ any j S, the boundary ∂(U C ) of U ∈C is the union a finite number of ∈ ∩ j ∩ j non-intersecting piecewise continuously differentiable Jordan curves in Cj. Theorem 2.14 (The Cauchy formulas). Let U Rn+1 be a slice Cauchy domain, let j S and ⊂ ∈ set ds = ds( j). If f is a (left) slice monogenic function on a set that contains U then j − 1 −1 f(x)= SL (s,x) dsj f(s), for any x U. (10) 2π C ∈ Z∂(U∩ j) If f is a right slice monogenic function on a set that contains U, then 1 −1 f(x)= f(s) dsj SR (s,x), for any x U. (11) 2π C ∈ Z∂(U∩ j) These integrals depend neither on U nor on the imaginary unit j S. ∈ Theorem 2.15 (Cauchy formulas on unbounded slice Cauchy domains). Let U Rn+1 be an S ⊂ unbounded slice Cauchy domain and let j . If f L(U) and f( ) := lim|x|→∞ f(x) exists and is finite, then ∈ ∈ SM ∞ 1 −1 f(x)= f( )+ SL (s,x) dsj f(s) for any x U. ∞ 2π C ∈ Z∂(U∩ j) If f R(U) and f( ) := lim f(x) exists and is finite, then ∈ SM ∞ |x|→∞ 1 −1 f(x)= f( )+ f(s) dsj SR (s,x) for any x U. ∞ 2π C ∈ Z∂(U∩ j ) 3. Poly slice monogenic functions We can now give the definition of poly slice monogenic functions and we develop the function theory. This function theory will be used to define the PS-functional calculus. This definition is the monogenic analog of Definition 3.17 in [11]. Definition 3.1 (Poly slice monogenic functions). Let M N and denote by M (U) the set of continuously differentiable functions with all their derivatives∈ up to order M on anC axially symmetric n+1 2 open set U R . We let = (u, v) R : u + Sv U . A function F : U Rn is called a left slice function,⊆ if it is of theU form{ ∈ ⊂ } → F (x)= F0(u, v)+jF1(u, v) for x = u +jv U 7 ∈ with the two functions F , F : Rn that satisfy the compatibility condition 0 1 U → F (u, v)= F (u, v), F (u, v)= F (u, v). (12) 0 − 0 1 − − 1 M If in addition F0 and F1 are in (U) and satisfy the poly Cauchy-Riemann equations of order M N C ∈ 1 M (∂u +j∂v) (F (u, v)+jF (u, v)) = 0, for all j S (13) 2M 0 1 ∈ then F is called left poly slice monogenic function of order M N. A function F : U Rn is called a right slice function if it is of the form ∈ → F (x)= F (u, v)+ F (u, v)j for x = u +jv U 0 1 ∈ n+1 M with two functions F0, F1 : R that satisfy (12). If in addition F0 and F1 are in (U) and satisfy the poly Cauchy-RiemannU → equations of order M N C ∈ 1 M (F (u, v)+ F (u, v)j) (∂u +j∂v) = 0, for all j S (14) 0 1 2M ∈ then F is called right poly slice monogenic of order M N. We will denote by M (U) and ∈ PSL M (U) the set of left and right poly slice monogenic functions on the open set U, respectively. PSR If F is a left (or right) slice function such that F0 and F1 are real-valued, then F is called intrinsic. By M (U) we denote the set of poly intrinsic slice monogenic functions. PN Remark 3.2. For the definition of slice monogenic functions we required that the pair (f0,f1) satisfies the Cauchy-Riemann system. In the case of poly slice monogenic functions we follow the same idea by observing that (for functions F of class M (U)) C M M 1 M M M−k k k M k (∂u +j∂v) F = (∂u) (∂v) j F = D j F 2M k k M−k,k Xk=0 Xk=0 M−k k where DM−k,k = (∂u) (∂v) . Due to the arbitrarity of j, the condition that F is left poly slice monogenic translates into a system of two differential equations of order M for the pair (F0, F1) of Rn-valued functions (which reduces to the Cauchy-Riemann system when M = 1): M M M M DM k,k DM k,k F (u, v) k − − k − 0 k=0(mod 4) k=2(mod 4) X X (15) M M M M + DM k,k + DM k,k F (u, v) = 0, − k − k − 1 k=1(Xmod 4) k=3(Xmod 4) M M M M DM k,k DM k,k F (u, v) k − − k − 0 k=1(Xmod 4) k=3(Xmod 4) M M M M + DM k,k DM k,k F (u, v) = 0. k − − k − 1 k=0(Xmod 4) k=2(Xmod 4) However, since this system is rather complicated to write, we prefer to use the above Definition 3.1. We point out that the definition of poly slice monogenic functions extends to functions with values in a Clifford Banach module in a very natural way. As in the quaternionic case, one can give the definition of strong and weak slice monogenicity, see [10]. 8 Definition 3.3 (Poly slice monogenic functions vector-valued). Let U Rn+1 be an axially sym- metric open set and let ⊆ = (u, v) R2 : u + Sv U . U { ∈ ⊂ } A function f : U XL with values in a left Clifford Banach module XL is called a left slice function, if is of the→ form F (x)= F (u, v)+jF (u, v) for x = u + jv U 0 1 ∈ with two functions F0, F1 : XL that satisfy the compatibility condition (12). If in addition M U → F0 and F1 are in (U) and satisfy the poly Cauchy-Riemann-equations (13), then F is called strongly left poly sliceC monogenic. A function f : U XR with values in a right Clifford-Banach module is called a right slice function if it is of the→ form F (x)= F (u, v)+ F (u, v)j for x = u + jv U 0 1 ∈ with two functions F0, F1 : XR that satisfy the compatibility condition (12). If in addition F0 M U → and F1 are in (U) and satisfy the poly Cauchy-Riemann-equations (14), then f is called strongly right poly sliceC monogenic. n+1 Definition 3.4. Let U R be an axially symmetric open set. A function F : U XL with ⊂ → values in a left Clifford-Banach module XL is called weakly left poly slice monogenic if ΛF is left ′ ′ poly slice monogenic for any Λ X , where X is the dual module. A function F : U XR with ∈ L L → values in a right Clifford-Banach module XR is called weakly right poly slice monogenic if ΛF is right poly slice monogenic for any Λ X′ . ∈ R We have the following proposition. N M M M Proposition 3.5. Let M and let F L (U) (resp. R (U) or (U)). Then F ′ ∈′ ∈ PS′ PS PN ∈ M+M (U) (resp. M+M (U) or M+M (U)) for all M ′ N. PSL PSR PN ∈ Proof. It is a direct consequences of the definition. Remark 3.6. The restriction of function F to the complex plane Cj is denoted by Fj. Theorem 3.7 (Poly splitting lemma and poly decomposition). Let U Rn+1 be an axially sym- metric open set. ⊆ M (Ia) (Poly splitting lemma). Let F (U). Then, for every j=j S let j ,..., jn be a ∈ PSL 1 ∈ 2 completion to a basis of Rn satisfying the defining relations jrjs +jsjr = 2δrs. Then, there exist n−1 − 2 polyanalytic functions A : U C C such that, for every z = u +jv, we have F ∩ j → j n−1 F (z)= A(z)jA, jA =ji . . . ji , j F 1 s |AX|=0 where A = i ...is is a subset of 2,...,n , with i <... Step (Ia). For every j = j1 S let j2,..., jn be a completion to a basis of Rn satisfying the ∈ n−1 defining relations jrjs +jsjr = 2δrs. Then there exist 2 functions A : U Cj Cj such that for every z = u +jv − F ∩ → n−1 F (z)= A(z)jA, jA =ji . . . ji , j F 1 s |AX|=0 where Fj(z) is the restriction to U Cj, and A = i1 ...is is a subset of 2,...,n , with i1 <... Definition 3.8. The functions f0, ..., fM−1 L(U) that appear in Theorem 3.7 in the poly decomposition (Ib) ∈ SM M−1 k F (x)= x fk(x), x U, ∀ ∈ k=0 X 10 are called the (left monogenic) components of the left poly slice monogenic function F . Similarly, we will call the functions f0, ..., fM−1 R(U) in (IIb) the components of the right poly slice monogenic function F . ∈ SM Proposition 3.9. Let U Rn+1 be an axially symmetric domain and let M N. Then we have: M ⊆ ∈ (I) F L (U) is intrinsic if and only if all its left slice monogenic components are also intrinsic.∈ PS M (II) F R (U) is intrinsic if and only if all its right slice monogenic components are also intrinsic. ∈ PS Proof. We prove (I). We use the poly decomposition in Theorem 3.7 to write M−1 k F (x)= x fk(x), Xk=0 where fk are the left slice monogenic components for all k = 0, .., M 1. First, we observe that if − all the functions fk are intrinsic, then F will preserve any complex plane U Cj that is to say that F (U C ) C , so it is intrinsic. ∩ ∩ j ⊆ j For the converse, let us assume that F = F0 +jF1 is intrinsic so that F0, F1 are real valued. This n means that in the decomposition F = |A|=0(F0,A +jF1,A)eA there is only the term corresponding M−1 k to e = 1 and expanding in the form F (u +jv) = (u jv) fk(u + jv) the functions fk ∅ P k=0 − are Cj-valued, see the proof of Theorem 3.7 (Ib), and so they take Cj into itself and so they are intrinsic. P Remark 3.10. Observe that since fk(x) is slice monogenic, for ℓ, k N, for k ℓ, we have that ∈ ≥ ℓ k ℓ k k−ℓ ∂ x fk(x) = ∂ x fk(x)= k(k 1)(k 2) . . . (k ℓ + 1)x fk(x), j j − − − ℓ k and ∂j x fk(x) = 0, fork<ℓ. The next result was already proved for quaternions, we revise its prove in more details here for the sake of completeness. Proposition 3.11. Let U Rn+1 be an axially symmetric open set and M N. Then, denoting by F g or gF the pointwise⊆ product, we have the following statements. ∈ M M (Ia) Let F (U) and g L(U). Then, F g belongs to L (U). ∈ PN M ∈ SM PMM (Ib) Let F L (U) and g (U). Then, gF belongs to L (U). ∈ PM M ∈N PM M (IIa) Let F (U) and g R(U). Then, gF belongs to (U). ∈ PN ∈ SM PMR (IIb) Let F M (U) and g (U). Then, F g belongs to M (U). ∈ PMR ∈N PMR (III) Let F M (U) and g (U). Then, gF = F g belongs to M (U). ∈ PN ∈N PN Proof. We prove just step (Ia). In much the same way we can prove the other points. So we assume M that F (U), g L(U) and j S and we set x = u +jv. We will prove that ∈ PN ∈ SM ∈ M ∂j (F g)(u + vj)=0. (18) Indeed, first we note that since F is intrinsic, we have F (U Cj) Cj. In particular, the Leibniz rule the equality ∩ ⊂ ∂j(F g)(u +jv)= F ∂j(g)(u + vj) + ∂j(F )g(u + vj) holds. We note that since F is poly slice monogenic of order M and g is slice monogenic we have ∂ (g)=0 and ∂ (F ) = 0. Thus, we obtain j j 6 ∂j(F g)(u +jv)= ∂j(F )g(u + vj). 11 Then, since F is intrinsic we can use the Leibniz rule M times and get M M−1 M ∂j (F g)(u +jv)= ∂j (F )∂j(g)(u + vj) + ∂j (F )g(u + vj). M Therefore, it follows that the formula (18) holds since F (U) and g L(U). Hence, the pointwise product F g is poly slice monogenic of order ∈M PNon U. ∈ SM In the case we have the explicit poly decomposition of a given poly slice monogenic function F of order M we can give it an integral representation using the Cauchy formula for the slice monogenic components fk(x) k ,...,M appearing in the poly decomposition formula. { } =0 −1 Corollary 3.12 (Slice monogenic integral representation of poly slice monogenic functions). Let U Rn+1 be a slice Cauchy domain. Let j S and set ds = ds( j). ⊆ ∈ j − (I) If F is a (left) poly slice monogenic function of order M on a set that contains U then M−1 1 k −1 F (x)= x SL (s,x) dsj fk(s), for any x U. (19) 2π C ∈ ∂(U∩ j) k Z X=0 (II) If F is a right slice monogenic function on a set that contains U, then M−1 1 −1 k F (x)= fk(s) dsj SR (s,x)x , for any x U. (20) 2π ∂(U∩Cj) ∈ Z Xk=0 The integrals in (19) and (20) depend neither on U nor on the imaginary unit j S. ∈ Proof. We consider the case (19) since (20) can be obtained with similar considerations. So (19) follows by replacing the Cauchy formula for the slice monogenic components fk(x) k=0,...,M−1 given by { } 1 −1 fk(x)= SL (s,x) dsj fk(s), for any x U 2π C ∈ Z∂(U∩ j) M−1 k into the poly decomposition formula F (x)= x fk(x). k X=0 Similarly we can prove the case of unbounded domains. Theorem 3.13 (Slice monogenic integral representation of poly slice monogenic functions on un- bounded slice Cauchy domains). Let U Rn+1 be an unbounded slice Cauchy domain and let j S. ⊂ ∈ Let F L(U) with poly slice decomposition ∈ PS M−1 k F (x)= x fk(x) Xk=0 and the components fk L(U) are such that the limits lim|x|→∞ fk(x) = fk( ) exist and are finite for all k = 0, ..., M∈ SM1. Then, we have ∞ − M−1 M−1 k 1 k −1 F (x)= x fk( )+ x SL (s,x) dsj fk(s) for any x U. ∞ 2π C ∈ k ∂(U∩ j) k X=0 Z X=0 Let F R(U) with poly slice decomposition ∈ PS M−1 k F (x)= fk(x)x k=0 X12 and the components fk R(U) are such that the limits lim|x|→∞ fk(x) = fk( ) exist and are finite for all k = 0, ..., M∈ SM1. Then, we have ∞ − M−1 M−1 k 1 −1 k F (x)= fk( )x + f(s) dsj SR (s,x)x for any x U. ∞ 2π ∂(U∩Cj ) ∈ Xk=0 Z Xk=0 Remark 3.14. Observe that we have two possibilities to write the Cauchy kernels, using the form I or the form II. The following result is a direct consequence of the form of the slice function, i.e., F (x)= F (u, v)+jF (u, v) for x = u +jv U 0 1 ∈ with the two functions F0, F1 : Rn that satisfy the compatibility condition (see the definition of poly slice monogenic functions).U → We recall it for the reader’s convenience. Theorem 3.15 (Poly Structure (or Poly Representation) Formula). Let U Rn+1 be an axially symmetric domain. ⊆ M (I) Let F (U). Then, for any vector x = u +jxv U, the following formula holds: ∈ PSL ∈ 1 1 F (x)= 1 jxj F (u +jv)+ 1+jxj F (u jv), for all u +jv U, j S. (21) 2 − 2 − ∈ ∈ h i h i M (II) Let F (U). Then, for any vector x = u +jxv U, the following formula holds: ∈ PSR ∈ 1 1 F (x)= F (u +jv) 1 jxj + F (u jv) 1+jxj , for all u +jv U, j S. (22) 2 − 2 − ∈ ∈ h i h i In the following we use the notation Rn+1 = Rn+1 . Then, we first recall the Runge’s theorem for slice monogenic functions which was proved∪ in {∞} [31]. We refer to this paper also for the terminology. Theorem 3.16. Let K be an axially symmetric compact set in Rn+1, and let A be a set having a point in each connected component of Rn+1 K. For any axially symmetric open set U K, for \ ⊃ every f L(U) and for every ε > 0 there exists a rational function r whose poles are spheres in A such∈ SMthat f(x) r(x) < ε, | − | for all x K. Similar considerations hold for every f R(U). ∈ ∈ SM Now, we are going to use Theorem 3.16 to prove the poly slice monogenic counterpart of the Runge’s theorem. First the definition of poly slice monogenic rational function. Definition 3.17. We say that a left poly slice monogenic function R(x) is rational if the left slice monogenic components rk(x), for k = 1, ..., M 1, are rational, i.e., − M−1 k R(x)= x rk(x), x U. (23) ∀ ∈ Xk=0 A right poly slice monogenic function R(x) is rational if the right slice monogenic components rk(x) are rational, i.e., M−1 k R(x)= rk(x)x , x U. (24) ∀ ∈ k=0 X 13 Theorem 3.18 (Poly slice Runge’s theorem). Let M N. Let K be an axially symmetric compact ∈ set in Rn+1, and let A be a set having a point in each connected component of Rn+1 K. For any M \ axially symmetric domain U K, for every F L (U) and for every ε > 0 there exists a left poly rational function R whose⊃ poles are spheres∈ in PSA such that F (x) R(x) < ε, | − | M for all x K. The same approximation holds for F R (U). In the case F is intrinsic the rational functions∈ R are also intrinsic. ∈ PS M Proof. We show just the case when F L (U), the other statements follows in much the same way. We note that K is a compact, in∈ particularPS K is bounded so that it is contained in some ball centered at the origin, i.e., K B(0, ρ), for some ρ > 0. Furthermore, since F is poly slice monogenic of order M on the domain⊂ U, we know the validity of the poly decomposition (Ib) in Theorem 3.7 with all fk 0≤k≤M−1 that are slice monogenic functions on U and fM−1 = 0. In particular, since K is contained{ } in U, we also have 6 M−1 k F (x)= x fk(x), x K. (25) ∀ ∈ Xk=0 Then, we can apply the Runge’s theorem 3.16 on each slice monogenic function fk. Thus, for all k = 0, ..., M 1, we know that, for every ε> 0, there exist a rational function rk whose poles are spheres in A−such that 1 fk(x) rk(x) < ε , (26) | − | M−1 j j=0 ρ ! for every x K. Now, we consider the poly slice monogenicP rational function given by ∈ M−1 k R(x)= x rk(x), x U. ∀ ∈ Xk=0 Let ε> 0. Then, we have M−1 M−1 k k F (x) R(x) = x (fk(x) rk(x) x fk(x) rk(x) , | − | − ≤ | | | − | k=0 k=0 X X for every x K. Therefore, we apply the inequality (26) to each slice monogenic component and get ∈ M−1 ε F (x) R(x) < x k , (27) | − | M−1 k | | k=0 ρ k ! X=0 for every x K. However, we know that K P B(0, ρ). So we have x ρ, for any x K. In particular, this∈ shows that ⊂ | | ≤ ∈ M−1 M−1 x k ρk, | | ≤ Xk=0 Xk=0 for every x K. Therefore, inserting this fact in the inequality (27) we obtain that F (x) R(x) < ε, for every∈x K. | − | ∈ 14 4. Cauchy formulas and product of poly slice monogenic functions In this section we develop the Cauchy formulas for poly slice monogenic functions. These will be used in the next section to define the PS-functional calculus for noncommuting operators. Lemma 4.1 (Poly Cauchy integral formula). Let U Rn+1 be a slice Cauchy domain, assume ⊆ that F M (U) and G M (U) for some M N. Then we have ∈ PSL ∈ PSR ∈ M−1 ℓ M−ℓ−1 ℓ ( 1) G(s)∂j dsj ∂j F (s) = 0, (28) ∂(U∩Cj) − Z Xℓ=0 1 where ds = ds( j) and ∂ := (∂u +j∂v) for j S. j − j 2 ∈ n n Proof. By writing G = |A|=0 eAGA, F = |A|=0 FAeA, see the proof of Theorem 3.7, (Ib), the result follows from the analogue theorem for polyanalytic functions of a complex variable. P P −1 −1 4.1. Cauchy formulas with kernels PℓSL and PℓSR . The poly slice monogenic Cauchy kernel are described in the next result: −1 −1 N Definition 4.2 (Cauchy kernels PℓSL and PℓSR ). Let ℓ . We define the left poly slice −1 −1 ∈ monogenic Cauchy kernels PℓSL and PℓSR of order ℓ +1 by (Re(s x))ℓ P S−1(s,x):= − S−1(s,x) ℓ L ℓ! L (Re(s x))ℓ = − (x2 2Re(s)x + s 2)−1(x s), s [x] − ℓ! − | | − 6∈ and right poly slice monogenic Cauchy kernels of order ℓ is defined by (Re(s x))ℓ P S−1(s,x):= − S−1 (s,x) ℓ R ℓ! R,ℓ (Re(s x))ℓ = − (x s¯)(x2 2Re(s)x + s 2)−1, s [x]. − ℓ! − − | | 6∈ Remark 4.3. Observe that we have used the slice monogenic Cauchy kernels written in form I so that with this observation the PS-functional calculus will work for paravector operators with noncommuting components, even though most of the properties hold true for paravector operators with commuting components. For the function theory it is also possible to use slice monogenic Cauchy kernels written in form II, in this case when we define the PS-functional calculus we are limited to commuting operators. Lemma 4.4. Let ℓ N and let x,s Rn+1 with s / [x]. The left poly slice monogenic Cauchy −1 ∈ ∈ ∈ kernel PℓSL (s,x) is left poly slice monogenic in x and right poly slice monogenic in s of order −1 ℓ + 1. The right poly slice monogenic Cauchy kernel PℓSR (s,x) is left poly slice monogenic in s and right poly slice monogenic in x of order ℓ + 1. Proof. For ℓ N. For all s,x Rn+1, we set ∈ ∈ (Re(s x))ℓ P (s,x)= − . ℓ ℓ! Then, the function Pℓ(s,x) is intrinsic and poly slice monogenic of order ℓ + 1 with respect to the −1 variables x and s. By a direct computation, for s / [x], the product Pℓ(s,x)SL (s,x) is left poly slice monogenic in x and right poly slice monogenic∈ in s of order ℓ + 1. Using similar arguments −1 we treat Pℓ(s,x)SR (s,x). 15 −1 −1 Rn+1 Theorem 4.5 (The poly slice Cauchy formulas with kernels PℓSL and PℓSR ). Let U be 1 ⊂ a slice Cauchy domain. For j S set ds = ds( j) and ∂ := (∂u +j∂v). ∈ j − j 2 (I) If F is a left poly slice monogenic function on a set that contains U then M−1 1 ℓ −1 ℓ F (x)= ( 2) PℓSL (s,x) dsj ∂j F (s), for any x U. (29) 2π C − ∈ ∂(U∩ j) ℓ Z X=0 (II) If F is a right poly slice monogenic function on a set that contains U, then M−1 1 ℓ ℓ −1 F (x)= ( 2) ∂j F (s) dsj PℓSR (s,x), for any x U. (30) 2π C − ∈ ∂(U∩ j) ℓ Z X=0 The integrals (29) and (30) depend neither on U nor on the imaginary unit j S. ∈ Proof. Consider (I). It is a consequence of the Lemma 4.1 and of Theorem 3.15. We now use M the poly splitting lemma (Ia). Let F (U), and for every j = j S let j ,..., jn be a ∈ PSL 1 ∈ 2 completion to a basis of Rn satisfying the defining relations jrjs +jsjr = 2δrs. Then there exist n−1 − 2 polyanalytic functions A : U C C such that, for every z = u +jv, we have F ∩ j → j n−1 F (z)= A(z)jA, jA =ji . . . ji , j F 1 s |AX|=0 where A = i1 ...is is a subset of 2,...,n , with i1 <... M−1 ℓ 1 ℓ 1 1 (Re(s z)) ℓ F (x)= ( 2) 1 jxj − dsj ∂j F (s) 2π ∂(U∩Cj) − 2 − s z ℓ! Z Xℓ=0 h i − M−1 ℓ 1 ℓ 1 1 (Re(s z)) ℓ + ( 2) 1+jxj − dsj ∂j F (s). 2π ∂(U∩Cj) − 2 s z ℓ! Z Xℓ=0 h i − Finally we get M−1 ℓ 1 ℓ (Re(s x)) −1 ℓ F (x)= ( 2) − SL (s,x) dsj ∂j F (s) 2π C − ℓ! ∂(U∩ j) ℓ Z X=0 where we have replaced the poly slice monogenic Cauchy kernel ℓ ℓ ℓ (Re(s x)) −1 1 (Re(s z)) 1 1 (Re(s z)) 1 − S (s,x)= 1 jxj − + 1+jxj − ℓ! L 2 − ℓ! s z 2 ℓ! s z h i − h i − written via the poly slice monogenic representation formula. Definition 4.6. We say that a poly slice monogenic function is poly slice monogenic function at infinity if its slice monogenic components are slice monogenic at infinity. Theorem 4.7 (Poly slice monogenic Cauchy formulas on unbounded slice Cauchy domains with −1 −1 Rn+1 S kernels PℓSL and PℓSR ). Let U be an unbounded slice Cauchy. For j set dsj = ds( j) 1 ⊂ ∈ − and ∂j := 2 (∂u +j∂v). (I)Let F L(U) with poly slice decomposition ∈ PS M−1 k F (x)= x fk(x) k X=0 and the components fk L(U) are such that the limits lim|x|→∞ fk(x) = fk( ) exist and are finite for all k = 0, ..., M∈ SM1. Then, we have ∞ − M−1 M−1 k 1 ℓ −1 ℓ F (x)= x fk( )+ ( 2) PℓSL (s,x) dsj ∂j F (s) for any x U. ∞ 2π ∂(U∩Cj) − ∈ Xk=0 Z Xℓ=0 (II) Let F R(U) with poly slice decomposition ∈ PS M−1 k F (x)= fk(x)x Xk=0 and the components fk R(U) are such that the limits lim|x|→∞ fk(x) = fk( ) exist and are finite for all k = 0, ..., M∈ SM1. Then we have ∞ − M−1 M−1 k 1 ℓ ℓ −1 F (x)= fk( )x + ( 2) ∂j F (s) dsj PℓSR (s,x) for any x U. ∞ 2π C − ∈ k=0 Z∂(U∩ j) ℓ=0 X X 17 Proof. Let us consider (I). For sufficiently large r > 0 , the set Ur := U Br(0) is a bounded slice ∩ Cauchy domain with x Ur and H Ur U. By ∈ \ ⊂ M−1 1 ℓ −1 ℓ F (x)= ( 2) PℓSL (s,x) dsj ∂j F (s) 2π ∂(Ur∩Cj) − Z Xℓ=0 M−1 1 ℓ −1 ℓ = ( 2) PℓSL (s,x) dsj ∂j F (s) 2π ∂(U∩Cj ) − Z Xℓ=0 M−1 1 ℓ −1 ℓ + ( 2) PℓSL (s,x) dsj ∂j F (s), for any x U. 2π C − ∈ ∂(Br(0)∩ j ) ℓ Z X=0 The Cauchy theorem for poly slice monogenic functions implies that we can vary r without changing the value of the second integral. Letting r tend to infinity, we find that the monogenic components converge to fk( ) and we obtain the statement, since ∞ M−1 M−1 1 ℓ −1 ℓ k lim ( 2) PℓSL (s,x) dsj ∂j F (s)= x fk( ). r→∞ 2π C − ∞ ∂(Br (0)∩ j ) ℓ k Z X=0 X=0 −1 Remark 4.8. There is a more direct proof of the Cauchy formulas with the Cauchy kernels PℓSL −1 and PℓSR . In fact, integrals k−1 1 ℓ −1 ℓ Jk(x) := ( 2) PℓSL (s,x) dsj ∂j F (s) 2π C − U∩ j ) ℓ Z X=0 can be computed directly. As an example consider k = 2. 1 −1 J2(x)= SL (s,x) dsj sf1(s)+ f0(s) 2π ∂(U∩C ) Z j 1 −1 + ( 2)(Re(s) x0)SL (s,x) dsj f1(s) 2π C − − Z∂(U∩ j ) using the relation S−1(s,x)s xS−1(s,x) = 1 and the Cauchy theorem we get L − L 1 −1 1 −1 J2(x)= x SL (s,x) dsj f1(s)+ SL (s,x) dsj f0(s) 2π C 2π C ZU∩ j ) Z∂(U∩ j ) so we obtain J2(x)= xf1(x)+ f0(x). −1 −1 4.2. Cauchy formulas with kernels ΠℓSL and ΠℓSR . −1 −1 N Definition 4.9 (The Cauchy kernels ΠℓSL and ΠℓSR ). Let ℓ . We define the left poly slice monogenic Cauchy kernels of order ℓ +1 by ∈ ℓ −1 1 ℓ k −1 ℓ−k ΠℓS (s,x):= x S (s,x)( s) L ℓ! k L − Xk=0 ℓ 1 ℓ = xk(x2 2Re(s)x + s 2)−1(x s)( s)ℓ−k, s [x] −ℓ! k − | | − − 6∈ k=0 X 18 and right poly slice monogenic Cauchy kernels of order ℓ + 1 is defined by ℓ −1 1 ℓ ℓ−k −1 k ΠℓS (s,x):= ( s) S (s,x)x R ℓ! k − R Xk=0 ℓ 1 ℓ = ( s)ℓ−k(x s¯)(x2 2Re(s)x + s 2)−1xk, s [x]. −ℓ! k − − − | | 6∈ k X=0 Lemma 4.10. Let ℓ N and let x,s Rn+1 with s / [x]. The left poly slice monogenic Cauchy −1 ∈ ∈ ∈ kernel ΠℓSL (s,x) is left poly slice monogenic in x and right poly slice monogenic in s of order −1 ℓ + 1. The right poly slice monogenic Cauchy kernel ΠℓSR (s,x) is left poly slice monogenic in s and right poly slice monogenic in x of order ℓ + 1. −1 Proof. Consider left poly slice monogenic Cauchy kernel ΠℓSL (s,x). The statement is a direct consequence of the definition of poly slice monogenicity and of the poly decomposition of poly slice −1 monogenic functions. In fact, ΠℓSL (s,x) is of the form ℓ −1 ℓ−k ΠℓSL (s,x)= ψℓ(s,x)s k X=0 where the components 1 ℓ k −1 ℓ−k ψℓ(s,x) := x S (s,x)( 1) ℓ! k L − are right poly slice monogenic function in the variable s for s / [x]. The order ℓ+1 is clear from the −1 ∈ definition. To see that the kernel ΠℓSL (s,x) is left poly slice monogenic in x it is more convenient −1 −1 to write SL (s,x) in the form II and reasoning in a similar way. The case of ΠℓSR (s,x) follows with similar considerations. −1 Lemma 4.11. For s / [x] the kernel ΠℓSL (s,x) is the unique left (resp. right) poly slice monogenic ∈ 1 ℓ −1 C extension in x (resp. s) of the kernel πℓ(z,s)= ℓ! (¯z s¯) (s z) , z,s j, for z = s. Analogously, −1 − − ∈ 6 for s / [x], the kernel ΠℓSR (s,x) is the unique right (resp. left) poly slice monogenic extension in ∈ 1 ℓ −1 x (resp. s) of the kernel πℓ(z,s)= (¯z s¯) (s z) , z,s C , z = s. ℓ! − − ∈ j 6 C −1 C Proof. When x = z j it is evident that the restriction of ΠℓSL (s,x) to j is πℓ(z,s). Since −1 1 ℓ ∈ ℓ k −1 ℓ−k ΠℓSL (s,x) = ℓ! k=0 k x SL (s,x)( s) and its slice monogenic components are unique, the assertion follows. − P −1 −1 Rn+1 Theorem 4.12 (Cauchy formulas with the kernels ΠℓSL and ΠℓSR ). Let U be a slice 1 ⊂ Cauchy domain. For j S set ds = ds( j) and ∂ := (∂u +j∂v). ∈ j − j 2 (I) If F is a left poly slice monogenic function on a set that contains U then M−1 1 −1 ℓ F (x)= ΠℓSL (s,x) dsj ∂j F (s), for any x U. (31) 2π ∂(U∩Cj) ∈ Z Xℓ=0 (II) If F is a right poly slice monogenic function on a set that contains U, then M−1 1 ℓ −1 F (x)= ∂j F (s) dsj ΠℓSR (s,x), for any x U. (32) 2π C ∈ ∂(U∩ j) ℓ Z X=0 The integrals (29) and (30) depend neither on U nor on the imaginary unit j S. 19 ∈ Proof. It follows in much the same way as the case of the poly slice Cauchy formulas with kernels −1 −1 PℓS and PℓS (see Theorem 4.2) using the poly splitting lemmas by taking x,s C . Then L R ∈ j the assertion follows by the poly representation formula, the polyanalytic Cauchy kernel πℓ used by Th´eodoresco see (2) and Lemma 4.11. Theorem 4.13 (Poly slice monogenic Cauchy formulas on unbounded slice Cauchy domains). Let n+1 1 U R be an unbounded slice Cauchy. For j S set ds = ds( j) and ∂ := (∂u +j∂v). ⊂ ∈ j − j 2 (I) Let F L(U) with poly slice decomposition ∈ PS M−1 k F (x)= x fk(x) Xk=0 and the components fk L(U) are such that the limits lim|x|→∞ fk(x) = fk( ) exist and are finite for all k = 0, ..., M∈ SM1. Then, we have ∞ − M−1 M−1 k 1 −1 ℓ F (x)= x fk( )+ ΠℓSL (s,x) dsj ∂j F (s) for any x U. ∞ 2π ∂(U∩Cj) ∈ Xk=0 Z Xℓ=0 (II) Let F R(U) with poly slice decomposition ∈ PS M−1 k F (x)= fk(x)x k X=0 and the components fk R(U) are such that the limits lim|x|→∞ fk(x) = fk( ) exist and are finite for all k = 0, ..., M∈ SM1. Then we have ∞ − M−1 M−1 k 1 ℓ −1 F (x)= fk( )x + ∂j F (s) dsj ΠℓSR (s,x) for any x U. ∞ 2π C ∈ k ∂(U∩ j) ℓ X=0 Z X=0 Proof. Consider (I). Recall that we assume the components fk L(U) of F R(U) are ∈ SM ∈ PS such that the limits lim fk(x) = fk( ) exist and are finite for all k = 0, ..., M 1. For |x|→∞ ∞ − sufficiently large r > 0 , the set Ur := U Br(0) is a bounded slice Cauchy domain with x Ur n+1 ∩ ∈ and R Ur U. By \ ⊂ M−1 1 −1 ℓ F (x)= ΠℓSL (s,x) dsj ∂j F (s) 2π C ∂(Ur∩ j) ℓ Z X=0 M−1 1 −1 ℓ = ΠℓSL (s,x) dsj ∂j F (s) 2π ∂(U∩Cj ) Z Xℓ=0 M−1 1 −1 ℓ + ΠℓSL (s,x) dsj ∂j F (s), for any x U. 2π ∂(U∩Cj ) ∈ Z Xℓ=0 The Cauchy theorem for poly slice monogenic functions implies that we can vary r without changing M−1 k the value of the second integral. Letting r tend to infinity, we find that it equals k=0 x fk( ) and we obtain the statement. ∞ P −1 −1 Remark 4.14. Also for the Cauchy formulas with the kernels ΠℓSL and ΠℓSR the integrals k−1 1 −1 ℓ Ik(r) := ΠℓSL (s,x) dsj ∂j F (s) 2π C Z∂(U∩ j ) ℓ=0 X20 can be computed directly. As an example consider the case k = 2, so we have 1 −1 I2(x)= SL (s,x) dsj sf1(s)+ f0(s) 2π ∂(U∩C ) Z j 1 −1 −1 + SL (s,x)( s)+ xSL (s,x) dsj f1(s) 2π ∂(U∩C ) − Z j and so we get I2(x)= xf1(x)+ f0(x) using the Cauchy formula of slice monogenic functions. 4.3. Poly ⊛L-product and ⊛R-product. We conclude this section with the product of poly slice monogenic functions. We note that given two poly slice monogenic functions F and G of orders N and M, we can use the poly decomposition formulas to define a natural product. This product take out of the class but it will be useful for the product rule of the PS-functional calculus. n+1 Definition 4.15 (Poly ⊛L-product and ⊛R-product and pointwise product ). Let U R be an axially symmetric open set and M,N 1. ∈ (I) Let F N (U) and G M (U)≥ and let ∈ PSL ∈ PSL N−1 M−1 k k F (x)= x fk(x) and G(x)= x gk(x), for all x U, ∈ k k X=0 X=0 be their poly decompositions for f , ..., fN and g , ..., gM L(U). We defined the poly 0 −1 0 −1 ∈ SM ⊛L-product of F and G by: N+M−2 ℓ (F ⊛L G)(x) := x (fk L gh)(x) , (33) ∗ ℓ k h ℓ ! X=0 +X= where L is the -product of left slice monogenic functions. (II)∗ Let F ∗ N (U) and G M (U) and let ∈ PSR ∈ PSR N−1 M−1 k k F (x)= fk(x)x and G(x)= gk(x)x , for all x U, ∈ Xk=0 Xk=0 be their poly decompositions for f , ..., fN and g , ..., gM R(U). We defined the poly 0 −1 0 −1 ∈ SM ⊛R-product of F and G by: N+M−2 ℓ (F ⊛R G)(x) := (fk R gh)(x) x , (34) ∗ ! Xℓ=0 k+Xh=ℓ where R is the star-product of right slice monogenic functions. (III)∗ Let F N (U) and G M (U) and let ∈ PN L ∈ PSL N−1 M−1 k k F (x)= x fk(x) and G(x)= x gk(x), for all x U, ∈ Xk=0 Xk=0 be their poly decompositions for f0, ..., fN−1 (U) and g0, ..., gM−1 L(U) . We defined the pointwise product of F and G by: ∈N ∈ SM N+M−2 ℓ (F G)(x) := x (fkgh)(x) . (35) ! Xℓ=0 k+Xh=ℓ N M Similarly we define the pointwise product for F R (U) and G R (U). 21∈ PS ∈ PN Remark 4.16. It is clear that the pointwise product of two poly slice monogenic functions does not preserve the poly slice monogenicity, but, as we will see, it does when we consider intrinsic functions. Theorem 4.17. Let U Rn+1 be an axially symmetric open set and M,N 1. (I) Let F N (U) ∈and G M (U) and let ≥ ∈ PSL ∈ PSL N+M−2 ℓ (F ⊛L G)(x) := x (fk L gh)(x) , (36) ∗ ℓ k h ℓ ! X=0 +X= M+N−1 the poly ⊛L-product of F and G. Then (F ⊛L G) (U). ∈ PSL (II) Let F N (U) and G M (U) and let ∈ PSR ∈ PSR N+M−2 ℓ (F ⊛R G)(x) := (fk R gh)(x) x , (37) ∗ ! Xℓ=0 k+Xh=ℓ M+N−1 be the poly ⊛R-product of F and G. Then (F ⊛R G) (U). ∈ PSR Proof. Let us prove (I). With similar computations we get (II). Observe that, in (36), the function hℓ(x) := (fk L gh)(x) ∗ k+Xh=ℓ is slice monogenic in U for all ℓ, by definition, so the function N+M−2 ℓ (F ⊛L G)(x) := x hℓ(x), Xℓ=0 belongs to M+N−1(U) thanks to the poly decomposition theorem. PSR As a consequence of the previous result we have the case of intrinsic functions which will be used for the product rule for the PS-functional calculus. Corollary 4.18. Let U Rn+1 be an axially symmetric open set and M,N 1. (I) Let F N (U) ⊆and G M (U). Then we have ≥ ∈ PN L ∈ PSL N+M−2 ℓ (F ⊛L G)(x)= F (x)G(x)= x fk(x)gh(x) , (38) ! Xℓ=0 k+Xh=ℓ and F G M+N−1(U). ∈ PSL (II) Let F N (U) and G M (U). Then we have ∈ PSR ∈ PN R N+M−2 ℓ (F ⊛R G)(x)= F (x)G(x)= fk(x)gh(x) x , (39) ! Xℓ=0 k+Xh=ℓ and F G M+N−1(U). ∈ PSR Proof. It is a direct consequence of the product theorem for slice monogenic functions, i.e., it is the case when the slice monogenic ⋆-product becomes the pointwise product. 22 5. Formulations of the PS-functional calculus via the ΠS-resolvent operators In the sequel, we will consider a Banach space V over R with norm . It is possible to endow k · k V with an operation of multiplication by elements of Rn which gives a two-sided module over Rn. A two-sided module V over Rn is called a Banach module over Rn, if there exists a constant C 1 ≥ such that va C v a and av C a v for all v V and a Rn. By Vn we denote k k ≤ k k| | k k ≤ | |k k ∈ ∈ V Rn over Rn that turns out to be a two-sided Banach module. An element in Vn is of the ⊗ type vA eA (where A = i ...ir, iℓ 1, 2,...,n , i < ... 0,1 n+1 Theorem 5.1. Let T (Vn) and let s R with T < s . ∈B ∈ k k | | (i) The left S-resolvent series equals +∞ T ms−m−1 = (T 2 2Re(s)T + s 2 )−1(T s ). − − | | I − I m=0 X (ii) The right S-resolvent series equals +∞ s−m−1T m = (T s )(T 2 2Re(s)T + s 2 )−1. − − I − | | I m=0 X 0,1 n+1 Definition 5.2. Let T (Vn). For s R , we set ∈B ∈ 2 2 s(T ) := T 2Re(s)T + s . Q − | | I We define the S-resolvent set ρS(T ) of T as n+1 ρS(T ) := s R : s(T ) is invertible in (Vn) { ∈ Q B } and we define the S-spectrum σS(T ) of T as n+1 σS(T ) := R ρS(T ). \ −1 For s ρS(T ), the operator s(T ) is called the pseudo S-resolvent operator of T at s. ∈ Q 23 0,1 Definition 5.3. Let T (Vn). For s ρS(T ), we define the left S-resolvent operator as ∈B ∈ −1 −1 S (s, T )= s(T ) (T s ), L −Q − I and the right S-resolvent operator as −1 −1 S (s, T )= (T s ) s(T ) . R − − I Q 0,1 Lemma 5.4. Let T (Vn). ∈B −1 (I) The left S-resolvent operator S (s, T ) is a (Vn)-valued right-slice mponogenic function of L B the variable s on ρS(T ). −1 (II) The right S-resolvent operator S (s, T ) is a (Vn)-valued left-slice monogenic function of R B the variable s on ρS(T ). 0,1 Definition 5.5. We denote by L(σS(T )), R(σS(T )) and (σS(T )), for T (Vn), the SM SM N ∈B set of all left, right and intrinsic slice monogenic functions with σS(T ) U, where U is a slice ⊂ Cauchy domain such that U dom(f) and dom(f) is the domain of the function f. ⊂ 0,1 Definition 5.6 (The S-functional calculus). Let T (Vn) for j S set dsj = ds( j). Then we have the formulations of the S-functional calculus. We∈B define ∈ − 1 −1 f(T ) := SL (s, T ) dsj f(s), for all f L(σS(T )), (40) 2π C ∈ SM Z∂(U∩ j) and 1 −1 f(T ) := F (s) dsj SR (s, T ), for all f R(σS (T )). (41) 2π C ∈ SM Z∂(U∩ j) Remark 5.7. The definition of the S-functional calculus is well posed because the integrals (40) and (41) depend neither on U nor on the imaginary unit j S. This fact is independent of the fact that the components of the operator T commute or not among∈ themselves. The S-resolvent equation is useful to prove several properties of the S-functional calculus. So it is natural to ask if it is possible to obtain an analog of the classical resolvent equation (λI A)−1(µI A)−1 = (λI A)−1 (µI A)−1 µ λ −1, λ, µ C σ(A), (42) − − − − − − ∈ \ where A is a complex operator on a Banach space. The generalization to this non commutative setting, involves both the left and the right S-resolvent operators and the analogue of the term (λI A)−1 (µI A)−1 µ λ −1, which is the difference of the resolvent operators (λI A)−1 − − − − − − (µI A)−1 multiplied by the Cauchy kernel (µ λ)−1 for the S-functional calculus becomes the − −1 − −1 difference of the S-resolvent operators SR (s, T ) SL (p, T ) entangled in a suitable way with the slice monogenic Cauchy kernels. In fact we have:− 0,1 Theorem 5.8 (The S-resolvent equation, see [8]). Let T (Vn) and let s,q ρS(T ) with −1 2 2 −1 ∈ B ∈ q / [s]. Set s(q) := (q 2Re(s)q + s ) . Then the equation ∈ Q − | | −1 −1 −1 −1 −1 −1 −1 S (s, T )S (q, T )= S (s, T ) S (q, T ) q s S (s, T ) S (q, T ) s(q) (43) R L R − L − R − L Q holds true. Equivalently, it can also be written as −1 −1 −1 −1 −1 −1 −1 S (s, T )S (q, T )= q(s) S (q, T ) S (s, T ) q s S (q, T ) S (s, T ) . (44) R L Q L − R − L − R The S-resolvent equation is a consequence of the left and the righ t S-resolvent equations: 0,1 Theorem 5.9. Let T (Vn) and let s ρS(T ). The left S-resolvent operator satisfies the left S-resolvent equation ∈B ∈ S−1(s, T )s TS−1(s, T )= (45) L − L I and the right S-resolvent operator satisfies the right S-resolvent equation −1 −1 sSR (s, T ) SR (s, T )T = . (46) − 24 I We point out that the equations (45) and (46) cannot be considered the generalizations of the classical resolvent equation. Only the equations in Theorem 5.8 have the properties of the classical resolvent equation (42). The product rule is a consequence of the S-resolvent equation. 0,1 Theorem 5.10 (Product rule). Let T (Vn) and let f (σS(T )) and g L(σS(T )) or ∈B ∈N ∈ SM let f R(σS(T )) and g (σS(T )). Then ∈ SM ∈N (fg)(T )= f(T )g(T ). In this section we give the formulations of the poly slice monogenic version of the S-functional based on the poly slice Cauchy formulas. This calculus that will be indicated by PS-functional calculus −1 −1 0,1 Definition 5.11 (The ΠS-resolvent operators ΠℓS (s, T ) and ΠℓS (s, T )). Let T (Vn) L R ∈ B and s ρS(T ). We define the poly left S-resolvent operator (for short ΠS-resolvent operator) of order ℓ∈+ 1, for ℓ N as ∈ ℓ −1 1 ℓ k −1 ℓ−k ΠℓS (s, T ):= T S (s, T )( s) L ℓ! k L − Xk=0 ℓ 1 ℓ k −1 ℓ−k = T s(T ) (T s )( s) , s [x] −ℓ! k Q − I − 6∈ Xk=0 and right poly slice monogenic Cauchy kernels of order ℓ + 1 is defined by ℓ −1 1 ℓ ℓ−k −1 k ΠℓS (s, T ):= ( s) S (s, T )T R ℓ! k − R k X=0 ℓ 1 ℓ ℓ−k −1 k = ( s) (T s ) s(T ) T , s [x]. −ℓ! k − − I Q 6∈ Xk=0 2 2 where s(T ) := T 2Re(s)T + s . Q − | | I 0,1 Lemma 5.12. Let T (Vn) and let ℓ N. Then we have: ∈B −1 ∈ (I) The poly left S-resolvent ΠℓS (s, T ) is a (Vn)-valued right poly slice monogenic function L B of the variable s on ρS(T ) of order ℓ + 1. −1 (II) The poly right S-resolvent ΠℓS (s, T ) is a (Vn)-valued left poly slice monogenic function R B of the variable s on ρS(T ) of order ℓ + 1. Proof. Consider case (I). It is a direct consequence of the definition because it is of the form ℓ −1 ℓ−k ΠℓSL (s, T )= ψℓ(s, T )s Xk=0 where the components 1 ℓ k −1 ℓ−k ψℓ(s, T ) := T S (s, T )( 1) ℓ! k L − are right (Vn)-valued slice monogenic function in the variable s on ρS(T ), recalling that Vn = B V Rn. Apply Theorem 3.7 which still holds for function (Vn)-valued we get the statement. The order⊗ ℓ + 1 is clear from the definition. Case (II) follows withB the same considerations. 0,1 We can define the left and the right PS-resolvent equations for T (Vn), that is when T has commuting components. This will have consequences on the product∈ BC rules. 25 N 0,1 M M Definition 5.13. Let M and let T (Vn). We denote by L (σS (T )), R (σS(T )) M ∈ ∈ B PS PS and (σS(T )) the set of all left, right and intrinsic poly slice monogenic functions F or order PN M, respectively, with σS(T ) U, where U is a slice Cauchy domain such that U dom(F ) and dom(F ) is the domain of the⊂ function F . ⊂ Using the Cauchy formula of poly slice monogenic functions we give the definition of the PS- functional calculus. 0,1 Definition 5.14 (The PS-functional calculus (I)). Let T (Vn), M N, for j S, set 1 −1 −1 ∈ B ∈ ∈ dsj = ds( j), ∂j := 2 (∂u +j∂v) let ΠℓSL (s, T ) and ΠℓSR (s, T ) be the ΠS-resolvent operators, for ℓ = 0, ..., M− 1. We define − M−1 1 −1 ℓ M F (T ) := ΠℓSL (s, T ) dsj ∂j F (s), for all F L (σS(T )), (47) 2π ∂(U∩Cj) ∈ PS Z Xℓ=0 and M−1 1 ℓ −1 M F (T ) := ∂j F (s) dsj ΠℓSR (s, T ), for all F R (σS(T )). (48) 2π ∂(U∩Cj) ∈ PS Z Xℓ=0 The following theorem shows that the definitions of the PS-functional calculus are well posed. 0,1 N S 1 Theorem 5.15. Let T (Vn), M , for j , set dsj = ds( j), ∂j := 2 (∂u +j∂v) and let −1 −1 ∈ B ∈ ∈ − ΠℓSL (s, T ) and ΠℓSR (s, T ) be the ΠS-resolvent operators, for ℓ = 0, ..., M 1. Then the integrals (47) and (48) depend neither on U nor on the imaginary unit j S. − ∈ Proof. The independence of the integrals (47) and (48) from the open set U is standard. The independence on the imaginary unit j S is based on the use of the poly slice monogenic Cauchy ∈ formula. We treat the case of F L(σS(T )). For functions in F R(σS(T )) the proof is ∈ PS ∈ PS similar with obvious changes. If U ′ U, then O := U U ′ is a slice Cauchy domain that contains 6⊂ ∩ ′′ ′′ ′′ ′ σS(T ). We can hence find a third slice Cauchy domain U with σS(T ) U and U O = U U . The above arguments show that the integrals over the boundaries of all⊂ three sets agree.⊂ ∩ So we choose two units i, j S and two slice Cauchy domains Uq,Us dom(F ) with σS(T ) Uq ∈ ⊂ ⊂ and Uq Us. (The subscripts q and s are chosen in order to indicate the respective variable of ⊂ c Rn+1 integration in the following computation). The set Uq := Uq is then an unbounded axially c \ symmetric domain with U ρS(T ). The left PS-resolvent operator is right poly slice monogenic q ⊂ on ρS(T ) and at infinity, since +∞ lim S−1(s, T ) = lim T ns−n−1 = 0 s→∞ L s→∞ n=0 X the left S-resolvent operator is slice monogenic. Now observe that the ΠS-resolvent operator ℓ −1 1 ℓ k −1 ℓ−k ΠℓS (s, T )= T S (s, T )( s) L ℓ! k L − k X=0 has operator valued slice monogenic components, and they go to zero, i.e., ℓ−k ( 1) ℓ k − T S−1(s, T ) 0, for s . ℓ! k L → →∞ So we can represent it with the Cauchy formula for unbounded slice Cauchy domains and it is right poly slice monogenic on the S-resolvent set ρS(T ), so we have M−1 −1 1 m −1 −1 ΠℓSL (s, T )= ∂i ΠℓSL (q, T ) dqi ΠmSR (q,s) 2π c C Z∂(Uq ∩ i) m=0 X26 −1 c −1 because ΠℓS (s, T ) is poly slice monogenic up to order M ℓ, for any s U . From S (q,s)= L ≥ ∈ q R S−1(s,q) we find that the Cauchy kernel − L m −1 1 m m−k −1 k ΠmS (s,q)= ( s) S (s,q)q R m! k − R Xk=0 can be written as m −1 1 m m−k −1 k ΠmS (q,s)= ( q) S (q,s)s R m! k − R Xk=0 ℓ 1 m = ( q)m−kS−1(s,q)sk. −m! k − L Xk=0 c So keeping in mind that ∂(U C )= ∂(Uq C ) we have the chain of equalities q ∩ i − ∩ i M−1 1 −1 ℓ F (T ) := ΠℓSL (s, T ) dsj ∂j F (s) 2π ∂(Us∩Cj) Z Xℓ=0 M−1 M−1 1 1 m −1 −1 ℓ = ∂i ΠℓSL (q, T ) dqi ΠmSR (q,s) dsj ∂j F (s) 2π C 2π c C ∂(Us∩ j) ℓ ∂(Uq ∩ i) m=0 Z X=0 Z X M−1 M−1 1 1 m −1 −1 ℓ = ∂i ΠℓSL (q, T ) dqi ΠmSL (s,q) dsj ∂j F (s) 2π C 2π C ∂(Us∩ j) ℓ ∂(Uq ∩ i) m=0 Z X=0 Z X M−1 M−1 1 m −1 1 −1 ℓ = ∂i ΠℓSL (q, T ) dqi ΠmSL (s,q) dsj ∂j F (s) 2π C 2π C ∂(Uq ∩ i) m ∂(Us∩ j) Z X=0 Z Xℓ=0 where we have used Fubini’s theorem and with some computations we obtain M−1 1 m −1 m F (T )= ∂i ΠmSL (q, T ) dqi ∂i F (q) 2π C ∂(Uq∩ i) m Z X=0 so we conclude that the poly slice monogenic Cauchy formula give the independence on the imagi- nary units in the sphere S because we chose Uq Us. ⊂ We start by proving some basic results related to the PS-functional calculus introduced in Definition 5.14 0,1 Proposition 5.16 (Linearity of the ΠS-functional calculus). Let T (Vn) and M 1. Then, we have ∈B ≥ M (1) If F, G (σS(T )), then we have: ∈ PSL (F + G)(T )= F (T )+ G(T ), (F λ)(T )= F (T )λ, for all λ Rn. ∈ M (2) If F, G (σS(T )), then we have: ∈ PSR (F + G)(T )= F (T )+ G(T ), (λF )(T )= λF (T ), for all λ Rn. ∈ Proof. This follows directly by construction of the PS-functional calculus in Definition 5.14. 27 Remark 5.17. We have proved in the previous section the poly decomposition (16) of a function M F (U) in terms of f , ..., fM L(U). So when the components are polynomials or ∈ PSL 0 −1 ∈ SM power series fℓ(x) for ℓ = 0, ..., M 1, then − ∞ u fℓ(x)= x Aℓ,u, with Aℓ,u Rn, for u N , ℓ = 0, ..., M 1. ∈ ∈ 0 − u=0 X So for fM = 0, we have the following decomposition −1 6 M−1 ∞ ℓ u F (x)= x x Aℓ,u, x U, (49) ∀ ∈ ℓ u=0 X=0 X where U is the set of convergence of all the series fℓ(x), for ℓ = 0, ..., M 1. Analogous considerations hold for the right case (17), and we have − M−1 ∞ u ℓ F (x)= Aℓ,ux x , x U. (50) ∀ ∈ u Xℓ=0 X=0 The following theorem shows the compatibility of the PS-functional calculus with the poly slice monogenic polynomials and series with respect to the slice monogenic components, that is when we consider for example the expansion (49), the PS-functional calculus gives M−1 ∞ ℓ u F (T )= T T Aℓ,u, ℓ u=0 X=0 X when F is defined on the S-spectrum of T . 0,1 n N S Theorem 5.18. Let T (Vn) where we set T = T0 j=1 ejTj. Let M , for j , set 1∈ B −1 − −1 ∈ ∈ ds = ds( j) and ∂ := (∂u +j∂v). Let ΠℓS (s, T ) and ΠℓS (s, T ) be the ΠS-resolvent operators, j − j 2 L PR for ℓ = 0, ..., M 1, and suppose σS(T ) U, where U is a slice Cauchy domain. (I) Then when− F is the series expansion⊂ in (49) we have M−1 M−1 ∞ 1 −1 ℓ ℓ u ΠℓSL (s, T ) dsj ∂j F (s)= T T Ak,u. 2π C ∂(U∩ j) u Z Xℓ=0 Xℓ=0 X=0 (II) Then when F is the series expansion in (50) we have M−1 M−1 ∞ 1 ℓ −1 u ℓ ∂j F (s) dsj ΠℓSR (s, T )= Ak,uT T . 2π C ∂(U∩ j) u=0 Z Xℓ=0 Xℓ=0 X Proof. It follows from the definition with some computations. Remark 5.19. From Theorem 5.18 we see a first difference with respect to the S-functional calculus for intrinsic functions. In fact, for the PS-functional calculus it is not enough that F is an intrinsic function to have that M−1 ∞ M−1 ∞ ℓ u u ℓ T T Ak,u = Ak,uT T ℓ u=0 ℓ u=0 X=0 X X=0 X but we have to require that T has commuting components. The following theorem is important to prove the product rule. 28 0,1 Theorem 5.20 (The intrinsic PS-functional calculus (I)). Let T (Vn) and set dsj = ds( j) 1 M −1 ∈ BC −1 − and ∂ := (∂u+j∂v). Let F (σS(T )). Let ΠℓS (s, T ) and ΠℓS (s, T ) be the ΠS-resolvent j 2 ∈ PN L R operators, for ℓ = 0, ..., M 1, and suppose σS(T ) U, where U is a slice Cauchy domain. Then the left and the right formulations− of the PS-functional⊂ calculus define the same operators; i.e., we have M−1 1 −1 ℓ F (T )= ΠℓSL (s, T ) dsj ∂j F (s) 2π C ∂(U∩ j) ℓ Z X=0 (51) M−1 1 ℓ −1 = ∂j F (s) dsj ΠℓSR (s, T ). 2π C ∂(U∩ j) ℓ Z X=0 Proof. It is a consequence of the Theorem 3.18, when we approximate the intrinsic function F by the rational function Fn we have M−1 M−1 1 −1 ℓ 1 ℓ −1 ΠℓSL (s, T ) dsj ∂j Fn(s)= ∂j Fn(s) dsj ΠℓSR (s, T ) 2π C 2π C ∂(U∩ j) ℓ ∂(U∩ j) ℓ Z X=0 Z X=0 0,1 for T (Vn) and from the continuity of the PS-functional calculus we get the statement. ∈ BC 6. Formulations of the PS-functional calculus via the modified S-resolvent operators There is an alternative way to define the PS-functional calculus: instead of using the Cauchy formula for poly slice monogenic functions, we can use the slice monogenic integral representation of poly slice monogenic function as in Corollary 4.2. In order to apply this strategy we need to define the modified S-resolvent operators and their related modified S-resolvent equation. As we have seen the PS-resolvent operators are poly slice monogenic, while the modified S-resolvent operators are slice monogenic. We define the modified S-resolvent operator series. 0,1 n+1 Definition 6.1. Let T (Vn) and let s R with T < s and B (Vn). We call ∈B ∈ k k | | ∈B +∞ +∞ T mBs−m−1, and s−m−1BT m m m X=0 X=0 the modified left (resp. right) S-resolvent operator series expansions. 0,1 n+1 Theorem 6.2. Let T (Vn) and let s R with T < s and let B (Vn). ∈B ∈ k k | | ∈B (i) The modified left S-resolvent operator series equals +∞ T mBs−m−1 = (T 2 2Re(s)T + s 2 )−1(TB Bs). − − | | I − m X=0 (ii) The modified right S-resolvent operator series equals +∞ s−m−1BT m = (BT sB)(T 2 2Re(s)T + s 2 )−1. − − − | | I m=0 X Proof. It follows directly from the relations +∞ qmBs−1−m = (q2 2Re(s)q + s 2)−1(qB Bs), (52) − − | | − m=0 X 29 and +∞ s−1−mBqm = (Bq sB)(q2 2Re(s)q + s 2)−1, (53) − − − | | m=0 X which hold for s,q Rn+1 with q < s , replacing q by T and T < s . See [8] for more details. ∈ | | | | k k | | 0,1 Definition 6.3 (The modified S-resolvent operators). Let T (Vn) and let B (Vn). For ∈ B ∈ B s ρS(T ), we define the modified left S-resolvent operator as ∈ −1 −1 S (s, T ; B) := s(T ) (TB Bs), L −Q − and the modified right S-resolvent operator as −1 −1 S (s, T ; B) := (BT sB) s(T ) , R − − Q 2 2 where s(T ) := T 2Re(s)T + s . Q − | | I 0,1 Lemma 6.4. Let T (Vn) and let B (Vn). ∈B ∈B −1 (I) The modified left S-resolvent operator S (s, T ; B) is a (Vn)-valued right-slice monogenic L B function of the variable s on ρS(T ). −1 (II) The modified right S-resolvent operator S (s, T ; B) is a (Vn)-valued left-slice monogenic R B function of the variable s on ρS(T ). Proof. It is can be proved by a direct computation as in the case of the S-resolvent operators. Some more interesting relation can be obtained when B (Vn) commutes with T and this fact will be crucial in the sequel. ∈B 0,1 Lemma 6.5. Let T (Vn) and suppose that B (Vn) commutes with T . Let s ρS(T ). Then we have ∈ B ∈ B ∈ −1 −1 SL (s, T ; B)= BSL (s, T ), and −1 −1 SR (s, T ; B)= SR (s, T )B. Proof. Since TB = BT the statement follows from the fact that (TB Bs) = B(T s ) and −1 −1 − − I s(T ) B = B s(T ) . Q Q 0,1 Theorem 6.6. Let T (Vn) and suppose that B (Vn) commutes with T . Let s ρS(T ). The modified left S-resolvent∈ B operator satisfies the modified∈ B left S-resolvent equation ∈ S−1(s, T ; B)s TS−1(s, T ; B)= BS−1(s, T )s TBS−1(s, T )= B (54) L − L L − L and the modified right S-resolvent operator satisfies the modified right S-resolvent equation sS−1(s, T ; B) S−1(s, T ; B)T = sS−1(s, T )B S−1(s, T )BT = B. (55) R − R R − R Proof. Since 2Re(s) and s 2 are real, they commute with the operator T . Therefore | | T s(T )= s(T )T Q Q and in turn −1 −1 s(T ) T = T s(T ) . Q Q Thus we have −1 −1 −1 −1 S (s, T ; B)s TS (s, T ; B)= s(T ) (TB Bs)s + T s(T ) (TB Bs) L − L − Q − Q − −1 = s(T ) ( (TB Bs)s + T (TB Bs)) Q − − − −1 = s(T ) B s(T )= B, Q 30 Q where we have used the fact that B and T commute. The modified right S-resolvent equation follows by similar computations. 0,1 Theorem 6.7 (The modified S-resolvent equation). Let T (Vn) and suppose that B (Vn) ∈B ∈B commutes with T and let s,q ρS(T ) with q / [s]. Then the equation ∈ ∈ −1 −1 −1 −1 −1 −1 −1 SR (s, T )BSL (q, T )= SR (s, T )B BSL (q, T ) q s SR (s, T )B BSL (q, T ) s(q) − − − Q (56) holds true. Equivalently, it can also be written as −1 −1 −1 −1 −1 −1 −1 SR (s, T )BSL (q, T )= q(s) BSL (q, T ) SR (s, T )B q s BSL (q, T ) SR (s, T )B , Q − − − (57) 2 2 where s(q) := q 2Re(s)q + s . Q − | | Proof. We show that −1 −1 −1 −1 −1 −1 SR (s, T )BSL (q, T ) s(q) = SR (s, T )B BSL (q, T ) q s SR (s, T )B BSL (q, T ) , Q − − − (58) which is equivalent to (56). The modified left S-resolvent equation (54) implies −1 −1 BSL (q, T )q = BTSL (q, T )+ B. Applying this identity in the third and fifth equality and using the fact the TB = BT : −1 −1 S (s, T )BS (q, T ) s(q) R L Q = S−1(s, T )BS−1(q, T )(q2 2s q + s 2) R L − 0 | | = S−1(s, T ) BS−1(q, T )q q 2s S−1(s, T ) BS−1(q, T )q + s 2S−1(s, T )BS−1(q, T ) R L − 0 R L | | R L = S−1(s, T ) BTS−1(q, T )+ B q 2s S−1(s, T ) BTS−1(q, T )+ B + s 2S−1(s, T )BS−1(q, T ) R L − 0 R L | | R L = S−1(s, T )T BS−1(q, T )q + S−1(s, T )Bq 2s S−1(s, T )BTS−1(q, T ) 2s S−1(s, T )B R L R − 0 R L − 0 R + s 2S−1(s, T )BS−1(q, T ) | | R L = S−1(s, T )T BTS−1(q, T )+ B + S−1(s, T )Bq 2s S−1(s, T )BTS−1(q, T ) 2s S−1(s, T )B R L R − 0 R L − 0 R + s 2S−1(s, T )BS−1(q, T ). | | R L So, the above relation becomes −1 −1 −1 −1 −1 −1 S (s, T )BS (q, T ) s(q)= S (s, T )TBTS (q, T )+ S (s, T )TB + S (s, T )Bq R L Q R L R R 2s S−1(s, T )BTS−1(q, T ) 2s S−1(s, T )B + s 2S−1(s, T )BS−1(q, T ). − 0 R L − 0 R | | R L −1 −1 Now we replace the term SR (s, T )BT by sSR (s, T )B B, two times, in the above formula using the right S-resolvent equation (55) written as − S−1(s, T )BT = sS−1(s, T )B B. R R − We get the equality −1 −1 −1 −1 −1 −1 S (s, T )BS (q, T ) s(q)= S (s, T )BT TS (q, T )+ S (s, T )TB + S (s, T )Bq R L Q R L R R 2s S−1(s, T )BT S−1(q, T ) 2s S−1(s, T )B + s 2S−1(s, T )BS−1(q, T ) − 0 R L − 0 R | | R L and replacing (55) in the above equation one more time in two places indicated in parenthesis we get −1 −1 −1 −1 −1 −1 S (s, T )BS (q, T ) s(q)= sS (s, T )B B TS (q, T )+ sS (s, T )B B + S (s, T )Bq R L Q R − L R − R −1 −1 −1 2 −1 −1 2s0 sSR (s, T )B B SL (q, T ) 2s0SR (s, T )B + s SR (s, T )BSL (q, T ) − − − 31 | | so we get −1 −1 −1 −1 −1 S (s, T )BS (q, T ) s(q)= s S (s, T )BT S (q, T ) BTS (q, T ) R L Q R L − L + sS−1(s, T )B B + S−1(s, T )Bq R − R 2s sS−1(s, T )B B S−1(q, T ) 2s S−1(s, T )B + s 2S−1(s, T )BS−1(q, T ). − 0 R − L − 0 R | | R L Replacing one more time we have −1 −1 −1 −1 S (s, T )BS (q, T ) s(q)= s sS (s, T )B B S (q, T ) R L Q R − L BTS−1(q, T )+ sS−1(s, T )B B + S−1(s, T )Bq − L R − R 2s sS−1(s, T )B B S−1(q, T ) 2s S−1(s, T )B − 0 R − L − 0 R + s 2S−1(s, T )BS−1(q, T ) | | R L so we obtain −1 −1 2 −1 −1 −1 S (s, T )BS (q, T ) s(q)= s S (s, T )BS (q, T ) sBS (q, T ) R L Q R L − L BTS−1(q, T )sS−1(s, T )B B + S−1(s, T )Bq − L R − R 2s sS−1(s, T )BS−1(q, T ) + 2s BS−1(q, T ) 2s S−1(s, T )B − 0 R L 0 L − 0 R + s 2S−1(s, T )BS−1(q, T ). | | R L Now we gather some terms in order to get −1 −1 2 2 −1 −1 S (s, T )BS (q, T ) s(q) = (s 2s s + s )S (s, T )BS (q, T ) R L Q − 0 | | R L sBS−1(q, T ) BTS−1(q, T )+ sS−1(s, T )B B + S−1(s, T )Bq − L − L R − R + 2s BS−1(q, T ) 2s S−1(s, T )B 0 L − 0 R and −1 −1 2 2 −1 −1 S (s, T )BS (q, T ) s(q) = (s 2s s + s )S (s, T )BS (q, T ) R L Q − 0 | | R L sBS−1(q, T ) BTS−1(q, T )+ sS−1(s, T )B B + S−1(s, T )Bq − L − L R − R + 2s BS−1(q, T ) 2s S−1(s, T )B. 0 L − 0 R Since sBS−1(q, T ) + 2s BS−1(q, T )= sBS−1(q, T ) we have − L 0 L L −1 −1 2 2 −1 −1 S (s, T )BS (q, T ) s(q) = (s 2s s + s )S (s, T )BS (q, T ) R L Q − 0 | | R L + sBS−1(q, T ) BTS−1(q, T )+ sS−1(s, T )B B + S−1(s, T )Bq L − L R − R 2s S−1(s, T )B. − 0 R Recalling the S-resolvent equation −1 −1 BSL (q, T )q = BTSL (q, T )+ B we obtain −1 −1 2 2 −1 −1 S (s, T )BS (q, T ) s(q) = (s 2s s + s )S (s, T )BS (q, T ) R L Q − 0 | | R L + sBS−1(q, T ) BS−1(q, T )q + sS−1(s, T )B + S−1(s, T )Bq L − L R R 2s S−1(s, T )B − 0 R and so −1 −1 2 2 −1 −1 S (s, T )BS (q, T ) s(q) = (s 2s s + s )S (s, T )BS (q, T ) R L Q − 0 | | R L −1 −1 −1 −1 + sBSL (q, T ) BSL (q, T )q sSR (s, T )B + SR (s, T )Bq. −32 − Finally we obtain −1 −1 2 2 −1 −1 S (s, T )BS (q, T ) s(q) = (s 2s s + s )S (s, T )BS (q, T ) R L Q − 0 | | R L + sBS−1(q, T ) BS−1(q, T )q sS−1(s, T )B + S−1(s, T )Bq L − L − R R = (s2 2s s + s 2)S−1(s, T )BS−1(q, T ) − 0 | | R L + (S−1(s, T )B BS−1(q, T ))q s(S−1(s, T )B BS−1(q, T ) R − L − R − L 2 2 and since s 2s0s + s = 0 we get (58). With similar computations we can show that also (57) holds. − | | 0,1 Definition 6.8. (The modified S-resolvent operators) Let T (Vn) and s ρS(T ). We define the modified left S-resolvent operator of order ℓ N as ∈B ∈ ∈ ℓ −1 ℓ −1 T S (s, T ) := T s(T ) (T s ), (59) L − Q − I and the modified right S-resolvent operator of order ℓ N as ∈ −1 ℓ −1 ℓ S (s, T )T := (T s ) s(T ) T , (60) R − − I Q 2 2 where s(T ) := T 2Re(s)T + s . Q − | | I ℓ Remark 6.9. Observe that when T has commuting components the operator B := T commutes with T , so in this case, by Lemma 6.5, we have −1 ℓ ℓ −1 SL (s, T ; T )= T SL (s, T ), and −1 ℓ −1 ℓ SR (s, T ; T )= SR (s, T )T . We are ready for an alternative, but equivalent, definition of the PS-functional calculus. 0,1 Definition 6.10 (The PS-functional calculus (II)). Let T (Vn) for j S set dsj = ds( j) and let M N. Then we have the formulations (II) of the PS∈-functional B calculus.∈ We define − ∈ M−1 1 ℓ −1 M F (T ) := T SL (s, T ) dsj fℓ(s), for all F L (σS(T )), (61) 2π ∂(U∩Cj) ∈ PS Z Xℓ=0 and M−1 1 −1 ℓ M F (T ) := fℓ(s) dsj SR (s, T )T , for all F R (σS(T )). (62) 2π ∂(U∩Cj) ∈ PS Z Xℓ=0 Theorem 6.11. The definition of the PS-functional calculus (II) in (61) and (62) is well posed because the integrals depend neither on U nor on the imaginary unit j S. ∈ Proof. The Definitions (61) and (62) are well posed since they are based on the S-functional calculus and Lemma 6.4 that assures that the modified S-resolvent operators preserve the slice monogenicity. The linearity of the PS-functional calculus (II) is a direct consequence of the definition. 0,1 Proposition 6.12. Let T (Vn) and M N. M ∈B ∈ (1) If F, G (σS(T )), then we have: ∈ PSL (F + G)(T )= F (T )+ G(T ), (F λ)(T )= F (T )λ, for all λ Rn. 33 ∈ M (2) If F, G (σS(T )), then we have: ∈ PSR (F + G)(T )= F (T )+ G(T ), (λF )(T )= λF (T ), for all λ Rn. ∈ 0,1 Theorem 6.13. Let T (Vn) and let M N. (I) Then, when F is the∈B series expansion in∈ (49), we have M−1 M−1 ∞ 1 ℓ −1 ℓ u T SL (s, T ) dsj fℓ(s)= T T Ak,u. 2π C ∂(U∩ j) ℓ ℓ u=0 Z X=0 X=0 X (II) Then, when F is the series expansion in (50), we have M−1 M−1 ∞ 1 −1 ℓ u ℓ fℓ(s) dsj SR (s, T )T = Ak,uT T . 2π C ∂(U∩ j) ℓ ℓ u=0 Z X=0 X=0 X n We recall that T = T ejTj. 0 − j=1 Proof. It is a consequenceP of the S-functional calculus. 0,1 Theorem 6.14 (The intrinsic PS-functional calculus (II)). Let T (Vn) and set dsj = ds( j) M ∈ BC − and let F be an intrinsic poly monogenic function F (σS(T )), for M N. Then the left and the right formulations of the PS-functional calculus∈ (II) PN define the same operator,∈ i.e., we have M−1 M−1 1 ℓ −1 1 −1 ℓ F (T )= T SL (s, T ) dsj fℓ(s)= fℓ(s) dsj SR (s, T )T . (63) 2π ∂(U∩Cj) 2π ∂(U∩Cj) Z Xℓ=0 Z Xℓ=0 0,1 Proof. It is a consequence of the fact that T (Vn) and of the S-functional calculus for intrinsic functions, i.e., ∈ BC 1 −1 1 −1 fℓ(T )= SL (s, T ) dsj fℓ(s)= fℓ(s) dsj SR (s, T ), 2π C 2π C Z∂(U∩ j) Z∂(U∩ j) holds. 7. Equivalence of the two definitions of the PS-functional calculus and the product rules According to Proposition 3.11 we have some possibilities to obtain the pointwise product of a poly slice monogenic function and a slice monogenic function that preserves the poly slice monogenicity. According to such proposition we obtain the related product rules. Using a product that takes out of the class of poly slice monogenic function of a given order, by Corollary 4.18, we obtain a more general product rule. First we establish the equivalence of the definitions of the PS-functional calculus (I) and (II). 0,1 Theorem 7.1 (Equivalence of the definitions of the PS-functional calculus). Let T (Vn), N S 1 −1 −1 ∈ B M , for j , set dsj = ds( j), ∂j := 2 (∂u +j∂v). Let ΠℓSL (s, T ) and ΠℓSR (s, T ) be the ΠS-resolvent∈ operators.∈ Then we− have M−1 1 −1 ℓ F (T )= ΠℓSL (s, T ) dsj ∂j F (s) 2π C ∂(U∩ j) ℓ Z X=0 (64) M−1 1 ℓ −1 M = T SL (s, T ) dsj fℓ(s), for all F L (σS(T )), 2π C ∈ PS Z∂(U∩ j) ℓ=0 X 34 and M−1 1 ℓ −1 F (T )= ∂j F (s) dsj ΠℓSR (s, T ) 2π C ∂(U∩ j) ℓ Z X=0 (65) M−1 1 −1 ℓ M = fℓ(s) dsj SR (s, T )T , for all F R (σS(T )). 2π ∂(U∩Cj) ∈ PS Z Xℓ=0 Proof. It is a consequence of Runge’s theorems for slice monogenic and poly slice monogenic func- tions and the continuity properties of the two formulations of the PS-functional calculus. As a corollary of the above theorem we have the four possible representations of the operator M f(T ) when F is an intrinsic poly slice monogenic function, i.e., when F (σS(T )). ∈ PN Theorem 7.2 (The formulations of PS-functional calculus for intrinsic functions). Let T 0,1 N S 1 N S∈ (Vn), M , for j , set dsj = ds( j), ∂j := 2 (∂u +j∂v) and Let M , for j , BC ∈ ∈ 1 − −1 −1 ∈ ∈ set dsj = ds( j) and ∂j := 2 (∂u +j∂v). Let ΠℓSL (s, T ) and ΠℓSR (s, T ) be the ΠS-resolvent − M operators. Then, for every F (σS(T )), we have ∈ PN M−1 1 −1 ℓ F (T )= ΠℓSL (s, T ) dsj ∂j F (s) 2π C ∂(U∩ j) ℓ Z X=0 M−1 1 ℓ −1 = T SL (s, T ) dsj fℓ(s) 2π C ∂(U∩ j) ℓ Z X=0 (66) M−1 1 ℓ −1 = ∂j F (s) dsj ΠℓSR (s, T ) 2π ∂(U∩Cj) Z Xℓ=0 M−1 1 −1 ℓ = fℓ(s) dsj SR (s, T )T . 2π ∂(U∩Cj) Z Xℓ=0 Proof. It is a consequence of Runge’s theorems for slice monogenic and poly slice monogenic func- tions and the continuity properties of the two formulations of the PS-functional calculus. The following lemma will be crucial in the proof of the product rules. n+1 Lemma 7.3 (See [8]). Let B (Vn). For any q,s R with q / [s] and let f be an intrinsic ∈ B ∈ ∈ slice monogenic function and let U be a slice Cauchy domain with U dom(f), setting ⊂ −1 2 2 −1 s(q) := (q 2Re(s)q + s ) Q − | | then 1 −1 f(s) dsj (sB Bq) s(q) = Bf(q) 2π C − Q Z∂(U∩ j) for any q that belongs to U and any j S. ∈ We start with Proposition 3.11 to obtain the first set of product rules for the PS-functional calculus. 0,1 Theorem 7.4 (Product rule (first case)). Let T (Vn) and M N. M ∈ BC ∈ (Ia) Let F (σS(T )) and g L(σS(T )). Then (F g)(T )= F (T )g(T ). ∈ PN M ∈ SM (Ib) Let F L (σS(T )) and g (σS(T )). Then (gF )(T )= g(T )F (T ). ∈ PM M ∈N (IIa) Let F (σS(T )) and g R(σS(T )). Then (gF )(T )= g(T )F (T ). ∈ PN M ∈ SM (IIb) Let F R (σS(T )) and g (σS(T )). Then (F g)(T )= F (T )g(T ). ∈ PM ∈N 35 M (III) Let F (σS (T )) and g (σS(T )). Then we have ∈ PN ∈N (gF )(T ) = (F g)(T )= F (T )g(T )= g(T )F (T ). Proof. We will show point (Ia). The other point follows in much the same way. So for F M ∈ (σS(T )) and g L(σS(T )) using the PS-functional calculus (II) for F and the S- PNfunctional calculus for∈g we SM have 1 −1 g(T ) := SL (s, T ) dsj g(s), for all f L(σS(T )). 2π C ∈ SM Z∂(U∩ j) Let Uq and Us be bounded slice Cauchy domains that contain σS(T ) such that Uq Us and ⊂ Us dom(F ) dom(g). The subscripts q and s refer to the respective variable of integration in ⊂ ∩ the following computation. We choose j S and we set Γs := ∂(Us Cj) and Γq := ∂(Uq Cj) for M ∈ ∩ ∩ neatness. So for F (σS(T )) by Theorem 6.14 we can represent F as ∈ PN M−1 1 −1 k F (T )= fk(s) dsj SR (s, T )T . 2π ∂(U∩Cj) Z Xk=0 When we consider the product M−1 1 −1 k 1 −1 F (T )g(T )= fk(s) dsj SR (s, T )T SL (q, T ) dqj g(q) 2π Γs 2π Γq Z Xk=0 Z so we write M−1 1 1 k F (T )g(T )= f (s) ds S−1(s, T )T S−1(q, T ) dq g(q). (67) 2π k j 2π R L j Γs k Γq Z X=0 Z k Now we use the modified S-resolvent equation with B = T that commute with T , since the paravector operator T has commuting components we have the modified S-resolvent equation −1 k −1 −1 k k −1 −1 k k −1 −1 S (s, T )T S (q, T )= S (s, T )T T S (q, T ) q s S (s, T )T T S (q, T ) s(q) R L R − L − R − L Q h i (68) we replace it (68) in (67) to get M−1 1 1 −1 k k −1 F (T )g(T )= fk(s) dsj SR (s, T )T T SL (q, T ) q 2π Γs 2π Γq − Z Xk=0 Z h −1 k k −1 −1 s S (s, T )T T S (q, T ) s(q) dq g(q) − R − L Q j but also i M−1 1 1 −1 k −1 F (T )g(T )= fk(s) ds S (s, T )T q s(q) dq g(q) 2π j 2π R Q j Γs k Γq Z X=0 Z M−1 1 1 k −1 −1 fk(s) dsj T SL (q, T )q s(q) dqj g(q) − 2π Γs 2π Γq Q Z Xk=0 Z M−1 1 1 −1 k −1 fk(s) dsj sSR (s, T )T s(q) dqj g(q) − 2π Γs 2π Γq Q Z Xk=0 Z M−1 1 1 k −1 −1 + fk(s) ds sT S (q, T ) s(q) dq g(q). 2π j 2π L Q j ZΓs k=0 ZΓq X 36 Now the integrals with the right S-resolvent operator are zero, in fact we can write them as M−1 1 1 −1 k −1 fk(s) ds S (s, T )T q s(q) dq g(q) 2π j 2π R Q j Γs k Γq Z X=0 Z M−1 1 −1 k 1 −1 = fk(s) ds S (s, T )T q s(q) dq g(q) = 0 2π j R 2π Q j Γs k Γq Z X=0 Z h i −1 by the Cauchy theorem because q s(q) and g(q) are slice monogenic. Similarly it is Q M−1 1 1 −1 k −1 fk(s) dsj sSR (s, T )T s(q) dqj g(q) = 0 −2π Γs 2π Γq Q Z Xk=0 Z so we remain with M−1 1 1 k −1 −1 F (T )g(T )= fk(s) dsj T SL (q, T )q s(q) dqj g(q) −2π Γs 2π Γq Q Z Xk=0 Z M−1 1 1 k −1 −1 + fk(s) dsj sT SL (q, T ) s(q) dqj g(q) 2π Γs 2π Γq Q Z Xk=0 Z and also we obtain M−1 1 1 k −1 k −1 −1 F (T )g(T )= fk(s) dsj sT SL (q, T ) T SL (q, T )q s(q) dqj g(q) 2π Γs 2π Γq − Q Z Xk=0 Z using Fubini’s theorem we get M−1 1 1 k −1 k −1 −1 F (T )g(T )= fk(s) dsj sT SL (q, T ) T SL (q, T )q s(q) dqj g(q) 2π Γq 2π Γs − Q Z h Z Xk=0 i k −1 and by Lemma 7.3, setting B := T SL (q, T ), we obtain M−1 1 k −1 F (T )g(T )= T SL (q, T )fk(q) dqj g(q) 2π Γq Z Xk=0 and finally M−1 1 k −1 F (T )g(T )= T SL (q, T ) dqj fk(q)g(q) 2π Γq Z Xk=0 M−1 k = T (fkg)(T ) = (F g)(T ), k X=0 and this concludes the proof. Based on the product of poly slice monogenic functions in Corollary 4.18 we prove the following product rule. This product rule requires that the paravector operator T has commuting compo- nents. 0,1 Theorem 7.5 (Product rule (second case)). Let T (Vn) and M,N N. N M ∈ BC ∈ (I) Let F (σS(T )) and G (σS(T )), then we have ∈NL ∈ PSL (F G)(T )= F (T )G(T ). (69) 37 N M (II) Let F (σS(T )) and G (σS(T )), then we have ∈ PSR ∈NR (F G)(T )= F (T )G(T ). (70) Proof. We reason as in Theorem 7.4 taking the same contours of integration and pointing out just the differences. We show point (I). The other point (II) follows in much the same way. Let Uq and Us be bounded slice Cauchy domains that contain σS(T ) such that Uq Us and ⊂ Us dom(F ) dom(G). The subscripts q and s refer to the respective variable of integration in ⊂ ∩ the following computation. We choose j S and we set Γs := ∂(Us Cj) and Γq := ∂(Uq Cj) for neatness. So by Theorem 6.14 we can represent∈ F as ∩ ∩ N−1 1 −1 k N F (T )= fk(s) ds S (s, T )T , for all F (σS(T )) 2π j R ∈ PN Γs k Z X=0 and M−1 1 ℓ −1 M G(T )= T S (q, T ) dq gℓ(q), for all F (σS(T )) 2π L j ∈ PSL Γq ℓ Z X=0 so when we consider the product N−1 M−1 1 k 1 ℓ F (T )G(T )= f (s) ds S−1(s, T )T T S−1(q, T ) dq g (q) 2π k j R 2π L j ℓ Γs k Γq ℓ Z X=0 Z X=0 and also. N−1 M−1 1 1 k+ℓ F (T )G(T )= f (s) ds S−1(s, T )T S−1(q, T ) dq g (q). 2π k j 2π R L j ℓ Γs k Γq ℓ Z X=0 Z X=0 We use the modified S-resolvent equation to get N−1 M−1 1 1 −1 k+ℓ −1 F (T )G(T )= fk(s) dsj SR (s, T )T q s(q) dqj gℓ(q) 2π Γs 2π Γq Q Z Xk=0 Z Xℓ=0 N−1 M−1 1 1 k+ℓ −1 −1 fk(s) dsj T SL (q, T )q s(q) dqj gℓ(q) − 2π Γs 2π Γq Q Z Xk=0 Z Xℓ=0 N−1 M−1 1 1 −1 k+ℓ −1 fk(s) ds sS (s, T )T s(q) dq gℓ(q) − 2π j 2π R Q j Γs k Γq ℓ Z X=0 Z X=0 N−1 M−1 1 1 k+ℓ −1 −1 + fk(s) dsj sT SL (q, T ) s(q) dqj gℓ(q). 2π Γs 2π Γq Q Z Xk=0 Z Xℓ=0 Also here the two integrals that contain the right S-resolvent operators are zero so we obtain N−1 M−1 1 1 k+ℓ −1 −1 F (T )G(T )= fk(s) dsj T SL (q, T )q s(q) dqj gℓ(q) − 2π Γs 2π Γq Q Z Xk=0 Z Xℓ=0 N−1 M−1 1 1 k+ℓ −1 −1 + fk(s) ds sT S (q, T ) s(q) dq gℓ(q). 2π j 2π L Q j ZΓs k=0 ZΓq ℓ=0 X 38 X Using Fubini’s theorem we finally have F (T )G(T )= N−1 M−1 1 1 k+ℓ −1 k+ℓ −1 −1 = fk(s) dsj sT SL (q, T ) T SL (q, T )q s(q) dqj gℓ(q) 2π Γq 2π Γs − Q Z h Z Xk=0 Xℓ=0 i k+ℓ −1 setting B := T SL (q, T ) in Lemma 7.3 we get N−1 M−1 1 k+ℓ −1 F (T )G(T )= T SL (q, T )fk(q) dqj gℓ(q) 2π Γq Z Xk=0 Xℓ=0 N−1 M−1 1 k+ℓ = T S−1(q, T ) dq (f g )(q) 2π L j k ℓ Γq k ℓ Z X=0 X=0 N−1 M−1 k+ℓ = T (fkgℓ)(T ) = (F G)(T ), Xk=0 Xℓ=0 and this concludes the proof. 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