Non commutative functional calculus: unbounded operators Fabrizio Colombo Graziano Gentili 8 Dipartimento di Matematica Dipartimento di Matematica 0 Politecnico di Milano Universit´a di Firenze 0 2 Via Bonardi, 9 Viale Morgagni, 67 A n 20133 Milano, Italy Firenze, Italy a [email protected] [email protected]fi.it J 5 Irene Sabadini Daniele C. Struppa 2 Dipartimento di Matematica Department of Mathematics ] Politecnico di Milano and Computer Sciences P Via Bonardi, 9 Chapman University S 20133 Milano, Italy Orange, CA 92866 USA, . h [email protected] [email protected] t a m [ Abstract 2 Inarecentwork,[3],wedevelopedafunctionalcalculusforboundedoperatorsdefinedon v 2 quaternionicBanachspaces. Inthispaperweshowhowthe resultsfrom[3]canbe extended 9 to the unbounded case, and we highlight the crucial differences between the two cases. In 5 particular,wededuceaneweigenvalueequation,suitablefortheconstructionofafunctional 3 calculus for operators whose spectrum is not necessarily real. . 8 0 AMS Classification: 47A10, 47A60, 30G35. 7 Key words: functional calculus, spectral theory, bounded and unbounded operators. 0 : v i 1 Introduction X ar Let V be a Banach space over the skew field H of quaternions, and let E be the Banach space of right linear operators acting on it. In a recent paper, [3], we have shown that if T is a bounded operator in E the standard eigenvalue problems (i.e. the search of the values λ ∈ H for which λI−T or Iλ−T arenot invertible) donot lead to a usefulfunctionalcalculus. Themain reason forthis difficulty consists inthefact thattheinverseof theoperator λI−T does notcorrespond, unlikewhathappensinthecomplexcase,toanintegralkernelofCauchytypefornon-commuting variables. In order to overcome this difficulty, we introduced a totally new eigenvalue problem, whichwecalled theS−eigenvalueproblem,andwhich,inthecommutative case, isidenticalwith the usual problem. The key observation which allowed the study of such an eigenvalue problem, was the recent development of slice regular functions, [6], [7]. These functions are likely to be the appropriate generalization of standard holomorphic functions, and they can be represented through a new, non-commutative, Cauchy kernel. It is by using this kernel, that one is able to develop a functional calculus for bounded operators in E. The specific techniques utilized in [3] did not allow us to tackle the important study of unbounded operators. These operators, however, are of great importance both in mathematics and in physics, because they can be 1 used to write functions of operators, such as the exponential of a closed operator, for which the corresponding power series expansion is not convergent and thus not suitable to define it. In particular, in quantum mechanics, the exponential function of an operator defines the evolution operator associated to Schro¨dinger equation. As it is well known, Quantum Mechanics can be formulated in the real, complex and quaternionic setting (see [1]), and for this reason it is important to introduce a quaternionic version of the functional calculus to allow the study of exponentials for quaternionic operators. Finally, we point out that the spectral theory can be extended to the case of n-tuples of operators. For the complex case, the reader is referred to [10], [11] and, by using the notion of monogenic functions with values in a Clifford algebra (see [2] for the case of one variable and [4] for the case of several variables), to [8] and the references therein. Acknowledgements The first and third authors are grateful to Chapman University for the hospitality during the period in which this paper was written. They are also indebted to G.N.S.A.G.A. of INdAM and the Politecnico di Milano for partially supporting their visit. 2 Preliminary results 2.1 Slice regular functions In this section we summarize the basic definitions from [6], [7], that we need in the sequel. LetHbetherealassociative algebraofquaternionswithrespecttothebasis{1,i,j,k} satisfying the relations i2 = j2 = k2 = −1,ij = −ji = k, jk = −kj = i, ki= −ik = j. We will denote a quaternion q as q = x + ix + jx + kx , x ∈ R, its conjugate as q¯ = 0 1 2 3 i x −ix −jx −kx , and we will write |q|2 = qq. 0 1 2 3 Let S be the sphere of purely imaginary unit quaternions, i.e. S= {q = ix +jx +kx | x2+x2+x2 = 1}. 1 2 3 1 2 3 Definition 2.1. Let U ⊆ H be an open set and let f : U → H be a real differentiable function. Let I ∈ S and let f be the restriction of f to the complex line L := R+IR passing through 1 I I and I. We say that f is a slice left regular function if for every I ∈S 1 ∂ ∂ +I f (x+Iy) = 0, I 2 (cid:18)∂x ∂y(cid:19) and we say it is slice right regular if for every I ∈ S 1 ∂ ∂ + I f (x+Iy) = 0. I 2 (cid:18)∂x ∂y (cid:19) Remark 2.2. Slice left regular functions on U ⊆ H form a right vector space R(U) and slice right regular functions on U ⊆ H form a left vector space. It is not true, in general, that the product of two regular functions is regular. In fact, every slice regular function can be represented as a power series, [7]: 2 Theorem 2.3. If B(0,R) is a ball centered in the origin with radius R > 0 and f : B :→ H is a slice left regular function, then f has a series expansion of the form +∞ 1 ∂nf f(q)= qn (0) n!∂xn nX=0 converging on B. Analogously, if f is slice right regular it can be expanded as +∞ 1 ∂nf f(q) = (0)qn. n!∂xn nX=0 Remark 2.4. An analogue statement holds for regular functions in an open ball centered in p ∈ R. 0 Definition 2.5. Let E be a bilateral quaternionic Banach space. A function f : H → E is said to be slice left regular if there exists an open ball B(0,r) ⊂ H and a sequence {a } ⊂ E n of elements of the Banach space such that, for every point q ∈ B(0,r) the function f(q) can be represented by the following series +∞ f(q)= qna , a ∈ E, (1) n n nX=0 converging in the norm of E for any q such that |q| < r. Analogously, f is said to be slice right regular if it can be expanded as +∞ f(q)= a qn, a ∈ E, (2) n n nX=0 Remark 2.6. From now on we will not specify whether we are considering left or right regular functions, since the context will clarify it. Definition 2.7. Let V be a bilateral vector space on H. A map T : V → V is said to be right linear if T(u+v) = T(u)+T(v), T(us) = T(u)s, for all s ∈ H and all u,v ∈ V. Remark 2.8. The set End(V) of right linear maps on V is a ring with respect to the usual sum of operators and with respect to the composition of operators defined in the usual way: for any two operators T,S ∈End(V) we have (TS)(u) = T(S(u)), ∀u∈ V. In particular, we have the identity operator I(u) = u, for all u ∈ V and setting T0 = I we can define powers of a given operator T ∈End(V): Tn = T(Tn−1) for any n ∈ N. An operator T is said to be invertible if there exists S such that TS = ST = I and we will write S = T−1. The set End(V) is a bilateral vector space on H with respect to the products by a scalar defined by (aT)(v) := aT(v) and (Ta)(v) := T(av). 3 2.2 The case of bounded operators Definition 2.9. Let V be a bilateral quaternionic Banach space. We will denote by B(V) the vector space of all right linear bounded operators on V. It is easy to verify that B(V) is a Banach space endowed with its natural norm. Definition 2.10. An element T ∈ B(V) is said to be invertible if there exists T′ ∈ B(V) such that TT′ = T′T = I. It is obvious that the set of all invertible elements of B(V) is a group with respect to the composition of operators defined in B(V). Let T be a linear quaternionic operator. There are two natural eigenvalue problems associated to T. The first, which one could call the left eigenvalue problem consists in the solution of equation T(v) = λv, and the second, which is called right eigenvalue problem, and consists in the solution of the equation T(v) = vλ. As it was shown in [3], none of them is useful to define a functional calculus, as another operator is the one associated to the notion of spectrum. Definition 2.11. (The S-resolvent operator series) Let T ∈ B(V) and let s ∈ H. We define the S-resolvent operator series as S−1(s,T):= Tns−1−n (3) Xn≥0 for kTk ≤ |s|. Theorem 2.12. (See [3]) Let T ∈ B(V) and let s ∈ H. Assume that s is such that T −sI is invertible. Then S(s,T) = (T −sI)−1s(T −sI)−T (4) is the inverse of S−1(s,T). Moreover, we have Tns−1−n = −(T2−2TRe[s]+|s|2I)−1(T −sI), (5) nX≥0 for kTk ≤ |s|. Definition 2.13. (The S-resolvent operator) Let T ∈ B(V) and let s ∈ H. We define the S-resolvent operator as S−1(s,T) := −(T2−2Re[s]T +|s|2I)−1(T −sI). (6) Definition 2.14. (The S-spectrum) Let T : V → V be a linear quaternionic operator on the Banach space V. We define the S-spectrum σ (T) of T related to the S-resolvent operator (6) S as: σ (T) = {s ∈ H : T2−2 Re[s]T +|s|2I is not invertible}. S The following theorem summarizes the properties of the S-spectrum (see [3]): Theorem 2.15. Let T ∈ B(V). Then: 1. The S-spectrum σ (T) is a compact nonempty set contained in { s ∈H : |s|≤ kTk }. S 2. If p = p + p I ∈ H \R is an S-eigenvalue of T with I ∈ S and p ,p ∈ R, then all the 0 1 0 1 elements of the sphere p +p S are S-eigenvalues of T. 0 1 Therefore, the S-spectrum σ (T) is a union of real points and 2-spheres. S 4 Let S3 = {q ∈H : |q|= 1} denote the unit sphere of H. For any set A ⊆ H, let us define the circularization of A as set circ(A) := x+yS3. x+[yI∈A The following definition will identify an important class of open sets. Definition 2.16. Let T ∈B(V). Let U ⊂ H be an open set such that (i) ∂(U ∩L ) is union of a finite number of rectifiable Jordan curves for every I ∈ S, I (ii) U contains the circularization of the S-spectrum σ (T). S A function f : H → H is said to be locally regular on σ (T) if there exists an open set U ⊂ H, S as above, on which f is regular. We will denote by R the set of locally regular functions on σ (T). σS(T) S Theorem 2.17. Let T ∈ B(V) and f ∈ R . Let U ⊂ H be an open set as in Definition 2.16 σS(T) and let U = U ∩L for I ∈ S. Then the integral I I 1 S−1(s,T) ds f(s) (7) I 2π Z ∂UI does not depend on the choice of the imaginary unit I and on the open set U. Theprecedingresultallowstogivethefollowingdefinitionwhichoffersanewfunctionalcalculus: Definition 2.18. Let T ∈ B(V) and f ∈ R . Let U ⊂ H be an open set as in Definition σS(T) 2.16, and set U = U ∩L for I ∈ S. We define I I 1 f(T)= S−1(s,T) ds f(s). (8) I 2π Z ∂UI 3 Functional calculus for unbounded operators We note thatif T is aclosed operator, theseries Tns−1−n does notconverge. Toovercome n≥0 thisdifficulty,weobservethattherighthandsidePofformula(5)containsthecontinuousoperator (T2−2TRe[s]+|s|2I)−1. From an heuristical point of view, the composition (T2−2TRe[s]+ |s|2I)−1(T −sI) gives a bounded operator, for suitable s. We will consider closed and densely defined operators. Definition 3.1. Let V be a quaternionic Banach space. We consider the linear closed densely defined operator T : D(T)⊂ V → V where D(T) denotes the domain of T. Let us assume that 1) D(T) is dense in V, 2) T −sI is densely defined in V, 3) D(T2) ⊂ D(T) is dense in V, 4) T2−2TRe[s]+|s|2I is one-to-one with range V. The S-resolvent operator is defined by S−1(s,T) = −(T2−2TRe[s]+|s|2I)−1(T −sI). (9) 5 Remark 3.2. We observe that, in principle, it is necessary also the following assumption: 5) the operator (T2 − 2TRe[s] + |s|2I)−1(T − sI) is the restriction to the dense subspace D(T) of V of a bounded linear operator. However this assumption is automatically fulfilled since it follows from the identity (T2−2TRe[s]+|s|2I)−1(T −sI)= T(T2−2TRe[s]+|s|2I)−1−(T2−2TRe[s]+|s|2I)−1sI, which is a consequence of (T2−2TRe[s]+|s|2I)−1T = T(T2−2TRe[s]+|s|2I)−1, that can be easily verified applying on the left to both sides the operator T2−2TRe[s]+|s|2I. Definition 3.3. Let T : D(T) ⊂ V → V be a linear closed densely defined operator as in Definition 3.1. We define the S-resolvent set of T to be the set ρ (T)= {s ∈H such that S−1(s,T) exists and it is in B(V)}. (10) S We define the S- spectrum of T as the set σ (T) = H\ρ (T). (11) S S Theorem 3.4. (Structure of the spectrum) Let T : V → V be a closed operator such that σ (T) 6= 0. If p = p +p I ∈ H\R is an S-eigenvalue of T with I ∈ S and p ,p ∈ R, then all S 0 1 0 1 the elements of the sphere p +p S are S-eigenvalues of T. The S-spectrum σ (T) is a union 0 1 S of real points and 2-spheres. Proof. The proof is analogous to the one of the bounded case. It immediately follows from the structure of the S-eigenvalue equation (T2−2TRe[s]+|s|2I)v = 0. Theorem 3.5. Let V be a quaternionic Banach space let T be a closed linear quaternionic operator on V. Let s ∈ ρ (T). Then the S-resolvent operator defined in (6) satisfies the S equation S−1(s,T)s−TS−1(s,T) = I. Proof. It follows by direct computation. Indeed, replacing (6) in the above equation we have −(T2−2Re[s]T +|s|2I)−1(T −sI)s+T(T2−2Re[s]T +|s|2I)−1(T −sI)= I. (12) Observe that T(T2−2Re[s]T +|s|2I)−1(T −sI) is a bounded operator because it is the sum of two bounded operators. Applying now (T2 −2Re[s]T +|s|2I) to both hands sides of (12), we get −(T −sI)s+(T2−2Re[s]T +|s|2I)T(T2−2Re[s]T +|s|2I)−1(T −sI)= T2−2Re[s]T +|s|2I. Since T and T2−2Re[s]T +|s|2I commute, we obtain the identity −(T −sI)s+T(T −sI)= T2−2Re[s]T +|s|2I which proves the statement. 6 Definition 3.6. Let s ∈ ρ (T). The equation S S−1(s,T)s−TS−1(s,T) = I (13) will be called the S-resolvent equation. Theorem 3.7. Let T be a closed linear operator on a quaternionic Banach space V. Assume that s ∈ ρ (T). Then the S-resolvent operator can be represented by S S−1(s,T) = (Re[s]I −T)−n−1(Re[s]−s)n (14) nX≥0 if and only if |Im[s]| k(Re[s]I −T)−1k < 1. (15) Proof. The equality −1 (T2−2TRe[s]+|s|2I)−1 = (T −Re[s]I)2+|Im[s]|2I (cid:16) (cid:17) −1 = (T −Re[s]I)2(I +|Im[s]|2(T −Re[s]I)−2) (cid:16) (cid:17) = I +|Im[s]|2(T −Re[s]I)−2 −1(T −Re[s]I)−2 (cid:0) (cid:1) = (−1)n|Im[s]|2n(T −Re[s]I)−2n(T −Re[s]I)−2, Xn≥0 yields S−1(s,T) = (−1)n+1|Im[s]|2n(T −Re[s]I)−2n−1(I +(T −Re[s]I)−1Im[s]) nX≥0 = (−1)n|Im[s]|2n(Re[s]I −T)−2n−1(I +(Re[s]I −T)−1(Re[s]−s)) nX≥0 and since (−1)n|Im[s]|2n = (Re[s]−s)2n we have S−1(s,T) = (Re[s]I −T)−2n−1(Re[s]−s)2n+ (Re[s]I −T)−2n−2(Re[s]−s)2n+1 nX≥0 nX≥0 = (Re[s]I −T)−1 (Re[s]I −T)−n(Re[s]−s)n nX≥0 which converges in B(V) if and only if (15) holds. Let V be a quaternionic Banach space and let T : D(T) ⊂ V → V be a linear operators. If at least one of its components is an unbounded operator then its resolvent is not defined at infinity. It is therefore natural to consider closed operators T for which the resolvent S−1(s,T) is not defined at infinity and to define the extended spectrum as σ (T):= σ (T)∪{∞}. S S Let us consider H = H∪{∞} endowed with the natural topology: a set is open if and only if it is union of open discs D(q,r) with center at points in q ∈ H and radius r, for some r, and/or union of sets the form {q ∈ H | |q| > r}∪{∞} = D′(∞,r)∪{∞}, for some r. 7 Definition 3.8. We say that f is regular function at ∞ if f(q) is an regular function in a set D′(∞,r) and lim f(q) exists and it is finite. We define f(∞) to be the value of this limit. q→∞ Definition 3.9. Let T : D(T) ⊂ V → V be a linear closed operator as in Definition 3.1 . Let U ⊂ H be an open set such that (i) ∂(U ∩L ) is union of a finite number of rectifiable Jordan curves for every I ∈ S, I (ii) U contains the circularization of the S-spectrum σ (T). S A function f is said to be locally regular on σ (T) if it is regular an open set U ⊂ H as above S and at infinity. We will denote by R the set of locally regular functions on σ (T). σS(T) S Consider α∈ H and the homeomorphism Φ : H → H for α ∈ H defined by p = Φ(s) =(s−α)−1, Φ(∞)= 0, Φ(α) = ∞. Definition 3.10. Let T : D(T) → V be a linear closed operator as in Definition 3.1 with ρ (T)∩R 6= ∅ and suppose that f ∈ R . Let us consider S σS(T) φ(p):= f(Φ−1(p)) and the operator A:= (T −kI)−1, for some k ∈ ρ (T)∩R. S We define f(T)= φ(A). (16) Remark 3.11. Observe that, if α= k ∈ R, we have that: i) the function φ is regular because it is the composition of the function f which is regular and Φ−1(p)= p−1+k which is regular with real coefficients; ii) in the case k ∈ ρ (T)∩R we have that (T −kI)−1 = −S−1(k,T). S To prove the fundamental Theorem 3.13 we need the following identities. Lemma 3.12. Let s, p ∈ H and k ∈ R such that p = (s−k)−1. Then the following identities hold s |p|2 = k|p|2+p , (17) 0 0 |p|2|s|2 = k2|p|2+2p k+1, (18) 0 p (2k−2s +p−1) = −p−2, (19) 0 |p|2 [s+(|s|2−k2−kp−1)(2k−2s +p−1)−1](2k−2s +p−1)p = 0. (20) 0 0 8 Proof. Identity (17) follows from Re[s−k]= Re[p−1]= Re[p|p|−2] from which we have s −k = p |p|−2. 0 0 Identity (18) follows from the chain of identities |s|2 = ss= (k+p−1)(k+p−1) =(k+p−1)(k+p−1) 2p 1 = k2+k(p−1+p−1)+p−1p−1 = k2+k 0 + . |p|2 |p|2 To prove (19) we consider the chain of identities p (2k−2s +p−1) = (2k−2s +p−1)p−1 0 |p|2 0 = (2k−2s +s−k)(s−k) = −(s−k)2 = −p−2. 0 Finally (20) follows from [s+(|s|2−k2−kp−1)(2k−2s +p−1)−1](2k−2s +p−1)p 0 0 = s(2(k−s )+p−1)p+(|s|2−k2−kp−1)p 0 now using (17) and (18) we get s(2(k−s )+p−1)p+(|s|2−k2−kp−1)p 0 p 2p k+1 0 0 = s −2 p+1 + p−k |p|2 |p|2 (cid:16) (cid:17) p 2p k+1 = (k+(p)−1) −2 0 p+1 + 0 p−k |p|2 |p|2 (cid:16) (cid:17) = −2p p−1p−1+p−1+p−1 = 0. 0 Theorem 3.13. If k ∈ ρ (T) ∩R 6= ∅ and Φ, φ are as above, then Φ(σ (T)) = σ (A) and S S S the relation φ(p) := f(Φ−1(p)) determines a one-to-one correspondence between f ∈ R and σS(T) φ∈ R . σS(A) Proof. First we consider the case p ∈ σ (A) and p 6= 0. Recall that S S−1(p,A) = −(A2−2ARe[p]+|p|2I)−1(A−pI), from which we obtain (A2−2ARe[p]+|p|2I)S−1(p,A) = −(A−pI). Let us apply the operator A−2 on the left to get (I −2Re[p]A−1+A−2|p|2)S−1(p,A) = −(A−1−A−2p). 9 Now we use the relations A−1 = T −kI, A−2 = T2−2kT +k2I (21) to get (I −2Re[p](T −kI)+(T2−2kT +k2I)|p|2)S−1(p,A) = −(T −kI −(T2−2kT +k2I)p). Using the identities (17) and (18) we have (T2|p|2−2s |p|2T +|s|2|p|2I)S−1(p,A) = −(T −kI −(T2−2kT +k2I)p). 0 So we get the equalities 1 S−1(p,A) = − (T2−2s T +|s|2I)−1(T −kI −(T2−2kT +k2I)p) |p|2 0 1 = − (T2−2s T +|s|2I)−1(Tp−1−kp−1I −T2+2kT −k2I)p |p|2 0 1 = − (T2−2s T +|s|2I)−1 −(T2−2s T +|s|2I)p |p|2 0 0 (cid:16) +(T(2k−2s +p−1)+(|s|2−k2−kp−1)I)p 0 (cid:17) p 1 = I− (T2−2s T +|s|2I)−1(T +(|s|2−k2−kp−1)(2k−2s +p−1)−1I)(2k−2s +p−1)p. |p|2 |p|2 0 0 0 With some calculation we get S−1(p,A) = p−1I 1 − (T2−2s T+|s|2I)−1(T−sI+[s+(|s|2−k2−kp−1)(2k−2s +p−1)−1]I)(2k−2s +p−1)p |p|2 0 0 0 and also 1 S−1(p,A) = p−1I +S−1(s,T)λ− (T2−2s T +|s|2I)−1Λ |p|2 0 where we have set p λ := (2k−2s +p−1) , 0 |p|2 Λ := [s+(|s|2−k2−kp−1)(2k−2s +p−1)−1](2k−2s +p−1)p. 0 0 Using the identities (19) and (20) we finally get S−1(p,A) = p−1I −S−1(s,T)p−2, but also S−1(s,T) = pI −S−1(p,A)p2. (22) So p ∈ ρ (A), p 6= 0 then s ∈ ρ (T). S S Now take s ∈ ρ (T) and observe that, with form the definitions of S−1(s,T) and of A, replacing S (21) in the S resolvent operator S−1(s,T) we obtain S−1(s,T)= −(A−2−2p |p|−2A−1+|p|−2I)−1(A−1+k−sI), 0 10