Spectral Theorem for quaternionic normal operators: Multiplication form

arXiv:1603.00697v1 [math.SP] 2 Mar 2016

SPECTRAL THEOREM FOR QUATERNIONIC NORMAL OPERATORS : MULTIPLICATION FORM G. RAMESH and P. SANTHOSH KUMAR

Abstract. In this article we prove the multiplication form of a spectral theorem for normal operators in quaternionic Hilbert spaces: Let T be a right linear normal operator in a quaternionic Hilbert space H with the domain D(T ) ⊂ H. We prove that for a fixed unit imaginary quaternion, say m, there exists a Hilbert basis Nm of H, a measure space (Ω, µ), a unitary operator U : H → L2 (Ω; H; µ) and a µ - measurable function φ : Ω → Cm , ( here Cm = {α + mβ; α, β ∈ R}) so that if T is expressed with respect to Nm then T x = U ∗ Mφ U x, for all x ∈ D(T ), where Mφ is the multiplication operator in L2 (Ω; H; µ) induced by φ with U (D(T )) ⊆ D(Mφ ). In the process, we prove that every complex Hilbert space is a slice Hilbert space of some quaternionic Hilbert space. Also, we prove that there exists an anti self-adjoint, unitary operator which commutes with an unbounded normal operator in a quaternionic Hilbert space. In proving all these results, we solve the problem by reducing the problem to the complex case and lift it to the quaternionic case.

1. INTRODUCTION The theory of quaternionic Hilbert spaces is found to be useful in quaternionic quantum mechanics. Particularly, in mathematical physics, finding the solution of the Schrodinger equation over the skew-field of quaternions is equivalent to the study of diagonalization of an anti self-adjoint, unitary operator on a quaternionic Hilbert space [6]. In quantum mechanics most of the operators which we come across are unbounded. In particular, this is the case in quaternionic quantum mechanics also. One of the most important operators in quantum mechanic is the position operator, which is nothing but a multiplication operator defined in a Hilbert space. This is a normal operator. In fact, it is well known in the classical theory of operators that every normal operator is a multiplication operator induced by a suitable function. One can ask whether the same is true or not in quaternionic setting. In this direction, upto our knowledge, only [12] dealt with such a question and the solution of this is given by using spectral system. In this article, the author used the simplectic image concept to obtain the result. This question is answered for normal operators in a real Hilbert spaces by S. H. Kulkarni and S. Agarwal in [1] using Banach algebra techniques. Both the above 1991 Mathematics Subject Classification. 47S10, 47B15, 35P05. Key words and phrases. quaternionic Hilbert space, normal operator, spherical spectrum, closed operator, multiplication operator, spectral measure, spectral theorem. 1

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G. RAMESH and P. SANTHOSH KUMAR

results do not use the concept of the spherical spectrum of a quaternionic normal operator. In our case we show that an unbounded normal operator on a quaternionic Hilbert space is unitarily equivalent to a multiplication operator induced by an appropriate function. The result exactly looks like the classical one. To our surprise, we observed that the function which determine the multiplication operator is Cm valued, for a fixed unit imaginary quaternion m. In proving our results we use the following approach: Given an unbounded normal operator in a quaternionic Hilbert space, we associate a unique operator in a complex Hilbert space, which have similar properties and by using the results of the complex operator theory, we obtain the results for quaternionic operator. Though this technique is available in the literature (see [7] for details ), to our knowledge we use this for the first time. Later, we extend this technique for quaternionic operators defined between two different Hilbert spaces. We organize this article in four sections. In the second section we recall basic properties of the ring of quaternions, quaternionic Hilbert spaces and operators in quaternionic Hilbert spaces. o In the third section, we prove two main results: • we show that every complex Hilbert space is a slice Hilbert space of a quaternionic Hilbert space for some anti self-adjoint, unitary operator and • a linear operator between two complex Hilbert spaces can be extended to a unique right linear operator between quaternionic Hilbert spaces. In the fourth section, we prove the spectral theorem (multiplication form) for right linear bounded operators on quaternionic Hilbert spaces. In the final section, we prove the similar results for unbounded normal operator in quaternionic Hilbert space by using the Z-transform. In this proof we restrict the operator to the slice Hilbert space, then associate it to the Z- transform to get the spectral theorem for the operator on the slice Hilbert space. Finally, we lift this result to the quaternionic case. 2. PRELIMINARIES Let H denotes the ring of all real quaternions. If q ∈ H, then q = q0 + q1 i + q2 j + q3 k, where qℓ ∈ R for each ℓ = 0, 1, 2, 3. Here i, j, k satisfy i2 = j 2 = k 2 = −1 and i · j · k = −1.

For q ∈ H, the conjugate of q is q = q0 − q1 i − q2 j − q3 k. The real part of H is denoted by Re(H) = {q ∈ H : q = q} and the imaginary part of H is denoted by Im(H) = {q ∈ H : q = −q} . The set S := {q ∈ Im(H) : |q| p = 1} is the imaginary unit sphere. For p, q ∈ H, we have p · q = q · p and |q| := q02 + q12 + q22 + q32 . For m ∈ S, the filed Cm := {α + mβ : α, β ∈ R} is called a slice of H. Here Cm is isomorphic to the complex field C through the mapping α S + mβ → α + iβ. For Cm . The positive m 6= ±n ∈ S, we have Cm ∩ Cn = R. We also have H = m∈S

upper half plane of Cm is defined by C+ m = {α + mβ; α ∈ R, β ≥ 0}.

Note 2.1. All the results in complex Hilbert spaces holds true in the slice Cm Hilbert spaces, for any m ∈ S. Relation on H: Define a relation on H by

p ∼ q ⇔ p = s−1 qs, for some 0 6= s ∈ H.

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Here ∼ is an equivalence relation on H. The equivalence class of q is given by [q] =:= {s−1 qs : 0 6= s ∈ H}.

If Im(q) 6= 0, then q = Re(q) +

Im(q) |Im(q)|

· |Im(q)|. It follows that

[q] = Re(q) + |Im(q)| · S. We have the following observations through the representation of the equivalence classes:   Im(q) for all m ∈ S, [q] ∩ Cm = Re(q) ± m · (1) , |Im(q)| (2) p ∼ q ⇔ Re(p) = Re(q) and |Im(p)| = |Im(q)|. Definition 2.1. [7, Definition 2.3] Let H be a right H− module. A map h·|·i : H × H −→ H satisfying: (1) If u ∈ H, then hu|ui = 0 ⇔ u = 0 (2) hu|v + w · qi = hu|vi + hu|wi · q, for all u, v ∈ H and q ∈ H (3) hu|vi = hv|ui, for all u, v ∈ H, is called an inner product on H. If we define kuk2 = hu|ui, for all u ∈ H, then k · k is a norm on H and is called the norm induced by h·|·i. If (H, k · k) is complete space then it is called a right quaternionic Hilbert space. Note 2.2. Throughout the article H denotes a right quaternionic Hilbert space. Definition 2.2. Let (Ω, µ) be a measure space and fix m ∈ S. If φ : Ω → Cm is measurable, then (1) essential supremum of φ is defined by 
ess sup(φ) = inf α : µ {x : φ(x) > α} = 0 (2) essential range of φ is defined by


ess ran(φ) := {λ : ∀ǫ > 0, µ {x : |φ(x) − λ| = 0} > ǫ}

(3) φ is said to be essentially bounded if ess sup(|φ|) is finite.

Proposition 2.3. [7, Proposition 2.5] Let N be a subset of a right quaternionic Hilbert space H such that, for z, z ′ ∈ N , hz|z ′ i = 0 if z 6= z ′ and hz|zi = 1. The following conditions are equivalent: (1) For every x, y ∈ H, it holds: X hx|yi = hx|zi hz|yi . z∈N

The above series converges absolutely (2) For every x ∈ H, it holds: X kxk2 = | hz|xi |2 z∈N

⊥

(3) N = {0} (4) span N = H.

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G. RAMESH and P. SANTHOSH KUMAR

Proposition 2.4. [7, Proposition 2.6] Every quaternionic Hilbert space H admits a subset N ,called Hilbert basis of H such that, for z, z ′ ∈ N , hz|z ′ i = 0 if z 6= z ′ and hz|zi = 1, and N satisfies equivalent conditions stated in proposition 2.3. Furthermore, if N is Hilbert basis of H then every x ∈ H can be uniquely decomposed as follows: X x= z hz|xi . z∈N

Definition 2.5. Let (Ω, µ) be a measure space. Then   Z   L2 (Ω; H; µ) := f : Ω → H | |f (x)|2 dµ(x) < ∞   Ω

is a right quaternionic Hilbert space with the inner product Z hf |gi = f (x) · g(x) dµ(x). Ω

For a fixed m ∈ S, L2 (Ω; Cm ; µ) =

  

f : Ω → Cm |

Z

Ω

  |f (x)|2 dµ(x) < ∞ 

is a Cm - Hilbert space with the inner product Z hf |gi = f (x) · g(x) dµ(x). Ω

Definition 2.6. [7, Definition 2.9] Let (H1 , kk1 ) and (H2 , kk2 ) be two Hilbert spaces. A map T : H1 → H2 is said to be a right H− linear operator if T (u · q + v) = T (u) · q + T (v),

for every u, v ∈ H1 and q ∈ H. We say that T is bounded or continuous, if there exists K > 0 such that If T is bounded, then

kT uk2 ≤ Kkuk1, for all u ∈ H1 . kT k = sup {kT uk2 : u ∈ H1 , kuk1 = 1}

is finite and it is called the norm of T .

We denote the set of all bounded right H - linear operators between H1 and H2 by B(H1 , H2 ) and B(H, H) = B(H). If T ∈ B(H1 , H2 ), the null space and the range space are denoted by N (T ) and R(T ) respectively. Definition 2.7. [7, Definition 2.12] Let T ∈ B(H). Then there exists a unique operator T ∗ ∈ B(H) such that hu|T vi = hT ∗ u|vi, for all u, v ∈ H. This operator T ∗ is called the adjoint of T. Definition 2.8. [7, Definition 2.12] Let T ∈ B(H). Then T is said to be (1) self-adjoint if T = T ∗ (2) positive if T = T ∗ and hx|T xi ≥ 0, for all x ∈ H (3) anti self-adjoint if T ∗ = −T (4) normal if T T ∗ = T ∗ T

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(5) unitary if T T ∗ = T ∗ T = I. Example 2.9. [12, Example 1.1] Let (Ω, µ) be σ- additive measure space and φ : Ω → Cm be µ-measurable. Then we define the multiplication operator Mφ : L2 (Ω; H; µ) → L2 (Ω; H; µ), by Then (1) (2) (3) (4) (5)

Mφ (g)(x) = φ(x) · g(x), for all g ∈ L2 (Ω; H; µ). Mφ is a right H- linear normal. Mφ∗ = Mφ Mφ ∈ B(L2 (Ω, H, µ)) if and only if φ is essentially bounded kMφ k = ess sup(|φ|) Mφ∗ = Mφ ⇔ φ = φ in µ- a.e.

In the forthcoming sections we show that every normal operator in quaternionic Hilbert space is some form of multiplication operator. Definition 2.10. Let T ∈ B(H). A closed subspace M of H is said to be invariant under T if T (M ) := {T u : u ∈ M } ⊆ M. Moreover, if M ⊥ is also invariant under T then we say that M is a reducing subspace for T . We recall the notion of the spherical spectrum of a right quaternionic linear operator in quaternionic Hilbert space. 2.3. Spherical spectrum: [7, Definition 4.1] Let T : H → H be a right linear operator with domain D(T ), a right linear subspace of H and q ∈ H. Define ∆q (T ) : D(T 2 ) → H by ∆q (T ) := T 2 − T (q + q) + I.|q|2 .

The spherical resolvent of T is denoted by ρS (T ) and is the set of all q ∈ H satisfying the following three properties: (1) N (∆q (T )) = {0} (2) R(∆q (T )) is dense in H (3) ∆q (T )−1 : R(∆q (T )) → D(T 2 ) is bounded. Then the spherical spectrum of T is defined by setting σS (T ) = H \ ρS (T ). Theorem 2.11. [7, Theorem 5.9] Let T ∈ B(H) be normal. Then there exists three mutually commuting bounded operators A, B and J on H such that T = A + J.B, where A =

T +T ∗ 2 ,B

=

|T −T ∗ | 2

and J is an anti self-adjoint, unitary operator.

Note 2.4. Throughout this article, J denotes an anti self-adjoint, unitary operator. Lemma 2.12. [8, Lemma 4.1] Let h·|·i : H × H → H be inner product on H and fix m ∈ S. Define Jm H± = {u ∈ H : J(u) = ±u · m}. Then Jm Jm (1) H± 6= {0} and the restriction of the inner product h·|·i to H± is Cm Jm valued. Therefore H± is Cm − Hilbert space, called the slice Hilbert space of H Jm Jm (2) H = H+ ⊕ H− .

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Jm Remark 2.13. If N is Hilbert basis of H+ , then N is also a Hilbert basis for H and it holds: X J(x) = z · m hz|xi . z∈N

(see [7, Proposition 3.8(f)] for details). Jm For every m ∈ S, H+ is Cm - Hilbert space. Our intension is to prove the converse statement: If K is Cm - Hilbert space for some m ∈ S, then there exists right quaternionic Hilbert space H and an anti self-adjoint, unitary J ∈ B(H) such Jm that K = H+ .

3. EXTENSION OF Cm -HILBERT SPACE TO A QUATERNIONIC HILBERT SPACE Jm It is known that a Cm - linear operator in a slice Hilbert space H+ can be extended uniquely to a right linear operator in H [7]. In this section we prove that a linear operator in any Cm - Hilbert space K can be extended uniquely to a quaternionic linear operator in some quaternionic Hilbert space H associated to K. This is possible by showing that every Cm - Hilbert space is a slice Hilbert space of some quaternionic Hilbert space, with respect to some anti self-adjoint, unity operator J.

Lemma 3.1. Let H be right quaternionic Hilbert space. Then H is separable if Jm and only if H+ is separable, for any m ∈ S and any anti self-adjoint and unitary operator J on H. Jm is a closed subset of H. If H is Proof. For any J and m ∈ S, the space H+ Jm Jm that separable, then H+ is also separable. It is clear from the definition of H± for any x ∈ H, we have Jm Jm x ∈ H+ ⇔ x · n ∈ H− . Jm If H+ has countable dense subset say D+ , then D− := {x·n : x ∈ D+ } is countable Jm and dense subset of H− . By [7, Lemma 3.10], D+ ⊕ D− is countable dense subset of H. This implies that H is separable.  Jm Jm Lemma 3.2. If x ∈ H+ and y ∈ H− , then hx|yi + hy|xi = 0.

Proof. It is clear from the proof of [7, Proposition 3.11(a)].



Jm We recall that every Cm - linear operator in H+ can be extended uniquely to a right linear operator in H. Also the converse is possible with some assumptions, which are given below. Jm Jm is a Cm − → H+ Proposition 3.3. [7, Proposition 3.11] If T : D(T ) ⊂ H+ linear operator, then there exists a unique right H− linear operator Te : D(Te) ⊂ T Jm H → H such that D(Te) H+ = D(T ), J(D(Te)) ⊂ D(Te) and Te(u) = T (u), for Jm every u ∈ H+ . The following facts holds: Jm (1) If T ∈ B(H+ ), then Te ∈ B(H) and kTek = kT k (2) J Te = TeJ.

e, On the other hand, let V : D(V ) → H be a right linear operator. Then V = U T Jm Jm for a unique bounded Cm − linear operator U : D(V ) H+ → H+ if and only if J(D(V )) ⊂ D(V ) and JV = V J. Furthermore,

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∗ Jm , then D(Te) = H and Te = Tf∗ (1) If D(T ) = H+ Jm Jm f = SeTe (2) If S : D(S) ⊂ H+ → H+ is Cm - linear, then ST (3) If S is the inverse of T , then Se is the inverse of Te.

Remark 3.4. In particular, if T ∈ B(H) is non self-adjoint, normal, there exist an anti self-adjoint, unitary J ∈ B(H) such that T J = JT (see [7, Theorem 5.9]). If T ∈ B(H) is self-adjoint, then the existence of such J is given in [7, Theorem 5.7(b)]. Hence all the results of Proposition 3.3 holds true for normal operators. Jm Jm Remark 3.5. Let q ∈ H and T : H+ → H+ be Cm - linear. Then by Proposition 3.3(2), we have

(3)

e q (T ), ∆q (Te) = ∆

e q (T ) denotes the extension of ∆q (T ) to H. where ∆

The Proposition 3.3 shows that for m ∈ S, the extension is possible only for the operators in slice Hilbert space of H. First we show that every Cm - Hilbert space is a slice Hilbert space of some quaternionic Hilbert space and for some anti self-adjoint, unitary J. Later we prove that the extension is also possible for linear operators between two Cm Hilbert spaces. Proposition 3.6. Let K be a Cm - Hilbert space. Then there is a quaternionic Hilbert space H and an anti self-adjoint, unitary J ∈ B(H) such that Jm K = H+ .

Proof. Define H := K × K. We define binary operations on H as follows: (x, y) + (z, w) = (x + z, y + w), for all (x, y), (z, w) ∈ H. For q ∈ H we can write q = α + β · n, for some α, β ∈ Cm , where n ∈ S is such that m · n = −n · m. Then (4)

(x, y) · (α + β · n) = (x · α − y · β , x · β − y · α).

Let us denote the inner product on K by h·|·iK . Define h·|·i : H × H → H by h(x, y)|(z, w)i = [hx|ziK + hw|yiK ] + [hx|wiK − hz|yiK ] · n. Using the fact that h·|·iK is an inner product on K, it can be shown that h·|·i is an inner product on H. The completeness of h·|·iK implies that H is complete with respect to the norm, k(x, y)k2 = kxk2 + kyk2 which is induced from the above inner product. There fore H is a right quaternionic Hilbert space. If we identify x of K by (x, 0) of H and y of K by (y, 0) of H, then the following identification holds: (x, y) = (x, 0) + (0, y) = (x, 0) + (y, 0) · n = x + y · n. Here we used Equation (4) to conclude (0, y) = (y, 0) · n. Define J : H → H by J(x + y · n) = (x − y · n) · m, for all x + y · n ∈ H.

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G. RAMESH and P. SANTHOSH KUMAR

We shall prove that J is anti self-adjoint and unitary. If x + y · n, z + w · n ∈ H then hx + y · n|J(z + w · n)i = hx + y · n|(z − w · n) · mi

= hx|(z − w · n) · mi + hy|(z − w · n) · mi

= hx · m|z + w · ni + hy · m · n|z + w · ni

∗

= h(−x + y · n) · m|z + w · ni .

Therefore J (x + y · n) = (−x + y · n) · m, for all x + y · n ∈ H. This implies J ∗ = −J and J ∗ J = JJ ∗ = I. Jm Jm It remains to show that H+ = K. If x + y · n ∈ H+ , then J(x + y · n) = (x + y · n) · m. By the definition of J, it follows that y = 0. Conversely, if x ∈ K Jm then J(x) = x · m. Hence K = H+ .  From Proposition 3.6 it is clear that for a fixed m ∈ S, a linear operator on any Cm - Hilbert space can be extended uniquely to the corresponding quaternionic Hilbert space. We prove the following lemma to give the relation between the spectrum of T and the spherical spectrum of Te.

Jm Jm Lemma 3.7. Let T : H+ → H+ be bounded Cm - linear and Te be the extension of T as given in Proposition 3.3. Then σS (Te) = {[λ] : λ ∈ σ(T )} .

Proof. If λ ∈ σ(T ), then (T − λ · I) is not invertible. We show that ∆λ (T ) is not invertible. Suppose ∆λ (T ) is invertible, there exists L ∈ B(H) such that L∆λ (T ) = ∆λ (T )L = I. This implies (T − λ · I)L = L(T − λ · I). Equivalently, (T − λ · I)[(T − λ · I)L] = [(T − λ · I)L](T − λ · I) = L.

This is contradiction. Therefore ∆λ (T ) is not invertible. We claim that ∆µ (Te) is not invertible for every µ ∈ [λ]. If ∆µ (Te) is invertible for some µ ∈ [λ], then by Equation (3) and Proposition 3.3(3), we have e µ (T )−1 = ∆ e λ (T )−1 . ∆µ (Te)−1 = ∆

This is contradiction to λ ∈ σ(T ). Hence

{[λ] : λ ∈ σ(T )} ⊆ σS (Te).

On the other hand if q ∈ σs (Te), then λ := Re(q) + m · |Im(q)| ∈ [q] ∈ σS (Te). By [7, Proposition 5.11(c)], we have λ ∈ σS (Te) ∩ C+ m = σ(T ). Thus σS (Te) = {[λ] : λ ∈ σ(T )}.



We generalize Proposition 3.3 for linear operators between two different Cm Hilbert spaces. Theorem 3.8. Let H, K be quaternionic Hilbert spaces. Let J1 , J2 be anti selfadjoint unitary operators on H and K, respectively. Fix m ∈ S. If T : D(T ) → J2 m J1 m K+ is Cm - linear with D(T ) ⊂ H+ , then there exists a unique right H- linear e e operator T : D(T ) → K such that T J1 m D(Te) H+ = D(T ), J1 (D(Te)) ⊂ D(Te) and Te(u) = T (u), for every u ∈ D(T ). Furthermore,

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J1 m J2 m (1) If T : H+ → K+ is bounded, then Te ∈ B(H, K) and kTek = kT k (2) J2 Te = TeJ1 on D(Te).

On the otherhand, let V : D(V ) → K be right H- linear with D(V ) ⊂ H. Then J1 m J2 m e , for U : D(V ) ∩ H+ , if and only if J2 V = V J1 on D(V ). V =U → K+

Proof. First we show that the extension is unique. Choose n ∈ S such that m · n = −n · m. Define Φ : H → H by (5)

Φ(x) = x · n, for all x ∈ H.

It is clear that Φ is anti Cm - linear isomorphism. Suppose that there exists an T J1 m extension Te of T such that D(Te) H+ = D(T )and J1 (D(Te)) ⊂ D(Te). Then J1 m e D(T ) ∩ H− = Φ(D(T )). This implies D(Te) = D(T ) ⊕ Φ(D(T )). If x ∈ D(Te), then x = x1 + x2 with x1 ∈ D(T ), x2 ∈ Φ(D(T )) and Te(x) = T (x1 ) − T (x2 · n) · n.

T J1 m If S : D(S) → K is another extension of T such that D(S) H+ = D(T ), J1 (D(S)) ⊂ D(S) and Sy = T y, for all y ∈ D(T ) then D(S) = D(T ) ⊕ Φ(D(T )) = D(Te) and we have S(x) = S(x1 ) + S(x2 ) = S(x1 ) − S(x2 · n) · n

= T (x1 ) − T (x2 · n) · n

= Te(x), for all x = x1 + x2 ∈ D(S).

This implies that extension Te of T is unique. J1 m J2 m J1 m J1 m Proof of (1): If T : H+ → K+ is bounded then D(Te) = H+ ⊕ H− = H. e e Since T is the extension of T , it follows that kT k ≤ kT k. If x = x1 + x2 ∈ H then by Lemma 3.2, we have kTexk2 = kT (x1 ) − T (x2 · n)k2 = kT x1 k2 + kT (x2 · n)k2 ≤ kT k2(kx1 k2 + kx2 k2 ) ≤ kT k2kxk2 .

This implies that kTek ≤ kT k. Hence kTek = kT k. Proof of (2): If x ∈ D(Te), then x = x1 + x2 with x1 ∈ D(T ), x2 ∈ Φ(D(T )) and J2 Te(x1 + x2 ) = J2 [T (x1 ) − T (x2 · n) · n]

= J2 (T (x1 )) − J2 (T (x2 · n)) · n = T (x1 ) · m − T (x2 · n) · m · n

= T (x1 · m) − T (−x2 · m · n) · n

= T (J1 x) − T (J1 x2 · n) · n

= Te(J1 x1 + J1 x2 )

= TeJ1 (x).

e , for some U : D(V ) T H J1 m → K J2 m , then J1 (D(V )) ⊂ D(V ). It is clear If V = U + + from the earlier proof that D(V ) = D(U ) ⊕ Φ(D(U )). For x = x1 + x2 ∈ D(V ), we

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have J2 V (x) = J2 (U x1 − U (x2 · n) · n)

= U (x1 ) · m − U (x2 · n) · m · n

= U (x1 · m) + U (x2 · n · m) · n = U (J1 x1 ) − U (J1 x2 · n) · n

= V J1 (x).

Conversely, assume that J2 V = V J1 . That is J2 V x = V J1 x, ∀ x ∈ D(V ). This T J1 m J2 m means that J1 x ∈ D(V ), for every x ∈ D(V ). Hence V (D(V ) H+ ) ⊆ K+ . T J1 m J2 m Define U : D(V ) H+ → K+ by U x = V x, for all x ∈ D(U ).

Here V is right H- linear extension of U such that J1 (D(V )) ⊂ D(V ), by the e. uniqueness of extension, we have V = U 

Our aim is to prove that
L2 (Ω; Cm ; µ) = L2 (Ω; H; µ)Jm + for some anti self-adjoint, unitary J ∈ B(L2 Ω; H; µ) . To establish this result, we need the following theorem.

Theorem 3.9. Let m, n ∈ S be such that m · n = −n · m. Let (Ω, µ) be a measure space. Then L2 (Ω; Cm ; µ) is closed in L2 (Ω; H; µ). More over, we represent L2 (Ω; H; µ) as {f + g · n|f, g ∈ L2 (Ω; Cm ; µ)}, where (f + g · n)(x) = f (x) + g(x) · n, for all x ∈ Ω. Proof. If f ∈ L2 (Ω; H; µ), then f (x) = αx + βx · n, for every x ∈ Ω, where αx , βx ∈ Cm . Define f1 (x) = αx and f2 (x) = βx . For l = 1, 2 the map fl : Ω → Cm satisfies Z Z 2 |fl (x)| dµ(x) ≤ |f (x)|2 dµ(x) < ∞. Ω

Ω

Hence f1 , f2 ∈ L2 (Ω; Cm ; µ). Define Ψ : L2 (Ω; H; µ) → {f + g · n|f, g ∈ L2 (Ω; Cm ; µ)} by Ψ(f ) = f1 + f2 · n, for every f ∈ L2 (Ω; H; µ). Here {f + g · n|f, g ∈ L2 (Ω; Cm ; µ)} is right quaternionic Hilbert space. Clearly, Ψ is well defined,right H- linear onto map. Next, we show that Ψ is an isometry. Let f ∈ L2 (Ω; H; µ). Then Z Z kΨ(f )k2 = |f1 (t) + f2 (t) · n|2 dµ(t) = |f1 (t)|2 + |f2 (t)|2 dµ(t) Ω

Ω

=

Z

2

|f1 (t)| dµ(t) +

Ω

Z

|f2 (t)|2 dµ(t)

Ω 2

= kf1 k + kf2 k

2

= kf k2 .

Thus Ψ is an isometrically isomorphism. It remains to show that L2 (Ω; Cm ; µ) is closed. Let {fk } be a sequence in L2 (Ω; Cm ; µ). If {fk } converges to f = f1 + f2 · n

SPECTRAL THEOREM: MULTIPLICATION FORM

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in L2 (Ω; H; µ), then kfk − f k2 = k(fk − f1 ) − f2 · nk2 = kfk − f1 k2 + kf2 k2 .

Since fk → f , it follows that kfk − f1 k2 → 0, as n → ∞ and f2 = 0. Therefore L2 (Ω; Cm ; µ) is closed in L2 (Ω; H; µ).  Corollary 3.10. There exists J ∈ B(L2 (Ω; H; µ)), an anti self-adjoint, unitary operator such that 2 L2 (Ω; H; µ)Jm + = L (Ω; Cm ; µ). Proof. It is clear from Proposition 3.6 and Theorem 3.9 that there exists J ∈ B(L2 (Ω; H; µ)) such that 2 L2 (Ω; H; µ)Jm + = L (Ω; Cm ; µ).



4. SPECTRAL THEOREM: BOUNDED OPERATORS In this section we prove the spectral theorem (multiplication form) for bounded normal operators on a quaternionic Hilbert space. We recall the spectral theorem for bounded normal operators on complex Hilbert spaces. Theorem 4.1. [4, Theorem 11.5] If N ∈ B(H) is a normal operator then there is a measure space (X, µ) and an essentially bounded µ- measurable function φ : X → C such that N is unitarily equivalent to Lφ , where Lφ is a left multiplication by φ acting on L2 (X; C; µ). More over, if H is separable then the measure space obtained above (X, µ) is σfinite. We prove the multiplication form of normal operator on quaternionic Hilbert Jm spaces by using Proposition3.3(1), the extension of Cm -linear operator on H+ to the linear operator on quaternionic Hilbert space H. MULTIPLICATION FORM:. Theorem 4.2. Let T ∈ B(H) be normal and fix m ∈ S. Then there exists (a) a Hilbert basis Nm of H (b) a measure space (Ω, µ) (c) a unitary operator U : H → L2 (Ω; H; µ) and (d) an essentially bounded µ- measurable function φ : Ω → Cm so that if T is expressed with respect to Nm , then T = U ∗ Mφ U,

where Mφ is a bounded multiplication operator on L2 (Ω; H; µ). Moreover, (1) kT k = ess sup(|φ|) (2) σS (T ) = {[λ] : λ ∈ ess ran (φ)}. Further more, if H is separable Hilbert space, then the obtained measure space (Ω, µ) is σ- finite.

12

G. RAMESH and P. SANTHOSH KUMAR

Jm Jm Proof. By Proposition 3.3, T+ : H+ → H+ is a Cm - linear bounded normal Jm e operator such that T+ = T . Let Nm be Hilbert basis for H+ then by Remark 2.13, Nm is Hilbert basis for H and moreover, X J(x) = z · m hz|xi . z∈N

If u = a + b, where a ∈

Jm H+ ,b

∈

Jm H− ,

then

T (u) = T+ (a) − T+ (b · n) · n.

By Theorem 4.1, there exist a measure space (Ω, µ), a Cm - valued µ- measurable Jm function φ on Ω and a unitary operator U+ : H+ → L2 (Ω; Cm ; µ) such that ∗ T + = U+ L φ U+ ,

where Lφ : L2 (Ω; Cm ; µ) → L2 (Ω; Cm ; µ) is defined by

Lφ (g)(t) = φ(t) · g(t), for all g ∈ L2 (Ω; Cm ; µ).

e φ : L2 (Ω; H; µ) → L2 (Ω; H; µ) given by By Theorem 3.3, we have L

e φ (g + h · n) = Lφ (g) + Lφ (h) · n, for all g, h ∈ L2 (Ω; Cm ; µ). L

e φ by Mφ . It is clear that Mφ is a right H- linear and Mφ |L2 (Ω;C ;µ) = Let us denote L m Lφ . For h = h1 + h2 · n ∈ L2 (Ω; H; µ) and x ∈ Ω, we have Mφ (h1 + h2 · n)(x) = Lφ (h1 )(x) + Lφ (h2 )(x) · n

= φ(x) · h1 (x) + φ(x) · h2 (x) · n

= φ(x)(h1 (x) + h2 (x) · n) = φ · (h1 + h2 )(x).

That is Mφ is a multiplication operator induced by φ. By Theorem 3.8, U+ has a unique extension U : H → L2 (Ω; H; µ) such that Jm Jm U (a + b) = U+ (a) − U+ (b · n) · n, for all a ∈ H+ , b ∈ H− .

Jm Jm Therefore, for all a ∈ H+ and b ∈ H− , we have

U ∗ Mφ U (a + b) = U ∗ Mφ (U+ (a) − U+ (b · n) · n)

= U ∗ [Lφ (U+ (a)) − Lφ (U+ (b · n)) · n]

∗ ∗ = U+ Lφ U+ (a) + U+ (Lφ U+ (b · n) · n · n) · n

∗ ∗ = U+ Lφ U+ (a) − U+ Lφ U+ (b · n) · n

= T+ (a) − T+ (b · n) · n

= T (a + b).

Hence U ∗ Mφ U = T . Proof of (1): It is clear from Proposition 3.3(1), that kT k = kT+ k and since kT+ k = ess sup(|φ|), we have kT k = ess sup(|φ|). Proof of (2): We know that σ(T+ ) = ess ran(φ). By Lemma 3.7, we have σS (T ) = {[λ] : λ ∈ ess ran(φ)}

Jm If H is separable, by Lemma 3.1 H+ is separable. Therefore by Theorem 4.1, (Ω, µ) is σ- finite. 

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Corollary 4.3. Let T ∈ B(H) be anti self-adjoint operator and fix m ∈ S. Then, there exists Hilbert basis Nm , a measure space (Ω; µ), a µ - measurable purely imaginary function ξ : Ω → Cm and a unitary operator U : H → L2 (Ω; H; µ) so that if T is expressed with respect to Nm then T = U ∗ Mξ U.

In particular, T is unitary ⇔ |ξ| = 1 in µ- a.e.

Proof. By Theorem 4.2, we have T = U ∗ Mξ U . Since T is anti self-adjoint, T ∗ = −T . It follows that U ∗ Mξ U = U ∗ M−ξ U . This implies ξ = −ξ. In particular, T is unitary ⇔ T ∗ T = T T ∗ = I ⇔ U ∗ M|ξ|2 U = I

⇔ U ∗ M|ξ|2 U = U ∗ M1 U (Here M1 is the identity on L2 (Ω; H; µ))

⇔ |ξ| = 1, µ - a.e.



Corollary 4.4. Let T ∈ B(H) be normal. Then T and T ∗ are unitarily equivalent. Proof. By Theorem 4.2, there exists a measure space (Ω, µ), µ- measurable Cm valued function ξ on Ω and a unitary U : H → L2 (Ω; H; µ) such that T = U ∗ Mξ U . This implies T ∗ = U ∗ Mξ U . For x ∈ Ω, define φ(x) = ξ(x)·n |ξ(x)| , for all x ∈ Ω. Here n ∈ S such that m·n = −n·m. Clearly, φ is non-zero a.e (µ), ess sup (|φ|) = 1 and Mφ is a right linear, unitary operator. Moreover, Mφ∗ Mξ Mφ = Mξ . Hence, T = U ∗ Mξ U = U Mφ∗ Mξ Mφ U = U Mφ∗ U T ∗ U ∗ Mφ U Let V = U Mφ∗ U . Then V is unitary and T = V T ∗ V ∗ . Hence T and T ∗ are unitarily equivalent.  Example 4.5. Let φ(t) = (i − j − k)t, for all t ∈ [0, 1] . Then φ is essentially bounded measurable function with the Lebesgue measure µ on [0, 1]. Define Mφ : L2 ([0, 1]; H; µ) → L2 ([0, 1]; H; µ) by Mφ (g)(t) = φ(t) · g(t) for all g ∈ L2 ([0, 1]; H; µ).

We show that Mφ is unitarily equivalent to a multiplication operator on L2 ([0, 1]; H; µ) induced by some complex valued measurable function. Define U : L2 ([0, 1]; H; µ) → L2 ([0, 1]; H; µ) by √ ( 3 + 1) − j + k p U (g)(t) = · g(t), for all g ∈ L2 ([0, 1]; H; µ). √ 6+2 3 It follows that √ ( 3 + 1) + j − k ∗ p U (h)(t) = , for all h ∈ L2 ([0, 1]; H; µ). √ 6+2 3 It can be easily verified that U is unitary. √ Define η(t) = 3 it for all t ∈ [0, 1]. Clearly, η is a complex valued essentially bounded measurable function. Also η induces a bounded multiplication operator

14

G. RAMESH and P. SANTHOSH KUMAR

Mη on L2 ([0, 1]; H; µ). We prove that Mφ is unitarily equivalent to Mη . For all g ∈ L2 ([0, 1]; H; µ), we have √ ( 3 + 1) − j + k ∗ ∗ p · g(t) U Mη U (g)(t) = U Mη √ 6+2 3 √ √ ( 3 + 1) − j + k ∗ p · g(t) = U 3it · √ 6+2 3 √ √ ( 3 + 1) + j − k √ ( 3 + 1) − j + k p p = · 3it · · g(t) √ √ 6+2 3 6+2 3 = (i − j − k)t · g(t) = Mφ (g)(t).

This example 4.5 infers that every multiplication operator induced by quaternion valued function is unitarily equivalent to a multiplication operator induced by some complex valued function. 5. SPECTRAL THEOREM: UNBOUNDED OPERATORS In this section we prove the spectral theorem ( multiplication form ) for unbounded normal operator in a quaternionic Hilbert space. For this, we restrict the operator to the slice Hilbert space, associate a bounded operator to the restricted operator through the Z- transform then by using Theorem 4.1, we obtain the result. Analogous to the complex Hilbert spaces, the theory of unbounded operators in quaternionic Hilbert space is studied in [7]. We recall some properties of unbounded operators in quaternionic Hilbert space, that we need for our purpose. Definition 5.1. [10, Definition 13.1] Let T : D(T ) ⊂ H1 → H2 be a right linear operator with domain D(T ), a right linear subspace of H1 . Then T is said to be densely defined, if D(T ) is dense in H1 . The graph of an operator T is denoted by G(T ) and is defined as G(T ) = {(x, T x)|x ∈ D(T )} . Definition 5.2. Let T : H1 → H2 be right linear with domain D(T ) ⊂ H1 . Then T is said to be closed, if the graph G(T ) is closed in H1 × H2 . Equivalently, if (xn ) ⊂ D(T ) with xn → x ∈ H1 and T xn → y, then x ∈ D(T ) and T x = y. Let S, T be densely defined closed operators with domains D(S), D(T ), respectively. Then S is said to be a restriction of T denoted by S ⊂ T , if D(S) ⊆ D(T ) and Sx = T x, for all x ∈ D(S). In this case, T is called an extension of S. Definition 5.3. An operator T with domain D(T ) is said to be normal, if T is densely defined, closed and T ∗ T = T T ∗. Definition 5.4. Let T : D(T ) → H be densely defined closed operator and S ∈ B(H), then we say that S commute with T if ST ⊆ T S. In other words, Sx ∈ D(T ) and ST x = T Sx, for all x ∈ D(T ). Recall that if T ∈ B(H) is normal, then by Remark 3.4 there exists a J ∈ B(H) anti self-adjoint, unitary such that T J = JT . We generalize this result to the case of densely defined, closed right linear operators.

SPECTRAL THEOREM: MULTIPLICATION FORM

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Theorem 5.5. [2, Theorem 6.1] Let T : D(T ) → H be densely defined closed, right 1 linear operator. Define the operator ZT := T (I + T ∗ T )− 2 . Then ZT has the following properties: 1 (1) ZT ∈
B(H), kZT k ≤ 1 and T = ZT (I − ZT∗ ZT )− 2 ∗ (2) ZT = ZT ∗ (3) If T is normal, then ZT is normal. Theorem 5.6. Let T : D(T ) → H be normal with D(T ) ⊂ H. Then there exists an anti self-adjoint unitary operator J ∈ B(H) such that J commutes with T . Proof. It is clear from the Theorem 5.5, that ZT is a bounded right linear normal operator. By Proposition 2.11, there exists an anti self-adjoint and unitary operator J ∈ B(H) such that ZT J = JZT and JZT∗ = ZT∗ J. This implies that J(I − ZT∗ ZT ) = (I − ZT∗ ZT )J. So J commutes with the square roof of bounded positive 1 1 operator (I − ZT∗ ZT ), that is J(I − ZT∗ ZT ) 2 = (I − ZT∗ ZT ) 2 J. 1 Now we show that J commutes with the inverse of (I − ZT∗ ZT ) 2 , which is an 1 1 unbounded operator. Let x ∈ D((I − ZT∗ ZT )− 2 ). Then x = (I − ZT∗ ZT ) 2 y, for 1 1 1 some y ∈ D((I − ZT∗ ZT ) 2 ) and Jx = J(I − ZT∗ ZT ) 2 y = (I − ZT∗ ZT ) 2 Jy. This 1 implies Jx ∈ D(I − ZT∗ ZT )− 2 . Moreover, 1

1

J(I − ZT∗ ZT )− 2 x = Jy = (I − ZT∗ ZT )− 2 Jx.

1

It is enough to show that JT ⊆ T J. Since T = ZT (I − ZT∗ ZT )− 2 and D(T ) = 1 D((I − ZT∗ ZT )− 2 ), for every x ∈ D(T ), we have Jx ∈ D(T ) and 1

1

JT x = JZT (I − ZT∗ ZT )− 2 x = ZT J(I − ZT∗ ZT )− 2 x 1

= ZT (I − ZT∗ ZT )− 2 Jx = T Jx.

Hence the result.



Lemma 5.7. Let H be a quaternionic Hilbert space and J is an anti self-adjoint Jm Jm → H+ is Cm - linear, then ZeT = ZTe . and unitary. If T : D(T ) ⊂ H+

Proof. By Theorem 5.6, there exists an anti self-adjoint, unitary operator J ∈ B(H) such that JT ⊆ T J. By Proposition 3.3, 1

− ZeT = Te(IeH+Jm + Te∗ Te) 2 1

− = Te(IH + Te∗ Te) 2

= ZTe .

^ It is enough to show that (IH+Jm + T ∗ T ) = (IH + Te∗ Te). We have

^ (IH+Jm + T ∗ T )(x + y) = (IH+Jm + T ∗ T )(x) − (IH+Jm + T ∗ T )(y · n) · n ∗ T (x + y) = (x + y) + Tg

= (IH + Te∗ Te)(x + y),

^ for all x + y ∈ D((IH+Jm + T ∗ T )).

Theorem 5.8. Let T : D(T ) → H be normal and fix m ∈ S. Then there exists



16

(a) (b) (c) (d) so that

G. RAMESH and P. SANTHOSH KUMAR

a a a a if

Hilbert basis Nm of H measure space (Ω, µ) unitary operator U : H → L2 (Ω; H; µ) and µ- measurable function φ : Ω → Cm T is expressed with respect to Nm then

T x = U ∗ Mη U x, for all x ∈ D(T ),

where Mη is right H- linear multiplication operator acting on L2 (Ω0 ; H; ν) with domain D(Mη ) = {g ∈ L2 (Ω0 ; H; ν)|η · g ∈ L2 (Ω0 ; H; ν)}. More over, if H is separable then the measure space obtained (Ω0 ; ν) is σ- finite. Jm Jm Proof. By Theorem 3.3, there exists a unique T+ : D(T ) ∩ H+ → H+ , a Cm Jm e linear operator such that T+ = T . Let Nm be Hilbert basis for H+ then by Remark 2.13, Nm is Hilbert basis for H and moreover, X J(x) = z · m hz|xi . z∈N

It is clear that ZeT+ = ZT and ZT+ is bounded Cm - linear operator. By Theorem Jm 4.1, there is a measure space (Ω; µ), a unitary operator U+ : H+ → L2 (Ω; Cm ; µ) and a µ- measurable function φ such that ∗ L φ U+ . ZT+ = U+

(6)

Here Ω = σ(ZT+ ). Define ξ : Ω → Cm by

1

ξ(p) = p(1 − |p|2 )− 2 , for all p ∈ Ω.

Then ξ is µ - measurable function such that


µ {x ∈ Ω : ξ(x) = ∞} = 0

By the Borel functional calculus for bounded Cm - linear operator ZT+ , we write (7)

1

ξ(ZT+ ) = ZT+ (I − ZT∗+ ZT+ )− 2 = T+ .

By Equations (6) and (7), we have 1

∗ ∗ T + = U+ Lφ U (I − U+ L|φ|2 U+ )− 2 ∗ = U+ L

1

φ(1−|φ|2 )− 2

U+ .

1

Let us denote ψ = φ(1 − |φ|2 )− 2 . Then

(8)

∗ T + x = U+ Lψ U+ x, for all x ∈ D(T+ ).

This implies that U+ (D(T+ )) ⊂ D(Lψ ). It is clear σ(T+ ) = ess ran(ψ) = Ω0 (say). Define a measure on Ω0 as ν(S) = µ(ξ −1 (S)), for every Borel subset S in Ω0 . If η(z) = z on Ω0 , then Lη : D(Lη ) → L2 (Ω0 ; Cm ; ν) defines a right linear operator in L2 (Ω; Cm ; µ) with the domain D(Lη ) = {g ∈ L2 (Ω0 ; Cm ; ν)|η · g ∈ L2 (Ω0 ; Cm ; ν)}. We establish a unitary between L2 (Ω; Cm ; µ) and L2 (Ω0 ; Cm ; ν) as follows: Define π : L2 (Ω; Cm ; µ) → L2 (Ω0 ; Cm ; ν) by π(g) = g ◦ ξ −1 , for all g ∈ L2 (Ω; Cm ; µ).

SPECTRAL THEOREM: MULTIPLICATION FORM

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We claim that π is a unitary. For g ∈ L2 (Ω; Cm ; µ), we have kπ(g)k2 =

Z

|(g ◦ ξ −1 )(s)|2 dν(s)

Z

|g(ξ −1 (s))|2 dν(s)

Z

|g(t)|2 dµ(t)

Ω0

=

Ω0

=

Ω

= kgk2 . This shows that π is one to one. If h ∈ L2 (Ω0 ; Cm ; ν), then h ◦ ξ ∈ L2 (Ω; Cm ; µ) and π(h ◦ ξ) = h. Let g ∈ L2 (Ω; Cm ; µ) and h ∈ L2 (Ω0 ; Cm ; ν). Then hπ(g)|hi =

Z

π(g)(x)h(x)dν(x)

Z

g(ξ −1 x)h(x)dν(x)

Z

g(y)h(ξ(y))dµ(y)

Ω0

=

Ω0

=

Ω

= hg|h ◦ ξi . This implies π ∗ (h) = h ◦ ξ, for all h ∈ L2 (Ω0 ; Cm ; ν). It can be verified that π π = ππ ∗ = I. First we express Lη in terms of Lψ . Later we construct unitary Jm V+ between H+ and L2 (Ω0 ; Cm ; ν). Consider ∗

(π ∗ Lη π)(g)(x) = π ∗ Lη (g ◦ ξ −1 )(x) = π(η · g ◦ ξ −1 )(x)

= η · (g ◦ ξ −1 ) ◦ ξ(x)

= (η ◦ ξ)(x) · g(x) = ψ(x) · g(x)

= Lψ (g)(x). This shows π ∗ Lη π = Lψ . Jm → L2 (Ω0 ; Cm ; ν) by Define V+ : H+ V+ = π ◦ U+ .

18

G. RAMESH and P. SANTHOSH KUMAR

The following diagram helps in understanding the construction of unitary operators. D(Lψ ) ⊆ L2 (Ω; Cm ; µ)

U+

Jm H+

V+

Lψ

π

D(Lη ) ⊆ L2 (Ω0 ; Cm ; ν)

L2 (Ω; Cm ; µ)

π∗

Lη

L2 (Ω0 ; Cm ; ν)

We claim that V+ is unitary and V+∗ Lη V+ u = T+ u for all u ∈ D(T+ ). It can be easily seen that V+∗ V+ = V+ V+∗ . Then by Equation (8), we have for all u ∈ D(T+ ), ∗ ∗ T + u = U+ π Lη πU+ u

= V+∗ Lη V+ u. Now extend the operator T+ to the operator T in H by using Theorem 3.3 and Theorem 3.8. The rest of the proof follows in the similar lines as in the case of fη = Mη and Vf bounded operators. Let L + = V . Then by extension of T+ we get T x = V Mη V x, for all x ∈ D(T ).

where Mη is the multiplication operator in L2 (Ω0 ; H; ν). If H is separable, then µ is σ- finite by Theorem 4.2. Therefore ν is also σ- finite.  The second author is thankful to INSPIRE (DST) for the support in the form of fellowship (No. DST/ INSPIRE Fellowship/2012/IF120551), Govt of India. References [1] S. Agrawal and S. H. Kulkarni, Spectral theorem for an unbounded normal operator in a real Hilbert space, J. Anal., 12(2004), 107–114. [2] Daniel Alpay, Fabrizio Colombo and David P. Kimsey, The spectral theorem for quaternionic unbounded normal operators based on the S-spectrum, arXiv:1409.7010. [3] J. B. Conway, A course in functional analysis, Graduate Texts in Mathematics,96, Springer, New York, 1985. [4] J. B. Conway, A course in operator theory, Graduate Studies in Mathematics,21, Amer. Math. Soc., Providence, RI, 2000. [5] N. Dunford and J. T. Schwartz, Linear operators. Part I, reprint of the 1958 original, Wiley Classics Library, Wiley, New York, 1988. [6] D. Finkelstein et al., Foundations of quaternion quantum mechanics, J. Mathematical Phys., 3 (1962), 207–220. [7] R. Ghiloni, V. Moretti and A. Perotti, Continuous slice functional calculus in quaternionic Hilbert spaces, Rev. Math. Phys. 25 (2013), no. 4, 1350006, 83 pp. [8] G. Ramesh and P. Santhosh Kumar, A note on the spectral theorem for compact normal operators on a Quaternionic Hilbert space.(Preprint). [9] G. Ramesh and P. Santhosh Kumar, Spectral theorem for normal operators in Quaternionic Hilbert spaces.(Preprint). [10] W. Rudin, Functional analysis, second edition, International Series in Pure and Applied Mathematics, McGraw-Hill, New York, 1991.

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¨ dgen, Unbounded self-adjoint operators on Hilbert space, Graduate Texts in Math[11] K. Schmu ematics, 265, Springer, Dordrecht, 2012. [12] K. Viswanath, Normal operators on quaternionic Hilbert spaces, Trans. Amer. Math. Soc. 162 (1971), 337–350. G. Ramesh, Department of Mathematics, Kandi, Sangareddy(M), Medak(Dist), Telangana, India 502285. E-mail address: [email protected] P. Santhosh Kumar, Department of Mathematics, Kandi, Sangareddy(M), Medak(Dist), Telangana, India 502285. E-mail address: [email protected]