Towards a theory of some unbounded linear operators on p-adic Hilbert spaces and applications

We are concerned with some unbounded linear operators on the so-called p-adic Hilbert space Eω. Both the Closedness and the selfadjointness of those unbounded linear operators are investigated. As applications, we shall consider the diagonal operator on Eω, and the solvability of the equation Au = v where A is a linear operator on Eω. AMS subject classification. 47S10; 46S10; 47A05; 47B25.


Introduction
The primary goal of this paper is to investigate upon some unbounded linear operators on the so-called p-adic Hilbert space E ω .For that, we first introduce and give the required background on the p-adic Hilbert space E ω .Next, we shall be dealing with natural issues such as the closedness and the self-adjointness of those unbounded linear operators.To do so, one equips the so-called direct sum E ω × E ω of E ω with itself, with both an ultrametric norm and a hilbertian structure.Afterwards, it goes back to introduce an unitary operator Γ on E ω × E ω which yields a remarkable description of A * , the adjoint of a linear operator A defined on E ω which does have an adjoint, in terms of A.
Let us mention that the p-adic Hilbert space E ω will play a key role throughout the paper.Apart from their intrinsic interests, p-adic Hilbert spaces have found extensive applications in theoretical physics.

T. Diagana
Let K be a complete ultrametric valued field.Classical examples of such a field include Q p the field of p-adic numbers where p ≥ 2 is a prime, C p the field of p-adic complex numbers, and the field of formal Laurent series( [4]).
An ultrametric Banach space E over K is said to be a free Banach space (see [2], [3], and [4]) if there exists a family (e i ) i∈I (I being an index set) of elements of E such that each element x ∈ E can be written in an unique fashion as: x i e i , lim i∈I x i e i = 0, and x = sup The family (e i ) i∈I is then called an orthogonal basis for E, and if e i = 1, for all i ∈ I, the family (e i ) i∈I is called an orthonormal basis.For a detailed description and properties of these spaces, we refer the reader to ( [2], [3], [4], and [11]) and the references therein.
Up to now, we shall suppose that the index set I is N, the set of all natural integers.
For a free Banach space E, let E * denote its (topological) dual and B(E) the Banach space of all bounded linear operators on E (see [2], [3], and [4]).Both E * and B(E) are equipped with their respective natural norms.
For (u, v) ∈ E × E * we define the linear operator (v ⊗ u) by setting: Let (e i ) i∈N be an orthogonal basis for E. We then define e i ∈ E * by It turns out that e i = 1 e i .Furthermore, every x ∈ E * can be expressed as a pointwise convergent series: x , e i e i .In addition to that, we have that: Now let us recall that every bounded linear operator A on E can be expressed as a pointwise convergent series, that is, there exists an infinite matrix (a ij ) (i,j)∈N×N with coefficients in K such that A = ij a ij (e j ⊗ e i ), and for any j ∈ N, lim i→∞ |a ij | e i = 0. (1.1) Moreover, for each j ∈ N, Ae j = i∈N a ij e i and its norm is defined by: In this paper we shall make extensive use of the p-adic Hilbert space E ω whose definition is given below.Again, for details, we refer the reader to ([2], [3], and [4]) and the references therein.
Let ω = (ω i ) i∈N be a sequence of non-zero elements in a complete non-Archimedean field K.We define the space E ω by i ω i = 0. Actually E ω is an ultrametric Banach space over K with the norm given by Let us also notice that E ω is a free Banach space (see [3]) and it has a canonical orthogonal basis.Namely, (e i ) i∈N , where e i is the sequence all of whose terms are 0 except the i-th term which is 1, in other words, e i = (δ ij ) j∈N where δ ij is the usual Kronecker symbol.We shall make extensive use of such a canonical orthogonal basis throughout the paper.It should be mentioned that for each i, e i = |ω i | 1/2 .Now if |ω i | = 1 we shall refer to (e i ) i∈N as the canonical orthonormal basis.
Let , : Then, , is a symmetric, non-degenerate form on E ω × E ω with value in K, and it satisfies the Cauchy-Schwarz inequality: Let us also mention that elements (e i ) i∈N of the canonical orthogonal basis for E ω satisfy The space E ω endowed with the bilinear form , defined above is called a p-adic Hilbert space.
It should also be observed that for every bounded linear operator A on E ω , the domain the whole of E ω .This is actually due to the fact that for every u = It is well-known ( [2] and [3]) that one can find bounded linear operators on E ω which do not have adjoints.Similarly, we shall see that the latter is still true for unbounded linear operators we shall deal with.Consequently, we denote by B 0 (E ω ) the space of all bounded linear operators which do have adjoints with respect to the non-degenerate form , defined above.It is known ( [2], [3] and [4]) that a bounded linear operator Let us recall that B 0 (E ω ) is stable under the operation of taking an adjoint and for any A ∈ B 0 (E ω ) : (A * ) * = A and A = A * .
Let ω = (ω i ) i∈N be a sequence of nonzero elements in a (complete) non-Archimedean field K. Consider the corresponding p-adic Hilbert space given by (E ω , , ).
In this paper, we initiate a theory for some unbounded linear operators within the p-adic framework, that is, on the p-adic Hilbert spaces E ω .As for bounded linear operators on E ω , some of the results go along the classical line and others deviate from it.For the most part, the statements of the results are inspired by their classical settings.However their proofs may depend heavily on the ultrametric nature of E ω and the ground field K.We especially emphasis on both the closedness and the self-adjointness of those unbounded linear operators on E ω (Theorems 3.3 and 3.4).To deal with those issues, we equip the direct sum E ω × E ω with both a complete ultrametric norm as well as a hilbertian structure ( it will be shown that this can be even done for a finite direct sum and E ωn where the sequences ω k = (ω k i ) i∈N for k = 1, 2, ..., n are respectively nonzero elements in K).Next we consider an unitary operator Γ on E ω × E ω which enables to describe the adjoint A * of a given operator A in terms of A.
To illustrate our abstract results, a few examples are discussed.In particular, a special attention is paid to the diagonal operator on E ω as well as the solvability of the equation Au = v where A is (possibly unbounded) a linear operator on E ω (Theorems 5.1 and 5.4).

Unbounded linear operators on E ω
Let ω = (ω i ) i∈N , = ( i ) i∈N be sequences of non-zero elements in a complete non-Archimedean field K, and let E ω , E be their corresponding p-adic Hilbert spaces.We suppose that (e i ) i∈N and (h j ) j∈N are respectively the canonical orthogonal bases associated to the p-adic Hilbert spaces E ω and E .
Let D ⊂ E ω be a subspace and let A : D ⊂ E ω → E be a linear transformation.As for bounded linear operators one can decompose A as a pointwise convergent series defined by:

Definition 2.1: An unbounded linear operator
Throughout the paper, we consider unbounded linear operators A whose domain We denote the collection of those unbounded linear operators by We have seen that if A ∈ B(E ω , E ) then its domain is the whole of E ω .We shall see that one can find elements of U (E ω , E ) whose domains differ from E ω (see Remark 4.2).As for bounded operators, there are elements of U (E ω , E ) which do not have adjoints (see Example 2.3 below).Actually in the next definition we state conditions which do guarantee the existence of the adjoint.Without lost of generality we shall suppose that E ω = E .As usual we denote In what follows, (K, |.|) denotes a complete non-Archimedean field.
Definition 2.2: Let ω = (ω i ) i∈N ⊂ K be a sequence of nonzero terms and let (E ω , , ) be the corresponding p-adic Hilbert space.An operator Furthermore, the adjoint A * of A is uniquely expressed by where a * i,j = w −1 i w j a j,i .We denote by U 0 (E ω ) the collection of operators in U (E ω ) which do have adjoint operators.Clearly, B 0 (E ω ) ⊂ U 0 (E ω ).
Example 2.3:(Unbounded operator with no adjoint).Set K = Q p the field of p-adic numbers endowed with the p-adic norm |.| p and let a i,j (e j ⊗ e i ) be the linear operator defined by its coefficients: and We have Proposition 2.4: The linear operator A = i,j a i,j (e j ⊗ e i ) defined above is in U (E ω ) and it does not have an adjoint.

Proof:
Since ∀j, lim Consequently, ∀i, j, λ i,j := Clearly, A := sup i,j To complete the proof we have to show that ∀i, lim = ∞, hence the adjoint of A does not exist.

Closed linear operators on E ω
Let A ∈ U (E ω ).As in the classical setting we define the graph of the linear operator A by The operator A is said to be closable if it has a closed extension.
As in the classical setting we characterize the closedness of an operator A ∈ U (E ω ) as follows: We denote the collection of closed linear operators A ∈ U (E ω ) by C(E ω ).In view of the above, B(E ω ) ⊂ C(E ω ).
Let us notice that in the classical context, the collection of closed operators is not a vector space.However if A is an unbounded linear operator acting in a Banach space X and if B ∈ B(X) then A + B their algebraic sum is closed.In the p-adic setting, there is no reason that the situation differs from the classical one.Similarly, if We shall prove the following important theorem: As previously mentioned, to deal with the closedness of an operator A ∈ U (E ω ) we first equip the direct sum E ω × E ω with both an ultrametric norm and a p-adic hilbertian structure.Actually, we shall show that this can be even done for a finite direct sum of Hilbert spaces E ω .
Throughout the rest of this section, we suppose that char(K) the characteristic of the field K is zero.Notice that examples of such fields include Q p (p ≥ 2 being prime) the field of p-adic numbers.
We now equip E n with the ultrametric norm defined by Now arguing that each (E ω k , . ) for k = 1, 2, ...n is an ultrametric Banach space it follows that (E n , ||.|| n ) is also an ultrametric Banach space.It also clear that E n (n fixed) is a free Banach space with orthogonal basis as {(L k i ) i∈N , k = 1, 2, ..., n} where L k i = (0, 0, ...e k i , ..0, 0, 0) whose terms are 0 except the i-th term which is e k i ((e k i ) i∈N being the canonical orthogonal basis for E ω k with k = 1, 2, .., n).Actually, for all i ∈ N and for each k = 1, 2, ..., n.We confer to {(L k i ) i∈N , k = 1, 2, ..., n} as the canonical orthogonal basis for E n .
Then, , n is a symmetric, non-degenerate form on E n × E n .
The space E n := E ω1 × E ω2 × ... × E ωn endowed with the K-bilinear form , n defined in (3.2) will be called a p-adic Hilbert space.
One can easily see that if n = 1 we retrieve the p-adic Hilbert space (E ω , , ) that we had introduced in Section 1.
Clearly, the norm (3.1) is not deduced from the K-bilinear form (3.2).However we have the Cauchy-Schwarz inequality: We shall prove the Cauchy-Schwarz inequality above only for n = 2 since the proof of the general case follows along the same lines.Proof: For all (u, v), (x, y) In view of the above, the direct sum E ω × E ω is a p-adic Hilbert space.
Let us mention that if M ⊂ E ω × E ω is a subspace, then its orthogonal complement M ⊥ with respect the K-bilinear form , 2 is defined by One can easily check that M ⊥ is closed.This is actually a consequence of the continuity of the bilinear form defined by Φ (x,y) : Let Γ be the bounded linear operator which goes from Clearly, the operator Γ defined above satisfies: From (i) it follows that Γ 2 = −I (Γ is an unitary operator) where I is the identity operator of E ω × E ω .
As we mentioned in Section 1, the unitary operator Γ enables us to describe yield A * (if A ∈ U 0 (E ω )) in terms of A.
We have

Theorem 3.4:
Proof: (Theorem 3.4).Since the adjoint A * of A does exist one defines its graph by: Thus for (x, y) ∈ D(A * ) × D(A) one has It follows that Az, x = z, y .Now by uniqueness of the adjoint, we obtain that: x ∈ D(A * ) and A * x = y, hence (x, y) ∈ G(A * ).
Proof: (Theorem 3.3).Since the adjoint of A * of A does exist and that ΓG(A) is a subspace of E ω × E ω , then using Theorem 3.4 it follows that Let A ∈ U (E ω ).We define the resolvent set ρ(A) of A as the set of all λ ∈ K such that the operator A λ := λI − A is one-to-one and that A −1 λ ∈ B(E ω ).In this event, the spectrum σ(A) of A is defined as the complement of ρ(A) in K.

The diagonal operator on E ω
Let ω = (ω i ) i∈N be a sequence of nonzero terms in K and let (E ω , , ) be the corresponding p-adic Hilbert space.Define the diagonal operator A ∈ U (E ω ) by its coefficients: a i,j = λ i δ i,j where δ i,j is the usual Kronecker symbols and Observe that: Proposition 4.1: Under previous assumptions, suppose that (4.1) holds.
Proof: First of all, let us make sure that the operator A is well-defined.For that note that |a i,i | = |λ i | and that |a i,j | = 0 if i = j.We have, lim Let us show that the adjoint A * of A does exist.This is actually obvious since ∀i ∈ N, lim b i,j e j ⊗ e i , where b i,j = w −1 i w j a j,i = a i,j for all i, j ∈ N. The latter yields A = A * .
To complete the proof one needs to compute ρ(A).For that, we have to solve the equation (A − λI)x = y, where x = i x i e i ∈ D(A) and y = i y i e i ∈ E ω .Considering the previous equation on (e i ) i∈N and using the fact A is self-adjoint it follows that: ∀i ∈ N, (λ i −λ).e i , x = e i , y .Equivalently, ∀i ∈ N, (λ i − λ).ω i x i = ω i y i .Now if ∀i ∈ N, λ i = λ the previous equation has a unique solution x.Moreover, x = (A − λ) Let us show that (A − λ)  5 The equation Ax = y on E ω

The bounded case
Let A ∈ B(E ω ) be a bounded linear operator on E ω .Consider the existence and uniqueness of a solution x = i∈N x i e i ∈ E ω to the equation: b ij e j ⊗ e i , where Similarly, Thus it is clear that Another application to Theorem 5.1 concerns the perturbation of bases for E ω .This is discussed in the next subsection.

Application to the perturbation of bases on E ω
We now apply Theorem 5.1 to the perturbation of bases.Let (e i ) i∈N be the canonical basis for E ω and let (f i ) i∈N ⊂ E ω be a family of vectors.
We require the following assumption: Corollary 5.3: Let (e i ) i∈N be the canonical basis for E ω and let (f i ) i∈N ⊂ E ω be a family of vectors.Suppose that (5.2) holds true.Then (f i ) i∈N is also a basis for E ω .
Proof: One defines a linear operator A : E ω → E ω by: ∀x = According to Theorem 5.1 the operator A is a bĳective and isometric.Since in the p-adic context, an isometric linear bĳection transforms an orthogonal basis into an orthogonal basis, then (f n ) n∈N is an orthogonal basis for E ω .

The unbounded case
Let ( n ) ⊂ K be sequence of nonzero terms and let (f i ) i∈N be another orthogonal basis for E ω .We shall suppose that there exists a nontrivial isometric linear bĳection T (see Subsection 5.2) such that T e i = f i for all i ∈ N, where , ∀i ∈ N, and f i , f j = i δ i,j (δ i,j being the usual Kronecker symbol).We shall study the existence and uniqueness of solutions to (5.1) in the particular case where A is the linear operator defined by where (µ i ) i∈N ⊂ K is a given sequence of nonzero terms.It is not hard to see that A is well-defined on D(A) and that it is an element of U (E ω ).Now since f m ∈ E ω it can be written as: ∀ m ∈ N, follows that a solution x to (5.1) is given by its coefficients: x m = 1 µ m m n∈N ω n y n a nm , ∀m ∈ N.
Since (f s ) s∈N is an orthogonal basis for E ω it is then clear that N (A) = {0} (N (A) being the kernel of A), hence the above solution is unique.Now from (5.3) it is clear that x m is well-defined.The remaining task now is to prove that the solution x ∈ D(A).Indeed from the expression of x m one has:

Remark 4 . 2 :
Let us notice that the domain D(A) of the diagonal operator A may not be equal to the whole of E ω .To see it, suppose that the ground field K contains a square of each of its elements and choose x = (x i ) i∈N where xi is given by: x2 i = 1 λ 2 i ω i for all i ∈ N. According to the assumption on the field K it is clear that for all i ∈ N, xi lies in K. Now, x ∈ E ω since lim i→∞ |x i | e i = lim i→∞ 1 |λ i | = 0, by (4.1).Meanwhile, one can easily see that x ∈ D(A) since lim i |x i | |λ i | e i = 1 = 0.

. 1 )
has a unique solution x given by x = n∈N B n y.Moreover, x = y , by Theorem 5.1.

k∈N x k e k , A( k∈N x k e k ) = k∈N x k f k .n e n = sup n∈N f n f n = 1 ,
In particular, Ae k = f k , ∀k ∈ N. Now from (5.2) one has e n − f n < e n , ∀n ∈ N, that is e n = f n , ∀n ∈ N. In view of the above, we have: (i) The decomposition Ax = k∈N x k f k is well-defined; (ii) A := sup n∈N Ae in particular A is a bounded linear operator on E ω .Furthermore, I − A := sup n∈N e n − Ae n e n < 1.
(e i ) i∈N is the canonical orthogonal basis for E ω .As a consequence, |ω i | = | i | for all i ∈ N.However, let us notice that the fact |ω i | = | i | for each i ∈ N, the sequences (ω i ) i∈N and ( i ) i∈N need not be equal.Since (f i ) i∈N is an orthogonal basis for E ω , for each x ∈ E ω , x = n∈N x n f n with lim n→∞ |x n | f n = 0, where f i =: | i | 1/2 = |ω i | 1/2

and Ax := i∈N µ i x i f
i , for each x = (x i ) i∈N ∈ D(A),

aTheorem 5 . 4 :
im e i with lim i |a im | e i = 0.Under assumptions (5.3)-(5.4), the equation (5.1) has a unique solution x = m∈N x m e m ∈ D(A) where x m = 1 µ m m n∈N ω n y n a nm , ∀m ∈ N. Proof: Let y = n∈N y n e n ∈ E ω ( lim n→∞ |y n | |ω n | 1/2 = 0).Now from (5.3) it is clear that Ax, f m = n∈N y n e n , f m , hence µ m m x m = n∈N ω n y n a nm .It −1 y given above is well-defined.For that it is sufficient to prove that lim i |y i | e i = 0 one has: ) and |a ij | e i e j ≤ 1 for all i, j = 1, 2, 3, ... Thus, A := sup hence A is bounded.Actually, one can easily show that the adjoint A * does exist.Nevertheless, A is not self-adjoint.Consider B = I − A, where I is the identity operator of E ω .It is then clear that B =