Characterization of compact and self-adjoint operators, and study of - - PDF document

Characterization of compact and self-adjoint

• perators, and study of positive operators on

a Banach space over a non-Archimedean field Khodr Shamseddine University of Manitoba Canada

1

Collaboration with Jos´ e Aguayo (Universidad de Concepcion) and Miguel Nova (Universidad Cat´

• lica de la

Sant´ ısima) Concepci´

• n, Chile
• Characterization of compact and self-adjoint
• perators on Free Banach spaces of count-

able type over the complex Levi-Civita field, Journal of Mathematical Physics, Volume 54 # 2, 2013.

• Inner product on B∗-algebras of operators
• n a free Banach space over the Levi-Civita

field, Indagationes Mathematicae, Volume 26 # 1, 2015.

• Positive operators on a Banach space over

the Levi-Civita field, in preparation.

Contents

• 1. The Levi-Civita Fields R and C
• 2. Normal Projections, Compact and Self-adjoint

Operators on the Space c0(C)

• 3. B∗-algebras of Operators on c0(C)
• 4. Positive Operators
• 5. Partial Order on Self-adjoint Operators

• 1. The Levi-Civita Fields R and C
• Let R = {f : Q → R|{q ∈ Q|f(q) ̸= 0} is left-

finite}.

• Notation: An element of R is denoted by x

and its function value at q ∈ Q by x[q].

• Supp(x) = {q ∈ Q| x[q] ̸= 0}.
• For x ∈ R, define

λ(x) = { min(supp(x)) if x ̸= 0 ∞ if x = 0 .

• Arithmetic on R: Let x, y ∈ R. We define

x + y and x · y as follows. For q ∈ Q, let (x + y)[q] = x[q] + y[q] (x · y)[q] = ∑

q1+q2=q

x[q1] · y[q2]. Then, x + y ∈ R and x · y ∈ R. Result: (R, +, ·) is a field. [Levi-Civita, 1892] Definition: C := R + iR. Then (C, +, ·) is also a field.

Order in R

• Define the relation ≤ on R × R as follows:

x ≤ y if x = y or (x ̸= y and (x − y)[λ(x − y)] < 0) .

• (R, +, ·, ≤) is an ordered field.
• R is real closed.

⇓ C is algebraically closed.

• The map E : R → R, given by

E(x)[q] = { x if q = 0 else , is an order preserving embedding.

• There are infinitely small and infinitely large

elements in R: The number d, given by d[q] = { 1 if q = 1 else , is infinitely small; while d−1 is infinitely large.

For x ∈ R, define |x|0 = { x if x ≥ 0 −x if x < 0 ; |x| = { e−λ(x) if x ̸= 0 if x = 0 . For z = x + iy ∈ C, define |z|0 = √ |x|2

0 + |y|2 0 =

√ x2 + y2; |z| = { e−λ(z) if z ̸= 0 if z = 0 = max{|x|, |y|} since λ(z) = min{λ(x), λ(y)}. Note that |·|o and |·| induce the same topology τv on R (or C). Moreover, C is topologically isomorphic to R2 provided with the product topology induced by |·| in R.

Properties of (R, τv)

• (R, τv) is a disconnected topological space.
• (R, τv) is Hausdorff.
• There are no countable bases.
• The topology induced to R is the discrete

topology.

• (R, τv) is not locally compact.
• τv is zero-dimensional (i.e.

it has a base consisting of clopen sets).

• τv is not a vector topology.
• For all x ∈ R (or C): x = ∑∞

n=1 x[qn] · dqn.

Uniqueness of R and C

• R is the smallest complete and real closed

non-Archimedean field extension of R. – It is small enough so that the R-numbers can be implemented on a computer, thus allowing for computational applications.

• C is the smallest complete and algebraically

closed non-Archimedean field extension of R (or C).

• 2. Normal Projections, Compact and

Self-adjoint Operators on c0(C) c0 (C) = { (xn)n∈N : xn ∈ C; lim

n→∞ xn = 0

} ≡ c0; c0 (R) = { (xn)n∈N : xn ∈ R; lim

n→∞ xn = 0

} ; L(c0) = {T : c0 → c0 : T linear & continuous}. We consider the following form: ⟨·, ·⟩ : c0 × c0 → C; ⟨z, w⟩ =

∞

∑

n=1

znwn. This is well-defined; and it satisfies: (1) ⟨z, z⟩ ≥ 0 and ⟨z, z⟩ = 0 if and only if z = 0; (2) ⟨ az1 + bz2, w ⟩ = a ⟨ z1, w ⟩ + b ⟨ z2, w ⟩ for a, b ∈ C and z1, z2, w ∈ c0; (3) ⟨z, w⟩ = ⟨w, z⟩ for z, w ∈ c0; (4) |⟨z, w⟩|2 ≤ |⟨z, z⟩| |⟨w, w⟩| (the Cauchy-Schwarz inequality). Let ∥z∥ := √ |⟨z, z⟩|. Then ∥·∥ is a non-Archimedean norm on c0. Moreover, ∥·∥ = ∥·∥∞.

Notation: If M is a subspace of c0, then M p will denote the subspace of all y ∈ c0 such that ⟨y, x⟩ = 0, for all x ∈ M. Definition: A sequence (zn) of non-null vec- tors of c0 has the Riemann-Lebesgue Property (RLP) if for all z ∈ c0, lim

n→∞ ⟨zn, z⟩ = 0.

Note: Any basis of c0 has the (RLP) property. Theorem: Let M be an infinite dimensional closed subspace of c0. Then, the following state- ments are equivalent: (1) M has a normal complement. That is c0 = M ⊕ M p. (2) M has an orthonormal base with the Riemann- Lebesgue Property. (3) There exists a normal projection P such that N (P) = M.

Any continuous linear operator u ∈ L (c0) can be identified with a matrix of the form [u] =            α11 α12 α13 · · · α1j · · · α21 α22 α23 · · · α2j · · · α31 α32 α33 · · · α3j · · ·

. . . . . .

αi1 αi2 αi3 · · · αij · · ·

. . . . . .

↓ ↓ ↓ · · · ↓ · · ·            where (1) limi→∞ αij = 0, for any j ∈ N, (2) supi,j∈N |αij| < ∞, (3) ∥u∥ = supi,j∈N |αij| = supn∈N ∥uen∥. Definition: An operator v : c0 → c0 is said to be an adjoint of a given operator u ∈ L (c0) if ⟨u (x) , y⟩ = ⟨x, v (y)⟩ , for all x, y ∈ c0. In that case, we will say that u admits an adjoint v. We will also say that u is self-adjoint if v = u.

Lemma: Let u ∈ L (c0) with associated matrix {αi,j}i,j≥1 . Then, u admits an adjoint operator v if and only if limj→∞ |αij| = 0, for each i ∈ N. So [u] =            α11 α12 α13 · · · α1j · · · → 0 α21 α22 α23 · · · α2j · · · → 0 α31 α32 α33 · · · α3j · · · → 0

. . . . . .

αi1 αi2 αi3 · · · αij · · · → 0

. . . . . .

↓ ↓ ↓ · · · ↓ · · ·            . In the classical Hilbert space theory, any con- tinuous linear operator admits an adjoint. This is not true in the non-Archimedean case. For example, the operator u ∈ L(c0) given by the matrix:         b b2 b3 · · · bj · · · 0 0 0 · · · 0 · · · 0 0 0 · · · 0 · · ·

. . . . . .

0 0 0 · · · 0 · · ·

. . . . . .

        , with 1 < |b| , doesn’t admit an adjoint.

The following two theorems provide charac- terizations for normal projections. Theorem: Let P ∈ L(c0). Then P is a normal projection if and only if P is self-adjoint and P 2 = P. Theorem: If P : c0 → c0 is a normal projection with R (P) = [{y1, y2, · · · }], where {y1, y2, · · · } is an

• rthonormal finite subset of c0 or an orthonor-

mal sequence with the Riemann-Lebesgue Prop- erty, then Px = ∑∞

i=1 ⟨x,yi⟩ ⟨yi,yi⟩yi.

Since C is not locally compact, convex com- pact sets of c0 are trivial. Definition: A subset C of c0 is called compactoid if for every ϵ > 0 there exists a finite subset S ⊂ c0 such that C ⊂ Bc0 (0, ϵ) + co (S) , where Bc0 (0, ϵ) = {x ∈ c0 : ∥x∥ ≤ ϵ} and co (S) is the ab- solutely (closed) convex hull of S. Definition: A linear operator T : c0 → c0 is said to be compact if T (Bc0) is compactoid, where Bc0 ={x ∈ c0 : ∥x∥ ≤ 1} is the unit ball of c0. Theorem: T ∈ L(c0) is compact if and only if, for each ϵ > 0, there exists a linear operator of finite-dimensional range S such that ∥T − S∥ ≤ ϵ.

The following theorem provides a way to con- struct compact and self-adjoint operators start- ing from an orthonormal sequence. Theorem: Let {y1, y2, · · · } be an orthonormal se- quence in c0. Then, for any λ = (λn) in c0 (R), the map T : c0 → c0 defined by T (·) =

∞

∑

n=1

λn ⟨·, yn⟩ ⟨yn, yn⟩yn ≡

∞

∑

n=1

λnPn (·) is a compact and self-adjoint operator. The converse is also true, as the following the-

• rem shows.

Theorem: If the linear operator T : c0 → c0 is compact and self-adjoint, then there exist λ = (λn) ∈ c0 (R) and an orthonormal sequence {yn} in c0 such that T =

∞

∑

n=1

λn ⟨·, yn⟩ ⟨yn, yn⟩yn.

• 3. B∗-algebras

Let A0 ≡ A0 (c0) := {T ∈ L (c0) : T has an adjoint}.

• A0 is a non-commutative Banach algebra

with unity.

• A0 contains normal projections.
• T ∈ A0 if and only if its associated matrix

has the form: [T] =           α11 α12 α13 α1j → 0 α21 α22 α23 α2j → 0 α31 α32 α33 α3j → 0 . . . . . . . . . ... . . . αi1 αi2 αi3 αij → 0 ↓ ↓ ↓ ↓ · · · · · ·          

• A0 can be rewritten as

A0 = { T ∈ L (c0) : ∀y ∈ c0, lim

j→∞ ⟨Tej, y⟩ = 0

} .

• For T ∈ A0, T ∗ ∈ A0 and (T ∗)∗ = T ∗∗ = T.

Therefore, the map ∗ : A0 → A0; T → T ∗, is an involution on A0.

• Altogether, we say A0 is a non-Archimedean

B∗−algebra. Let A1 = { T ∈ L (c0) : lim

n→∞ Ten = 0

} .

• From |⟨Ten, y⟩| ≤ ∥Ten∥ ∥y∥ , we have that

A1 ⊂ A0. But A1 ̸= A0 since Id / ∈ A1.

• A1 is a closed subalgebra of A0.
• T ∈ A1 ⇔ T is compact and T ∈ A0.

Let A2 = {T ∈ A1 : T = T ∗}.

• Note that the operator

S (·) =

∞

∑

i=1

ai ⟨·, ei⟩ ei+1, where a = (ai) ∈ c0, is in A1, but it is not self-adjoint; therefore A2 is a proper subset

• f A1.
• A2 is a closed subset of A1.
• T ∈ A2 if and only if there exist an element

(λn) ∈ c0(R) and an orthonormal sequence {yn}n∈N in c0 such that T =

∞

∑

n=1

λn ⟨·, yn⟩ ⟨yn, yn⟩yn. – ∥T∥ = ∥ (λn) ∥. – A2 ∼ = c0(R).

Inner Product in A1 Since limn→∞ Sen = 0 and limn→∞ Ten = 0 for S, T ∈ A1, the mapping ⟨·, ·⟩ : A1×A1 → C; (S, T) → ⟨S, T⟩ =

∞

∑

i=1

⟨Sei, Tei⟩ , is well-defined, linear in the first variable and linear conjugate in the second variable.

• ⟨S, T⟩ = ⟨T, S⟩ for all S, T ∈ A1.
• ⟨S, S⟩ ≥ 0 and ⟨S, S⟩ = 0 ⇔ S = 0.
• √

|⟨S, S⟩| = ∥S∥ for all S ∈ A1.

• |⟨S, T⟩| ≤ ∥S∥∥T∥ for all S, T ∈ A1.
• c0 is isometrically isomorphic to a closed

subspace S of A1. Moreover, the restriction

• f the inner product in A1 to S coincides

with the inner product defined in c0.

• 4. Positive operators

Definition: For T ∈ A1, we say that T is positive and write T ≥ 0 if ⟨Tx, x⟩ ≥ 0 for all x ∈ c0(C). Proposition: Let S, T ≥ 0 in A1 and α ≥ 0 in R be given. Then

• αS + T ≥ 0.
• T is self-adjoint; that is T ∈ A2.
• For all x, y ∈ c0, we have that

|⟨Tx, y⟩|2 ≤ |⟨Tx, x⟩| |⟨Ty, y⟩| . Proposition: Let T ∈ A1. Then both TT ∗ and T ∗T are positive.

Theorem: For T ∈ A1, the following are equiv- alent: (1) T ≥ 0. (2) T is self-adjoint; and all of its eigenvalues are in R and non-negative. (3) There exists S ≥ 0 in A1 such that T = S2. (4) There exists S ∈ A1 such that T = S∗S. Proof: (1) ⇒ (2): Assume T ≥ 0. Then T is self-

• adjoint. Now let λ be an eigenvalue of T and let

v ∈ c0(C) be a corresponding eigenvector. Then 0 ≤ ⟨Tv, v⟩ = ⟨λv, v⟩ = λ⟨v, v⟩. Since ⟨v, v⟩ > 0, it follows that λ ∈ R and λ ≥ 0. (2) ⇒ (3): Assume (2) is true. Since T is com- pact and self-adjoint, there exist (λn) ∈ c0(R) and an orthonormal sequence {yn} in c0(C) such that T =

∞

∑

n=1

λn ⟨·, yn⟩ ⟨yn, yn⟩yn. For each m ∈ N, we have that Tym =

∞

∑

n=1

λn ⟨ym, yn⟩ ⟨yn, yn⟩ yn = λmym.

Thus, λn is an eigenvalue of T for each n and hence λn ∈ R and λn ≥ 0 for all n ∈ N. Let S : c0(C) → c0(C) be given by S =

∞

∑

n=1

√ λn ⟨·, yn⟩ ⟨yn, yn⟩yn. Then S ∈ A1. For all x ∈ c0(C), we have that ⟨Sx, x⟩ =

∞

∑

n=1

√ λn ⟨x, yn⟩ ⟨yn, yn⟩⟨yn, x⟩ =

∞

∑

n=1

√ λn ⟨x, yn⟩⟨x, yn⟩ ⟨yn, yn⟩ ≥ 0. Hence S ≥ 0. Also, for all x ∈ c0(C): S2x = S(Sx) = S ( ∞ ∑

n=1

√ λn ⟨x, yn⟩ ⟨yn, yn⟩yn ) =

∞

∑

n=1

√ λn ⟨x, yn⟩ ⟨yn, yn⟩S (yn) =

∞

∑

n=1

λn ⟨x, yn⟩ ⟨yn, yn⟩yn = Tx. Hence S2 = T. (3) ⇒ (4): Assume there exists S ≥ 0 in A1 such that T = S2. Then S is self-adjoint. Thus, S = S∗ and hence T = S2 = SS = S∗S. (4) ⇒ (1): Follows from previous proposition.

Remark: Let T and S be as in the previous

• theorem. Then
• S is unique. We say S is the positive square

root of T and write S = √ T.

• Moreover,

∥S∥ = ∥( √ λn)∥ = max

n∈N

{ | √ λn| } = max

n∈N

{ |λn|1/2} = [ max

n∈N {|λn|}

]1/2 = ∥(λn)∥1/2 = ∥T∥1/2. Proposition: Let T ≥ 0 in A1, let S = √ T, and let R ∈ A1 be given. Then TR = RT ⇔ SR = RS.

Proposition: Let T ≥ 0 in A1 and x ∈ c0(C) be

• given. Then ⟨Tx, x⟩ = 0 if and only if Tx = 0.

Proof: If Tx = 0 then ⟨Tx, x⟩ = 0 by definition

• f the inner product. Now assume ⟨Tx, x⟩ = 0.

Then, since T ≥ 0, there exists S ∈ A1 such that T = S∗S. Thus, ⟨S∗Sx, x⟩ = 0, and hence ⟨Sx, Sx⟩ = 0, from which we get Sx = 0. It fol- lows that Tx = S∗Sx = S∗0 = 0. Corollary: Let T ≥ 0 in A1. Then ⟨Tx, x⟩ = 0 for all x ∈ c0(C) if and only if T = 0. Proposition: Let S, T ∈ A1 be positive. Then ST ≥ 0 ⇔ ST = TS. Proof: (⇒): Easy. (⇐): Assume ST = TS. Let N = √

• T. Applying

a previous proposition, we have that NS = SN. Now let x ∈ c0(C) be given. Then ⟨STx, x⟩ = ⟨S(NN)x, x⟩ = ⟨(SN)Nx, x⟩ = ⟨(NS)Nx, x⟩ = ⟨N(SN)x, x⟩ = ⟨SNx, N ∗x⟩ = ⟨SNx, Nx⟩ ≥ 0.

Proposition: Let T ∈ A2 be given. Then there exist unique positive operators A and B such that T = A − B and AB = BA = 0. Proof: Since T is compact and self-adjoint, there exist (λn) ∈ c0(R) and an orthonormal sequence {yn} in c0(C) such that T =

∞

∑

n=1

λn

⟨·,yn⟩ ⟨yn,yn⟩yn. Thus,

T =

∞

∑

n = 1 λn > 0

λn ⟨·, yn⟩ ⟨yn, yn⟩yn +

∞

∑

n = 1 λn < 0

λn ⟨·, yn⟩ ⟨yn, yn⟩yn =

∞

∑

n = 1 λn > 0

λn ⟨·, yn⟩ ⟨yn, yn⟩yn −

∞

∑

n = 1 λn < 0

(−λn) ⟨·, yn⟩ ⟨yn, yn⟩yn = A − B. For all x ∈ c0(C), we have that ABx =

∞

∑

n = 1 λn < 0

(−λn) ⟨x, yn⟩ ⟨yn, yn⟩A(yn) =

∞

∑

n = 1 λn < 0

(−λn) ⟨x, yn⟩ ⟨yn, yn⟩   

∞

∑

l = 1 λl > 0

λl ⟨yn, yl⟩ ⟨yl, yl⟩ yl    = 0. Hence AB = 0. It follows that BA = 0 too.

Uniqueness: Assume that T = A1 − B1 with A1 and B1 positive operators in A2 and A1B1 = B1A1 = 0. Write A1 =

∞

∑

l=1

αl ⟨·, xl⟩ ⟨xl, xl⟩xl and B1 =

∞

∑

j=1

βj ⟨·, zj⟩ ⟨zj, zj⟩zj. Fix l0 ∈ N. Then Txl0 = (A1 − B1) xl0 = A1xl0 − B1xl0 = αl0xl0 − B1A1 ( 1 αl0 xl0 ) = αl0xl0, since B1A1 = 0. Thus, αl0 is an eigenvalue of T; and hence αl0 is equal to some λn > 0. Similarly we show that, for each j ∈ N, −βj is equal to some λn < 0. Hence {αl : l ∈ N} = {λn : n ∈ N, λn > 0} and {−βj : j ∈ N} = {λn : n ∈ N, λn < 0} . It then follows that A1 = A and B1 = B.

Proposition: Let T, A and B be as above. Then (1) ⟨A, B⟩ = 0; and (2) ∥T∥ = max {∥A∥ , ∥B∥}. Proof: (1) First note that since A and B are positive

• perators, they are both self-adjoint. Thus,

⟨A, B⟩ =

∞

∑

n=1

⟨Aen, Ben⟩ =

∞

∑

n=1

⟨B∗Aen, en⟩ =

∞

∑

n=1

⟨BAen, en⟩ =

∞

∑

n=1

⟨0, en⟩ = 0. (2) Also ⟨B, A⟩ = ⟨A, B⟩ = 0. Thus, ⟨T, T⟩ = ⟨A − B, A − B⟩ = ⟨A, A⟩ + ⟨B, B⟩ ; and hence |⟨T, T⟩| = |⟨A, A⟩ + ⟨B, B⟩| = max {|⟨A, A⟩| , |⟨B, B⟩|}

• r, equivalently,

∥T∥2 = max { ∥A∥2 , ∥B∥2} = (max {∥A∥ , ∥B∥})2 . It follows that ∥T∥ = max {∥A∥ , ∥B∥}.

• 5. Partial Order on A2

Definition: For S, T ∈ A2, we say that S ≥ T (or T ≤ S) if S − T ≥ 0. Theorem: The relation ≥ defines a partial or- der on A2.

• Reflexivity: For all T ∈ A2, T ≥ T.
• Antisymmetry: Let S, T ∈ A2 be such that

S ≥ T and T ≥ S. Then S − T ≥ 0 and T − S ≥ 0. Thus, for all x ∈ c0(C): ⟨(S − T)x, x⟩ ≥ 0 and ⟨(T − S)x, x⟩ ≥ 0, from which we get ⟨(S − T)x, x⟩ = 0 for all x ∈ c0(C). It follows that S − T = 0 and hence S = T.

• Transitivity: Let R, S, T ∈ A2 be such that

R ≥ S and S ≥ T. Then R − S ≥ 0 and S − T ≥ 0. It follows that R − T = (R − S) + (S − T) ≥ 0, and hence R ≥ T.

Example: Let S, T ∈ A2 be the operators given by the matrix representations [S] =     1 0 0 · · · 0 1 0 · · · 0 0 0 · · · . . . . . . . . . ...     and [T] =     0 0 0 · · · 0 2 0 · · · 0 0 0 · · · . . . . . . . . . ...     . Then [S−T] =     1 0 · · · 0 −1 0 · · · 0 · · · . . . . . . . . . ...     ; [T −S] =     −1 0 0 · · · 1 0 · · · 0 0 · · · . . . . . . . . . ...     . Since both S − T and T − S have a negative eigenvalue (-1), it follows that neither is ≥ 0. Proposition: If S ≥ T and U ≥ V in A2 then S + U ≥ T + V . Moreover, if S ≥ T in A2 and α ≥ 0 in R then αS ≥ αT. Proof: Let S, T, U, V ∈ A2 be such that S ≥ T and U ≥ V ; and let α ≥ 0 in R be given. Then (S + U) − (T + V ) = (S − T) + (U − V ) ≥ 0 αS − αT = α(S − T) ≥ 0

For R, S, T ∈ A2: R ≥ 0 and S ≥ T ̸⇒ SR ≥ TR. Example: Let R, S, T in A2 be given by [R] =     0 0 0 · · · 0 1 0 · · · 0 0 0 · · · . . . . . . . . . ...     ; [S] =     2 −1 0 · · · −1 1 0 · · · 0 · · · . . . . . . . . . ...     ; [T] =     1 0 0 · · · 0 0 0 · · · 0 0 0 · · · . . . . . . . . . ...     . Then R ≥ 0 and [S − T] =     1 −1 0 · · · −1 1 0 · · · 0 · · · . . . . . . . . . ...     ; so S − T ≥ 0 since, for all x ∈ c0(C), we have that ⟨(S − T)x, x⟩ = x1(x1 − x2) + x2(x2 − x1) = |x1|2

• − x1x2 − x2x1 + |x2|2
• = |x1|2
• − 2R (x1x2) + |x2|2
• ≥ |x1|2
• − 2|x1|o|x2|o + |x2|2
• = (|x1|o − |x2|o)2 ≥ 0,

where, for z = α + iβ ∈ C, R(z) = α denotes the R-part of the C-number z.

However, [SR] =     0 −1 0 · · · 1 0 · · · 0 · · · . . . . . . . . . ...     and [TR] = 0. Thus [SR − TR] =     0 −1 0 · · · 1 0 · · · 0 · · · . . . . . . . . . ...     . Therefore SR−TR ̸≥ 0 since it is not self-adjoint; and hence SR ̸≥ TR. Proposition: Let S, T ∈ A2 be given. Then S ≥ T ⇔ ⟨Sx, x⟩ ≥ ⟨Tx, x⟩ for all x ∈ c0(C). Proposition: Let S ≥ T in A2 and let R ∈ A1 be

• given. Then R∗SR ≥ R∗TR.

Proof: Note that R∗SR, R∗TR ∈ A2. Let x ∈ c0(C) be given. Then ⟨(R∗SR − R∗TR)x, x⟩ = ⟨R∗(S − T)Rx, x⟩ = ⟨(S − T)Rx, Rx⟩ ≥ 0 since S − T ≥ 0. Thus, R∗SR − R∗TR ≥ 0, and hence R∗SR ≥ R∗TR.