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1 Definitions and Notation

Let B1, B2 be Banach spaces with the norms k·k1 and k·k2 such that B1 ⊂B2.

We will write k·k1 ≻k·k2 if

+ ∃C ∈ R ∀x ∈B1 kxk2 ≤ Ckxk1. Throughout this paper we adhere to the following notation. By M we denote a von Neumann algebra that acts on a Hilbert space H with the scalar product h·, ·i. We denote its selfadjoint part by Msa, and the set of all projections in M pr pr by M . Let p ∈M and x ∈M, then by xp we denote the restriction of pxp to pH (i.e. xp := pxp|pH ), also we denote the reduction of M to pH by Mp. By h C(M) we denote the center of M. By M∗ and M∗ we denote the predual of M and its Hermitian part, respectively. If an operator x is affiliated with M then we write xηM. We denote the domain of an operator x by D(x). The adjoint operator is denoted by x∗. We denote the identity operator, the zero operator and the zero vector by 1,0 and 0 , respectively. We use standard notation for multiplication of a functional ϕ ∈ M∗ by an operator x ∈M, namely, xϕ, ϕx and xϕx denote the linear functionals y 7→ ϕ(xy), y 7→ ϕ(yx) and y 7→ ϕ(xyx), respectively. We also consider partial order for positive selfadjoint operators affiliated with M. If x is affiliated with M we denote it as xηM. For positive selfadjoint x, y η M we write x ≤ y if and only if

1 1 1 1 1 D(y 2 ) ⊂ D(x 2 ) and kx 2 fk2 ≤ky 2 fk2 for all f ∈ D(y 2 ).

If for an increasing net (xj )j∈J of operators affiliated with M there exists x = sup xj , then we write xj ր x. For a positive selfadjoint operator xηM we use j∈J −1 + xλ to denote λx(λ + x) with λ ∈ R \{0}. From the Spectral theorem it + follows that the mapping λ 7→ xλ ∈M is monotone operator-valued function 1 1 1 2 and lim x f = x 2 f for all f ∈ D(x 2 ), therefore xλ ր x. For an unbounded λ→+∞ λ + x and ϕ ∈M∗ we define ϕ(x) as ϕ(x) := lim ϕ(xλ). λ→+∞ From now on a stands for a positive selfadjoint operator affiliated with M. We consider D+ a ≡{ϕ ∈M∗ | ϕ(a) < +∞},

2 Dh D+ D+ D Dh a ≡ a − a and a ≡ linC a. D+ + Dh h D Note that if operator a is bounded, then a = M∗ , a = M∗ and a = M∗. Dh We define a seminorm k·ka on a as

+ kϕka := inf{ϕ1(a)+ ϕ2(a) | ϕ = ϕ1 − ϕ2; ϕ1, ϕ2 ∈ D }.

Also, Theorem 2 from [7] states, that if operator a is bounded, then

1 1 2 2 kϕka = ka ϕa k.

If k·ka is a norm, then we call it the a-norm. Note that the 1-norm coincides h with the restriction of the standard norm in M∗ onto M∗ .

h definition 1 ([16]). By L1 (a) we denote the completion of the real normed Dh D+ space a = linR a with the norm D+ ra(ϕ) = inf{ϕ1(a)+ ϕ2(a) | ϕ = ϕ1 − ϕ2, ϕ1, ϕ2 ∈ a },

D+ + where a = {ϕ ∈M∗ | ϕ(a) < +∞}.

h sa The dual of L1 (a) is (L∞(a), k·ka), where

al L∞(a) ≡{x ∈ (Da) | λ ∈ R, −λa ≤ x ≤ λa}

R Dh and kxka ≡ inf{λ ∈ | − λa ≤ x ≤ λa}. We identify the elements of a h with the corresponding elements in L1 (a). Further for an injective operator a sa we always assume that L∞(a) is equiped with the a-norm. 1 1 1 1 For x ∈ M we define the sesquilinear form a a 2 xa 2 on D(a 2 ) × D(a 2 ) by \1 1 1 1 the equality a 2 xa 2 (f,g) := hxa 2 f,a 2 gi. The set of all such sesquilinear forms is denoted by \1 1 Sa(M) ≡{a 2 xa 2 | x ∈ M}.

sa We consider partial order on Sa(M ), such that

\1 1 \1 1 a 2 xa 2 ≤ a 2 ya 2

\1 1 \1 1 1 sa if and only if a 2 xa 2 (f,f) ≤ a 2 ya 2 (f,f) for all f ∈ D(a 2 ). By Sa(M ) we 1 1 denote the seminormed space of sesquilinear forms {a 2 xa 2 | x ∈Msa} equiped \1 1 + \1 1 \1 1 \1 1 with the seminorm pa(a 2 xa 2 ) := inf{λ ∈ R | −λa 2 1a 2 ≤ a 2 xa 2 ≤ λa 2 1a 2 }. definition 2 ([1]). Let (X0,X1) be the pair of Banach spaces. For t > 0 and x ∈ X0 + X1 let

K(x, t; X0,X1) = inf kx0kX0 + t kx1kX1 | x = x0 + x1, x0 ∈ X0, x1 ∈ X1 .  3 By K-method of interpolation we call the construction of the space Kθ,q(X0,X1) as the linear subspace of the sum X0 + X1 such that

∞ 1/q q dt t−θK(x, t; X ,X ) < ∞. 0 1 t Z0  h
definition 3. By Lp,q(a) we denote the noncommutative Lorentz space with h h p, q ∈ (1, +∞) which is the interpolation space K(p−1)/p,q(L1 (a),L∞(a))

h definition 4. By Lp (a) we denote the noncommutative Lebesgue space which h is interpolation space Lp,p(a).

2 Preliminaries

For ϕ ∈ Da the equality

1 1 1 1 a 2 ϕa 2 (x) = lim ϕ(a 2 xa 2 ) with x ∈M λ→+∞ λ λ

1 1 defines the normal functional a 2 ϕa 2 ∈M∗. If an operator a is injective then

\1 1 \1 1 \1 1 inf{λ | − λa 2 1a 2 ≤ a 2 xa 2 ≤ λa 2 1a 2 } = kxk for any x ∈Msa

\1 1 and the latter implies that the mapping u1 : x 7→ a 2 xa 2 is an isometrical isomorphism of M onto Sa(M). For an injective operator a the mapping

\1 1 u : x ∈ M 7→ a 2 xa 2 ∈ L∞(a) is an isometrical isomporphism of M onto L∞(a). Thus, L∞(a) is isometricaly −1 isomorphic to Sa(M). Further we call the isomorphism u1 u the canonical iso- morphism of Sa() 7→ M onto L∞(a) and identify the corresponding elements. t ∗ Moreover, the adjoint mapping u is an isometrical isomorphism of L∞(a) onto M∗. For an injective operator a the continue of the mapping

1 1 2 2 v : ϕ ∈ Da(⊂ L1(a)) 7→ a ϕa ∈M∗ is an isometrical isomorphism of L1(a) onto M∗

Let B1 and B2 be Banach spaces and k·k1 ≻ k·k2, then the embedding

id : x ∈B1 7→ x ∈B2 is continuous.

4 3 Mixed Limits of Banach Spaces

By Bk,n we denote two-indexed family of Banach spaces. Let τ k,n be the topol- k,n ogy of the norm k·kk,n which is natural norm of the Banach space B .

Let k1 < k2 and n1

k1,n1 k1,n2 B ⊂ B k·kk1,n1 ≻ k·kk1,n2 ∪ ∪ and f f . k2,n1 k2,n2 B ⊂ B k·kk2,n1 ≻ k·kk2,n2 Note that k1,n1 k1,n2 τ |k2 ⊃ τ |k2,n1 ∩ ∩ , k2 ,n1 k2,n2 τ ⊃ τ |n1

k2,n1 with τ|k2 ≡ τ|n1 ≡ τ|k2 ,n1 := {X ∩B | X ∈ τ}. Consider the limits

k k,n k,n B = B и Bn = B n>0 [ k>\0 k k with the topology τ and the topology τn, respectively. The topology τ is the strongest topology on Bk such that the mappings

k k,n Bk ϕn : x ∈ (B , k·kk,n) 7→ x ∈ (1) are continuous and the topology τn is the weakest on Bn such that the mappings

n B k,n ψk : x ∈ n 7→ x ∈ (B , k·kk,n) (2) are coninuous.

Lemma 1. Let k

(i) Bk ⊃ Bn;

(ii) Bk ⊂ Bn.

Proof. (i) We have

Bn = Bn,γ, Bk = Bk,γ и Bn,γ ⊂Bk,γ . γ>0 γ>0 [ [

n n,γ0 k,γ0 Evidently, if x ∈ B , then there exists γ0 such that x ∈B ⊂B . Therefore, x ∈ Bk.

5 γ,k γ,n (ii) If x ∈ Bk, then ∀γ > 0 x ∈B ⊂B . Thus,

x ∈ Bγ,n = Bn. γ>0 \ definition 5. We call

L = limBk,k ≡ Bn,n ≥ k>\0 n[k an upper limit. definition 6. And we call

L = limBk,k ≡ Bn,n ≥ k>[0 n\k a lower limit.

Proposition 1. L ⊃ L.

k1,n k2,n Proof. If k1 < k2, then B ⊃B . Therefore,

L = Bk,n = Bk,n ⊃ Bn,n. n>0 ≥ ≥ k>\0 [ k>\0 n[k k>\0 n[k

k,n1 k,n2 If n1

Bn,n ⊃ Bn,k = Bn,k = L. ≥ ≥ n>0 k>[0 n\k k>[0 n\k k>[0 \ definition 7. The family Bk,n is called converging if L = L.

Lemma 2. Let k

k n k τ |n := {X ∩ B | X ∈ τ }

k (i.e. the topology induced by the τ on Bn), then

k n τ |n ⊂ τ .

6 k Bk k k −1 Proof. Let X0 ∈ τ |n, i.e. X0 = X ∩ , where X ∈ τ . Then (ϕγ ) (X) is k,γ open in (B , k·kk,γ). The embedding

n,k n,γ k,γ mγ : x ∈ (B , k·kn,γ) 7→ x ∈ (B , k·kk,γ ) is continuous, therefore

k n,k −1 n,k −1 k −1 (ϕγ mγ ) (X) = (mγ ) (ϕγ ) (X)

n,γ is open in (B , k·kn,γ) for any γ > 0. n n The topology τ is the strongest topology, such that any embedding ϕγ is n continuous. If X0 ∈/ τ , then there exists topology

n n n τ = τ ∪{X0 ∩ Y | Y ∈ τ }∪{X0 ∪ Y | Y ∈ τ }

n stronger, then the topology τ , X0 ∈ τ. We show that for any A ∈ τ the n −1 preimage (ϕγ ) (A) is open. n n Consider three cases A ∈ τ , A = X0 ∩ Y , A = X0 ∪ Y (Y ∈ τ ).

n β −1 1. If A ∈ τ , then (ϕγ ) (A) is open.

2. If A = X0 ∩ Y , then

n −1 Bn n −1 n,γ n −1 (ϕγ ) (X ∩ ∩ Y ) = (ϕγ ) X ∩ B ∩ (ϕγ ) (Y )= γ>0 ! [

n −1 n,γ n −1 k n,k −1 n −1 = (ϕγ ) (X ∩B ) ∩(ϕγ ) (Y )= ϕγ mγ (X) ∩(ϕγ ) (Y ). γ>0 ! γ>0 ! [ [
k n,k −1 n −1 is open, since ϕγ mγ (X) and ϕγ (Y ) are open.

3. If A = X0 ∪ Y , then

n −1 Bn n −1 n,γ n −1 (ϕγ ) ((X ∩ ) ∪ Y ) = (ϕγ ) X ∩ B ∪ (ϕγ ) (Y )= γ>0 ! [

n −1 n,γ n −1 k n,k −1 n −1 = (ϕγ ) (X ∩B ) ∪(ϕγ ) (Y )= ϕγ mγ (X) ∪(ϕγ ) (Y ) γ>0 ! γ>0 ! [ [
k n,k −1 n −1 is open, since ϕγ mγ (X) and ϕγ (Y ) are open.

Thus, we get the contradiction with the maximality of the topology τ n, n therefore X0 ∈ τ .

7 Lemma 3. Let k

τn|k := {X ∩ Bk | X ∈ τn} be the topology induced by the topology τn on Bk, then

τk ⊃ τn|k.

∞ Proof. The topology τk is determinded by the family of semi-norms {k·kγ,k}γ=1, ∞ and the topology τn|k is determined by the family {k·kγ,n}γ=1. Evidently, that k·kγ,n ≺k·kγ,k.

For L define the topology τ the strongest topology, such that any embedding

Φk : x ∈ (Bk, τk) 7→ x ∈ (L, τ) (3) is continuous. Also, determine the topology τ, such that it will be the weakest topology on L with the embedings

k k Ψk : x ∈ (L, τ) 7→ x ∈ (B , τ ) (4) being continuous.

Theorem 1. The embedding

Λ: x ∈ (L, τ) 7→ x ∈ (L, τ) (5) is continuous.

Proof. Note that

k ∀n0 > 0 τ|n0 = τn|n0 и τ = τ |∞. ≥ 0 n\n k>[0 At the same time

k,n k k,n τn = τ |∞ и ∀n0 > 0 τ |n0 = τ |n0 . ≥ 0 k>[0 n\n

It is sufficient to prove that for any n0 > 0 we have the inclusion

B τ|n0 ⊃ τ|n0 , where τ|n0 = {X ∩ n0 | X ∈ τ}.

Thus k,n τ|n0 = τ |∞ |n0 ≥ 0 ! n\n k>[0

8 и k τ|n0 = τ |∞ |n0 . ! k>[0 B Note, that by reduction we obtain n0 space with the topology:

k,n k τ|n0 = τ |n0 и τ|n0 = τ |n0 . n≥n0 k>0 ! k>0 \ [
[
k k,n But then τ |n0 = n≥n0 τ |n0 , thus T τ| = τ k,n| ⊃ τ k,n| = τ| . n0 n0  n0  n0 n≥n0 k>0 ! k>0 n≥n0 \ [
[ \
 

4 Limits of noncommutative L∞ spaces

We split this section into four parts. Firstly, we consider general properties for

L∞(a) spaces and its norms. Secondly, we consider case of bounded a, in the third case we consider unbounded a such that a−1 is bounded, and at last we consider the general case of unbounded a.

4.1 Case of bounded operator

Lemma 4. Let aηM (a is affiliated with M, a ≥ 0), then Sa(M) = Sλa(M) for any λ> 0.

\1 1 1 1 1 1 Proof. Let a 2 xa 2 ∈ Sa(M). Evidently, D(a 2 )= D((λa) 2 )= D(λ 2 a 2 ), thus

\1 1 1 1 1 1 1 1 2 2 2 2 2 2 a xa (f,g)= hxa f,a gi = hx 1 (λa) f, 1 (λa) gi = λ 2 λ 2

1 1 1 1 \1 1 1 = hx(λa) 2 f, (λa) 2 gi = (λa) 2 x(λa) 2 (f,g) for any f,g ∈ D(a 2 ). λ λ \1 1 \1 1 Hence, λa 2 xa 2 = (λa) 2 x(λa) 2 , and since Sa(M), Sλa(M) are linear spaces, we have Sa(M)= Sλa(M).

Particularly, if a is bounded, then Sa(M) = Sa0 (M), where a0 = a/kak, if −1 −1 a is such that a is bounded, then Sa(M)= Sa∞ (M), where a∞ = ka ka.

Proposition 2. Let aηM, a ≥ 0, then k·ka in L∞(a) is equivalent to k·kλa, where λ> 0.

9 Proof. Since, L∞(a) is isometrically isomorphic to Sa(M) for x ∈ L∞(a) we \1 1 may consider x = a 2 ya 2 , where y ∈M, then kxka = kyk. On the other hand,

1/2 1/2 \1 1 λ 1 1 λ kxk = ka 2 ya 2 k = a 2 ya 2 = λa λa λ1/2 λ1/2 λa

1 1 1 1 1 1 1 1 2 2 2 2 = (λa) y(λa) = (λa ) y(λa) = kyk = kxka λ λ λa λ λ λa

+ Particularly, for the bounded a ∈M we have k·ka is equivalent to k·ka0 , −1 −1 where a0 = a/kak. If a is bounded and we consider a∞ = ka ka, then k·ka∞ is equivalent to k·ka. α Now, consider L∞(a ).

Lemma 5. For a bounded a ∈M+ and α, β ∈ R+, if α<β, then

α β L∞(a ) ⊃ L∞(a ).

β β/\2 β/2 Proof. Let y ∈ L∞(a ), then we consider y = a xa , where x ∈M. Thus, \ y(f,g)= aβ/2xaβ/2(f,g)= hxaβ/2f,aβ/2gi =

1 = h(a(β−α)/2xa(β−α)/2)aα/2f,aα/2gi for any f,g ∈ D(a 2 ). Note, that x′ := a(β−α)/2xa(β−α)/2 ∈M, thus

α/\2 ′ α/2 α y = a x a ∈ L∞(a ).

The following definition is standard. definition 8. Let X ⊃ Y be normed spaces with the norms k · kX , k · kY , + respectively. We write k·kX ≺k·kY if and only if there exists C ∈ R , such that for any x ∈ Y the inequality kxkX ≤ CkxkY holds. Lemma 6. For a bounded a ∈M+ and α, β ∈ R+. If α<β, then

k·kaα ≺k·kaβ .

β β/\2 β/2 Proof. Let y ∈ L∞(a ), then y = a xa , where x ∈M, then

β/\2 β/2 (β−α)/2 (β−α)/2 kykaα = ka xa kaα = ka xa k≤

(β−α)/2 2 (β−α) (β−α) ≤ka k kxk = ka kkxk = ka kkykaβ

10 Now, let us consider the limit space.

+ definition 9. For the bounded a ∈M we define the topological space (L∞(a), τ(a)) α α α as the limit space for L∞(a ), where L∞(a) := L∞(a ) and τ(a) := τ∞(a ), α>1 α>1 α α α α τ∞(a ) = {X ∩ L∞(a) | X ∈ τ(a )}, τ(a ) isT the topology on L∞(a S) of the norm k·kaα .

The latter definition essentially means, that τ(a) is the initial topology on the L∞(a) for the family of mapping

α ϕα : x ∈ L∞(a) 7→ x ∈ (L∞(a ), k·kaα ).

Also, we can describe this topology as the topology on L∞(a) defined by the family of the seminorms {k·kaα }α>1. The familiy of the seminorms {k·kan }n∈N describes the same topology, thus we get the following theorem.

+ Theorem 2. For the bounded a ∈ M the space (L∞(a), τ(a)) is metriziable locally-convex space (Frechet space).

For the more detailed proof of the latter theorem see [14][Lemma 2].

4.2 Case of unbounded operator with bounded inverse

Now, consider case, when aηM,a ≥ 0, a is not bounded, but a−1 is bounded.

Lemma 7. For a aηM, a ≥ 0 and α, β ∈ R+. If α<β, then

α β L∞(a ) ⊂ L∞(a )

α α/\2 α/2 Proof. Let y ∈ L∞(a ), then we consider y = a xa , where x ∈M. Thus, \ y(f,g)= aα/2xaα/2(f,g)= hxaα/2f,aα/2gi =

1 = h(a(α−β)/2xa(α−β)/2)aβ/2f,aβ/2gi for any f,g ∈ D(a 2 ).

Note, that x′ := (a−1)(β−α)/2x(a−1)(β−α)/2 ∈ M, therefore we have that y = β/\2 ′ β/2 β a x a ∈ L∞(a ).

Lemma 8. For aηM, a ≥ 0 such that a−1 is bounded and α, β ∈ R+, such that α<β we have

k·kaα ≻k·kaβ .

11 α α/\2 α/2 Proof. Let y ∈ L∞(a ), then y = a xa , where x ∈M, then

α/\2 α/2 (α−β)/2 (α−β)/2 kykaβ = ka xa kaβ = ka xa k≤

≤k(a−1)(β−α)/2k2kxk =

−1 (β−α) −1 (β−α) = k(a ) kkxk = k(a ) kkykaα

definition 10. For the unbounded a ∈ M+ with bounded inverse a−1 we α define the topological space (L∞(a), τ∞(a)) as the limit space for L∞(a ), where α L∞(a) := L∞(a ) and τ(a) is the strongest topology such that for all α> 0 α>1 α the mappingsS mα : x ∈ L∞(a ) 7→ x ∈ (L∞(a), τ∞(a)) are continuous.

Evidently, (L∞(a), τ∞(a)) is an (LB)-space.

4.3 Case of unbounded operator with unbounded inverse

Now, consider case, when aηM, a ≥ 0, a and a−1 are unbounded, simultane- ously. Then we take spectra gecomposition

+∞ a = λdPλ Z0 and determine

1 +∞ 1 +∞ a0 := λdPλ, a∞ := λdPλ; p0 := dPλ, p∞ := dPλ. Z0 Z1 Z0 Z1 Note, that

a0 = ap0 = p0a = p0ap0,

a∞ = p∞a∞ = a∞p∞ = p∞a∞p∞,

p∞p0 = p0p∞ = 0 = a0a∞ = a∞a0.

Evidently, H = p0H ⊕ p∞H, thus we represent M as a subalgebra of the algebra of matrices M2(M):

x p xp M p Mp x = p0 0 ∞ ∈ p0 0 ∞ ; p∞xp0 xp∞ ! p∞Mp0 Mp∞ ! meaning that if x acts on h ∈ H, then

xh = x(p0h ⊕ p∞h)=

12 x p xp p h p xp h + p xp h p xh = p0 0 ∞ 0 = 0 0 0 ∞ = 0 = p∞xp0 xp∞ ! p∞h! p∞xp0h + p∞xp∞h! p∞xh!

= p0(xh) ⊕ p∞(xh).

1 1 Evidently, a 2 Ma 2 in this notation is represented as

L∞(a0) S 1/2 1/2 (M) a0 ,a∞ , S 1/2 1/2 (M) L∞(a∞) a∞ ,a0 ! where S 1/2 1/2 (M) and S 1/2 1/2 (M) are defined by the following definitions a0 ,a∞ a∞ ,a0 definition 11. Let a,bηM, x ∈M, then the sesquilinear form

axb : (f,g) ∈ D(b) × D(a) 7→ R is defined by the equalityd

axb(f,g) := hxbf,agi. definition 12. Let a,bηM.d We define the linear space of sesqulinear forms

Sa,b(M) := {axb | x ∈ M} endowed with the norm kaxbka,b := kxkd.

Let a be injective, thenda0|p0H is injective bounded operator in Mp0 act- ing on p0H and a∞|p∞H is injective operator affiliated with Mp∞ (acting on p∞H) with bounded inverse. Moreover, L∞(a0) is isometrically isomorphic to

L∞(a0|p0H ) as well as L∞(a∞) is isometrically isomorphic to L∞(a∞|p∞H ) by construction. Also, it is evident, that there exists the isomorphism axb 7→ bxa between the spaces Sa,b(M) and Sb,a(M). Thus, we only need to consider the d d limits for S α β (M) for α, β → +∞. a0 ,a∞ Further we also use the notation

α,β S := S α β (M) 0,∞ a0 ,a∞ and α,β S := S α β (M). ∞,0 a∞,a0

Let α1 < α2 and β1 <β2, αi,βi ∈ R, then note, that the following schemes

α1,β1 α1,β2 α1,β1 α1,β2 S0,∞ ⊂ S0,∞ S∞,0 ⊃ S∞,0 ∪ ∪ and ∩ ∩ hold. α2,β1 α2,β2 α2,β1 α2,β2 S0,∞ ⊂ S0,∞ S∞,0 ⊃ S∞.0

13 As for the norms,

k·k α1 β1 ≻ k·k α1 β2 k·k α1 β1 ≺ k·k α1 β2 a0 ,a∞ a0 ,a∞ a∞ ,a0 a∞ ,a0 f f and g g .

k·k α2 β1 ≻ k·k α2 β2 k·k α2 β1 ≺ k·k α2 β2 a0 ,a∞ a0 ,a∞ a∞ ,a0 a∞ ,a0

α,β α,β α β By τ0,∞ we denote the topology of the norm k·ka0 ,a0∞ on the space S0,∞. Note, that α1,β1 α1,β2 τ0,∞ |α2 ⊃ τ0,∞ |α2,β1 ∩ ∩ , α2,β1 α2 ,β2 τ0,∞ ⊃ τ0,∞ |β1 where α1,β2 τ|α2 ≡ τ|β1 ≡ τ|α2,β1 := {X ∩ S0,∞ | X ∈ τ}. α,β α,β Sine there exists the isomorphism between S0,∞ and S∞,0, we will only consider one case. Consider the limits

Sα α,β Sβ α,β 0 = S0,∞ and ∞ = S0,∞ α>0 β>[0 \ α β α with the topologies τ0 and τ∞, respectively. Topology τ0 is the strongest topol- Sα ogy on 0 , such that all the mappings

α α,β α ϕ : x ∈ (S , k·k α β ) 7→ x ∈ S (6) β 0,∞ a0 ,a∞ 0

β β are continuous and the topology τ∞ is the weakest topology on S∞ that all the mappings β β α,β ψ : x ∈ S 7→ x ∈ (S , k·k α β ) (7) α ∞ 0,∞ a0 ,a∞ are continuous. α,β α α Remark Note, that any (S , k·k α β ) is Banach space, thus (S , τ ) is 0,∞ a0 ,a∞ 0 0 an (LB)-space. Sα α Remark Note, that ( ∞, τ∞) is a Frechet space, since its topology is de- termined by the countable set of the seminorms.

Lemma 9. Let α<β, α, β ∈ R+, then

Sα Sβ (i) 0 ⊃ 0 ; Sα Sβ (ii) ∞ ⊂ ∞.

Proof. (i) Since

Sβ β,γ Sα α,γ β,γ α,γ 0 = S0,∞, 0 = S0,∞ and S0,∞ ⊂ S0,∞, γ>0 γ>0 [ [

14 Sβ β,γ0 α,γ0 it follows that if x ∈ 0 , then there exists γ0 such that x ∈ S0,∞ ⊂ S0,∞ , thus Sα x ∈ 0 . Sα γ,α γ,β (ii) If x ∈ ∞, then for all γ > 0 x ∈ S0,∞ ⊂ S0,∞, thus γ,β Sβ x ∈ S0,∞ = ∞. γ>0 \

We consider the following constructions:

L Sα L Sβ 0,∞ = 0 , 0,∞ = ∞. α>0 \ β>[0 Note that

L Sβ α,β α,β Sα L 0,∞ = ∞ = S0,∞ ⊂ S0,∞ = 0 = 0,∞. α>0 α>0 α>0 β>[0 β>[0 \ \ β>[0 \ Proposition 3.

L α,α β,β β,β α,α L 0,∞ ⊃ limS0,∞ ≡ S0,∞ ⊃ S0,∞ ≡ limS0,∞ ⊃ 0,∞. α>0 ≥ α>0 ≥ \ β[α [ β\α α1,β α2,β Proof. Note, that if α1 < α2, then S0,∞ ⊃ S0,∞ , thus

L α,β α,β β,β 0,∞ = S0,∞ = S0,∞ ⊃ S0,∞ α>0 α>0 ≥ α>0 ≥ \ β>[0 \ β[α \ β[α α,β1 α,β2 Note, that if β1 <β2, then S0,∞ ⊂ S0,∞ , thus β,β β,α β,α L S0,∞ ⊃ S0,∞ = S0,∞ = 0,∞. α>0 ≥ α>0 ≥ α>0 [ β\α [ β\α [ β>\0

α,β L L definition 13. We call the family S0,∞ converging if 0,∞ = 0,∞ and denote α,β α,β L it as S0,∞ −−→ 0,∞. α,β Corollary 1. If the family S0,∞ is converging, then α,α α,α L limS0,∞ = limS0,∞ = 0,∞. Lemma 10. Let α<β, α,β ∈ R+ and

α Sβ α τ0 |β := {X ∩ 0 | X ∈ τ0 } α Sβ i.e. the topology induced by τ0 on 0 , then

α β τ0 |β ⊂ τ0 .

15 α Sβ α α −1 Proof. Let X0 ∈ τ0 |β, i.e. X0 = X ∩ 0 with X ∈ τ0 . Then (ϕγ ) (X) is open α,γ α γ in (S0,∞, k·ka0 ,a∞ ). Note, that the embedding

β,α β,γ α,γ m : x ∈ (S , k·k β γ ) 7→ x ∈ (S , k·k α γ ) γ 0,∞ a0 ,a∞ 0,∞ a0 ,a∞ is also continuous, thus

α β,α −1 β,α −1 α −1 (ϕγ mγ ) (X) = (mγ ) (ϕγ ) (X)

β,γ is open in (S , k·k β γ ) for any γ > 0. 0,∞ a0 ,a∞ β The topology τ0 is the strongest topology, such that all of the embeddings β β ϕγ are continuous. If X0 ∈/ τ0 , then there exists the toplogy

β β β τ = τ0 ∪{X0 ∩ Y | Y ∈ τ0 }∪{X0 ∪ Y | Y ∈ τ0 }

β such that it is stronger, then τ0 , X0 ∈ τ and for any A ∈ τ the preimages β −1 (ϕγ ) (A) are open. Further we explain why it is open. β β Consider three cases A ∈ τ0 , A = X0 ∩ Y , A = X0 ∪ Y (Y ∈ τ0 ). β β −1 1) Let A ∈ τ0 , then evidently (ϕγ ) (A) is open. 2) Let A = X0 ∩ Y , then

β −1 Sβ β −1 β,γ β −1 (ϕγ ) (X ∩ 0 ∩ Y ) = (ϕγ ) X ∩ S0,∞ ∩ (ϕγ ) (Y )= γ>0 ! [

β −1 β,γ β −1 α β,α −1 β −1 = (ϕγ ) X ∩ S0,∞ ∩(ϕγ ) (Y )= ϕγ mγ (X) ∩(ϕγ ) (Y ). γ>0 ! γ>0 ! [   [
α β,α −1 β −1 is open, since ϕγ mγ (X) is open and ϕγ (Y ) is open. 3) Let A =X0 ∪ Y ,
then

β −1 Sβ β −1 β,γ β −1 (ϕγ ) (X ∩ 0 ) ∪ Y = (ϕγ ) X ∩ S0,∞ ∪ (ϕγ ) (Y )= γ>0 !   [

β −1 β,γ β −1 α β,α −1 β −1 = (ϕγ ) X ∩ S0,∞ ∪(ϕγ ) (Y )= ϕγ mγ (X) ∪(ϕγ ) (Y ). γ>0 ! γ>0 ! [   [
α β,α −1 β −1 is open, since ϕγ mγ (X) is open and ϕγ (Y ) is open. β Thus, we get a contradiction
with the maximality
of τ0 , therefore X0 ∈ β τ0 .

Lemma 11. Let α<β,α,β ∈ R+ and

β Sα β τ∞|α := {X ∩ ∞ | X ∈ τ∞}

16 β Sα i.e. the topology induced by τ∞ on ∞, then

α β τ∞ ⊃ τ∞|α.

α ∞ γ α Proof. The topology τ∞ is determined with the set of the seminorms {k·ka0 ,a∞ }γ=1 β ∞ and the topology τ | is determined by the system {k·k γ β } . It is sufficient ∞ α a0 ,a∞ γ=1 to note, that k·k γ β ≺k·k γ α . a0 ,a∞ a0 ,a∞

For L0,∞ it is natural to define the topology τ0,∞ which is determined as the strongest topology such that all of the embeddings

Sα α L Φα : x ∈ ( ∞, τ∞) 7→ x ∈ ( 0,∞, τ0,∞) (8) are continuous.

On the other side, it is natural to define the topology τ0,∞ which would be the weakest topology on L0,∞ such that all of the embeddings

L Sα α Ψα : x ∈ ( 0,∞, τ0,∞) 7→ x ∈ ( 0 , τ0 ) (9) are continuous.

Theorem 3. The embedding

Λ: x ∈ (L0,∞, τ0,∞) 7→ x ∈ (L0,∞, τ0,∞) (10) is continuous.

Proof. Note, that

β α ∀β0 > 0 τ0,∞|β0 = τ∞|β0 and τ0,∞ = τ0 |∞. ≥ 0 α>0 β\β [ At the same time,

β α,β α α,β τ∞ = τ0,∞|∞ and ∀β0 > 0 τ0 |β0 = τ0,∞|β0 . α>0 ≥ 0 [ β\β

It is sufficient to prove that for any β0 > 0 we have the inclusion

Sβ0 τ0,∞|β0 ⊃ τ0,∞|β0 , where τ|β0 = {X ∩ ∞ | X ∈ τ}.

Thus,

α,β τ0,∞|β0 = τ0,∞|∞ |β0 ≥ 0 α>0 ! β\β [

17 and α τ0,∞|β0 = τ0 |∞ |β0 . α>0 ! [ β0 Note, that the reductions always lead to the space S∞ and may be rewrited as

α,β α τ0,∞|β0 = τ0,∞|β0 and τ0,∞|β0 = (τ0 |β0 ) . ≥ 0 α>0 ! α>0 β\β [   [ α α,β But then τ0 |β0 = β≥β0 τ0,∞|β0 , therefore T α,β α,β τ ∞| 0 = τ | 0 ⊃ τ | 0 = τ ∞| 0 . 0, β 0,∞ β  0,∞ β  0, β ≥ 0 α>0 ! α>0 ≥ 0 β\β [   [ β\β    

Remark Note, that τ0,∞ is a topology of (LF )-space and τ0,∞ is a topology of the projective limit of (LB)-spaces. Remark All the constructions of this section may be applied to the system α,β L L L S∞,0. We will distinguish such constructions by the notation ∞,0, ∞,0, ∞,0, α,β τ∞,0 and so on.

4.4 Bring it all together

α Evidently we may consider different limits of L∞(a ) spaces, using the con- structions above, particularly, we may define definition 14. Let a be a positive operator affiliated with von Neumann algebra M acting on the Hilbert space H, such that not a nor a−1 are necessarily α bounded. Then we define the lower limit of the spaces limL∞(a ) as a vector space of sesquilinear forms formally wriiten as

(L∞(a0), τ∞(a0)) (L0,∞, τ0,∞) L L ( ∞,0, τ∞,0) ( ∞(a∞), τ∞(a∞))! definition 15. Let a be a positive operator affiliated with von Neumann algebra M acting on the Hilbert space H, such that not a nor a−1 are necessarily α bounded. Then we define the upper limit of the spaces limL∞(a ) as a vector space of sesquilinear forms formally wriiten as

(L∞(a0), τ∞(a0)) (L0,∞, τ0,∞) L L ( ∞,0, τ∞,0) ( ∞(a∞), τ∞(a∞))!

18 Theorem 4. Let L0,∞ = L0,∞, τ0,∞ = τ0,∞, then

α α L∞(a); = limL∞(a ) = limL∞(a ) is (LF )-space.

Corollary 2. If a ≥ 0 is affiliated with the center C(M) of the von Neumann α algebra M, then the limits space L∞(a) = lim L∞(a ) is an (LF)-space. α

5 Lp-spaces

Let M be a von Neumann algebra and a ∈M be a positive bounded injective linear operator. We consider L1(a) and L∞(a), descirbed previously (also see [16]). It is easy to see that for any α<β,α,β ∈ R

h α h β h α h β .( L1 (a ) ֒→ L1 (a ); L∞(a ) ←֓L∞(a

h h We also obtain, that L1 (a) is isometrically isomorphic to L1 (a/kak). Note, that for any a such that kak≤ 1 and any pair α<β the inequalities

β α α α K(x, t; L1(a ),L∞(a ))

α α α β K(x, t; L1(a ),L∞(a ))

α α β α ( L1(a )+ L∞(a ) ֒→ L1(a )+ L∞(a

α β β β ( L1(a )+ L∞(a ) ֒→ L1(a )+ L∞(a hold.

Theorem 5. Let a be positive injective bounded operator from the von Neumann algebra M and kak≤ 1.

α h β h α (( Lp(a ) ֒→ K(p−1)/p,p(L1 (a ),L∞(a

h α β β ( K(p−1)/p,p(L1 (a ),L∞(a )) ֒→ Lp(a

α h α h α with Lp(a ) being the interpolation space K(p−1)/p,p(L1 (a ),L∞(a )).

19 Proof. The inclusions are proved analogously. Consider as the example the h α β α embedding K(p−1)/p,p(L1 (a ,a )) в Lp(a ). First of all note, that

α β α α L1(a )+ L∞(a ) ⊂ L1(a )+ L∞(a ).

After that note, that

∞ 1/q −θ α α q dt t K(x, t; L (a ),L∞(a )) ≤ 1 t Z0 
∞ 1/q −θ α β q dt ≤ t K(x, t; L (a ),L∞(a )) < ∞, 1 t Z0  h α β
α thus if x ∈ K(p−1)/p,p(L1 (a ,a )), then x ∈ Lp(a ). At the same time the inequality also implies the continuity.

From the latter Theorem we obtain that we, also, can build the following construction for the fixed p ∈ (0, 1). L α h β h α Let p(a ) be the inductive limit (by β → ∞) of K(p−1)/p,p(L1 (a ),L∞(a )) R α h α h β and p(a ) be the projective limit (by β → ∞) of K(p−1)/p,p(L1 (a ),L∞(a )). α α The projective limit of Lp(a ) will be the upper limit limLp(a ) and the induc- α α tive limit of Rp(a ) will be the lower limit limLp(a ). If they coincide tas the α topological spaces, then Lp(a ) has the mixed limit for α converging to infinity.

Acknowledgments

This work was supported by Russian Foundation for Basic Research Grant 18- 31-00218.

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