International Journal of Analysis and Applications
ISSN 2291-8639
Volume 5, Number 2 (2014), 123-135
http://www.etamaths.com

ON SOME ISOMORPHISMS BETWEEN BOUNDED LINEAR

MAPS AND NON-COMMUTATIVE Lp-SPACES

E. J. ATTO1,∗, V.S.K. ASSIAMOUA1 AND Y. MENSAH1,2

Abstract. We define a particular space of bounded linear maps using a Von

Neumann algebra and some operator spaces. By this, we prove some isomor-

phisms, and using interpolation in some particular cases, we get analogue of
non-commutative Lp spaces.

1. Introduction

In the fifties, many authors have studied on non-commutative Lp-spaces like
Segal [13], Kunze [9], Dixmier [4], Stinespring [14]. But the recent emergency of
the theory of operator spaces from the late 80’s to the early 90’s in the works of
Effros and Ruan [5],[6], [7],[12], Blecher and Paulsen [2],[3] allowed Gilles Pisier
since the mid 90’s to expose the general theory of non-commutative Lp-spaces [11],
using for instance a Von Neumann algebra M equipped with a particular type of
trace ϕ. As the complex interpolation method contribute to define a new Banach
space using a compatible pair of Banach spaces, this method was also used to define
non-commutative Lp-spaces.

Our aim in this paper is to define for each 1 ≤ p ≤ ∞ a particular space of
bounded linear maps denoted Lp(M,ϕ,E; F), using some operator spaces E, F and
the non-commutative spaces L1(M,ϕ), L∞(M,ϕ), such that those particular spaces
have some properties with the non-commutative Lp spaces like the isomorphism.
We firstly view those types of spaces as Banach spaces and secondly give them an
operator space structure.

Before stating our results, we shall recall the concept of operator space and the
complex interpolation method to make the paper more comprehensive.

2. Preliminary Notes

2.1. Operator spaces. H being an Hilbert space, we denote by B(H) the Banach
space of all bounded operators from H into H, endowed with the operator norm

‖T‖∞ = sup{‖Tξ‖H,ξ ∈ H,‖ξ‖≤ 1}.

A closed subset E ⊂ B(H) is called an operator space. But there exist an abstract
characterization of an operator space given by Ruan (see [7] and [12] for more
details):

2010 Mathematics Subject Classification. 46B03; 46L07.
Key words and phrases. Operator spaces, Banach spaces, linear maps, non-commutative Lp-

spaces.

c©2014 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

123



124 ATTO, ASSIAMOUA AND MENSAH

Theorem 2.1 (Ruan theorem). A complex vector space E is an operator space
if and only if for each n ≥ 1, there is a complete norm ‖.‖n on Mn(E), the space
of n×n matrices with entries in E, such that the following properties are satisfied:

(1) ‖αxβ‖n ≤‖α‖‖x‖n‖β‖
(2) ‖x⊕y‖n+m = max{‖x‖n,‖y‖m}

for x ∈ Mn(E),y ∈ Mm(E),α,β ∈ Mn.

Let M be a Von Neumann algebra on the Hilbert space H, that is, M ⊂ B(H)
is a C∗-algebra closed in the weak operator topology and contains the identity
operator. We denote by M+ the set of all positive elements of M. Then, we recall
the following definitions concerning the notion of trace:

Definition 2.2. A trace ϕ on M+ is a function ϕ : M+ →]0, +∞], such that
(1) ϕ(x + y) = ϕ(x) + ϕ(y) for any x,y ∈ M+.
(2) ϕ(λx) = λϕ(x) for any 0 ≤ λ ≤∞ and x ∈ M+ with the usual convention

0.∞ = 0.
(3) ϕ(xy) = ϕ(yx) for any x,y ∈ M+.

Definition 2.3. A trace ϕ is called:

i) faithful if x ∈ M+,ϕ(x) = 0 =⇒ x = 0.
ii) finite if ϕ(x) < ∞ for any x ∈ M+.

iii) normal if for any x ∈ M+ and any increasing net (xα) converging to x in
the strong operator topology, ϕ(xα) → ϕ(x).

iv) semi-finite if for any x ∈ M+ , there exist y ∈ M+such that ϕ(y) < ∞ and
y ≤ x.

In the following, the Von Neumann algebra M is assumed to be equipped with
a faithful, normal and semi-finite trace ϕ.

2.2. The complex interpolation method and the non-commutative Lp
spaces. A couple of (complex) Banach spaces (X0; X1) is said to be compatible
if they are both embedded by continuous injective linear maps into a Hausdorff
topological vector space X. In this case, X0 and X1 are viewed as vector subspaces
of X. Then their intersection X0 ∩X1 is equipped with the norm

‖x‖X0∩X1 = max{‖x‖X0,‖x‖X1}
and their sum is defined by

X0 + X1 = {x0 + x1 : xk ∈ Xk, k = 0, 1}

with the norm

‖x‖X0+X1 = inf{‖x0‖X0 + ‖x1‖X1 : x = x0 + x1,xk ∈ Xk, k = 0, 1}.

It is easy to check that X0 ∩X1 and X0 + X1 are again Banach spaces. Then we
have

X0 ∩X1 ⊂ X0, X1 ⊂ X0 + X1, contractiveinjections.
Let B = {z ∈ C : 0 ≤ Re(z) ≤ 1}. Let F(X0,X1) be the family of all functions
f : B → X0 + X1 satisfying the following conditions:

(1) f is continuous on B and analytic in the interior of B;



ON SOME ISOMORPHISMS 125

(2) f(k + it) ∈ Xk for t ∈ R and the function t 7→ f(k + it) is continuous from
R into Xk, k = 0, 1;

(3) lim
|t|→∞

‖f(k + it)‖Xk = 0, k = 0, 1.

We equip F(X0,X1) with the norm:

‖f‖F(X0,X1) = {sup
t∈R
‖f(it)‖X0, sup

t∈R
‖f(1 + it)‖X1}.

Then it is easy to check that F(X0,X1) becomes a Banach space. Let 0 < θ < 1,
the complex interpolation space (X0,X1)θ is defined as the space of all those x ∈
X0 + X1 for which there exists f ∈F(X0,X1) such that f(θ) = x. Equipped with

‖x‖θ = inf{‖f‖F(X0,X1) : f(θ) = x, f ∈F(X0,X1)},

(X0,X1)θ becomes a Banach space. Indeed, by the maximum principle, the map
f 7→ f(θ) is a contraction from F(X0,X1) to X0 + X1. Then (X0,X1)θ can be
isometrically identified with the quotient of F(X0,X1) by the kernel of this map.

Remark 2.1. The following properties are easy:

i) (X0,X1)θ = (X1,X0)1−θ isometrically.
ii) X0 ∩X1 is dense in (X0,X1)θ.
iii) Let (X0,X1) and (Y0,Y1) be two compatible couples.

Let T : X0 + X1 → Y0 + Y1 be a linear map which is bounded from Xk to
Yk for k = 0 and k = 1. Then T is bounded from (X0,X1)θ to (Y0,Y1)θ for
any 0 < θ < 1; moreover

‖T : (X0,X1)θ → (Y0,Y1)θ‖≤‖T : X0 → Y0‖1−θ‖T : X1 → Y1‖θ.

This statement is usually called interpolation theorem.

Note that by tradition in interpolation theory, the assumption on T in the state-
ment (iii) above means that T maps Xk into Yk and its restriction to Xk belongs
to B(Xk,Yk) (k = 0, 1).

Set Ip = {x ∈ M : ϕ(|x|p) < ∞}, (1 ≤ p < ∞), equipped with the norm
‖x‖p = (ϕ(|x|p)1/p. The completion of Ip under this norm is a Banach space and is
denoted Lp(M,ϕ) by G. Pisier in [11] where it is called non-commutative Lp space.

Since all Lp(M,ϕ), 1 ≤ p ≤∞, are injected into L1(M,ϕ), (Lp0 (M,ϕ),Lp1 (M,ϕ))
is a compatible couple for any p0,p1 ∈ [1;∞]. The following is the complex inter-
polation theorem on non-commutative Lp-spaces.

Theorem 2.4. Let 1 ≤ p0 < p1 ≤ ∞, 0 < θ < 1 and let p be determined by
1
p

= 1−θ
p0

+ θ
p1

. Then

(Lp0 (M,ϕ),Lp1 (M,ϕ))θ = Lp(M,ϕ)

with equal norms.

For more details, see [1] and [8].
Particularly, for p0 = 1 and p1 = ∞ we get

(L1(M,ϕ),L∞(M,ϕ))θ = Lp(M,ϕ)

whith θ = 1
p



126 ATTO, ASSIAMOUA AND MENSAH

Remark 2.2. (i) E ⊂ B(H) being an operator space, the non-commutative
vector valued Lp spaces for 1 ≤ p ≤∞ are defined as follow:

L1(M,ϕ,E) = L1(M,ϕ)⊗̂E

L0∞(M,ϕ,E) = M ⊗min E

Lp(M,ϕ,E) = (L1(M,ϕ,E),L
0
∞(M,ϕ,E))θ 1 < p < ∞

whith θ = 1
p
.

(ii) The dual space of Lp(M,ϕ) is Lq(M,ϕ) for 1 ≤ p < ∞( 1p +
1
q

= 1) with

respect to the following duality

〈x,y〉 = ϕ(xy), x ∈ Lp(M,ϕ), y ∈ Lq(M,ϕ).

In over words

(Lp(M,ϕ))
′ = Lq(M,ϕ) isometrically.

(iii) In the Lebesgue-Bochner theory, if E is a Banach space, it is well known that
the dual of Lp(Ω,µ; E) is not in general Lq(Ω,µ; E

′) unless E′ possesses the
Radon Nikodym property (in short the RNP). In [11], it was introduced an
operator space analogue of the RNP which is called the ORNP. Thus, if E is
an operator space such that is dual E′ has the ORNP, then the dual theorem
is confirmed. Namely the dual of Lp(M,ϕ,E) is completely isometric to
Lq(M,ϕ,E

′). One must also note that the ORNP of an operator space
implies the RNP of the underlying Banach space, but the converse is not
true.

The following theorem has been proved by G. Pisier in [11] pages 49,50.

Theorem 2.5 (Pisier). Let (M,ϕ) be any non-commutative probability space
(in short n.c.p. space). Let E be an operator space. If E′ has the ORNPp′ with
1 < p < ∞ and q = p/(p− 1), then we have a completely isometric identity

Lp(M,ϕ,E)
′ = Lq(M,ϕ,E

′).

3. Main Results

3.1. The spaces L1(M,ϕ,E; F) and L∞(M,ϕ,E; F).. Let E,F ⊂ B(H) two
operator spaces. We denote by ‖.‖E and ‖.‖F the operator norm on E and F
inherited from B(H) and L(E,F) the space of all bounded linear maps from E into
F equipped with the norm ‖f‖L(E,F) = sup{‖f(x)‖F : ‖x‖E ≤ 1}.

Set L1(M,ϕ,E; F) the space of all bounded linear maps from L1(M,ϕ) into
L(E,F).

Theorem 3.1. The mapping

u 7−→‖u‖ϕ,1 = sup{‖u(x)‖L(E,F) : ϕ(|x|) ≤ 1}

is a norm on L1(M,ϕ,E; F) . L1(M,ϕ,E; F) equipped with this norm is a Banach
space.

Proof. It is obvious that ‖.‖ϕ,1 is a norm. Since L1(M,ϕ), (E,‖.‖E) and (F,‖.‖F )
are Banach spaces, then (L1(M,ϕ,E; F),‖.‖ϕ,1) is a Banach space. �



ON SOME ISOMORPHISMS 127

Moreover, L1(M,ϕ,E; F) is a Banach algebra if endowed with the inner product
denoted · as follow: for all u,v ∈ L1(M,ϕ,E; F), u ·v = w such that
w(x)(y) = u(x)(y) ◦ v(x)(y), with x ∈ L1(M,ϕ) and y ∈ E. Here, the product
between the two elements u(x)(y),v(x)(y) of F is the natural product of operator
inherited from the Banach algebra B(H). More precisely,

∀T1,T2 ∈ B(H),T1 ◦T2(ξ) = T1(T2(ξ))

with ξ ∈ H. And we obtain

‖u ·v‖ϕ,1 = sup{‖(u ·v)(x)‖L(E,F) : ϕ(|x|) ≤ 1}
= sup{sup{‖(u ·v)(x)(y)‖F : ‖y‖E ≤ 1} : ϕ(|x|) ≤ 1}
= sup{sup{‖u(x)(y) ◦v(x)(y)‖F : ‖y‖E ≤ 1} : ϕ(|x|) ≤ 1}
≤ sup{sup{‖u(x)(y)‖F‖v(x)(y)‖F : ‖y‖E ≤ 1} : ϕ(|x|) ≤ 1}
≤ sup{ sup

‖y‖E≤1
{‖u(x)(y)‖F‖v(x)(y)‖F} : ϕ(|x|) ≤ 1}

≤ sup{( sup
‖y‖E≤1

‖u(x)(y)‖F )( sup
‖y‖E≤1

‖v(x)(y)‖F ) : ϕ(|x|) ≤ 1}

≤ sup{
(
‖u(x)‖L(E,F)

)(
‖v(x)‖L(E,F)

)
: ϕ(|x|) ≤ 1}

≤ sup
ϕ(|x|)≤1

{
(
‖u(x)‖L(E,F)

)(
‖v(x)‖L(E,F)

)
}

≤ ( sup
ϕ(|x|)≤1

‖u(x)‖L(E,F))( sup
ϕ(|x|)≤1

‖v(x)‖L(E,F))

≤ ‖u‖ϕ,1‖v‖ϕ,1

Definition 3.2. Set L∞(M,ϕ,E; F) the space of all bounded linear maps from
M into L(E,F) equipped with the norm

‖u‖ϕ,∞ = sup{‖u(x)‖L(E,F) : ‖x‖∞ ≤ 1},

where ‖.‖∞ is the operator norm on M ⊂ B(H).

Theorem 3.3. L∞(M,ϕ,E; F) is a Banach space.

Proof. The proof of this theorem is in the same spirit of the one of the previous
theorem . �

It is also easy to check that L∞(M,ϕ,E; F) endowed with the same inner prod-
uct used for L1(M,ϕ,E; F) is a Banach algebra.

Theorem 3.4 (duality). The topological dual of L1(M,ϕ,E; F) is isometrically
isomorph to L∞(M,ϕ,E; F

′) where F ′ is the dual of F :

(L1(M,ϕ,E; F))
′ ' L∞(M,ϕ,E; F ′)

Proof. Let us consider the mapping

T : L∞(M,ϕ,E; F
′) −→ (L1(M,ϕ,E; F))′

u 7−→ Tu,

such that

〈Tu,v〉 = sup
ϕ(|x|)≤1,‖y‖E≤1

|〈u(x)(y),v(x)(y)〉|,

where u ∈ L∞(M,ϕ,E,F ′), v ∈ (L1(M,ϕ,E; F))′, and so:
∀x ∈ L1(M,ϕ),y ∈ E we have u(x)(y) ∈ F ′, v(x)(y) ∈ F.



128 ATTO, ASSIAMOUA AND MENSAH

(1) Linearity and boundedness of T: The linearity of T is trivial. Let us prove
T is bounded. We have:

|〈Tu,v〉| = sup
ϕ(|x|)≤1,‖y‖E≤1

|〈u(x)(y),v(x)(y)〉|

≤ sup
ϕ(|x|)≤1,‖y‖E≤1

‖u(x)(y)‖F′‖v(x)(y)‖F

≤ sup
ϕ(|x|)≤1,‖y‖E≤1

‖u(x)(y)‖F′ sup
ϕ(|x|)≤1,‖y‖E≤1

‖v(x)(y)‖F

≤ sup
‖x‖∞≤1,‖y‖E≤1

‖u(x,y)‖F′ sup
ϕ(|x|)≤1,‖y‖E≤1

‖v(x)(y)‖F

≤ ‖u‖ϕ,∞‖v‖ϕ,1
Thus, ‖Tu‖≤‖u‖ϕ,∞ and T is bounded with ‖T‖≤ 1.

(2) We prove now that ‖T‖ = 1. Since ‖T‖ ≤ 1, all we have to do is to prove
‖T‖≥ 1.
Let a ∈ F such that ‖a‖F = 1. Since a 6= 0, there exist b∗ ∈ F

′
such that

‖b∗‖ = 1 and 〈b∗,a〉 = ‖a‖F = 1. For (x0,y0) fixed in L1(M,ϕ) ×E, one
define u ∈ L∞(M,ϕ,E; F ′) as follow:

u(x)(y) =

{
b∗ if (x,y) = (x0,y0)

0 if not

and v ∈ L1(M,ϕ,E; F) by

v(x)(y) =

{
a if (x,y) = (x0,y0)

0 if not

Afterwards,

〈Tu,v〉 = sup
ϕ(|x|)≤1,‖y‖E≤1

|〈u(x)(y),v(x)(y)〉|

= 〈u(x0)(y0),v(x0)(y0)〉
= 〈b∗,a〉
= 1

Since 〈Tu,v〉 ≤ ‖T‖‖u‖‖v‖, with ‖u‖ = ‖v‖ = 1, then ‖T‖ ≥ 1 and
‖T‖ = 1.

(3) subjectivity of T
Suppose f ∈ (L1(M,ϕ,E; F))′ and (x0,y0) fixed in L1(M,ϕ) ×E.
Let φ ∈ L1(M,ϕ,E; F) such that φ(x)(y) = 0 if (x,y) 6= (x0,y0).
We set a(x0,y0) = φ(x0)(y0) ∈ F,
then there exist a linear functional b(x0,y0) ∈ F

′ such that

〈f,φ〉 = 〈b(x0,y0),a(x0,y0〉 > 0
Set Φ ∈ L∞(M,ϕ,E; F) as follow:

Φ(x)(y) =

{
b(x0,y0) if (x,y) = (x0,y0)

0 if not

So we have:

〈f,φ〉 = 〈Φ(x0)(y0),φ(x0)(y0)〉
= sup

ϕ(|x|)≤1,‖y‖E≤1
|〈Φ(x)(y),φ(x)(y)〉|

= 〈TΦ,φ〉.
Hence the linear functional f and TΦ coincide on L∞(M,ϕ,E; F

′). In over
words f = TΦ and T is subjective.



ON SOME ISOMORPHISMS 129

In conclusion, T is isometric.
�

3.2. The spaces L 0p (M,ϕ,E; F) with 1 < p < ∞.

Definition 3.5. Let L 0p (M,ϕ,E; F) be the space of all bounded linear maps
from Lp(M,ϕ) into L(E,F) for 1 < p < ∞.

Theorem 3.6. For each 1 < p < ∞,the space L 0p (M,ϕ,E; F) endowed with the
norm

‖u‖ϕ,p = sup{‖u(x)‖L(E,F) : ϕ(|x|p) ≤ 1}
is a Banach space.

Proof. In the same spirit as in theorem 3.1 �

Theorem 3.7 (Duality). The topological dual of L 0p (M,ϕ,E; F) is isometrically

isomorph to L 0q (M,ϕ,E; F
′), (

1

p
+

1

q
= 1) where F ′ is the dual of F :

(L 0p (M,ϕ,E; F))
′ ' L 0q (M,ϕ,E,F

′)

Proof. Let us consider the mapping

T : Lq(M,ϕ,E; F
′) −→ (Lp(M,ϕ,E; F))′

u 7−→ Tu,
such that

〈Tu,v〉 = sup
ϕ(|x|)≤1,‖y‖E≤1

|〈u(x)(y),v(x)(y)〉|,

where u ∈ Lq(M,ϕ,E,F ′), v ∈ (Lp(M,ϕ,E; F))′, and so:
∀x ∈ L1(M,ϕ),y ∈ E we have u(x)(y) ∈ F ′, v(x)(y) ∈ F.

(1) Linearity and boundedness of T: The linearity of T is trivial. Let us prove
T is bounded. We have:

|〈Tu,v〉| = sup
ϕ(|x|)≤1,‖y‖E≤1

|〈u(x)(y),v(x)(y)〉|

≤ sup
ϕ(|x|)≤1,‖y‖E≤1

‖u(x)(y)‖F′‖v(x)(y)‖F

≤ sup
ϕ(|x|)≤1,‖y‖E≤1

‖u(x)(y)‖F′ sup
ϕ(|x|)≤1,‖y‖E≤1

‖v(x)(y)‖F

≤ sup
ϕ(|x|q)≤1,‖y‖E≤1

‖u(x)(y)‖F′ sup
ϕ(|x|p)≤1,‖y‖E≤1

‖v(x)(y)‖F

≤

(
sup

ϕ(|x|q)≤1
‖u(x)‖L(E,F′)

)(
sup

ϕ(|x|p)≤1
‖v(x)‖L(E,F

)
≤ ‖u‖ϕ,q‖v‖ϕ,p

Thus, ‖Tu‖≤‖u‖ϕ,q and T is bounded with ‖T‖≤ 1.
(2) We prove now that ‖T‖ = 1. Since ‖T‖ ≤ 1, all we have to do is to prove
‖T‖≥ 1.
Let a ∈ F such that ‖a‖F = 1. Since a 6= 0, there exist b∗ ∈ F

′
such that

‖b∗‖ = 1 and 〈b∗,a〉 = ‖a‖F = 1. For (x0,y0) fixed in Lp(M,ϕ) ×E, one
define u ∈ Lq(M,ϕ,E; F ′) as follow:

u(x)(y) =

{
b∗ if (x,y) = (x0,y0)

0 if not



130 ATTO, ASSIAMOUA AND MENSAH

and v ∈ Lp(M,ϕ,E; F) by

v(x)(y) =

{
a if (x,y) = (x0,y0)

0 if not

Afterwards,

〈Tu,v〉 = sup
ϕ(|x|)≤1,‖y‖E≤1

|〈u(x)(y),v(x)(y)〉|

= 〈u(x0)(y0),v(x0)(y0)〉
= 〈b∗,a〉
= 1

Since 〈Tu,v〉 ≤ ‖T‖‖u‖‖v‖, with ‖u‖ = ‖v‖ = 1, then ‖T‖ ≥ 1 and
‖T‖ = 1.

(3) subjectivity of T
Suppose f ∈ (Lp(M,ϕ,E; F))′ and (x0,y0) fixed in Lp(M,ϕ) ×E.
Let φ ∈ Lp(M,ϕ,E; F) such that φ(x)(y) = 0 if (x,y) 6= (x0,y0).
We set a(x0,y0) = φ(x0)(y0) ∈ F ,
then there exist a linear functional b(x0,y0) ∈ F

′ such that

〈f,φ〉 = 〈b(x0,y0),a(x0,y0〉 > 0

Set Φ ∈ Lq(M,ϕ,E; F) as follow:

Φ(x)(y) =

{
b(x0,y0) if (x,y) = (x0,y0)

0 if not

So we have:

〈f,φ〉 = 〈Φ(x0)(y0),φ(x0)(y0)〉
= sup

ϕ(|x|)≤1,‖y‖E≤1
|〈Φ(x)(y),φ(x)(y)〉|

= 〈TΦ,φ〉.

Hence the linear functional f and TΦ coincide on Lq(M,ϕ,E; F
′). In over

words f = TΦ and T is subjective.

In conclusion, T is isometric.
�

Corollary 3.8. for 1 ≤ p < ∞ L 0p (M,ϕ,C; C) is isomorphic to Lq(M,ϕ):

L 0p (M,ϕ,C; C) ' Lq(M,ϕ)

where q is such that
1

p
+

1

q
= 1 (called the conjugate of p).

Proof.

L 0p (M,ϕ,C; C) = L(Lp(M,ϕ),L(C,C))
' L(Lp(M,ϕ),C)
' (Lp(M,ϕ))′
' Lq(M,ϕ)

�



ON SOME ISOMORPHISMS 131

3.3. The spaces Lp(M,ϕ,E; F) by using the interpolation method.

Theorem 3.9. There is a contractive injection from L∞(M,ϕ,E; F) into L1(M,ϕ,E; F)

Proof. Since L1(M,ϕ,E; F) denotes the space of all bounded linear maps from
L1(M,ϕ) into L(E,F) and L∞(M,ϕ,E; F) the space of all bounded linear maps
from M into L(E,F), with M ⊂ L1(M,ϕ), it is obvious to claim that
L∞(M,ϕ,E; F) ⊂ L1(M,ϕ,E; F). Let z0 be an element of F.

Now, we consider the injection:

f : L∞(M,ϕ,E; F) −→ L1(M,ϕ,E; F)
u 7−→ f(u)

such that ∀(x,y) ∈ L∞(M,ϕ) ×E ,

f(u)(x)(y) =

{
u(x)(y) if x ∈ L1(M,ϕ)

z0 if not

Then we have:

‖f(u)‖ϕ,1 = sup{‖f(u)(x)(y)‖F : ϕ(|x|) ≤ 1,‖y‖E ≤ 1}
≥ sup{‖f(u)(x)(y)‖F : ‖x‖∞ ≤ 1,‖y‖E ≤ 1}
≥ sup{‖u(x)(y)‖F : ‖x‖∞ ≤ 1,‖y‖E ≤ 1}
≥ ‖u‖ϕ,∞

�

This theorem allows as to view the pair (L∞(M,ϕ,E; F), L1(M,ϕ,E; F)) as a
compatible couple of Banach spaces and so we can apply the complex interpolation
method to define a new Banach space.

Definition 3.10. Let ϕ be a semi-finite normal faithful trace on an injective
Von Neumann algebra M ⊂ B(H) and let E,F ⊂ B(H) two operators spaces. If
1 < p < ∞, we define

Lp(M,ϕ,E; F) = (L∞(M,ϕ,E; F), L1(M,ϕ,E; F))θ

where θ =
1

p
.

Theorem 3.11. L1(M,ϕ,E; F) is isomorphic to L(L1(M,ϕ)⊗E,F) and L∞(M,ϕ,E; F)
is isomorphic to L(M ⊗E,F):

L1(M,ϕ,E; F) 'L(L1(M,ϕ) ⊗E,F),

L∞(M,ϕ,E; F) 'L(M ⊗E,F).

Proof. We want to prove firstly that

L1(M,ϕ,E; F) 'L(L1(M,ϕ) ⊗E,F).

Let us consider the map:

F : L1(M,ϕ,E; F) −→L(L1(M,ϕ) ⊗E,F)
u 7−→F(u)

such that

F(u)(x⊗y) = u(x)(y)



132 ATTO, ASSIAMOUA AND MENSAH

i) Linearity:
We have

F(λu)(x⊗y) = (λu)(x)(y)
= λu(x)(y)
= λFu(x⊗y)

and

F(u + v)(x⊗y) = (u + v)(x)(y)
= u(x)(y) + v(x)(y)
= F(u)(x⊗y) + F(v)(x⊗y)
= [F(u) + F(v)](x⊗y)

ii) F is bijective
For all v ∈L(L1(M,ϕ) ⊗E,F) and x ∈ L1(M,ϕ), set:
G(v)(x) the mapping y 7→ v(x ⊗ y), where y ∈ E. This mapping is an
element of L(E,F). In fact for x fixed, y 7→ v(x ⊗ y) is linear, and is
bounded since v is a bounded linear map. Thus, we get a map:

G(v) : L1(M,ϕ) −→L(E,F).

Now we want to prove that this map is linear. For all x,x′ ∈ L1(M,ϕ), y ∈
E,

G(v)(x + λx′)(y) = v((x + λx′) ⊗y)
= v(x⊗y + λ(x′ ⊗y))
= v(x⊗y) + λv(x′ ⊗y)
= G(v)(x)(y) + λG(v)(x′)(y)
= [G(v)(x) + λG(v)(x′)](y)

Since ∀y,G(v)(x + λx′)(y) = [G(v)(x) + λG(v)(x′)](y), then

G(v)(x + λx′) = G(v)(x) + λG(v)(x′),

and G(v) is linear. By this, we claim that G is a map from
L(L1(M,ϕ) ⊗E,F) into L1(M,ϕ,E; F).

Finally we prove that G is the inverse of F:

(G ◦F)(u)(x⊗y) = G[F(u)(x⊗y)]
= G(u(x)(y))
= u(x⊗y)

Since (G ◦F)(u)(x⊗y) = u(x⊗y), then

(G ◦F)(u) = u

(F ◦G)(v)(x)(y) = F[G(v)(x)(y)]
= F(v(x⊗y))
= v(x)(y)

Since (F ◦G)(v)(x)(y) = v(x)(y), then

(G ◦F)(v) = v.

So L1(M,ϕ,E; F) 'L(L1(M,ϕ) ⊗E,F).
The proof of the second isomorphism use the same method by replacing for

instance L1(M,ϕ) by M and L1(M,ϕ,E; F) by L∞(M,ϕ,E; F). �



ON SOME ISOMORPHISMS 133

Corollary 3.12.

L1(M,ϕ,E; F) 'L(L1(M,ϕ),E),F)
and

L∞(M,ϕ,E; F) 'L(L∞(M,ϕ,E),F)

Proof.
L1(M,ϕ,E; F) ' L(L1(M,ϕ) ⊗E,F)

' L(L1(M,ϕ)⊗̂E,F)
' L(L1(M,ϕ,E),F)

and
L∞(M,ϕ,E; F) ' L(M ⊗E,F)

' L(M ⊗min E,F)
' L(L∞(M,ϕ,E),F)

�

Corollary 3.13. For all 1 < p < ∞, Lp(M,ϕ,E; F) is isomorphic to
(L(L∞(M,ϕ,E),F),L(L1(M,ϕ,E),F))θ:

Lp(M,ϕ,E; F) ' (L(L∞(M,ϕ,E),F),L(L1(M,ϕ,E),F))θ

with θ =
1

p

Proof. This is a direct consequence of Definition 3.10 and Theorem 3.11. �

Corollary 3.14. for 1 ≤ p < ∞ Lp(M,ϕ,C; C) is isomorphic to Lq(M,ϕ):
Lp(M,ϕ,C; C) ' Lq(M,ϕ)

where q is such that
1

p
+

1

q
= 1 (called the conjugate of p).

Proof.

Lp(M,ϕ,C; C) = (L∞(M,ϕ,C; C), L1(M,ϕ,C; C))θ
= (L(L∞(M,ϕ),L(C; C)),L(L1(M,ϕ),L(C; C)))θ
' (L(L∞(M,ϕ),C)),L(L1(M,ϕ),C)))θ
' ((L∞(M,ϕ))′, (L1(M,ϕ))′)θ
' ((L1(M,ϕ)), (L∞(M,ϕ)))θ
' ((L∞(M,ϕ)), (L1(M,ϕ)))1−θ
' ((L∞(M,ϕ)), (L1(M,ϕ)))1/q
' Lq(M,ϕ)

�

3.4. Operator Space structure. For any integer n ∈ N∗, we identify Mn(L1(M,ϕ,E; F))
with L1(M,ϕ,E; Mn(F)):

Mn(L1(M,ϕ,E; F)) ≈ L1(M,ϕ,E; Mn(F)),
via the correspondence

Mn(L1(M,ϕ,E; F)) → L1(M,ϕ,E; Mn(F))
(ϕkl)1≤k,l≤n 7→ ϕn

where ϕn is an element of L1(M,ϕ,E; Mn(F)) defined by
∀(x,y) ∈ L∞(M,ϕ) ×E, ϕn(x,y) = (ϕkl(x,y))1≤k,l≤n .



134 ATTO, ASSIAMOUA AND MENSAH

Now, by identifying Mn(L1(M,ϕ,E; F)) with L1(M,ϕ,E; Mn(F)), we can set
an operator space structure on L1(M,ϕ,E; F). Thus, the norm ‖·‖n in Mn(L1(M,ϕ,E; F))
is the one defined on L1(M,ϕ,E; Mn(F)). In fact, Mn(F) being an operator space,
L1(M,ϕ,E; Mn(F)) is well-defined as a Banach space and so the norm ‖·‖n is com-
plete for all n ∈ N∗ . Let us prove that the two conditions of Ruan theorem are
realised:
Let n,m ∈ N∗, u ∈ Mn(L1(M,ϕ,E; F)), v ∈ Mm(L1(M,ϕ,E; F)) and α,β ∈ Mn.

First condition:

‖u⊕v‖n+m = ‖u⊕v‖L1(M,ϕ,E;Mn+m(F))
= sup{‖(u⊕v)(x,y)‖Mn+m(F) : ϕ(|x|) ≤ 1,‖y‖E ≤ 1}
= sup{max

{
‖u(x,y)‖Mn(F),‖v(x,y)‖Mm(F)

}
: ϕ(|x|) ≤ 1,‖y‖E ≤ 1}

= sup
ϕ(|x|)≤1,‖y‖E≤1

{
max

{
‖u(x,y)‖Mn(F),‖v(x,y)‖Mm(F)

}}
= max

{
sup

ϕ(|x|)≤1,‖y‖E≤1

{
‖u(x,y)‖Mn(F)

}
, sup
ϕ(|x|)≤1,‖y‖E≤1

{
‖v(x,y)‖Mm(F)

}}
= max

{
‖u‖L1(M,ϕ,E;Mn(F)),‖v‖L1(M,ϕ,E;Mm(F))

}
Second condition:

‖αuβ‖n = ‖αuβ‖L1(M,ϕ,E;Mn(F))
= sup

ϕ(|x|)≤1,‖y‖E≤1
‖(αuβ)(x,y)‖Mn(F)

= sup
ϕ(|x|)≤1,‖y‖E≤1

‖α(u(x,y))β‖Mn(F)

u(x,y) being in Mn(F), where F is an operator space, according to the second
condition of Ruan theorem, we have obviously

‖α(u(x,y))β‖Mn(F) ≤‖α‖‖u(x,y)‖Mn(F)‖β‖.
So

‖αuβ‖ ≤ sup
ϕ(|x|)≤1,‖y‖E≤1

‖α‖‖u(x,y)‖Mn(F)‖β‖

≤ ‖α‖

(
sup

ϕ(|x|)≤1,‖y‖E≤1
‖u(x,y)‖Mn(F)

)
‖β‖

≤ ‖α‖‖u‖n‖β‖.
Using the same method, we give to L∞(M,ϕ,E; F) an operator space structure

by identifying Mn(L∞(M,ϕ,E; F)) with L∞(M,ϕ,E; Mn(F)).
In the following, L1(M,ϕ,E; F) and L∞(M,ϕ,E; F) are viewed as operator

spaces and by interpolation, we define for 1 < p < ∞, the operator space:
Lp(M,ϕ,E; F) = (L∞(M,ϕ,E; F), L1(M,ϕ,E; F))θ

where θ =
1

p
.

The following theorem is the analogous of Pisier’s Theorem that we’ve recalled
in the previous sequence (Theorem 2.5).

Theorem 3.15. Let (M,ϕ) be any n.c.p. space. Let E be an operator space.
If E′ has the ORNPq with 1 < p < ∞ and q = pp−1 , then we have a completely
isometric identity

Lp(M,ϕ,E; C) = Lq(M,ϕ,E′).



ON SOME ISOMORPHISMS 135

Proof.
Lp(M,ϕ,E; C) = L(Lp(M,ϕ,E),C)

= (Lp(M,ϕ,E))
′

= (Lq(M,ϕ,E
′)

�

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Lectures in the Summer School on Banach spaces and Operator spaces, Nankai University

-China, July 16 - July 20, 2007.

1Department of Mathematics, University of Lomé, POBox 1515 Lomé, Togo

2International Chair in Mathematical Physics and Applications (ICMPA-Unesco Chair),University

of Abomey-Calavi, Benin

∗Corresponding author