Cb-frames for operator spaces

0.2 MB·

by  Rui Liu

CB-FRAMES FOR OPERATOR SPACES
6 RUI LIU AND ZHONG-JIN RUAN
1
0
2 Abstract. In this paper, we introduce the concept of cb-frames for operator
n spaces. We show that there is a concrete cb-frame for the reduced free group
a C∗-algebra Cr∗(F2), which is derived from the infinite convex decomposition of the
J biorthogonal system (λs,δs)s∈F2. We show that, in general, a separable operator
3 spaceX hasacb-frameifandonlyifithasthe completelyboundedapproximation
2
property if and only if it is completely isomorphic to a completely complemented

subspace of an operator space with a cb-basis. Therefore, a discrete group Γ is
]
A weaklyamenableifandonlyifthereducedgroupC*-algebraC∗(Γ)hasacb-frame.
r
O Finally,weshowthat,incontrasttoBanachspacecase,thereexistsaseparableop-
erator space,which can not be completely isomorphicto a subspace of anoperator
.
h space with a cb-basis.
t
a
m
[

1. Introduction
1
v
8 In [10], Junge, Nielsen, Ruan and Xu introduced the notion of cb-basis for operator
1 spaces. Let us recall that a separable operator space X has a cb-basis if X has a
2
Schauder basis (e ) and the natural projections
6 n
0
m
. ∞
1 P α e = α e (1.1)
0 m(cid:0)X n n(cid:1) X n n
6 n=1 n=1
1 satisfy sup P < . It is shown in [10] that every separable nuclear C -algebra
: mk mkcb ∞ ∗
v has a cb-basis. In particular, the reduced group C -algebra C (Γ) of an amenable
∗ r∗
i
X group Γ has a cb-basis.
r It is clear from the definition that if a separable operator space X has a cb-basis,
a
then it must have the completely bounded approximation property (CBAP), i.e. there
exists a sequence of finite-rank maps Φ : X X such that sup Φ < and
k → kk kkcb ∞
Φ (x) x for every x X. It is natural to ask whether CBAP implies cb-basis.
k
→ ∈
This is not true for general operator spaces since there exist a separable Banach space
X which has the bounded approximation property (BAP), but has no Schauder basis.
Then MIN(X), the space X equipped with the MIN-operator space structure, is an
operator space with CBAP, but no cb-basis. However the problem is still open for
Key words and phrases. completely bounded approximationproperty, cb-basis, cb-frame, opera-
tor space, reduced group C∗-algebra.
The first author was partially supported by the NSFC 11101220 and 11201336. The second
author was partially supported by the Simons Foundation.
1
2 RUILIU ANDZHONG-JINRUAN
separable C*-algebras. In particular, it is not known whether C (Γ) has a cb-basis if
r∗
Γ is a weakly amenable discrete group.
In [8], Han and Larson introduced the concept of frames as a compression of a
basis. It is a generalization of dual frame pairs from Hilbert spaces to Banach spaces.
In [2], Casazza, Han and Larson showed that a separable Banach space has the BAP
if and only if it has a frame. Motivated by these results, we can consider cb-frames
for operator spaces. Here is the definition. Let X be an operator space and X be
∗
its canonical operator dual. A sequence (x ,f ) X X is a cb-frame for X if
n n ∗
⊂ ×
(x ,f ) is a frame, i.e.
n n
∞ ∞
x = f (x)x = (x f )(x)
X n n X n ⊗ n
n=1 n=1
for all x X, and the initial sums
∈
m
S (x) = x f
m X n ⊗ n
n=1
define completely bounded maps on X with sup S < .
In section 2, we show that there is a natural cbm-kframmkecbfor C∞(F ), which is derived
r∗ 2
fromtheinfiniteconvex decompositionofthebiorthogonalsystem (λs,δs)s F2. Hereλ
is the left regular representation of F and δ B (F ) = C (F ) is the cha∈racteristic
function at s F . 2 s ∈ r 2 r∗ 2 ∗
2
∈
In Section 3, we prove some equivalent conditions for general operator spaces. We
show in Theorem 3.1 and Theorem 3.2 that a separable operator space X has a cb-
frame if and only if it has the CBAP if and only if it is completely isomorphic to a
completely complemented subspace of an operator space with a cb-basis. These are
natural operator space analogues of corresponding Banach space results by Casazza,
Han and Larson [2], Johnson, Rosenthal and Zippin [9], and Pe lczyn´ski [14]. We
also show in Remark 3.4 that the cb-basis constructed in the proof of Theorem 3.2
is minimal.
It is known from the Banach-Mazur theorem that any separable Banach space
can be isometrically embedded into C[0,1], which is a separable Banach space with
a Schauder basis. So it is natural to ask whether the corresponding result holds
for general operator spaces. Using the Hilbertian operator space X constructed by
0
Oikhberg and Ricard [13], we show in Theorem 4.5 that this is false for operator
spaces.
2. Cb-frame for C (F )
r∗ 2
Let F be the free group of two generators and let λ be the left regular representa-
2
tion of F . The reduced group C -algebra C (F ) is defined to be the norm-closure of
2 ∗ r∗ 2
CB-FRAMES FOR OPERATOR SPACES 3
span λ s F in B(ℓ (F )). It is well-known (see [6, 7]) that C (F ) has the com-
{ s| ∈ 2} 2 2 r∗ 2
pletely contractive approximation property (CCAP). The main result in this section
is the following theorem, in which we show that C (F ) has a cb-frame.
r∗ 2
Theorem 2.1. There exists a cb-frame (x ,f ) for C (F ), which is derived from the
n n r∗ 2
infinite convex decomposition of the biorthogonal system (λs,δs)s F2. More precisely,
there is a surjective map ϕ : N F and a sequence of positive sc∈alars (a ) satisfying
2 n
→
(i) x = λ ,f = a δ for all n N;
n ϕ(n) n n ϕ(n)
∈
(ii) a = 1 for all s F .
X n ∈ 2
n ϕ 1(s)
∈ −
The proof will be carried out in the following arguments and lemmas. Let us first
recall from [6] that the word length function s F s [0, ) is conditionally
2
∈ 7→ | | ∈ ∞
negative definite on F , and thus by Schoenburg’s theorem the map ϕ (s) = e ts is a
2 t − | |
positive definite function (with ϕ (e) = 1) on F for any t > 0. This gives us a family
t 2
of unital completely positive maps (Φ ) on C (F ) such that
t t>0 r∗ 2
Φ (λ ) = ϕ (s)λ = e tsλ (2.1)
t s t s − | | s
for all s F . It is clear from (2.1) that lim Φ (λ ) λ = 0 for all s F .
2 t 0+ t s s 2
∈ → k − k ∈
Then (Φ ) is a family of unital completely positive maps converging to the identity
t t>0
on C (F ) in the point-norm topology.
r∗ 2
Let W = s F : s = d be the set of all words in F with length d, and let
d 2 2
{ ∈ | | }
χ be the characteristic function on W . Then
Wd d
P : λ C (F ) λ χ (s) C (F )
d s ∈ r∗ 2 7→ s Wd ∈ r∗ 2
is the finite-rank projection onto the subspace E = span λ : s = d . It is known
d s
{ | | }
(see [1, 15]) that this projection is completely bounded with P 2d. For each
d cb
t > 0 and m N, we get a completely bounded finite-rank makp k ≤
∈
Φ = Φ (P + +P )
t,m t 0 m
···
from C (F ) onto m E = span λ : s m . We can easily obtain the following
r∗ 2 Pd=0 d { s | | ≤ }
lemma, which is known by experts in the fields. We include a calculation for the
convenience of readers.
Lemma 2.2. For each t > 0 and m N, we have
∈
∞
Φ Φ 2 e tdd 0 (as m ).
k t − t,mkcb ≤ X − → → ∞
d=m+1
4 RUILIU ANDZHONG-JINRUAN
Proof. Let x = P∞d=0Ps=da(s)⊗λs be an element in B(ℓ2)⊗Cr∗(F2) with finitely
| |
many a(s) nonzero. Then we have
∞
(id Φ id Φ )(x) = a(s) Φ (λ )
k B(ℓ2) ⊗ t − B(ℓ2) ⊗ t,m k k X X ⊗ t s k
d=m+1 s=d
| |
∞ ∞
a(s) e tdλ = e td a(s) λ
≤ X k X ⊗ − sk X − k X ⊗ sk
d=m+1 s=d d=m+1 s=d
| | | |
∞ ∞
= e td (id P )(x) e td P x .
X − k B(ℓ2) ⊗ d k ≤ X − k dkcbk k
d=m+1 d=m+1
Since P 2d, we can conclude that
d cb
k k ≤
∞ ∞
Φ Φ e td P 2 e tdd.
k t − t,mkcb ≤ X − k dkcb ≤ X −
d=m+1 d=m+1
Since P∞d=1e−tdd is a convergent positive infinite series, its remainder part converges
to 0. Therefore, we can conclude that
∞
Φ Φ 2 e tdd 0.
k t − t,mkcb ≤ X − →
d=m+1
(cid:3)
According to Lemma 2.2, for each t > 0, we can find m N such that Φ Φ
t t,m cb
∈ k − k
is sufficiently small. Therefore, we can carefully choose a sequence of (monotone
decreasing) t 0 and a sequence of (monotone increasing) m such that
k k
→ → ∞
kl→im∞kΦtk−Φtk,mkkcb ≤ kl→im∞d=Xm∞k+1e−tkdd = kl→im∞ (e1−−tk(me−kt+k2))2+(mk +11−)ee−−ttkk(mk+1) = 0.
For example, we can choose t = 1/√k and m = k, and we get
k k
∞
lim e d/√kd = 0.
−
X
k
→∞d=k+1
Itno tthhiesicdaesnet,ittyhemfianpitoen-raCnk(Fm)apins {thΦe1/p√oki,kn}tk-n∈Nor(mwittohpsoulopgkyk.Φ1/√k,kkcb < ∞) converge
r∗ 2
Let δ B (F ) = C (F ) be the biorthogonal functionals of λ . For k = 1,
{ s} ∈ r 2 r∗ 2 ∗ { s}
we set
Ψ (x) = Φ (x) = e sδ (x)λ = λ e sδ (x). (2.2)
1 1,1 X −| | s s X (cid:0) s ⊗ −| | s(cid:1)
s 1 s 1
| |≤ | |≤
CB-FRAMES FOR OPERATOR SPACES 5
There are 5 terms in (2.2). If we list these 5 terms by the index 1 j 5 and use
≤ ≤
y (respectively, g ) for the corresponding λ (respectively, e sδ ) in each term,
1,j 1,j s −| | s
we can write
5
Ψ = λ e sδ = (y g ). (2.3)
1 X (cid:0) s ⊗ −| | s(cid:1) X 1,j ⊗ 1,j
s 1 j=1
| |≤
For k 2, we set
≥
Ψ (x) = (Φ Φ )(x)
k 1/√k,k − 1/√k−1,k−1
= (e s/√k e s/√k 1)δ (x)λ + e s/√kδ (x)λ
X −| | − −| | − s s X −| | s s
s k 1 s=k
| |≤ − | |
= λ (e s/√k e s/√k 1)δ (x)+ λ e s/√kδ (x()2..4)
X (cid:0) s ⊗ −| | − −| | − s(cid:1) X (cid:0) s ⊗ −| | s(cid:1)
s k 1 s=k
| |≤ − | |
There are 2 3k 1 = 1+4+4 3+ +4 3k 1 terms in (2.4). So if we list these terms
−
· − · ··· ·
by the index 1 j 2 3k 1 andwe use y (respectively, g ) forthe corresponding
k,j k,j
≤ ≤ · −
λ (respectively, (e s/√k e s/√k 1)δ or e s/√kδ ) in each term, then we can write
s −| | −| | − s −| | s
−
23k 1
· −
Ψ = λ (e s/√k e s/√k 1)δ + λ e s/√kδ = (y g ).
k X (cid:0) s⊗ −| | − −| | − s(cid:1) X (cid:0) s⊗ −| | s(cid:1) X k,j⊗ k,j
s k 1 s=k j=1
| |≤ − | |
(2.5)
This is a sequence of completely bounded maps on C (F ) with
r∗ 2
Ψ = Φ Φ 2sup Φ (2.6)
k kkcb k 1/√k,k − 1/√k−1,k−1kcb ≤ k k √1k,kkcb
and for each x C (F ), we have
∈ r∗ 2
∞ ∞
x = lim Φ (x) = Φ (x)+ (Φ (x) Φ (x)) = Ψ (x). (2.7)
k 1/√k,k 1,1 X 1/√k,k − 1/√k−1,k−1 X k
→∞ k=2 k=1
Now to get a frame, we need to further modify the terms in (2.3) and (2.5) by
defining x = y and f = gk,j when i = p(2 3k 1) + j with 0 p
k,i k,j k,i (23k 1)2 · − ≤ ≤
(2 3k 1)2 1 and 1 j 2 3k· −1. In this case, we can write
· − − ≤ ≤ · −
23k 1 (23k 1)3
· − · −
Ψ = (y g ) = (x f ) (2.8)
k X k,j ⊗ k,j X k,i ⊗ k,i
j=1 i=1
for all k N, and thus for each x C (F ), we have
∈ ∈ r∗ 2
23k 1 (23k 1)3
∞ ∞ · − ∞ · −
x = Ψ (x) = (y g )(x) = (x f )(x) .
X k X(cid:0) X k,j ⊗ k,j (cid:1) X(cid:0) X k,i ⊗ k,i (cid:1)
k=1 k=1 j=1 k=1 i=1
6 RUILIU ANDZHONG-JINRUAN
Now we need to set up an appropriateorder to relate each pair (k,i) with a positive
integer n. For k = 1, we have 53 terms related to (1,i). So we simply let n = i. For
k 2, we let n = k−1(2 3r 1)3 +i with 1 i (2 3k 1)3.
≥ Pr=1 · − ≤ ≤ · −
Lemma 2.3. If we let x = x and f = f , then (x ,f ) is a frame for C (F ).
n k,i n k,i n n r∗ 2
Pcornovoef.rgWesetnoeexd. tSoinscheowxt=haPtf∞ko=r1eΨvekr(yx)xi∈s aCcr∗o(nFv2e),rgtehnetinsfierniietseisnerCier∗s(FP2)∞n,=f1o(rxna⊗rbfintr)a(xry)
ǫ > 0 there exists k 2 such that for any k k
0 0
≥ ≥
∞ x
Ψ (x) + Ψ (x) + k k < ǫ.
kX r k k k k 2 3k 1
r=k · −
For any m > m0 = Pkr=0−11(2·3r−1)3, we can write m = Prk=−11(2·3r−1)3+j for some
k k and 1 j (2 3k 1)3. In this case, there exists 0 p (2 3k 1)2 1
0
≥ ≤ ≤ · − ≤ ≤ · − −
and 1 q 2 3k 1 such that
≤ ≤ · −
m
x (x f )(x)
k −X n ⊗ n k
n=1
k 1(23r 1)3 j
− · −
= x (x f )(x) (x f )(x)
k −X X r,i ⊗ r,i −X k,t ⊗ k,t k
r=1 i=1 t=1
k 1 q
∞ − p 1
= Ψ (x) Ψ (x) Ψ (x) (y g )(x)
kX r −X r − (2 3k 1)2 k − (2 3k 1)2 X k,i ⊗ k,i k
r=1 r=1 · − · − i=1
∞ x
Ψ (x) + Ψ (x) + k k < ǫ.
≤ kX r k k k k 2 3k 1
r=k · −
This shows that x = P∞n=1(xn ⊗fn)(x) for every x ∈ Cr∗(F2). (cid:3)
Lemma 2.4. The sequence (x ,f ) is a cb-frame for C (F ).
n n r∗ 2
Proof. We need to show that the initial sums S = m (x f ) are completely
bounded maps on C (F ) with sup S < . mLet mPn=N1 bena⊗ponsitive integer. As
r∗ 2 k mkcb ∞ ∈
we have seen in the proof of Lemma 2.3 that we can write m = k−1(2 3r 1)3+j
for some k N and 1 j (2 3k 1)3, and there exists 0 Pp r=1(2 ·3k −1)2 1
∈ ≤ ≤ · − ≤ ≤ · − −
CB-FRAMES FOR OPERATOR SPACES 7
and 1 q 2 3k 1 such that
≤ ≤ · −
m k 1(23r 1)3 j
− · −
S = x f = (x f )+ (x f )
k mkcb kX n ⊗ nkcb kX X r,i ⊗ r,i X k,t ⊗ k,t kcb
n=1 r=1 i=1 t=1
k 1 q
− p 1
= Ψ + Ψ + (y g )
kX r (2 3k 1)2 k (2 3k 1)2 X k,i ⊗ k,i kcb
r=1 · − · − i=1
p 2 3k 1
Φ + Ψ + · −
≤ k √k1 1,k−1kcb (2 3k 1)2k kkcb (2 3k 1)2
− · − · −
3 sup Φ +1.
≤ · k k √1k,kkcb
Finally, we let us recall from (2.3), (2.5) and (2.8) that we can write
53 53 52 5 gl
Ψ = (x f ) = (x f ) = ( (yl 1,j))
1 X n ⊗ n X 1,i ⊗ 1,i X X 1,j ⊗ 52
n=1 i=1 l=1 j=1
with yl = y = λ and gl = g = e−|s|δ for corresponding s 1. For k 2, we
1,j 1,j s 1,j 1,j 52 s | | ≤ ≥
can write
Pkr=1(2·3r−1)3 (2·3k−1)3 (2·3k−1)2 2·3k−1 gl
Ψ = (x f ) = (x f ) = ( (yl k,j ))
k X n ⊗ n X k,i ⊗ k,i X X k,j ⊗ (2 3k 1)2
n=Pkr=−11(2·3r−1)3+1 i=1 l=1 j=1 · −
with ykl,j = yk,j = λs and gkl,j = gk,j = (e−|s|/√(2k3−ke−1|s)|2/√k−1)δs or (e2−3|ks|/√1)k2δs for corre-
sponding s k. So for each s F , there are· m−any positive int·eg−ers n N such
2
that x =|λ|.≤We let ϕ : n N ∈ s F be the map such that x = λ , a∈nd let a
n s 2 n s n
∈ → ∈
be the coefficient for the corresponding δ . Then it is easy to see that statement (i)
s
(cid:3)
and (ii) in theorem hold true.
Remark 2.5. The cb-frame (x ,f ) for C (F ) in Theorem 2.1 is not unconditional,
n n r∗ 2
ethacaht ixs,inthCe s(eFrie)s(sxee=[2P]).∞n=If1iftni(sxu)xnncodnodeitsionnoatl,cothnevnerfgoerueancchonxditioCna(lFly)inthneoinrmfinfioter
r∗ 2 ∈ r∗ 2
series
∞ ∞
x = (x f )(x) = a δ (x)λ
X n ⊗ n X n ϕ(n) ϕ(n)
n=1 n=1
converges unconditionally. In this case, we can rearrange its order such that
∞ ∞
x = a δ (x)λ = ( ( a )δ (x)λ ). (2.9)
X n ϕ(n) ϕ(n) X X X n s s
n=1 k=0 s=k ϕ(n)=s
| |
We note that for any s = e in F the summation a in the last term of (2.9) is
6 2 Pϕ(n)=s n
a positive infinite series with a = 1. The last equality makes sense since we
Pϕ(n)=s n
8 RUILIU ANDZHONG-JINRUAN
can apply an ε-argument to replace such an infinite sum by a finite sum if necessary.
So we can conclude from (2.9) that
∞ ∞ ∞
x = ( ( a )δ (x)λ ) = ( δ (x)λ ) = P (x).
X X X n s s X X s s X k
k=0 s=k ϕ(n)=s k=0 s=k k=0
| | | |
This implies that for any x C(T) = C (Z) ֒ C (F )
∈ ∼ r∗ → r∗ 2
K K
∞
x = P (x) = lim P (x) = lim a zj
X k X k X j
K K
k=0 →∞k=0 →∞j= K
−
converges uniformly in C(T). This is impossible. At this moment, it is not known
whether there is any unconditional frame for C (F ).
r∗ 2
3. Cb-frame, CBAP and complemented embedding property
In this section, we prove that a separable operator space X has a cb-frame if and
only if it has the CBAP if and only if it is completely isomorphic to a completely
complemented subspace of an operator space with a cb-basis. We separate it to two
results.
Theorem 3.1. A separable operator space X has a cb-frame if and only if X has the
CBAP.
Proof. The“onlyif”partisobvious. Weonlyneedtoprovethe“if”part. Theproofis
motivated by Pe lczyn´ski’s decomposition technique. Suppose that X has the CBAP.
Then there is a sequence of finite-rank maps (Φ ) on X such that sup Φ K <
k kk kkcb ≤
and x = lim Φ (x) for all x X. Let Ψ = Φ = Φ Φ (with Φ = 0)
k k 1 1 1 0 0
∞ →∞ ∈ −
and Ψ = Φ Φ for k 2. Then (Ψ ) is a sequence of finite-rank maps on X
k k k 1 k
− − ≥
such that supkkΨkkcb ≤ 2K < ∞ and x = P∞k=1Ψk(x) for all x ∈ X. Let m(k)
be the dimension of Ψ (X). It is known from Auerbach theorem that there exists
k
a biorthogonal basis (y ,y ) for Ψ (X) such that y 1, y 1
k,j k∗,j 1≤j≤m(k) k k k,jk ≤ k k∗,jk ≤
and y ,y = δ . For each 1 j m(k), g = y Ψ is a bounded linear
h k,i k∗,ji ij ≤ ≤ k,j k∗,j ◦ k
functionalonX such that g 2K. Weget (y ,g ) inX X such that
k,j k,j k,j 1 j m(k) ∗
k k ≤ ≤ ≤ ×
Ψ = m(k)y g satisfying y 1 and g 2K. As we have discussed
k Pj=1 k,j ⊗ k,j k k,jk ≤ k k,jk ≤
before Lemma 2.3, we define x = y and f = gk,j when i = p m(k)+j with
k,i k,j k,i m(k)2 ·
0 p m(k)2 1 and 1 j m(k). Then we can use a similar argument as that
≤ ≤ − ≤ ≤
given in Lemma 2.3 and Lemma 2.4 to show that if we let x = x and f = f
n k,i n k,i
when n = Prk=−01m(r)3 +i and 1 ≤ i ≤ m(k)3, then (xn,fn) is a cb-frame for X. (cid:3)
Theorem 3.2. An operator space X has a cb-frame if and only if X is completely
isomorphic to a completely complemented subspace of an operator space with a cc-
basis.
CB-FRAMES FOR OPERATOR SPACES 9
Proof. The “if” part is obvious since every completely complemented subspace of an
operator space with the CBAP always has the CBAP. We only need to prove the
“only if” part. Let (x ,f ) X X be a cb-frame of X with x = 1 for all i N.
i i ∗ i
∈ × k k ∈
Let c be the linear space of all sequences of complex numbers with finitely many
00
nonzeros, and (e ) be the canonical basis of c . For any u M (c ) = c (M ),
i 00 n 00 ∼ 00 n
there is a unique linear expression u = u e with finitel∈y many u = 0 in M .
i i i n
We define a norm on M (c ) as fPollow⊗s: 6
n n 00
|k·|k
m m
u = sup u x = max u x < . (3.1)
|k |kn m 1kX i ⊗ ikn m 1 kX i ⊗ ikn ∞
≥ i=1 ≥ i=1
More precisely for any u,v M (c ), we have
n 00
∈
m m m
u+v = max (u +v ) x max( u x + v x ) u + v .
|k |kn m 1 kX i i ⊗ ikn ≤ m 1 kX i ⊗ ikn kX i ⊗ ikn ≤ |k kn |k kn
≥ i=1 ≥ i=1 i=1
If u = 0, then m u x = 0 for all m N. By induction, we can conclude
u =|k0|.knThis shows kthPati=e1achi⊗ ikn is a norm on∈M (c ). Moreover, satisfies
n n 00 n
|k·|k |k·|k
the following properties:
(N1) u e = u x = u for all u M and i N.
i n i n n n
|k ⊗ |k k ⊗ k k k ∈ ∈
(N2) For any u M and m l, we have m u e l u e .
i ∈ n ≤ |kPi=1 i ⊗ i|kn ≤ |kPi=1 i ⊗ i|kn
(N3) For any x X, ( n f (x)e ) is a Cauchy sequence in (c , ).
∈ Pi=1 i i ∞n=1 00 |k·|k1
Since (N1) and (N2) are obvious, we only need to prove (N3). Since (x ,f ) is a
i i
fisranmesufocrhXth,afto,rfoarnyanxy∈mX>, nwe hanve, wxe=haPve∞i=1fi(mx)xi.fT(xh)exn fo<r aǫn.yNǫow>i0f,wteheleret
y =0 n f (x)e for each n N≥, we0 get kPi=n+1 i ik
n Pi=1 i i ∈
m k
y y = f (x)e = max f (x)x < ǫ.
|k m − n|k1 |k X i i|k1 n+1 k mk X i ik
i=n+1 ≤ ≤ i=n+1
This shows (N3).
10 RUILIU ANDZHONG-JINRUAN
Now we prove that is actually an operator space matrix norm. For u =
n
u e M (c ),w|k=· |kw e M (c ), and α M ,β M , we have
i i n 00 i i m 00 n n
P ⊗ ∈ P ⊗ ∈ ∈ ∈
l
u w = (u w ) e = max (u w ) x
|k ⊕ |kn+m |kX i ⊕ i ⊗ i|kn+m l 1 kX i ⊕ i ⊗ ikn+m
≥ i=1
l l
= max ( u x ) ( w x )
l 1 k X i ⊗ i ⊕ X i ⊗ i kn+m
≥ i=1 i=1
l l
= maxmax u x , w x
l 1 nkX i ⊗ ikn kX i ⊗ ikmo
≥ i=1 i=1
l l
= max max u x ,max w x
n l 1 kX i ⊗ ikn l 1 kX i ⊗ ikmo
≥ i=1 ≥ i=1
= max u , w , (3.2)
n m
{|k |k |k |k }
and
αuβ = α( u e )β = (αu β) e
|k |kn |k X i ⊗ i |kn |kX i ⊗ i|kn
l l
= max (αu β) x = max α( u x )β
l 1 kX i ⊗ ikn l k X i ⊗ i kn
≥ i=1 i=1
l
max α u x β = α u β . (3.3)
≤ l 1 k knkX i ⊗ iknk kn k kn|k |knk kn
≥ i=1
Thus, by the abstract characterization theorem given in [16], this newly defined
matricial norm determines an operator space structure on c . We let
n 00
{|k · |k }
Y = c denote the completion. It is known from [5, Fact 6.3] that (e ) is
00−|k·|k i
a basis for Y. According to (N2), (e ) is actually a cc-basis for Y since for any
i
u = u e M (c ), the natural projections P satisfy
i i n 00 m
P ⊗ ∈
m
(P ) (u) = u e u .
k m n kn |kX i ⊗ i|kn ≤ |k |kn
i=1
Now let us define a linear map
Q : α e c α x X. (3.4)
X i i ∈ 00 → X i i ∈
For any u = u e M (c ), we get
i i n 00
P ⊗ ∈
(Q) (u) = (Q) u e = u x
n n(cid:0)X i ⊗ i(cid:1) X i ⊗ i
and
(Q) (u) = u x u
k n kn kX i ⊗ ikn ≤ |k |kn