# Double duals and Hilbert modules

Huaxin Lin

###### Abstract

Let $A$ be a $C^{*}$-algebra, $H$ be a Hilbert $A$-module and $K(H)$ be the
closure of the set of finite rank module maps. We show that the
$W^{*}$-algebra of all bounded $A^{**}$-module maps on the smallest self-dual
Hilbert $A^{**}$-module containing $H$ is isomorphic to $K(H)^{**}$ as
$W^{*}$-algebras. We also show that the unit ball of $H$ is closed in
$H^{\sharp},$ the dual of $H,$ in an $A$-weak topology of $H^{\sharp}$ as well
as dense in the unit ball of $H^{\sharp}$ in a weak*-topology and some
versions of Kaplansky density theorem for Hilbert $C^{*}$-modules.

##  1 Introduction

Hilbert $C^{*}$-modules as a generalization of Hilbert spaces were first
introduced by I. Kaplansky ([6]) in 1953 in special cases and later by W.
Paschke ([13]) for general $C^{*}$-algebras in 1973. Hilbert $C^{*}$-modules
are crucial to Kasparov’s formulation of $KK$-theory ([7]). Early applications
also include $C^{*}$-algebraic quantum group theory (see [16]). Later in the
study of Cuntz -semigroups in connection with the classification of amenable
$C^{*}$-algebras, Hilbert $C^{*}$-modules play an important role (see, for
example, [1], [2], [5], and [12]).

Let $A$ be a $C^{*}$-algebra. Unlike Hilbert spaces, bounded module maps on a
Hilbert $A$-module $H$ may not have adjoints and the dual module $H^{\sharp},$
i.e., the Banach $A$-module of all bounded module maps from $H$ to $A$ may not
be identified as elements in $H.$ Moreover, the $C^{*}$-algebra $L(H)$ of all
bounded module maps with adjoints may not be a $W^{*}$-algebras. If
$H_{0}\subset H$ is a Hilbert $A$-submodule, a bounded module map
$\varphi:H_{0}\to A$ may not be extended to a bounded module map from $H$ to
$A.$ In general, one should not expect that $H$ to be decomposed to an
orthogonal direct sum of $H_{0}$ and its orthogonal complement. In fact,
$H_{0}$ may not even have an orthogonal complement. Study of these phenomena
may be found, for eaxmples, in [9], [10] and more recently in [1].

However, Paschke found (see [13]) out that, if $A$ is a $W^{*}$-algebra, then
the dual module $H^{\sharp}$ of a Hilbert $A$-module $H$ can be made into a
Hilbert $A$-module in a natural way which extends $H,$ and $H^{\sharp}$ is a
self-dual Hilbert $A$-module. Even if $A$ is not a $W^{*}$-algebra, one can
extend $H$ into an $A^{**}$-module $H\bullet A^{**}$ naturally. Then its dual
$H^{\sim}:=(H\bullet A^{**})^{\sharp}$ becomes a self-dual Hilbert
$A^{**}$-module containing $H.$ In fact, $H^{\sim}$ is the smallest self-dual
Hilbert $A^{**}$-module containing $H$ as a Hilbert $A$-submodule (see
Proposition 3.2). Paschke showed the Banach algebra of all bounded module maps
on $H^{\sim}$ becomes a $W^{*}$-algebra.

For a Hilbert $A$-module $H,$ the rank one module maps are the module maps $T$
of the form $T(h)=x\langle y,h\rangle$ for all $h\in H$ (and fixed $x,y\in H,$
where $\langle\cdot,\cdot\rangle$ is the $A$-valued inner product). Denote by
$F(H)$ the linear span of rank one module maps and denote by $K(H)$ the norm
closure of $F(H).$ $K(H)$ is a $C^{*}$-algebra and an important algebra
related to the Hilbert module $H.$ It was proved by Kasparov (see Theorem 1 of
[7]) that the $C^{*}$-algebra $L(H)$ may be identified with $M(K(H)),$ the
multiplier algebra of $K(H),$ and it was proved in [9] that the Banach algebra
of all bounded module maps on $H$ is identified with the left multipliers of
$K(H)$ (all Hilbert $A$-modules considered in this paper are right
$A$-modules). Over the decades, we eventually realized that it is rather
convenient to work in $B(H^{\sim})$ in many occasions as we study module maps
on a Hilbert module $H.$ It is not difficult to establish a natural normal
homomorphism $\Psi:K(H)^{**}\to B(H^{\sim})$ which extends beyond $M(K(H))$
and $LM(K(H)).$ It remained unknown for many years whether $\Psi$ is an
isomorphism. The original motivation of this paper is to show that indeed
$\Psi$ is an isomorphism between $W^{*}$-algebras $K(H)^{**}$ and
$B(H^{\sim}).$

As we study the relation among Hilbert modules $H,$ $H\bullet A^{**}$ and
$H^{\sim},$ naturally, we ask how dense of $H$ in $H\bullet A^{**}$ and in
$H^{\sim}$ is? Since $H^{\sim}=(H\bullet A^{**})^{\sharp},$ the dual of
$H\bullet A^{**},$ one may also ask the density of $H$ in $H^{\sharp}$ in
general.

We first note that it was shown (Theorem 6.1 of [1]) that $H$ is dense in
$H^{\sharp}$ in an $A$-weak topology. More precisely, for any $\xi\in
H^{\sharp},$ there is a net $\\{x_{\alpha}\\}$ in $H$ with
$\|x_{\alpha}\|\leq\|\xi\|$ for all ${\alpha}$ such that
$\lim_{\alpha}\|\xi(x)-\langle x_{\alpha},x\rangle\|=0$ for all $x\in H.$
However, we show here that the unit ball of $H$ is closed in $H^{\sharp}$ in
the topology where $x_{\alpha}\to\xi$ if and only if
$\lim_{\alpha}\|\langle\xi-x_{\alpha},\zeta\rangle\|=0$ for all $\zeta\in
H^{\sharp},$ and where the inner product is extended to $H^{\sim}.$

On the other hand, it is easy to see that for any $\xi\in H\bullet A^{**}$
there is a net $\\{x_{\lambda}\\}$ in $H$ such that
$\lim_{\lambda}\pi_{U}(\langle
x_{\lambda},y\rangle)(v)=\pi_{U}(\langle\xi,y\rangle)(v)$ for all $y\in
H\bullet A^{**}$ and $v\in H_{U},$ where $H_{U}$ is the Hilbert space
corresponding to the universal representation $\pi_{U}$ of $A.$ To be a more
useful approximation, one may ask whether the net can be chosen to be bounded
(by $\|\xi\|$). We will present a Kaplansky style density theorem. Perhaps
more interesting question is how dense $H$ is in $H^{\sim}=(H\bullet
A^{**})^{\sharp}.$ Since $H^{\sim}$ is the dual of $H\bullet A^{**},$ it is
relatively easy to show that, for any $\zeta\in H^{\sim},$ there is a net
$\\{z_{\alpha}\\}$ in $H$ such that $\lim_{\lambda}f(\langle
z_{\alpha},y\rangle)=f(\langle\zeta,y\rangle)$ for all $y\in H\bullet A^{**}$
and $f\in A^{*}.$ It is more challenging to show that $y$ can be replaced by
any element in $H^{\sim}=(H\bullet A^{**})^{\sharp}.$ We show that the unit
ball of $H$ is actually dense in the unit ball of $H^{\sim}$ in the
weak*-topology (as $H^{\sim}$ is a conjugate space), another Kaplansky style
density theorem. In fact, we show a stronger density theorem that, for any
$\xi\in H^{\sim},$ there is a net $\\{x_{\alpha}\\}$ in $H$ with
$\|x_{\alpha}\|\leq\|\xi\|$ such that

$\displaystyle\lim_{\alpha}f(\langle\xi-x_{\alpha},\xi-
x_{\alpha}\rangle)=0\,\,\,{\rm for\,\,\,all}\,\,\,f\in A^{*}.$ (e 1.1)

##  2 Self-duals

######  Definition 2.1.

Let $A$ be a $C^{*}$-algebra. Denote by $\tilde{A}$ the minimum unitization of
$A.$ We will use the following convention: If $A$ is a $C^{*}$-subalgebra of a
unital $C^{*}$-algebra $B,$ we write $1_{\tilde{A}}=1_{B},$ if either $A$ is
unital and $1_{A}=1_{\tilde{A}}=1_{B},$ or $A^{\perp}=\\{b\in
B:ba=ab=0\\}=\\{0\\},$ and we unitize $A$ by adjoint $1_{B}$ to form
$\tilde{A}\subset B.$

######  Definition 2.2.

Let $X$ be a Hilbert space and $B(X)$ be the $C^{*}$-algebra of all bounded
linear operators on $X.$ Suppose that $A\subset B(X).$ Then
$\overline{A}^{SOT}$ is the closure of $A$ in the strong operator topology.
Note that if $\\{e_{\alpha}\\}$ is an approximate identity for $A,$ then
$e_{\alpha}\nearrow 1_{M},$ i.e., $e_{\alpha}$ increasingly converges to the
identity of $M=\overline{A}^{SOT}$ in the strong operator topology as well as
in the weak*-topology (of $M$). In particular, we may write
$1_{\tilde{A}}=1_{M}.$

This works particularly for the pair $A$ and $A^{**}$ (where $X$ is $H_{u},$
the Hilbert space corresponding to the universal representation of $A$).

In general, if $M$ is a $W^{*}$-algebra, denote by $M_{*}$ the pre-dual of
$M.$

######  Definition 2.3.

Let $A$ be a $C^{*}$-algebra. In this paper, we use the formal definition of
Hilbert modules in [13] and consider only right $A$-modules. Recall that a
linear space $H$ is a pre-Hilbert module if it is also a right $A$-module with
an inner product $H\times H\to A$ satisfying the following properties: for any
$x,y,z\in H,$ $a\in A$ and $\lambda\in\mathbb{C},$

(i) $\langle x,\lambda y+z\rangle=\langle x,y\rangle+\lambda\langle
x,y\rangle;$

(ii) $\langle x,ya\rangle=\langle x,y\rangle a;$

(iii) $\langle x,y\rangle^{*}=\langle y,x\rangle;$

(iv) $\langle x,x\rangle\geq 0;$ if $\langle x,x\rangle=0,$ then $x=0.$

Define $\|x\|=\|\langle x,x\rangle\|^{1/2}$ for $x\in H.$ Then $H$ becomes a
normed space. $H$ is a Hilbert $A$-module if $H$ is complete with this norm.

Denote by $H^{\sharp}$ the Banach space of all bounded module maps from $H$
into $A.$ A Hilbert $A$-module is said to be self-dual, if, for every $f\in
H^{\sharp},$ there is $x\in H$ such that

$f(y)=\langle x,y\rangle\,\,\,{\rm for\,\,\,all}\,\,\,y\in H.$

Denote by $B(H)$ the Banach algebra of all bounded module maps from $H$ into
itself, and by $L(H),$ the $C^{*}$-algebra of all those bounded module maps
$T$ with an adjoint $T^{*}$ in $L(H),$ defined by

$\langle T(x),y\rangle=\langle x,T^{*}(y)\rangle\,\,\,{\rm
for\,\,\,all}\,\,\,x,y\in H.$

Let $F(H)$ be the algebra of all finite rank module maps, i.e., the linear
span of all bounded module maps of the form: $\theta_{x,y}:H\to H,$ defined by

$\theta_{x,y}(\xi)=x\langle y,\xi\rangle$

for all $\xi\in H$ and $x,y\in H.$ Denote by $K(H)$ the $C^{*}$-algebra of all
bounded module maps which are (uniform) norm closure of $F(H).$

By Theorem 1 of [7], we identify $L(H)$ with $M(K(H))$ the multiplier algebra
of $K(H)$ and, by Theorem 1.5 of [9], $B(H)$ with $LM(K(H))$ the Banach
algebra of left multipliers of $K(H)$ ( in $K(H)^{**}$). If $H$ is self-dual,
then $B(H)=L(H).$

We refer to [7], [13], [9] and [10] for common terminologies related to
Hilbert $C^{*}$-modules.

######  Definition 2.4.

Let $A$ be a $C^{*}$-algebra and $H$ a Hilbert $A$-module. Let us give the
definition of self-dual Hilbert $A^{**}$-module $H^{\sim}$ (see Definition 1.3
of [9]).

We may view $H$ as a Hilbert $\tilde{A}$-module. Let $B$ be a unital
$C^{*}$-algebra containing $A$ and $1_{\tilde{A}}=1_{B}$ (see the convention
in 2.1). The algebraical tensor product $H\otimes B$ becomes a right
$B$-module if we set $(h\otimes a)\cdot b=h\otimes ab$ for any $h\in H$ and
$a,b\in B.$ Define $\langle-,-\rangle:H\otimes B\times H\otimes B\to B$ by

$\langle\sum_{i}h_{i}\otimes a_{i},\,\sum_{j}x_{j},\otimes
b_{j}\rangle=\sum_{i,j}a_{i}\langle h_{i},x_{j}\rangle b_{j},$

and $N=\\{z\in H\otimes A^{**}:\langle z,z\rangle=0\\}.$ Then $(H\otimes B)/N$
becomes a per-Hilbert $B$-module (see 4 of [13] –but exchange $B$ with $A$).
Denote by $H\bullet B:=(H\otimes B)/N)^{-}$ (the completion) the Hilbert
$B$-module.

We are particularly interested in the case that $B=A^{**}.$ We view
$\tilde{A}$ as a $C^{*}$-subalgebra of $A^{**}.$ Then $H^{\sim}:=(H\bullet
A^{**})^{\sharp}$ is a self-dual Hilbert $A^{**}$-module.

Note that $\tilde{A}$ is ultraweakly dense in $A^{**}$ (since $A$ is). By
applying Theorem 4.2 of [13] to the pair $A^{**}$ (as $A$ in Theorem 4.2 of
[13]) and $\tilde{A}$ (as $B$ in Theorem 4.2 of [13], see also the remark
right after the proof of Theorem 4.2 of [13]), we obtain an isometric
(surjective) isomorphism $\iota:H^{\sim}:=(H\bullet A^{**})^{\sharp}\to
B(H,A^{**}),$ the Banach space of all bounded $A$-module maps from $H$ to
$A^{**}$ (as written $M(H,A^{**})$ in Theorem 4.2 of [13]).

Let $x\in H$ and $b\in B.$ Then

$\|(x\otimes b)/N\|^{2}=\|b^{*}\langle x,x\rangle b\|\leq\|x\|^{2}\|b^{*}b\|.$

Hence

$\displaystyle\|(x\otimes b)/N\|\leq\|x\|\|b\|.$ (e 2.2)

In what follows, for $x\in H$ and $b\in B,$ we write $x\bullet b:=(x\otimes
b)/N.$

In general, if $E$ is a self-dual Hilbert module, then $B(E)=L(E)$ (see 3.5 of
[13]). If in addition, $A$ is a $W^{*}$-algebra, $B(E)$ is also a $W^{*}$
-algebra (see Proposition 3.11 of [13]).

Let us recall the description of the pre-dual of $B(E)$ in this case. Denote
by $E_{\sim}$ the linear space $E$ with the “twisted” scalar multiplication
(i.e., $\lambda x=\bar{\lambda}x$ for $x\in E$ and $\lambda\in\mathbb{C}$) and
consider $E\otimes E_{\sim}\otimes A_{*}$ with the greatest cross-norm, where
$A_{*}$ is the usual pre-dual of $W^{*}$-algebra $A.$ For each $T\in B(E),$
define a linear functional $\check{T}$ on $E_{\sim}\otimes E\otimes A_{*}$ by

$\displaystyle\check{T}(\sum_{j=1}^{n}x_{j}\otimes y_{j}\otimes
g_{j})=\sum_{j=1}^{n}g_{j}(\langle T(x_{j}),y_{j}\rangle)$ (e 2.3)

for $x_{j},y_{j}\in E$ and $g_{j}\in A_{*},$ $1\leq j\leq n.$ The map
$T\to\check{T}$ is a linear isometry of $B(E)=L(E)$ into $(E_{\sim}\otimes
E\otimes A_{*})^{*}.$ It was shown (Proposition 3.10 of [13]) that
$B(E{\check{)}}$ is weak*-closed in $E_{\sim}\otimes E\otimes A_{*}.$ A
bounded net $\\{T_{\alpha}\\}$ in $B(E)$ converges to $T\in B(E)$ in the
weak*-topology if and only if $f(\langle T_{\alpha}(x),y\rangle)\to f(\langle
T(x),y\rangle)$ for all $x,y\in E$ and $f\in A_{*}$ (Remark 3.9 and
Proposition 3.10 of [13]). In particular, $B(H^{\sim})$ is a $W^{*}$-algebra.

######  Definition 2.5.

Keep the notation in 2.4. Then $\iota:H\to H\bullet B$ defined by $x\to
x\otimes 1$ is an injective map. Note that

$\displaystyle\langle(x\cdot b)\otimes 1-x\otimes b,(x\cdot b)\otimes
1-x\otimes b\rangle$ (e 2.4) $\displaystyle=\langle x\cdot b,x\cdot
b\rangle-\langle x\cdot b,x\rangle b-b^{*}\langle x,x\cdot
b\rangle+b^{*}\langle x,x\rangle b$ (e 2.5) $\displaystyle=b^{*}\langle
x,x\rangle b-b^{*}\langle x,x\rangle b-b^{*}\langle x,x\rangle b+b^{*}\langle
x,x\rangle b=0.$ (e 2.6)

Hence $\iota(x\cdot b)=x\otimes b/N$ for all $b\in B.$ In the case $B=A^{**},$
we then extend $\iota$ from $H^{\sharp}$ to $(H\bullet A^{**})^{\sharp}$ by

$\iota(f)(x\otimes b)=f(x)b\,\,\,{\rm for\,\,\,all}\,\,\,x\in H\,\,\,{\rm
and}\,\,\,b\in A^{**}$

and $f\in H^{\sharp}.$ Note that the map is a module map from $H^{\sharp}$ to
$(H^{\sim})^{\sharp}$ (which is conjugate module isomorphic to $H^{\sim}$).

From now on, we may view $H$ as a submodule of $H^{\sim}$ and, sometime, omit
the map $\iota.$

The following is a convenient and easy fact that $H\bullet B$ is smallest
Hilbert $B$-module containing $H$ as a Hilbert $A$-module.

######  Proposition 2.6.

Let $A$ and $B$ be a pair of $C^{*}$-algebras such that $A\subset B,$ $B$ is
unital and $1_{\tilde{A}}=1_{B}.$ Suppose that $H$ is a Hilbert $A$-module,
$H_{1}$ is a Hilbert $B$-module and there is an embedding $\iota:H\to H_{1}$
as Hilbert modules, i.e., $\iota$ is a linear and $A$-module map such that

$\displaystyle\langle\iota(x),\iota(y)\rangle=\langle
x,y\rangle\,\,\,\text{for\,\,\,all}\,\,\,x,y\in H.$ (e 2.7)

Then there is a unique $B$-module embedding $\tilde{\iota}:H\bullet B\to
H_{1}$ (such that

$\displaystyle\tilde{\iota}(x\bullet
b)=\iota(x)b\,\,\,\text{for\,\,\,all}\,\,\,x\in H\,\,\,{\rm and}\,\,\,b\in
B\,\,\,\text{and}\,\,\,$ (e 2.8)
$\displaystyle\langle\tilde{\iota}(\xi),\tilde{\iota}(\zeta)\rangle=\langle\xi,\zeta\rangle\,\,\,\text{for\,\,\,all}\,\,\,\xi,\zeta\in
H\bullet B).$ (e 2.9)

###### Proof.

For any $\sum_{i=1}^{n}x_{i}\bullet a_{i},$ where $x_{i}\in H$ and $a_{i}\in
B$ ($1\leq i\leq n$), define

$\displaystyle\tilde{\iota}(\xi)=\sum_{i=1}^{n}\iota(x_{i})a_{i}.$ (e 2.10)

Then, for $\zeta=\sum_{i=1}^{n}y_{i}\bullet b_{i},$

$\displaystyle\langle\tilde{\iota}(\xi),\tilde{\iota}(\zeta)\rangle=\sum_{i,j}^{n}a_{i}^{*}\langle
x_{i},y_{j}\rangle b_{j}=\langle\xi,\zeta\rangle.$ (e 2.11)

In particular,

$\displaystyle\|\tilde{\iota}(\xi),\tilde{\iota}(\xi)\rangle\|=\|\sum_{i,j}^{n}a_{i}^{*}\langle
x_{i},x_{j}\rangle b_{j}\|=\|\xi\|^{2}.$ (e 2.12)

Therefore $\|\tilde{\iota}\|\leq 1$ on $(H\otimes B)/N.$ So $\tilde{\iota}$ is
uniquely extended to a contractive linear map from $H\bullet B$ into $H_{1}.$
It is a $B$-module map. Since $(H\otimes B)/N$ is dense in $H\bullet B,$

$\langle\tilde{\iota}(x),\tilde{\iota}(y)\rangle=\langle x,y\rangle\,\,\,{\rm
for\,\,\,all}\,\,\,x,y\in H\bullet B.$

To see this embedding is unique, let $\tilde{\iota}_{1}$ be another such
embedding. Then $(\tilde{\iota}-\tilde{\iota}_{1})|_{H}=0.$ For any
$\xi=\sum_{i=1}^{n}x_{i}\bullet a_{i},$ where $x_{i}\in H$ and $a_{j}\in B,$

$\displaystyle(\tilde{\iota}-\tilde{\iota}_{1})(\xi)=\sum_{i=1}^{n}(\iota(x_{i})-\iota(x_{i}))\bullet
a_{i}=0.$ (e 2.13)

In other words, $\tilde{\iota}_{1}=\tilde{\iota}.$ ∎

######  Definition 2.7.

Keep the notations in 2.3, 2.4 and 2.5. Recall that $F(H)$ is the algebra of
all finite rank module maps. Define $\Psi_{0}:F(H)\to F(H\bullet B)\subset
B(H\bullet B)$ by

$\displaystyle\Psi_{0}(\theta_{x,y})(\zeta)=\iota(x)\langle\iota(y),\zeta\rangle$
(e 2.14)

for all $\zeta\in H\bullet B,$ $x,y\in H.$ $\Psi$ is a $*$-preserving
homomorphism from the $*$-algebra $F(H)$ into $F(H\bullet B).$ Moreover,
$\Psi_{0}$ is an isometry on $F(H).$ In particular, $\|\Psi_{0}\|=1.$
Therefore it extends uniquely to a $C^{*}$-algebra homomorphism from $K(H)$ to
$K(H\bullet B)$ (which preserves the norm). It has to be an isometry as $F(H)$
is dense in $K(H).$

In the case that $B=A^{**},$ we may define $\tilde{\Psi}_{0}:F(H)\to
F(H^{\sim})\subset B(H^{\sim})$ by

$\displaystyle\tilde{\Psi}_{0}(\theta_{x,y})(\zeta)=\iota(x)\langle\iota(y),\zeta\rangle$
(e 2.15)

for all $\zeta\in H^{\sim},$ $x,y\in H.$ Then $\tilde{\Psi}_{0}$ is a
$*$-preserving homomorphism from the $*$-algebra $F(H)$ into $F(H^{\sim})$ and
it extends uniquely to a $C^{*}$-algebra homomorphism $\tilde{\Psi}_{0}$ from
$K(H)$ to $K(H^{\sim})$ which preserves the norm. Recall that
$\iota(H^{\sharp})\subset H^{\sim}.$

######  Proposition 2.8.

Let $A\subset B$ be a pair of $C^{*}$-algebras, where $B$ is unital and
$1_{B}=1_{\tilde{A}}.$ Let $T\in K(H).$ Then $\Psi_{0}(T)(x\bullet
b)=T(x)\bullet b$ for all $x\in H$ and $b\in B.$

###### Proof.

From the definition, for any $S\in F(H),$ any $x\in H$ and any $b\in B,$

$\displaystyle\Psi_{0}(S)(x\otimes b)=S(x)\otimes b\,\,\,({\rm mod}\,N).$ (e
2.16)

Fix $T\in K(H)$ and let $\epsilon>0.$ There exists $S\in F(H)$ such that

$\displaystyle\|T-S\|<\epsilon/2(1+\|x\bullet b\|+\|x\|\|b\|).$ (e 2.17)

Then

$\displaystyle\|\Psi_{0}(T)-\Psi_{0}(S)\|<\epsilon/4(1+\|x\otimes
b\|+\|x\|\|b\|)\,\,\,{\rm and}\,\,\,\|T(x)\bullet b-S(x)\bullet
b\|\leq\epsilon/2.$ (e 2.18)

Hence

$\displaystyle\|\Psi_{0}(T)(x\bullet b)-T(x)\bullet b\|<\epsilon.$ (e 2.19)

Since this holds for all $\epsilon>0,$ we conclude that

$\Psi_{0}(x\bullet b)=T(x)\bullet b.$

∎

######  Lemma 2.9.

Let $A$ and $B$ be as in Proposition 2.8, and $H$ be a Hilbert $A$-module.
Suppose that $\\{E_{\lambda}\\}$ is an approximate identity for $K(H).$ Then
$\\{\Psi_{0})(E_{\lambda})\\}$ forms an approximate identity for $K(H\bullet
B).$ Moreover

$\displaystyle\lim_{\lambda}\|\Psi_{0}(E_{\lambda})(x)-x\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,x\in H\bullet B.$ (e 2.20)

###### Proof.

By Lemma 3.1 of [1],

$\displaystyle\lim_{\lambda}\|E_{\lambda}(x)-x\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,x\in H.$ (e 2.21)

Let $S=\sum_{i=1}^{n}\theta_{x_{i},y_{i}},$ where $x_{i},y_{i}\in(H\otimes
B)/N,$ $1\leq i\leq n.$ Write $x_{i}=\sum_{j=1}^{k(i)}\xi_{j,i}\bullet
b_{j,i},$ where $\xi_{j,i}\in H$ and $b_{j,i}\in B,$ $j=1,2,...,k(i),$
$i=1,2,...,n.$ By Proposition 2.8,

$\displaystyle\Psi_{0}(E_{\lambda})(\xi_{j,i}\bullet
b_{j,i})=E_{\lambda}(\xi_{j,i})\bullet b_{j,i}.$ (e 2.22)

By (e 2.21),

$\displaystyle\lim_{\lambda}\|\Psi_{0}(E_{\lambda})(\xi_{j,i}\bullet
b_{j,i})-(\xi_{j,i}\bullet b_{j,i})\|=0$ (e 2.23)

for $j=1,2,...,k(i),$ $i=1,2,...,n.$ It follows that

$\displaystyle\lim_{\lambda}\|\Psi_{0}(E_{\lambda})(x_{i})-x_{i}\|=0,\,\,\,i=1,2,...,n.$
(e 2.24)

For any $z\in H\bullet B,$

$\displaystyle\Psi_{0}(E_{\lambda})\theta_{x_{i},y_{i}}(z)$ $\displaystyle=$
$\displaystyle(\Psi_{0}(E_{\lambda})x_{i})\langle y_{i},z\rangle$ (e 2.25)
$\displaystyle=$ $\displaystyle E_{\lambda}(x_{i})\langle y_{i},z\rangle.$ (e
2.26)

It follows that (for $1\leq i\leq n$)

$\displaystyle\lim_{\lambda}\|\Psi_{0}(E_{\lambda})\theta_{x_{i},y_{i}}-\theta_{x_{i},y_{i}}\|=0.$
(e 2.27)

Hence

$\displaystyle\lim_{\lambda}\|\Psi_{0}(E_{\lambda})S-S\|=0.$ (e 2.28)

The set of those module maps with the form of $S$ is norm dense in $K(H\bullet
B).$ Therefore we conclude that

$\displaystyle\lim_{\lambda}\|\Psi_{0}(E_{\lambda})S-S\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,S\in K(H\bullet B).$ (e 2.29)

It follows that $\\{\Psi_{0}(E_{\lambda})\\}$ forms an approximate identity
for $K(H\bullet B).$

∎

######  2.10.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then
$H^{\sharp}$ is a Banach $A$-module in general. Recall that, for each $T\in
B(H),$ one may define a bounded conjugate module map $T^{*}:H\to H^{\sharp}$
as follows. For $x,y\in H,$ define

$\displaystyle T^{*}(x)(y)=\langle x,T(y)\rangle.$ (e 2.30)

So, for a fixed $x,$ $T^{*}(x)$ gives an element in $H^{\sharp}.$ Moreover,
$T^{*}$ is a bounded conjugate module map from $H$ to $H^{\sharp}$ with
$\|T^{*}\|=\|T\|.$ However, if we view $H$ as a submodule of $H^{\sharp},$
then $T^{*}$ is a bounded module map. Note that, if $T\in L(H),$ then
$T^{*}\in L(H)$ and $T^{*}(H)\subset H.$

If $A$ is a $W^{*}$-algebra, by Theorem 3.2 of [13], $H^{\sharp}$ becomes a
Hilbert $A$ module in a natural way. For $T\in B(H)$ and $f\in H^{\sharp},$
define, for each $x\in H,$

$\displaystyle\tilde{T}(f)(x)=\langle f,T^{*}(x)\rangle,$ (e 2.31)

where $T^{*}$ is defined above. Thus $\tilde{T}(f)$ is a bounded linear module
map from $H$ to $A$ with $\|\tilde{T}(f)\|\leq\|T\|\|f\|.$ Hence we extend $T$
to a bounded (conjugate) module map from $H^{\sharp}$ to $H^{\sharp}.$ As we
view $H^{\sharp}$ as a Hilbert $A$-submodule in this case, $T$ is in fact a
bounded module map on $H^{\sharp}$ (we will take the conjugate as Hilbert
space cases). By Corollary 3.7 of [13], such extension is unique.

By Lemma 3.7 of [10], one may ease the assumption that $A$ is a
$W^{*}$-algebra to the assumption that $A$ is a monotone complete
$C^{*}$-algebra.

######  Proposition 2.11.

Let $A$ and $B$ be as in 2.8, $H$ be a Hilbert $A$-module and
$\\{E_{\lambda}\\}$ an approximate identity for $K(H).$ Then

$\displaystyle\lim_{\lambda}(\sup\\{\|\tilde{\Psi}_{0}(E_{\lambda})(f)(x)-f(x)\|:f\in(H\bullet
B)^{\sharp},\,\|f\|\leq 1\\})=0\,\,\,{\rm for\,\,\,all}\,\,\,x\in H\bullet B.$
(e 2.32)

Moreover, suppose that $(H\bullet B)^{\sharp}$ extends $H\bullet B$ as Hilbert
$B$-module, then, for any $T\in B((H\bullet B)^{\sharp}),$

$\displaystyle\lim_{\lambda}\|\langle\tilde{\Psi}_{0}(E_{\lambda})T\tilde{\Psi}_{0}(E_{\lambda})(x),y\rangle-\langle
T(x),y\rangle\|=0\,\,\,{\rm for\,\,\,all}\,\,\,x,y\in H\bullet B.$ (e 2.33)

###### Proof.

Fix $f\in H^{\sharp}.$ For any $x\in H\bullet B,$ by Lemma 2.9,

$\displaystyle\|\tilde{\Psi}_{0}(E_{\lambda})(f)(x)-f(x)\|=\|f(E_{\lambda}(x))-f(x)\|\leq\|f\|\|E_{\lambda}(x)-x\|\to
0.$ (e 2.34)

Hence (e 2.32) holds.

To see the “Moreover” part of the lemma, let $T\in B((H\bullet B)^{\sharp}).$
Then, for any $x,y\in H\bullet B,$

$\displaystyle\|\langle\tilde{\Psi}_{0}(E_{\lambda})T\tilde{\Psi}_{0}(E_{\lambda})(x),y\rangle-\langle
T(x),y\rangle\|$ $\displaystyle\leq\|\langle
T\tilde{\Psi}_{0}(E_{\lambda})(x),\tilde{\Psi}_{0}(E_{\lambda})(y)\rangle-\langle
T(x),\tilde{\Psi}_{0}(E_{\lambda})(y)\rangle\|$ (e 2.35) $\displaystyle\hskip
36.135pt+\|\langle T(x),\tilde{\Psi}_{0}(E_{\lambda})(y)\rangle-\langle
T(x),y\rangle\|$ (e 2.36)
$\displaystyle\leq\|y\|\|T\|\|\Psi_{0}(E_{\lambda})(x)-x\|+\|T\|\|x\|\|\Psi_{0}(E_{\lambda})(y)-y\|$
(e 2.37)

By applying Lemma 2.9 to the two terms of the last inequality above, we
conclude that

$\displaystyle\lim_{\lambda}\|\langle\tilde{\Psi}_{0}(E_{\lambda})T\tilde{\Psi}_{0}(E_{\lambda})(x),y\rangle-\langle
T(x),y\rangle\|=0\,\,\,{\rm for\,\,\,all}\,\,\,x,y\in H\bullet B.$ (e 2.38)

∎

######  Definition 2.12.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Recall (Theorem
1.5 of [9]) that we identify $B(H)$ with $LM(K(H)),$ the Banach algebra of
left multipliers of $K(H)$ (in $K(H)^{**}$).

By 2.9, $\Psi_{0}$ maps $K(H)$ into $K(H\bullet B)$ which maps approximate
identities to approximate identities. It follows that we may extend a
homomorphism $\Psi_{0}:B(H)=LM(K(H))\to LM(K(H\bullet B))=B(H\bullet B)$ by

$\Psi_{0}(T)=\lim_{\lambda}\Psi_{0}(TE_{\lambda}),$

where the convergence is in the left strict topology of $LM(K(H\bullet B)).$
Since ${\Psi_{0}}|_{K(H)}$ is an isometry, so is $\Psi_{0}.$

We are mostly interested in the case that $B=A^{**}.$ By Theorem 3.2 of [13],
$(H\bullet A^{**}))^{\sharp}$ is a self-dual Hilbert $A^{**}$-module.
Therefore, by 2.10, for each $T\in B(H),$ the extension
${\widetilde{\Psi_{0}(T)}}$ is unique. Hence $\Psi_{0}$ may be extended to a
Banach algebra isomorphism $\tilde{\Psi}_{0}$ from $B(H)$ into $B(H^{\sim})$
such that

$\displaystyle\tilde{\Psi}_{0}(T)|_{H\bullet A^{**}}=\Psi_{0}(T)\,\,\,{\rm
for\,\,\,all}\,\,\,T\in B(H).$ (e 2.39)

We will visualize the map $\Psi_{0}$ a bit more.

######  Proposition 2.13.

Let $A$ and $B$ be a pair of $C^{*}$-algebras as in Proposition 2.8 and $H$ be
a Hilbert $A$-module. Then, for any $T\in B(H),$

$\displaystyle\lim_{\lambda}\|\Psi_{0}(T)\Psi_{0}(E_{\lambda})(x)-\Psi_{0}(T)(x)\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,x\in H\bullet B.$ (e 2.40)

Moreover

$\displaystyle\Psi_{0}(T)(x\bullet b)=T(x)\bullet b\,\,\,{\rm
for\,\,\,all}\,\,\,x\in H\,\,\,{\rm and}\,\,\,b\in B.$ (e 2.41)

Consequently, $\tilde{\Psi}_{0}({\rm id}_{H})={\rm id}_{H^{\sim}}.$

###### Proof.

The identity (e 2.40) follows from immediately from Lemma 2.9.

Since

$\displaystyle\Psi_{0}(TE_{\lambda})(x\bullet b)=TE_{\lambda}(x)\bullet b,$ (e
2.42)

by (e 2.40) and by Lemma 3.1 of [1],

$\Psi_{0}(T)(x\bullet b)=T(x)\bullet b$

for all $x\in H$ and $b\in B.$

For the last part of the proposition, we note that, by considering the pair
$A$ and $A^{**},$ and by the “Moreover part” of the proposition,
$\Psi_{0}({\rm id}_{H})={\rm id}_{H\bullet A^{**}}.$ Therefore, since the
extension $\tilde{\Psi}_{0}({\rm id}_{H\bullet A^{**}})$ is unique (Corollary
3.7 of [13], see 2.10 for convenience), we must have that
$\tilde{\Psi}_{0}({\rm id}_{H})={\rm id}_{H^{\sim}}.$ ∎

The following is a slightly strengthened restatement of Proposition 2.3 of
[1].

######  Proposition 2.14.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then there is a
homomorphism $\Psi$ from $K(H)^{**}$ into $B(H^{\sim})$ such that
$\Psi|_{B(H)}=\tilde{\Psi}_{0}.$ Moreover, if $T\in K(H)^{**}$ and
$T_{\lambda}\in K(H)^{**}$ such that $T_{\lambda}\to T$ in the weak*-topology,
then

$\displaystyle\lim_{\lambda}f(\langle\Psi(T_{\lambda})(x),y\rangle)=f(\langle\Psi(T)(x),y\rangle)\,\,\,{\rm
for\,\,\,all}\,\,\,x,y\in H^{\sim}\,\,\,{\rm and}\,\,\,f\in A^{*}.$ (e 2.43)

###### Proof.

By 2.4, $B(H^{\sim})=L(H^{\sim})$ is a $W^{*}$-algebra (see Proposition 3.11
of [13]). Let $\pi:B(H^{\sim})\to B(H_{\pi})$ be a faithful normal
representation such that $\pi(B(H^{\sim}))$ is weakly closed in $B(H_{\pi}).$
Then, by, for example, Theorem 3.7.7 of [14] and Cor. 46.5 of [4], there is a
normal homomorphism $\Phi:K(H)^{**}\to B(H_{\pi})$ such that
$\Phi|_{K(H)}=\pi\circ{\tilde{\Psi}_{0}}|_{K(H)}$ and
$\pi\circ\tilde{\Psi}_{0}(K(H))$ is weakly dense in $\Phi(K(H)^{**}).$ Since
$\pi(B(H^{\sim}))$ is a von Neumann algebra,
$\Phi(K(H)^{**})\subset\pi(B(H^{\sim})).$ Since $\pi$ is injective, we may
define $\Psi=\pi^{-1}\circ\Phi.$ Recall that $\pi^{-1}$ is an isomorphism
between $W^{*}$-algebras $\pi(B(H^{\sim}))$ and $B(H^{\sim}).$ It follows that
$\Psi$ is weak *-continuous. Then,
$\Psi|_{K(H)}=\pi^{-1}\circ{\pi\circ\tilde{\Psi}_{0}}|_{K(H)}={\tilde{\Psi}_{0}}|_{K(H)}.$

Let $V=B(H^{\sim})_{*}$ be the pre-dual (as Banach spaces). Then $\Psi$
induces a map $\Psi^{*}:V\to K(H)^{*},$ the pre-dual of $K(H)^{**},$ by
$L(\Psi^{*}(v))=\Psi(L)(v)$ for all $L\in(K(H)^{*})^{*}$ and $v\in V.$ Thus if
$T_{\lambda}\in K(H)^{**}$ such that $T_{\lambda}\to T$ in the weak *-topology
in $K(H)^{**},$ then $\Psi(T_{\lambda})(v)=T_{\lambda}(\Psi^{*}(v))$ converges
to $T(\Psi^{*}(v))=\Psi(T)(v)$ for all $v\in V.$ In other words,
$\Psi(T_{\lambda})\to\Psi(T)$ in the weak *-topology in $V^{*}=B(H^{\sim}).$
By 2.4 (see Remark 3.9 and proof of Theorem 3.10 of [13]), this implies, in
particular, for any $f\in A^{*},$ $x,y\in H^{\sim},$
$f(\langle\Psi(T_{\lambda}((x),y\rangle)\to f(\langle\Psi(T)(x),y\rangle).$

By Theorem 1.5 of [9], $B(H)=LM(K(H)).$ Let $\\{E_{\lambda}\\}$ be an
approximate identity for $K(H).$ Then $TE_{\lambda}\in K(H)$ for all $T\in
B(H).$ It follows from Proposition 2.13 that, for $T\in B(H),$

$\displaystyle\lim_{\lambda}\|\Psi(TE_{\lambda})(f)(x)-\Psi(T)(f)(x)\|=\lim_{\lambda}\|\Psi(T)\Psi(E_{\lambda})(f)(x)-\Psi(T)(f)(x)\|=0$
(e 2.44)

for all $x\in H\bullet A^{**}$ and $f\in(H\bullet A^{**})^{\sharp}.$ On the
other hand, by Proposition 2.9,

$\displaystyle\lim_{\lambda}\|\Psi_{0}(TE_{\lambda})(x)-\Psi_{0}(T)(x)\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,x\in H\bullet A^{**}.$ (e 2.45)

However, we have shown
$\Psi(TE_{\lambda})(y)=\tilde{\Psi}_{0}(TE_{\lambda})(y)=\Psi_{0}(TE_{\lambda})(y)$
for all $y\in H\bullet A^{**}$ (see also (2.12)). Therefore, combining these
three facts, for $x,y\in H\bullet A^{**},$ we obtain

$\displaystyle\langle\Psi(T)(x),y\rangle=\langle\Psi_{0}(T)(x),y\rangle.$ (e
2.46)

It follows that $\Psi(T)|_{H\bullet A^{**}}=\Psi_{0}(T).$ Since the extension
of $\Psi_{0}(T)$ to a bounded module map on $(H\bullet A^{**})^{\sharp}$ is
unique (see the end of 2.10, also Lemma 3.5 of [10]), we have
$\Psi(T)=\tilde{\Psi}_{0}(T)$ for all $T\in B(H).$ Hence

$\Psi|_{B(H)}=\tilde{\Psi}_{0}.$

∎

######  Definition 2.15.

Let $M$ be a $W^{*}$-algebra and $H$ be a Hilbert $M$-module. Then, by 3.2 of
[13], $H^{\sharp}$ is a self-dual Hilbert $M$-module. Let $F_{0}:F(H)\to
F(H^{\sharp})$ be the homomorphism defined by

$F_{0}(\theta_{x,y})(z)=x\langle y,z\rangle\,\,\,{\rm for\,\,\,all}\,\,\,z\in
H^{\sharp}\,\,\,{\rm and}\,\,\,x,y\in H.$

Clearly $F_{0}$ is an isometry. It extends uniquely a homomorphism
$F_{0}:K(H)\to K(H^{\sharp}).$ We further extend
$F:\widetilde{K(H)}\to\widetilde{K(H^{\sharp})}$ by $F({\rm id}_{H})={\rm
id}_{H^{\sharp}}.$

######  Proposition 2.16.

Let $M$ be a $W^{*}$-algebra and $H$ be a Hilbert $M$-module. Then there
exists a unital normal homomorphism $F:K(H)^{**}\to B(H^{\sharp})$ such that
$F|_{K(H)}=F_{0}$ and, if $T_{\lambda}\to T$ in weak*-topology of $W,$ then

$\displaystyle\lim_{\lambda}f(\langle F(T_{\lambda})(x),y\rangle)=f(\langle
F(T)(x),y\rangle)$ (e 2.47)

for all $x,y\in H^{\sharp}$ and $f\in M_{*},$ the pre-dual of $M.$ Moreover,
$F(T)=\tilde{T}$ for all $T\in B(H)$ as defined in the later part of 2.10.

###### Proof.

Recall that $B(H^{\sharp})$ is a $W^{*}$-algebra. We may assume that
$B(H^{\sharp})$ acting on a Hilbert space $X$ as a von Neumann algebra with
$1_{B(H^{\sharp})}={\rm id}_{X}.$ Then, by Theorem 1.8.2 of [11] (see also
Theorem 3.7.7 of [14]), there is a unital normal homomorphism
$F:K(H)^{**}\to\overline{F_{0}(K(H))}^{SOT}\subset B(H^{\sharp})$ such that
$F|_{K(H)}=F_{0}.$ So $F$ is weak*-continuous (see, for example, Cor. 46.5 in
[4]).

Suppose that $T_{\lambda}\to T$ in the weak*-topology of $K(H)^{**},$ then
$F(T_{\lambda})\to F(T)$ in the weak*-topology of $B(H^{\sharp}).$ Therefore
(see the later part of Definition 2.4, also, Remark 3.9 and the proof of
Proposition 3.9 of [13]),

$\displaystyle f(\langle F(T_{\lambda})(x),y\rangle)\to f(\langle
F(T)(x),y\rangle)\,\,\,{\rm for\,\,\,all}\,\,\,x,y\in H^{\sharp}\,\,\,{\rm
and}\,\,\,f\in M_{*}.$ (e 2.48)

Let $\\{E_{\lambda}\\}$ be an approximate identity for $K(H).$ Then, for any
$T\in B(H),$ by Lemma 3.1 of [1],

$\displaystyle\lim_{\lambda}\|F(T)F(E_{\lambda})(x)-F(T)(x)\|=\lim_{\lambda}\|F(T)E_{\lambda}(x)-F(T)(x)\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,x\in H.$ (e 2.49)

On the other hand, since
$F(T)F(E_{\lambda})|_{H}=F(TE_{\lambda})|_{H}=TE_{\lambda},$ and

$\displaystyle\lim_{\lambda}\|TE_{\lambda}(x)-T(x)\|=0,$ (e 2.50)

we conclude that

$\displaystyle T(x)=F(T)(x)\,\,\,{\rm for\,\,\,all}\,\,\,x\in H.$ (e 2.51)

Since the extension of $T$ to $H^{\sharp}$ is unique (by Proposition 3.6 of
[13], see also Lemma 3.5 of [10]), $\tilde{T}=F(T).$ ∎

##  3 Isomorphism of $B(H^{\sim})$ and $K(H)^{**}$

Let $A$ be a monotone $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then,
by Lemma 3.7 of [10], $H^{\sharp}$ becomes a self-dual Hilbert $A$-module such
that $\langle\tau,x\rangle=\tau(x)$ for all $x\in H$ and $\tau\in H^{\sharp}.$
Note that, if $E$ is self-dual, we map $E^{\sharp}$ onto $E$ conjugately just
as in the case of Hilbert spaces.

We will apply the following lemma several times.

######  Proposition 3.1.

Let $A$ be a monotone $C^{*}$-algebra, $H_{1}\subset H_{2}$ be Hilbert
$A$-modules such that $H_{2}$ is self-dual. Then $H_{1}^{\sharp}$ is an
orthogonal summand of $H_{2}^{\sharp}$ and the embedding $H_{1}^{\sharp}\to
H_{2}^{\sharp}$ extends the embedding $H_{1}\subset H_{2}.$

###### Proof.

Define $P_{0}:H_{2}\to H_{1}^{\sharp}$ by

$\displaystyle P_{0}(y)(x)=\langle y,x\rangle\,\,\,{\rm
for\,\,\,all}\,\,\,y\in H_{2}\,\,\,{\rm and}\,\,\,x\in H_{1}.$ (e 3.52)

It is a bounded module map (by viewing $H_{1}^{\sharp}$ as a Hilbert module
instead of the dual to avoid the conjugation) with $\|P_{0}\|=1.$ Note that
${P_{0}}|_{H_{1}}=\operatorname{id}_{H_{1}}.$

Let $\tau\in H_{1}^{\sharp}.$ Since $A$ is monotone complete, by Theorem 3.8
of [10], there is $\tilde{\tau}\in H_{2}^{\sharp}$ such that
$\tilde{\tau}|_{H_{1}}=\tau$ and $\|\tilde{\tau}\|=\|\tau\|.$ This implies
that $P_{0}$ is surjective.

Define $j:H_{1}^{\sharp}\to H_{2}^{\sharp}=H_{2}$ by

$\displaystyle j(x)(y)=\langle x,P_{0}(y)\rangle\,\,\,{\rm
for\,\,\,all}\,\,\,x\in H_{1}^{\sharp}\,\,\,{\rm and}\,\,\,y\in H_{2}.$ (e
3.53)

Then $j$ extends the embedding $H_{1}\subset H_{2}.$ Now, for $x\in
H_{1}^{\sharp}$ and $y\in H_{2},$ by (e 3.52) and (e 3.53),

$\displaystyle P_{0}\circ j(x)(y)$ $\displaystyle=$ $\displaystyle
P_{0}(j(x))(y)=\langle j(x),y\rangle=\langle x,P_{0}(y)\rangle$ (e 3.54)
$\displaystyle=$ $\displaystyle\langle
P_{0}(y),x\rangle^{*}=(P_{0}(y)(x))^{*}$ (e 3.55) $\displaystyle=$
$\displaystyle\langle y,x\rangle^{*}=\langle x,y\rangle.$ (e 3.56)

It follows that $P_{0}\circ j=\operatorname{id}|_{H_{1}^{\sharp}}.$ Thus
$j:H_{1}^{\sharp}\to H_{2}$ is an emebeding. With identification of
$H_{1}^{\sharp}$ and $j(H_{1}^{\sharp}),$
${P_{0}}|_{H_{1}^{\sharp}}=\operatorname{id}|_{H_{1}^{\sharp}}.$ It follows
that $P_{0}$ is a projection and $H_{1}^{\sharp}$ is an orthogonal summand of
$H_{2}.$ ∎

Applying Lemma 3.1 and Proposition 2.6, we obtain the following
characterization of $H^{\sim}.$

######  Proposition 3.2.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then $H^{\sim}$
is the smallest self-dual Hilbert $A^{**}$-module containing $H$ as a Hilbert
$A$\- submodule.

###### Proof.

Let $H_{1}$ be a self-dual Hilbert $A^{**}$-module containing $H$ as a Hilbert
$A$-submodule. Then, by Proposition 2.6,

$\displaystyle H\subset H\bullet A^{**}\subset H_{1}.$ (e 3.57)

Applying Lemma 3.1, since $H_{1}$ is self-dual,

$\displaystyle H^{\sim}=(H\bullet A^{**})^{\sharp}\subset
H_{1}^{\sharp}=H_{1}.$ (e 3.58)

The proposition follows. ∎

######  3.3.

In the next proposition, let $A$ be a $C^{*}$-algebra, $H_{1}\subset H$ be
Hilbert $A$-modules. Then, by Proposition 2.6, $H_{1}\bullet A^{**}\subset
H\bullet A^{**}.$ Since $A^{**}$ is monotone complete, and $(H\bullet
A^{**})^{\sharp}=H^{\sim}$ and $(H_{1}\bullet A^{**})^{\sharp}=H_{1}^{\sim},$
by Lemma 3.1, we may write
$H^{\sim}=H_{1}^{\sim}\oplus(H_{1}^{\sim})^{\perp}.$ Denote by $P:H^{\sim}\to
H_{1}^{\sim}$ the projection. Note that $P\in L(H^{\sim}).$ By Lemma 3.2 of
[10], $K(H_{1})$ is a hereditary $C^{*}$-subalgebra of $K(H).$ Let
$\Psi_{H}:K(H)^{**}\to B(H^{\sim})$ and $\Psi_{1}:K(H_{1})^{**}\to
B(H_{1}^{\sim})$ be the homomorphisms given by Proposition 2.14, respectively.

######  Proposition 3.4.

Use the notation above, we have that

$\Psi_{1}={\Psi_{H}}|_{K(H_{1})^{**}}=P\Psi_{H}P|_{K(H_{1})^{**}},$

in particular,
$\Psi_{1}(T)=\Psi_{H}(T)|_{H_{1}^{\sim}}=P\Psi_{H}(T)P|_{H_{1}^{\sim}}$ for
$T\in K(H_{1})^{**}.$ Moreover,

$P\Psi_{H}(L)P|_{K(H_{1})^{**}}\subset\Psi_{1}(K(H_{1})^{**})\,\,\,{\rm
for\,\,\,all}\,\,\,L\in K(H)^{**}.$

Furthermore, $\Psi(Q)=P,$ where $Q$ is the open projection in $K(H)^{**}$
corresponding to the hereditary $C^{*}$-subalgebra $K(H_{1}).$

###### Proof.

Denote by $\Psi_{K(H),0}$ the injective homomorphism from $K(H)$ into
$K(H\bullet A^{**}),$ $\Psi_{K(H_{1}),0}$ the injective homomorphism from
$K(H_{1})$ into $K(H_{1}\bullet A^{**})$ described in 2.7, respectively.

Fix $S\in K(H_{1}).$ For each $x\in H_{1}$ and $b\in A^{**},$ by 2.8,

$\displaystyle\Psi_{K(H),0}(S)(x\bullet b)=S(x)\bullet b\,\,\,{\rm and}\,\,\,$
(e 3.59) $\displaystyle P\Psi_{K(H),0}(S)P(x\bullet b)=P(S(x\bullet
b))=S(x)\bullet b=\Psi_{K(H_{1}),0}(S)(x\bullet b).$ (e 3.60)

It follows that

$\Psi_{K(H),0}(S)|_{H_{1}\bullet A^{**}}=P\Psi_{K(H),0}(S)P|_{H_{1}\bullet
A^{**}}=\Psi_{K(H_{1}),0}(S).$

Since the extensions of $\Psi_{K(H),0}(S)|_{H_{1}\bullet A^{**}}$ and
$\Psi_{K(H_{1}),0}(S)$ to bounded module maps on $H_{1}^{\sim}$ are unique,
and, $\Psi(S)|_{H_{1}^{\sim}}$ and $\Psi(S)$ are corresponding extensions, by
Corollary 3.7 of [13], we conclude that
$\Psi(S)|_{H_{1}^{\sim}}=P\Psi_{H}(S)P|_{H_{1}^{\sim}}=\Psi_{1}(S).$

Let $T\in K(H_{1})^{**}$ and $\\{T_{\lambda}\\}\subset K(H_{1})$ be a net such
that $T_{\lambda}\to T$ in the weak*-topology. By Proposition 2.14, for any
$g\in A^{*},$

$\displaystyle\lim_{\lambda}|g(\langle\Psi_{H}(T_{\lambda})(x),y\rangle)-g(\langle\Psi_{H}(T)(x),y\rangle)|=0\,\,\,{\rm
and}\,\,\,$ (e 3.61)
$\displaystyle\lim_{\lambda}|g(\langle\Psi_{1}(T_{\lambda})(x),y\rangle)-g(\langle\Psi_{1}(T)(x),y\rangle)|=0$
(e 3.62)

for all $x,y\in H_{1}^{\sim}.$ Since we have shown that
$\Psi_{H}(T_{\lambda})|_{H_{1}^{\sim}}=P\Psi_{H}(T_{\lambda})P|_{H_{1}^{\sim}}=\Psi_{1}(T_{\lambda}),$
we conclude that

$\displaystyle\Psi_{H}(T)|_{H_{1}^{\sim}}=P\Psi_{H}(T)P|_{H_{1}^{\sim}}=\Psi_{1}(T).$
(e 3.63)

Hence

$\displaystyle\Psi_{1}=P\Psi_{H}P|_{K(H_{1})^{**}}=\Psi_{H}|_{K(H_{1})^{**}}.$
(e 3.64)

Let $\\{q_{\lambda}\\}$ be an approximate identity for $K(H_{1}).$ Then
$q_{\lambda}\nearrow{\rm id}_{H_{1}}\in K(H_{1})^{**}.$ It follows from
Proposition 2.14 that

$\displaystyle\lim_{\lambda}f(\langle\Psi_{1}(q_{\lambda}(y)),z\rangle)=f(\langle
y,z\rangle)\,\,\,{\rm for\,\,\,all}\,\,\,y,z\in H_{1}^{\sim}\,\,\,{\rm
and}\,\,\,f\in A^{*}.$ (e 3.65)

On the other hand, we also have that $q_{\lambda}\nearrow Q$ in $K(H)^{**}.$
By Lemma 3.1, $H^{\sim}=H_{1}^{\sim}\oplus(H_{1}^{\sim})^{\perp}.$ Note that
$q_{\lambda}(x)\in H_{1}$ for all $x\in H.$ Then, for $x\in H,$ $b\in A^{**},$
and $g\in(H_{1}^{\sim})^{\perp},$ by Proposition 2.8,

$\displaystyle\langle\Psi_{H}(q_{\lambda})(x\bullet
b),g\rangle=g(q_{\lambda}(x\bullet b))^{*}=g(q_{\lambda}(x)\bullet b)^{*}=0.$
(e 3.66)

It follows that, for any $y\in H\bullet A^{**}$ and
$g\in(H_{1}^{\sim})^{\perp},$

$\displaystyle\langle\Psi_{H}(q_{\lambda})(y),g\rangle=0.$ (e 3.67)

Hence, for $g\in(H_{1}^{\sim})^{\perp},$

$\displaystyle\langle y,\Psi_{H}(q_{\lambda})(g)\rangle=0\,\,\,{\rm
for\,\,\,all}\,\,\,y\in H\bullet A^{**}.$ (e 3.68)

It follows that $\Psi_{H}(q_{\lambda})(g)=0$ and

$\displaystyle\langle\Psi_{H}(q_{\lambda})(z),g\rangle=\langle
z,\Psi_{H}(q_{\lambda})(g)\rangle=0\,\,\,{\rm for\,\,\,all}\,\,\,z\in
H^{\sim}.$ (e 3.69)

In other words, $\Psi_{H}(q_{\lambda})(z)\in H_{1}^{\sim}$ for all $z\in H$
and $\lambda.$ Therefore

$\displaystyle
P\Psi_{H}(q_{\lambda})=\Psi_{H}(q_{\lambda})=\Psi_{H}(q_{\lambda})P.$ (e 3.70)

Note that $Pz\in H_{1}^{\sim}$ for any $z\in H^{\sim}.$ Thus, by (e 3.63) and
(e 3.62),

$\displaystyle\lim_{\lambda}f(\langle\Psi(q_{\lambda})(y),z\rangle)$
$\displaystyle=$
$\displaystyle\lim_{\lambda}f(\langle\Psi(q_{\lambda})(P(y)),P(z)\rangle$ (e
3.71) $\displaystyle=$
$\displaystyle\lim_{\lambda}f(\langle\Psi_{1}(q_{\lambda})(P(y)),P(z)\rangle$
(e 3.72) $\displaystyle=$ $\displaystyle f(\langle P(y),P(z)\rangle)=f(\langle
P(y),z\rangle).$ (e 3.73)

By (e 3.61) and (e 3.63),
$\lim_{\lambda}f(\langle\Psi(q_{\lambda})(y),z\rangle)=f(\langle\Psi(Q)(y),z\rangle).$
Therefore

$\displaystyle\Psi(Q)=P.$ (e 3.74)

This proves the “Furthermore” part. In what follows we will identify $Q$ with
$P$ as well as $\Psi(Q)$ and $\Psi(P).$

Now let $L\in K(H)^{**}$ and $\\{L_{\lambda}\\}\subset K(H)$ be a net such
that $L_{\lambda}\to L$ in weak *-topology. By Proposition 2.14, for any $g\in
A^{*},$ $x,y\in H_{1}^{\sim},$

$\displaystyle\lim_{\lambda}|g(\langle\Psi_{H}(T_{\lambda})(x),y\rangle)-g(\langle\Psi_{H}(T)(x),y\rangle)|=0\,\,\,{\rm
and}\,\,\,$ (e 3.75)
$\displaystyle\lim_{\lambda}|g(\langle\Psi_{1}(T_{\lambda})(x),y\rangle)-g(\langle\Psi_{1}(T)(x),y\rangle)|=0$
(e 3.76)

(note that $\Psi_{1}(T_{\lambda})=P\Psi_{1}(T_{\lambda})P$). We also have, for
any $x,y\in H_{1}^{\sim},$

$\displaystyle\langle\Psi_{H}(PT_{\lambda}P)(x),y\rangle=\langle\Psi_{H}(T_{\lambda})(x),y\rangle,$
(e 3.77) $\displaystyle\langle
P\Psi_{H}(T)P(x),y\rangle=\langle\Psi_{H}(T)(x),y\rangle,$ (e 3.78)

Since $PT_{\lambda}P\in K(H_{1})^{**},$ by the first part of the lemma,
$\Psi_{H}(PT_{\lambda}P)(x)=\Psi_{1}(PT_{\lambda}P)(x)$ for $x\in
H_{1}^{\sim}.$ It follows that $P\Psi_{H}(T)P(x)=\Psi_{1}(PTP)(x)$ for all
$x\in H_{1}^{\sim}.$ Then
$P\Psi_{H}(T)P=\Psi_{1}(PTP)\in\Psi_{1}(K(H_{1})^{**}).$ ∎

######  3.5.

Let $A$ be a $C^{*}$-algebra and let, for $n\in\mathbb{N},$
$H_{n}=A^{(n)}=\\{(a_{1},a_{2},...,a_{n}):a_{j}\in A,1\leq j\leq n\\},$ the
direct sum of $n$-copies of $A,$ where $\langle
a,b\rangle=\sum_{j=1}^{n}a_{j}^{*}b_{j},$ if $a=(a_{1},a_{2},...,a_{n})$ and
$b=(b_{1},b_{2},...,b_{n}).$ Let

$\displaystyle H_{A}=\\{\\{a_{n}\\}:a_{n}\in A\,\,\,{\rm
and}\,\,\,\sum_{i=1}^{n}a_{k}^{*}a_{k}\,\,\,{\rm converges\,\,in\,\,norm}\\}$
(e 3.80)

be the standard countably generated Hilbert (right) $A$-module. Note that

$\displaystyle\langle\\{a_{n}\\},\\{b_{n}\\}\rangle=\sum_{n=1}^{\infty}a_{n}^{*}b_{n}.$
(e 3.81)

We note that $H_{A}$ is the closure of $\cup_{n}A^{(n)}.$ We may also view
$H_{n}=A^{(n)}$ as an orthogonal summand of $H_{A}.$ Then

$\displaystyle
H_{A}^{\sharp}=\\{\\{a_{n}\\}:\\{\|\sum_{k=1}^{n}a_{k}^{*}a_{k}\|\\}\,\,{\rm
bounded}\\}.$ (e 3.82)

If $g=\\{a_{n}\\}\in H^{\sharp},$ then

$\displaystyle g(x)=\sum_{n=1}^{\infty}a_{n}^{*}b_{n}\,\,\,{\rm
for\,\,\,all}\,\,\,x=\\{b_{n}\\}\in H_{A},$ (e 3.83)

where the sum converges in norm. Moreover
$\|g\|=\lim_{n\to\infty}\|\sum_{k=1}^{n}b_{k}^{*}b_{k}\|.$

If $A$ is a $W^{*}$-algebra, as mentioned earlier, $H_{A}^{\sharp}$ becomes a
Hilbert $A$-module in a natural way (see Theorem 3.2 of [13]). In fact, we may
define

$\displaystyle\langle x,y\rangle=\sum_{n=1}^{\infty}a_{n}^{*}b_{n}\,\,\,{\rm
for\,\,\,all}\,\,\,x=\\{a_{n}\\},\,\,y=\\{b_{n}\\}\in H_{A}^{\sharp}.$ (e
3.84)

To see the right side converges in the weak*-topology, we first let $f\in
A^{*}.$ Note that, if $\\{a_{n}\\}\in H_{A}^{\sharp},$

$\displaystyle|\sum_{k=1}^{N}f(a_{k}^{*}a_{k})|=|f(\sum_{k=1}^{N}a_{k}^{*}a_{k})|\leq\|f\|\|\sum_{k=1}^{N}a_{k}^{*}a_{k}\|$
(e 3.85)

for any integer $N.$ Hence $\\{\sum_{k=1}^{n}f(a_{k}^{*}a_{k})\\}$ is bounded,
increasing and converges for any positive linear functional $f.$ Hence, for
any $m>n,$

$\displaystyle\sum_{k=n}^{m}f(a_{k}^{*}a_{k})\to 0\,\,\,{\rm
as}\,\,n\to\infty\,\,\,{\rm for\,\,\,all}\,\,\,f\in A^{*}.$ (e 3.86)

For any positive linear functional $f$ and for any $m>n$ in $\mathbb{N},$

$\displaystyle|f(\sum_{k=n}^{m}a_{k}^{*}b_{k})|$ $\displaystyle=$
$\displaystyle|\sum_{k=n}^{m}f(a_{k}^{*}b_{k})|\leq\sum_{k=n}^{m}|f(a_{k}^{*}b_{k})|$
(e 3.87) $\displaystyle\leq$
$\displaystyle\sum_{k=n}^{m}|f(a_{k}^{*}a_{k})|^{1/2}|f(b_{k}^{*}b_{k})|^{1/2}$
(e 3.88) $\displaystyle\leq$
$\displaystyle((\sum_{k=n}^{m}|f(a_{k}^{*}a_{k})|)(\sum_{k=n}^{m}|f(b_{k}^{*}b_{k})|))^{1/2}$
(e 3.89) $\displaystyle\leq$
$\displaystyle\|f\|^{1/2}\|\\{b_{k}\\}\||(\sum_{k=n}^{m}|f(a_{k}^{*}a_{k})|)^{1/2}\to
0\,\,{\rm as}\,\,n\to\infty.$ (e 3.90)

It follows that $f(\sum_{k=1}^{n}a_{k}^{*}b_{k})$ converges for all $f\in
A^{*}$ as $n\to\infty.$ Let us write the limit as
$f(\sum_{k=1}^{\infty}a_{k}^{*}b_{k}).$ Then, by the above inequalities (with
$n=1$), we also have

$\displaystyle|f(\sum_{k=1}^{\infty}a_{k}^{*}b_{k})|\leq\|f\|M_{b}M_{a},$ (e
3.91)

where

$\displaystyle
M_{a}=\sup\\{\|\sum_{k=1}^{n}a_{k}^{*}a_{k}\|\\}^{1/2}\,\,\,{\rm
and}\,\,\,M_{b}=\sup\\{\|\sum_{k=1}^{n}b_{k}^{*}b_{k}\|\\}^{1/2}.$ (e 3.92)

Thus $\sum_{k=1}^{\infty}a_{k}^{*}b_{k}$ defines a bounded linear functional
on $A^{*}.$ Its restriction on $A_{*}$ gives an element in $A$ (recall that
$A$ is assumed to be a $W^{*}$-algebra). This shows the infinite series in the
right side of (e 3.84) converges in the weak*-topology. It is then standard to
verify that (e 3.84) defines an inner product which extends the inner product
on $H_{A}.$

Let $A$ act on Hilbert space $X$ (as a $W^{*}$-algebra). Consider $l^{2}(X)$
the Hilbert space direct sum of countably many copies of $X.$ Suppose that
$b=\\{b_{n}\\}\in H_{A}^{\sharp}.$ Then the infinite matrix
${\bar{b}}=(b_{i,j})$ with $b_{i,1}=b_{i},$ $i\in\mathbb{N},$ and $b_{i,j}=0,$
if $j\geq 2,$ defines a bounded linear operator on $l^{2}(X),$ by
${\bar{b}}(v)=(b_{1}(v_{1}),b_{2}(v_{1}),...,b_{n}(v_{1}),...),$ where
$v=(v_{1},v_{2},...,v_{n},...)\in l^{2}(X).$ Moreover

$\displaystyle\|{\bar{b}}\|^{2}=\|{\bar{b}}^{*}{\bar{b}}\|=\|\sum_{i=1}^{\infty}b_{i}^{*}b_{i}\|=\sup\\{\|\sum_{i=1}^{n}b_{i}^{*}b_{i}\|:n\in\mathbb{N}\\}$
(e 3.93)

(some of these details in this subsection may be found in [8]).

######  Proposition 3.6.

Let $C$ be a unital $C^{*}$-algebra and $A\subset C$ be a $C^{*}$-subalgebra
such that $1_{\tilde{A}}=1_{C}.$ Denote by $R=\overline{AC},$ the closed right
ideal of $C$ generated by $A.$ Then

(1) $H_{A}\bullet C=\\{\\{b_{n}\\}\in H_{C}:b_{n}\in R\\}.$

(2) If $C$ is a $W^{*}$-algebra and $e_{\alpha}\nearrow 1_{C},$ where
$\\{e_{\alpha}\\}$ is an approximate identity for $A,$ then

$(H_{A}\bullet C)^{\sharp}=H_{C}^{\sharp}.$

###### Proof.

To see (1), we first note that $A\bullet C=R$ as Hilbert $C$-modules. Hence
$A^{(n)}\bullet C=R^{(n)}.$ Clearly, $H_{A}\bullet C\subset H_{C}.$ We note
that $\\{\\{r_{n}\\}\in H_{C}:r_{n}\in R\\}$ is closed in $H_{C}.$ Since both
$\cup_{n}A^{(n)}\bullet C$ and $\cup_{n=1}^{\infty}R^{(n)}$ are dense in
$\\{\\{r_{n}\\}\in H_{C}:r_{n}\in R\\},$ and $\cup_{n}A^{(n)}\bullet C$ is
dense $H_{A}\bullet C,$ we obtain that

$\displaystyle\\{\\{r_{n}\\}\in H_{C}:r_{n}\in R\\}=H_{A}\bullet C.$ (e 3.94)

This proves (1).

For (2) we may assume that $A\subset C\subset B(X),$ where $X$ is a Hilbert
space, $1_{C}={\rm id}_{X},$ and $C(X)=X.$ Otherwise, we replace $X$ by
$1_{C}(X).$

Claim 1: $\overline{R(X)}=C(X)=X.$ Since $e_{\alpha}\nearrow 1_{C}={\rm
id}_{X},$ for any $v\in X,$ $e_{\alpha}(v)\to v.$ This proves the claim.

Claim 2: $R^{\sharp}=C,$ where $R^{\sharp}$ is the dual of Hilbert $C$-module
$R$ (as we now assume that $C$ is a $W^{*}$-algebra).

Let $f\in R^{\sharp}.$ Then $f(e_{\alpha})r=f(e_{\alpha}r)\to f(r)$ for all
$r\in R$ in norm as $e_{\alpha}r\to r$ in norm. Hence $f(e_{\alpha})r(v)\to
f(r)(v)$ for all $r\in R$ and $v\in X.$ Define $T$ on $R(X)$ by
$T(r(v))=\lim_{\alpha}f(e_{\alpha})r(v)$ for all $v\in X$ and $r\in R.$ One
checks that it is a well-defined linear map on $R(X).$ Moreover,
$\|T\|\leq\sup\\{\|f(e_{\alpha})\|:{\alpha}\\}\leq\|f\|.$ Since, by Claim 1,
$\overline{R(X)}=X,$ $T$ extends uniquely a bounded operator (denote by $T$
again) on $X.$ Moreover, $f(e_{\alpha})$ converges to $T$ on $X.$ Since $C$ is
closed in the weak operator topology, $T\in C.$ Moreover $Tr(v)=f(r)(v)$ for
all $v\in X.$ It follows that $Tr=f(r)$ for all $r\in R.$

For each $c\in C,$ define $f_{c}\in R^{\sharp}$ by

$\displaystyle f_{c}(r)=c^{*}r\,\,\,{\rm for\,\,\,all}\,\,\,r\in R.$ (e 3.95)

For the above $T,$ we note that $f_{T^{*}}(r)=Tr$ for all $r\in R.$ Hence the
map $c\to f_{c}$ is surjective. To see it is injective, suppose that
$c^{*}r=0$ for all $r\in R.$ Then

$\displaystyle c^{*}e_{\alpha}c=0\,\,\,{\rm for\,\,\,all}\,\,\,{\alpha}.$ (e
3.96)

Since $c^{*}e_{\alpha}c\nearrow c^{*}c,$ this implies that $c^{*}c=0.$ Thus
the map $c\mapsto f_{c}$ is injective which extends the identity map on $R.$
It follows that $R^{\sharp}=C.$ Claim 2 is proved.

By Claim 2, we obtain that $((A^{(n)})\bullet C)^{\sharp}=C^{(n)}.$ By (1),
$(A^{(n)})\bullet C$ is a direct summand of $H_{A}\bullet C.$ Hence we may
write $((A^{(n)})\bullet C)^{\sharp}\subset(H_{A}\bullet C)^{\sharp}.$
Together with (1), we obtain that

$\displaystyle H_{A}\bullet C\subset H_{C}\subset(H_{A}\bullet C)^{\sharp}.$
(e 3.97)

Note $H_{C}$ is a Hilbert $C$-submodule of the self-dual Hilbert $C$ module
$(H_{A}\bullet C)^{\sharp}.$ It follows from Lemma 3.1 that

$\displaystyle(H_{A}\bullet C)^{\sharp}\subset
H_{C}^{\sharp}\subset(H_{A}\bullet C)^{\sharp}.$ (e 3.98)

consequently, $H_{C}^{\sharp}=(H_{A}\bullet C)^{\sharp}.$

∎

######  3.7.

Note that, if $A$ is unital, $H_{A}\bullet C=H_{C}.$

From the above discussion, we obtain the following:

######  Lemma 3.8.

Let $A$ be a $C^{*}$-algebra, $H_{n}=(A^{**})^{(n)}$ and
$P_{n}:H_{A^{**}}^{\sharp}\to H_{n}$ be the projection.

(1) Let $S\subset H_{A^{**}}^{\sharp}$ be a bounded subset. Then, for any
$f\in A^{*}$ and $x\in H_{A^{**}}^{\sharp},$

$\displaystyle\lim_{n\to\infty}\sup\\{|f(\langle P_{n}(x),y\rangle)-f(\langle
x,y\rangle)|:y\in S\\}=0\,\,\,{\rm and}\,\,\,$ (e 3.99)
$\displaystyle\lim_{n\to\infty}\sup\\{|f(\langle y,P_{n}(x)\rangle)-f(\langle
y,x\rangle)|:y\in S\\}=0.$ (e 3.100)

(2) Moreover

$\displaystyle\lim_{n\to\infty}|f(\langle P_{n}(x),P_{n}(x)\rangle)-f(\langle
x,x\rangle)|=0\,\,\,{\rm for\,\,\,all}\,\,\,x\in H_{A^{**}}^{\sharp}\,\,\,{\rm
and}\,\,\,f\in A^{*}.$ (e 3.101)

###### Proof.

Set $M=\sup\\{\|y\|:y\in S\\}+1.$ Let $f$ be a positive linear functional in
$A^{*}$ and $x=\\{a_{n}\\}\in H_{A^{**}}^{\sharp}.$ For each $y=\\{b_{n}\\}\in
S,$

$\displaystyle|f(\langle P_{n}(x),y\rangle)-f(\langle x,y\rangle)|$
$\displaystyle=$ $\displaystyle|\sum_{k=n+1}^{\infty}f(a_{k}^{*}b_{k})|$ (e
3.102) $\displaystyle\leq$
$\displaystyle(\sum_{k=n+1}^{\infty}f(a_{k}^{*}a_{k}))^{1/2}(\sum_{k=n+1}^{\infty}f(b_{k}^{*}b_{k}))^{1/2}$
(e 3.103) $\displaystyle\leq$
$\displaystyle\|f\|\|y\|(\sum_{k=n+1}^{\infty}f(a_{k}^{*}a_{k}))^{1/2}$ (e
3.104) $\displaystyle\leq$ $\displaystyle
M\|f\|(\sum_{k=n+1}^{\infty}f(a_{k}^{*}a_{k}))^{1/2}.$ (e 3.105)

By what has been discussed in 3.5,

$\displaystyle\lim_{n\to\infty}(\sum_{k=n+1}^{\infty}f(a_{k}^{*}a_{n}))^{1/2}=0.$
(e 3.106)

Thus, for this $f$ and $x,$ $|f(\langle P_{n}(x),y\rangle)-f(\langle
x,y\rangle)|$ converges uniformly on $S.$ Almost the identical estimates shows
that $|f(\langle y,P_{n}(x)\rangle)-f(\langle y,x\rangle)|$ converges
uniformly on $S.$

Since any $f\in A^{*}$ can be written as a linear combination of four positive
linear functionals in $A^{*},$ the first part of the statement holds.

For the second part, we note that, for any $f\in A^{*}$ and $x\in
H_{A^{**}}^{\sharp},$ by the first part of the lemma (since
$\|P_{n}(x)\|\leq\|x\|$),

$\displaystyle\lim_{n\to\infty}|f(\langle P_{n}(x),P_{n}(x)\rangle)-f(\langle
x,P_{n}(x)\rangle)|=0.$ (e 3.107)

We also have

$\displaystyle\lim_{n\to\infty}|f(\langle P_{n}(x),x\rangle)-f(\langle
x,x\rangle)|=0.$ (e 3.108)

Hence the second part of the lemma also follows. ∎

The following are two easy facts: We present here for convenience.

######  Lemma 3.9.

Let $A$ be a $C^{*}$-algebra.

(1) Let $H$ be a Hilbert $A$-module, and $\\{E_{\lambda}\\}$ be an approximate
identity for $K(H).$ Suppose $T\in K(H)^{**}$ is a non-zero positive element.
Then there is $\lambda_{0}$ such that

$\displaystyle E_{\lambda}TE_{\lambda}\not=0\,\,\,{\rm
for\,\,\,all}\,\,\,\lambda\geq\lambda_{0}.$ (e 3.109)

(2) Let $T\in K(H_{A})^{**}$ be a non-zero positive element and
$P_{n}:H_{A}\to H_{n}=A^{(n)}$ be the projection ($n\in\mathbb{N}$). Then,
there exists $n_{0}\in\mathbb{N}$ such that

$\displaystyle P_{n}TP_{n}\not=0\,\,\,{\rm for\,\,\,all}\,\,\,n\geq n_{0}.$ (e
3.110)

###### Proof.

Let $f\in A^{*}$ be a positive linear functional. Then

$\displaystyle|f(T^{1/2}(1-E_{\lambda}))|^{2}\leq
f(T)f((1-E_{\lambda})^{2})\leq f(T)f((1-E_{\lambda}))\to 0.$ (e 3.111)

It follows that $f(T^{1/2}E_{\lambda})\to f(T^{1/2})$ for all positive linear
functionals, whence for all $f\in A^{*}.$ Since $T^{1/2}\not=0,$ for some
$\lambda_{0},$ $T^{1/2}E_{\lambda}\not=0$ for all $\lambda\geq\lambda_{0}.$ It
follows that

$E_{\lambda}TE_{\lambda}\not=0$

for all $\lambda\geq\lambda_{0}.$ This proves (1).

There are several easy proofs for (2). Let us use part (1). Choose an
approximate identity $\\{e_{\alpha}\\}$ for $A.$ Let $\lambda=({\alpha},n)$
and $\lambda_{1}=({\beta}_{1},n)\leq\lambda_{2}=({\beta}_{2},m)$ if
${\beta}_{1}\leq{\beta}_{2}$ and $n\leq m.$ Define

$\displaystyle E_{{\beta},n}={\rm
diag}(\overbrace{e_{\beta},e_{\beta},...,e_{\beta}}^{n},0,...).$ (e 3.112)

Then $\\{E_{{\beta},n}\\}$ forms an approximate identity for $K(H_{A})\cong
A\otimes{\cal K}.$ Let $T\in K(H)^{**}_{+}$ be a non-zero positive element. By
(1), there is ${\beta}_{0}$ and $n_{0}\in\mathbb{N}$ such that

$\displaystyle E_{{\beta},n}TE_{{\beta},n}\not=0\,\,\,{\rm
for\,\,\,all}\,\,\,({\beta},n)\geq({\beta}_{0},n_{0}).$ (e 3.113)

Hence
$\|T^{1/2}E_{{\beta},n}^{2}T^{1/2}\|=\|E_{{\beta},n}TE_{{\beta},n}\|\not=0$
for all $({\beta},n)\geq({\beta}_{0},n_{0}).$ Since

$\displaystyle T^{1/2}P_{n}T^{1/2}\geq T^{1/2}E_{{\beta},n}^{2}T_{1/2}\not=0,$
(e 3.114)

we have $T^{1/2}P_{n}T^{1/2}\not=0.$ It follows that

$\displaystyle P_{n}TP_{n}\not=0\,\,\,{\rm for\,\,\,all}\,\,\,n\geq n_{0}.$ (e
3.115)

∎

######  Lemma 3.10.

Let $A$ be a $C^{*}$-algebra and $H$ be a countably generated Hilbert
$A$-module. Then the homomorphism $\Psi$ from $K(H)^{**}$ into $B(H^{\sim})$
(give by Proposition 2.16) is injective.

###### Proof.

Let $H_{n}=A^{(n)}=\\{(a_{1},a_{2},...,a_{n}):a_{j}\in A\\}$ be the Hilbert
$A$-module whose inner product is defined by $\langle
x,y\rangle=\sum_{j=1}^{n}a_{j}^{*}b_{j},$ where $x=\\{a_{j}\\}_{1\leq j\leq
n}$ and $y=\\{b_{j}\\}_{1\leq j\leq n}.$ One identifies $K(H_{n})$ with
$M_{n}(A).$

Claim: The map $\Psi:K(H_{n})^{**}\to B(H_{n}^{\sim})$ is a
$W^{*}$-isomorphism.

Since we identify $K(H_{n})$ with $M_{n}(A),$ we have
$K(H_{n})^{**}=M_{n}(A^{**}).$

By (2) of Proposition 3.6 (and Claim 2 of the proof), $(H_{n}\bullet
A^{**})^{\sharp}=(A^{**})^{(n)}.$ So $H_{n}^{\sim}=(A^{**})^{(n)}.$ Note that
$B(H_{n}^{\sim})=M_{n}(A^{**}).$ One then easily checks that
$\Psi:K(H_{n})^{**}\to B(H_{n}^{\sim})$ is a $W^{*}$-isomorphism. This proves
the claim.

Let us consider the map $\Psi_{H_{A}}:K(H_{A})^{**}\to B(H_{A}^{\sim})$ given
by Proposition 2.14. Put $T\in K(H)^{**}_{+}\setminus\\{0\\}$

By (2) of Lemma 3.9, there exists $n_{0}\in\mathbb{N}$ such that
$P_{n}TP_{n}\not=0.$ Recall that $H_{n}$ is a direct summand of $H_{A}.$ Hence
by the claim and applying Proposition 3.4, we conclude that have
$\Psi_{H_{A}}(P_{n}TP_{n})\not=0$ for all $n\geq n_{0}.$ There must be an
element $x\in H_{n}$ such that

$\displaystyle\langle\Psi_{H_{A}}(P_{n}TP_{n})(x),x\rangle\not=0.$ (e 3.116)

It follows that $\langle\Psi_{H_{A}}(T)x,x\rangle\not=0.$ Hence
$\Psi_{H_{A}}(T)\not=0.$ This implies that ${\rm ker}\Psi_{H_{A}}=\\{0\\}.$

In general, since $H$ is countably generated, by Kasparov’s absorbing theorem
(Theorem 2 of [7]), we may write $H_{A}=H\oplus H^{\perp}.$ To show $\Psi$ is
injective, let $T\in B(H)^{**}$ be a non-zero element. Then
$K(H)^{**}=PK(H_{A})^{**}P,$ where $P:H_{A}\to H$ is the projection. Hence
$PTP=T$ in $K(H_{A})^{**}.$ We have shown that $\Psi_{H_{A}}(PTP)\not=0.$
Since, by Lemma 3.4, $\Psi(T)=P\Psi_{H_{A}}(T)P|_{H^{\sim}}\not=0.$ This
implies that $\Psi$ is injective.

∎

######  Lemma 3.11.

Let $A$ be a $C^{*}$-algebra and $H$ be a countably generated Hilbert
$A$-module. Then there is an isomorphism $\Psi$ from $K(H)^{**}$ onto
$B(H^{\sim})$ as $W^{*}$-algebras.

###### Proof.

By Lemma 3.10 (and by Proposition 2.14), it suffices to show that $\Psi$ is
surjective. Let us first consider the case $H=H_{A}$ (even though $H_{A}$ is
not countably generated when $A$ is not $\sigma$-unital). By the end of 2.4
(see also Remark 3.9 (and 3.10) of [13]), to show that $T\in
B(H^{\sim})=B(H_{A^{**}}^{\sharp})$ is in $\Psi(K(H_{A})^{**}),$ it suffices
to show that, for any $\epsilon>0,$ any finite subsets $X\subset
H_{A^{**}}^{\sharp}$ and a finite subset ${\cal F}\subset A^{*},$ there exists
$S\in K(H)^{**}$ such that

$\displaystyle|f(\langle\Psi(S)(x),y\rangle)-f(\langle
T(x),y\rangle)|<\epsilon\,\,\,{\rm for\,\,\,all}\,\,\,x,y\in X\,\,\,{\rm
and}\,\,\,f\in{\cal F}.$ (e 3.117)

For any $T\in B(H_{A}^{\sim})=B(H_{A}^{\sharp}),$

$\displaystyle|f(\langle P_{n}TP_{n}(x),y\rangle-f(\langle T(x),y\rangle)|$
$\displaystyle\leq$ $\displaystyle|f(TP_{n}(x),P_{n}(y)\rangle)-f(\langle
T(x),P_{n}(y)\rangle)|$ (e 3.119) $\displaystyle+|f(\langle
T(x),P_{n}(y)\rangle)-f(\langle T(x),y\rangle)|$

for any $x,y\in H_{A^{**}}^{\sharp}$ and $f\in A^{*}.$ However,
$\|P_{n}(y)\|\leq\|y\|$ for all $n\in\mathbb{N}.$ By (1) of Lemma 3.8,

$\displaystyle|f(TP_{n}(x),P_{n}(y)\rangle)-f(\langle
T(x),P_{n}(y)\rangle)|\to 0,$ (e 3.120)

and by (2) of Lemma 3.8,

$\displaystyle|f(\langle T(x),P_{n}(y)\rangle)-f(\langle T(x),y\rangle)|\to
0.$ (e 3.121)

It follows that

$\displaystyle\lim_{n\to\infty}|f(\langle P_{n}TP_{n}(x),y\rangle-f(\langle
T(x),y\rangle)|=0$ (e 3.122)

for all $x,y\in H_{A}^{\sharp}$ and $f\in A^{*}.$

We then choose $n_{0}\in\mathbb{N}$ such that for all $n\geq n_{0}$ (recall
$P_{n}$ is a projection)

$\displaystyle|f(\langle
P_{n}TP_{n}(x),P_{n}(y)\rangle)-f(T(x),y\rangle)|<\epsilon$ (e 3.123)

for all $x,y\in X$ and $f\in{\cal F}.$

Now fix $n\geq n_{0}.$ Then we have $P_{n}(x),P_{n}(y)\in(H_{n})^{\sim}$ for
all $x,y\in X,$ and $P_{n}TP_{n}\subset B(H_{n}^{\sim}).$ By the claim for
$H_{n}$ in the proof of Lemma 3.10, we obtain an element $S\in K(H_{n})^{**}$
such that $\Psi_{n}(S)=(P_{n}TP_{n})|_{H_{n}^{\sim}},$ where
$\Psi_{n}:K(H_{n})^{**}\cong M_{n}(A^{**})\to B(H_{n}^{\sim})=M_{n}(A^{**})$
is the isomorphism given by the claim. Note, by Lemma 3.4, that
$\Psi(S)=P_{n}\Psi(S)=\Psi(S)P_{n}=\Psi_{n}(S).$ Hence it follows that, for
all $x,y\in X$ and $f\in{\cal F}$ (and $n\geq n_{0}$),

$\displaystyle|f(\langle\Psi(S)(x),y\rangle)-f(\langle T(x),y\rangle)|$
$\displaystyle=$ $\displaystyle|f(\langle
P_{n}\Psi(S)P_{n}(x),y\rangle)-f(T(x),y\rangle)|$ (e 3.124) $\displaystyle=$
$\displaystyle|(f(\langle P_{n}\Psi(S)P_{n}(x),P_{n}(y)\rangle)-f(\langle
P_{n}TP_{n}(x),P_{n}(y))|$ $\displaystyle+|f(\langle
P_{n}TP_{n}(x),P_{n}(y)\rangle)-f(T(x),y\rangle)|$
$\displaystyle<0+\epsilon=\epsilon.$

As mentioned above, this implies that $\Psi$ is surjective.

For a general countably generated Hilbert $A$-module $H,$ by Kasparov’s
absorbing theorem (Theorem 2 of [7]), we may write $H_{A}=H\oplus H^{\perp}.$
By 3.4, $H_{A}^{\sim}=H^{\sim}\oplus(H^{\perp})^{\sim}.$ Let $S\in
B(H^{\sim})\setminus\\{0\\}.$ Define $T\in B(H_{A}^{\sim})$ by
$T|_{H^{\sim}}=S$ and $S|_{(H^{\perp})^{\sim})}=\\{0\\}.$ We have shown that
there is $L\in B(H_{A})^{**}$ such that $\Psi_{H_{A}}(L)=S.$ Then $PSP=S,$ by
3.4, $\Psi(L)=P\Psi_{H_{A}}(L)P|_{H^{\sim}}=T.$ Hence $\Psi$ is surjective. ∎

######  Theorem 3.12.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then there is an
isomorphism $\Psi$ (given by Proposition 2.14) from $K(H)^{**}$ onto
$B(H^{\sim})$ as $W^{*}$-algebras. Moreover,

$\displaystyle\Psi|_{B(H)}=\tilde{\Psi}_{0}.$ (e 3.125)

###### Proof.

By Proposition 2.14, it suffices to show that $\Psi$ is bijective. If $K(H)$
is unital, by Proposition 2.8 of [1], $H$ is finitely generated. The theorem
then follows from Lemma 3.11. So we will assume that $K(H)$ is not unital.

Let $\\{E_{\lambda}\\}$ be an approximate identity for $K(H)$ and
$H_{\lambda}=\overline{E_{\lambda}(H)}.$ Then
$K(H_{\lambda})=\overline{E_{\lambda}K(H)E_{\lambda}}$ is $\sigma$-unital. By
Proposition 3.2 of [1], $H_{\lambda}$ is countably generated.

Denote by $P_{\lambda}:H^{\sim}\to H_{\lambda}^{\sim}$ the projection given by
3.1 and $\Psi_{\lambda}:K(H_{\lambda})^{**}\to B(H_{\lambda}^{\sim})$ be the
map given by Proposition 2.14.

To see $\Psi$ is injective, let $T\in K(H)^{**}_{+}\setminus\\{0\\}.$ It
follows from Lemma 3.9 that $E_{\lambda}TE_{\lambda}\not=0$ for all
$\lambda\geq\lambda_{0}$ and some $\lambda_{0}.$ Since $H_{\lambda}$ is
countably generated, by Lemma 3.11,
$\Psi_{\lambda}(E_{\lambda}TE_{\lambda})\not=0$ (for
$\lambda\geq\lambda_{0}$). By Proposition 3.4,

$\displaystyle\Psi(E_{\lambda}TE_{\lambda})|_{H_{\lambda}^{\sim}}=\Psi_{\lambda}(E_{\lambda}TE_{\lambda}).$
(e 3.126)

It follows that $\Psi(E_{\lambda}TE_{\lambda})|_{H_{\lambda}^{\sim}}\not=0$
for all $\lambda\geq\lambda_{0}.$ For $\lambda\geq\lambda_{0},$ there are
$x,y\in H_{\lambda}$ such that

$\displaystyle\langle\Psi(T)(E_{\lambda}(x)),E_{\lambda}(y)\rangle=\langle\Psi(E_{\lambda}TE_{\lambda})(x),y\rangle\not=0.$
(e 3.127)

Hence $\Psi(T)\not=0.$ This shows that $\Psi$ is injective.

To see that $\Psi$ is surjective, let $L\in B(H^{\sim}).$ Since, by
Proposition 2.14, $\Psi(K(H)^{**})$ is weak*-closed in $W^{*}$-algebra
$B(H^{\sim}),$ it suffices to show the following: for any $\epsilon>0,$ any
finite subsets $X,Y\subset H^{\sim},$ and finite subset ${\cal F}\subset
A^{*},$ there exists $T\in K(H)^{**}$ such that

$\displaystyle|f(\langle\Psi(T)(x),y\rangle)-f(L(x),y\rangle)|<\epsilon\,\,\,{\rm
for\,\,\,all}\,\,\,x\in X,\,\,y\in Y\,\,\,{\rm and}\,\,\,f\in{\cal F}$ (e
3.128)

(see the last part of 2.4). We now fix $\epsilon,$ $X,Y$ and ${\cal F}.$ By
Proposition 2.14 (since $E_{\lambda}\nearrow 1_{K(H)^{**}}$),

$\displaystyle\lim_{\lambda}f(\langle
x,\Psi(E_{\lambda})(y)\rangle)=\lim_{\lambda}f(\langle\Psi(E_{\lambda})(x),y\rangle)=f(\langle
x,y\rangle)$ (e 3.129)

for all $x,y\in H^{\sim}$ and $f\in A^{*}.$ It follows that there is
$\lambda_{0}$ such that, for all $\lambda\geq\lambda_{0},$

$\displaystyle|f(\langle\Psi(E_{\lambda})(x),L^{*}(y)\rangle)-f(\langle
x,L^{*}(y)\rangle)|<\epsilon/2,$ (e 3.130) $\displaystyle{\rm or}\,\,\,$
$\displaystyle|f(\langle L\Psi(E_{\lambda})(x),y\rangle)-f(\langle
L(x),y\rangle)|<\epsilon/2$ (e 3.131)

for all $x\in X,$ $y\in Y$ and $f\in{\cal F}.$ (The proof would be shorter if
we know
$\lim_{\lambda}f(\langle\Psi(E_{\lambda})L\Psi(E_{\lambda})(x),y\rangle)=f(\langle
L(x),y\rangle).$) However, we may also assume that, for fixed $\lambda_{0},$
there is $\lambda_{1}\geq\lambda_{0}$ such that

$\displaystyle|f(\langle
L\Psi(E_{\lambda_{0}})(x),\Psi(E_{\lambda})(y)\rangle)-f(\langle
L\Psi(E_{\lambda_{0}})(x),y\rangle)|<\epsilon/2,$ (e 3.132) $\displaystyle{\rm
or}\,\,\,$
$\displaystyle|f(\langle\Psi(E_{\lambda})L\Psi(E_{\lambda_{0}})(x),y\rangle)-f(\langle
L\Psi(E_{\lambda_{0}})(x),y\rangle)|<\epsilon/2$ (e 3.133)

for all $x\in X,$ $y\in Y$ and $f\in{\cal F},$ and $\lambda\geq\lambda_{1}.$
It follows that, for all $x\in X,$ $y\in Y$ and $f\in{\cal F},$ if
$\lambda\geq\lambda_{1},$

$\displaystyle|f(\langle
L(x),y\rangle)-f(\langle\Psi(E_{\lambda})L\Psi(E_{\lambda_{0}})(x),y\rangle)|$
(e 3.134) $\displaystyle\leq|f(\langle L(x),y\rangle)-f(\langle
L\Psi(E_{\lambda_{0}})(x),y\rangle)|$ (e 3.135) $\displaystyle\hskip
28.90755pt+|f(\langle
L\Psi(E_{\lambda_{0}})(x),y\rangle)-f(\langle\Psi(E_{\lambda})L\Psi(E_{\lambda_{0}})(x),y\rangle)|$
(e 3.136) $\displaystyle<\epsilon/2+\epsilon/2=\epsilon.$ (e 3.137)

Fix $\lambda\geq\lambda_{1}\geq\lambda_{0}.$ Then
$H_{\lambda}=\overline{E_{\lambda}(H)}\supset H_{\lambda_{0}}.$ We also note
that $\Psi(E_{\lambda})L\Psi(E_{\lambda_{0}})|_{H_{\lambda}^{\sim}}\in
B(H_{\lambda}).$ Since $H_{\lambda}$ is countably generated, by Lemma 3.11,
there is $T_{\lambda}\in K(H_{\lambda})^{**}$ such that

$\displaystyle\Psi_{\lambda}(T_{\lambda})=\Psi(E_{\lambda})L\Psi(E_{\lambda_{0}})|_{H_{\lambda}^{\sim}}.$
(e 3.138)

However, by Proposition 3.4,

$\displaystyle
P_{\lambda}\Psi(T_{\lambda})P_{\lambda}|_{H_{\lambda}^{\sim}}=\Psi(T_{\lambda})|_{H_{\lambda}^{\sim}}=\Psi_{\lambda}(T_{\lambda}).$
(e 3.139)

Fix $\lambda\geq\lambda_{1}\geq\lambda_{0}.$ Then, for any $x\in X,y\in Y$ and
$f\in A^{**},$ by (e 3.139), (e 3.138) and (e 3.137),

$\displaystyle|f(\langle\Psi(T_{\lambda})(x),(y)\rangle)-f(\langle
L(x),y\rangle)|$ $\displaystyle=$
$\displaystyle|f(\langle\Psi(T_{\lambda})P_{\lambda}(x),P_{\lambda}(y)\rangle)-f(\langle
L(x),y\rangle)|$ $\displaystyle=$
$\displaystyle|f(\langle\Psi(E_{\lambda})L\Psi(E_{\lambda_{0}})P_{\lambda}(x),P_{\lambda}(y)\rangle)-f(\langle
L(x),y\rangle)|$ $\displaystyle=$ $\displaystyle|f(\langle
P_{\lambda}\Psi(E_{\lambda})L\Psi(E_{\lambda_{0}})P_{\lambda}(x),y\rangle)-f(\langle
L(x),y\rangle)|$ $\displaystyle=$
$\displaystyle|f(\langle\Psi(E_{\lambda})L\Psi(E_{\lambda_{0}})(x),y\rangle)-f(\langle
L(x),y\rangle)|<\epsilon.$

As mentioned above, this implies that $\Psi$ is surjective.

∎

######  Corollary 3.13.

Let $A$ be a $W^{*}$-algebra and $H$ be a Hilbert $A$-module. Then
$F:K(H)^{**}\to B(H^{\sharp}),$ the map given by Proposition 2.16, is a
surjective map.

###### Proof.

Consider the pair $A$ and $A^{**}$ and $H^{\sim}=(H\bullet A^{**})^{\sharp}.$
By Corollary 4.3 of [13], $H^{\sim}=B(H,A^{**}),$ the $A^{**}$-module of all
bounded $A^{**}$-valued $A$-module maps from $H$ into $A^{**}.$ It follows
that $H^{\sharp}\subset H^{\sim}$ as an $A$-submodule. It then follows from
Proposition 2.6 that $H^{\sharp}\bullet A^{**}\subset H^{\sim}$ as Hilbert
$A^{**}$-modules. Then, by applying Lemma 3.1,

$(H^{\sharp}\bullet A^{**})^{\sharp}\subset H^{\sim}.$

However, $H\bullet A^{**}\subset H^{\sharp}\bullet A^{**}.$ By applying Lemma
3.1 again, we obtain

$H^{\sim}=(H\bullet A^{**})^{\sharp}\subset(H^{\sharp}\bullet
A^{**})^{\sharp}.$

Hence $(H^{\sharp}\bullet A^{**})^{\sharp}=H^{\sim}.$ Denote by
$\tilde{\Psi}:K(H)^{**}\to B(H^{\sim})$ the isomorphism given by Theorem 3.12
and by $\tilde{\Psi}_{H^{\sharp}}:B(H^{\sharp})\to B((H^{\sharp}\bullet
A^{**})^{\sharp})=B(H^{\sim})$ the map given by Theorem 3.12, respectively.

Now let $T\in B(H^{\sharp}).$ Then, by applying Theorem 3.12, we obtain $a\in
K(H)^{**}$ such that $\tilde{\Psi}(a)=\tilde{\Psi}_{H^{\sharp}}(T).$ It
follows that

$\displaystyle\tilde{\Psi}(a)|_{H^{\sharp}}=T.$ (e 3.140)

Since $a\in K(H)^{**},$ there exists a net $\\{a_{\alpha}\\}$ in $K(H)$ such
that $a_{\alpha}\to a$ in the weak*-topology. Therefore, by Proposition 2.14,
for any $f\in A^{*}$ and any $\xi,\zeta\in H^{\sim},$

$\displaystyle\lim_{\alpha}f(\langle(\tilde{\Psi}(a)-\tilde{\Psi}(a_{\alpha}))(\xi),\zeta\rangle)=0.$
(e 3.141)

On the other hand, by Proposition 2.16, for any $g\in A_{*}$ and any $x,y\in
H,$

$\displaystyle\lim_{\alpha}g(\langle(F(a)-a_{\alpha})(x),y\rangle)=0.$ (e
3.142)

Hence

$\displaystyle g(\langle(F(a)-\tilde{\Psi}(a))(x),y\rangle)=0\,\,\,{\rm
for\,\,\,all}\,\,\,x,y\in H\,\,\,{\rm and}\,\,\,g\in A_{*}.$ (e 3.143)

Since $\tilde{\Psi}(a)|_{H^{\sharp}}=T,$ we actually have

$\displaystyle g(\langle(F(a)-T)(x),y\rangle)=0\,\,\,{\rm
for\,\,\,all}\,\,\,x,\,y\in H\,\,\,{\rm and}\,\,\,g\in A_{*}.$ (e 3.144)

Note that $F(a),\,T\in B(H^{\sharp}).$ So $F(a)(x),T(x)\in H^{\sharp}$ for all
$x\in H.$ It follows that

$\displaystyle\langle(F(a)-T)(x),y\rangle\in A\,\,\,{\rm
for\,\,\,all}\,\,\,x,\,y\in H.$ (e 3.145)

By (e 3.144),

$\langle(F(a)-T)(x),y\rangle=0\,\,\,{\rm for\,\,\,all}\,\,\,x,y\in H.$

Hence $F(a)=T.$ In other words, $F$ is surjective. ∎

##  4 A Kaplansky density theorem in Hilbert modules

As mentioned in the introduction, in this section we study the density of $H$
in $H\bullet A^{**}.$

######  Definition 4.1.

Let $X$ be a Hilbert space and $A\subset B(X)$ be a $C^{*}$-subalgebra of
$B(H).$ Let $M={\bar{A}}^{{}^{SOT}},$ the strong operator closure of $A,$ and
$H$ be a Hilbert $A$-module. Recall, by Proposition 2.6, $H\bullet M$ is the
smallest Hilbert $M$-module containing $H$ as Hilbert $A$-module. We consider
the question how large $H$ in $H\bullet M$ as a submodule.

Let $\epsilon>0,$ and $V$ be a finite subset of $X.$ For each $\xi\in H\bullet
M,$ define

$\displaystyle N_{\xi,\epsilon,V}=\\{z\in H\bullet
M:\|\langle\xi-z,\xi-z\rangle(v)\|<\epsilon,\,\,\,v\in V\\}.$ (e 4.146)

Let ${\cal T}_{s}$ be the topology generated by $N_{\xi,\epsilon,V}$ for all
$\xi\in H\bullet M,$ $\epsilon\in\mathbb{R}_{+}\setminus\\{0\\},$ any finite
subset $Y\in H\bullet M$ and any finite subset $V\subset X.$ In other words,
in ${\cal T}_{s},$ a net $\\{z_{\alpha}\\}$ converges to $\xi$ in $H\bullet M$
if and only if

$\displaystyle\lim_{\alpha}\|\langle\xi-z_{\alpha},\xi-
z_{\alpha}\rangle(v)\|=0\,\,\,{\rm for\,\,\,all}\,\,\,v\in X.$ (e 4.147)

In the special case that $X=H_{U}$ is the Hilbert space corresponding to the
universal representation $\pi_{U}$ of $A$ and $M=A^{**},$ we use ${\cal
T}_{su}$ for the topology generated by $N_{\xi,\epsilon,V}$ for all $\xi\in
H\bullet A^{**},$ $\epsilon\in\mathbb{R}_{+}\setminus\\{0\\},$ and any finite
subset $V\subset H_{U}.$

It is easy to see that $H$ is dense in $H\bullet M$ in topology ${\cal
T}_{s}.$ However, to be more useful, we will show in Theorem 4.4 that the unit
ball of $H$ is dense in the unit ball of $H\bullet M$ in ${\cal T}_{s},$ a
Kaplansky style density theorem.

######  Lemma 4.2.

Suppose that $x\in H\bullet M$ and $\\{x_{\alpha}\\}\subset H\bullet M$ is a
bounded net. Then $x_{\alpha}\to x$ in ${\cal T}_{s}$ if and only if, for any
$v\in X,$

$\displaystyle\lim_{\alpha}\sup\\{\|\langle y,x_{\alpha}-x\rangle(v)\|:y\in
H\bullet M,\,\,\|y\|\leq 1\\}=0.$ (e 4.148)

Moreover, if $x_{\alpha}\to x$ in ${\cal T}_{s},$ then, for any $f\in M_{*},$

$\displaystyle\lim_{\alpha}\sup\\{|f(\langle y,x_{\alpha}-x\rangle)|:y\in
H\bullet M,\,\,\|y\|\leq 1\\}=0.$ (e 4.149)

###### Proof.

Suppose that $x_{\alpha}\to x$ in ${\cal T}_{s}.$ We have (see (ii) of
Proposition 2.3 of [13]), for any $y\in H\bullet M,$

$\displaystyle\langle x_{\alpha}-x,y\rangle\cdot\langle
y,x_{\alpha}-x\rangle\leq\|y\|^{2}\langle x_{\alpha}-x,x_{\alpha}-x\rangle.$
(e 4.150)

Then, for any $v\in X,$ and any $y\in H\bullet M$ with $\|y\|\leq 1,$

$\displaystyle\|\langle y,x_{\alpha}-x\rangle(v)\|^{2}$ $\displaystyle=$
$\displaystyle\langle\langle x_{\alpha}-x,y\rangle\cdot\langle
y,x_{\alpha}-x\rangle v,v\rangle_{X}$ (e 4.151) $\displaystyle\leq$
$\displaystyle\|y\|^{2}\langle\langle x_{\alpha}-x,x_{\alpha}-x\rangle
v,v\rangle_{X}$ (e 4.152) $\displaystyle\leq$ $\displaystyle\|\langle
x_{\alpha}-x,x_{\alpha}-x\rangle v\|\|v\|\to 0$ (e 4.153)

(where $\langle\cdot,\cdot\rangle_{X}$ is the inner product in the Hilbert
space $X$). Conversely, let $K=\sup_{\alpha}\\{\|x_{\alpha}\|+\|x\|\\}+1.$
Then

$\displaystyle\|\langle x_{\alpha}-x,x_{\alpha}-x\rangle(v)\|\leq
K\sup\\{\|\langle y,x_{\alpha}-x\rangle(v)\|:y\in H\bullet M,\,\,\|y\|\leq
1\\}\to 0$ (e 4.154)

For the “Moreover” part of the lemma, suppose that $\langle
x_{\alpha}-x,x_{\alpha}-x\rangle\to 0$ in the strong operator topology. Then
it converges in the weak operator topology. However $\\{\langle
x_{\alpha}-x,x_{\alpha}-x\rangle\\}$ is bounded, this also implies that it
converges to zero in $\sigma$-weak topology and in weak*-topology. In other
words,

$\displaystyle\lim_{\alpha}f(\langle
x_{\alpha}-x,x_{\alpha}-x\rangle)=0\,\,\,{\rm for\,\,\,all}\,\,\,f\in M_{*}.$
(e 4.155)

Let $f\in M_{*}$ be a positive normal functional. Then,
$f(\langle\cdot,\cdot\rangle)$ defines a pseudo inner product on $H\bullet M.$
Hence, for any $y\in H\bullet M,$ we have, by Cauchy-Bunyakovsky-Schwarz
inequality,

$\displaystyle|f(\langle y,x_{\alpha}-x\rangle)|^{2}\leq f(\langle
y,y\rangle)f(\langle
x_{\alpha}-x,x_{\alpha}-x\rangle)\leq\|f\|^{2}\|y\|^{2}f(\langle
x_{\alpha}-x,x_{\alpha}-x\rangle).$ (e 4.156)

Thus

$\displaystyle\lim_{\alpha}\sup\\{|f(\langle y,x_{\alpha}-x\rangle)|:y\in
H\bullet M,\,\,\|y\|\leq 1\\}=0.$ (e 4.157)

∎

######  Lemma 4.3.

Let $X$ be a Hilbert space, $A\subset B(X)$ be a $C^{*}$-subalgebra and
$M=\overline{A}^{{SOT}}$ such that ${\rm id}_{X}\in M.$ Then the unit ball of
$H_{A}$ is dense in the unit ball of $H_{A}\bullet M$ in ${\cal T}_{s}.$

###### Proof.

Let $\xi\in H_{A}\bullet M$ with $\|\xi\|\leq 1.$ We will show that there is a
net $\\{x_{\alpha}\\}\in H$ such that $\|x_{\alpha}\|\leq\|\xi\|$ and
$\lim_{\alpha}\|\langle x_{\alpha}-\xi,x_{\alpha}-\xi\rangle(v)\|=0$ for all
$v\in X.$ From the inequality

$\|\langle x_{\alpha}-\xi,x_{\alpha}-\xi\rangle(v)\|\leq\|\langle
x_{\alpha}-\xi,x_{\alpha},\xi\rangle^{1/2}\|\|\langle
x_{\alpha}-\xi,x_{\alpha},\xi\rangle^{1/2}(v)\|\leq 2\|\|\langle
x_{\alpha}-\xi,x_{\alpha},\xi\rangle^{1/2}(v)\|,$

we conclude that it is enough to show that there is net $\\{x_{\alpha}\\}\in
H$ such that $\|x_{\alpha}\|\leq\|\xi\|$ and $\lim_{\alpha}\|\langle
x_{\alpha}-\xi,x_{\alpha}-\xi\rangle^{1/2}v\|=0$ for all $v\in X.$ Therefore
it suffices to show that, for any $\epsilon>0$ and any finite subset $V\subset
X,$ there exists $z\in H$ with $\|z\|\leq 1$ such that

$\displaystyle\|(\langle\xi-z,\xi-z\rangle)^{1/2}(v)\|<\epsilon\,\,\,{\rm
for\,\,\,all}\,\,\,v\in V.$ (e 4.158)

To simplify notation, we may also assume that $\|v\|\leq 1$ for all $v\in V.$

Denote by $R={\overline{AM}},$ the closed right ideal of $M$ generated by $A.$
Note, by Proposition 3.6,

$\displaystyle H_{A}\bullet M=\\{\\{b_{n}\\}\in H_{B}:b_{n}\in R\\}.$ (e
4.159)

We write $\xi=\\{b_{n}\\}\in H_{A}\bullet M.$ There exists
$n_{0}\in\mathbb{N}$ such that, for all $n\geq n_{0},$

$\displaystyle\|\sum_{k=n}^{\infty}b_{n}^{*}b_{n}\|<\epsilon/2.$ (e 4.160)

Fix an integer $n\geq n_{0}.$ Let $P_{n}:H_{A}\bullet M\to
R^{(n)}=\\{(c_{1},c_{2},...,c_{n}):c_{i}\in R\\}$ be the projection. Put

$\displaystyle S=\begin{pmatrix}b_{1}&0&0\cdots&0\\\ b_{2}&0&0\cdots&0\\\
\vdots&&&\vdots\\\ b_{n}&0&0\dots&0\end{pmatrix}.$ (e 4.161)

For any $v\in V,$ put

$\displaystyle u_{v}=\begin{pmatrix}v\\\ 0\\\ \vdots\\\ 0\end{pmatrix}.$ (e
4.162)

By the Kaplansky density theorem, there is $L\in M_{n}(A)$ such that

$\displaystyle\|L\|\leq\|S\|\,\,\,{\rm
and}\,\,\,\|L(u_{v})-S(u_{v})\|<\epsilon/2$ (e 4.163)

for all $v\in V.$ Hence, denoting by $\langle\cdot,\cdot\rangle_{X}$ the inner
product in $X,$

$\displaystyle\langle(L-S)^{*}(L-S)u_{v},u_{v}\rangle_{X}<\epsilon/2\,\,\,{\rm
for\,\,\,all}\,\,\,v\in V.$ (e 4.164)

Define $q={\rm diag}(1,0,...,0)\in M_{n}(M).$ Then $S=Sq.$ Replacing $L$ by
$Lq,$ we may write

$\displaystyle L=\begin{pmatrix}a_{1}&0&0\cdots&0\\\ a_{2}&0&0\cdots&0\\\
\vdots&&&\vdots\\\ a_{n}&0&0\dots&0\end{pmatrix},$ (e 4.165)

where $a_{i}\in A,$ $i=1,2,...,n.$ Then

$\displaystyle\|\sum_{i=1}^{n}a_{i}^{*}a_{i}\|=\|L^{*}L\|=\|L\|^{2}\leq\|S\|^{2}=\|\sum_{i=1}^{n}b_{i}^{*}b_{i}\|\leq\|\xi\|^{2}.$
(e 4.166)

It follows from (e 4.164) that

$\displaystyle\langle\sum_{i=1}^{n}(b_{i}-a_{i})^{*}(b_{i}-a_{i})(v),v\rangle_{X}<\epsilon/2.$
(e 4.167)

Put $x=(a_{1},a_{2},...,a_{n},0,0,\cdots)\in H_{A}.$ Then, by (e 4.166),
$\|x\|\leq\|\xi\|,$ and

$\displaystyle\langle\langle\xi-x,\xi-x\rangle(v),v\rangle_{X}$
$\displaystyle=$
$\displaystyle\langle\sum_{i=1}^{n}(b_{i}-a_{i})^{*}(b_{i}-a_{i})(v),v\rangle_{X}+\langle\sum_{i=n+1}^{\infty}b_{i}^{*}b_{i}(v),v\rangle_{X}$
(e 4.168) $\displaystyle<$
$\displaystyle\epsilon/2+\|\sum_{i=n+1}^{\infty}b_{i}^{*}b_{i}\|<\epsilon/2+\epsilon/2=\epsilon.$
(e 4.169)

In other words, for any $v\in V,$

$\displaystyle\|(\langle\xi-x_{\alpha},\xi-
x_{\alpha})^{1/2}(v)\|=\langle\langle\xi-x,\xi-x\rangle(v),v\rangle_{X}<\epsilon.$
(e 4.170)

The lemma then follows. ∎

######  Theorem 4.4.

Let $X$ be a Hilbert space, $A\subset B(X)$ be a $C^{*}$-subalgebra and
$M=\overline{A}^{{SOT}}$ and ${\rm id}_{X}\in M.$ Let $H$ be a Hilbert
$A$-module. Then the unit ball of $H$ is dense in the unit ball of $H\bullet
M$ in ${\cal T}_{s}.$

###### Proof.

Let $\xi\in H\bullet M$ with $\|\xi\|\leq 1.$

Let us first assume that $H$ is countably generated $A$-module. By Lemma 4.2,
it suffices to show that, for any $\epsilon>0$ and any finite subset $V\subset
X,$ there exists $z\in H$ with $\|z\|\leq 1$ such that

$\displaystyle\|\langle y,\xi-z\rangle)(v)\|<\epsilon\,\,\,{\rm
for\,\,\,all}\,\,\,y\in H\,\,{\rm with}\,\,\|y\|\leq 1\,\,\,{\rm
and}\,\,\,v\in V.$ (e 4.171)

To simplify notation, we may also assume that $\|v\|\leq 1$ for all $v\in V.$

By Kasparov’s absorbing theorem (Theorem 2 of [7]), we may write
$H_{A}=H\oplus H^{\perp}.$ It follows that

$H_{A}\bullet M=H\bullet M\oplus H^{\perp}\bullet M.$

Define $Q:H_{A}\to H$ to be the projection. Then $Q\in L(H_{A})=M(K(H_{A})).$
We identify $Q$ with $\Psi_{0}(Q)$ in the sense that $Q\in L(H_{M})$ which
extends $Q|_{H_{A}}.$ In particular, $H\bullet M=Q(H_{A}\bullet M).$

By applying Lemma 4.3 and Lemma 4.2, we obtain $z\in H_{A}$ with
$\|z\|\leq\|\xi\|$ such that

$\displaystyle\|\langle y,\xi-z\rangle(v)\|<\epsilon\,\,\,{\rm
for\,\,\,all}\,\,\,y\in H\bullet M,\,\,\|y\|\leq 1\,\,\,{\rm and}\,\,\,v\in
V.$ (e 4.172)

Note $Q(\xi)=\xi$ and $Q(y)=y$ for all $y\in H.$ Put $x=Q(z)\in H.$ We have

$\displaystyle\|\langle y,\xi-x\rangle(v)\|=\|\langle
y,Q(\xi)-Q(z))\rangle(v)\|=\|\langle Q(y),\xi-z\rangle(v)\|=\|\langle
y,\xi-z\rangle(v)\|<\epsilon.$ (e 4.173)

This proves the case that $H$ is countably generated.

Next we let $H$ be a general Hilbert $A$-module. We will show that, for any
$\epsilon>0$ and any finite subset $V\subset X,$ there exists $z\in H$ with
$\|z\|\leq 1$ such that

$\displaystyle\|\langle\xi-z,\xi-z\rangle)(v)\|<\epsilon\,\,\,{\rm
for\,\,\,all}\,\,\,v\in V.$ (e 4.174)

Again, we may also assume that $\|v\|\leq 1$ for all $v\in V.$

Let $\\{E_{\lambda}\\}$ be an approximate identity for $K(H).$ Then, as in the
proof of Theorem 3.12, $H_{\lambda}=\overline{E_{\lambda}(H)}$ is countably
generated for each $\lambda.$ It follows from Lemma 2.9 that there is
$\lambda$ such that

$\displaystyle\|\Psi_{0}(E_{\lambda})(\xi)-\xi\|<\epsilon/4.$ (e 4.175)

Fix such a $\lambda.$ Note that, by Proposition 2.8,
$\Psi_{0}(E_{\lambda})(\xi)\in H_{\lambda}\bullet M\subset H\bullet M.$ Since
$H_{\lambda}$ is countably generated, by the first part of the proof, we
obtain $x\in H_{\lambda}$ with
$\|x\|\leq\|\Psi_{0}(E_{\lambda})(\xi)\|\leq\|\xi\|$ such that

$\displaystyle\sup\\{\|\langle y,\Psi_{0}(E_{\lambda})(\xi)-x\rangle(v)\|:y\in
H\bullet M,\,\,\|y\|\leq 1\\}<\epsilon/4.$ (e 4.176)

Then, applying (e 4.175) and then (e 4.176),

$\displaystyle\|\langle\xi-x,\xi-x\rangle(v)\|$ $\displaystyle\leq$
$\displaystyle\|\langle\xi-x,\xi-\Psi_{0}(E_{\lambda})(\xi)\rangle(v)\|+\|\langle\xi-x,\Psi_{0}(E_{\lambda})(\xi)-x\rangle(v)\|$
(e 4.177) $\displaystyle<$ $\displaystyle
2\|\xi-\Psi_{0}(E_{\lambda})(\xi)\|+2\|\langle{\xi-x\over{2}},\Psi_{0}(E_{\lambda}(\xi)-x\rangle(v)\|$
(e 4.178) $\displaystyle<$ $\displaystyle\epsilon/2+\epsilon/2=\epsilon.$ (e
4.179)

∎

We then obtain the following corollary as a Kaplansky density theorem:

######  Theorem 4.5.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then the unit
ball of $H$ is dense in the unit ball of $H\bullet A^{**}$ in ${\cal T}_{su}.$

##  5 Closeness of $H$

Let $H$ be a Hilbert $A$-module., Then, by Theorem 6.1 of [1], the unit ball
of $H$ is $A$-weakly dense (see Definition 3.3 of [1]) in the unit ball of
$H^{\sharp},$ i.e., for any $f\in H^{\sharp},$ there is a net
$\\{x_{\alpha}\\}$ in $H$ with $\|x_{\alpha}\|\leq\|f\|$ such that
$\lim_{\alpha}\|\langle f-x_{\alpha},y\rangle\|=0$ for all $y\in H.$ In the
case that $A$ is a $W^{*}$-algebra, $H^{\sharp}$ is a Hilbert $A$-module. One
may ask whether one can find the net $\\{x_{\alpha}\\},$ in addition,
$\lim_{\alpha}\|\langle f-x_{\alpha},\xi\rangle\|=0$ for all $\xi\in
H^{\sharp}?$

We begin with the following example.

######  Example 5.1.

Let $M$ be a $W^{*}$-algebra which contains a self-adjoint element $a$ with
infinite spectrum. Then, by the spectral theory, one obtains a sequence of
mutually orthogonal non-zero projections $p_{1},p_{2},...,p_{n},....$ Let
$H=H_{M}$ and let $\xi=\\{p_{n}\\}\in H_{M}^{\sharp}.$ Note that
$\|\xi\|=\|\sum_{n=1}^{\infty}p_{n}\|=1$ (the convergence is in the strong
operator topology and weak*-topology of $M$). We claim that there is no net
$\\{x_{\alpha}\\}$ in $H_{M}$ such that

$\displaystyle\lim_{\alpha}\|\langle\xi-x_{\alpha},\xi\rangle\|=0.$ (e 5.180)

Otherwise, there would be $x\in H_{M}$ such that

$\displaystyle\|\langle\xi-x,\xi\rangle\|<1/4.$ (e 5.181)

Since $x=\\{a_{n}\\}\in H_{M},$ there is $N\in\mathbb{N}$ such that

$\displaystyle\|\sum_{N+1}a_{n}^{*}a_{n}\|<(1/16(1+\|x\|))^{2}.$ (e 5.182)

Choose $q=\sum_{n=N+1}^{\infty}p_{n}\in M.$ Define $P_{N}:H_{M}^{\sharp}\to
M^{(N)}=\\{(b_{1},b_{2},...,b_{N}):b_{i}\in M\\}$ to be the projection. Then

$\displaystyle\|\langle\xi-P(x),\xi\rangle\|$ $\displaystyle\leq$
$\displaystyle\|\langle\xi-x,\xi\rangle\|+\|\langle(1-P_{N})(x),\xi\rangle\|$
(e 5.183) $\displaystyle<$ $\displaystyle
1/4+\|(1-P_{N})(x)\|\|\xi\|<1/4+1/16=5/16.$ (e 5.184)

On the other hand

$\displaystyle 5/16$ $\displaystyle\geq$
$\displaystyle\|\langle\xi-P(x),\xi\rangle\|\geq\|\langle\xi-P(x),\xi\rangle
q\|$ (e 5.186)
$\displaystyle=\|(\sum_{N+1}^{\infty}p_{n}-\sum_{i=1}^{N}(p_{i}-a_{i})^{*}p_{i})q\|$
$\displaystyle=$ $\displaystyle\|\sum_{N+1}^{\infty}p_{n}q\|=1.$ (e 5.187)

A contradiction. In other words, the question at the beginning of this section
is negative. This also follows from Corollary 5.7 below. However, we think
that the example above might also be helpful.

######  Lemma 5.2.

Let $A$ be a $C^{*}$-algebra. Suppose that $\xi\in H_{A}^{\sharp}$ and
$\\{x_{\alpha}\\}$ is a bounded net such that

$\displaystyle\lim_{\alpha}\|\xi(x)-x_{\alpha}(x)\|=0\,\,\,\text{for\,\,\,all}\,\,\,x\in
H_{A}$ (e 5.188)

and $\xi(x_{\alpha}):=\langle\xi,x_{\alpha}\rangle$ converges in norm. Then
$\xi\in H_{A}$ and
$\langle\xi,\xi\rangle=\lim_{\alpha}\langle\xi,x_{\alpha}\rangle.$

###### Proof.

Write $\xi=\\{b_{n}\\}$ and $x_{\alpha}=\\{a_{{\alpha},n}\\},$ where
$\\{b_{n}\\}\in H_{A}^{\sharp},$ $a_{{\alpha},n}\in A$ and, for each
${\alpha},$ $\\{a_{{\alpha},n}\\}\in H_{A}.$

Put

$\displaystyle M=1+\sup\\{\|x_{\alpha}\|:{\alpha}\\}+\|\xi\|<\infty\,\,\,{\rm
and}\,\,\,a=\lim_{\alpha}\langle x_{\alpha},\xi\rangle.$ (e 5.189)

Note $\xi(x_{\alpha})=\langle\xi,x_{\alpha}\rangle\in A$ for all ${\alpha}.$
Hence $a\in A.$

Put $P_{n}:H_{A}^{\sharp}\to H_{n}:=A^{(n)}$ be the projection to the first
$n$ copies of $A.$ Then $P_{n}\xi\in H_{n}\subset H_{A}.$ It follows that

$\displaystyle\lim_{\alpha}\langle
x_{\alpha},P_{n}(\xi)\rangle=\langle\xi,P_{n}(\xi)\rangle=\sum_{j=1}^{n}b_{j}^{*}b_{j}.$
(e 5.190)

Fix $f\in A^{*}.$ Let $\epsilon>0.$ By Lemma 3.8, since $\\{x_{\alpha}\\}$ is
bounded, there is an integer $N\in\mathbb{N}$ such that, for all $n\geq N,$

$\displaystyle|f(\langle
x_{\alpha},\xi\rangle)-f(x_{\alpha},P_{n}(\xi)\rangle)|<\epsilon/3\,\,\,{\rm
for\,\,\,all}\,\,\,{\alpha}.$ (e 5.191)

Fix any $n\geq N.$ Choose ${\alpha}_{0}$ such that, for all
${\alpha}\geq{\alpha}_{0},$

$\displaystyle\|\langle
x_{\alpha},P_{n}(\xi)\rangle-\sum_{j=1}^{n}b_{j}^{*}b_{j}\|<\epsilon/3(1+\|f\|)\,\,\,{\rm
and}\,\,\,$ (e 5.192) $\displaystyle\|\langle
x_{\alpha},\xi\rangle-a\|<\epsilon/3(1+\|f\|).$ (e 5.193)

It follows that, for all $n\geq N,$

$\displaystyle|f(a)-f(\sum_{j=1}^{n}b_{j}^{*}b_{j})|$ $\displaystyle\leq$
$\displaystyle|f(a-\langle x_{\alpha},\xi\rangle)|+|f(\langle
x_{\alpha},\xi\rangle)-f(x_{\alpha},P_{n}(\xi)\rangle)|$ (e 5.195)
$\displaystyle+\|f\|\|\langle
x_{\alpha},P_{n}(\xi)\rangle-\sum_{j=1}^{n}b_{j}^{*}b_{j}\|<\epsilon/3+\epsilon/3+\epsilon/3=\epsilon.$

Hence on the state space $S(A)$ of $A,$

$\displaystyle\lim_{n\to\infty}f(\sum_{j=1}^{n}b_{j}^{*}b_{j})=f(a).$ (e
5.196)

On the compact space $S(A)$ (in the weak *-topology), $\hat{a}(f)=f(a)$ for
all $f\in S(A)$ is a continuous function, and
$\widehat{\sum_{j=1}^{n}b_{j}^{*}b_{j}}$ is increasing. By the Dini theorem,
$\widehat{\sum_{j=1}^{n}b_{j}^{*}b_{j}}$ converges uniformly to $\hat{a}$ on
$S(A).$ It follows that

$\displaystyle\sum_{j=1}^{n}b_{j}^{*}b_{j}\to a$ (e 5.197)

in norm. This implies that $\xi=\\{b_{n}\\}\in H_{A}$ and
$\langle\xi,\xi\rangle=a=\lim_{\alpha}\langle\xi,x_{\alpha}\rangle.$ ∎

######  Proposition 5.3.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then, for any
$T\in K(H),$ one has $\Psi_{0}(T)(H^{\sharp})\subset H,$ where $\Psi_{0}$ is
Definition 2.7.

###### Proof.

Suppose that $T\in F(H)$ and $T=\sum_{i=1}^{m}\theta_{x_{i},y_{i}}$ for some
$x_{i},y_{i}\in H,$ $i=1,2,...,m.$ Then, for any $\xi\in H^{\sharp},$

$\displaystyle\Psi_{0}(T)(\xi)=\sum_{i=1}^{m}x_{i}\langle
y_{i},\xi\rangle=\sum_{i=1}^{m}\xi(y_{i})^{*}x_{i}\in H.$ (e 5.198)

Since $F(H)$ is dense in $K(H),$ this implies that
$\Psi_{0}(T)(H^{\sharp})\subset H.$ ∎

######  Lemma 5.4.

Let $A$ be a $C^{*}$-algebra, $H$ be a Hilbert $A$-module and
$\\{E_{\lambda}\\}$ an approximate identity for $K(H).$ Then, for any $\xi\in
H^{\sim}$ and any $f\in A^{*},$

$\displaystyle\lim_{\alpha}\sup\\{f(\langle\xi-\Psi_{0}(E_{\lambda})(\xi),y\rangle)|:y\in
H^{\sim},\|y\|\leq 1\\}=0.$ (e 5.199)

###### Proof.

By Lemma 2.9, $\\{\Psi_{0}(E_{\lambda})\\}$ is an approximate identity for
$K(H\bullet A^{**}).$ In the universal representation of $K(H\bullet A^{**}),$
$1-\Psi_{0}(E_{\lambda})$ converges to zero in the strong operator topology.
Note that $\|1-\Psi_{0}(E_{\lambda})\|\leq 1.$ Therefore
$(1-\Psi_{0}(E_{\lambda}))(1-\Psi_{0}(E_{\lambda}))$ also converges to zero in
the strong operator topology. Hence it converges to zero in the weak operator
topology. Since $\\{(1-\Psi_{0}(E_{\lambda}))^{2}\\}$ is bounded, it also
converges to zero in the weak*-topology of $K(H\bullet A^{**}).$ Recall that
$(H\bullet A^{**})^{\sharp}=H^{\sim}.$ It follows from Proposition 2.16, for
any $\xi\in H^{\sim},$

$\displaystyle\lim_{\alpha}|f(\langle\xi-\Psi_{0}(E_{\lambda})(\xi),\xi-\Psi_{0}(E_{\lambda})(\xi)\rangle)|$
$\displaystyle=$
$\displaystyle\lim_{\alpha}|f(\langle\xi-F\circ\Psi_{0}(E_{\lambda})(\xi),\xi-F\circ\Psi_{0}(E_{\lambda})(\xi)\rangle)$
(e 5.200) $\displaystyle=$
$\displaystyle\lim_{\alpha}|f(\langle(1-F\circ\Psi_{0}(E_{\lambda}))^{2}(\xi),\xi\rangle)|=0.$

Suppose that $y\in H^{\sim}$ and $\|y\|\leq 1.$ Then, for any positive linear
functional $f\in A^{*},$

$\displaystyle f(\langle\xi-\Psi_{0}(E_{\lambda})(\xi),y\rangle)^{2}$
$\displaystyle\leq$ $\displaystyle
f(\langle\xi-\Psi_{0}(E_{\lambda})(\xi),\xi-\Psi_{0}(E_{\lambda})(\xi)\rangle)f(y^{*}y)$
(e 5.201) $\displaystyle\leq$
$\displaystyle\|f\|f(\langle\xi-\Psi_{0}(E_{\lambda})(\xi),\xi-\Psi_{0}(E_{\lambda})(\xi)\rangle).$
(e 5.202)

It follows that, for any $f\in A^{*},$

$\displaystyle\lim_{\alpha}\sup\\{f(\langle\xi-\Psi_{0}(E_{\lambda})(\xi),y\rangle)|:y\in
H^{\sim},\|y\|\leq 1\\}=0.$ (e 5.203)

∎

######  Theorem 5.5.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Suppose that
$\xi\in H^{\sharp}$ and there is a bounded net $\\{x_{\alpha}\\}$ in $H$ such
that

$\displaystyle\lim_{\alpha}\|\xi(x)-\langle
x_{\alpha},x\rangle\|=0\,\,\,\text{for\,\,\,all}\,\,\,x\in H$ (e 5.204)

and $\xi(x_{\alpha}):=\langle\xi,x_{\alpha}\rangle$ converges in norm. Then
$\xi\in H$ and
$\langle\xi,\xi\rangle=\lim_{\alpha}\langle\xi,x_{\alpha}\rangle\in A.$

###### Proof.

First let us assume $H$ is countably generated. Then, by Kasparov’s absorbing
theorem (Theorem 2 of [7]), we may write $H_{A}=H\oplus H^{\perp}.$ Then
$\xi\in H^{\sharp}\subset H_{A}^{\sharp}.$ By applying Lemma 5.2, we obtain
that

$\displaystyle\xi\in H_{A}\,\,\,{\rm
and}\,\,\,\langle\xi,\xi\rangle=\lim_{\alpha}\langle\xi,x_{\alpha}\rangle.$ (e
5.205)

Since $\xi(x_{\alpha})\in A,$ then $a=\langle\xi,\xi\rangle\in A.$ Let
$P:H_{A}\to H$ be the projection. Then $P\in L(H_{A}).$ Put $\eta=P(\xi)\in
H.$ Note that $\langle P(\xi)-\xi,x\rangle=0$ for all $x\in H$ and $\xi\in
H^{\sharp}.$ Hence $\xi=\eta.$ Therefore this case follows.

In what follows we will work in $H^{\sim}$ and use the inner product in
$H^{\sim}$ whenever it is convenient.

In general, let $a=\lim_{\alpha}\langle\xi,x_{\alpha}\rangle.$ Since
$\langle\xi,x_{\alpha}\rangle=\xi(x_{\alpha})\in A$ for all ${\alpha},$ we
have $a\in A.$

Claim: $a=\langle\xi,\xi\rangle$ (in the inner product of $H^{\sim}$).

Let $\\{E_{\lambda}\\}$ be an approximate identity for $K(H).$ Let
$\epsilon>0$ and $f\in A^{*}$ with $\|f\|\leq 1.$ By applying Lemma 5.4, have
(since $\\{\|\xi-x_{\alpha}\|\\}$ is bounded)

$\displaystyle\lim_{\lambda}\big{(}\sup_{\alpha}\\{|f(\langle\xi-\Psi_{0}(E_{\lambda})(\xi),\xi-
x_{\alpha}\rangle)|\\}\big{)}=0.$ (e 5.206)

Thus, by applying Lemma 5.4 and (e 5.206), we obtain $\lambda_{0}$ such that,
for all $\lambda\geq\lambda_{0},$

$\displaystyle|f(\langle\xi-\Psi_{0}(E_{\lambda})(\xi),\xi)|<\epsilon/3\,\,\,{\rm
and}\,\,\,$ (e 5.207)
$\displaystyle|f(\langle\xi-\Psi_{0}(E_{\lambda})(\xi),\xi-
x_{\alpha}\rangle)|<\epsilon/3\,\,\,{\rm for\,\,\,all}\,\,\,{\alpha}.$ (e
5.208)

Recall that, by Proposition 5.3, $\Psi_{0}(E_{\lambda})(\xi)\in H.$ Fix any
$\lambda\geq\lambda_{0},$ choose ${\alpha}_{0}$ such that, for any
${\alpha}\geq{\alpha}_{0},$

$\displaystyle\|\langle\xi,x_{\alpha}\rangle-a\|<\epsilon/3\,\,\,{\rm
and}\,\,\,|f(\langle\Psi_{0}(E_{\lambda})(\xi),\xi-
x_{\alpha}\rangle)|<\epsilon/3.$ (e 5.209)

Now, by the first inequality of (e 5.209), (e 5.208) and then the second
inequality of (e 5.209),

$\displaystyle|f(\langle\xi,\xi\rangle-a)|$ $\displaystyle<$
$\displaystyle\epsilon/3+|f(\langle\xi,\xi\rangle-\langle\xi,x_{\alpha}\rangle)|$
(e 5.210) $\displaystyle=$ $\displaystyle\epsilon/3+|f(\langle\xi,\xi-
x_{\alpha}\rangle)|$ (e 5.211) $\displaystyle\leq$
$\displaystyle\epsilon/3+|f(\langle\xi-\Psi_{0}(E_{\lambda})(\xi),\xi-
x_{\alpha}\rangle)|+|f(\langle\Psi_{0}(E_{\lambda})(\xi),\xi-
x_{\alpha}\rangle)|$ (e 5.212) $\displaystyle<$
$\displaystyle\epsilon/3+\epsilon/3+\epsilon/3=\epsilon.$ (e 5.213)

Since this holds for any $\epsilon,$ we conclude that

$\displaystyle f(\langle\xi,\xi\rangle)=f(a)\,\,\,{\rm for\,\,\,all}\,\,\,f\in
A^{*}.$ (e 5.214)

By the Hahn-Banach theorem, we obtain that $\langle\xi,\xi\rangle=a.$ The
claim is proved.

There exists $x_{1}\in\\{x_{\alpha}\\}$ such that

$\displaystyle\|\langle\xi-x_{\alpha},x_{1}\rangle\|<1/2\,\,\,{\rm
and}\,\,\,\|\langle x_{1},\xi\rangle-a\|<1/2.$ (e 5.215)

Suppose that we have found $x_{1},x_{2},...,x_{n}$ such that

$\displaystyle\|\langle\xi-x_{j},x_{i}\rangle\|<1/2^{j}\,\,\,{\rm
and}\,\,\,\|\langle x_{j},\xi\rangle-a\|<1/2^{j},\,\,i=1,2,...,j-1,$ (e 5.216)

and $j=1,2,...,n.$ Then choose $x_{n+1}\in\\{x_{\alpha}\\}$ such that

$\displaystyle\|\langle\xi-x_{n+1},x_{i}\rangle\|<1/2^{n+1}\,\,\,{\rm
and}\,\,\,\|\langle x_{n+1},\xi\rangle-a\|<1/2^{n+1},\,\,i=1,2,...,n.$ (e
5.217)

Thus, by the induction, we obtain a subsequence $\\{x_{n}\\}$ in
$\\{x_{\alpha}\\}$ such that

$\displaystyle\lim_{n\to\infty}\|\langle x_{n},\xi\rangle-a\|=0\,\,\,{\rm
and}\,\,\,\lim_{n\to\infty}\|\langle\xi-x_{n},x_{i}\rangle\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,i\in\mathbb{N}.$ (e 5.218)

Denote by $H_{0}$ the Hilbert $A$-submodule generated by
$\\{x_{1},x_{2},...,x_{n},...,\\}.$ In particular, $x_{n}\in H_{0},$
$n\in\mathbb{N}.$ Let $\eta=\xi|_{H_{0}}.$

Now $H_{0}$ is countably generated and $x_{n}\in H_{0}$ such that

$\displaystyle\lim_{n\to\infty}\|\eta(x_{n})-a\|=\lim_{n\to\infty}\|\xi(x_{n})-a\|=0.$
(e 5.219)

Moreover, if $y=\sum_{i=1}^{m}x_{i}\cdot a_{i},$ where $a_{i}\in A,$ then

$\displaystyle\lim_{n\to\infty}\|\eta(y)-\langle x_{n},y\rangle\|=0.$ (e
5.220)

Since $\\{x_{n}\\}$ is bounded (since $\\{x_{\alpha}\\}$ is bounded), this
implies that

$\displaystyle\lim_{n\to\infty}\|\eta(y)-\langle x_{n},y\rangle\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,y\in H_{0}.$ (e 5.221)

Applying what has been proved, we conclude that $\eta\in H_{0}$ and
$\lim_{n\to\infty}\langle\eta,x_{n}\rangle=\langle\eta,\eta\rangle=a.$

Considering Hilbert $A^{**}$-modules $H_{0}\bullet A^{**}\subset H\bullet
A^{**}.$ By Lemma 3.1, we obtain a projection $P:H^{\sim}\to H_{0}^{\sim}$
such that $P|_{H_{0}\bullet A^{**}}={\rm id}_{H_{0}\bullet A^{**}}.$ Then
$\eta=P(\xi).$ Hence, by the claim,

$\displaystyle\|(1-P)\xi\|^{2}$ $\displaystyle=$
$\displaystyle\|\langle(1-P)(\xi),(1-P)(\xi)\|\leq\|\langle(1-P)(\xi),\xi\rangle\|+\|\langle(1-P)(\xi),P(\xi)\rangle\|$
(e 5.222) $\displaystyle=$ $\displaystyle\|\langle\xi,\xi\rangle-\langle
P(\xi),\xi\rangle\|+0=\|a-\langle
P(\xi),P(\xi)\rangle\|=\|a-\langle\eta,\eta\rangle\|=0.$

In other words, $P(\xi)=\eta=\xi.$ The theorem follows. ∎

######  Definition 5.6.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then
$H^{\sharp}\subset H^{\sim}.$

For each $\xi\in H^{\sharp},$ $\epsilon>0$ and a finite subset $Y\subset
H^{\sharp},$ define

$\displaystyle O_{\xi,\epsilon,Y}=\\{\zeta\in
H^{\sharp}:\|\langle\xi-\zeta,y\rangle\|<\epsilon:y\in H^{\sharp}\\},$ (e
5.223)

where the inner product is taken from $H^{\sharp},$ if $H^{\sharp}$ is a
Hilbert $A$-module, or from $H^{\sim}$ (with value in $A^{**}$).

Denote by ${\cal T}_{NW}$ the topology in $H^{\sharp}$ generated by
$O_{\xi,\epsilon,Y}$ for all $\xi\in H^{\sharp},$
$\epsilon\in\mathbb{R}_{+}\setminus\\{0\\},$ and finite subsets $Y\subset
H^{\sharp}.$ Note that a net $\\{\zeta_{\alpha}\\}$ converges to $\xi$ in
$H^{\sharp}$ in ${\cal T}_{NW},$ if and only if

$\displaystyle\lim_{\alpha}\|\langle\xi-\zeta_{\alpha},y\rangle\|=0$ (e 5.224)

for all $y\in H^{\sharp}.$ where the inner product is the one defined above.

######  Corollary 5.7.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then, with
${\cal T}_{NW},$ the unit ball of $H$ is closed in $H^{\sharp}.$

###### Proof.

Let $\xi\in H^{\sharp}.$ Suppose that there is a net $\\{x_{\alpha}\\}$ in $H$
with $\|x_{\alpha}\|\leq 1$ such that

$\displaystyle\lim_{\alpha}\|\langle\xi-x_{\alpha},\eta\rangle\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,\eta\in H^{\sharp},$ (e 5.225)

where the inner product is in $H^{\sim}.$ Then, for each $x\in H,$
$\lim_{\alpha}\|\langle\xi-x_{\alpha},x\rangle\|=0$ and
$\\{\xi(x_{\alpha})\\}=\\{\langle\xi,x_{\alpha}\rangle\\}$ converges in norm
to $\langle\xi,\xi\rangle.$ By Theorem 5.5, $\xi\in H.$ ∎

######  Corollary 5.8.

Let $A$ be a monotone complete $C^{*}$-algebra and $H$ be a Hilbert
$A$-module. Then the unit ball of $H$ is closed in $H^{\sharp}$ in the
topology ${\cal T}_{NW},$ where we view $H^{\sharp}$ as a self-dual Hilbert
$A$-module.

######  Lemma 5.9.

Let $X$ be a Hilbert space, $A\subset B(X)$ be a $C^{*}$-subalgebra and
$M=\overline{A}^{{SOT}}$ and ${\rm id}_{X}\in M.$ Let $H$ be a Hilbert
$A$-module. Suppose that $\xi\in H\bullet M$ and $\langle\xi,x\rangle\in A$
for all $x\in H.$ Then $\xi\in H.$

###### Proof.

First let us consider the case that $H=H_{A}.$ Then, by Proposition 3.6,

$\displaystyle H_{A}\bullet M=\\{\\{a_{n}\\}:a_{n}\in\overline{AM}\,\,\,{\rm
and}\,\,\,\sum_{k=1}^{n}a_{k}^{*}a_{k}\,\,{\rm converges\,\,in\,\,norm}\\}.$
(e 5.226)

Write $\xi=\\{b_{n}\\}\in H_{A}\bullet M.$ The condition that
$\langle\xi,x\rangle\in A$ for all $x\in H_{A}$ implies that $\xi\in
H_{A}^{\sharp}.$ It follows that $b_{n}\in A.$ Hence $\xi\in H_{A}.$

Next, let us assume that $H$ is countably generated. Let $\xi\in H\bullet M$
and $\langle\xi,x\rangle\in A$ for all $x\in H.$ By Kasparov’s absorbing
theorem, we may write $H_{A}=H\oplus H^{\perp}.$ It follows from what has been
proved that $\xi\in H_{A}.$ Let $P:H_{A}\to H$ be the projection. Then
$P(\xi)\in H.$ However, $\langle\xi-P(\xi),x\rangle=0$ for all $x\in H.$ For
any $y\in H^{\perp},$ Since $\xi\in H\bullet M,$ $\langle\xi,y\rangle=0$ for
all $y\in H.$ Hence $\xi=P(\xi)\in H.$

In general, since $\xi\in H\bullet M,$ there are $x_{n,i}\in H,$
$i=1,2,...,k(n),$ $b_{n,i}\in M,$ $i=1,2,...,k(n),$ $n\in\mathbb{N}$ such that

$\displaystyle\lim_{n\to\infty}\|\xi-\sum_{i=1}^{k(n)}x_{n,i}\bullet
b_{n,i}\|=0.$ (e 5.227)

Let $H_{0}$ be the Hilbert $A$-submodule generated by $\\{x_{n,i}:1\leq i\leq
k(n),\,n\in\mathbb{N}\\}.$ Then $\xi\in H_{0}\bullet M$ and $\xi|_{H_{0}}\in
H_{0}^{\sharp}$ as $\langle\xi,h\rangle\in A$ for all $h\in H_{0}\subset H.$
From what has just been proved, $\xi\in H_{0}\subset H.$ ∎

We end this section with the following result.

######  Theorem 5.10.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then the unit
ball of $H$ is closed in $H^{\sim}$ in the topology ${\cal T}_{NW}$ of
$H^{\sim}=(H\bullet A^{**})^{\sharp}.$

###### Proof.

Let $\\{x_{\alpha}\\}$ be a net in the unit ball of $H$ and $\xi\in H^{\sim}$
such that

$\displaystyle\lim_{\alpha}\|\langle\xi-x_{\alpha},\zeta\rangle\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,\zeta\in H^{\sim}.$ (e 5.228)

Since $H^{\sim}=(H\bullet A^{**})^{\sharp}$ and $H\subset H\bullet A^{**},$ by
applying Corollary 5.8, we conclude that $\xi\in H\bullet A^{**}.$

We also have, for all $y\in H,$

$\displaystyle\lim_{\alpha}\|\langle\xi-x_{\alpha},y\rangle\|=0.$ (e 5.229)

Since $\langle x_{\alpha},y\rangle\in A,$ it follows that
$\langle\xi,y\rangle\in A.$ By Lemma 5.9, $\xi\in H.$ ∎

##  6 A Kaplansky style density theorem in the self-dual Hilbert modules

In the last section, we show that $H$ is closed in $H^{\sharp}$ and in
$H^{\sim}$ in the topology ${\cal T}_{NW}$ of $H^{\sharp},$ and that of
$H^{\sim},$ respectively. In this section, however, we will show that $H$ is
dense in $H^{\sim}$ in a weaker topology. In fact, by Theorem 4.5, it is easy
to show that $H$ is dense in $H^{\sharp}$ in ${\cal T}_{0},$ the topology
defined below (see 6.1). A similar question is whether one can replace $x$ in
(e 6.230) by any element in $H^{\sharp}.$

######  Definition 6.1.

Let $A$ be a $W^{*}$-algebra and $H$ be a Hilbert $A$-module.

Let $\epsilon>0$ and $Y\subset H$ and ${\cal F}\subset A_{*}$ be finite
subsets. Let $\xi\in H^{\sharp}.$ Denote

$\displaystyle O_{\xi,\epsilon,Y,{\cal F}}=\\{\zeta\in
H^{\sharp}:|f(\langle\xi-\zeta,x\rangle)|<\epsilon,\,\,x\in Y,\,\,f\in{\cal
F}\\}\subset H^{\sharp}.$ (e 6.230)

Let ${\cal T}_{0}$ be the topology of $H^{\sharp}$ generated by the subsets
$O_{\xi,\epsilon,Y,{\cal F}}.$

Let $\epsilon>0$ and $Y\subset H^{\sharp}$ and ${\cal F}\subset A_{*}$ be
finite subsets. Let $\xi\in H^{\sharp}.$ Denote

$\displaystyle O_{\xi,\epsilon,Y,{\cal F}}=\\{\zeta\in
H^{\sharp}:|f(\langle\xi-\zeta,x\rangle)|<\epsilon,\,\,x\in Y\\}\subset
H^{\sharp}.$ (e 6.231)

Let ${\cal T}_{w}$ be the topology of $H^{\sharp}$ generated by the subsets
$O_{\xi,\epsilon,Y,{\cal F}}.$

In fact, by Proposition 3.8 of [13] and the definition before it, ${\cal
T}_{w}$ is the weak*-topology of $H^{\sharp}$ as a conjugate space. So a
natural question is whether $H$ is dense in $H^{\sharp}$ in ${\cal T}_{w}.$ To
be more useful (perhaps may not be used twice on Sundays– -cf. 2.3.4. of
[14]), we will also prove a Kaplansky style density theorem as Theorem 6.4.

Let us also consider another topology. Let $\epsilon>0,$ $\xi\in H^{\sharp}$
and ${\cal F}\subset A_{*}$ be a finite subset. Denote by

$\displaystyle O_{\epsilon,\xi,{\cal F}}=\\{\zeta\in
H^{\sharp}:|f(\langle\xi-\zeta,\xi-\zeta\rangle)|<\epsilon,\,\,f\in{\cal
F}\\}.$ (e 6.232)

Let ${\cal T}_{ws}$ be the topology generated by $O_{\epsilon,\xi,{\cal F}}$
for all $\epsilon>0,$ $\xi\in H^{\sharp}$ and finite subsets ${\cal F}\subset
A_{*}.$ Note that ${\cal T}_{ws}$ is stronger than ${\cal T}_{w}$ which is
stronger than ${\cal T}_{0}.$

######  Lemma 6.2.

Let $X$ be a Hilbert space and $A\subset B(X)$ be a $C^{*}$-subalgebra.
Suppose that $M=\overline{A}^{SOT}$ with ${\rm id}_{X}\in M$ and
$b=\\{b_{k}\\}\in H_{M}^{\sharp}.$ There is a net
$a_{\alpha}=\\{(a_{1,{\alpha}},a_{2,{\alpha}},...,a_{n,{\alpha}},...\\}\in
H_{A}$ such that

$\displaystyle\|\sum_{j=1}^{\infty}a_{j,{\alpha}}^{*}a_{j,{\alpha}}\|^{1/2}\leq\|b\|\,\,\,\text{and}\,\,\,$
(e 6.233)
$\displaystyle\lim_{\alpha}f(\sum_{j=1}^{\infty}(b_{j}-a_{j,{\alpha}})^{*}(b_{j}-a_{j,{\alpha}}))=0$
(e 6.234)

for all $f\in M_{*}.$

###### Proof.

Let $Y=l^{2}(X),$ the Hilbert space direct sum of countably many copies of
$X.$ Let ${\bar{b}}=(c_{i,j})\in B(Y),$ where $c_{i,1}=b_{i},$
$i\in\mathbb{N},$ and $c_{i,j}=0,$ if $j\geq 2$ (see (e 3.93)). Denote by
$P_{n}:Y\to X^{(n)}$ the projection, where $X^{(n)}$ is the direct sum of
(first) $n$ copies of $X.$ Let $\epsilon>0$ and $V\in L^{2}(X)$ be a finite
subset. Then there is $n_{0}\in\mathbb{N}$ such that

$\displaystyle\|(1-P_{n_{0}})(v)\|<\epsilon/2(1+\|b\|)\,\,\,{\rm
for\,\,\,all}\,\,\,v\in V.$ (e 6.235)

There is $d\in M_{n_{0}}(A)$ such that

$\displaystyle\|({\bar{b}}-d)(P_{n_{0}}(v))\|<\epsilon/4\,\,\,{\rm
for\,\,\,all}\,\,\,v\in V.$ (e 6.236)

We have

$\displaystyle\|({\bar{b}}-dP_{n_{0}})(v)\|$ $\displaystyle\leq$
$\displaystyle\|({\bar{b}}-dP_{n_{0}})(1-P_{n_{0}})(v)\|+\|({\bar{b}}-d)P_{n_{0}}(v)\|$
(e 6.237) $\displaystyle=$
$\displaystyle\|{\bar{b}}(1-P_{n_{0}})(v)\|+\epsilon/4<\epsilon\,\,\,{\rm
for\,\,\,all}\,\,\,v\in V.$ (e 6.238)

Let $B$ be the self-adjoint algebra of those bounded operators on $Y$ which
can be expressed as infinite matrices with entries in $A$ and all are zero
except finitely many of them. Then, by what has been proved, we conclude that,
in the strong operator topology (of $B(Y)$), operator ${\bar{b}}$ is in the
closure of operators in $B$ in the strong operator topology.

Then, by the Kaplansky density theorem, there is a net $\\{d_{\alpha}\\}\in B$
with $\|d_{\alpha}\|\leq\|{\bar{b}}\|$ such that

$\displaystyle\lim_{\alpha}\|({\bar{b}}-d_{\alpha})v\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,v\in Y.$ (e 6.239)

Since $\\{\|{\bar{b}}-d_{\alpha}\|\\}$ is bounded, we also have

$\displaystyle\lim_{\alpha}\|({\bar{b}}-d_{\alpha})^{*}({\bar{b}}-d_{\alpha})v\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,v\in Y.$ (e 6.240)

We further note that

$\displaystyle\|{\bar{b}}\|^{2}=\|({\bar{b}})^{*}{\bar{b}}\|=\|\sum_{j=1}^{\infty}b_{j}^{*}b_{j}\|\leq\|b\|.$
(e 6.241)

Then

$\displaystyle\lim_{\alpha}\|({\bar{b}}-d_{\alpha})^{*}({\bar{b}}-d_{\alpha})P_{1}v\|=0\,\,\,{\rm
for\,\,\,all}\,\,\,v\in Y.$ (e 6.242)

Note ${\bar{b}}P_{1}={\bar{b}}.$ Let
$d_{\alpha}^{\prime}=d_{\alpha}P_{1}=(d_{i,j,{\alpha}}),$ where
$d_{i,j,{\alpha}}=0$ if $j\geq 2.$ Put $a_{j,{\alpha}}=d_{1,j,{\alpha}},$
$j\in\mathbb{N}.$ Then, for all $n\in\mathbb{N},$

$\displaystyle\|\sum_{j=1}^{n}a_{j,{\alpha}}^{*}a_{j,{\alpha}}\|\leq\|(d_{\alpha}^{\prime})^{*}d_{\alpha}^{\prime}\|=\|d_{\alpha}^{\prime}\|^{2}\leq\|{\bar{b}}\|^{2}\leq\|b\|^{2}.$
(e 6.243)

Put $a_{\alpha}=\\{a_{j,{\alpha}}\\}.$ Since $d_{\alpha}\in B,$ for each
${\alpha},$ there are only finitely many of $a_{j,{\alpha}}$ which are not
zero. Hence $a_{\alpha}\in H_{A}.$ Then $\|a_{\alpha}\|\leq\|b\|.$ Thus (e
6.233) holds. On the other hand, by (e 6.242),

$\displaystyle\lim_{\alpha}\|({\bar{b}}-d_{\alpha}^{\prime})^{*}({\bar{b}}-d_{\alpha}^{\prime})P_{1}v\|=0.$
(e 6.244)

Let $h\in X.$ By (e 6.244),

$\displaystyle\lim_{\alpha}\,\,\,\|\sum_{j=1}^{\infty}(b_{j}-a_{j,{\alpha}})^{*}(b_{j}-a_{j,{\alpha}})h\|=0.$
(e 6.245)

In other words,
$\sum_{i=1}^{\infty}(b_{j}-a_{j,{\alpha}})^{*}(b_{j}-a_{j,{\alpha}})=\langle
b-a_{\alpha},b-a_{\alpha}\rangle\to 0$ in the strong operator topology.
However,

$\displaystyle\|\sum_{j=1}^{n}(b_{j}-a_{j,{\alpha}})^{*}(b_{j}-a_{j,{\alpha}})\|=\|({\bar{b}}-d_{\alpha}^{\prime})\|^{2}\leq(\|\bar{b}\|+\|d_{\alpha}\|)^{2}\leq
4\|b\|^{2}.$ (e 6.246)

Therefore $\sum_{i=1}^{n}(b_{j}-a_{j,{\alpha}})^{*}(b_{j}-a_{j,{\alpha}})\to
0$ in the $\sigma$-weak operator topology and hence in weak*-topology (see,
for example, 4.6.13 of [15]). Therefore (e 6.234) holds. ∎

######  Lemma 6.3.

Let $A\subset B(X)$ be a $C^{*}$-subalgebra and let $M={\bar{A}}^{SOT}$ with
$1_{X}\in M.$ Suppose that $H$ is a countably generated Hilbert $A$-module.
Then $H$ is dense in $(H\bullet M)^{\sharp}$ in the following sense: For any
$\xi\in(H\bullet M)^{\sharp},$ there is a net $x_{\alpha}\in H$ with
$\|x_{\alpha}\|\leq\|\xi\|$ such that

$\displaystyle\lim_{\alpha}\sup\\{|f(\langle\xi-
x_{\alpha},\zeta\rangle)|:\zeta\in(H\bullet M)^{\sharp},\|\zeta\|\leq
1\\}=0\,\,\,{\rm for\,\,\,all}\,\,\,f\in M_{*}.$ (e 6.247)

###### Proof.

Let us first prove this for $H=H_{A},$ even though when $A$ is not
$\sigma$-unital, $H_{A}$ is not countably generated. Lemma 6.2 provides a net
$\\{x_{\alpha}\\}$ in $H_{A}$ with $\|x_{\alpha}\|\leq\|\xi\|$ such that

$\displaystyle\lim_{\alpha}f(\langle\xi-x_{\alpha},\xi-
x_{\alpha}\rangle)=0\,\,\,{\rm for\,\,\,all}\,\,\,f\in M_{*}.$ (e 6.248)

Recall that, for any positive linear functional $f,$ $H_{M}^{\sharp}\times
H_{M}^{\sharp}\to\mathbb{R}$ defined by $[x,y]_{f}=f(\langle x,y\rangle)$ (for
all $x,y\in H_{M}^{\sharp}$) is a pseudo inner product. Therefore, by the
Cauchy-Bunyakovsky-Schwarz inequality,

$\displaystyle f(\langle x,y\rangle)^{2}\leq f(\langle x,x\rangle)f(\langle
y,y\rangle)\,\,\,{\rm for\,\,\,all}\,\,\,x,y\in H_{M}^{\sharp}.$ (e 6.249)

It follows that, for any positive normal linear functional $f,$

$\displaystyle\sup\\{|f(\langle\xi-x_{\alpha},\zeta\rangle)|:\zeta\in
H_{M}^{\sharp},\|\zeta\|\leq 1\\}^{2}$ $\displaystyle\leq$
$\displaystyle\sup\\{f(\langle\zeta,\zeta\rangle)f(\langle\xi-x_{\alpha},\xi-
x_{\alpha}\rangle):\zeta\in H_{M}^{\sharp},\,\|\zeta\|\leq 1\\}$ (e 6.250)
$\displaystyle=$ $\displaystyle\|f\|f(\langle\xi-x_{\alpha},\xi-
x_{\alpha}\rangle)\to 0.$

Thus we proved (e 6.247) holds for $H=H_{A}.$

Now let $H$ be a countably generated Hilbert $A$-module. Then, by Kasparov’s
absorbing theorem, we may write $H_{A}=H\oplus H^{\perp}.$ Hence $H_{A}\bullet
M=H\bullet M\oplus(H^{\perp}\bullet M).$ It follows that
$H_{M}^{\sharp}=(H_{A}\bullet M)^{\sharp}=(H\bullet
M)^{\sharp}\oplus(H^{\perp}\bullet M)^{\sharp}.$ Let
$P:H_{M}^{\sharp}\to(H\bullet M)^{\sharp}$ be the projection such that
$P|_{H}={\rm id}_{H}.$ Then, by what has been proved for $H_{A},$ there is a
net $y_{\alpha}\in H_{A}$ such that $\|y_{\alpha}\|\leq\|\xi\|$ and, for any
$f\in M_{*},$

$\displaystyle\lim_{\alpha}\sup\\{f(\langle\xi-
y_{\alpha},\zeta\rangle):\zeta\in H_{M}^{\sharp},\|\zeta\|\leq 1\\}=0.$ (e
6.251)

Put $x_{\alpha}=P(y_{\alpha})\in(H\bullet M)^{\sharp}.$ Note that
$P(\xi)=\xi.$ Then, for any $f\in M_{*},$

$\displaystyle\lim_{\alpha}\sup\\{f(\langle\xi-
x_{\alpha},\zeta\rangle):\zeta\in(H\bullet M)^{\sharp},\|\zeta\|\leq 1\\}$
$\displaystyle=$ $\displaystyle\lim_{\alpha}\sup\\{f(\langle\xi-
x_{\alpha},P(\zeta)\rangle):\zeta\in(H\bullet M)^{\sharp},\|\zeta\|\leq 1\\}$
$\displaystyle=$ $\displaystyle\lim_{\alpha}\sup\\{f(\langle\xi-
y_{\alpha},\zeta\rangle):\zeta\in(H\bullet M)^{\sharp},\|\zeta\|\leq 1\\}$
$\displaystyle\leq$ $\displaystyle\lim_{\alpha}\sup\\{f(\langle\xi-
y_{\alpha},\zeta\rangle):\zeta\in H_{M}^{\sharp},\|\zeta\|\leq 1\\}=0.$

This proves the lemma. ∎

######  Theorem 6.4.

Let $X$ be a Hilbert space, $A\subset B(X)$ a $C^{*}$-subalgebra and
$M=\overline{A}^{SOT}$ with $1_{M}={\rm id}_{X},$ and let $H$ be a Hilbert
$A$-module. Then the unit ball of $H$ is dense in the unit ball of $(H\bullet
M)^{\sharp}$ in ${\cal T}_{ws}$ (of $(H\bullet M)^{\sharp}$).

###### Proof.

Let $\xi\in(H\bullet M)^{\sharp}$ with $\|\xi\|\leq 1.$ It suffices to show
that, for any $\epsilon>0,$ any finite subset $Y\subset(H\bullet M)^{\sharp}$
and any finite subset ${\cal F}\subset M_{*},$ there is $x\in H$ such that

$\displaystyle\|x\|\leq\|\xi\|\,\,\,{\rm
and}\,\,\,|f(\langle\xi-x,y\rangle)|<\epsilon\,\,\,{\rm
for\,\,\,all}\,\,\,y\in(H\bullet M)^{\sharp},\|y\|\leq 1,\,\,\,{\rm
and}\,\,\,f\in{\cal F}.$ (e 6.253)

Let us fix $\epsilon$ and ${\cal F}.$ Choose an approximate identity
$\\{E_{\lambda}\\}$ for $K(H).$ It follows that $E_{\lambda}\nearrow{\rm
id}_{H}.$ Note that ${\rm id}_{H}\in M(K(H)).$ By the last part of Proposition
2.13, $\Psi_{0}({\rm id}_{H})={\rm id}_{H\bullet M}.$ By 3.17 of [13],
$F\circ\Psi_{0}({\rm id}_{H})={\rm id}_{(H\bullet M)^{\sharp}},$ where $F$ is
the map given by Proposition 2.16. Note also that, by Lemma 2.9,
$\\{\Psi_{0}(E_{\lambda})\\}$ is an approximate identity for $K(H\bullet M).$
In the universal representation of $K(H\bullet M),$ $1-\Psi_{0}(E_{\lambda})$
converges to zero in the strong operator topology. Note that
$\|1-\Psi_{0}(E_{\lambda})\|\leq 1.$ Therefore
$(1-\Psi_{0}(E_{\lambda}))^{*}(1-\Psi_{0}(E_{\lambda}))$ also converges to
zero in the strong operator topology. Hence (since
{$\|(1-\Psi_{0}(E_{\lambda}))^{*}(1-\Psi_{0}(E_{\lambda}))\|\\}$ is bounded,)
it converges to zero in the weak*-topology. By Proposition 2.16, we have, for
all $f\in M_{*},$

$\displaystyle
f(\langle\xi-F\circ\Psi_{0}(E_{\lambda})(\xi),\xi-F\circ\Psi_{0}(E_{\lambda})(\xi)\rangle)$
(e 6.254)
$\displaystyle=f(\langle(1-F\circ\Psi_{0}(E_{\lambda}))^{*}(1-F\circ\Psi_{0}(E_{\lambda}))(\xi),\xi\rangle)\to
0.$ (e 6.255)

Next let $g$ be a positive normal linear functional in $M_{*}.$ Then, for any
$y\in(H\bullet M)^{\sharp}$ with $\|y\|\leq 1,$

$\displaystyle|g(\langle\xi-F\circ\Psi_{0}(E_{\lambda})(\xi),y\rangle)|^{2}$
$\displaystyle\leq$ $\displaystyle
g(\langle\xi-F\circ\Psi_{0}(E_{\lambda})(\xi),\xi-F\circ\Psi_{0}(E_{\lambda})(\xi)\rangle)g(\langle
y,y\rangle)$ (e 6.256) $\displaystyle\leq$
$\displaystyle\|g\|\|y\|^{2}g(\langle\xi-F\circ\Psi_{0}(E_{\lambda})(\xi),\xi-F\circ\Psi_{0}(E_{\lambda})(\xi)\rangle).$
(e 6.257)

Hence, by (e 6.255),

$\displaystyle\lim_{\alpha}\big{(}\sup\\{|g(\langle\xi-F\circ\Psi_{0}(E_{\lambda})(\xi),y\rangle)|:y\in(H\bullet
M)^{\sharp},\,\|y\|\leq 1\\}\big{)}=0.$ (e 6.258)

It follows that, for any $f\in M_{*},$

$\displaystyle\lim_{\alpha}\big{(}\sup\\{|f(\langle\xi-F\circ\Psi_{0}(E_{\lambda})(\xi),y\rangle)|:y\in(H\bullet
M)^{\sharp},\,\|y\|\leq 1\\}\big{)}=0.$ (e 6.259)

Thus we obtain $\lambda_{0}$ such that, for all $\lambda\geq\lambda_{0},$

$\displaystyle|f(\langle\xi-F\circ\Psi_{0}(E_{\lambda})(\xi),y\rangle)|<\epsilon/2\,\,\,{\rm
for\,\,\,all}\,\,\,y\in(H\bullet M)^{\sharp},\,\|y\|\leq 1,\,\,\,{\rm
and}\,\,\,f\in{\cal F}.$ (e 6.260)

Let $H_{\lambda}=\overline{E_{\lambda}(H)}.$ As in the proof of Theorem 3.12,
$H_{\lambda}$ is countably generated. Moreover, by Lemma 3.1,

$(H\bullet M)^{\sharp}=(H_{\lambda}\bullet
M)^{\sharp}\oplus((H_{\lambda}\bullet M)^{\sharp})^{\perp}.$

Let $P_{\lambda}:(H\bullet M)^{\sharp}\to(H_{\lambda}\bullet M)^{\sharp}$ be
the projection. Note that

$\Psi_{\lambda}(E_{\lambda})(\xi),\Psi_{\lambda}(E_{\lambda})(y)\in
P_{\lambda}((H\bullet M)^{\sharp})=(H_{\lambda}\bullet M)^{\sharp}$

for all $y\in(H\bullet M)^{\sharp}.$

It follows from Lemma 6.3 that there is $x\in H_{\lambda}$ with
$\|x\|\leq\|\Psi(E_{\lambda})(\xi)\|\leq\|\xi\|$ such that

$\displaystyle|f(\langle\Psi(E_{\lambda})(\xi)-x,P_{\lambda}(y)\rangle)|<\epsilon/2\,\,\,{\rm
for\,\,\,all}\,\,\,y\in(H_{\lambda}\bullet M)^{\sharp}\,\,\,{\rm
and}\,\,\,\|y\|\leq 1.$ (e 6.261)

Since $P_{\lambda}\Psi(E_{\lambda})=\Psi(E_{\lambda})$ and $x\in H_{\lambda},$
we have, for all $y\in(H_{\lambda}\bullet M)^{\sharp}\,\,\,{\rm
and}\,\,\,\|y\|\leq 1,$

$\displaystyle|f(\langle\Psi(E_{\lambda})(\xi)-x,y\rangle)|$ $\displaystyle=$
$\displaystyle|f(\langle
P_{\lambda}\Psi(E_{\lambda})(\xi)-P_{\lambda}(x),y\rangle)|$ (e 6.262)
$\displaystyle=$
$\displaystyle|f(\langle\Psi(E_{\lambda})(\xi)-x,P_{\lambda}(y)\rangle)|<\epsilon/2.$
(e 6.263)

Thus, for all $y\in(H\bullet M)^{\sharp}$ with $\|y\|\leq 1$ and $f\in{\cal
F},$

$\displaystyle|f(\langle\xi-x,y\rangle)|\leq|f(\langle\xi-\Psi(E_{\lambda})(\xi),y\rangle)|+|f(\langle\Psi(E_{\lambda})(\xi)-x,y\rangle)|<\epsilon/2+\epsilon/2=\epsilon.$
(e 6.264)

∎

The next two statements are the main results of this section:

######  Corollary 6.5.

Let $A$ be a $W^{*}$-algebra and $H$ be a Hilbert $A$-module. Then the unit
ball of $H$ is dense in $H^{\sharp}$ in ${\cal T}_{ws}.$

###### Proof.

Let $M=A$ and then apply Theorem 6.4. ∎

######  Theorem 6.6.

Let $A$ be a $C^{*}$-algebra and $H$ be a Hilbert $A$-module. Then the unit
ball of $H$ is dense in $H^{\sim}$ in ${\cal T}_{ws}$ (as $H^{\sim}=(H\bullet
A^{**})^{\sharp}$).

###### Proof.

We choose the universal representation $\pi_{U}$ and its strong operator
closure $A^{\prime\prime}=A^{**}.$ Then apply Theorem 6.4. ∎

## References

  * [1] L. G. Brown and H. Lin, Projective Hilbert modules and sequential approximations, Science China Math., (2024), https://doi.org/10.1007/s11425-023-2194-x.
  * [2] N. Brown and A. Ciuperca, Isomorphism of Hilbert modules over stably finite $C^{*}$-algebras, J. Funct. Anal. 257 (2009), 332–339.
  * [3] A. Ciuperca and G. A. Elliott, A remark on invariants for $C^{*}$-algebras of stable rank one, Internat. Math. Res. Notices 5 (2008), Art. ID rnm 158, 33 pp.
  * [4] J. B. Conway, A course in operator theory, Grad. Stud. Math., 21, Amer Math. Soc., Providence, RI, 2000. xvi+372 pp. ISBN:0-8218-2065-6.
  * [5] K. Coward, G. Elliott and C. Ivanescu, The Cuntz semigroup as an invariant for $C^{*}$-algebras, J. Reine Angew. Math. 623 (2008), 161–193.
  * [6] I. Kaplansky, Modules over operator algebras, Amer. J. Math. 75 (1953), 839–858.
  * [7] G. Kasparov, Hilbert $C^{\ast}$-modules: theorems of Stinespring and Voiculescu, J. Operator Theory 4 (1980), 133–150.
  * [8] H. Lin, Hilbert $C^{*}$-modules and their bounded module maps, Sci. China Ser. A 34 (1991), 1409–1421.
  * [9] H. Lin, Bounded module maps and pure completely positive maps, J. Operator Theory 26  (1991), 121–138.
  * [10] H. Lin, Injective Hilbert $C^{\ast}$-modules, Pacific J. Math. 154 (1992), 131–164.
  * [11] H. Lin, An introduction to the classification of amenable $C^{*}$-algebras, World Scientific Publishing Co., Inc., River Edge, NJ, 2001. xii+320 pp. ISBN: 981-02-4680-3.
  * [12] E. Ortega, M. Rørdam and H. Thiel, The Cuntz semigroup and comparison of open projections, J. Funct. Aanal., 260 (2011), 3474–3493.
  * [13] W. Paschke, Inner product modules over $B^{*}$-algebras, Trans. Amer. Math. Soc, 182 (1973), 443–468.
  * [14] G. K. Pedersen, $C^{*}$-algebras and their Automorphism Group, L.M.S monographs, Academic Press, London, New York, San Francisco, 1979.
  * [15] G. K. Pedersen, Analysis Now, Graduate Text Book in Mathematics, Springer -Verlag New York 1989, ISBN 0-387-96788-5, ISBN 3-540-96788-5.
  * [16] S. Baaj and G. Skandalis, Unitaires multiplicatifs et dualité pour les produits croisés de C*-algèbres, Annales Scientifiques de l’École Normale Supérieure. 26 (4) (1993), 425–488.