Taken from C*-algebras and Operator Theory by Gerald Murphy.
Preparatory to our introduction of the weak and ultraweak topologiesm we show now that is the dual of
, and
is the dual of
.
Let be a Hilbert space, and suppose that
. It follows from Theorem 2.4.16 (https://jpmccarthymaths.wordpress.com/2011/01/18/c-algebras-and-operator-theory-2-4-compact-hilbert-space-operators/) that the function
,
,
is linear and bounded, and . We therefore have a map
,
,
which is clearly linear and norm-decreasing. We call this map the canonical map from to
.
Theorem 4.2.1
If is a Hilbert space, then the canonical map from
to
is an isometric linear isomorphism.
Proof
Let denote this map. If
, then
, so
(a quick calculation). Thus,
is injective.
Now suppose that . Then the function
,
,
is a sesquilinear form on and
. Hence, there is a unique operator
on
such that
(
) (Theorem 2.3.6). Also,
. Let
be an orthonormal basis for
and let
be the polar decomposition of
. If
is finite, set
(so
is a projection). Then
.
Hence, , so
. If
, then
, so
equals
on
and therefore on
; that is,
. Therefore,
is an isometric linear isomorphism
Corollary 4.2.2
is a Banach *-algebra under the trace-class norm.
Proof
It is a dual space and therefore complete
Suppose again that is a Hilbert space and suppose that
. Then the function
,
,
is linear and bounded, and , by Theorem 2.4.16 (https://jpmccarthymaths.wordpress.com/2011/01/18/c-algebras-and-operator-theory-2-4-compact-hilbert-space-operators/). We call the norm-decreasing linear map
,
, the canonical map from
to
.
Theorem 4.2.3
If is a Hilbert space, then the canonical map from
to
is an isometric linear isomorphism.
Proof
Let denote this map. If
, then
, that is
, so
, for all
. Hence,
, and therefore
is injective.
Now let . The map
,
,
is a sesquilinear form on , and
. Hence, there is a unique operator
such that
for all
. Also,
. Now
, so
equals
on
, and therefore on
(by Theorem 2.4.17 https://jpmccarthymaths.wordpress.com/2011/01/18/c-algebras-and-operator-theory-2-4-compact-hilbert-space-operators/). Therefore,
is an isometric linear isomorphism
If is a Hilbert space, the Hausdorff locally convex topology on
generated be the separating family of semi-norms (A family of semi-norms for
is a set of semi-norms
such that for all non-zero
, there exists a
such that
.):
,
, for
is called the weak operator topology on . If
is a net in
, then
converges weakly to an operator
if and only if
for all
.
The ultraweak or –weak topology on
is the Hausdorff locally convex topology on
generated by the semi-norms
,
, where
.
The weak topology is weaker than the ultraweak topology (!). For if is a net converging ultraweakly to and operator
, then for each
the net
converges to
, so
converges to
weakly (
). Clearly, the weak topology is also weaker than the strong topology (
).
The operations of addition and scalar multiplication are of course () weakly and ultraweakly continuous, but it is easy to see that the involution operation is
is also continuous for these topologies (
converges to
if and only if
for all
. So then
and hence
.). We showed in Example 4.1.1 (https://jpmccarthymaths.wordpress.com/2011/06/24/von-neumann-algebras-the-double-commutant-theorem/) that the involution is not strongly continuous in general, so the weak and strong topologies do not coincide in general.
Continuity of multiplication in the weak topology does not hold in general. Let be a Hilbert space with an orthonormal basis
and let
be the unilateral shift on this basis. Then the sequences
and
both converge weakly to zero, but the product sequence
is the constant
.
We have seen that for an arbitrary Hilbert space , the Banach space
is the dual of
. It is clear from this that the ultraweak topology is just the weak* topology on
. Hence, the closed unit ball of
is ultraweakly compact, by the Banach-Alaoglu theorem (http://en.wikipedia.org/wiki/Banach-Alaoglu_theorem#Generalization:_Bourbaki.E2.80.93Alaoglu_theorem).
Theorem 4.2.4
If is a Hilbert space, then the relative weak and ultraweak topologies on the closed unit ball of
coincide, and hence the ball is weakly compact.
Proof
The identity map from the ball with the ultraweak topology to the ball is a continuous bijection from a compact space to a Hausdorff space and is therefore a homeomorphism
Remark 4.2.1
The closed unit ball of is not strongly compact in general. For if we suppose the contrary, then the identity map of the ball with the relative strong topology to the ball with the relative weak topology is a continuous bijection from a compact space to a Hausdorff space, and therefore a homeomorphism. This means that the relative strong and weak topologies on the ball coincide. The involution operation
is weakly continuous, but it follows from Example 4.1.1 that in general this operation is not strongly continuous when restricted to the ball. We therefore have a contradiction.
If is a set of operators on a Hilbert space
, then the weak closure of
is contained in
. For if
is in this weak closure, there is a net
in
converging weakly to
, and therefore if
, we have
,
so and
commute so
.
Theorem 4.2.5
Suppose that is a *-algebra on a Hilbert space
containing
.
- The weak closure of
is
.
is a von Neumann algebra if and only if it is weakly closed.
Proof
This is immediate from the preceding remark and Theorem 4.1.5 (https://jpmccarthymaths.wordpress.com/2011/06/24/von-neumann-algebras-the-double-commutant-theorem/). In fact, this result is basically just the completion of that theorem
Theorem 4.2.6
Let be a Hilbert space and
a linear functional on
. The following are equiavlent:
is weakly continuous.
is strongly continuous.
- There are vectors
and
such that, for all
Proof
The implications 3. implies 1. and 1. implies 2. are clear (Let be a net in
weakly convergent to
. Now
,
as addition is weakly continuous. So is weakly continuous.). To show that 2. implies 3., suppose that
is strongly continuous. Then by Theorem A.1 (same as Excercise 5.1 from http://www.maths.lancs.ac.uk/~belton/www/notes/fa_notes.pdf), there is a positive number
ans there exist
such that
for all
. We may suppose that
. Then, for any
:
Let be the subspace of
consisting of all
, as
runs over
, and let
be its norm closure. The function
,
,
is well defined (If , do we have
? By linearity this is equivalent to asking if we have
, do we have
? But
so we are done.),
linear, and bounded, with , so it extends to a linear norm-decreasing functional on
which we also denote by
. By the Riesz representation theorem for linear functionals on Hilbert spaces, there is a unique element
such that
for all
. Hence, for all
This shows 2. implies 3
Theorem 4.2.7
Let be a Hilbert space and
a convex subset of
. Then
is strongly closed if and only if it is weakly closed.
Proof
Since the weak topology is weaker than the strong topology, a weakly closed set is strongly closed. Suppose, therefore, that is strongly closed, and let
be a point in the weak closure. Then there is a net of operators
in
converging to
weakly, and hence, for every weakly continuous linear functional
on
, we have
. By Theorem 4.2.6, the weakly continuous functionals on
are the same as the strong continuous, so by Corollary A.8 (Let
be a convex set in a locally convex space
. Then for any point
,
(norm closure) if and only if there is a net
in
such that
converges to
for all continuous linear functionals
on
.),
is in the strong closure of
; that is
. Hence
is weakly closed
Corollary 4.2.8
If is a *-algebra on
, then
is a von Neumann algebra if and only if it is weakly closed.
Proof
Immediate, since is convex
We show now that a von Neumann algebra is the dual space of a Banach space. This is not true for arbitrary C*-algebras. In fact, by a theorem of Sakai (Theorem 1.16.7), every C*-algebra that is a dual space is isomorphic to a von Neumann algebra.
Suppose that is a von Neumann algebra on a Hilbert space
. We set
.
This is a vector subspace of , closed with respect to the trace-class norm (because it is the intersection of the kernels of the continuous linear maps
,
?). Set
. Then
is a Banach space when endowed with the quotient norm (corresponding to the trace-class norm
). If
, we have a well-defined bounded linear functional
,
.
The map
,
,
is clearly norm-decreasing and linear. We call it the canonical map from to
.
Theorem 4.2.9
Let be a von Neumann algebra on a Hilbert space
. Then the canonical map from
to
is an isometric linear isomorphism.
Proof
Let and
be the canonical maps. By Theorem 4.2.3,
is an isometric linear isomorphism. If
, then
(If
then
for all
which implies that
.) Thus
is injective. If
, and
is the quotient map from
to
, then
, so
for some
. To show that
, we need only show that
for all
, using the fact that
is strongly closed, the characterisation of strongly continuous linear functionals on
given in Theorem 4.2.6, and Corollary A.9 (“Let
be a closed vector subspace of a locally convex space
and
. Then there is a continuous linear functional
on
such that
and
.” Suppose
for all
and that
. Then there exists a strong operator topology continuous
such that
but
. But there exist
,
such that
.
So then
and
for all
.
However let and by definition in
. Now we claim that
We have that
Now we can write and
. As
is an orthonormal basis
and
and the result follows.
Now
,
a contradiction so as required.) If
, then
(
). , so
. Therefore,
is a bijection. Observe also that if
, then there exists
such that
and
, so
. Since
was arbitrary, this shows that
. It follows that
is isometric
It is easy to check that the weak* topology on is just the relative ultraweak topology on
.
Theorem 4.2.10
Let be a von Neumann algebra on a Hilbert space
, and let
be a linear functional. Then
is ultraweakly continuous if and only if there exists
such that
for all
.
Proof
This follows from the identification of , the remark preceding this theorem, and Theorem A.2 (Let
be a normed vector space. Then a linear functional
is weak* continuous if and only if
for some
.)
Remark
Let be a von Neumann algebra on a Hilbert space
containing
. If
is a normal element of
, then for each
the element
is in
. This is because
is in the closed linear span of the characteristic functions on
, and if
is a Borel set of
, then the spectral projection
belongs to
by Theorem 4.1.11. From this and the proof of Theorem 2.5.8, it follows that if
is a unitary, then
for some hermitian element
.
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July 11, 2011 at 12:38 pm
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October 5, 2011 at 2:13 pm
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