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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