Taken from C*-algebras and Operator Theory by Gerald Murphy.
We prepare the way for the density theorem with some useful results on strong convergence.
Theorem 4.3.1
If is a Hilbert space, the involution is strongly continuous when restricted to the set of normal operators of .
Proof
Let and suppose that are normal operators in . Then
If is a net of normal operators strongly convergent to a normal operator , then the net is convergent to and the net is convergent to , so is convergent to . Therefore, is strongly convergent to
Remark 4.3.1
If is a bounded subset of , where is a Hilbert space, then the map
, ,
is strongly continuous. The proof of this is the inequality
.
We say that a continuous function from to is strongly continuous if for every Hilbert space and each net of Hermitian operators on converging strongly to a Hermitian operator , we have converging strongly to .
Theorem 4.3.2
If is a continuous bounded function, then is strongly continuous.
Proof
Let denote the set of strongly continuous functions. This is clearly a vector space (with the pointwise-defined operations), and it follows from Remark 4.3.1 that if belong to and one of them is bounded, then .
We show first that : Let . It is easy to verify that is a closed subalgebra of , and by Theorem 4.3.1 it is self-adjoint. If is the inclusion function, then
,
and belong to and , . We show that . Let be a Hilbert space and suppose that are hermitian operators on . Then
,
,
.
Therefore, if ,
,
,
,
since and . Hence, is strongly continuous, and therefore . Since , we have (Remark 4.3.1), so , and therefore . The set separates the points of (Let . If either are , we have no problem as attains a unique max at . If are both positive/ negative, then separates them because is strictly increasing/ decreasing for positive/ negative numbers. If differ in sign, then separates them because is odd.), and for all , so by the Stone-Weierstrass theorem (Louis De Branges) the C*-subalgebra generated by and is . Hence, .
Now suppose that . Then , so , and therefore also. Consequently
If is a convex set of operators on a Hilbert space , then its strong and its weak closures coincide, by Theorem 4.2.7. If is a *-subalgebra of , then its weak closure is a von Neumann algebra. These observations are used in the proof of the following theorem, which is known as the Kaplansky Density Theorem.
The Kaplansky Density Theorem
Let be a Hilbert space and a C*-subalgebra of with strong closure .
- is strongly dense in .
- The closed unit ball of is strongly dense in the closed unit ball of .
- The closed unit ball of is strongly dense in the closed unit ball of .
- If contains , then the unitaries of are strongly dense in the unitaries of .
Proof
If , then there is a net in strongly convergent to , so converges weakly to , and therefore is weakly convergent to . Hence, is in the weak closure of , and therefore in the strong closure of this set, since its weak and strong closures coincide (As is convex: Theorem 4.2.7). This proves condition 1.
Suppose now that is in the closed unit ball of . Then by condition 1. there is a net in strongly convergent to . The function defined by setting for and elsewhere belongs to , and therefore by Theorem 4.3.2, it is strongly continuous. Hence, is strongly convergent to . But clearly, , since . Moreover, is in the closed unit ball of is in the closed unit ball of for all indices , since and . This proves condition 2.
The algebra is a C*-subalgebra of , and is strongly dense in the von Neumann algebra
(Let
Let ordered by reverse inclusion. Having is equivalent to the existence of a net such that in the strong operator topology. This is the case because for all , is a strongly open neighbourhood of which implies that , so just choose any in here. Then consider
.
Now for all ;
).
If is in the closed unit ball of , then
is a hermitian operator on lying in the strong closure of , and since ; it follows from condition 2. that there is a net in the closed unit ball of that strongly converges to . Hence, is strongly convergent on to , and is in the closed unit ball of for all indices . Thus condition 3. is proved.
Suppose now that contains and let and denote the unitaries of and respectively. If , then by Remark 4.2.2 there is a hermitian element of such that . By condition 1. there is a net in strongly convergent to . However, the function
, ,
is strongly continuous by Theorem 4.3.2, so converges strongly to . Since and , condition 4. is proved
Theorem 4.3.4
Let and be Hilbert spaces, a von Neumann algebra on , and a weakly continuous *-homomorphism. Then is a von Neumann algebra on .
Proof
Observe first that is a C*-algebra on . By applying Remark 4.1.2, we may suppose that contains .
Let ans suppose that , so there is a number such that . Write for some element and let be the polar decomposition of . By Theorem 4.1.10 . Let be the spectral resolution of the identity for and . Then ,
(A spectral resolution of the identity for a self-adjoint are such that for all
is a (probability) measure on such that
.
Now is as follows:
Now
.
In turn, with ;
.
),
and . Hence , so
.
Therefore, , so , so , since is a projection. Consequently, . Also
.
Set
.
We have shown that (). The closed unit ball of , is weakly compact by Theorem 4.2.4, so is weakly compact. Observe that is the weak closure of (To say this we must see that that is a weakly closed set such that . Then if is another weakly closed set such that , then we have . This follows from the fact that the weak topology is weaker than the norm topology.). We claim that is the closed unit ball of . Let be an element of the closed unit ball of and take a sequence converging to . Now for some , so and . Hence, for some . Thus, is a sequence in converging in norm to , so . This shows that the closed unit ball of is contained in , and the reverse inclusion is obvious.
Now let be a non-zero element of the weak closure of in . By the Kaplansky density theorem, is in the weak closure of the unit ball of (), and therefore , so . This shows that the weak closure of is equal to , so is a von Neumann algebra
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