Proof
Let H be a Hilbert space and L(H) the bounded operators on H. Consider a self-adjoint subalgebra M of L(H). Suppose also, M contains the identity operator on H.
As stated above, the theorem claims the following are equivalent:
- i) M = M′′.
- ii) M is closed in the weak operator topology.
- iii) M is closed in the strong operator topology.
The adjoint map T → T* is continuous in the weak operator topology. So the commutant S’ of any subset S of L(H) is weakly closed. This gives i) ⇒ ii). Since the weak operator topology is weaker than the strong operator topology, it is also immediate that ii) ⇒ iii). What remains to be shown is iii) ⇒ i). It is true in general that S ⊂ S′′ for any set S, and that any commutant S′ is strongly closed. So the problem reduces to showing M′′ lies in the strong closure of M.
For h in H, consider the smallest closed subspace Mh that contains {Mh| M ∈ M}, and the corresponding orthogonal projection P.
Since M is an algebra, one has PTP = TP for all T in M. Self-adjointness of M further implies that P lies in M′. Therefore for any operator X in M′′, one has XP = PX. Since M is unital, h ∈ Mh, hence Xh∈ Mh and for all ε > 0, there exists T in M with ||Xh - Th|| < ε.
Given a finite collection of vectors h1,...hn, consider the direct sum
The algebra N defined by
is self-adjoint, closed in the strong operator topology, and contains the identity operator. Given a X in M′′, the operator
lies in N′′, and the argument above shows that, all ε > 0, there exists T in M with ||Xh1 - Th1||,...,||Xhn - Thn|| < ε. By definition of the strong operator topology, the theorem holds.
Read more about this topic: Von Neumann Bicommutant Theorem
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