Von Neumann Bicommutant Theorem - Proof

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,...h_n, 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||,...,||Xh_n - Th_n|| < ε. By definition of the strong operator topology, the theorem holds.