Proof of (i)

By definition of the weak operator topology, for any x and y in H, the map T → <Tx, y> is continuous in this topology. Therefore, for any operator O (and by substituting once y → O^∗y and once x → Ox), so is the map

${\displaystyle T\to \langle Tx,O^{*}y\rangle -\langle TOx,y\rangle =\langle OTx,y\rangle -\langle TOx,y\rangle .}$

Let S be any subset of L(H), and S′ its commutant. For any operator T not in S′, <OTx, y> - <TOx, y> is nonzero for some O in S and some x and y in H. By the continuity of the abovementioned mapping, there is an open neighborhood of T in the weak operator topology for which this is nonzero, therefore this open neighborhood is also not in S′. Thus S′ is closed in the weak operator, i.e. S′ is weakly closed. Thus every commutant is weakly closed, and so is M′′; since it contains M, it also contains its weak closure.

Proof of (ii)

This follows directly from the weak operator topology being coarser than the strong operator topology: for every point x in cl_S(M), every open neighborhood of x in the weak operator topology is also open in the strong operator topology and therefore contains a member of M; therefore x is also a member of cl_W(M).

Proof of (iii)

Fix X ∈ M′′. We will show X ∈ cl_S(M).

Fix an open neighborhood U of X in the strong operator topology. By definition of the strong operator topology, U contains a finite intersection U(h₁,ε₁) ∩...∩U(h_n,ε_n) of subbasic open sets of the form U(h,ε) = {O ∈ L(H): ||Oh - Xh|| < ε}, where h is in H and ε > 0.

Fix h in H. Consider the closure cl(Mh) of Mh = {Mh : M ∈ M} with respect to the norm of H and equipped with the inner product of H. It is a Hilbert space (being a closed subspace of a Hilbert space H), and so has a corresponding orthogonal projection which we denote P. P is bounded, so it is in L(H). Next we prove:

Lemma. P ∈ M′.

Proof. Fix x ∈ H. Then Px ∈ cl(Mh), so it is the limit of a sequence O_nh with O_n in M for all n. Then for all T ∈ M, TO_nh is also in Mh and thus its limit is in cl(Mh). By continuity of T (since it is in L(H) and thus Lipschitz continuous), this limit is TPx. Since TPx ∈ cl(Mh), PTPx = TPx. From this it follows that PTP = TP for all T in M.

By using the closure of M under the adjoint we further have, for every T in M and all x, y ∈ H:

${\displaystyle \langle x,TPy\rangle =\langle x,PTPy\rangle =\langle Px,TPy\rangle =\langle T^{*}Px,Py\rangle =\langle PT^{*}Px,y\rangle =\langle T^{*}Px,y\rangle =\langle Px,Ty\rangle =\langle x,PTy\rangle }$

thus TP = PT for all T ∈ M, meaning P lies in M′.

Since M is unital, h ∈ Mh, hence XPh = Xh. By definition of the bicommutant, XP = PX, so XPh = PXh ∈ cl(Mh). We thus see that Xh ∈ cl(Mh). Thus for every ε > 0, there exists T in M with ||Xh − Th|| < ε. Then T lies in U(h,ε).

Thus in every open neighborhood U of X in the strong operator topology there is a member of M, and so X is in the strong operator topology closure of M.

Non-unital case

A C*-algebra M acting on H is said to act non-degenerately if for h in H, Mh = {0} implies h = 0. In this case, it can be shown using an approximate identity in M that the identity operator I lies in the strong closure of M. Therefore, the conclusion of the bicommutant theorem holds for M.