proof of von Neumann double commutant theorem

Lemma - Let $H$ be a Hilbert space and $B(H)$ its algebra of bounded operators. Let $𝒩$ be a *-subalgebra of $B(H)$ that contains the identity operator and is closed in the strong operator topology. If $T\in {𝒩}^{\prime \prime }$, the double commutant of $𝒩$, then for each $x\in H$ there is an operator $A\in 𝒩$ such that .

: Let $\overline{𝒩x}\subseteq H$ be the closure of the subspace $𝒩x:=\{Sx:S\in 𝒩\}$. It is clear that $𝒩x$ is an invariant subspace for $𝒩$, hence so is its closure $\overline{𝒩x}$ (see this entry (http://planetmath.org/InvariantSubspacesForSelfAdjointAlgebrasOfOperators), Proposition 5).

Let $P$ be the orthogonal projection onto $\overline{𝒩x}$. Since $\overline{𝒩x}$ is invariant for $𝒩$, we have that $P\in {𝒩}^{\prime }$ (see this entry (http://planetmath.org/InvariantSubspacesForSelfAdjointAlgebrasOfOperators), last theorem). Since $𝒩$ contains the identity operator, we know that $x$ belongs to $\overline{𝒩x}$. Hence,

$Tx=TPx=PTx$

where the last equality comes from the fact that $T\in {𝒩}^{\prime \prime }$ and $P\in {𝒩}^{\prime }$. Thus, we see that $Tx\in \overline{𝒩x}$, which implies that there exists an $A\in 𝒩$ such that . $\mathrm{\square }$

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$(1)⟹(2)$ Since ${ℳ}^{\prime \prime }$ is the commutant of some set, namely it is the commutant of ${ℳ}^{\prime }$, it follows that ${ℳ}^{\prime \prime }$ is closed in the weak operator topology (see this entry (http://planetmath.org/CommutantIsAWeakOperatorClosedSubalgebra)). But we are assuming that $ℳ={ℳ}^{\prime \prime }$, hence $ℳ$ is closed in the weak operator topology.

$(2)⟹(3)$ This part is obvious since the weak operator topology is weaker than the strong operator topology.

$(3)⟹(1)$ Suppose $ℳ$ is closed in the strong operator topology.

A subset of $B(H)$ is always contained in its double commutant, thus $ℳ\subseteq {ℳ}^{\prime \prime }$. So it remains to prove the opposite inclusion.

Let $T\in {ℳ}^{\prime \prime }$. We are going to prove that $T$ belongs to the strong operator closure of $ℳ$, and since $ℳ$ is closed under this topology, it will follow that $T\in ℳ$.

Recall that the strong operator topology is the topology in $B(H)$ generated by the family of seminorms $\parallel \cdot {\parallel }_x,x\in H$ defined by ${\parallel S\parallel }_x:=\parallel Sx\parallel$. A local base around $T$, in this topology, consists of sets of the form

$V(x_1,\mathrm{\dots },x_n;ϵ):=\{S\in B(H):\parallel (S-T)x_i\parallel \le ϵ,i=1,\mathrm{\dots },n\},x_1,\mathrm{\dots },x_n\in H,ϵ>0$

We can however consider $ϵ$ to be $1$, since $V(x_1,\mathrm{\dots },x_n;ϵ)=V({ϵ}^{-1}{x}_1,\mathrm{\dots },{ϵ}^{-1}{x}_n;1)$.

For every $x_1,\mathrm{\dots },x_n\in H$ we want to find $A\in ℳ$ such that $A\in V(x_1,\mathrm{\dots },x_n;\mathrm{\hspace{0.17em}1})$, i.e. such that , for each $i$.

Let $\tilde{H}$ be the direct sum of Hilbert spaces $\tilde{H}:={\oplus }_{i=1}^{n}H$. For every $A\in B(H)$ let $\tilde{A}\in B(\tilde{H})$ be the direct sum of bounded operators (http://planetmath.org/DirectSumOfBoundedOperatorsOnHilbertSpaces) $\tilde{A}:={\oplus }_{i=1}^{n}A$, i.e.

$\tilde{A}(y_1,\mathrm{\dots },y_n)=(A{y}_1,\mathrm{\dots },A{y}_n),y_1,\mathrm{\dots },y_n\in H$

We have that $𝒩:=\{\tilde{A}:A\in ℳ\}$ is a *-algebra of bounded operators in $\tilde{H}$.

Claim 1 - $\tilde{T}\in {𝒩}^{\prime \prime }$.

The algebra $B(\tilde{H})$ can be canonically identified with the algebra of $n\times n$ matrices with entries in $B(H)$, and $𝒩$ corresponds to the diagonal matrices with an element $A\in ℳ$ in the diagonal. Thus, it is easy to check that ${𝒩}^{\prime }$ is precisely the set of matrices whose entries belong to ${ℳ}^{\prime }$.

Since the unit matrices (http://planetmath.org/UnitMatrix) belong to ${𝒩}^{\prime }$, it follows that ${𝒩}^{\prime \prime }$ consists solely of diagonal matrices with one element on the diagonal (see this entry (http://planetmath.org/CentralizerOfMatrixUnits)). It is easy to check that ${𝒩}^{\prime \prime }$ is precisely the set of diagonal matrices with one element of ${ℳ}^{\prime \prime }$ in the diagonal. Hence, we conclude that $\tilde{T}\in {𝒩}^{\prime \prime }$, and Claim 1 is proved.

Now, we observe that $𝒩$ is a *-subalgebra of $B(\tilde{H})$ that contains the identity operator. Since $ℳ$ is closed in the strong operator topology, it follows easily that $𝒩$ is also closed in the strong operator topology. Since $\tilde{T}\in {𝒩}^{\prime \prime }$, Lemma 1 that for each $\tilde{x}:=(x_1,\mathrm{\dots },x_n)\in \tilde{H}$ there exists an operator $\tilde{A}\in 𝒩$ such that . But this is implies that for each $1\le i\le n$.

Thus, $T\in V(x_1,\mathrm{\dots },x_n;\mathrm{\hspace{0.17em}1})$. Hence we conclude that $T$ belongs to the operator closure of $ℳ$, but since $ℳ$ is closed under this topology, $T\in ℳ$.

We conclude that ${ℳ}^{\prime \prime }=ℳ$. $\mathrm{\square }$