Proof of Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation

Each result cited below is universally quantified over the data in its own statement. The following facts are used throughout.

(Ops) By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations, sums, scalar multiples and composites of elements of $\mathcal{L}(H)$ lie in $\mathcal{L}(H)$, and so does $I$. Composition distributes over sums and commutes with scalar multiples. Composition of maps is associative.

(Adj) Let $A,B\in\mathcal{L}(H)$. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint, $A$ has an adjoint $A^{*}\in\mathcal{L}(H)$, so $\langle A^{*}\eta,\xi\rangle=\langle\eta,A\xi\rangle$ for all $\xi,\eta\in H$ (Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint). By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, $A$ is the adjoint of $A^{*}$. Hence $(A^{*})^{*}=A$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique, and $\langle A\eta,\xi\rangle=\langle\eta,A^{*}\xi\rangle$. Again by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, $B^{*}A^{*}$ is the adjoint of $AB$.

(Def) If $w\in H$ and $\langle w,w\rangle=0$, then $w=0$, by condition 4 of Complex Inner Product Space.

Proof of clause 1 (Algebra). We have $I\in\mathcal{L}(H)$ and $IA=A=AI$ for every $A$, so $I\in\mathcal{S}'$. Let $S,T\in\mathcal{S}'$, $c\in\mathbb{C}$ and $A\in\mathcal{S}$. By (Ops), $S+T$, $cS$ and $ST$ lie in $\mathcal{L}(H)$, and

Now assume $\mathcal{S}^{*}\subseteq\mathcal{S}$, and let $S\in\mathcal{S}'$ and $A\in\mathcal{S}$. Then $A^{*}\in\mathcal{S}$, so $A^{*}S=SA^{*}$. By (Adj), the adjoint of $A^{*}S$ is $S^{*}(A^{*})^{*}=S^{*}A$ and the adjoint of $SA^{*}$ is $(A^{*})^{*}S^{*}=AS^{*}$. These are adjoints of the same map, so $S^{*}A=AS^{*}$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique. Since $S^{*}\in\mathcal{L}(H)$, this shows $S^{*}\in\mathcal{S}'$.

Proof of clause 2 (Order). Let $\mathcal{S}\subseteq\mathcal{T}$ and $T\in\mathcal{T}'$. Then $T\in\mathcal{L}(H)$ commutes with every element of $\mathcal{T}$, in particular with every element of $\mathcal{S}$, so $T\in\mathcal{S}'$.

For $A\in\mathcal{S}$ we have $A\in\mathcal{L}(H)$, and $AT=TA$ for every $T\in\mathcal{S}'$ by the definition of $\mathcal{S}'$. So $A\in(\mathcal{S}')'=\mathcal{S}''$, which proves $\mathcal{S}\subseteq\mathcal{S}''$.

Applying this inclusion to the set $\mathcal{S}'$ gives $\mathcal{S}'\subseteq((\mathcal{S}')')'=(\mathcal{S}'')'=\mathcal{S}'''$. Applying order reversal to $\mathcal{S}\subseteq\mathcal{S}''$ gives $\mathcal{S}'''=(\mathcal{S}'')'\subseteq\mathcal{S}'$. Hence $\mathcal{S}'''=\mathcal{S}'$.

Proof of clause 3 (Weak limits). Let $A\in\mathcal{S}$ and $\xi,\eta\in H$. Apply the hypothesis to the vectors $A\xi,\eta$: the sequence $\langle\eta,T_kA\xi\rangle$ converges to $\langle\eta,TA\xi\rangle$. Apply it to $\xi,A^{*}\eta$: the sequence $\langle A^{*}\eta,T_k\xi\rangle$ converges to $\langle A^{*}\eta,T\xi\rangle=\langle\eta,AT\xi\rangle$, where the equality is (Adj). Since $T_k\in\mathcal{S}'$, (Adj) gives, for every $k$,

So the two sequences are the same sequence in $\mathbb{C}$. By Uniqueness of Limits in a Metric Space in the metric space $(\mathbb{C},|\cdot-\cdot|)$ (a metric by claim 9 of Properties of Complex Conjugation and Modulus), $\langle\eta,TA\xi\rangle=\langle\eta,AT\xi\rangle$. Now put $w=TA\xi-AT\xi$. Conditions 2 and 3 of Complex Inner Product Space give $\langle\eta,w\rangle=0$ for every $\eta$, and $\eta=w$ gives $w=0$ by (Def). Hence $TA=AT$ for every $A\in\mathcal{S}$, and $T\in\mathcal{S}'$ because $T\in\mathcal{L}(H)$.

For the final sentence, suppose $T_k\to T$ in operator norm, and let $\xi,\eta\in H$. Conditions 2 and 3 of Complex Inner Product Space, claim 1 of The Induced Norm is a Norm, and Induces a Metric and Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound give

Given $\varepsilon>0$, put $M=\lVert\eta\rVert\,\lVert\xi\rVert+1$. By Limit of a Sequence of Real Numbers there is $N$ with $\lVert T_k-T\rVert_{\mathrm{op}}<\varepsilon/M$ for $k\ge N$, and then the left-hand side is below $\varepsilon$. So $\langle\eta,T_k\xi\rangle\to\langle\eta,T\xi\rangle$ for all $\xi,\eta$, and $T\in\mathcal{S}'$ by what was just proved.

Proof of clause 4 (Conjugation). By Conjugation of a Complex Hilbert Space §conjugation, $J$ is additive, $J(c\xi)=\overline{c}\,J\xi$, $J(J\xi)=\xi$ and $\langle J\xi,J\eta\rangle=\langle\eta,\xi\rangle$. Let $X,Y:H\to H$ be maps. For every $\xi\in H$, by Conjugation of a Complex Hilbert Space §conjugated, $(JXJ)(JYJ)\xi=J(X(J(J(Y(J\xi)))))=J(X(Y(J\xi)))=J(XY)J\xi$ and $J(JXJ)J\xi=J(J(X(J(J\xi))))=X\xi$, because $J(J\eta)=\eta$ for every $\eta\in H$ (Conjugation of a Complex Hilbert Space §conjugation). Hence every pair of maps $X,Y:H\to H$ satisfies

Also $\lVert J\xi\rVert^{2}=\langle J\xi,J\xi\rangle=\langle\xi,\xi\rangle=\lVert\xi\rVert^{2}$, so $\lVert J\xi\rVert=\lVert\xi\rVert$ by the uniqueness of nonnegative square roots in Norm Induced by a Complex Inner Product.

Let $A,B\in\mathcal{L}(H)$ and $c\in\mathbb{C}$. For $\xi,\eta\in H$,

using claim 1 of Properties of Complex Conjugation and Modulus. So $JAJ$ is linear. Moreover $\lVert JAJ\xi\rVert=\lVert AJ\xi\rVert\le\lVert A\rVert_{\mathrm{op}}\lVert J\xi\rVert=\lVert A\rVert_{\mathrm{op}}\lVert\xi\rVert$. Hence $JAJ\in\mathcal{L}(H)$ and $\lVert JAJ\rVert_{\mathrm{op}}\le\lVert A\rVert_{\mathrm{op}}$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound. The identity $J(JAJ)J=A$ is ($\ast$). Applying the inequality just proved to $JAJ$ in place of $A$ gives $\lVert A\rVert_{\mathrm{op}}=\lVert J(JAJ)J\rVert_{\mathrm{op}}\le\lVert JAJ\rVert_{\mathrm{op}}$. So $\lVert JAJ\rVert_{\mathrm{op}}=\lVert A\rVert_{\mathrm{op}}$.

For the adjoint, let $\xi,\eta\in H$. Use $\langle Ju,Jv\rangle=\langle v,u\rangle$ twice, together with $J\circ J=I$ and (Adj):

Since $A^{*}\in\mathcal{L}(H)$, the map $JA^{*}J$ is linear by the above, so it is an adjoint of $JAJ$ (Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint). By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique, $(JAJ)^{*}=JA^{*}J$.

The identity $(JAJ)(JBJ)=J(AB)J$ is ($\ast$). Pointwise, $J(A+B)J\xi=J(AJ\xi+BJ\xi)=JAJ\xi+JBJ\xi$ and $J(cA)J\xi=J(c\,AJ\xi)=\overline{c}\,JAJ\xi$.

Finally, $J\mathcal{S}J\subseteq\mathcal{L}(H)$ by the above, so its commutant is defined. Let $T\in\mathcal{L}(H)$, so that $JTJ\in\mathcal{L}(H)$. For $A\in\mathcal{S}$, ($\ast$) gives

The map $X\mapsto JXJ$ is injective on maps $H\to H$, by ($\ast$). So $T(JAJ)=(JAJ)T$ if and only if $(JTJ)A=A(JTJ)$. Consequently $T\in(J\mathcal{S}J)'$ if and only if $JTJ\in\mathcal{S}'$. If $JTJ\in\mathcal{S}'$, then $T=J(JTJ)J\in J\mathcal{S}'J$. Conversely, if $T=JRJ$ with $R\in\mathcal{S}'$, then $T\in\mathcal{L}(H)$ and $JTJ=R\in\mathcal{S}'$. Hence $(J\mathcal{S}J)'=J\mathcal{S}'J$.