Taken from C*-algebras and Operator Theory by Gerald Murphy.

We prepare the way for the density theorem with some useful results on strong convergence.

Theorem 4.3.1

If H is a Hilbert space, the involution T\mapsto T^* is strongly continuous when restricted to the set of normal operators of B(H).

Proof

Let x\in H and suppose that T,S are normal operators in B(H). Then

\|(S^*-T^*)(x)\|^2=\langle S^*x-T^*x,S^*x-T^*x\rangle

=\|Sx\|^2-\|Tx\|^2+\langle TT^*x,x\rangle-\langle ST^*x,x\rangle

+\langle TT^*x,x\rangle-\langle TS^*x,x\rangle

=\|Sx\|^2-\|Tx\|^2+\langle (T-S)T^*x,x\rangle+\langle x,(T-S)T^*x\rangle

\leq \|Sx\|^2-\|Tx\|^2+2\|(T-S)T^*x\|\|x\|.

If \{T_\lambda\}_{\lambda\in\Lambda} is a net of normal operators strongly convergent to a normal operator T, then the net \|T_\lambda x\|^2 is convergent to \|Tx\|^2 and the net \{(T-T_\lambda)T^*x\} is convergent to 0, so \{T_\lambda^*x-T^*x\} is convergent to 0. Therefore, \{T_\lambda^*\} is strongly convergent to T^* \bullet

Remark 4.3.1

If B is a bounded subset of B(H), where H is a Hilbert space, then the map

B\times B(H)\rightarrow B(H)(S,T)\mapsto ST,

is strongly continuous. The proof of this is the inequality

\|STx-S_1T_1x\|\leq \|S\|\|Tx-T_1x\|+\|(S-S_1)T_1x\|.

We say that a continuous function from \mathbb{R} to \mathbb{C} is strongly continuous if for every Hilbert space H and each net \{T_\lambda\}_{\lambda\in\Lambda} of Hermitian operators on H converging strongly to a Hermitian operator T, we have \left\{f\left(T_\lambda\right)\right\} converging strongly to f(T).

Theorem 4.3.2

If f:\mathbb{R}\rightarrow\mathbb{C} is a continuous bounded function, then f is strongly continuous.

Proof

Let A denote the set of strongly continuous functions. This is clearly a vector space (with the pointwise-defined operations), and it follows from Remark 4.3.1 that if f,g belong to A and one of them is bounded, then fg\in A.

We show first that \mathcal{C}_0(\mathbb{R})\subset A: Let A_0=A\cap \mathcal{C}_0(\mathbb{R}). It is easy to verify that A_0 is a closed subalgebra of \mathcal{C}_0(\mathbb{R}), and by Theorem 4.3.1 it is self-adjoint. If z:\mathbb{R}\rightarrow\mathbb{C} is the inclusion function, then

f(z)=\frac{1}{1+z^2},

and g=zf belong to \mathcal{C}_0(\mathbb{R}) and \|f\|_\infty\|g\|_\infty\leq 1. We show that f,g\in A_0. Let H be a Hilbert space and suppose that S,T are hermitian operators on H. Then

g(T)-g(S)=T(1+T^2)^{-1}-S(1+S^2)^{-1},

=(1+T^2)^{-1}\left[T(1+S^2)-(1+T^2)S\right](1+S^2)^{-1},

=(1+T^2)^{-1}[T-S+T(S-T)S](1+S^2)^{-1}.

Therefore, if x\in H,

\|g(T)(x)-g(S)(x)\|\leq \|(1+T^2)^{-1}(T-S)(1+S^2)^{-1}(x)\|,

+\|(1+T^2)^{-1}(T(S-T)S)(1+S^2)^{-1}(x)\|,

\leq\|(T-S)(1+S^2)^{-1}(x)\|+\|(S-T)S(1+S^2)^{-1}(x)\|,

since \|(1+T^2)^{-1}\| and \|(1+T^2)^{-1}T\|\leq1. Hence, g is strongly continuous, and therefore g\in A_0. Since z\in A, we have zg\in A (Remark 4.3.1), so f=1-zg\in A, and therefore f\in A_0. The set \{f,g\} separates the points of \mathbb{R} (Let x,y\in \mathbb{R}. If either are  0, we have no problem as f attains a unique max at 0. If x,y are both positive/ negative, then f separates them because f is strictly increasing/ decreasing for positive/ negative numbers. If x,y differ in sign, then g separates them because g is odd.), and f(t)>0 for all t, so by the Stone-Weierstrass theorem (Louis De Branges) the C*-subalgebra generated by f and g is \mathcal{C}_0(\mathbb{R}). Hence, A_0=\mathcal{C}(\mathbb{R}).

Now suppose that h\in \mathcal{C}_b(\mathbb{R}). Then hf,hg\in\mathcal{C}_0(\mathbb{R}), so hf,hg\in A, and therefore zhg\in A also. Consequently h=hf+zhg\in A \bullet

If C is a convex set of operators on a Hilbert space H, then its strong and its weak closures coincide, by Theorem 4.2.7. If A is a *-subalgebra of B(H), then its weak closure is a von Neumann algebra. These observations are used in the proof of the following theorem, which is known as the Kaplansky Density Theorem.

The Kaplansky Density Theorem

Let H be a Hilbert space and A a C*-subalgebra of B(H) with strong closure B.

  1. A_{\text{sa}} is strongly dense in B_{\text{sa}}.
  2. The closed unit ball of A_{\text{sa}} is strongly dense in the closed unit ball of B_{\text{sa}}.
  3. The closed unit ball of A is strongly dense in the closed unit ball of B.
  4. If A contains I_H, then the unitaries of A are strongly dense in the unitaries of B.

Proof

If T\in B_{\text{sa}}, then there is a net \{T_\lambda\}_{\lambda\in\Lambda} in A strongly convergent to T, so \{T_\lambda^*\} converges weakly to T^*, and therefore \{\text{Re}(T_\lambda)\} is weakly convergent to T. Hence, T is in the weak closure of A_{\text{sa}}, and therefore in the strong closure of this set, since its weak and strong closures coincide (As A is convex: Theorem 4.2.7). This proves condition 1.

Suppose now that T is in the closed unit ball of B_{\text{sa}}. Then by condition 1. there is a net \{T_\lambda\}_{\lambda\in\Lambda} in A_{\text{sa}} strongly convergent to T. The function f:\mathbb{R}\rightarrow \mathbb{C} defined by setting f(t)=t for t\in [-1,1] and f(t)=1/t elsewhere belongs to C_0(\mathbb{R}), and therefore by Theorem 4.3.2, it is strongly continuous. Hence, \{f(T_\lambda)\} is strongly convergent to f(T). But clearly, f(T)=T, since \sigma(T)\subset [-1,1]. Moreover, f(T_\lambda) is in the closed unit ball of A_{\text{sa}} is in the closed unit ball of A_{\text{sa}} for all indices \lambda, since \overline{f}=f and \|f\|_\infty\leq 1. This proves condition 2.

The algebra M_2(A) is a C*-subalgebra of M_2(B(H))=B(H^{(2)}), and is strongly dense in the von Neumann algebra M_2(B)

(Let 

B=\left(\begin{array}{cc}b_{11}&b_{12}\\ b_{21}&b_{22}\end{array}\right)\in M_2(B).

Let \Lambda=\{\text{strongly open neighbourhoods of }0\} ordered by reverse inclusion. Having b_{ij}\in\overline{A} is equivalent to the existence of  a net \{a(\lambda)_{ij}\}_{\lambda\in\Lambda} such that a(\lambda)_{ij}\rightarrow b_{ij} in the strong operator topology. This is the case because for all \lambda\in\Lambdab_{ij}+\lambda is a strongly open neighbourhood of b_{ij} which implies that (b_{ij}+\lambda)\cap A\neq\emptyset, so just choose any a(\lambda)_{ij} in here. Then consider

A(\lambda)=\left(\begin{array}{cc}a(\lambda)_{11}&a(\lambda)_{12}\\ a(\lambda)_{21}&a(\lambda)_{22}\end{array}\right).

Now for all \mathbf{x}\in H^2;

\|(B-A(\lambda))\mathbf{x}\|=\left\|\left(\begin{array}{cc}b_{11}-a(\lambda)_{11}&b_{12}-a(\lambda)_{12}\\ b_{21}-a(\lambda)_{21}&b_{22}-a(\lambda)_{22}\end{array}\right)\left(\begin{array}{c}u\\ v\end{array}\right)\right\|

=\|(b_{11}-a(\lambda)_{11})u+(b_{12}-a(\lambda)_{12})v\|+\|(b_{21}-a(\lambda)_{21})u+(b_{22}-a(\lambda)_{22})v\|

\leq \|b_{11}-a(\lambda)_{11}\|\,\|u\|+\|b_{12}-a(\lambda)_{12}\|\,\|v\|+

\|b_{21}-a(\lambda)_{21}\|\,\|u\|+\|b_{22}-a(\lambda)_{22}\|\,\|v\|\rightarrow0).

If T is in the closed unit ball of B, then

S=\left(\begin{array}{cc}0&T\\ T^*&0\end{array}\right)

is a hermitian operator on H^{(2)} lying in the strong closure of M_2(A), and since \|S\|\leq 1; it follows from condition 2. that there is a net \{S_\lambda\}_{\lambda\in\Lambda} in the closed unit ball of M_2(A)_{\text{sa}} that strongly converges to S. Hence, \{[S_\lambda]_{(1,2)}\} is strongly convergent on H to T, and [S_\lambda]_{(1,2)} is in the closed unit ball of A for all indices \lambda. Thus condition 3. is proved.

Suppose now that A contains I_H and let (A) and U(B) denote the unitaries of A and B respectively. If T\in U(B), then by Remark 4.2.2 there is a hermitian element S of B such that T=e^{iS}. By condition 1. there is a net \{S_\lambda\}_{\lambda\in\Lambda} in A_{\text{sa}} strongly convergent to S. However, the function

f:\mathbb{R}\rightarrow\mathbb{C}t\mapsto e^{it},

is strongly continuous by Theorem 4.3.2, so \{f(S_\lambda)\} converges strongly to f(S). Since f(S_\lambda)=e^{iS_\lambda}\in U(A) and f(S)=T, condition 4. is proved \bullet

Theorem 4.3.4

Let H_1 and H_2 be Hilbert spaces, A a von Neumann algebra on H_1, and \varphi(A)\rightarrow B(H_2) a weakly continuous *-homomorphism. Then \varphi(A) is a von Neumann algebra on H_2.

Proof

Observe first that \varphi(A) is a C*-algebra on H_2. By applying Remark 4.1.2, we may suppose that A contains I_{H_1}.

Let S\in \varphi(A) ans suppose that \|S\|<1, so there is a number \alpha such that \|S\|<\alpha<1. Write S=\varphi(T) for some element T\in A and let T=U|T| be the polar decomposition of T. By Theorem 4.1.10 U\in A. Let E be the spectral resolution of the identity for |T| and G=\{\lambda\in\sigma(|T|): \lambda\geq\alpha\}. Then E(G)\in A\alpha E(G)\leq |T|E(G)

(A spectral resolution of the identity for a self-adjoint R\in B(H) are E:\mathbb{B}(\mathbb{R})\rightarrow \{\text{projections on }H\} such that for all x\in H

\mu_x(G)=\langle x,E(G)x\rangle

is a (probability) measure on \mathbb{B}(\mathbb{R}) such that

\langle x,f(R)x\rangle=\int_\mathbb{R} f(t)\,d\mu_x(t).

Now G is as follows:

Now 

\langle x,\alpha E(G)x\rangle=\alpha\int \mathbf{1}_G(t)\,d\mu_x(t)=\int_G\alpha \,d\mu_{x}(t)\leq \int_G|t|\,d\mu_x(t).

In turn, with |t|=t;

\int_Gt\,d\mu_x(t)=\int_\mathbb{R}\mathbf{1}_G(t)t\,d\mu_x(t)

=\langle x,\mathbf{1}_G(|T|)\text{id}(|T|)x\rangle=\langle x,E(G)|T|x\rangle.

),

and |T|(I_{H_1}-E(G))\leq \alpha(I_{H_1}-E(G)). Hence 0\leq \alpha\,\varphi(E(G))\leq \varphi(|T|)\varphi(E(G)), so

\alpha\|\varphi(E(G))\|\leq\|\varphi(|T|)\|\|\varphi(E(G))\|\leq \|\varphi(T)\|

=\|U^*U|T|\|\leq\|\varphi(T)\|=\|S\|<\alpha.

Therefore, \|\varphi(E(G))\|<1, so \varphi(E(G))<1, so \varphi(E(G))=0, since \varphi(E(G)) is a projection. Consequently, S=\varphi(T-E(G)). Also

\|T(I_{H_1}-E(G))\|\leq \||T|(I_{H_1}-E(G))\|\leq \alpha\|I_{H_1}-E(G)\|\leq\alpha<1.

Set

R=\{T_1\in A:\|T_1\|<1\}.

We have shown that \varphi(R)=\{S\in \varphi(A):\|T_1\|<1\} (\checkmark). The closed unit ball of AA_1 is weakly compact by Theorem 4.2.4, so \varphi(A_1) is weakly compact. Observe that A_1 is the weak closure of R (To say this we must see that that A_1 is a weakly closed set such that R\subset A_1 \checkmark. Then if B is another weakly closed set such that R\subset B, then we have A_1\subset B. This follows from the fact that the weak topology is weaker than the norm topology.). We claim that \varphi(A_1) is the closed unit ball of \varphi(A). Let S be an element of the closed unit ball of \varphi(A) and take a sequence \{\varepsilon_n\}\subset (0,1) converging to 1. Now S=\varphi(T) for some T\in A, so \varepsilon_n S\in\varphi(A) and \|\varepsilon_n S\|<1. Hence, \varepsilon_n S=\varphi(T_n) for some T_n\in R. Thus, \varepsilon_n S is a sequence in \varphi(A_1) converging in norm to S, so S\in\varphi(A_1). This shows that the closed unit ball of \varphi(A) is contained in \varphi(A_1), and the reverse inclusion is obvious.

Now let T be a non-zero element of the weak closure of \varphi(A) in B(H_2). By the Kaplansky density theorem, T/\|T\| is in the weak closure of the unit ball of \varphi(A) (\checkmark), and therefore T/\|T\|\in\varphi(A_1), so T\in\varphi(A). This shows that the weak closure of \varphi(A) is equal to \varphi(A), so \varphi(A) is a von Neumann algebra \bullet

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