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\Lambda$$b_{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 $A$$A_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 $\bullet$