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

Preparatory to our introduction of the weak and ultraweak topologiesm we show now that $L^1(H)$ is the dual of $K(H)$, and $B(H)$ is the dual of $L^1(H)$.

Let $H$ be a Hilbert space, and suppose that $T\in L^1(H)$. It follows from Theorem 2.4.16 (https://jpmccarthymaths.wordpress.com/2011/01/18/c-algebras-and-operator-theory-2-4-compact-hilbert-space-operators/) that the function

$\text{tr}(T\cdot):K(H)\rightarrow\mathbb{C}$$S\mapsto \text{tr}(TS)$,

is linear and bounded, and $\|\text{tr}(T\cdot)\|\leq \|T\|$. We therefore have a map

$L^1(H)\rightarrow K(H)^\star$$T\mapsto \text{tr}(T\cdot)$,

which is clearly linear and norm-decreasing. We call this map the canonical map from $L^1(H)$ to $K(H)^\star$.

## Theorem 4.2.1

If $H$ is a Hilbert space, then the canonical map from $L^1(H)$ to $K(H)^\star$ is an isometric linear isomorphism.

### Proof

Let $\theta$ denote this map. If $\theta(T)=0$, then $\text{tr}(T^\star T)=$, so $T=0$ (a quick calculation). Thus, $\theta$ is injective.

Now suppose that $\tau\in K(H)^\star$. Then the function

$\sigma: H^2\rightarrow \mathbb{C}$$(x,y)\mapsto \tau(|x\rangle\langle y|)$,

is a sesquilinear form on $H$ and $\|\sigma\|\leq\|\tau\|$. Hence, there is a unique operator $T$ on $H$ such that $\langle Tx,y\rangle=\sigma(x,y)=\tau(|x\rangle\langle y|)$ ($x,y\in H$) (Theorem 2.3.6). Also, $\|T\|=\|\sigma\|$. Let $E$ be an orthonormal basis for $H$ and let $T=U|T|$ be the polar decomposition of $T$. If $F\subset E$ is finite, set $P_F=\sum_{e\in F}|e\rangle\langle e|$ (so $P_F$ is a projection). Then

$\sum_{e\in F}\langle |T|e,e\rangle=\sum_{e\in F}\langle Te,Ue\rangle$

$=\sum_{e\in F}\sigma(e,Ue)$

$=\sum_{e\in F}\tau(|e\rangle\langle Ue)$

$=\tau\left(\left(\sum_{e\in F}|e\rangle\langle e|\right)U^*\right)$

$=\tau (P_FU^\star)\leq \|\tau\|$.

Hence, $\|T\|_1\leq\|\tau\|$, so $T\in L^1(H)$. If $x,y\in H$, then $\text{tr}(T(|x\rangle\langle y|))=\langle Tx,y\rangle=\tau(|x\rangle\langle y|)$, so $\text{tr}(T\cdot)$ equals $\tau$ on $F(H)$ and therefore on $K(H)$; that is, $\theta(T)=\tau$. Therefore, $\theta$ is an isometric linear isomorphism $\bullet$

### Corollary 4.2.2

$L^1(H)$ is a Banach *-algebra under the trace-class norm.

### Proof

It is a dual space and therefore complete $\bullet$

Suppose again that $H$ is a Hilbert space and suppose that $S\in B(H)$. Then the function

$\text{tr}(\cdot S):L^1(H)\rightarrow \mathbb{C}$$T\mapsto \text{tr}(TS)$,

is linear and bounded, and $\|\text{tr}(\cdot S)\|\leq\|S\|$, by Theorem 2.4.16 (https://jpmccarthymaths.wordpress.com/2011/01/18/c-algebras-and-operator-theory-2-4-compact-hilbert-space-operators/). We call the norm-decreasing linear map

$B(H)\rightarrow L^1(H)^\star$, $S\mapsto \text{tr}(\cdot S)$, the canonical map from $B(H)$ to $L^1(H)^\star$.

## Theorem 4.2.3

If $H$ is a Hilbert space, then the canonical map from $B(H)$ to $L^1(H)^\star$ is an isometric linear isomorphism.

### Proof

Let $\theta$ denote this map. If $\theta(S)=0$, then $\theta(S)(|x\rangle\langle y|)=0$, that is $\text{tr}(|x\rangle\langle S^\star y|)=0$, so $\langle x,S^*y\rangle=0$, for all $x,y\in H$. Hence, $S=0$, and therefore $\theta$ is injective.

Now let $\tau\in L^1(H)^*$. The map

$\sigma:H^2\rightarrow \mathbb{C}$$(x,y)\mapsto \tau(|x\rangle\langle y|)$,

is a sesquilinear form on $H$, and $\|\sigma\|\leq\|\tau\|$. Hence, there is a unique operator $S\in B(H)$ such that $\langle Sx,y\rangle=\tau(|x\rangle\langle y|)$ for all $x,y\in H$. Also, $\|S\|=\|\sigma\|\leq\|\tau\|$. Now $\theta(S)(|x\rangle\langle y|)=\text{tr}(|x\rangle\langle S^*y)=\langle Sx,y\rangle=\tau(|x\rangle\langle y|)$, so $\theta(S)$ equals $\tau$ on $F(H)$, and therefore on $L^1(H)$ (by Theorem 2.4.17 https://jpmccarthymaths.wordpress.com/2011/01/18/c-algebras-and-operator-theory-2-4-compact-hilbert-space-operators/). Therefore, $\theta$ is an isometric linear isomorphism $\bullet$

If $H$ is a Hilbert space, the Hausdorff locally convex topology on $B(H)$ generated be the separating family of semi-norms (A family of semi-norms for $B(H)$ is a set of semi-norms $S=\{\rho_i\}_{i\in I}$ such that for all non-zero $T\in B(H)$, there exists a $\rho\in S$ such that $\rho(T)>0$.):

$B(H)\rightarrow\mathbb{R}^+$$T\mapsto |\langle Tx,y\rangle|$, for $x,y\in H$

is called the weak operator topology on $B(H)$. If $\{T_\lambda\}_{\lambda\in\Lambda}$ is a net in $B(H)$, then $\{T_\lambda\}$ converges weakly to an operator $T$ if and only if $\langle Tx,y\rangle=\lim_\lambda\langle T_\lambda x,y\rangle$ for all $x,y\in H$.

The ultraweak or $\sigma$weak topology on $B(H)$ is the Hausdorff locally convex topology on $B(H)$ generated by the semi-norms

$B(H)\rightarrow\mathbb{R}^+$$T\mapsto|\text{tr}(TS)|$, where $S\in L^1(H)$.

The weak topology is weaker than the ultraweak topology (!). For if $\{T_\lambda\}_{\lambda\in\Lambda}$ is a net converging ultraweakly to and operator $T$, then for each $x,y\in H$ the net $\{\langle T_\lambda x-Tx,y\rangle\}_\lambda=\{\text{tr}((T_\lambda-T)(|x\rangle\langle y|))\}^\lambda$ converges to $0$, so $\{T_\lambda\}$ converges to $T$ weakly ($\checkmark$). Clearly, the weak topology is also weaker than the strong topology ($\checkmark$).

The operations of addition and scalar multiplication are of course ($\checkmark$) weakly and ultraweakly continuous, but it is easy to see that the involution operation is $T\mapsto T^*$ is also continuous for these topologies ($\{T_\lambda\}$ converges to $T$ if and only if $\langle x,T_\lambda y\rangle\rightarrow \langle x,Ty\rangle$ for all $x,\,y\in H$. So then $\overline{\langle T_\lambda^*x,y\rangle}\rightarrow\overline{\langle T^*x,y\rangle}$ and hence $T_\lambda^*\rightarrow T$.).  We showed in Example 4.1.1 (https://jpmccarthymaths.wordpress.com/2011/06/24/von-neumann-algebras-the-double-commutant-theorem/) that the involution is not strongly continuous in general, so the weak and strong topologies do not coincide in general.

Continuity of multiplication in the weak topology does not hold in general. Let $H$ be a Hilbert space with an orthonormal basis $\{e_n\}_{n\geq 1}$ and let $U$ be the unilateral shift on this basis. Then the sequences $\{U^{*n}\}$ and $\{U^n\}$ both converge weakly to zero, but the product sequence $\{U^{*n}U^n\}$ is the constant $I_H$.

We have seen that for an arbitrary Hilbert space $H$, the Banach space $B(H)$ is the dual of $L^1(H)$. It is clear from this that the ultraweak topology is just the weak* topology on $B(H)$. Hence, the closed unit ball of $B(H)$ is ultraweakly compact, by the Banach-Alaoglu theorem (http://en.wikipedia.org/wiki/Banach-Alaoglu_theorem#Generalization:_Bourbaki.E2.80.93Alaoglu_theorem).

## Theorem 4.2.4

If $H$ is a Hilbert space, then the relative weak and ultraweak topologies on the closed unit ball of $B(H)$ coincide, and hence the ball is weakly compact.

### Proof

The identity map from the ball with the ultraweak topology to the ball is a continuous bijection from a compact space to a Hausdorff space and is therefore a homeomorphism $\bullet$

### Remark 4.2.1

The closed unit ball of $B(H)$ is not strongly compact in general. For if we suppose the contrary, then the identity map of the ball with the relative strong topology to the ball with the relative  weak topology is a continuous bijection from a compact space to a Hausdorff space, and therefore a homeomorphism. This means that the relative strong and weak topologies on the ball coincide. The involution operation $T\mapsto T^*$ is weakly continuous, but it follows from Example 4.1.1 that in general this operation is not strongly continuous when restricted to the ball. We therefore have a contradiction.

If $C$ is a set of operators on a Hilbert space $H$, then the weak closure of $C$ is contained in $C^{\prime\prime}$. For if $T$ is in this weak closure, there is a net $\{T_\lambda\}_{\lambda\in\Lambda}$ in $C$ converging weakly to $T$, and therefore if $S\in C^\prime$, we have

$\langle TSx,y\rangle=\lim_\lambda \langle T_\lambda Sx,y \rangle=\lim_\lambda\langle ST_\lambda x,y\rangle$

$=\lim_\lambda\langle T_\lambda x,S^*y\rangle=\langle Tx,S^*y\rangle=\langle STx,y\rangle$,

so $S$ and $T$ commute so $T\in C^{\prime\prime}$.

## Theorem 4.2.5

Suppose that $A$ is a *-algebra on a Hilbert space $H$ containing $I_H$.

1. The weak closure of $A$ is $A^{\prime\prime}$.
2. $A$ is a von Neumann algebra if and only if it is weakly closed.

### Proof

This is immediate from the preceding remark and Theorem 4.1.5 (https://jpmccarthymaths.wordpress.com/2011/06/24/von-neumann-algebras-the-double-commutant-theorem/). In fact, this result is basically just the completion of that theorem  $\bullet$

## Theorem 4.2.6

Let $H$ be a Hilbert space and $\rho$ a linear functional on $B(H)$. The following are equiavlent:

1. $\rho$ is weakly continuous.
2. $\rho$ is strongly continuous.
3. There are vectors $x_1,\dots,x_n$ and $y_1,\dots,y_n\in H$ such that, for all $T\in B(H)$
$\rho(T)=\sum_{j=1}^n\langle Tx_j,y_j\rangle$.
Remarks below show that we could include another equivalent condition, namely that there exists a $R\in F(H)$ such that $\rho(T)=\text{tr}(TR)$.

### Proof

The implications 3. implies 1. and 1. implies 2. are clear (Let $\{T_\lambda\}_\lambda$ be a net in $B(H)$ weakly convergent to $T\in B(H)$. Now

$\rho(T_\lambda)=\sum_{j=1}^n\langle T_\lambda x_j,y_j\rangle\rightarrow\tau(T)$,

as addition is weakly continuous. So $\rho$ is weakly continuous.). To show that 2. implies 3., suppose that $\rho$ is strongly continuous. Then by Theorem A.1 (same as Excercise 5.1 from http://www.maths.lancs.ac.uk/~belton/www/notes/fa_notes.pdf), there is a positive number $M$ ans there exist $x_1,\dots,x_n\in H$ such that $|\rho(T)|\leq M\max_{1\leq j\leq n}\|Tx_j\|$ for all $T\in B(H)$. We may suppose that $M=1$. Then, for any $T\in B(H)$:

$|\rho(T)|\leq \left(\sum_{j=1}^n\|Tx_j\|^2\right)^{1/2}$

Let $K_0$ be the subspace of $H^{(n)}$ consisting of all $(Tx_1,\dots,Tx_n)$, as $T$ runs over $B(H)$, and let $K$ be its norm closure. The function

$\sigma:K_0\rightarrow\mathbb{C}$$(Tx_1,\dots,Tx_n)\mapsto \rho(T)$,

is well defined (If $(Tx_1\,Tx_2\,\cdots\,Tx_n)^t=(Sx_1\,Sx_2\,\cdots\,Sx_n)^t$, do we have $\rho(T)=\rho(S)$? By linearity this is equivalent to asking if we have $(Tx_1\,Tx_2\,\cdots\,Tx_n)^t=\mathbf{0}$, do we have $\rho(T)=0$? But

$|\rho(T)|\leq\left(\sum_{i=1}^n\|Tx_i\|^2\right)^{1/2}=0$ so we are done.),

linear, and bounded, with $\|\sigma\|<1$, so it extends to a linear norm-decreasing functional on $K$ which we also denote by $\sigma$. By the Riesz representation theorem for linear functionals on Hilbert spaces, there is a unique element $y=(y_1,\dots,y_n)\in K$ such that $\sigma(z)=\langle z,y\rangle$ for all $z\in K$. Hence, for all $T\in H$

$\rho(T)=\sigma(Tx_1,\dots,Tx_n)=\sum_{j=1}^n\langle Tx_j,y_j\rangle.$

This shows 2. implies 3 $\bullet$

## Theorem 4.2.7

Let $H$ be a Hilbert space and $C$ a convex subset of $B(H)$. Then $C$ is strongly closed if and only if it is weakly closed.

### Proof

Since the weak topology is weaker than the strong topology, a weakly closed set is strongly closed. Suppose, therefore, that $C$ is strongly closed, and let $T$ be a point in the weak closure. Then there is a net of operators $\{T_\lambda\}_{\lambda\in\Lambda}$ in $C$ converging to $T$ weakly, and hence, for every weakly continuous linear functional $\rho$ on $B(H)$, we have $\rho(T)=\lim_\lambda \rho(T_\lambda)$. By Theorem 4.2.6, the weakly continuous functionals on $B(H)$ are the same as the strong continuous, so by Corollary A.8 (Let $C$ be a convex set in a locally convex space $X$. Then for any point $x\in X$$x\in\overline{C}$ (norm closure) if and only if there is a net $\{x_\lambda\}_{\lambda\in\Lambda}$ in $C$ such that $\{\rho(x_\lambda)\}_{\lambda\in\Lambda}$ converges to $\rho(x)$ for all continuous linear functionals $\rho$ on $X$.), $T$ is in the strong closure of $C$; that is $T\in C$. Hence $C$ is weakly closed $\bullet$

## Corollary 4.2.8

If $A$ is a *-algebra on $H$, then $A$ is a von Neumann algebra if and only if it is weakly closed.

### Proof

Immediate, since $A$ is convex $\bullet$

We show now that a von Neumann algebra is the dual space of a Banach space. This is not true for arbitrary C*-algebras. In fact, by a theorem of Sakai (Theorem 1.16.7), every C*-algebra that is a dual space is isomorphic to a von Neumann algebra.

Suppose that $A$ is a von Neumann algebra on a Hilbert space $H$. We set

$A^\perp=\left\{S\in L^1(H):\text{tr}(TS)=0;T\in A\right\}$.

This is a vector subspace of $L^1(H)$, closed with respect to the trace-class norm (because it is the intersection of the kernels of the continuous linear maps $\varphi_T: L^1(H)\mapsto\mathbb{C}$$S\mapsto \text{tr}(TS)$?). Set $A_*=L^1(H)/A^\perp$. Then $A_*$ is a Banach space when endowed with the quotient norm (corresponding to the trace-class norm $\checkmark$). If $T\in A$, we have a well-defined bounded linear functional

$\varphi(T):A_*\rightarrow\mathbb{C}$$S+A^\perp\mapsto\text{tr}(TS)$.

The map

$\varphi:A\mapsto (A_*)^*$$T\mapsto \varphi(T)$,

is clearly norm-decreasing and linear. We call it the canonical map from $A$ to $(A_*)^*$.

## Theorem 4.2.9

Let $A$ be a von Neumann algebra on a Hilbert space $H$. Then the canonical map from $A$ to $(A_*)^*$ is an isometric linear isomorphism.

### Proof

Let $\theta: B(H)\rightarrow L^1(H)^*$ and $\varphi:A\rightarrow (A_*)^*$ be the canonical maps. By Theorem 4.2.3, $\theta$ is an isometric linear isomorphism. If $\varphi(T)=0$, then $\theta(T)=0$ (If $\varphi(T)=0$ then $\text{tr}(TS)=0$ for all $S\in L^1(H)$ which implies that $\theta(T)=0$.) Thus $\varphi$ is injective. If $\rho\in(A_*)^*$, and $\pi$ is the quotient map from $L^1(H)$ to $L^1(H)/A^\perp$, then $\rho\pi\in L^1(H)^*$, so $\rho\pi=\theta(T)$ for some $T\in B(H)$. To show that $T\in A$, we need only show that $\text{tr}(TR)=0$ for all $R\in A^\perp$, using the fact that $A$ is strongly closed, the characterisation of strongly continuous linear functionals on $B(H)$ given in Theorem 4.2.6, and Corollary A.9 (“Let $Y$ be a closed vector subspace of a locally convex space $X$ and $x\in X\backslash Y$. Then there is a continuous linear functional $\rho$ on $X$ such that $\rho(Y)=0$ and $\rho(X)=1$.” Suppose $\text{tr}(TR)=0$ for all $R\in A^\perp\subset L^1(H)$ and that $T\not\in A$. Then there exists a strong operator topology continuous $\chi:B(H)\rightarrow \mathbb{C}$ such that $\chi_{|A}=0$ but $\chi(T)=1$. But there exist $x_1,\,x_2,\,\dots,\,x_n$$y_1,\,y_2,\,\dots,\,y_n\in H$ such that

$\chi(S)=\sum_{i=1}^n\langle x_i,Sy_i\rangle$.

So then

$\chi(T)=\sum_{i=1}^n\langle x_i,Ty_i\rangle=1$ and

$\chi(L)=0$ for all $L\in A$.

However let $R=\sum_{i=1}^n|x_i\rangle\langle y_i|\in F(H)\subset L^1(H)$ and by definition in $A^\perp$. Now we claim that

$\text{tr}(|x\rangle\langle y|)=\langle y,x\rangle.$

We have that

$\text{tr}(|x\rangle\langle y|)=\sum_{i\in I}\langle e_i,x\rangle\langle y,e_i\rangle.$

Now we can write $x=\sum_{i\in I}\alpha_ie_i$ and $y=\sum_{i\in I}\beta_ie_i$. As $\{e_i\}_{i\in I}$ is an orthonormal basis $\langle e_i,x\rangle=\overline{\alpha_i}$ and $\langle y,e_i\rangle=\beta_i$ and the result follows.

Now

$\text{tr}(RT)=\text{tr}\left(\sum_{i=1}^n|x_i\rangle\langle y_i|T\right)=\text{tr}\left(\sum_{i=1}^n|x_i\rangle\langle Ty_i|\right)$

$=\sum_{i=1}^n\langle x_i,Ty_i\rangle=\chi(T)=1$

a contradiction so $T\in A$ as required.) If $S\in L^1(H)$, then $\varphi(T)(\pi(S))=\text{tr}(TS)=\theta(T)(S)=\rho(\pi(S))$ ($\checkmark$). , so $\varphi(T)=\rho$. Therefore, $\varphi$ is a bijection. Observe also that if $\varepsilon>0$, then there exists $S\in L^1(H)$ such that $\|S\|_1\leq 1$ and $|\theta(T)(S)|>\|\theta(T)\|-\varepsilon$, so $\|\rho\|\geq |\rho(\pi(S))|>\|T\|-\varepsilon$. Since $\varepsilon$ was arbitrary, this shows that $\|\rho\|\geq\|T\|$. It follows that $\varphi$ is isometric $\bullet$

It is easy to check that the weak* topology on $A$ is just the relative ultraweak topology on $A$.

## Theorem 4.2.10

Let $A$ be a von Neumann algebra on a Hilbert space $H$, and let $\rho:A\rightarrow\mathbb{C}$ be a linear functional. Then $\rho$ is ultraweakly continuous if and only if there exists $T\in L^1(H)$ such that $\rho(S)=\text{tr}(TS)$ for all $S\in A$.

### Proof

This follows from the identification of $A=(A_*)^*$, the remark preceding this theorem, and Theorem A.2 (Let $X$ be a normed vector space. Then a linear functional $\rho:X^*\rightarrow\mathbb{K}$ is weak* continuous if and only if $\rho=\hat{x}$ for some $x\in X$.$\bullet$

### Remark

Let $A$ be a von Neumann algebra on a Hilbert space $H$ containing $I_H$. If $T$ is a normal element of $A$, then for each $f\in B_\infty(\sigma(T))$ the element $f(T)$ is in $A$. This is because $f$ is in the closed linear span of the characteristic functions on $B_\infty(\sigma(T))$, and if $S$ is a Borel set of $\sigma(T)$, then the spectral projection $\chi_S(T)=E(S)$ belongs to $A$ by Theorem 4.1.11. From this and the proof of Theorem 2.5.8, it follows that if $T\in A$ is a unitary, then $T=e^{iS}$ for some hermitian element $S\in A$.