Introduction to von Neumann algebras, Lecture 2 (Definitions, the double commutant theorem, etc.)

by Orr Shalit

In this second lecture we start a systematic study of von Neumann algebras.

1. Some topologies on $B(H)$

On $B(H)$ we have the topology generated by the operator norm. There are other natural topologies that may be used.

Definition 1: The weak operator topology (WOT, sometimes also called the weak topology) is the topology on $B(H)$ generated by the sets

$V(A, x, y, \epsilon) := \{T : |\langle (T-A)x,y \rangle | < \epsilon\}$ .

where $A \in B(H)$ , $x,y \in H$ and $\epsilon > 0$ (recall that this means that a basis for this topology is given by finite intersections of the above sets). In fact, it suffices to consider sets of the form $V(A,x,y,1)$ (why?) Equivalently, a net $T_n$ converges to $T$ in the weak operator topology if and only if $\langle T_n x, y \rangle \to \langle T x, y \rangle$ for all $x,y \in H$ .

Remark: It is common to call the weak operator topology “the weak topology”, for short (and also because that’s how von Neumann called it). Since $B(H)$ is a Banach space, the term “weak topology” can also mean the topology generated on $B(H)$ by the space $B(H)^*$ of bounded functionals on $B(H)$ ; however, that weak topology is rarely used, so this abuse of terminology hardly ever causes confusion .

Definition 2: The strong operator topology (SOT, sometimes also called the strong topology) is the topology on $B(H)$ generated by the sets

$U(A, x, \epsilon) := \{T : \|(T-A)x\| < \epsilon\}$ .

where $A \in B(H)$ , $x \in H$ and $\epsilon > 0$ . As above, it suffices to consider sets of the form $U(A,x,1)$ . Equivalently, a net $T_n$ converges to $T$ in the strong operator topology if and only if $T_n x \to T x$ (in the norm) for all $x \in H$ .

Remark: Once again, the terminology “strong topology” is used in Banach space theory to refer to the norm topology, in contrast to the weak topology. However, in our setting, since “weak topology” does not mean the topology on $B(H)$ induced by the Banach space dual $B(H)^*$ , “strong topology” is usually not used for the norm topology, but for the strong operator topology.

A beginner might be tempted to think that, since the norm topology comes from a norm and in particular is metric space topology, it should be more convenient to work with than the above two topologies (it turns out that the weak and strong topologies are not metrizable). Why would one want to work with a weaker topology? One reason is that a closed set has better chances of being compact in a weaker topology . Another reason is that closures are bigger, hence richer. For example, we have already seen that the spectral projections of a selfadjoint operator $T$ , are not in the norm closed algebra generated by $T$ , but rather in the weak/strong closed algebra generated by $T$ (these turn out to be the same, see below).

There are other natural topologies, we shall encounter them later.

It is easy to see that the weak and strong topologies are Hausdorff, and that addition and scalar multiplication are jointly continuous. If $T \in B(H)$ , then the maps $A \mapsto AT$ and $A \mapsto TA$ are WOT and SOT continuous. The adjoint operation is WOT continuous. Indeed, if $T_n \to T$ wot, then

$\langle T_n^* x,y \rangle = \langle x, T_n y \rangle \to \langle x, Ty \rangle = \langle T^* x,y \rangle$ .

However, the adjoint operation is not SOT continuous. For example, if $S$ denotes the unilateral shift on $\ell^2(\mathbb{N})$ (that is, $S (a_1, a_2, \ldots) = (0, a_1, a_2, \ldots)$ ), then $T_k := (S^*)^k$ converges SOT to $0$ as $k \to \infty$ , but $T_k^* = S^k$ does not.

Exercise A: Prove that the strong topology is strictly stronger than the weak topology.

Multiplication is not jointly continuous, WOT nor SOT, but it is SOT continuous when restricted to bounded sets.

Exercise B: Show that multiplication is not jointly SOT continuous (this requires finding a tricky example). However, show that the map $B(H)_1 \times B(H) \to B(H)_1$ given by $(A, B) \mapsto AB$ is SOT continuous (here, $B(H)_1$ denotes the closed unit ball of $B(H)$ ). What about joint WOT continuity of multiplication?

Exercise C: Explain why the previous exercise implies that the strong topology is not metrizable. What about the weak topology – is it metrizable?

Theorem 3: The closed unit ball $B(H)_1$ of $B(H)$ is WOT compact.

Proof: The proof is similar to the proof of Alaoglu’s theorem. For every $x,y \in H$ , define

$D_{x,y} = \{z \in \mathbb{C} : |z| \leq \|x\|\|y\|\}$ .

Let ${\bf X} = \Pi_{x,y \in H} D_{x,y}$ , equipped with the product topology. By Tychonoff’s theorem, ${\bf X}$ is compact. We consider ${\bf X}$ as a subspace of the space of all functions $b: H \times H \to \mathbb{C}$ , equipped with the topology of pointwise convergence, which is the same thing as the product topology.

Now let $\Phi: B(H)_1 \to {\bf X}$ be defined by

$\Phi(T)(x,y) = \langle Tx, y \rangle$ .

$\Phi$ is injective, and by definition, $\Phi$ is a homeomorphism between $B(H)_1$ with the weak operator topology and $\Phi(B(H)_1)$ with the topology of pointwise convergence. To show that $B(H)_1$ is compact, it suffices to show that $\Phi(B(H)_1)$ is closed.

Now, if $b \in \overline{\Phi(B(H)_1}$ , then it is not hard to show that $b$ must be a sesqui-linear form on $H$ and it is bounded in the sense that $|b(x,y)| \leq \|x\|\|y\|$ . It follows from a familiar consequence of the Riesz representation theorem that there is some $B \in B(H)_1$ , such that $b(x,y) = \langle Bx, y \rangle$ for all $x,y \in H$ , therefore $b \in \Phi(B(H)_1)$ , as required.

Exercise D: Show that $B(H)_1$ with the strong topology is not compact.

Exercise E: If $H$ is separable, then $B(H)_1$ is metrizable in both the weak and the strong operator topologies. The metric for $B(H)_1$ with the weak topology is

$d_w(A,B) = \sum_{m,n=1}^\infty 2^{-m-n} |\langle (A-B)x_m,x_n \rangle |$ ,

where $\{x_n\}$ is a dense sequence in the unit ball of $H$ . Fill in the rest of the details.

2. The double commutant theorem

We recall some definitions, and give a couple of new ones.

Definition 4: A $*$ -subalgebra of $B(H)$ that is SOT closed and contains $1 = I_H$ is called a von Neumann algebra.

Definition 5: For a set $\mathcal{S} \subseteq B(H)$ , the commutant of $\mathcal{S}$ is the set $\mathcal{S}'$ defined

$\mathcal{S}' = \{T \in B(H) : TS = ST$ for all $S \in \mathcal{S}\}$ .

Exercise F: Prove that $\mathcal{S} \subseteq \mathcal{S}''$ and that $\mathcal{S}' = \mathcal{S}'''$ .

Definition 6: The null space of a subset $\mathcal{S} \subseteq B(H)$ is the set

$null \mathcal{S} = \cap_{T \in \mathcal{S}} ker T$ .

Definition 7: A subset $\mathcal{S} \subseteq B(H)$ is said to be non-degenerate if $[\mathcal{S} H] = H$ (where $[\mathcal{A}]$ denote the closed linear span of a set $\mathcal{A} \subseteq H$ ).

Exercise G: A $*$ -algebra $A \subseteq B(H)$ is non-degenerate if and only if $null(A) = \{0\}$ .

Theorem 8 (von Neumann’s double commutant theorem): Let $A \subseteq B(H)$ be a $*$ -subalgebra with a trivial null space. Then

$A'' := (A')' = \overline{A}^{WOT} = \overline{A}^{SOT}$ .

In particular, a $*$ -algebra $A$ is a von Neumann algebra if and only if $A'' = A$ .

Before the proof, we will prove a basic lemma, which will also emphasize an important role that the commutant plays.

Lemma: Let $A \subseteq B(H)$ be a $*$ -subalgebra, let $p \in B(H)$ be an orthogonal projection, and set $M \subseteq pH$ . Then $M$ is invariant for $A$ if and only if $p \in A'$ .

Proof: For $a \in B(H)$ , $aM \subseteq M$ if and only if $pa x = ax$ for all $x \in M$ , which happens if and only if $pap = ap$ . Therefore, $a^*M \subseteq M$ if and only if $pa^* p = a^* p$ , or $pap = pa$ . Therefore, if $M$ is invariant for $A$ , then for every $a \in A$ , the adjoint $a^* \in A$ too, so $ap = pap = pa$ , therefore $p \in A'$ . This argument works backwards as well.

Proof of the double commutant theorem: As $A''$ is WOT closed, it holds that

$\overline{A}^{SOT} \subseteq \overline{A}^{WOT} \subseteq A''$ .

Now let $t \in A''$ . To show that $t \in \overline{A}^{SOT}$ , we have to find, for every $x_1, \ldots, x_n \in H$ , an operator $a \in A$ such that $\|t x_i - a x_i \| < 1$ for all $i$ .

Case 1: $n=1$ . Let $p$ be the orthogonal projection onto the subspace $[Ax_1]$ . Then by the lemma, $p \in A'$ . Now, $A p^{\perp} x_1 = p^{\perp} A x_1 = 0$ , so the assumption $null A = \{0\}$ implies that $p^{\perp} x_1 = 0$ , or $x_1 \in [A x_1]$ .

Now since $t \in A''$ and $p \in A'$ , we have

$t x_1 = t p x_1 = p t x_1 \in [A x_1]$ .

This means that there is some $a \in A$ such that $\|a x_1 - t x_1\| < 1$ .

Case 2: $n \geq 2$ . Define $H^{(n)} = H \oplus \cdots \oplus H$ ( $n$ times), and for every operator $a \in B(H)$ , define $a^{(n)} = a \oplus \cdots \oplus a$ in $B(H^{(n)}) = M_n(B(H))$ (the last equality should be clear, if it is not STOP NOW!).

Put $A^{(n)} = \{a^{(n)} : a \in A\}$ . Then $A^{(n)}$ is a $*$ subalgebra of $B(H^{(n)})$ . A calculation shows that $(A^{(n)})' = M_n(A')$ , and that $(A^{(n)})'' = (M_n(A'))' = (A'')^{(n)}$ .

Now put $x = (x_1, \ldots, x_n)$ , where $x_1, \ldots, x_n \in H$ . Applying Case 1 to the algebra $A^{(n)}$ and the vector $x$ , we find $a^{(n)} \in A^{(n)}$ such that $\|t^{(n)} x - a^{(n)} x\| <1$ , and this implies $\|t x_i - a x_i \| < 1$ for all $i$ .

Corollary 9: A unital $*$ -subalgebra $A \subseteq B(H)$ is a von Neumann algebra if and only if $A = A''$ .

Exercise H: Prove that every von Neumann algebra is the commutant of the image of a unitary representation.

Corollary 10: Let $T \in B(H)$ , and suppose that $T = U|T|$ is its polar decomposition. Then both $|T|$ and $U$ are in $W^*(T) =\{T,T^*\}''$ (the von Neumann algebra generated by $T$ ).

Proof: We have already seen that $|T| \in C^*(T)$ . To show that $U \in W^*(T)$ , we will show that $U$ commutes with every $S \in \{T,T^*\}'$ (here we are using the double commutant theorem). Given such $S$ , we have

$S U|T| = ST = TS = U|T|S = U S |T|$ .

It follows that $S Ux = U Sx$ for every $x \in Im |T|$ . To finish the proof we need to show that $SUx = USx$ for every $x \in ker T = ker |T| = Im |T|^\perp$ . By the definition of $U$ , $Ux = 0$ for $x \in ker T = Im|T|^\perp$ , so $S Ux = 0$ . On the other hand, if $x \in ker T$ , then $T Sx = ST x = 0$ , so $S (ker T) \subseteq ker T$ . Thus $U S x = 0$ for $x \in ker T$ , and we are done.

Example: If $A = M_n(\mathbb{C})$ , then $A' = \mathbb{C} I_n$ , and $A'' = (\mathbb{C}I_n)' = M_n(\mathbb{C})$ . More generally, if $A = B(H)$ , then $A' = \mathbb{C} I_H$ , and $A'' = (\mathbb{C}I_H)' = B(H)$ .

The first equality follows from the fact that $A$ has no invariant subspaces, thus the only projections in $A'$ are $0$ and $I_H$ . Since a von Neumann algebra is generated by its projections, $A' =\mathbb{C} I_H$ .

Similarly, if $A = K(H)$ (the algebra of compact operators), then $K(H)' = \mathbb{C} I_H$ and $\overline{K(H)}^{SOT} = K(H)'' = (\mathbb{C} I_H)' = B(H)$ (that the SOT/WOT closure of $K(H)$ is $B(H)$ is also easy to prove directly, but note how it just falls out of the air here).

Example: Let $(X, \mu)$ be a probability space, and consider the algebra $A = \{M_f : f \in L^\infty(X, \mu)\}$ , where we are using the notation introduced in the previous lecture:

$M_f : h \mapsto fh$ for $h \in L^2(X,\mu)$ .

If you check carefully the notes of the previous lecture, you will see that we never really proved that this algebra is SOT closed. We shall now see that $A' = A$ , and it will follow that $A = A''$ is a von Neumann algebra. It is common to abuse notation and write $(L^\infty)' = L^\infty = (L^\infty)''$ .

Of course, $A \subseteq A'$ because $A$ is commutative. Let $T \in A'$ . If $T$ was in $A$ , then $T = M_f$ and we would be able to recover $f$ as $f = M_f 1 = T1$ . Here, $1$ denotes the constant function with value $1$ everywhere on $X$ , which is in $L^2(X, \mu)$ because $\mu$ is a probability measure. We therefore define $f := T1$ , and hope to be able to show that $f$ – originally only known to be in $L^2$ – is bounded, and that $T = M_f$ . But this is easy: for every function $h \in L^\infty \cap L^2$ ,

$fh = M_h f = M_h T 1 = T M_h 1 = Th$ ,

therefore $\|fh\|_2 \leq \|T\| \|h\|$ . It follows (for example, by considering characteristic functions of sets of finite measure) that $f \in L^\infty$ , and the equation above shows that $T$ agrees with $M_f$ on the space $h \in L^\infty \cap L^2$ , which is dense in $L^2$ . Since both operators are bounded, they are equal.

Surely one can prove directly that $A$ is a strongly closed algebra, but the above simple computation shows the beauty and power of the double commutant theorem: an analytical problem (involving, perhaps, the limits of convergent unbounded nets) is reduced to a rather simple minded algebraic problem: computing the commutant.

Example: Suppose that $X$ is a locally compact Hausdorff space, and that $\mu$ is a regular Borel measure on $X$ . Identify the algebra $C_c(X)$ of continuous, compactly supported functions on $X$ , with the algebra of operators $\{M_f : f \in C_c(X)\} \subseteq B(L^2)$ . Using an argument similar to the one above, one can show that $C_c(X)' = C_c(X)'' = L^\infty(X, \mu)$ .

Exercise I: Prove that the von Neumann algebra $W^*(T)$ generated by a selfadjoint operator $T$ that has a cyclic vector, is unitarily equivalent to $L^\infty(X, \mu)$ , where $\mu$ is a probability measure on $X:=\sigma(T) \subseteq \mathbb{R}$ . Give an example of an operator (necessarily, non-cyclic) such that the von Neumann algebra that it generates is not unitarily equivalent to $L^\infty(X, \mu)$ (for any probability space $(X, \mu)$ ). Your example is still related to $L^\infty$ – how so?

3. Existence of limits of monotone nets

Theorem 11: Let $T_n$ be an increasing net of selfadjoints in a von Neumann algebra $A \subseteq B(H)$ . Suppose that this net is bounded above, in the sense that there is some some operator $M \in B(H)$ such that $T_n \leq M$ for all $n$ . Then $T_n$ is SOT convergent to some $T \in A_{sa}$ , and $A$ is the least upper bound for $T_n$ in $B(H)$ . In case where every $T_n$ is a projection, then $T$ is the projection onto the closure of the union of the ranges of $T_n$ .

Proof: For every $x \in H$ ,

$\langle T_n x , x \rangle \leq \langle M x, x \rangle$ ,

and the left hand side is a bounded increasing net of real numbers; it therefore has a limit , which we write as

$\Lambda(x) = \lim_n \langle T_n x,x \rangle$ .

Now if we define

$B(x,y) := \frac{1}{4} \left(\Lambda(x+y) - \Lambda(x-y) +i\Lambda(x+iy) - i\Lambda(x-iy) \right)$ .

Then $B(x,y) = \lim_n \langle T_n x, y \rangle$ , so it is a bounded sesquilinear form. It follows from a familiar consequence of Riesz theorem that there exists a $T \in B(H)$ such that $B(x,y) = \langle T x, y \rangle$ for all $x, y \in H$ . It easily follows that $T = WOT \lim_n T_n \in A_{sa}$ , and that $T \geq T_n$ for all $n$ . Moreover, if $S \geq T_n$ for all $n$ , then $\langle S x,x \rangle \geq \lim_n \langle T_n x,x \rangle = \langle Tx,x \rangle$ , for all $x \in H$ , so $S \geq T$ . We therefore have that $T = \sup_n T_n$ .

SOT convergence is trickier. Without loss of generality, let us assume that $T_n \geq 0$ for all $n$ (if not, then just consider the net $T_n - T_{n_0} \geq$ for some fixed index $n_0$ ).

Now $0 \leq T - T_n \leq T$ , so $\|T- T_n\| \leq \|T\|$ .

$\|(T - T_n)x\|^2 = \langle (T-T_n)^2 x, x \rangle \leq \|T-T_n\| \langle (T-T_n) x, x \rangle \to 0$ ,

where the inequality is justified by the following claim.

Claim: For every positive operator $B \geq 0$ , it holds that $B^2 \leq \|B\| B$ , in the sense that $\langle B^2 x, x \rangle \leq \langle \|B\|Bx, x \rangle$ for all $x \in H$ .

Proof of claim: This follows immediately from the functional calculus, because for a bounded nonnegative function $f$ , it clearly holds that $f^2 \leq \|f\|_\infty f$ almost everywhere.

That concludes the proof of the claim, so we have that $T_n \to T$ in the SOT.

The final assertion (regarding projections) is left as an exercise.

Exercise J: Prove that if $T_n$ is a bounded increasing net of projections, then $\sup_n T_n$ is the projection onto the union of the ranges of the $T_n$ .

Remark: Here are another two ways to see that $B^2 \leq \|B\|B$ . First, recall that if $S \leq T$ and if $W$ is any operator, then $W^* SW \leq W^* T W$ . Indeed:

$\langle W^*SW x,x \rangle = \langle S (Wx), Wx \rangle \leq \langle T (Wx) , Wx \rangle = \langle W^*TW x , x \rangle$ .

Now, we simply apply this to $W = B^{1/2}$ , $S = B$ and $T = \|B\|$ :

$B^2 = B^{1/2} B B^{1/2} \leq B^{1/2} \|B\| B^{1/2} = \|B\|B$ .

Alternatively,

$\langle B^2 x, x \rangle = \|Bx\|^2 = \|B^{1/2}(B^{1/2})x\|^2 \leq \|B^{1/2}\|^2\langle Bx, x \rangle$ .

But, by our characterization of the norm of a selfadjoint operator, together with the spectral mapping theorem

$\|B^{1/2}\|^2 = (\max \sigma(B^{1/2}))^2 = \max \sigma((B^{1/2})^2) = \|B\|$ .

4. Kaplansky’s density theorem(s)

By the double commutant theorem, if $T \in A''$ , then there is a net $T_n \in A$ such that $T_n \to T$ .

Technical problem: Convergent nets may be unbounded.

Technical solution: Kaplansky’s density theorem below shows that one may assume that the net is bounded by $\|T\|$ .

Exercise K: Use Kaplansky’s density theorem (Theorem 14 below) to prove directly that $L^\infty[0,1] \subseteq B(L^2[0,1])$ is SOT closed (“directly” means without making use of the double commutant theorem).

Proposition 12: Let $K \subseteq B(H)$ be a convex set. Then, the WOT and SOT closures of $K$ coincide.

Remark: Recall that in a normed space, a convex set is weakly closed if and only if it is closed in the norm, hence the weak and norm closures of a convex set are the same (short explanation: every weakly closed set is strongly closed, and conversely, if a strongly closed set $K$ is convex, then, by the Hahn-Banach theorem, $K$ is equal to the intersection of closed half spaces, hence it is weakly closed as the intersection of weakly closed sets).

Proof of Proposition 12: Clearly, $\overline{K}^{SOT} \subseteq \overline{K}^{WOT}$ . Let $A \in \overline{K}^{WOT}$ . To show that $A \in \overline{K}^{SOT}$ , we fix $x_1, \ldots, x_n \in H$ , and find $T \in K$ such that $\|T x_i - A x_i \|<1$ for all $i$ – this will show that there is a point $T \in K \cap(\cap_{i=1}^n U(A, x_i, 1))$ for every open set of the form $\cap_{i=1}^n U(A,x_i, 1)$ , whence $A \in \overline{K}^{SOT}$ .

Assume that $n=1$ , otherwise repeat the trick from the proof of the double commutant theorem. Write $x = x_1$ . Now, $A \in \overline{K}^{WOT}$ , so it follows that

$A x \in \overline{K x}^{weak} = \overline{K x}^{\|\cdot\|}$ ,

where the last equality follows from the remark preceding the proof. Thus, by definition, there is some $T \in K$ such that $\|A x - T x\|<1$ , as required.

From general functional analysis (in a locally convex topological vector space, a linear functional is continuous if and only if its kernel is closed) we obtain:

Corollary 13: The SOT and the WOT have the same continuous linear functionals.

Theorem 14 (Kaplansky’s density theorem): Let $A$ be a non-degenerate *-subalgebra of $B(H)$ . Then

$\overline{A_1}^{SOT} = (A'')_1$ (i.e., the closed unit ball of $A$ is SOT dense in the closed unit ball of $A''$ ).
$\overline{(A_{sa})_1}^{SOT} = ((A'')_{sa})_1$ .
$\overline{A_+ \cap A_1}^{SOT} = (A'')_+ \cap (A'')_1$ .
If $A$ is a unital C*-algebra, then $\overline{U(A)}^{SOT} \supseteq U(A'')$ .

Proof: The heart of the matter is to prove item 2: that is, to prove $\overline{(A_{sa})_1}^{SOT} = ((A'')_{sa})_1$ . This done, the rest of the assertions follow by various tricks or by adapting the arguments. For example, suppose that we know that $\overline{(A_{sa})_1}^{SOT} = ((A'')_{sa})_1$ holds for every $A$ . To prove that every $T \in (A'')_1$ is the strong limit of a net of elements $T_n \in A_1$ , we define

$\tilde{T} = \begin{pmatrix}0 & T \\ T^* & 0 \end{pmatrix} \in M_2(A) \subseteq B(H \oplus H)$ .

Then we know that there is some bounded net of selfadjoints

$\tilde{T}_n = \begin{pmatrix}* & T_n \\ T_n^* & * \end{pmatrix} \in ((M_2(A))_{sa})_1$

converging to $\tilde{T}$ , and it follows that $\|T_n\| \leq 1$ and that $SOT \lim_n T_n = T$ .

Thus we turn to the heart of the matter, which is proving that $\overline{(A_{sa})_1}^{SOT} = ((A'')_{sa})_1$ . We may assume that $A$ is closed in the norm. Using Proposition 12, it is easy to prove that $\overline{(A_{sa})_1}^{SOT} \subseteq ((A'')_{sa})_1$ (Hey there! If you don’t see why Proposition 12 is needed, then make a note to return to this point at the end of the proof) . We therefore assume that $T \in (A'')_{sa}$ and $\|T\| \leq 1$ , and we will find a net of selfadjoint contractions in $A$ that converges to $T$ .

By the double commutant theorem, there is a net $a_n \in A$ such that $a_n \to T$ in the SOT, and therefore WOT. Since the adjoint is WOT continuous, $a_n^* \to T^* = T$ in the WOT, and therefore $\frac{a_n + a_n^*}{2} \to T$ weakly. The elements of this sequence are selfadjoint, thus,

$T \in \overline{A_{sa}}^{WOT} = \overline{A_{sa}}^{SOT}$ ,

where the last equality follows from Proposition 12. It follows that $T$ is the SOT limit of a net in $A_{sa}$ , say $SOT - \lim_n T_n = T$ . The net $T_n$ might still be missing the crucial property $\|T_n\| \leq 1$ . To fix this, we need the following technical lemma.

Lemma: For every $f \in C_0(\mathbb{R})$ , and every net $a_n \in B(H)_{sa}$ , if $SOT - \lim_n a_n = a$ , then $SOT - \lim_n f(a_n) = f(a)$ .

The lemma will be proved below; in the meanwhile, let $f \in C_0(\mathbb{R})$ be defined by $f(t) = t$ if $t$ is in the interval $[-1,1]$ , and otherwise $f(t) = t^{-1}$ . Then by the functional calculus, $f(T_n)$ are selfadjoint contractions, and $f(T)= T$ . By the lemma, on the other hand, $SOT-\lim_n f(T_n) = f(T) = T$ . This shows that $T \in \overline{(A_{sa})_1}^{SOT}$ , and the proof of assertion 2 is complete.

Exercise L: Complete the proof of Theorem 14.

Proof of the lemma: For the span of this proof, let us agree that all functions are real valued. Let us say that a function $f \in C(\mathbb{R})$ is strongly continuous if $SOT-\lim_n a_n = a$ implies $SOT-\lim_n f(a_n) = f(a)$ for every net of selfadjoint operators $a_n$ . Let us define

$V = \{f \in C(\mathbb{R}) : f$ is strongly continuous $\}$ .

If $f \in V$ and $g \in C_0(\mathbb{R}) \cap V$ , then $fg \in C_0(\mathbb{R}) \cap V$ , because multiplication is SOT continuous on $B(H)_1 \times B(H)$ . In Exercise K below, you will be asked to prove that if $f_k$ is a sequence of strongly continuous functions, and if $f_k$ converge uniformly to $f \in C(\mathbb{R})$ , then $f$ is also strongly continuous. It follows that $C_0(\mathbb{R}) \cap V$ is a norm closed subalgebra of $C_0(\mathbb{R})$ . Thus, to prove that $C_0(\mathbb{R}) \subseteq V$ , all that we have to do is to find enough functions in $C_0(\mathbb{R}) \subseteq V$ to separate points, and then apply the (locally compact version of the) Stone-Weierstrass theorem.

We will finish the proof by showing that $\psi(t) = \frac{t}{1+t^2}$ and that $\phi(t) = \frac{1}{1+t^2}$ are in $V$ . Since $\phi(t) = 1 - t \cdot \psi(t)$ , and using our observation above on the product of strongly continuous functions, we just have to deal with $\psi$ . If $a,b \in B(H)_{sa}$ , then

$\psi(a) - \psi(b) = (1+a^2)^{-1}a - b(1+b^2)^{-1} = (1+a^2)^{-1} \left[a - b + a(b-a)b \right](1+b^2)^{-1}$ .

Now let $a_n$ converge SOT to $a$ . Then for every $x \in H$ ,

$\psi(a_n)x - \psi(a)x = (1+a_n^2)^{-1} (a_n - a)(1+a^2)^{-1}x + (1+a_n^2)^{-1} a_n (a-a_n)a (1+a^2)^{-1}x$ ,

and both summands on the RHS tend to zero. For example, to see that

$(1+a_n^2)^{-1} a_n (a-a_n)a (1+a^2)^{-1}x \to 0$ ,

we set $y = a (1+a^2)^{-1}x$ , and then we obtain that $(a-a_n)y \to 0$ because $a_n \to a$ SOT. But then $\|(1+a_n^2)^{-1} a_n\| = \|\psi(a_n)\| \leq \|\psi\|_\infty$ , so the whole expression converges to $0$ .

Exercise M: Prove that if $f_k$ is a sequence of strongly continuous functions, and if $f_k$ converge uniformly to $f \in C(\mathbb{R})$ , then $f$ is also strongly continuous.

Corollary 15: A non-degenerate *-algebra $A \subseteq B(H)$ is a von Neumann algebra if and only if the unit ball $A_1$ of $A$ is WOT compact.

5. Additional exercises

Definitions: A family of operators $\mathcal{S} \subseteq B(H)$ is said to be topologically irreducible when $0$ and $H$ are the only closed subspaces invariant under the action of all $a \in \mathcal{S}$ . The family $\mathcal{S}$ is said to be algebraically irreducible if for every $0 \neq x \in H$ and every $y \in H$ , there exists $a \in \mathcal{S}$ such that $a \in \mathcal{S}$ such that $ax = y$ .

Exercise N: Let $\mathcal{S} = \mathcal{S}^* \subseteq B(H)$ . Prove that $\mathcal{S}$ is topologically irreducible if and only if $\mathcal{S}' = \mathbb{C}I$ (equivalently, if and only if $\mathcal{S}'' = B(H)$ ).

Exercise O: Prove that a C*-algebra $A \subseteq B(H)$ is topologically irreducible if and only if it is algebraically irreducible.

Thanks to the result of the above exercise, neither the terminology “topologically irreducible” nor “algebraically irreducible” is used; a C*-algebra satisfying either requirement is simply said to be irreducible.

Exercise P: Generalize the result in Exercise O in the following way: if a C*-algebra $A$ is irreducible, then for any $x_1, \ldots, x_n \in H$ and $y_1, \ldots, y_n \in H$ there exists $a \in A$ such that $ax_i = y_i$ for all $i$ . Moreover if there exists a selfadjoint (respectively, unitary) operator $t \in B(H)$ such that $t x_i = y_i$ for all $i$ , then there is a selfadjoint (respectively, unitary) operator $a \in A$ such that $ax_i = y_i$ for all $i$ .

Exercise O: Prove Corollary 15.

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Noncommutative Analysis