# Operator Topologies

We previously defined the notion of positive linear maps and states on *-algebras, and noted that there always exists seminorms defining the ${L^2}$ and ${L^\infty}$ topologies. However, for applications to noncommutative probability theory, these are often not the most convenient modes of convergence to be using. Instead, the weak, strong, ultraweak and ultrastrong operator topologies can be used. This, rather technical post, is intended to introduce these concepts and prove their first properties.

Weak convergence on a *-probability space ${(\mathcal A,p)}$ is straightforward to define. A net ${a_\alpha\in\mathcal A}$ tends weakly to the limit ${a}$ if and only if ${p(xa_\alpha y)\rightarrow p(xay)}$ for all ${x,y\in\mathcal A}$. Before going into details, I note that the operator topologies are usually applied to the algebra of bounded operators on a Hilbert space so, in this post, I work in more generality than just for *-probability spaces. Let us call a pair ${(\mathcal A,V)}$ a *-algebra representation, where ${\mathcal A}$ is a *-algebra, ${V}$ is a semi-inner product space, and ${\mathcal A}$ acts on ${V}$ by left multiplication in a fashion consistent with the algebra operations,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle (a+b)x=ax+bx,\smallskip\\ &\displaystyle (\lambda a)x=\lambda(ax),\smallskip\\ &\displaystyle (ab)x=a(bx),\smallskip\\ &\displaystyle \langle x,a^*y\rangle=\langle ax,y\rangle, \end{array}$

for all ${a,b\in\mathcal A}$, ${x,y\in V}$ and ${\lambda\in{\mathbb C}}$. Define the operator seminorm ${\lVert\cdot\rVert}$ on ${\mathcal A}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert a\rVert&\displaystyle=\inf\left\{K\in{\mathbb R}^+\colon\lVert ax\rVert\le K\lVert x\rVert\right\}\smallskip\\ &\displaystyle=\sup\left\{\lVert ax\rVert\colon x\in V,\ \lVert x\rVert\le1\right\}. \end{array}$

We will say that an element ${a\in\mathcal A}$ is bounded if ${\lVert a\rVert}$ is finite, and ${(\mathcal A,V)}$ is bounded if ${\lVert a\rVert}$ is finite for all ${a\in\mathcal A}$.

For a *-probability space ${(\mathcal A,p)}$, then we have the ${L^2(p)}$ semi-inner product on ${\mathcal A}$ given by ${\langle x,y\rangle=p(x^*y)}$, so that considering ${\mathcal A}$ as acting on itself by left-multiplication, ${(\mathcal A,\mathcal A)}$ defines a *-algebra representation. The operator semi-norm on ${\mathcal A}$ is then the same as the ${L^\infty(p)}$ seminorm.

We define weak and strong topologies on ${\mathcal A}$. Given a collection ${\mathcal F}$ of maps ${f\colon\mathcal A\rightarrow X}$ from set ${\mathcal A}$ to topological space ${X}$, they generate a topology on ${\mathcal A}$ defined as the weakest topology making each ${f\in\mathcal F}$ continuous. It can be seen that this has a base consisting of the sets ${f_1^{-1}(U_1)\cap\cdots\cap f_n^{-1}(U_n)}$ for ${f_1,\ldots,f_n\in\mathcal F}$ and open ${U_1,\ldots,U_n\subseteq X}$. A net ${a_\alpha\in\mathcal A}$ converges to a limit ${a}$ in the topology generated by ${\mathcal F}$ iff ${f(a_\alpha)\rightarrow f(a)}$ for all ${f\in\mathcal F}$. In the following, we use the standard topology on ${{\mathbb C}}$ and the (strong) topology on ${V}$ defined by its seminorm.

Definition 1 Let ${(\mathcal A,V)}$ be a *-algebra representation. Then, define the following topologies on ${\mathcal A}$.

• The weak topology generated by the maps ${\mathcal A\rightarrow{\mathbb C}}$, ${a\mapsto\langle x,ay\rangle}$, for ${x,y\in V}$.
• The strong topology generated by the the maps ${\mathcal A\rightarrow V}$, ${a\mapsto ax}$, for ${x\in V}$.

So, a net ${a_\alpha\in\mathcal A}$ converges to a limit ${a}$ in the weak topology iff ${\langle x,a_\alpha y\rangle\rightarrow\langle x,ay\rangle}$ for all ${x,y\in V}$, and in the strong topology iff ${\lVert(a_\alpha-a)x\rVert\rightarrow0}$ for all ${x\in V}$. As these are vector topologies, ${a_\alpha\rightarrow a}$ iff ${a_\alpha-a\rightarrow0}$, so that the topology is characterised by the nets converging to zero. Convergence in the weak and strong topologies are related in straightforward way.

Lemma 2 A net ${a_\alpha\in\mathcal A}$ tends strongly to zero iff ${a_\alpha^*a_\alpha}$ tends weakly to zero.

Proof: If ${a_\alpha\rightarrow0}$ strongly then, for all ${x,y\in V}$, ${a_\alpha x}$ and ${a_\alpha y}$ tend to zero and, hence,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lvert\langle x,a_\alpha^*a_\alpha y\rangle\rvert&\displaystyle=\lvert\langle a_\alpha x,a_\alpha y\rangle\rvert\smallskip\\ &\displaystyle\le\lVert a_\alpha x\rVert\lVert a_\alpha y\rVert\rightarrow0. \end{array}$

Therefore, ${a_\alpha^*a_\alpha\rightarrow0}$ weakly. Conversely, if ${a_\alpha^*a_\alpha\rightarrow0}$ weakly then, for all ${x\in V}$,

$\displaystyle \lVert a_\alpha x\rVert^2=\langle x,a_\alpha^*a_\alpha x\rangle\rightarrow0$

and, hence, ${a_\alpha\rightarrow0}$ strongly. ⬜

I look at continuity of the algebra operations with respect to the weak and strong topologies.

Lemma 3 Let ${(\mathcal A,V)}$ be a *-algebra representation. Then, the vector space operations

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle {\mathbb C}\times\mathcal A\rightarrow\mathcal A,\ (\lambda,a)\mapsto\lambda a,\smallskip\\ &\displaystyle \mathcal A\times\mathcal A\rightarrow\mathcal A,\ (a,b)\mapsto a+b \end{array}$ (1)

are jointly continuous, using either the weak or strong topology for ${\mathcal A}$. The involution

 $\displaystyle \mathcal A\rightarrow\mathcal A,\ a\mapsto a^*$ (2)

is weakly continuous. The operator product

 $\displaystyle \mathcal A\times\mathcal A\rightarrow\mathcal A,\ (a,b)\mapsto ab$ (3)

is individually continuous in each of ${a,b}$, using either the weak or strong topology for ${\mathcal A}$. Furthermore, it is jointly strongly continuous if ${a}$ is restricted to a norm-bounded subset of ${\mathcal A}$.

Proof: For each ${x,y\in V}$, the maps

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle {\mathbb C}\times\mathcal A\rightarrow{\mathbb C},\ (\lambda,a)\mapsto\langle x,(\lambda a)y\rangle=\lambda \langle x,ay\rangle,\smallskip\\ &\displaystyle \mathcal A\times\mathcal A\rightarrow{\mathbb C},\ (a,b)\mapsto\langle x,(a+b)y\rangle=\langle x,ay\rangle+\langle x,by\rangle \end{array}$

are jointly continuous using the weak topology on ${\mathcal A}$, as multiplication and addition are continuous in ${{\mathbb C}}$. Hence (1) are jointly weakly continuous. Similarly, the maps

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle {\mathbb C}\times\mathcal A\rightarrow V,\ (\lambda,a)\mapsto(\lambda a)x=\lambda (ax),\smallskip\\ &\displaystyle \mathcal A\times\mathcal A\rightarrow V,\ (a,b)\mapsto(a+b)x=(ax)+(bx) \end{array}$

are jointly continuous using the strong topology on ${\mathcal A}$, as scalar multiplication and addition are continuous in ${V}$. Hence (1) are jointly strongly continuous. Next, the map

$\displaystyle \mathcal A\rightarrow{\mathbb C},\ a\mapsto\langle x,(a^*)y\rangle = \overline{\langle y,ax\rangle}$

is weakly continuous, as complex conjugation is continuous in ${{\mathbb C}}$. So, (2) is weakly continuous. Also,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \mathcal A\times\mathcal A\rightarrow{\mathbb C},\ (a,b)\mapsto\langle x,(ab)y\rangle&\displaystyle=\langle x,a(by)\rangle\smallskip\\ &\displaystyle=\langle a^*x,by\rangle \end{array}$

is weakly continuous in ${a}$ and in ${b}$ separately, so (3) is weakly continuous in ${a}$ and ${b}$ separately. Similarly,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \mathcal A\times\mathcal A\rightarrow V,\ (a,b)\mapsto(ab)x =a(bx) \end{array}$

is strongly continuous in ${a}$ for each fixed ${b}$, so (3) is strongly continuous in ${a}$. Finally, if ${a_\alpha,b_\alpha\in\mathcal A}$ are nets converging strongly to ${a}$ and ${b}$ respectively, with ${\lVert a_\alpha\rVert\le K}$ for some real ${K}$ then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert a_\alpha b_\alpha x-abx\rVert &\displaystyle \le\lVert a_\alpha(b_\alpha -b)x\rVert+\lVert (a_\alpha-a)bx\rVert\smallskip\\ &\displaystyle \le K\lVert(b_\alpha-b)x\rVert+\lVert(a_\alpha-a)bx\rVert\smallskip\\ &\displaystyle \rightarrow0. \end{array}$

Hence, (3) is jointly strongly continuous when restricted to ${\lVert a\rVert\le K}$. ⬜

Notably, involution is not strongly continuous. For example, taking ${V=\ell^2}$ to be the space of sequences ${x=x_0,x_1,\ldots}$ of complex numbers with ${\sum_n\lvert x_n\rvert^2 < \infty}$, this has inner product ${\langle x,y\rangle=\sum_n\bar x_ny_n}$. The shift maps ${a_n\colon\ell^2\rightarrow\ell^2}$ given by ${(a_nx)_k=x_{k+n}}$ converge strongly to zero. However, their adjoints are isometries, ${\lVert a_n^*x\rVert=\lVert x\rVert}$, so do not converge strongly. Sometimes, an additional operator topology is used, the *-strong topology, under which a net converges, ${a_\alpha\rightarrow\alpha}$ if and only if it converges in the strong topology and the adjoints ${a_\alpha^*\rightarrow a^*}$ converge strongly. This forces the adjoint map to be *-strong continuous. I will not make use of this topology here.

When applied to unbounded sets, the weak and strong topologies are actually slightly too weak for our purposes. So, we define a couple of extra topologies — the ultraweak and ultrastrong topologies — by enlarging the vector space ${V}$. For any semi-inner product space, define

$\displaystyle V^\infty=V\oplus V\oplus V\oplus\cdots=\bigoplus_{n=1}^\infty V.$

An element ${x}$ of ${V^\infty}$ is a sequence ${x_1,x_2,\ldots\in V}$ such that ${\sum_n\lVert x_n\rVert^2}$ is finite. For elements ${x=(x_n)_{n}}$ and ${y=(y_n)_n}$ of ${V^\infty}$, define the semi-inner product,

$\displaystyle \langle x,y\rangle=\sum_{n=1}^\infty\langle x_n,y_n\rangle.$

This sum is absolutely convergent since,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \sum_{n=1}^\infty\lvert\langle x_n,y_n\rangle\rvert&\displaystyle\le\sum_{n=1}^\infty\lVert x_n\rVert\lVert y_n\rVert\smallskip\\ &\displaystyle\le\sqrt{\sum_n\lVert x_n\rVert^2\sum_n\lVert y_n\rVert^2}. \end{array}$

This makes ${V^\infty}$ into a semi-inner product space which is, respectively, a true inner product space or a Hilbert space, whenever ${V}$ is. For any bounded linear map ${a\colon V\rightarrow V}$, then we can also define ${a}$ to act diagonally on ${V^\infty}$ by ${(ax)_n=ax_n}$. It can be seen that the operator norm of ${a}$ acting on ${V^\infty}$ is the same as its action on ${V}$. Hence, if ${(\mathcal A,V)}$ is a bounded *-algebra representation, then ${(\mathcal A,V^\infty)}$ also defines a *-algebra representation.

If ${(\mathcal A,V)}$ is a bounded *-algebra representation, then the ultraweak and ultrastrong topologies on ${\mathcal A}$ are just the weak and strong topologies defined with respect to the representation ${(\mathcal A,V^\infty)}$.

Definition 4 Let ${(\mathcal A,V)}$ be a bounded *-algebra representation. Then, define the following topologies on ${\mathcal A}$.

• The ultraweak topology generated by the maps ${\mathcal A\rightarrow{\mathbb C}}$, ${a\mapsto\langle x,ay\rangle}$, for ${x,y\in V^\infty}$.
• The ultrastrong topology generated by the maps ${\mathcal A\rightarrow V^\infty}$, ${a\mapsto ax}$, for ${x\in V^\infty}$.

So, a net ${a_\alpha\in\mathcal A}$ converges ultraweakly to a limit ${a}$ iff

$\displaystyle \sum_n\langle x_n,ay_n\rangle\rightarrow\sum_n\langle x_n,ay_n\rangle$

for all ${x,y\in V^\infty}$, and converges ultrastrongly to ${a}$ iff

$\displaystyle \sum_n\lVert(a_\alpha-a)x_n\rVert^2\rightarrow0$

for all ${x\in V^\infty}$.

These topologies, and the use of ${V^\infty}$, may seem a bit mysterious at first sight. Other than the simple fact that these definitions turn out to be useful, I will give a brief justification. The collection of maps ${F\colon\mathcal A\rightarrow{\mathbb C}}$ of the form ${F(a)=\langle x,ay\rangle}$, for ${x,y\in V}$, is not closed under linear combinations. Consider the set of all linear combinations of such maps, denoted for now by ${\mathcal X}$. This is a vector space of bounded linear functionals ${F\colon\mathcal A\rightarrow{\mathbb C}}$ of the form

$\displaystyle F(a)=\sum_{n=1}^N\langle x_n,ay_n\rangle$

for finite sequences ${x_n,y_n\in V}$ so, under the operator norm, ${\mathcal X}$ is a normed space. The weak topology on ${\mathcal A}$ is then generated by ${\mathcal X}$. The completion, ${\bar{\mathcal X}}$, of ${\mathcal X}$ is a Banach space of bounded linear functions ${F\colon\mathcal A\rightarrow{\mathbb C}}$, which can be shown to be of the form ${F(a)=\langle x,ay\rangle}$ for ${x,y\in V^\infty}$. It is natural to want all functions in ${\bar{\mathcal X}}$ to be continuous, for which we would use the topology generated by ${\bar{\mathcal X}}$, but this is precisely the ultraweak topology. See, also, lemma 12 below which, for classical probability spaces, gives a simple description of the maps ${a\mapsto\langle x,ay\rangle}$ for ${x,y\in V^\infty}$ and identifies the ultraweak topology with ${\sigma(L^\infty,L^1)}$.

Applying lemma 2 to ${(\mathcal A,V^\infty)}$ relates the ultraweak and ultrastrong topologies.

Lemma 5 A net ${a_\alpha\in\mathcal A}$ tends ultrastrongly to zero iff ${a_\alpha^*a_\alpha}$ tends ultraweakly to zero.

Similarly, applying lemma 3 to the action of ${\mathcal A}$ on ${V}$ extends it to the ultraweak and ultrastrong topologies.

Lemma 6 Let ${(\mathcal A,V)}$ be a bounded *-algebra representation. Then, the vector space operations

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle {\mathbb C}\times\mathcal A\rightarrow\mathcal A,\ (\lambda,a)\mapsto\lambda a,\smallskip\\ &\displaystyle \mathcal A\times\mathcal A\rightarrow\mathcal A,\ (a,b)\mapsto a+b \end{array}$

are jointly continuous, using either the ultraweak or ultrastrong topology for ${\mathcal A}$. The involution

$\displaystyle \mathcal A\rightarrow\mathcal A,\ a\mapsto a^*$

is ultraweakly continuous. The operator product

$\displaystyle \mathcal A\times\mathcal A\rightarrow\mathcal A,\ (a,b)\mapsto ab$

is individually continuous in each of ${a,b}$, using either the ultraweak or ultrastrong topology for ${\mathcal A}$. Furthermore, it is jointly ultrastrongly continuous if ${a}$ is restricted to a norm-bounded subset of ${\mathcal A}$.

The ordering between topologies is straightforward to establish. Note that, despite the terminology, the ultraweak topology is stronger than the weak topology.

Lemma 7 The topologies on ${\mathcal A}$ are ordered as follows, where the arrows point from stronger to weaker topologies.

 ultrastrong ⇒ ultraweak ⇓ ⇓ strong ⇒ weak

Proof: Strong ⇒ Weak: Suppose that the net ${a_\alpha\in\mathcal A}$ tends strongly to ${0}$. For any ${x,y\in V}$, ${a_\alpha y\rightarrow0}$ and, hence, ${\langle x,a_\alpha y\rangle\rightarrow0}$, so ${a_\alpha\rightarrow0}$ weakly.

Ultrastrong ⇒ Ultraweak: Apply strong ⇒ weak’ to ${(\mathcal A,V^\infty)}$.

Ultraweak ⇒ Weak: Suppose that the net ${a_\alpha\in\mathcal A}$ tends ultraweakly to ${0}$. For ${x,y\in V}$, define ${\tilde x,\tilde y\in V^\infty}$ by ${\tilde x=(x,0,0,\ldots)}$ and ${\tilde y=(y,0,0,\ldots)}$. Then,

$\displaystyle \langle x,a_\alpha y\rangle=\langle\tilde x,a_\alpha\tilde y\rangle\rightarrow0$

and, hence, ${a_\alpha\rightarrow0}$ weakly.

Ultrastrong ⇒ Strong: Suppose that the net ${a_\alpha\in\mathcal A}$ tends ultrastrongly to ${0}$. By lemma 5, ${a_\alpha^*a_\alpha\rightarrow0}$ ultraweakly so, by what we have just shown, ${a_\alpha^*a_\alpha\rightarrow0}$ weakly and, by lemma 2 , ${a_\alpha\rightarrow0}$ strongly. ⬜

The closed unit ball of ${\mathcal A_1}$ is defined with respect to the operator seminorm,

$\displaystyle \mathcal A_1=\left\{a\in\mathcal A\colon\lVert a\rVert\le1\right\},$

which is a closed subset of ${\mathcal A}$ under the operator topologies.

Lemma 8 The unit ball ${\mathcal A_1}$ is closed under the weak, strong, ultraweak, and ultrastrong topologies.

Proof: As the weak topology is weaker than all the others, it is sufficient to show that ${\mathcal A_1}$ is weakly closed. Consider a net ${a_\alpha\in\mathcal A_1}$ converging to a limit ${a\in\mathcal A}$. Then, for ${x\in V}$,

$\displaystyle \lVert ax\rVert^2=\langle ax,ax\rangle=\lim_\alpha\langle ax,a_\alpha x\rangle\le\lVert ax\rVert\lVert x\rVert$

showing that ${\lVert ax\rVert\le\lVert x\rVert}$, so ${a\in\mathcal A_1}$. ⬜

On bounded sets, the weak and strong topologies coincide with the ultra’ topologies. Note that a vector topology restricted to ${\mathcal A_1}$ is uniquely determined by the convergence of nets ${a_\alpha\in\mathcal A_1}$ to zero, since ${a_\alpha\in\mathcal A_1}$ converges to a limit ${a\in\mathcal A_1}$ iff ${(a_\alpha-a)/2\in\mathcal A_1}$ tends to zero.

Lemma 9 If ${(\mathcal A,V)}$ is a bounded *-algebra representation then,

• the weak and ultraweak topologies coincide on ${\mathcal A_1}$.
• the strong and ultrastrong topologies coincide on ${\mathcal A_1}$.

Proof: Let ${a_\alpha\in\mathcal A_1}$ be a net converging weakly to zero. For any ${x,y\in V^\infty}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \langle x,a_\alpha y\rangle&\displaystyle=\sum_{n=1}^\infty\langle x_n,a_\alpha y_n\rangle\smallskip\\ &\displaystyle\le\sum_{n=1}^\infty\lvert\langle x_n,a_\alpha y_n\rangle\rvert\smallskip\\ &\displaystyle\le\sum_{n=1}^\infty\lVert x_n\rVert\lVert y_n\rVert < \infty. \end{array}$

For each ${n}$, by weak convergence, ${\langle x_n,a_\alpha y_n\rangle\rightarrow0}$ so, by dominated convergence, ${\langle x,a_\alpha y\rangle\rightarrow0}$. Hence ${a_\alpha\rightarrow 0}$ ultraweakly.

Next, suppose that ${a_\alpha\in\mathcal A_1}$ tends strongly to ${0}$. By lemma 2, ${a_\alpha^* a_\alpha\in\mathcal A_1}$ tends weakly to zero so, by what we have just shown, it also converges ultraweakly. By lemma 5, ${a_\alpha\rightarrow0}$ ultrastrongly. ⬜

We note that, to show that a linear map ${F\colon\mathcal A\rightarrow{\mathbb C}}$ is of the form ${F(a)=\langle x,ay\rangle}$ for some ${x,y\in V^\infty}$, it is enough to express it as

$\displaystyle F(a)=\sum_{n=1}^\infty\langle x_n,ay_n\rangle$

for sequences ${x_n,y_n\in V}$ such that ${\sum_n\lVert x_n\rVert\lVert y_n\rVert}$ is finite. In this case, by scaling, we can find ${\tilde x_n,\tilde y_n\in V}$ satisfying ${\lVert \tilde x_n\rVert^2=\lVert\tilde y_n\rVert^2=\lVert x_n\rVert\lVert y_n\rVert}$ and ${\langle\tilde x_n,a\tilde y_n\rangle=\langle x_n,ay_n\rangle}$. Then,

$\displaystyle \sum_n\lVert\tilde x_n\rVert^2=\sum_n\lVert\tilde y_n\rVert^2=\sum_n\lVert x_n\rVert\lVert y_n\rVert < \infty.$

So, defining ${\tilde x=(\tilde x_n)_n}$ and ${\tilde y=(\tilde y_n)_n}$ in ${V^\infty}$, then ${F(a)=\langle\tilde x,a\tilde y\rangle}$.

If we have a bounded *-probability space ${(\mathcal A,p)}$ then, as previously noted, ${\mathcal A}$ has semi-innner product ${\langle x,y\rangle=p(x^*y)}$. So, ${(\mathcal A,\mathcal A)}$ is a *-algebra representation, with ${\mathcal A}$ acting on itself by left-multiplication. Hence, the definitions and results here apply to ${\mathcal A}$ which is, of course, the whole point of this post.

Lemma 10 Let ${(\mathcal A,p)}$ be a bounded *-probability space. Then, there exists ${x,y\in\mathcal A^\infty}$ such that ${p(a)=\langle x,ay\rangle}$.

Proof: If ${\mathcal A}$ is unitial, then the result is trivial, as we can write ${p(a)=\langle 1,a1\rangle}$. More generally, by lemma 10 of the post on states, there exists a sequence ${u_n\in\mathcal A}$ with ${\lVert u_n\rVert_2=1}$ and ${p(u_n)\rightarrow1}$, in which case, taking ${u_0=0}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle p(a)&\displaystyle=\lim_{n\rightarrow\infty}p(u_n^*au_n)\smallskip\\ &\displaystyle=\sum_{n=1}^\infty\left(\langle u_n,au_n\rangle-\langle u_{n-1},au_{n-1}\rangle\right)\smallskip\\ &\displaystyle=\sum_{n=1}^\infty\left(\langle(u_n-u_{n-1},au_n\rangle+\langle u_{n-1},a(u_n-u_{n-1})\rangle\right). \end{array}$

However, again by lemma 10, ${u_n}$ is ${L^2}$-Cauchy, so we can pass to a subsequence with ${\lVert u_n-u_{n-1}\rVert < 2^{-n}}$ for all ${n >1}$. So,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle \sum_{n=1}^\infty\left(\lVert u_n-u_{n-1}\rVert_2\lVert u_n\rVert_2+\lVert u_{n-1}\rVert_2\lVert u_n-u_{n-1}\rVert_2\right)\smallskip\\ &\displaystyle \qquad\le1+\sum_{n=2}^\infty\left(2^{-n}+2^{-n}\right) < \infty. \end{array}$

Next, for a *-probability space, the fact that it has a semi-inner product means that we have further topologies on ${\mathcal A}$. First, there is the ${L^2(p)}$ seminorm topology, also called the strong ${L^2(p)}$ topology, so that ${x_\alpha\in\mathcal A}$ tends to a limit ${a}$ iff ${\lVert a_\alpha-a\rVert_2\rightarrow0}$. There is also a weak ${L^2(p)}$ topology, where ${x_\alpha\rightarrow x}$ iff ${\langle y,x_\alpha\rangle\rightarrow\langle y,x\rangle}$ for all ${y\in\mathcal A}$. These topologies are related to the ones described above as follows.

Lemma 11 Let ${(\mathcal A,p)}$ be a bounded *-probability space. Then,

• the ultraweak topology on ${\mathcal A}$ is stronger than the weak ${L^2(p)}$ topology.
• the ultrastrong topology on ${\mathcal A}$ is stronger than the strong ${L^2(p)}$ topology.

If ${\mathcal A}$ is unitial then,

• the weak topology on ${\mathcal A}$ is stronger than the weak ${L^2(p)}$ topology.
• the strong topology on ${\mathcal A}$ is stronger than the strong ${L^2(p)}$ topology.

If ${p}$ is tracial, so that ${p(ab)=p(ba)}$ for all ${a,b\in\mathcal A}$, then,

• the weak ${L^2(p)}$ topology on ${\mathcal A}$ is stronger than the weak topology.
• the strong ${L^2(p)}$ topology on ${\mathcal A}$ is stronger than the strong topology.
• the weak, ultraweak and weak ${L^2(p)}$ topologies coincide on ${\mathcal A_1}$.
• the strong, ultrastrong and strong ${L^2(p)}$ topologies coincide on ${\mathcal A_1}$.

If ${p}$ is tracial and ${\mathcal A}$ is unitial then, the weak and weak ${L^2(p)}$ topologies coincide, and the strong and strong ${L^2(p)}$ topologies coincide.

Proof: Suppose that net ${a_\alpha\in\mathcal A}$ tends to zero weakly. If ${\mathcal A}$ is unitial then this implies that ${\langle y,a_\alpha\rangle=\langle y,a_\alpha1\rangle}$ tends to zero for all ${y\in\mathcal A}$, so ${a_\alpha\rightarrow0}$ ${L^2}$-weakly.

Suppose that ${a_\alpha\rightarrow0}$ strongly. If ${\mathcal A}$ is unitial then this implies that ${a_\alpha=a_\alpha1}$ tends to zero ${L^2}$-strongly.

Suppose that ${a_\alpha\rightarrow0}$ ultraweakly, then ${y^*a_\alpha\rightarrow0}$ ultraweakly for any ${y\in\mathcal A}$. By lemma 10, ${p}$ is ultraweakly continuous, so ${\langle y,a_\alpha\rangle=p(y^*a_\alpha)\rightarrow0}$, and ${a_\alpha\rightarrow0}$ ${L^2}$-weakly.

Suppose that ${a_\alpha\rightarrow0}$ ultrastrongly. Lemma 5 says that ${a_\alpha^*a_\alpha\rightarrow0}$ ultraweakly and, as ${p}$ is ultraweakly continuous, ${\lVert a_\alpha\rVert_2^2=p(a_\alpha^*a_\alpha)\rightarrow0}$, so ${a_\alpha\rightarrow0}$ ${L^2}$-strongly.

Suppose that ${a_\alpha\rightarrow0}$ in the weak ${L^2(p)}$ topology. If ${p}$ is tracial then, for ${x,y\in\mathcal A}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \langle x,a_\alpha y\rangle&\displaystyle=p(x^*a_\alpha y)=p(yx^*a_\alpha)\smallskip\\ &\displaystyle =\langle xy^*, a_\alpha \rangle\rightarrow0, \end{array}$

showing that ${a_\alpha\rightarrow0}$ weakly.

Suppose that ${a_\alpha\rightarrow0}$ in the strong ${L^2(p)}$ topology. If ${p}$ is tracial then, for ${x\in\mathcal A}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert a_\alpha x\rVert_2^2&\displaystyle=p(x^*a_\alpha^* a_\alpha x)=p(a_\alpha xx^* a_\alpha^*)\smallskip\\ &\displaystyle=\lVert x^* a_\alpha^*\rVert_2^2\le\lVert x^*\rVert_\infty^2\lVert a_\alpha^*\rVert_2^2\smallskip\\ &\displaystyle=\lVert x^*\rVert_\infty^2p(a_\alpha a_\alpha^*)=\lVert x^*\rVert_\infty^2 p(a_\alpha^*a_\alpha)\smallskip\\ &\displaystyle=\lVert x^*\rVert_\infty^2\lVert a_\alpha\rVert_2^2\rightarrow0, \end{array}$

so ${a_\alpha\rightarrow 0}$ strongly.

Finally, if ${p}$ is tracial then, from the above, the weak ${L^2}$ topology lies between the weak and ultraweak so, as lemma 9 says that the weak and ultraweak topologies coincide on ${\mathcal A_1}$, all three topologies coincide on ${\mathcal A_1}$. Similarly, the strong, strong ${L^2}$, and ultrastrong topologies coincide on ${\mathcal A_1}$. ⬜

So, for tracial states, the weak, ultraweak and weak ${L^2}$ topologies coincide on ${\mathcal A_1}$, as do the strong, ultrastrong, and strong ${L^2}$ topologies. For this reason, when dealing only with tracial *-probability spaces, it is possible to get by with just the ${L^2}$ topology. On the other hand, for general states, the ${L^2}$ topologies can be strictly weaker. Example 5 of the post on states gives a *-probability space ${(\mathcal A,p)}$ and self adjoint ${a\in\mathcal A_1}$ such that ${p(a^n)=0}$ for all ${n\ge1}$. Hence, ${a^n\rightarrow0}$ trivially in the strong (and weak) ${L^2}$ topologies. However, there exists (unitary) ${u\in\mathcal A}$ with ${p(u^*a^nu)=1}$ for all ${n}$, so that ${a^n}$ does not tend to zero in the operator topologies.

These additional topologies can be added to the ordering described by lemma 7.

 strong ${L^2(p)}$ ⇒ weak ${L^2(p)}$ ⇑ ⇑ ${L^\infty(p)}$ ⇒ ultrastrong ⇒ ultraweak ⇓ ⇓ strong ⇒ weak

To demonstrate the definitions given above, I apply them to the case of a classical probability space ${(\Omega,\mathcal F,{\mathbb P})}$. Use ${L^\infty(\Omega,\mathcal F,{\mathbb P})}$ to denote the space of uniformly bounded (and measurable) complex-valued random variables ${X\colon\Omega\rightarrow{\mathbb C}}$, ${L^1(\Omega,\mathcal F,{\mathbb P})}$ for the integrable random variables, and ${L^2(\Omega,\mathcal F,,{\mathbb P})}$ for the square-integrable random variables. It is usual to identify any two random variables which are almost surely equal. Then, it is standard theory that ${L^1}$ is a Banach space under the norm ${\lVert X\rVert_1={\mathbb E}[\lvert X\rvert]}$, with dual space ${L^\infty}$ under the essential supremum norm, ${\lVert X\rVert_\infty}$, and under the pairing ${(X,Y)\mapsto{\mathbb E}[XY]}$ for ${X\in L^\infty}$ and ${Y\in L^1}$. The ${\sigma(L^\infty,L^1)}$ topology on ${L^\infty}$ is the weakest topology such that ${X\mapsto{\mathbb E}[XZ]}$ is continuous for each ${Z\in L^1}$. This is also called the weak topology on ${L^\infty}$ although, to avoid confusion with the weak topology defined above, I do not use this terminology . On ${L^2(\Omega,\mathcal F,{\mathbb P})}$, we have an associated norm ${\lVert X\rVert_2={\mathbb E}[\lvert X\rvert^2]^{\frac12}}$.

Lemma 12 Let ${(\Omega,\mathcal F,{\mathbb P})}$ be a probability space. With respect to the *-probability space ${(L^\infty(\Omega,\mathcal F,{\mathbb P}),{\mathbb P})}$, the following are equivalent for any map ${F\colon L^\infty\rightarrow{\mathbb C}}$,

1. ${F(A)=\langle x,Ay\rangle}$, for some ${x,y\in\mathcal A^\infty}$.
2. ${F(A)={\mathbb E}[AZ]}$, for some ${Z\in L^1(\Omega,\mathcal F,{\mathbb P})}$.

Furthermore,

• the ultraweak topology on ${L^\infty}$ coincides with the ${\sigma(L^\infty,L^1)}$ topology.
• the strong topology on ${L^\infty}$ coincides with the ${L^2(\Omega,\mathcal F,{\mathbb P})}$ norm topology.
• the ultrastrong and strong topologies on the unit ball of ${L^\infty}$ coincide with convergence in probability.

Proof: Suppose that ${F(A)=\langle x,Ay\rangle}$ for some ${x,y\in V^\infty}$, with ${V=L^\infty\subseteq L^2}$ as an inner product space. Then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \sum_n{\mathbb E}[\lvert x_ny_n\rvert]&\displaystyle\le\sum_n\lVert x_n\rVert_2\lVert y_n\rVert_2\smallskip\\&\displaystyle\le\lVert x\rVert\lVert y\rVert < \infty. \end{array}$

Hence, setting ${Z=\sum_n x_n^*y_n}$, by monotone convergence,

$\displaystyle {\mathbb E}[\lvert Z\rVert]\le\sum_n{\mathbb E}[\lvert x_ny_n\rvert] < \infty.$

So, ${Z\in L^1}$ and, by dominated convergence,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle F(A)&\displaystyle=\sum_n\langle x_n,Ay_n\rangle=\sum_n{\mathbb E}\left[x_n^*Ay_n\right]\smallskip\\ &\displaystyle={\mathbb E}\left[A\sum_nx_n^*y_n\right]={\mathbb E}[AZ] \end{array}$

as required.

Conversely, suppose that ${F(A)={\mathbb E}[AZ]}$ for some ${Z\in L^1}$. Write ${Z=U\lvert Z\rvert}$ for ${U\in{\mathbb C}}$ with ${\lvert U\rvert=1}$. Then, set ${x_n=1_{\{n-1\le\lvert Z\rvert < n\}}\sqrt{\lvert Z\rvert}}$ and ${y_n=Ux_n}$. As these are uniformly bounded, they are in ${L^\infty}$ and, by monotone convergence,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \sum_{n=1}^\infty\lVert x_n\rVert_2^2= \sum_{n=1}^\infty\lVert y_n\rVert_2^2= &\displaystyle\sum_{n=1}^\infty{\mathbb E}\left[1_{\{n-1\le \lvert Z\rvert < n\}}\lvert Z\rvert\right]\smallskip\\ &\displaystyle={\mathbb E}[\lvert Z\rvert] < \infty. \end{array}$

So, we can define ${x=(x_n)_n}$ and ${y=(y_n)_n}$ in ${V^\infty}$. By dominated convergence,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \langle x,Ay\rangle&\displaystyle=\sum_n\langle x_n,Ay_n\rangle=\sum_n{\mathbb E}[x_n^*Ay_n]\smallskip\\ &\displaystyle=\sum_n{\mathbb E}\left[A1_{\{n-1\le\lvert Z\rvert < n\}}Z\right]={\mathbb E}[AZ] \end{array}$

as required.

We have shown equivalence of the two statements for ${F}$. Hence, from the definition of the ultraweak topology, it is the weakest topology on ${L^\infty}$ such that ${A\mapsto{\mathbb E}[AZ]}$ is continuous for all ${Z\in L^1}$. By definition, this is the same as the ${\sigma(L^\infty,L^1)}$ topology.

Next, as ${L^\infty(\Omega,\mathcal F,{\mathbb P})}$ is unitial and commutative, the strong and strong ${L^2}$ topologies coincide by lemma 11.

Finally, we show that the strong and ultrastrong topologies and convergence in probability agree on the unit ball of ${L^\infty}$. First, the dominated convergence theorem says that, on the unit ball, the ${L^2}$-norm topology coincides with convergence in probability and, as we showed above, also agrees with the strong topology. Also, the strong and ultrastrong topologies coincide by lemma 9. ⬜

Lastly, I will note some further basic properties relating the operator topologies, the proofs of which are slightly out of scope here but which will be included in more detail in a later post. Given a bounded *-algebra representation ${(\mathcal A,V)}$, use ${S^{\rm w},S^{\rm s},S^{\rm uw},S^{\rm us}}$ to denote the closure of ${S\subseteq\mathcal A}$ under, respectively, the weak, strong, ultraweak, and ultrastrong topologies. For example, lemma 9 above can be expressed by the identities

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle S^{\rm w}=S^{\rm uw},\smallskip\\ &\displaystyle S^{\rm s}=S^{\rm us}, \end{array}$ (4)

for all ${L^\infty}$-bounded sets ${S}$. Now, for the special case where ${V}$ is a Hilbert space, then it is well-known that the weak and strong topologies are compatible. This means that a linear map ${F\colon\mathcal A\rightarrow{\mathbb C}}$ is weakly continuous if and only if it is strongly continuous. By the Hahn–Banach theorem, this is equivalent to both topologies having the same closed convex sets, so ${S^{\rm w}=S^{\rm s}}$ for convex ${S}$. Considering the action on ${V^\infty}$, it is similarly true that ${S^{\rm uw}=S^{\rm us}}$. Using Hilbert space completions, this last statement can be extended to the general situation where ${V}$ is only a semi-inner product space. So, we have the identity

 $\displaystyle S^{\rm uw}=S^{\rm us}$ (5)

for all convex ${S\subseteq\mathcal A}$. Next, as lemma 9 states that the unit ball is closed under each of the operator topologies, we have ${(S\cap\mathcal A_1)^\tau\subseteq S^\tau\cap\mathcal A_1}$ where ${\tau}$ is any of the operator topologies. For the special case where ${V}$ is a Hilbert space, the Kaplansky density theorem states that, if ${S}$ is a *-subalgebra, then the reverse inequality holds for the strong topology and, by considering the action on ${V^\infty}$, for the ultrastrong topology. In this form, the statement generalizes to arbitrary semi-inner product spaces ${V}$. Combining with (4) and (5), we have the following satisfying string of equalities for any *-subalgebra ${\mathcal B}$ of ${\mathcal A}$,

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle \mathcal B^{\rm uw}\cap\mathcal A_1=\mathcal B^{\rm us}\cap\mathcal A_1=(\mathcal B\cap\mathcal A_1)^{\rm w}\smallskip\\ &\displaystyle =(\mathcal B\cap\mathcal A_1)^{\rm s}=(\mathcal B\cap\mathcal A_1)^{\rm uw}=(\mathcal B\cap\mathcal A_1)^{\rm us}. \end{array}$ (6)