Initial Topology and Weak and Strong Operator Topology

Question

Given a set $X$ and a family of topological spaces $(Y_i)_{i \in I}$ and functions $f_i : X \to Y_i$, we can define a topology on $X$ called the initial topology on $X$ with respect to the topological spaces and functions $(Y_i, f_i)_{i \in I}$. This topology is defined to be the coarsest/weakest/smallest topology with respect to which all the maps $f_i$ are continuous. In terms of open sets, all sets of the form $f_i^{-1}(U)$ for $U$ open in $Y_i$ and some $i \in I$ forms a subbasis of this (the initial) topology on $X$. Also, a net $(x_j) \subseteq X$ converges to $x \in X$ with respect to this topology if and only if $f_i(x_j) \to f_i(x)$ in $Y_i$ for every $i \in I$.

Using this procedure, we can define the weak and strong operator topology on $B(\mathcal{H})$, the $C^*$-algebra of all bounded linear operators on the Hilbert space $\mathcal{H}$, via certain families of seminorms. According to the book An introduction to $II_1$ factors, the weak operator topology is the initial topology generated by the seminorms $p_{\xi , \eta}(x) := |\langle \xi , x \eta \rangle |$ for all $\xi , \eta \in \mathcal{H}$, while the strong operator topology is the initial topology generated by the seminorms $p_{\xi}(x) = \|x \xi\|$ for all $\xi \in \mathcal{H}$. Hence, $(x_i)_{i \in I} \subseteq B(\mathcal{H})$ converges to $x \in B(\mathcal{H})$ with respect to the weak topology if and only if $|\langle \xi , x_i \eta \rangle | \to |\langle \xi , x \eta \rangle |$ for every $\xi , \eta \in \mathcal{H}$, while it converges with respect to the strong operator topology if and only if $\|x_i \xi\| \to \|x \xi\|$ for every $\xi \in \mathcal{H}$.

However, in many other sources (wiki, Conway's book on functional analysis, etc.), the weak and strong operator topology seem to be defined slightly different. According to these other sources, $x_i \to x$ in the weak operator topology if and only if $\langle \xi , x_i \eta \rangle \to \langle \xi , x \eta \rangle$ for all $\xi ,\eta \in \mathcal{H}$, while $x_i \to x$ in the strong operator topology if and only if $\|(x_i - x) \xi \| \to 0$ for all $\xi \in \mathcal{H}$. This is slightly different from what I wrote above. Perhaps I am being dumb, but I don't quite see the equivalence between the way the weak and strong operator topology are defined in...compared to other sources.

If $\langle \xi , x_i \eta \rangle \to \langle \xi , x \eta \rangle$, then, because the absolute value function is continuous, we have that $|\langle \xi , x_i \eta \rangle| \to | \langle \xi , x \eta \rangle |$, but I don't see how the converse holds. Again, if $\|(x_i - x) \xi \| \to 0$, then, using the reverse triangle inequality, we have

$$|~ \|x_i \xi \| - \|x \xi \| ~ | \le \|x_i \xi - x \xi \| = \|(x_i -x) \xi \| \to 0$$

But I don't see why the converse holds.

Answer 1

Indeed there are many equivalent characterizations of the strong and the weak operator topology. Following Chapter 2.1.4 / Proposition 2.1.20 of my PhD thesis (arXiv version here) --- and recalling that the initial topology is defined as the smallest topology such that all maps from a specified family are continuous --- the following holds (your setting is recovered via $X=Y=\mathcal H$):

Proposition. Let normed spaces $X,Y$ and a topology $\tau$ on $\mathcal B(X,Y)$ be given. The following statements are equivalent:

1. $\tau$ is the initial topology on $\mathcal B(X,Y)$ with respect to the family $\{T\mapsto \|Tx-Sx\|\}_{x\in X,S\in\mathcal B(X,Y)}$. We say that $\tau$ is the topology induced by the seminorms $T\mapsto \|Tx\|$.
2. $\tau$ is the initial topology with respect to the family $\{T\mapsto Tx\}_{x\in X}$.
3. For all nets $(T_i)_{i\in I}$ in $ \mathcal B(X,Y)$ and all $T\in \mathcal B(X,Y)$ one has $T_i\to T$ in $\tau$ if and only if $T_ix\to Tx$ for all $x\in X$.
4. $\tau$ is the topology generated by the basis $$\mathsf S:=\{N(T,A,\varepsilon)\,:\,T\in\mathcal B(X,Y),A\subset X\text{ finite, }\varepsilon>0\}$$ where $N(T,A,\varepsilon):= \{S\in\mathcal B(X,Y)\,:\,\|Tx-Sx\|<\varepsilon\text{ for all }x\in A\}$, that is, $\tau$ is the collection of arbitrary unions of elements of $\mathsf S$.

If $\tau$ satisfies one (and thus all) these conditions we call it the strong operator topology. Moreover, $\tau$ in general is not the initial topology with respect to the family $\{T\mapsto\|Tx\|\}_{x\in X}$.

For the weak operator topology one gets an analogous result & proof which I, however, will omit for now (if you're interested you can find all the details in the document I linked).

Either way the proof of this becomes quite simple upon learning that comparing topologies can be done, equivalently, via nets/generalized sequences:

Given two topologies $\tau_1,\tau_2$ on an arbitrary set $X$ the following are equivalent:

• $\tau_1\subseteq\tau_2$ (i.e. $\tau_1$ is weaker than $\tau_2$)
• ${\rm id}_X:(X,\tau_2)\to(X,\tau_1)$ is continuous (definition of continuity)
• For every net $(x_i)_{i\in I}$ in $X$ which converges to $x\in X$ with respect to $\tau_2$ one has $x_i\to x$ with respect to $\tau_1$. (characterization of continuity via nets, cf. also Theorem 11.8 in "General Topology" (1970) by Willard)

In particular, given a set $X$ and a family $\mathcal F:=(f_j)_{j\in J}$ of functions $f_j:X\to Y_j$ (each $Y_j$ being a topological space) a net $(x_i)_{i\in I}$ in $X$ converges to $x\in X$ in the initial topology w.r.t. $\mathcal F$ iff $f_j(x_i)\to f_j(X)$ for all $j\in J$ (cf. Theorem 8.10 in Willard's "General Topology")

With this let us come to the

Proof idea (of the above proposition). First note that by "(2) $\Leftrightarrow$ (3)" is precisely the characterization of the initial topology in terms of nets given above. Then "(1) $\Leftrightarrow$ (3)" boils down to two inequalities, one of which you have stated already (the reverse triangle inequality):

• If $T_ix\to Tx$, i.e. $\|T_ix-Tx\|\to 0$, then for all $S,x$ $$ \big|\|T_ix-Sx\|-\|Tx-Sx\|\big|\leq\|T_ix-Sx-Tx+Sx\|=\|T_ix-Tx\|\to 0 $$
• If $\|T_ix-Sx\|\to 0$ for all $x,S$, then setting $S=T$ yields $\|T_ix-Tx\|\to 0$, i.e. $T_ix\to Tx$. This is also where your observation comes in that it is not enough to know that $\|T_ix\|\to\|Tx\|$ in order to deduce that $\|T_ix-Tx\|\to 0$. In other words the initial topology with respect to $T\mapsto \|Tx\|$ is, in general, not a characterization of the strong operator topology

(4) $\Rightarrow$ (3): This follows from a characterization of a topology in terms of a basis of the topology. More precisely, a net $(x_i)_{i\in I}$ converges to $x$ if and only if for all basis elements $B$ there exists $i_0\in I$ such that $x_i\in B$ for all $i\succeq i_0$. From this it is straightforward to relate (3) and (4).