Strong Operator Topology vs Weak Operator Topology

Question

Let $X$ and $Y$ be Banach spaces. I want to make sure I understand these topologies very well before moving on with my studies.

For definitions I have that (source: these notes from Tao):

The strong operator topology on $L(X,Y)$ is the map $T \mapsto \|Tx\|_Y$ is continuous for $T \in L(X,Y)$, $x \in X$, and where $\|\cdot\|_Y$ is the norm in $Y$.

The weak operator topology on $L(X,Y)$ is the map $T \mapsto |f(Tx)|$ is continuous for $T \in L(X,Y)$, $x \in X$, and $f \in Y^*$.

From these definitions I have three questions:

1. I have seen different texts and notes instead define the strong operator topology in terms of the map $T \mapsto |Tx|$, where instead of a norm there is an absolute value. Is this simply notation or is there more to this?

2. I am trying to visualize what a neighborhood basis of the two topologies look like, say centered around some operator $S$. Based on my current understanding, I believe for the strong operator topology they should be of the form, (where $x \in X$ and $\epsilon > 0$ is fixed): $$\{T \in L(X,Y) : \|Tx - Sx\|_Y < \epsilon\}$$ and for the weak operator topology they will be of the form (where, additionally, $f \in Y^*$ is fixed): $$\{T \in L(X,Y) : |f(Tx - Sx)| < \epsilon\}.$$ Is this correct? Instead of a neighborhood, how would one define the basic open sets of these topologies? For completeness, what do the open balls look like in the operator norm topology?

3. I have read that, as the name suggests, the weak operator topology is weaker than the strong operator topology. However it is not clear to me why this is? How does the inclusion of a linear functional in some way relax the number of open sets?

Answer 1

1: It is merely notational.

2: The sets you’ve written are neighbourhoods, but you’ve written a subbasis, not a basis. What I mean by that is that you need to allow non-fixed $x$ to get the “finite intersection of open sets is open” criterion.

Let $S\in L(X,Y)$. A basic strong open neighbourhood of $S$ is of the form: $$\{T\in L(X,Y):\|(S-T)x_i\|\lt\varepsilon_i,\forall1\le i\le n\}$$For $\{x_i\}_{i=1}^n\subset X$ and $\varepsilon_i\gt0$. Notice that the expression $\|(S-T)\|$ is correct but your original description of the SOT was wrong. It is the coarsest topology in which $T\mapsto Tx$ is continuous for each $x$, not $T\mapsto|Tx|$. In the latter case the basic neighbourhoods would feature the expression $|\|Tx_i\|-\|Sx_i\||$ which is ... weird, and unhelpful (no longer the topology of pointwise convergence). The above set is an intersection of subbasic open sets (where $x$ is fixed).

Likewise we need to change the expression for the weak basis:

A basic weak open neighbourhood of $S$ is of the form: $$\{T\in L(X,Y):|\psi_i((S-T)x_i)|\lt\varepsilon_i,\forall1\le i\le n\}$$Where $\{\psi_i\}_{i=1}^n\subset Y^\ast$ and $\{x_i\}_{i=1}^n\subset X$ and $\{\varepsilon_i\}_{i=1}^n\subset\Bbb R_+$. Again, the above expression is a finite intersection of subbasic open neighbourhoods, in order to create a full basis. Also, the description of the WOT was wrong: we want the maps $T\mapsto\psi(Tx)$ to be continuous.

3: Linear functionals have a kernel with codimension $1$. If I want to squeeze $S,T$ into the same weak neighbourhood where particular $\psi,x$ are fixed, I need to make sure that $S-T$ has small image along the support of $\psi$ - they may create a huge vector along the kernel. The same goes for multiple $\psi$: we intersect their kernels, we will still (when $Y$ is infinite dimensional) have an enormous vector subspace on which they are all $0$. I need only require $S-T$ is small outside of this subspace, but within the shared kernel? It can be huge. This is similar to the fact that every weak open neighbourhood in the weak topology on $X$ is unbounded when $X$ is infinite dimensional.