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Given two normed spaces $(X, ||\cdot||_X)$ and $(Y,||\cdot||_Y)$ the space of bounded linear maps $\mathcal{B}(X,Y)$ can be equipped with the strong operator topology (SOT) as follows:

The initial topology with respect to[*] The topology induced by the
family $S := \{s_x : x \in X\}$, $$s_x : \begin{cases} &\mathcal{B}(X,Y) &\to &[0,\infty) \\ &T &\mapsto &||T(x)||_Y \end{cases} $$

(Definition taken from my lecture notes, but seem to coincide with Operator topologies - Wikipedia) $S$ is a separating family of seminorms, so this yields a locally-convex topological vector space.

In a similar fashion, we can define the weak operator topology (WOT) as

The initial topology with respect to $W :=\{w_{x,y'} : x \in X, y' \in Y'\}$. $$w_{x,y'} : \begin{cases} &\mathcal{B}(X,Y) &\to &\mathbb{C} \\ &T &\mapsto &y'(T(x)) \end{cases} $$ $Y'$ representing the topological dual space of $Y$.

Now my questions are:

Why do we have to go all the way to $[0,\infty)$ and $\mathbb{C}$, respectively? Could these topologies be defined as the initial topologies with respect to the family $P:= \{ p_x : x \in X\}$ of pointwise evaluation $$p_x : \begin{cases} &\mathcal{B}(X,Y) &\to &(Y,\mathcal{T}) \\ &T &\mapsto &T(x) \end{cases} $$ only this time we equip $Y$ with either the norm topology $\mathcal{T} := \mathcal{T}_{||\cdot||_Y}$ or the weak topology $\mathcal{T} := \mathcal{T}_w$ instead?

and also

Wikipedia describes the weak-topology in a slightly different way, namely using extra absolute values $$|W| :=\{|w_{x,y'}| : x \in X, y' \in Y'\}.$$ Are there any benefits in doing this? (I guess this has something to do with the connection of separating families of seminorms and point separating subspaces of the algebraic dual.)

*Edit: As Daniel pointed out the "topology induced by a family of seminorms" is different from "the initial topology with respect to the seminorms".

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Yes, we could also define the strong resp. weak operator topologies as the initial topology with respect to the evaluation maps $p_x$, where $Y$ is endowed with the strong resp. weak topology.

For the weak operator topology that is obvious from the transitivity of initial topologies, since the weak topology is just the initial topology with respect to the family $\{ y' : y' \in Y'\}$.

For the strong operator topology, we can't use such an abstract argument, unfortunately, because the strong topology on $Y$ is not the initial topology with respect to $\lVert\,\cdot\,\rVert_Y \colon Y \to [0,\infty)$.

At this point, let me remark that the strong operator topology also is (generally) not the initial topology with respect to the family $\{ s_x \colon T \mapsto \lVert Tx\rVert_Y \mid x \in X\}$. A neighbourhood of $S \in B(X,Y)$ with respect to that initial topology is of the form

$$\left\lbrace T : \lvert \lVert Tx_i\rVert_Y - \lVert Sx_i\rVert_Y\rvert < \varepsilon, \; 1\leqslant i \leqslant n \right\rbrace\tag{1}$$

for some finite family $\{x_1,\dotsc,x_n\} \subset X$, while the neighbourhoods in the strong operator topology are of the form

$$\left\lbrace T : \lVert Tx_i - Sx_i\rVert_Y < \varepsilon, \; 1\leqslant i \leqslant n \right\rbrace,\tag{2}$$

and such a neighbourhood rarely contains a set of the form $(1)$. The initial topology with respect to $\{s_x\}$ cannot distinguish between operators that differ by an isometry of $Y$.

The strong operator topology is the topology induced by the seminorms $s_x$, not the initial topology with respect to these.

Back to the possibility to define the strong operator topology as the initial topology with respect to the evaluation maps $p_x$: The evaluation maps are continuous with respect to the strong operator topology, hence that is finer than the initial topology. Conversely, a neighbourhood of $S\in B(X,Y)$ of the form $(2)$ can be written as

$$\bigcap_{i=1}^n p_{x_i}^{-1}\left(B_\varepsilon(p_{x_i}(S))\right),$$

which is open in the initial topology, so the initial topology is also finer than the strong operator topology, hence the two are equal.

Wikipedia describes the weak-topology in a slightly different way, namely using extra absolute values $$\lvert W\rvert := \{\lvert w_{x,y'}\rvert : x\in X, y' \in Y'\}.$$ Are there any benefits in doing this? (I guess this has something to do with the connection of separating families of seminorms and point separating subspaces of the algebraic dual.)

Yes, in that way you have seminorms that define the topology. On the other hand, that means the weak operator topology is (generally) not the initial topology with respect to $\lvert W\rvert$.

There are benefits to both approaches/definitions. Defining the topologies as initial topologies gives direct access to the general topological facts about initial topologies, and defining them as the topologies induced by a family of seminorms gives direct access to the theorems about locally convex topologies.

The theorems about locally convex topologies are more important (and, I dare say, also better known in the target audience) for analysts working with the various operator topologies, so altogether, defining the topologies by the seminorms seems preferable.

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  • $\begingroup$ As you point out the norm topology is not induced by the map $y \mapsto ||y||_Y$. I think I can understand why, but still it would be nice if it were in some way. Would it be possible to iron out this unevenness by for example using the initial topology with respect to the family of "translated norms" $\{||.-y||_Y : y \in Y\}$? Or by saying: The initial topology with respect to the norm and the "+"-operation? Could this then be applied to the operator-topolgy situation, so I could say "The operator topology is the initial topology with respect to ..." instead of the "induced by"-construction? $\endgroup$ – knedlsepp Feb 21 '14 at 22:19
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    $\begingroup$ Yes, the norm topology is the initial topology with respect to the family $n_y \colon z \mapsto \lVert z-y\rVert_Y$. Then we can say that the strong operator topology is the initial topology with respect to the $n_y\circ p_x$. $\endgroup$ – Daniel Fischer Feb 21 '14 at 22:29

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