I'm not sure this is an appropriate question for this site but I've tried math stack exchange and got no answers. Also, this problem arose in one of my research problems, so I'm stating it here.

The strong operator topology is defined on Simon and Reed's book as follows. It is the weakest topology on $\mathcal{L}(X,Y)$ such all the maps $E_{x}: \mathcal{L}(X,Y) \to Y$ defined by: $$E_{x}(T) := Tx $$ are continuous for all $x \in X$. Here, $X$ and $Y$ are supposed to be Banach spaces and $\mathcal{L}(X,Y)$ is the space of all bounded linear operators from $X$ to $Y$. A neighboorhood basis for this topology, in Simon's words, is given by the sets of the form: $$ \{S: \hspace{0.1cm} S\in \mathcal{L}(X,Y), \hspace{0.1cm} ||Sx_{i}||_{Y}<\epsilon, \hspace{0.2cm} i=1,...n\}$$ where $x_{1},...,x_{n}$ is any finite collection of elements of $X$ and $\epsilon > 0$.

I know the notion of strong topology can be extended to more general spaces such as topological vector spaces, but I don't want to get too deep into the theory. However, I'm interested in the case where $X$ is not Banach but $Y = \mathbb{C}$ is Banach.

**My question is:** In my setup, if $X$ is a Fréchet space and $Y=\mathbb{C}$ is Banach, the above definition seems to work just fine if I replace $\mathcal{L}(X,Y)$ the space of bounded linear operators to its analogue, the space of all continuous linear maps. The same properties seem to hold in this case. Is it a correct definition of a strong topology to my particular case? In other words, if I was to consider $X$ as a topological vector space and $X^{*}$ its topological dual, would the strong topology defined on $X$ be the same topology I'm proposing?

strong operator topologyas well asweak$^*$ topologyquite unfortunate (although they are of course standard). A good name, IMHO, would betopology of pointwise convergence. $\endgroup$ – Jochen Wengenroth Jun 8 '20 at 13:502more comments