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](https://rads.stackoverflow.com/amzn/click/com/0125850506) 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 topology on a topological vector space
This is a clash of two cultures which use the adjective "strong" for a topology with completely different meanings. I agree with Jochen's that this choice of terminology is quite unfortunate. I believe the question the OP is after is what is the correct topology on spaces of distributions like $\mathscr{D}'$, $\mathscr{S}'$ and their sequence space concrete realization like $s'$, etc. The answer is the strong topology in the sense of the topological vector spaces literature. The more precise (Jochen comment compliant) terminology would be the *topology of uniform convergence on bounded sets*.


First review the basic definitions given in


<https://math.stackexchange.com/questions/3510982/doubt-in-understanding-space-d-omega/3511753#3511753>


to which one can add the following. For a LCTVS $V$, and a subset $A\in V$, we say that $A$ is bounded iff for every continuous seminorm $\rho$ on $V$,
$$
\sup\_{v\in A}\rho(v)\ <\ \infty\ .
$$
The strong dual $V'$ is the space of continuous linear forms $L:V\rightarrow\mathbb{C}$
with locally convex topology defined by the collection of seminorms
$$
\rho\_A(L)=\sup\_{v\in A}|L(v)|
$$
indexed by the (nonempty) bounded sets $A$ in $V$.


Take the space of sequences $s$. To a nonempty subset $A\subset s$, one can associate an envelope ${\rm env}(A)$ which is the sequence $(a\_n)$ given by
$$
a\_n=\sup\_{x\in A}|x\_n|\ .
$$


**Exercise 1:** Show that a $A$ is bounded iff ${\rm env}(A)\in s\_+$ (sequences in $s$ with nonnegative entries).


**Exercise 2:** Consider $s'$ realized as a space of sequences. Show that the previous strong topology is the same as the locally convex topology defined by the seminorms
$$
||y||\_{\omega}=\sum\_n \omega\_n\ |y\_n|
$$
indexed by $\omega\in s\_+$.


Note that one can take $\ell^{\infty}$ or $\ell^p$ norms instead of $l^1$ with the same result. The relation to the envelope is more immediate with the $\ell^{\infty}$ variant. However, the $\ell^1$ choice allows a more immediate comparison with the weak-$\ast$ topology
which is defined by the seminorms
$$
||y||\_x=\left| \sum\_n x\_n y\_n \right|
$$
indexed by $x\in s$. A glance at the last formula, versus the one for $||y||\_{\omega}$,
should be enough to see that the $||y||\_x$ are very bad seminorms to work with.


**Exercise 3:** Repeat the process for $V=s'$ instead of $s$ and show that the strong dual is $s$ with its original topology. Namely, $s$ is reflexive, just like a finite-dimensional space.


**Moral of the story:** The strong and weak-$\ast$ topologies are not that far apart because a bounded set "behaves like a single vector", however, it allows one to put the absolute values where they should be, i.e., "inside" the sum for the duality pairing. For more on this, see my answer to the MO question: [Is the tensor product of distributions a continuous bilinear map with respect to the weak topology?](https://mathoverflow.net/questions/393207/is-the-tensor-product-of-distributions-a-continuous-bilinear-map-with-respect-to)


