Bounded-open topology vs norm on $L\left(X,Y\right)$

Question

In general topology there is two ways of introducing a topology on the space of (continuous) maps between, say, metric spaces: set-open topology and uniform topology (it is a uniformity of uniform convergence on a certain family of subsets of the domain). In general they do not coincide; they do if the family consists only of compacts.

Now let $X,Y$ be normed space. The usual norm on the space of continuous linear maps from $X$ into $Y$ defines the topology of uniform convergence on bounded subsets of $X$. My question is whether this topology coincides with the bounded-open topology.

Often in the linear case bounded sets behave like compacts in general, which sort of motivates this question. In fact, the question is valid for any pair of TVS.

The answer is affirmative in the case, when $X$ is reflexive and $Y$ is finite-dimensional. As continuity of a linear map implies the weak continuity, the topology (and hence the uniformity) on $Y$ stays the same, if we replace the original one with the weak, and bounded sets in $X$ become relatively compact, we can rely on the general-topological version.

Thank you.

Answer 1

In case, anyone is interested, my question happened to be rather simple.

Set-open topology coincides with the uniform topology if the target space is homogeneous enough. More precisely:

Let $X$ be a set and let $Y$ be an abelian topological group. Let $\mathcal{S}$ be a collection of subsets of $X$ and let $\mathcal{F}\left(X,Y\right)$ be the set of all maps from $X$ into $Y$ endowed with a pointwise addition. Then $\mathcal{S}$-open topology is compatible with the group structure and $\mathcal{F}\left(X,Y\right)$ equipped with this topology is isomorphic (as a uniform space) to $\mathcal{F}\left(X,Y\right)$ with $\mathcal{S}$-uniformity.

Maybe we can discard the commutativity in the statement, or even replace the "external" algebraic condition by something else, but I don't want to think about this now =)