let $X$ be some Banach space. Let $L(X)$ be the set of continuous operators on $X$ into $X$.
Let $(\tau_i)_i$ be a set of topologies on $L(X)$ s.t. $L(X)$ is topological vector space (i.e. addition and scalar multiplication continuous, not necessarily Hausdorff).
Let $\tau$ be the initial topology on $L(X)$ induced by the linear identities $f_i : L(X) \longrightarrow (L(X),\tau_i)$. Then $(L(X),\tau)$ is a topological vector space, too.
For the sake of illustration: if we restrict $(\tau_i)_i$ to the topologies that are coarser than the operator norm topology on $X$, $\tau_{op}$, then obviously $\tau = \tau_{op}$.
Question: Does $\tau = \tau_{op}$ hold even in general? Alternatively, is there at least a "nice" set of restrictions on $(\tau_i)_i$, or a characterization of "reasonable" topologies, for which we have $\tau = \tau_{op}$?
Assuming that your $f_i$ is just the identity map with $\tau_i$ on the codomain, $\tau$ is just the coarsest topology finer than all $\tau_i$. This explains the assertion that restricting to those $\tau_i$ coarser than $\tau_{op}$ yields $\tau=\tau_{op}$. But then to ask whether $\tau=\tau_{op}$ in general is to ask whether every TVS topology on $L(X)$ is coarser than $\tau_{op}$. To contradict it you would just need an example of a TVS topology that is not comparable to $\tau_{op}$ or one that is strictly finer than $\tau_{op}$. If $X$ is infinite dimensional, then there are Banach space norms on $L(X)$ that are not equivalent to the operator norm, which by the open mapping theorem means the topologies are not comparable. So the answer is no.