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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 $\tau = \tau_{op}$?

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What is $f_i$? Is it just the identity map? I initially read it as taking all linear bijections, but that would contradict other assertions in your question. –  Jonas Meyer Dec 22 '10 at 18:28
    
Thanks, i clarified this. –  Martin Dec 22 '10 at 19:13
    
"linear identities"? So the answer is yes, they are each the identity map? –  Jonas Meyer Dec 22 '10 at 21:14
    
yes, that's it. Further rewritten. –  Martin Dec 22 '10 at 22:30
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up vote 3 down vote accepted

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.

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Further, on any vector space at all, there is a unique finest locally convex topology, namely, the (locally convex) colimit of all the finite-dimensional subspaces (with their canonical topologies). –  paul garrett Jul 22 '11 at 15:16
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