# $c^\infty$ topology on $L(E, F)$

In Kriegl/Michor's "Convenient Setting for Global Analysis", they put on the set $L(E, F)$ of bounded linear operators between locally convex spaces $E$, $F$ the subspace topology induced by the inclusion $L(E, F) \subset C^\infty(E, F)$, where smooth maps are those that map smooth curves in $E$ to smooth curves in $F$.

Question: Is this topology somehow related to one of the common topologies on $L(E, F)$, at least in special cases (i.e. when $E$, $F$ satisfy special assumptions)?

I am interested in this, because smooth maps from $M \rightarrow L(E, F)$ ($M$ a manifold) are needed to define vector bundles and vector bundle maps with locally convex spaces as fibers.

Lemma 5.3. in the book states that a subset is bounded in $$L(E, F)$$ if and only if it is uniformly bounded on bounded subsets of E, i.e. the inclusion $$L(E, F) \rightarrow C^\infty (E, F)$$ is initial when we endow $$L(E,F)$$ with the topology of uniform convergence on bounded sets. Though this does nothing for you to establish that the topologies are the same one can read this as a statement on smoothness into the space $$L(E,F)$$ with respect to the uniform topology. As the concepts of differentiability discussed in the book do not depend on continuity but only on bornology, the above statement means that the smooth curves into (and thus the smooth mappings defined on) the spaces with both structures coincide. Hence to test the differentiability it makes no difference with respect to which structure you test. Moreover, you can even get continuity if you assume that $$E,F$$ are Banach spaces and $$M$$ is a compact manifold. Then $$C^\infty (M,L(E,F))$$ is a complete metrisable space and being convenient smooth implies that the mapping is also continuous with respect to the original topologies (which then coinicde with the $$c^\infty$$-topology of the function space.
Remark: I suspect that for $$E,F$$ are finite dimensional, the induced topology on $$L(E,F)$$ should be the same as the operator norm topology. Though I was not able to show it yet, it suffices to prove that the topology turns $$L(E,F)$$ into a locally convex space (which is finite-dimensional in this case, whence it must be the standard topology by uniqueness of the norm topology)
• As to the last remark, in fact it suffices to prove a bit less, that is, that the topology makes $L(E,F)$ a Hausdorff TVS (the uniqueness result still holds) – Pietro Majer Apr 29 '19 at 20:06