Given a continuous map of TVS $u:E\to F$, we can associate to it the transpose map $u^t:F'\to E'$. Here, $E',F'$ are the continuous duals of $E,F$, respectively. The map $u^t$ is continuous, so long as $E', F'$ carry a reasonable topology (the weak dual topology, the compact convex convergence topology, the compact convergence topology, or the strong dual topology).
We assume the duals carry the strong dual topology. Moreover, Let $L(E,F)$ denote the TVS of continuous linear maps $E\to F$, equipped with the bounded-convergence topology. In particular, $L(E,\mathbb R)$ is topologically isomorphic to $E'$.
The transposition operator gives rise to a map $t:L(E,F)\to L(F',E')$. My question is: Under which circumstances is this operator continuous?
I cannot find any results online or in books. I scanned through Treves "Topological Vector Spaces, Distributions and Kernels" but failed to find any reference to it.
You may assume $E,F$ are nuclear reflexive spaces, and in fact consider $$t:L(E,F')\to L(F,E').$$
My (clueless) attempt:
A basic neighborhood of $0$ in $L(F,E')$ is $$ U=\{v:F\to E'|v(B)\subset A\}$$ for some $B\subset F$ bounded and $A\subset E'$ basic. However, $A=C^0$ is a polar of some bounded set $C\subset E$. $u\in L(E,F')$ is in $t^{-1}(U)$, that is, $u^t\in U$, if and only if $$u^t(B)\subset C^0.$$ This happens if and only if $$\forall b\in B,\qquad \sup_{c\in C}|\langle u^t(b),c\rangle|\leq 1$$ which happens if and only if $$\forall b\in B,\qquad \sup_{c\in C}|\langle b,u(c)\rangle|\leq 1$$ which happens if and only if $$\forall b\in B,\qquad \sup_{x\in u(C)}|\langle b,x\rangle|\leq 1.$$ This happens if and only if $$ B\subset (u(C))^0$$ which happens if and only if $$u(C)\subset B^0.$$ Thus, $$t^{-1}(U)=\{u:E\to F'|u(C)\subset B^0\}$$ is open in $L(E,F')$.
