Given $(X,s)$ a (real) barreled locally convex space (that is, every closed convex and absorbing set in $(X,s)$ is a neighborhood of the origin), is there a (strictly) finer, non-barreled linear locally convex topology $\tau$ on $X$ that is compatible with the duality $(X,X^*)$?
In other words $s\preceq\tau$, $(X,s)^*=(X,\tau)^*$, and $\tau$ is not barreled.
On
A slightly different argument: Any barrelled space $X$ with continuous dual $X'$ carries the strong topology $\beta(X,X')$ as the latter is the finest polar topology on $X$ of the dual pair $(X,X')$ and - as the space is barrelled - every closed polar is a neighbourhood of $0$.
So you have $\beta(X,X')$ compatible with the duality, hence by Mackey-Arens $\beta(X,X') = \tau(X,X')$, the Mackey topology, which in turn is the finest topology compatible with the duality. So there is no finer topology compatible with the duality.
The answer is negative. Every linear locally convex topology $\tau$ on $X$ that is compatible with the duality has the same closed convex sets as $s$. Hence, every $\tau-$closed convex and absorbing set $M$ is $s-$closed convex and absorbing, and since $s$ is barreled, $M$ is an $s-$neighborhood of $0$. But because $s\preceq\tau$, $M$ is also a $\tau-$neighborhood of $0$. Therefore every $\tau-$closed convex and absorbing set is a $\tau-$neighborhood of $0$, that is, $\tau$ is barreled.
Conclusion: -all topologies compatible with the duality and finer that a given barreled topology are barreled.