In some current work, my co-authors and I had wanted in a certain argument to appeal to $\kappa^{\lt\kappa}=\kappa$ in $L[A]$, in a situation where $A\subset\kappa$ and $\kappa$ was weakly inaccessible in $V$, but $\kappa$ was below the continuum in $V$ (and so $\kappa^{\lt\kappa}\neq\kappa$ in $V$). But we have lost confidence in this statement. Is it consistent that there is a counterexample?
Question. Is it (relatively) consistent that $\kappa$ is weakly inaccessible, but there is $A\subset\kappa$ with $L[A]\models\kappa^{\lt\kappa}>\kappa$?
The model $L[A]$ here is the relative constructible universe, defined as in the constructible universe, but with a predicate for the set $A$.
We had at first thought a condensation argument might show $\kappa^{\lt\kappa}=\kappa$ in every $L[A]$, but upon looking at it closely, I can't quite make it work. Condensation works fine above $\kappa$, but I don't see why, for example, every real number in $L[A]$ must be added by a stage before $\kappa$. It would be interesting to me even to resolve the question of how big $2^\omega$ must be or can be in comparison with $\kappa$ in $L[A]$, where $A\subset\kappa$ and $\kappa$ is a regular limit cardinal in $V$.
(Meanwhile, we've saved our application by finding another route for our argument that does not rely on this issue.)
