Does $\Sigma_1$-separation over $L_\alpha$ implies admissibility? Inspired by the following tweet, I tried to prove the following statement:

If $\alpha>\omega$ is a limit ordinal and $L_\alpha\models \Sigma_1\text{-Separation}$. Then $L_\alpha\models \mathsf{KP}$.

The only axiom we need to verify is $\Delta_0$-Collection. Since $\langle L_\beta\mid\beta<\alpha\rangle$ is $\Delta_1$-definable over $L_\alpha$, the problem reduces to proving the following one:

If $a\in L_\alpha$ and $f\colon a\to \alpha$ is $\Sigma_1$-definable over $L_\alpha$, then $f$ is bounded.

Unfortunately, I have no idea how to proceed from there. There might be some similar proof in textbooks about admissible set theory or higher computability (or fine structure theory), but I do not know which reference would be relevant. I would appreciate your help!
 A: In fact we can get $\Sigma_1$-collection directly! The following argument is pulled from a set of handwritten notes by Ronald Jensen, which seems to be called "Admissible Sets". Jensen proves that failure of a strengthened (not just $\Sigma_1$-collection) version of admissibility implies failure of $\Sigma_1$-separation, so this is modified into the contrapositive direction.
Let $\alpha$ be such that $L_\alpha$ satisfies $\Sigma_1$-separation. Then assume there exists such a function $f:a\to L_\alpha$ whose range is not bounded by some member of $L_\alpha$. We may then construct the set $X=\{x\in a\mid x\notin f(x)\}$, which is a $\Sigma_1(L_\alpha)$ set. Since $f$'s range is unbounded in $L_\alpha$, $X$ is not necessarily present in any earlier $\textrm{Def}(L_\beta)$ for $\beta<\alpha$. Let $z\in a$ be such that $f(z)=X$, this should be possible to choose since $f$'s range is not restricted to any member of $X\subseteq L_\alpha$. Then using the usual diagonal argument, $z\in X\iff z\notin f(z)\iff z\notin X$, which is a contradiction. So $f$ couldn't have existed, and $L_\alpha$ satisfies $\Sigma_1$-collection.
