I am trying to show $(x\text{ is constructible})^L\iff(x\text{ is constructible})$. That is, I am trying to show that the statement "$x$ is constructible" is absolute for $L$.
I am following Jech, and here are my thoughts:
To show this, I want to have the following implications:
$$(x\text{ is constructible})^L\iff\exists\alpha\in L(x\in L_\alpha)^L\iff \exists\alpha(x\in L_\alpha^L)\iff \exists\alpha(x\in L_{\alpha})\iff(x\text{ is constructible})$$
Now $L$ is an inner model, so certainly every possible $\alpha$ is a member of $L$ giving us the second iff. It's the third iff that gives me trouble. We want to show that $x\in L_{\alpha}^L$ is absolute. The idea is to express it in the Godelian hierarchy Jech builds up using the Godel Normal Form Theorem.
I think we can express $y=L_{\alpha}^L$ iff $(\forall x\in y)(\exists\beta(\beta<\alpha\wedge x\in L_{\beta}^L)$. Now we want to express $x\in L_{\beta}^L$ in a nice way. I think the way we should do this is is by saying $\exists\gamma(\gamma<\beta\wedge x\in\text{cl}(L_{\gamma}^L)$. Here $\text{cl}$ is closure under the Godel operations given in the normal form theorem. I think this follows by Corollary 13.9 in Jech. Now, we want to write out the closure of $L_{\gamma}^L$. I suppose the question though is that I don't know what $L_{\gamma}^L$ should be represented as let alone its closure.
I do have some thoughts, however. I think that what I'm missing here is that $L_{\epsilon}$ is defined using transfinite recursion. Now Lemma 13.12 in Jech tells us that functions defined in this way are $\Sigma_1$ if the class function defining them is $\Sigma_1$ (i.e. absolute upwards). So I think if I can find the class function defining this recursion, and show it is $\Sigma_1$, I can show at least that $L_{\epsilon}$ is absolute upwards. I am still confused on what to do for the $\Pi_1$ case and wondering if I am on the right track.
Sorry for the very long post; I'm trying to work out all the little details of how this stuff works.
Yes, you are on the right track. A function defined by a transfinite recursion on a $\Sigma_1$ function is $\Sigma_1,$ as can be shown by writing out the defining formula directly. To show that it is also $\Pi_1,$ observe that any $\Sigma_1$ class function on the ordinals is also $\Pi_1$ since we can express $F(\alpha)=y$ as $$\forall z(F(\alpha)=z\to z=y).$$ (More generally, any $\Sigma_n$ function with a $\Delta_n$ domain is $\Delta_n$.)