Is there any reference where I can find something on approximation of analytic functions on a domain in complex plane by $L^{p}$ analytic functions of the same domain?



Approximation of analytic functions by Lp functions
First of all I am not sure that the domains allowing nonzero analytic $L^p$ functions are fully understood even in dimension $1$. I think that much of the cases are covered in the booklet ,,Selected problems on exceptional sets" by Carleson in the last chapter (I do not have access to it right now), where a characetization is given with the help of Hausdorff measure.


I remember that the $L^2$ case (the most studied) is fully understood in dimension 1 and the characterization is that the complement should not be a polar set.


Lemma 4.3.1 in ,,An introduction to complex analysis in several variables" by Hormander says: Let p be a strictly plurisubharmonic smooth function in $\Omega$ (this is equivalent to pseudoconvexity in several variables, in $\mathbb C^{1}$ such a function always exists). Then every function which is analytic in a neighborhood of a sublevel set of p (in particular every function analytic in $\Omega$) can be approximated in $L^2$ norm over the sublevelset by functions analytic in $\Omega$. With little glueing or directly by estimates on $\bar \partial$ one can ensure that the approximating function is $L^2$ on $\Omega$.


If the domain is bounded (or more generally of bounded Lebesgue measure) then $L^2$ gives $L^p$ by Jensen's inequality. In the other cases I believe your question boils down to $L^p$ estimates on the $\bar\partial$ equation on specific domains.


Cauchy estimates yield that $L^{p}$ convergence gives one locally uniform convergence.


Now it depends what kind of approximation you want. If you want locally uniform approximation by $L^p$ functions- you have it. If You want to have global uniform convergence, then I fear that You have to assume that your initial function is $L^p$ which makes your question void. The same applies if you want global $L^p$ convergence. 


Also bear in mind that the sublevelsets are not arbitrary. Consider the function $\frac{1}{z}$ on $\mathbb D\setminus\{0\}$. Any $L^2$ function on $\mathbb D\setminus\{0\}$ will however extend through zero, because it is a removable set for such functions.


