Regarding basis of holomorphic Hardy space Let $\Omega\subset\mathbb{C}^n$ be any $C^{\infty}$ bounded domain and let $H^2(\partial\Omega)$ denotes a holomorphic Hardy space which is a $L^2(\partial\Omega)$ closure of $A^{\infty}(\Omega)(=\mathscr{O}(\Omega)\cap C^{\infty}(\overline{\Omega})).$ Because $H^2(\partial\Omega)$ is a closed subspace of  $L^2(\partial\Omega)$ there exists a countable dense subset {$\phi_j$}  where $ j\in\mathbb{N}$ such that
$$
f=\sum_{j=1}^{\infty}\langle f,\phi_j\rangle\phi_j\qquad
\forall f\in H^2(\partial\Omega)
$$
My Question: can we choose $ \forall j\in\mathbb{N},\;\; \phi_j\in A^{\infty}(\Omega)?$
 A: One should be able to find a countable linearly independent subset of $A^\infty(\Omega)$ whose linear span is dense in $H^2(\partial\Omega)$ (it would suffice to show that the orthogonal complement of this subset inside $H^2$ is zero).
Once you have such a subset, order it in any ways as a sequence and then apply the Gram--Schmidt orthogonalization process.

Some more details on the second step. Suppose that we have a sequence $(f_n)_{n=1}^\infty$ in an inner product space, which is linearly independent. Equivalently, if we define $V_n$ to be the linear span of the set $\{f_1,\dots, f_n\}$, then we require that $0 \subset V_1 \subset V_2 \subset \dots$ where each inclusion is strict.
The Gram-Schmidt orthogonalization process takes this sequence and produces an orthonormal sequence $(u_n)_{n=1}^\infty$, with the property that the span of $\{u_1,\dots, u_n\}$ is equal to $V_n$ for each $n\geq 1$. In particular, if $\bigcup_{n=1}^\infty V_n$ is dense in the original inner product space, then we have produced an orthonormal basis.
In particular, if we start in the inner product space that is $A^\infty(\Omega)$ with the $L^2$-inner product, then we automatically get an orthonormal basis for $H^2(\partial \Omega)$ that is contained in $A^\infty(\Omega)$.
