Let $\Omega\subset\mathbb{C}^n$ be a $C^{\infty}$ bounded domain. Let $H^2(\partial\Omega)$ denote the Hardy space, and $S(.,.)$ denote its Szego Kernel. We know that $$ \forall f\in H^2(\partial\Omega),\text{ for each }z\in\Omega,\;\; F(z)=\langle f,S(.,z)\rangle_{L^2(\partial\Omega)}\in\mathcal{O}(\Omega) $$
My Question is: for $f\in L^2(\partial\Omega)$ and $F(z)=\langle f,S(.,z)\rangle_{L^2(\partial\Omega)} \implies f\in H^2(\partial\Omega)$?
Note that by definition of $F$, $F\in\mathcal{O(\Omega)}\implies\;F$ is a harmonic function on $\Omega.$
$$
\begin{split}
F(z) & = \langle f,S(.,z)\rangle_{L^2(\partial\Omega)} \\
& =\langle f,PS(.,z)\rangle_{L^2(\partial\Omega)} \\
& = \langle P^{*}f,S(.,z)\rangle_{L^2(\partial\Omega)} \\
&= \langle Pf,S(.,z)\rangle_{L^2(\partial\Omega)} = Pf(z)
\end{split}
$$
for $z\in\Omega$, where $P$ denotes the Szego projection.
Therefore, $Pf=F$ on $\Omega$ and this implies
$$
\sup_{\epsilon>0}\int_{\partial\Omega}|(F)_{\epsilon}(z)|^2 dS = \sup_{\epsilon>0}\int_{\partial\Omega} |(Pf)_{\epsilon}(z)|^2 dS<{\infty}$$
since $$Pf\in H^{2}(\partial\Omega) \implies F\in H^{2}(\partial\Omega).$$
I think that from this it follows that $f\in H^2(\partial\Omega)$ but I don't know how it follows.
I appreciate any help you can provide.
