This is an extension of a previously asked question:
A function $f\in L^2(D)$ is weakly holomorphic if, for every $\phi\in \mathcal{C}^{\infty}_c(D)$, $$\int_D f\partial_{\bar{z}}\phi = 0.$$ I'm trying to show that each such $f$ is smooth on the interior of $D$ and is in fact a strong solution to $\partial_{\bar{z}}f=0$; i.e., $f$ is holomorphic in the usual sense.
Here's what I've proven so far:
Let $B$ be a bounded open set in $\mathbb{R}^n$ and $f\in L^p(D)$ for $1<p<\infty$. Let $g$ be a smooth, non-negative function supported in the unit ball with Lebesgue integral 1 and consider the mollifier $$f_{\epsilon}(x)=\epsilon^{-n}\int_{B}g(\frac{x-y}{\epsilon})f(y)dy$$ where $x\in B$ and $\epsilon<|x,\partial B|$. Then, $f_{\epsilon}\rightarrow f$ uniformly in the $L^p$ sense as $\epsilon\rightarrow 0$. Consequently, $f$ can be approximated by smooth, compactly supported functions in $B$.
Now, supposing this is true in the complex case if we just replace $B$ with the unit disk $D$ (I haven't proven this, but I think it's correct...) then I'm essentially done if I can demonstrate that my mollifying functions are holomorphic since I already know they're smooth. I'm not quite sure where the hypothesis would come into play, however. Anyway, is this the correct route?
Thanks.