Let $f=(u,v)\in \mathscr{D}'(U,\mathbb{C})$ be a distribution, where $U\subset\mathbb{C}=\mathbb{R}^2$ is an open set and $u$ and $v$ are the projection of $f$ onto the real and imaginary axis (ie $\langle f,\phi\rangle=\langle u,\phi\rangle+i\langle v,\phi\rangle$). Suppose that
$$
\frac{\partial}{\partial \overline{z}}f=0\qquad\text{in U,}
$$
where $\frac{\partial}{\partial \overline{z}}=\frac{1}{2}\bigg(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\bigg)$ and the derivatives are in distributional sense.
Does it follow that $f$ is holomorphic in the classical sense, ie $f\in C^\infty(U,\mathbb{C})$ and the Cauchy-Riemann equations are satisfied?


The obvious idea would be to mollify, get holomorphic functions and then take the limit. But how can we conclude that the limit is still holomorphic?



$\frac{\partial f}{\partial \overline{z}}=0$ in distributional sense implies $f$ is holomorphic
I've just realized that, if $f$ is in $L^1\_{loc}$ and not just in $\mathscr{D}'$, my question can be answered using Weyl's lemma for harmonic functions. Indeed, from $\frac{\partial f}{\partial \overline{z}}=0$ it easily follows that $\Delta u=\Delta v=0$, and then Weyl's lemma implies that $u$ and $v$ are smooth. But then $f$ is smooth and satisfies the Cauchy-Riemann equations, thus is holomorphic.


EDIT: As pointed out by Ben McKay in the comments, the hypothesis that $f\in L^1\_{loc}$ is not necessary.


