Existence of specific weak derivative Suppose there exits a sequence $(\phi_n)_n\subset C_0^\infty(\Omega)$, where $\Omega\subset\mathbb{R}^2$ is a bounded domain with $C^\infty$ boundary, such that $(\partial_1+\partial_2)\phi_n$ converges to some function $\phi$ in $L^2$, does there exist some function $\psi\in L^2$ with $(\partial_1+\partial_2)\psi=\phi$ at least in distributional sense, i.e.
\begin{equation}
\int_{\Omega} \psi\cdot ((\partial_1+\partial_2)\alpha)dx=-\int_{\Omega}\phi\cdot\alpha dx
\end{equation}
for all $\alpha\in C_0^\infty(\Omega)$?
 A: Let's suppose $\Omega$ is an axis-aligned square; the general case can be handled by a partition of unity subordinate to a locally finite cover by such squares (Whitney squares). After translation/scaling, $\Omega=(-1,1)^2$.  Define $$I[\phi](x,y) = \int_0^{(x+y)/2} \phi\left( \frac{x-y}{2} + t,\  \frac{y-x}{2} + t\right)\,dt$$
Notice that if $\phi$ is smooth, the function $\psi=I[\phi]$ satisfies $\psi_x+\psi_y=\phi$. Also, if $\phi$ is approximated in the $L^2$ sense by some functions $f_n$, Fubini's theorem implies $I[f_n]\to I[f]$ uniformly on a.e. line with slope $1$; in particular $I[f_n]\to I[f]$ in $L^2$.
Since convergence of distributions implies distributional convergence of their derivatives, it follows that 
$$
(\partial_x+\partial_y)I[\phi]= \lim (\partial_x+\partial_y)I[f_n] = \lim \phi_n = \phi
$$

I remark that individual partials of $I[\phi]$ can be quite singular; in general, $I[\phi]$ is not a Sobolev function. For example, consider $\phi(x,y)= \operatorname{sign}(x-y)$, which can be approximated by using $\phi_n(x,y)=\frac12(x+y) \chi_n(x-y)$, where $\chi_n$ is a smooth approximation to $\operatorname{sign}$. For this $\phi$, there is no $W^{1,1}$ function $f$ such that $f_x+f_y=\phi$.
