I have also posted the question here. Let me explain what difficulties I have. In fact, one may write \begin{equation} \partial_1(f-\partial_1 u)=0 \end{equation} in $\Omega$. Then one may have the formula $(f-\partial_1 u)(x,y)=g(y)$ for some one dimensional function $g$. If one wants to construct a counter example, then $g$ should have no derivative. But since $u$ is approximated by smooth functions with compact support, it looks like very impossible to construct such example.

Update: So far I have proved that if $f\in H_0^1(\Omega)$, then $\partial_1\alpha$ has to be in $H^1_{loc}(\Omega)$. The trick is to use the discrete Fourier transform.

Let me explain the trick I used (I am also not quite sure if I was mistaken, since I am not very familiar with the Fourier transform method). Since $f$ and $\alpha$ can be approximated by smooth functions with compact support, we can extend them to $\mathbb{R}^2$ by setting their value equal to zero outside of the region and they belong to $H^1(\mathbb{R}^2)$ and $M(\mathbb{R}^2)$. In particular, they and their derivatives can be written as the discrete Fourier sum (I will write also scalar components by $c$, since they don't essentially influence the result), for example, \begin{align} f(x)&=c\sum_{n\in\mathbb{Z}^2}Ff(n)e^{in\cdot x},\\ \partial_2f(x)&=c\sum_{n\in\mathbb{Z}^2}n_2Ff(n)e^{in\cdot x},\\ \partial_1\alpha(x)&=c\sum_{n\in\mathbb{Z}^2}n_1F\alpha(n)e^{in\cdot x} \end{align}

and so on. In fact, using the standard differnce quotient method one derive \begin{equation} f(x)-\partial_1\alpha(x)=G(x_2) \end{equation} for some $G\in L_2(\Omega)$. Now for $g(x):=G(x_2)$ we also have its Fourier sum \begin{equation} G(x)=c\sum_{n_2\in\mathbb{Z}}Fg(n_2)e^{in_2\cdot x_2}. \end{equation}

Sum up all, we have \begin{equation} \partial_2f(x)=c\sum_{n\in\mathbb{Z}^2}n_1n_2F\alpha(n)e^{in\cdot x}+c\sum_{n_2\in\mathbb{Z}}n_2Fg(n_2)e^{in_2x_2}. \end{equation} Since $\partial_2 f\in L_2$, we have \begin{equation} \sum_{n\in\mathbb{Z}^2}|n_1n_2F\alpha(n)|^2<\infty. \end{equation} Then define $v(x)=c\sum_{n\in\mathbb{Z}^2}n_1n_2F\alpha(n)e^{in\cdot x}$, and $v$ is in $L_2$ and using the definition of weak derivative, it is actually equal to $\partial_2\alpha$.

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    $\begingroup$ Normally, it's better to wait for a while before reposting here. $\endgroup$ – Alex Degtyarev Jul 22 '15 at 13:08
  • $\begingroup$ @AlexDegtyarev Ok, sorry for this. I just don't know if every one visit both websites usually, since I am also new here and I used to use the MS website. $\endgroup$ – Peter Jul 22 '15 at 13:10

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