Let $u:\Omega\subset\mathbb{R}^2\to\mathbb{R}$ (with $\Omega$ a bounded Lipschitz domain) be a function from $H^1(\Omega)$ with $\Delta u\in L^2(\Omega)$. Is it true that $\dfrac{\partial^2 u}{\partial x_1^2}\in L^2(\Omega)$ and $\dfrac{\partial^2 u}{\partial x_2^2}\in L^2(\Omega)$?
By $\Delta u=f\in L^2(\Omega)$ in the sense of distributions we mean that:
$$\int_{\Omega} u(x)\Delta\phi(x)\ dx=\int_{\Omega} f(x)\phi(x)\ dx,\ \forall\ \phi\in C^{\infty}_{c}(\Omega)$$
So I wonder if there are $f_1,f_2\in L^2(\Omega)$ such that:
$$\int_{\Omega} u(x)\dfrac{\partial^2\phi}{\partial x_1^2}(x)\ dx=\int_{\Omega} f_1(x)\phi(x)\ dx\ \text{and}\ \int_{\Omega}u(x)\dfrac{\partial^2\phi}{\partial x_2^2}(x)\ dx=\int_{\Omega} f_2(x)\phi(x)\ dx\ \forall\ \phi\in C^{\infty}_{c}(\Omega)$$
???
$\bullet$ I tried to obtain a counterexample for $\Omega=(0,1)\times (0,1)$ by constructing a function $h(x)=\dfrac{2}{3}\sqrt{x^3}$. It has the property that $h\in H^1(0,1)$ but $\dfrac{d^2 h}{dx^2}=\dfrac{1}{2\sqrt{x}}$ which is not in $L^2(0,1)$. But it's hard to contruct the function $u$ from that.