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Let $\Omega \subset R^2$ is a ball. Consider the equation [ \begin{cases} -\triangle u = f(x), \quad x \in \Omega \\ u \big|_{\partial \Omega} = 0. \end{cases} ] It suffices to prove that for $p \geq 2$ [ |D^2u|{L^p(\Omega)} \leq C |f|{L^p(\Omega)}. ] At first, as you know, using integration by parts we have \begin{equation}\label{equ1} \|u\|_{H^1(\Omega)} \leq C \|f\|_{L^2} \leq C \|f\|_{L^p(\Omega)}. \end{equation} Then consider a cutoff function $\eta \in C^\infty_0(\Omega)$, denote by $v = \eta u$, then $v$ satisfies the equation [ -\triangle v = \eta f - 2\nabla u \cdot \nabla \eta - \triangle \eta u, \quad x \in R^2. ] It's known that $\xi_i\xi_j/|\xi|^2$ is an $L^{R^2}$ multiplier, that is, [ |\partial_i\partial_j u|{L^q(R^2)} \leq C |\triangle u|{L^q(R^2)}, \quad q \in (1, \infty). ] Using this fact and \eqref{equ1}, notice the support of $\eta$ we obtain [ |u|{H^2{(\Omega)}} \leq C |f|{L^p}(\Omega). ] Then the Sobolev embedding theorem yields that [ |u|{W^{1,q}{(\Omega)}} \leq C |f|{L^p(\Omega)}, \quad 2 \leq q < \infty. ] Proceed the above argument again, we find [ |D^2u|{L^p(\Omega)} \leq C |f|{L^p(\Omega)} ] as desired.