A "straightforward" inequality.

Question

In section 5.5 page page 117 of Fanghua, Quing's book, in order to obtain $W^{2,p}$ estimates the idea is to show that for $p \in (1,\infty]$ the condition $\theta \in L^p(\Omega)$ implies $D^2u \in L^p(\Omega)$ and \begin{equation} \| D^2u\|_{L^p(\Omega)} \le 2 \| \theta \|_{L^p(\Omega)} \end{equation} where $\Omega$ is a bounded domain and $u$ is a continuous function in $\Omega$. As in section 2 we define for $M>0$ \begin{eqnarray*} G_M(u,\Omega) & =& \{ x_0 \in \Omega: \mbox{there is an affine function} \ L \ \mbox{such that} \ \\ & &L(x) - \dfrac{M}{2}|x -x_0 |^2 \le u(x) \le L(x) + \dfrac{M}{2}|x- x_0|^2\\ & & \mbox{for}\ x \in \Omega \ \mbox{ with equality at} \ x_0 \} \\ A_M(u,\Omega)& =& \Omega \backslash G_M(u,\Omega). \end{eqnarray*} Finally the function $\theta$ is given by \begin{equation} \theta(x) = \theta(u,\Omega)(x) = \inf \{ M: x \in G_M(u,\Omega)\} \in [0, \infty] \end{equation} for $x \in \Omega.$ The author said that this is straightforward but I can't see this so easy. I'd like that you do this for me, Thank you.