In the proof of theorem2 in: https://www.math.ualberta.ca/~xinweiyu/527.1.11f/lec14.pdf , I am struggling with the estimate of the cutoff function.
Theorem: Let $u \in W^{1,2}(\Omega) $ be a weak solution of $\Delta u = f $ with $f \in L^2(\Omega)$. For any $\Omega' \subset \subset \Omega$, we have $u \in W^{2,2}(\Omega') $, and \begin{equation} \label{eq:main_estimate} \|u\|_{W^{2,2}(\Omega')} \leq C \left( \|u\|_{L^2(\Omega)} + \|f\|_{L^2(\Omega)} \right), \end{equation} where the constant depends on the distance between $ \Omega' $ and $ \partial \Omega $. Furthermore, $ \Delta u = f $ almost everywhere in $ \Omega $.
To prove this result, we first establish the following intermediate gradient estimate: \begin{equation} \label{eq:gradient_estimate} \|\nabla u\|_{L^2(\Omega')}^2 \leq \frac{17}{\delta^2} \|u\|_{L^2(\Omega)}^2 + \delta^2 \|f\|_{L^2(\Omega)}^2, \end{equation} without any additional assumptions.
Proof. Let $\eta(x)$ be a cut-off function defined by \begin{equation} \label{eq:cut_off} \eta(x) = \begin{cases} 1 & x \in \Omega', \\ 1 - \frac{1}{\delta} \text{dist}(x, \Omega') & 0 < \text{dist}(x, \Omega') \leq \delta, \\ 0 & \text{dist}(x, \Omega') > \delta \end{cases} \end{equation} and set the test function \begin{equation} \label{eq:test_function} v = \eta^2 u \in W_0^{1,2}(\Omega). \end{equation}
Some calculation yields \begin{equation} \label{eq:integration_by_parts} \int_\Omega \eta^2 |\nabla u|^2 + 2 \int_\Omega (\eta \nabla \eta) \cdot (u \nabla u) = -\int_\Omega \eta^2 f u. \end{equation}
Using Young's inequality: \begin{equation} \label{eq:youngs_inequality} |a b| \leq \frac{\epsilon}{2} a^2 + \frac{1}{2\epsilon} b^2, \quad a, b \in \mathbb{R}, \, \epsilon > 0, \end{equation} on the second term on the LHS ($\epsilon = 1/2$) and on the RHS ($\epsilon = \frac{1}{\delta^2}$), we arrive at \begin{equation} \label{eq:lhs_rhs_bound} \int_\Omega \eta^2 |\nabla u|^2 \leq \frac{1}{2} \int_\Omega \eta^2 |\nabla u|^2 + 2 \int_\Omega u^2 |\nabla \eta|^2 + \frac{1}{2\delta^2} \int_\Omega \eta^2 u^2 + \frac{\delta^2}{2} \int_\Omega \eta^2 f^2. \end{equation}
By moving the first term on the RHS to the left, we obtain \begin{equation} \frac{1}{2}\int_{\Omega'}|\nabla u|^2 \leq \frac{1}{2}\int_\Omega \eta^2 |\nabla u|^2 \leq 2 \int_\Omega u^2 |\nabla \eta|^2 + \frac{1}{2\delta^2} \int_\Omega u^2 + \frac{\delta^2}{2} \int_\Omega f^2. \end{equation}
Assuming $ \eta^2 \leq 1 $ and $|\nabla \eta |^2 \leq \frac{4}{\delta^2}$ (to be proved), we conclude: \begin{equation} \frac{1}{2}\int_{\Omega'}|\nabla u|^2 \leq \frac{8}{\delta^2}\int_\Omega u^2 + \frac{1}{2\delta^2} \int_\Omega u^2 + \frac{\delta^2}{2} \int_\Omega f^2. \end{equation}
Thus, \begin{equation} \label{eq:final_estimate} \int_{\Omega'} |\nabla u|^2 \leq \frac{17}{\delta^2}\int_\Omega u^2 + \delta^2 \int_\Omega f^2. \end{equation}
However, it seems that the estimate $ |\nabla \eta |^2 \leq \frac{4}{\delta^2}$ is needed, but I am unable to prove this. Could someone help with this step?
