i consider a function $u\in W^{2,2}(B^+_R)$ where $B_R^+:=B_R\cap \mathbb{R}^m_+$ is the upper ball with radius $0<R\leq 1$. The function $u$ shall additionally satisfy the following boundary conditions $u=g$ and $Du=Dg$ on the flat boundary $T_R$ of $B_R^+$. Now, i need to approximate u by smooth functions because i have to use integration by parts and there appear weak derivatives of third order. I choose $u^{\varepsilon}(x):=g(x)+[\zeta(x^m)\cdot (u-g)(x)]\ast\eta_{\varepsilon}$ with $\zeta=0$ on $[0,2\varepsilon]$ and $\zeta=1$ on $[3\varepsilon,\infty)$. $\zeta$ is a cut-off function and $\eta_{\varepsilon}$ a mollifier. Now i want to show that $\lim_{\varepsilon \searrow 0}\dfrac{\partial u^{\varepsilon}}{\partial x^m}=\dfrac{\partial u}{\partial x^m}$ for $x^m>2\varepsilon$. First i calculate $$ \dfrac{\partial u^{\varepsilon}}{\partial x^m}=\dfrac{\partial g}{\partial x^m}-\int_{B_{\varepsilon}(x)}\zeta(z^m)\cdot (u-g)(z)\cdot\dfrac{\partial \eta_{\varepsilon}}{\partial x^m}(z-x)dz=\dfrac{\partial g}{\partial x^m}-\int_{B_{\varepsilon}(x)}\zeta(z^m)\cdot (u-g)(z)\cdot\dfrac{\partial \eta_{\varepsilon}}{\partial z^m}(z-x)dz $$ And by integration by parts with respect to $z^m$ I get $$ \dfrac{\partial u^{\varepsilon}}{\partial x^m}=\dfrac{\partial g}{\partial x^m}+\int_{B_{\varepsilon}(x)}\zeta(z^m)\cdot \dfrac{\partial (u-g)(z)}{\partial z^m}\cdot\eta_{\varepsilon}(z-x)dz+\int_{B_{\varepsilon}(x)}\zeta'(z^m)\cdot (u-g)(z)\cdot\eta_{\varepsilon}(z-x)dz. $$ Let $z^m>2\varepsilon$. On the open interval $(2\varepsilon,3\varepsilon)$ I choose $\zeta=1/\varepsilon\cdot (z^m-2\varepsilon)$. Then $\zeta'=1/\varepsilon$ on $(2\varepsilon,3\varepsilon)$ and 0 else. So, i get $$ \dfrac{\partial u^{\varepsilon}}{\partial x^m}=\dfrac{\partial g}{\partial x^m}+\int_{B_{\varepsilon}(x)}\zeta(z^m)\cdot \dfrac{\partial (u-g)(z)}{\partial z^m}\cdot\eta_{\varepsilon}(z-x)dz+\dfrac{1}{\varepsilon}\int_{B_{\varepsilon}(x)} (u-g)(z)\cdot\eta_{\varepsilon}(z-x)dz. $$ I do: $$ \vert \dfrac{\partial u^{\varepsilon}}{\partial x^m}-\dfrac{\partial u}{\partial x^m}\vert\leq \int_{B_{\varepsilon}(x)}\vert \zeta(z^m)\cdot \dfrac{\partial (u-g)(z)}{\partial z^m}-\dfrac{\partial (u-g)(x)}{\partial x^m}\vert\cdot\eta_{\varepsilon}(z-x)dz +\dfrac{1}{\varepsilon}\int_{B_{\varepsilon}(x)} \vert (u-g)(z)\vert\cdot\eta_{\varepsilon}(z-x)dz $$ How can i show that the left-hand side tends to 0 for $\varepsilon \searrow 0$. I have a troubles with cut-off function. How to handle with it. Can anyone help me?
Best regards