Approximation by convolution and cut-off function

Question

let u be a $L^p$-function on the upper half-ball with radius $R$ and center at 0 $B_R^+\subset \mathbb{R}^m_+$ with boundary condition u=g on flat boundary of the half-ball. I construct the following approximation: $$ u^{\varepsilon}(x):=g(x)+\int_{B^+_{\varepsilon}(x)}\zeta(z^m)\cdot (u-g)(z)\cdot \eta_{\varepsilon}(x-z)dz, $$ where $0\leq\zeta\leq1$ is a cut-off function with $\zeta=0$ on $[0,2\varepsilon]$ and $\zeta=1$ on $[3\varepsilon,\infty]$. Now, let $x^m>2\varepsilon$. I want to show that $u^{\varepsilon}$ tends to $u$ for $\varepsilon\searrow0$ for a.e. $x$. My attempt: \begin{align*} \vert u^{\varepsilon}(x)-u(x)\vert&=\bigg\vert g(x)-u(x)+\int_{B^+_{\varepsilon}(x)}\zeta(z^m)\cdot (u-g)(z)\cdot \eta_{\varepsilon}(x-z)dz\bigg\vert\\ &=\bigg\vert\int_{B^+_{\varepsilon}(x)}[\zeta(z^m)\cdot (u-g)(z)-(u-g)(x)]\cdot \eta_{\varepsilon}(x-z)dz\bigg\vert\\ &\leq \Vert \eta\Vert_{\infty}\varepsilon^{-m} \int_{B^+_{\varepsilon}(x)}\vert\zeta(z^m)\cdot (u-g)(z)-(u-g)(x)\vert dz\\ &\leq \Vert \eta\Vert_{\infty}\varepsilon^{-m} \int_{B^+_{\varepsilon}(x)}\vert(\zeta(z^m)-1)\cdot (u-g)(z)\vert dz\\ &\quad+\Vert \eta\Vert_{\infty}\varepsilon^{-m} \int_{B^+_{\varepsilon}(x)}\vert (u-g)(z)-(u-g)(x)\vert dz\\ &\leq \Vert \eta\Vert_{\infty}\varepsilon^{-m} \int_{B^+_{\varepsilon}(x)}\vert\zeta(z^m)-1\vert \vert (u-g)(z)\vert dz\\ &\quad+\Vert \eta\Vert_{\infty}\vert B^+_{1}(0)\vert\dfrac{1}{\vert B^+_{\varepsilon}(x)\vert} \int_{B^+_{\varepsilon}(x)}\vert (u-g)(z)-(u-g)(x)\vert dz \end{align*} For $\varepsilon\searrow0$ the second term tends to 0 due to Lebesgue Differentiation theorem. What about the first term? How can i argue that this converges also to 0?

Answer 1

If $u$ is just an $L^p$ function, then it is defined up to sets of Lebesgue measure zero, so specifying $u=g$ on the flat part of the semi-ball does not mean anything. You will not be able to make the first term converge. Lebesgue differentiation theorem is for every point up to a set of Lebesgue measure zero. What you need is $u$ to be in some Sobolev space and then use the theory of traces.