Let $B_1=\{x\in\mathbb{R}^N:\ |x|<1\}$. Let $u\in L^p(B_1)$ with $p\in (1,\infty)$ and suppose that $u$ is also defined in the boundary of $B_1$ and satisfies $u_{|\partial B_1}\in L^p(\partial B_1)$, where I'm assuming the surface measure in $\partial B_1$.
I want to find a sequence $u_n\in C^\infty(\overline{B_1})$ such that $u_n\to u$ in $L^p(B_1)$ and $(u_n)_{|\partial B_1}\to u_{|\partial B_1}$ in $L^p(\partial B_1)$
I was trying to do something like this: Let $\epsilon>0$ be a small number and consider the ball $B_{1+\epsilon}$. Extend $u$ in $B_{1+\epsilon}$ by reflecting it along the radial axis and call it $v$. Let $h_\delta$ be a mollifier sequence and define $$u_\delta(x)=(h_\delta\ast u)(x)=\int_{B_{1+\epsilon}}h_\delta(x-y)v(y)dy$$
The last sequence does converge to $u$ in $L^p(B_1)$ but I dont know how to argue with the boundary. For example, by using Fubini's theorem, I can show that for almost $t\in (0,1+\epsilon)$, there exist a subsequence of $u_\delta$ (not relabaled) such that $(u_n)_{|\partial B_t}\to u_{|\partial B_t}$ in $L^p(\partial B_t)$.
Any to how to go further or even another idea is appreciated.
Thank you
Extend $u$ to $B_2$ so that $u(x)=u(x/|x|)$ when $1<|x|<2$. Show that the sequence $v_n(x)=u(x/(1-n^{-1}))$ converges to $u$ in $L^p(B_1)$. (Same proof as for continuity of translations in $L^p$.) Then convolve $v_n$ with a mollifier $\varphi_n$ whose support contained in $B_{1/n}$. On the boundary $|x|=1$, the value of $v_n*\varphi_n$ is obtained by averaging the boundary values of $u$, and therefore will be close to $u|_{\partial B_1}$.
For example, consider $u=0$ on $(-1,1)$, $u(\pm 1)=1$.
Step 1: Extension
Step 2: Scaling
Step 3: Mollification