Suppose $\Omega:=B(0,1)\subset \mathbb R^2$. Let $u\in BV(\Omega)$ be a radially symmetric, i.e., $u(x)=u(Rx)$ for all $R\in SO(2)$. In addition, suppose $w$ to be an affine function,. i.e., $\partial_1 w=c_1$, $\partial_2w=c_2$ where $c_1$ $c_2$ are two constant.
I am wondering could we obtain a sequence of function $(u_n)\subset C^\infty(\bar\Omega)$ with $u_n$ is radially symmetric and $$ \|\nabla u_n\|_{L^1(\Omega)}\to \|Du\|_{\mathcal M(\Omega)}\tag 1$$ as well as $$ \|\nabla u_n+\nabla w\|_{L^1(\Omega)}\to \|Du+\nabla w\|_{\mathcal M(\Omega)}\tag 2$$ where $\|\cdot\|_{\mathcal M(\Omega)}$ means the total variation of radon measure.\
Thank you!
Let $\phi$ be a standard bump function (radially symmetric, and in particular even: $\phi(-x)=\phi(x)$). I claim that $w*\phi=w$. Indeed, write $w(x) = a+\langle b,x\rangle$ and compute $$ (w*\phi)(x) = \int \phi(y)(a+\langle b,x-y\rangle)\,dy = a + \langle b,x\rangle - \int \phi(y)\langle b,y\rangle\,dy $$ Here the integrand $\phi(y)\langle b,y\rangle$ changes sign when $y$ is replaced by $-y$; therefore, $\int \phi(x-y)\langle b,y\rangle\,dy=0$ by symmetry. We conclude with $$ (w*\phi)(x) = w(x) $$
Let $u_n $ be $u$ convolved with $n^{-1} \phi(x/n)$. Then (1) holds. And since $u_n+w$ is equal to the convolution of $u+w$ with $n^{-1} \phi(x/n)$, (2) holds as well.