# A special approximation of BV functions

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!

• Is there any reason why mollification of $u$ would not work for this? – user147263 Mar 5 '15 at 4:17
• @FamousBlueRaincoat If we just use mollification, then $(1)$ would be easy but I can not prove $(2)$. I can only have $\geq$ but not $=$ by lower semi continuous. – spatially Mar 5 '15 at 4:19
• Somehow I don't see what makes the difference here. Mollification commutes with adding an affine function; that is, $u_n+w$ would be the same as mollification of $u+w$. So, (2) is just (1) applied to $u+w$. – user147263 Mar 5 '15 at 4:22
• @FamousBlueRaincoat Sorry I still confused. If you mollify $u+w$ then we will have $u_n+w_n$ but not $u_n+w$. I probably missed something here. Could just write me a simple answer? Thank you! – spatially Mar 5 '15 at 4:25
• Convolution of an affine function with a standard mollifier is precisely that affine function again. Try it out: the linear term cancels out in the integral because it's odd while the mollifier is even. – user147263 Mar 5 '15 at 4:26

## 1 Answer

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.