0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Is there any reason why mollification of $u$ would not work for this? $\endgroup$ – user147263 Mar 5 '15 at 4:17
  • $\begingroup$ @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. $\endgroup$ – spatially Mar 5 '15 at 4:19
  • $\begingroup$ 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 $u+w$. $\endgroup$ – user147263 Mar 5 '15 at 4:22
  • $\begingroup$ @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! $\endgroup$ – spatially Mar 5 '15 at 4:25
  • $\begingroup$ 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. $\endgroup$ – user147263 Mar 5 '15 at 4:26
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.