We know in one dimension that $BV(I)$ is an algebra, for both $I$ is bounded or unbounded. Now suppose $\Omega\subset R^N$ is open bounded with Lipschitz boundary. My question is: is it $BV(\Omega)$ is an algebra? i.e., if $u\in BV(\Omega)$ and $v\in BV(\Omega)$, do we have $uv\in BV(\Omega)$?
The only thing I can think so far is to use definition of $BV$, so to say, take $\varphi\in C_c^\infty(\Omega,R^N)$ and we write $$\int_\Omega uv\,\text{div}\varphi\,dx$$ and we try to estimate this with the norm of $u$ and $v$. However, so far I only can do this...: $$\int_\Omega uv\,\text{div}\varphi\,dx \leq \left(\int_{\Omega}u^2\,|{\text{div}\varphi}|\,dx\right)^{\frac{1}{2}}\left(\int_{\Omega}v^2\,|{\text{div}\varphi}|\,dx\right)^{\frac{1}{2}}$$ which is not really helpful...
Any ideas?
For $n > 1$, $BV(\Omega)$ is no longer an algebra. To come up with a counterexample, consider the imbeddings $$ W^{1,1}(\Omega) \subset BV(\Omega) \subset L^{n/(n-1)}(\Omega) \, . $$ Now pick a function $u \in W^{1,1}(\Omega)$ such that $u \notin L^{2n/(n-1)}(\Omega)$. Then $u^2 \notin L^{n/(n-1}(\Omega)$ and therefore $u^2 \notin BV$. For example, for $n = 2$, you can choose $u(x) = |x|^{-\alpha}$ for any $\alpha \in [1/4, 1)$.