For $u\in L^1(\Omega)$, its <total variation seminorm on a domain> is
$$
\operatorname{TV}(u)
=\sup\left\{
\int_\Omega u\,\operatorname{div}\varphi\,dx:
\varphi\in C_c^1(\Omega;\mathbb R^n),\ 
\|\varphi\|_\infty\leq1
\right\}.
$$
The corresponding <function of bounded variation on a domain>[bounded-variation space] and its zero-mean subspace are
$$
BV(\Omega)=\{u\in L^1(\Omega):\operatorname{TV}(u)<\infty\},
\qquad
BV_0(\Omega)=\left\{u\in BV(\Omega):\int_\Omega u\,dx=0\right\}.
$$
If $u_\Omega=|\Omega|^{-1}\int_\Omega u\,dx$, the <Poincaré inequality for total variation> is
$$
\boxed{\|u-u_\Omega\|_{L^1(\Omega)}\leq C_\Omega\operatorname{TV}(u)}.
$$
In particular, $\|u\|_{L^1}\leq C_\Omega\operatorname{TV}(u)$ for $u\in BV_0(\Omega)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 326 / 2 / 1 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-326/2/1/solution