BV Sobolev–Poincaré inequality for Neumann TV

First there is $C_1<\mathrm{\infty }$ such that every zero-mean $v\in BV(\mathrm{\Omega })$ satisfies $\mathrm{\parallel }v{\mathrm{\parallel }}_1\le C_{1}TV(v;\mathrm{\Omega })$. If no such constant existed, [A1] would select $v_k$ with ${\int }_{\mathrm{\Omega }}{v}_k=0$, $\mathrm{\parallel }{v}_k{\mathrm{\parallel }}_1=1$, and $TV(v_k;\mathrm{\Omega })<1\mathrm{/}k$. Their BV norms are bounded, so [F1] gives a subsequence converging in $L^1$ to $v$. The limit has mean zero and $\mathrm{\parallel }v{\mathrm{\parallel }}_1=1$, while lower semicontinuity gives $TV(v;\mathrm{\Omega })=0$. Hence every distributional derivative of $v$ is zero; [F2] makes $v$ constant a.e., and its mean forces $v=0$, contradicting $\mathrm{\parallel }v{\mathrm{\parallel }}_1=1$.

For any $u\in BV(\mathrm{\Omega })$ set $v=u-u_{\mathrm{\Omega }}$. Then $v$ has mean zero, $TV(v;\mathrm{\Omega })=TV(u;\mathrm{\Omega })$ by [F5], and step 1.1 gives $\mathrm{\parallel }v{\mathrm{\parallel }}_1\le C_{1}TV(u;\mathrm{\Omega })$. By [F3], $\tilde{v}\in BV({\mathbb{R}}^2)$ and $TV(\tilde{v};{\mathbb{R}}^2)\le C_{\mathrm{e}\mathrm{x}\mathrm{t}}(\mathrm{\parallel }v{\mathrm{\parallel }}_1+TV(v;\mathrm{\Omega }))\le C_{\mathrm{e}\mathrm{x}\mathrm{t}}(C_1+1)TV(u;\mathrm{\Omega })\mathrm{.}$ Applying [F4] and using $\mathrm{\parallel }\tilde{v}{\mathrm{\parallel }}_{L^2({\mathbb{R}}^2)}=\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}$ proves the Statement with $C_{\mathrm{\Omega }}=C_{\mathrm{e}\mathrm{x}\mathrm{t}}(C_1+1)$. The same argument also shows every BV function on this bounded domain belongs to $L^2(\mathrm{\Omega })$.