Test-field formula and L1 lower semicontinuity of spatial TV

Suppose $u\in BV(\mathrm{\Omega })$ and write $\mu =Du$. The distributional derivative identity gives ${\int }_{\mathrm{\Omega }}u\mathrm{div}\phi dx=-{\int }_{\mathrm{\Omega }}\phi \cdot d\mu$. The continuous-field dual formula [F2] bounds the supremum over smooth fields by $\mathrm{\mid }\mu \mathrm{\mid }(\mathrm{\Omega })$. Conversely, for any $f\in C_c(\mathrm{\Omega };{\mathbb{R}}^n)$ with $\mathrm{\mid }f\mathrm{\mid }\le 1$, for every $\epsilon >0$, density [F3] gives one smooth compactly supported $g$ with $\mathrm{\parallel }g-f{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$; replacing $g$ by $g\mathrm{/}\max (1,\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }})$ keeps the unit bound and changes this error by at most another $\epsilon$. Finiteness of the coordinate measures bounds the integral error by a fixed constant times $\epsilon$. Letting $\epsilon \downarrow 0$ (without selecting a sequence of approximants) shows that the smooth supremum is at least the continuous-field supremum $\mathrm{\mid }\mu \mathrm{\mid }(\mathrm{\Omega })$. Thus $V(u)=TV(u;\mathrm{\Omega })$.

Assume [A1], let real-valued $u\in L^1(\mathrm{\Omega })$, and suppose $M:=V(u)<\mathrm{\infty }$. For each coordinate $i$, define $L_i(\psi ):=-{\int }_{\mathrm{\Omega }}u{\mathrm{\partial }}_i\psi dx$ on real $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$. Taking the test vector with only its $i$th component equal to $\psi$ and also its negative shows $\mathrm{\mid }{L}_i(\psi )\mathrm{\mid }\le M\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}$. By density [F3], $L_i$ extends uniquely to a bounded real linear functional on real $C_0(\mathrm{\Omega })$. Apply [F4] directly to this real functional to obtain a finite signed Radon measure ${\mu }_i$ representing $D_{i}u$. Applying [F2] to $\mu =({\mu }_1,\dots ,{\mu }_n)$ and using [F3] to pass between continuous and smooth unit fields gives $\mathrm{\mid }\mu \mathrm{\mid }(\mathrm{\Omega })=M$. Hence $u\in BV(\mathrm{\Omega })$ and $TV(u;\mathrm{\Omega })=M$.

If the $L^1$ limit $u$ is already in $BV(\mathrm{\Omega })$, boundedness of $\mathrm{div}\phi$ gives $\int u\mathrm{div}\phi ={\lim }_k\int u_k\mathrm{div}\phi$ for each admissible $\phi$. Each approximating integral is at most $TV(u_k;\mathrm{\Omega })$ by step 1.1, so taking the supremum gives $TV(u;\mathrm{\Omega })\le {\mathrm{lim\hspace{0.17em}inf}}_{k}TV(u_k;\mathrm{\Omega })$; this branch uses no choice axiom.

If only $L:={\mathrm{lim\hspace{0.17em}inf}}_{k}TV(u_k;\mathrm{\Omega })<\mathrm{\infty }$ is known, choose a subsequence whose variations converge to $L$. The same test-integral limit gives $V(u)\le L$. Under [A1], step 1.2 implies $u\in BV(\mathrm{\Omega })$ and $TV(u;\mathrm{\Omega })=V(u)\le L$. If $L=+\mathrm{\infty }$, the extended inequality is automatic.