Finite L² distributional variation has a finite Radon derivative

Every $L^2$ function is integrable on compact subsets by Cauchy–Schwarz, so its distributional derivatives are defined. For $i=1,2$, set $L_i(\psi ):=-{\int }_{\mathrm{\Omega }}u{\mathrm{\partial }}_i\psi dx$ on $C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{R})$. If $\psi \mathrm{\ne }0$, the vector field $\phi =\psi e_i\mathrm{/}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}$ and its negative are admissible in [F1]; hence $\mathrm{\mid }{L}_i(\psi )\mathrm{\mid }\le M\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}$. The same bound is immediate for $\psi =0$. By [F3], $L_i$ extends uniquely to a bounded real linear functional on $C_0(\mathrm{\Omega })$.

Countable Choice and [F4] represent each extension by a finite signed Radon measure ${\mu }_i$. Formula [F2] identifies ${\mu }_i$ with $D_{i}u$, so $\mu =({\mu }_1,{\mu }_2)$ has finite vector Radon variation by [F5]. For every continuous compactly supported $g$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, approximate uniformly by a smooth compactly supported $h$ with $\mathrm{\parallel }h-g{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$, and replace $h$ by $h\mathrm{/}\max \{1,\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}\}$. The replacement has norm at most one and differs from $g$ by at most $2\epsilon$. Since ${\sum }_i\mathrm{\mid }{\mu }_i\mathrm{\mid }(\mathrm{\Omega })<\mathrm{\infty }$, its integral against $\mu$ approaches that of $g$ as $\epsilon \downarrow 0$. Thus the supremum over smooth unit fields equals the continuous-field supremum in [F5]. By [F2], $\int u\mathrm{div}\phi =-\int \phi \cdot d\mu$; symmetry of the unit test class removes this minus sign. Consequently $\mathrm{\mid }Du\mathrm{\mid }(\mathrm{\Omega })=M=T{V}_2(u;\mathrm{\Omega })$, proving the assertion.