Strict smooth approximation in BV

If $\mathrm{\Omega }=\mathrm{\varnothing }$, take the zero sequence. If $n=0$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, its singleton supports $u$ itself, which is smooth, has zero gradient and variation, and gives the required constant sequence. Assume henceforth $n\ge 1$. By [F6], take a countable locally finite nonnegative smooth partition of unity $({\eta }_j{)}_{j\ge 1}$ on $\mathrm{\Omega }$, with each ${\eta }_j$ compactly supported in $\mathrm{\Omega }$.

Fix $\delta >0$. Extend each $w_j:={\eta }_{j}u$ by zero to ${\mathbb{R}}^n$. Fix a nonnegative unit-mass smooth bump $\rho$ and its rescalings ${\rho }_{\epsilon }$. For each $j$, the approximate-identity property [F2] gives $w_j\ast {\rho }_{\epsilon }\to w_j$ in $L^1$, and also $(u\mathrm{\nabla }{\eta }_j)\ast {\rho }_{\epsilon }\to u\mathrm{\nabla }{\eta }_j$ in $L^1$. Choose the least nonnegative integer $m_j$ for which ${\epsilon }_j:=2^{-m_j}$ is small enough that ${\epsilon }_j<2^{-j}$, the convolution supports remain in $\mathrm{\Omega }$, and $\mathrm{\parallel }{w}_j\ast {\rho }_{{\epsilon }_j}-w_j{\mathrm{\parallel }}_1+\mathrm{\parallel }(u\mathrm{\nabla }{\eta }_j)\ast {\rho }_{{\epsilon }_j}-u\mathrm{\nabla }{\eta }_j{\mathrm{\parallel }}_1<2^{-j}\delta$. Set $u_{\delta }:={\sum }_j{w}_j\ast {\rho }_{{\epsilon }_j}$. This sum is locally finite: for a compact $C⋐\mathrm{\Omega }$, choose $r>0$ so its closed $r$-neighborhood $C_r$ is compactly contained in $\mathrm{\Omega }$. For all sufficiently large $j$, ${\epsilon }_j<r$; if the thickened support of $w_j$ meets $C$, then $\mathrm{supp}{\eta }_j$ meets $C_r$, and local finiteness of $({\eta }_j)$ leaves only finitely many such indices. Thus $u_{\delta }$ is smooth by [F3]. Summing the first error terms and using ${\sum }_j{\eta }_j=1$ gives $\mathrm{\parallel }{u}_{\delta }-u{\mathrm{\parallel }}_{L^1(\mathrm{\Omega })}<\delta$.

The product rule follows by testing against $C_c^{\mathrm{\infty }}$: $D({\eta }_{j}u)={\eta }_{j}Du+u\mathrm{\nabla }{\eta }_{j}dx$. Convolution commutes with distributional derivatives by the defining test pairing and Tonelli [F4] applied to positive and negative parts. These interchanges are product-measurable: each localized $L^1$ density depends only on its own integration variable, and all remaining kernel and smooth test factors are Borel on the product. Absolute integrability follows from the compact supports, the $L^1$ density norms and the finite coordinate variations. Thus [F4] applies before making the interchange. Since ${\sum }_j{\eta }_j=1$, local finiteness gives ${\sum }_{j}u\mathrm{\nabla }{\eta }_j=0$. Therefore $\mathrm{\nabla }{u}_{\delta }={\sum }_j({\eta }_{j}Du)\ast {\rho }_{{\epsilon }_j}+{\sum }_j((u\mathrm{\nabla }{\eta }_j)\ast {\rho }_{{\epsilon }_j}-u\mathrm{\nabla }{\eta }_j)$. For the first sum, $\mathrm{\mid }({\eta }_{j}Du)\ast {\rho }_{{\epsilon }_j}\mathrm{\mid }\le ({\eta }_j\mathrm{\mid }Du\mathrm{\mid })\ast {\rho }_{{\epsilon }_j}$ pointwise, so Tonelli and ${\sum }_j{\eta }_j=1$ bound its $L^1$ norm by $\mathrm{\mid }Du\mathrm{\mid }(\mathrm{\Omega })$. The second sum has $L^1$ norm below $\delta$ by the choices in step 2.1. Hence ${\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{\nabla }{u}_{\delta }\mathrm{\mid }dx\le TV(u;\mathrm{\Omega })+\delta$.

Take ${\delta }_k=1\mathrm{/}(k+1)$ for $k\in \mathbb{N}$ and set $u_k=u_{{\delta }_k}$. Step 2.1 gives $u_k\to u$ in $L^1$, while step 3.1 gives ${\mathrm{lim\hspace{0.17em}sup}}_k{\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{\nabla }{u}_k\mathrm{\mid }dx\le TV(u;\mathrm{\Omega })$. For each $k$, integration by parts against compactly supported smooth tests identifies $D{u}_k=\mathrm{\nabla }{u}_{k}dx$. Since $\mathrm{\nabla }{u}_k$ is continuous and integrable by step 3.1, [F8] gives $u_k\in BV(\mathrm{\Omega })$ and $TV(u_k;\mathrm{\Omega })={\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{\nabla }{u}_k\mathrm{\mid }dx$. The choice-free lower-semicontinuity clause of [F5], now applied to $u_k,u\in BV(\mathrm{\Omega })$, gives the reverse inequality. Thus the variations converge. For $\mathrm{\Omega }={\mathbb{R}}^n$, choose a smooth cutoff ${\chi }_R$ equal to one on $B(0,R)$, supported in $B(0,2R)$, and satisfying $\mathrm{\mid }\mathrm{\nabla }{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$, using [F7]. Since each $u_{1\mathrm{/}(2(k+1))}$ and its gradient are integrable, choose $R_k$ so its $L^1$ tail, gradient tail, and $C\mathrm{\parallel }{u}_{1\mathrm{/}(2(k+1))}{\mathrm{\parallel }}_1\mathrm{/}{R}_k$ are all below $1\mathrm{/}(k+1)$. The products ${\chi }_{R_k}{u}_{1\mathrm{/}(2(k+1))}$ are compactly supported and smooth, converge to $u$ in $L^1$, and have gradient integrals bounded above by $TV(u;{\mathbb{R}}^n)+O(1\mathrm{/}(k+1))$; [F8] puts them in $BV({\mathbb{R}}^n)$ and lower semicontinuity gives convergence of their variations as well.