Proof. The proof of the equality and of the fact that $E^{f,g}$ (a)-(f) of Definition 3.8 may be deduced from the proof of [Reference Dal Maso and Toader20, Proposition 3.11], with minor changes.

We are left with showing that $E^{f,g}$ enjoys properties (g) and (h) of Definition 3.8. We begin proving (g). Since $E^{f,g}$ satisfies properties (a) and (b), it is enough to prove (3.6) for every $A\in\mathcal{A}_c({{\mathbb R}}^d)$. Without loss of generality, we may assume that $u|_A\in GBV_\star(A;{{\mathbb R}}^k)$. Let $w\in W^{1,1}_{\rm loc}({{\mathbb R}}^d;{{\mathbb R}}^k)$, let R > 0, let $m\in{\mathbb N}$, and let $i\in\{1,...,m\}$. We set $v^i_R:=w+\psi^i_R\circ (u-w)$ and note that $v_R^i|_A\in GBV_\star(A;{{\mathbb R}}^k)$, thanks to Proposition 2.3 and to the fact that $GBV_\star(A;{{\mathbb R}}^k)$ is a vector space.

Consider the set $A_{\rm reg}:=\{x\in A\colon \widetilde{u}(x),\,\widetilde{w}(x)\,\,\, \text{exist}\}$. We observe that by Proposition 2.4 A _reg and J_u are Borel sets and that ${\mathcal H}^{d-1}(A\setminus (A_{\rm reg}\cup J_{u}))=0$. We set

\begin{eqnarray*} &&\displaystyle A_{\rm in}^i:=\{x\in A_{\rm reg}\colon|\widetilde{u}(x)-\widetilde{w}(x)|\leqslant \sigma^{i-1}R\},\\ &&\displaystyle A^i:=\{x\in A_{\rm reg}\colon\sigma^{i-1}R \lt |\widetilde{u}(x)-\widetilde{w}(x)| \lt \sigma^{i}R\},\\ &&\displaystyle A^i_{\rm out}:=\{x\in A_{\rm reg}\colon|\widetilde{u}(x)-\widetilde{w}(x)|\geqslant \sigma^{i}R\}. \end{eqnarray*}

Thanks to (i) of Proposition 2.5, for every $i\in\{1,...,m\}$, we have that

\begin{eqnarray*} &\nonumber \displaystyle \nabla v^i_R= \nabla u \quad {\mathcal{L}}^d\text{-} \text{a.e. in } A^i_{\rm in}, \end{eqnarray*}

(3.8)\begin{eqnarray} &\displaystyle\nabla v^i_R=\nabla w +\nabla\psi^i_R\circ(u-w)(\nabla u-\nabla w)\quad {\mathcal{L}}^d\text{-} \text{a.e. in } A^i, \end{eqnarray}

\begin{eqnarray*} &\nonumber \displaystyle \nabla v^i_R=\nabla w \quad {\mathcal{L}}^d\text{-} \text{a.e. in } A_{\rm out}^i. \end{eqnarray*}

Therefore,

(3.9)\begin{equation} \int_Af(x,\nabla v^i_R)\,dx=\int_{A^i_{\rm in}}f(x,\nabla u)\,dx+\int_{A^i}f(x,\nabla v^i_R)\,dx+\int_{A^i_{\rm out}}f(x,\nabla w)\,dx. \end{equation}

Exploiting (f2 $'$), (f3 $'$), and (3.8), and recalling that $\text{Lip}(\psi_R^i)= 1$, we obtain

\begin{align*} \int_{A^i}f(x,\nabla v^i_R)\,dx&\leqslant 2c_3k^{1/2}\int_{A^i}|\nabla w|\, dx+c_3k^{1/2}\int_{A^i}|\nabla u|\, dx+c_4{\mathcal{L}}^d(A^i)\\ &\leqslant 2c_3k^{1/2}\int_{A^i}|\nabla w|\, dx+\frac{c_3k^{1/2}}{c_1}\int_{A^i}f(x,\nabla u)\,dx+c_4{\mathcal{L}}^d(A^i),\\ \int_{A^i_{\rm out}}f(x,\nabla w)\,dx&\leqslant c_3k^{1/2}\int_{A^R_{u,w}}|\nabla w|\, dx+c_4{\mathcal{L}}^d(A_{u,w}^R). \end{align*}

These inequalites and (3.9) lead to

\begin{eqnarray*} \nonumber \sum_{i=1}^m\int_{A}f(x,\nabla &v^i_R)\, dx\leqslant m\Big(\int_A f(x,\nabla u)\, dx +c_3k^{1/2}\int_{A_{u,w}^R}|\nabla w|\, dx+c_4{\mathcal{L}}^d(A_{u,w}^R)\Big) \end{eqnarray*}

(3.10)\begin{eqnarray} &+\frac{c_3k^{1/2}}{c_1}\int_{A}f(x,\nabla u)\, dx+ 2c_3\int_{A}|\nabla w|\, dx+c_4{\mathcal{L}}^d(A). \end{eqnarray}

To estimate the term depending on D ^cu in $E^{f,g}$, we use Proposition 2.5(ii) to rewrite

\begin{align*} \int_{A}f^\infty\Big(x,\frac{dD^cv^i_R}{d|D^cv^i_R|}\Big)\,d|D^cv^i_R|=\int_{A_{\rm in}^i}f^\infty\Big(x,\frac{dD^cu}{d|D^cu|}\Big)\,d|D^cu| \nonumber\\ +\int_{A^i}f^\infty\Big(x,\frac{\nabla\psi_R^i(\widetilde{u}-\widetilde{w})}{|\nabla\psi_R^i(\widetilde{u}-\widetilde{w})|}\frac{dD^cu}{d|D^cu|}\Big)\,|\nabla\psi_R^i(\widetilde{u}-\widetilde{w})| d|D^cu|+\int_{A^i_{\rm out}}f^\infty(x,0)\,d|D^cu| \end{align*}

\begin{equation*} \leqslant \int_{A}\!\!f^\infty\!\Big(x,\frac{dD^cu}{d|D^cu|}\Big)d|D^cu|+\int_{A^i}\!\!f^\infty\!\Big(x,\frac{\nabla\psi_R^i(\widetilde{u}{-}\widetilde{w})}{|\nabla\psi_R^i(\widetilde{u}{-}\widetilde{w})|}\frac{dD^cu}{d|D^cu|}\Big)\!|\nabla\psi_R^i(\widetilde{u}{-}\widetilde{w})| d|D^cu|, \end{equation*}

where we have used that $w\in W^{1,1}_{\rm loc}({{\mathbb R}}^d;{{\mathbb R}}^k)$ and that $f^\infty(x,0)=0.$ Taking advantage of (3.3) and of $\text{Lip}(\psi_R^i)=1$, we infer that

\begin{eqnarray*} \int_{A^i}f^\infty\Big(x,\!\!\!&&\frac{\nabla\psi_R^i(\widetilde{u}-\widetilde{w})}{|\nabla\psi_R^i(\widetilde{u}-\widetilde{w})|}\frac{dD^cu}{d|D^cu|}\Big)\,|\nabla\psi_R^i(\widetilde{u}-\widetilde{w})| d|D^cu|\\ && \leqslant c_3k^{1/2}\int_{A^i}|\nabla\psi_R^i(\widetilde{u}-\widetilde{w})|d|D^cu|\leqslant \frac{c_3k^{1/2}}{c_1}\int_{A^i}f^\infty\Big(x,\frac{dD^cu}{d|D^cu|}\Big)\,d|D^cu|. \end{eqnarray*}

Together with the previous inequalities, this leads to

\begin{eqnarray*} \nonumber\sum_{i=1}^m\int_{A}f^\infty\Big(x,\frac{dD^cv^i_R}{d|D^cv^i_R|}\Big)\,d|D^cv^i_R|&\leqslant & m\int_{A}f^\infty\Big(x,\frac{dD^cu}{d|D^cu|}\Big)\,d|D^cu| \end{eqnarray*}

(3.11)\begin{eqnarray} & + \frac{c_3k^{1/2}} {c_1}\int_{A}f^\infty\Big(x,\frac{dD^cu}{d|D^cu|}\Big)\,d|D^cu|. \end{eqnarray}

We now estimate the surface term in $E^{f,g}$. To this scope, we first remark that for ${\mathcal H}^{d-1}$-a.e. $x\in J_{v^i_R}$, we have the inclusion $J_{v^i_R}\subset A\setminus A_{\rm reg}$ and that for ${\mathcal H}^{d-1}$-a.e $x\in J_{v^i_R}$, we have $\nu_{v^i_R}(x)=\nu_{u}(x)$. Moreover, since $w\in W^{1,1}_{\rm loc}({{\mathbb R}}^d;{{\mathbb R}}^k)$, for ${\mathcal H}^{d-1}$-a.e. $x\in A$ the approximate limit $\widetilde{w}(x)$ exists, so that for ${\mathcal H}^{d-1}$-a.e. every $x\in J_{v^i_R}$, we have

(3.12)\begin{eqnarray} &&\displaystyle (v^i_R)^+(x)=\widetilde{w}(x)+\psi_R^i(u^+(x)-\widetilde{w}(x)), \end{eqnarray}

(3.13)\begin{eqnarray} &&\displaystyle (v^i_R)^-(x)=\widetilde{w}(x)+\psi_R^i(u^-(x)-\widetilde{w}(x)). \end{eqnarray}

For $i\in\{1,...,m\}$ we introduce a partition of $\widetilde{J}_{v^i_R}:= \{{x\in J_{v^i_R}}\colon \widetilde w(x) \text{exists}\} $:

\begin{eqnarray*} &\displaystyle S^i_1:=\{x\in \widetilde{J}_{v^i_R}\colon |u^+(x)-\widetilde{w}(x)| \lt \sigma^{i-1}R,\,|u^{-}(x)-\widetilde w(x)| \lt \sigma^{i-1}R\},\\ &\displaystyle S^i_2:=\{x\in \widetilde{J}_{v^i_R}\colon |u^+(x)-\widetilde{w}(x)|,|u^{-}(x)-\widetilde{w}(x)|\in[\sigma^{i-1}R,\sigma^{i}R]\},\\ &\displaystyle S^i_3:=\{x\in \widetilde{J}_{v^i_R}\colon |u^+(x)-\widetilde{w}(x)| \gt \sigma^{i}R,\,|u^{-}(x)-\widetilde{w}(x)| \gt \sigma^{i}R\}\},\\ &\displaystyle S^i_4:=\{x\in \widetilde{J}_{v^i_R}\colon |u^+(x)-\widetilde{w}(x)| \lt \sigma^{i-1}R,\,|u^{-}(x)-\widetilde{w}(x)| \gt \sigma^{i}R\},\\ & \displaystyle S^i_5:=\{x\in \widetilde{J}_{v^i_R}\colon |u^+(x)-\widetilde{w}(x)| \gt \sigma^{i}R,\,|u^{-}(x)-\widetilde{w}(x)| \lt \sigma^{i-1}R\}, \\ &\displaystyle S^i_6:=\{x\in \widetilde{J}_{v^i_R}\colon |u^+(x)-\widetilde{w}(x)|\in[\sigma^{i-1}R,\sigma^iR],\,\, |u^-(x)-\widetilde{w}(x)|\notin[\sigma^{i-1}R,\sigma^iR]\},\\ &\displaystyle S^i_7:=\{x\in \widetilde{J}_{v^i_R}\colon \,|u^{+}(x)-\widetilde{w}(x)|\notin[\sigma^{i-1}R,\sigma^iR], \,\, |u^-(x)-\widetilde{w}(x)|\in [\sigma^{i-1}R,\sigma^iR]\}. \end{eqnarray*}

Note that for $\ell\in\{2,6,7\}$

\begin{equation*} \displaystyle \text{if } S^i_\ell\cap S^j_\ell\neq \emptyset \text{\,\,\,\,then\,\,}\ |i-j|\leqslant 1, \end{equation*}

so that

(3.14)\begin{equation}\sum_{i=1}^m(\chi_{S^i_2}+\chi_{S^i_6}+\chi_{S^i_7})\leqslant 9. \end{equation}

By definition of S ⁱ₁ and of S ⁱ₃, recalling (3.12) and (3.13), we have

(3.15)\begin{eqnarray} &\displaystyle\int_{S_1^i}g(x,[v^{i}_R],\nu_{v^i_R})\,d{\mathcal H}^{d-1}=\int_{S_1^i}g(x,[u],\nu_u)\,d{\mathcal H}^{d-1}, \end{eqnarray}

(3.16)\begin{eqnarray} &\displaystyle\int_{S_3^i}g(x,[v^{i}_R],\nu_{v^i_R})\,d{\mathcal H}^{d-1}=\int_{S_3^i}g(x,0,\nu_u)\,d{\mathcal H}^{d-1}=0. \end{eqnarray}

Since $\text{Lip}(\psi_R^i)=1$, we have $|[\psi_R^i(u-w)]|\leqslant |[u-w]|=|[u]|$ in $\widetilde{J}_{{v^i_R}}$. Hence, by (g2) and (g3) we deduce that

\begin{eqnarray*} \nonumber \int_{S_2^i\cup S_6^i\cup S_7^i}g(x,[v^{i}_R],\nu_{v^i_R})\,d{\mathcal H}^{d-1}&\leqslant& c_3k \int_{S_2^i\cup S_6^i\cup S_7^i}|[\psi_R^i(u-w)]|\land 1\,d{\mathcal H}^{d-1} \end{eqnarray*}

(3.17)\begin{eqnarray} \leqslant c_3k \int_{S_2^i\cup S_6^i\cup S_7^i}|[u]|\land 1\,d{\mathcal H}^{d-1} &\leqslant& \frac{c_3k}{c_1} \int_{S_2^i\cup S_6^i\cup S_7^i}g(x,[u],\nu_u)\,d{\mathcal H}^{d-1}. \end{eqnarray}

We are left with estimating the surface integral over $S_4^i$ and S ⁱ₅. By definition, if $x\in S^i_4$ then $|u^+(x)-\widetilde{w}(x)| \lt \sigma^{i-1}R$ and $|u^{-}(x)-\widetilde{w}(x)| \gt \sigma^{i}R.$ Therefore,

(3.18)\begin{equation} \begin{aligned} |[u](x)|=|u^+(x)-u^-(x)|&\geqslant &|u^-(x)-\widetilde{w}(x)|-|u^+(x)-\widetilde{w}(x)|\\ &\geqslant &\sigma^iR-\sigma^{i-1}R=\sigma^{i-1}R(\sigma-1)\geqslant c_6k\sigma^{i-1}R, \end{aligned} \end{equation}

where we have used that thanks to (3.4) we have $\sigma \gt c_6k+1$. On the other hand, $[v^i_R](x)=u^+(x)-\widetilde{w}(x)$ so that

\begin{equation*} c_6k|[v^i_R]|= c_6k|u^+(x)-\widetilde{w}(x)|\leqslant c_6k\sigma^{i-1}R. \end{equation*}

From these inequalities, we deduce that $c_6k|[v^i_R]|\leqslant |[u]|$, which by (g6) implies

\begin{equation*} g(x,[v^i_R],\nu_{v^i_R})\leqslant g(x,[u],\nu_{v^i_R})=g(x,[u],\nu_u) \quad {\mathcal H}^{d-1}\text{-a.e. in }S^{i}_4. \end{equation*}

The same argument shows that ${\mathcal H}^{d-1}$-a.e. in S ⁱ₅ it holds

\begin{equation*} g(x,[v_R^i],\nu_{v^i_R})\leqslant g(x,[u],\nu_u). \end{equation*}

From these last two inequalities, we get

(3.19)\begin{equation} \int_{S_4^i\cup S^i_5}g(x,[v^{i}_R],\nu_{v^i_R})\,d{\mathcal H}^{d-1}\leqslant \int_{S_4^i\cup S^i_5}g(x,[u],\nu_{u})\,d{\mathcal H}^{d-1}. \end{equation}

Combining (3.15)-(3.17) and (3.19), we obtain

\begin{align*} \int_{J_{v^i_R}}g(x,[v^i_R],\nu_{v^{i}_R})\,d{\mathcal H}^{d-1}\leqslant &\int_{J_u}g(x,[u],\nu_u)\,d{\mathcal H}^{d-1}\\ &+\frac{c_3k}{c_1}\int_{S^i_2\cup S^i_6\cup S^i_7 }g(x,[u],\nu_u)\,d{\mathcal H}^{d-1}, \end{align*}

which, in light of (3.14), implies that

(3.20)\begin{equation} \sum_{i=1}^m\int_{J_{v^i_R}}g(x,[v^i_R],\nu_{v^{i}_R})\,d{\mathcal H}^{d-1}\leqslant \Big(m+\frac{9c_3k}{c_1}\Big)\int_{J_u}g(x,[u],\nu_u)\,d{\mathcal H}^{d-1}. \end{equation}

Finally, from (3.10), (3.11), and (3.20) we deduce that

\begin{eqnarray*} \frac1m\sum_{i=1}^mE^{f,g}(v^i_R,A) &\leqslant& E^{f,g}(u,A) + c_3k^{1/2}\int_{A^R_{u,w}}|\nabla w|\, dx+c_4{\mathcal{L}}^d(A^R_{u,w})\\ &&+\frac{C}{m}\left( E^{f,g}(u,A) + {\mathcal{L}}^d(A) +\int_A|\nabla w|\, dx \right), \end{eqnarray*}

where $C:=\max\{9c_3k/c_1,2c_3k^{1/2},c_4\}$. This shows that $E^{f,g}$ satisfies (g).

To see that $E^{f,g}$ satisfies (h) it is enough to verify it for $A\in\mathcal{A}_c({{\mathbb R}}^d)$ and for $u|_A\in GBV_\star(A;{{\mathbb R}}^k)$. Let $\lambda \leqslant 1/c_6$ and let $R\in SO(k)$. Note that by (3.1) we also have $\lambda \leqslant (c_1/c_3)k^{-1/2}$. We set $v:=\lambda Ru$. By (f2 $'$), (f3), and (g6) we can estimate

\begin{eqnarray*} E^{f,g}(v,A)&=&\int_{A}\!\!f(x,\nabla v)\, dx+\int_{A}\!\!f^\infty\!\Big(x,\frac{dD^cv}{d|D^cv|}\Big) d|D^cv|+\int_{A\cap J_{u}}g(x,[v],\nu)\, d{\mathcal H}^{d-1}\\ &\leqslant& c_3k^{1/2}\lambda\Big( \int_A|\nabla u|\, dx+|D^cu|(A)\Big)+\int_{A}g(x,[u],\nu)\,d{\mathcal H}^{d-1}+c_4{\mathcal{L}}^d(A)\\ &\leqslant& E^{f,g}(u,A)+(c_4+c_2){\mathcal{L}}^d(A), \end{eqnarray*}

concluding the proof.

Proof. For every sequence $R_m\to+\infty$ as $m\to+\infty$, we have that $\psi_{R_m}\circ u\to u_j$ in $L^0(A;{{\mathbb R}}^k)$ as $m\to+\infty$, so that the lower semicontinuity of $E(\cdot,A_j)$ with respect to the topology of $L^0(A_j;{{\mathbb R}}^k)$ implies that

(3.22)\begin{equation} E(u_j,A_j)\leqslant \liminf_{m\to+\infty}E(\psi_{R_m}\circ u_j,A_j) \end{equation}

for $j=1,2.$

Assume now that $E(u_1,A_1)$ and $E(u_2,A_2)$ are both finite. We can choose a sequence r_m converging to $+\infty$ as $m\to+\infty$ such that

\begin{equation*} c_4{\mathcal{L}}^d(A_{u_j,0}^{r_m})\leqslant \frac{1}{m}\quad \text{for every }m\in{\mathbb N}\ \text{and for}\ j=1,2, \end{equation*}

where $A^{r_m}_{u_j,0}:=\{x\in A: |u_j(x)|\leqslant r_m \}$, in agreemeent with (g) of Defintion 3.8, with w = 0. From that property applied with w = 0, we obtain that there exists $i(m)\in\{1,...,m\}$ such that

\begin{align*} \nonumber E(\psi_{r_m}^{i(m)}\circ u_1,A_1)&+ E(\psi_{r_m}^{i(m)}\circ u_2,A_2)\leqslant E(u_1,A_1)+E(u_2,A_2)\\ &\nonumber \end{align*}

\begin{align*} +\frac{2+C(E(u_1,A_1)+E(u_2,A_2))+{\mathcal{L}}^d(A_1)+{\mathcal{L}}^d(A_2))}{m}. \end{align*}

We now set $R_m:=\sigma^{i(m)}r_m$, so that $\psi^{i(m)}_{r_m}=\psi_{R_m}$. Hence, the previous inequality gives

\begin{equation*} \limsup_{m\to+\infty}(E(\psi_{R_m}\circ u_1,A_1)+E(\psi_{R_m}\circ u_2,A_2))\leqslant E(u_1,A_1)+E(u_2,A_2). \end{equation*}

Combining this last inequality with (3.22), we obtain (3.21) in the case where $E(u_1,A_1)$ and $E(u_2,A_2)$ are both finite. A simpler proof yields the result in the other cases.

Proof. The strategy adopted here is that of [Reference Dal Maso and Toader20, Lemma 3.24], which is based on the classical localization method for Γ-convergence (see [Reference Dal Maso17, Chapters 15,16,18,19]). As already mentioned, the main difference in the proof of this compactness result in the vectorial case lies in the different type of truncation used, which is dealt with in Lemma 4.5 below.

Given $A\in\mathcal{A}({{\mathbb R}}^d)$, we set

(4.1)\begin{gather} \displaystyle E'(\cdot,A):=\Gamma\text{-}\liminf_{n\to+\infty}E_n(\cdot,A)\,\,\text{and }\,\,E^{\prime\prime}(\cdot,A):=\Gamma\text{-}\limsup_{n\to+\infty}E_n(\cdot,A), \end{gather}

(4.2)\begin{gather}\displaystyle E'_-(\cdot,A):=\sup_{A'\in\mathcal{A}_c(A)}E'(\cdot,A)\,\,\text{and }\,\, E^{\prime\prime}_-(\cdot,A):=\sup_{A'\in\mathcal{A}_c(A)}E^{\prime\prime}(\cdot,A'), \end{gather}

where the $\Gamma\text{-}\liminf$ and the $\Gamma\text{-}\limsup$ are computed with respect to the topology of $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$. Given $u\in L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$, it is immediate to check that $E'(u,\cdot)$, $E^{\prime\prime}(u,\cdot)$, $E'_-(u,\cdot),$ and $E_-^{\prime\prime}(u,\cdot)$ are all increasing set functions.

By a compactness theorem for Γ-convergence of increasing functionals (see [Reference Dal Maso17, Theorem 16.9]), there exists a subsequence of $(E_n)_n$, not relabelled, such that

(4.3)\begin{equation} E(u,A):=E'_-(u,A)=E^{\prime\prime}_-(u,A)\quad \text{for}\ u\in L^0({{\mathbb R}}^d;{{\mathbb R}}^k)\ \text{and } A\in\mathcal{A}({{\mathbb R}}^d) .\\ \end{equation}

Thus, we can define $E\colon L^0({{\mathbb R}}^d;{{\mathbb R}}^k)\times\mathcal{B}({{\mathbb R}}^d)\to[0+\infty]$ as

(4.4)\begin{equation} E(u,B)=\inf\{E(u,A):\,\,A\in\mathcal{A}({{\mathbb R}}^d),\,\,B\subset A \}\quad \text{for every } B\in\mathcal{B}({{\mathbb R}}^d). \end{equation}

In Lemma 4.9 we shall prove that

(4.5)\begin{equation} E(u,A):=E'(u,A)=E^{\prime\prime}(u,A)\quad \text{for}\ u\in L^0({{\mathbb R}}^d;{{\mathbb R}}^k)\ \text{and }A\in\mathcal{A}_c({{\mathbb R}}^d) .\\ \end{equation}

This implies that for every $A\in \mathcal{A}_c({{\mathbb R}}^d)$ the sequence $(E_n(\cdot,A))_n\ \Gamma$-converges to $E(\cdot,A)$ in the topology of $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$. We are left with proving that $E\in\mathfrak{E}_\textit{sc}$.

From some general Γ-convergence results (see [Reference Dal Maso17, Proposition 16.15]), it follows that E satisfies property (a) of Definition 3.8. Property (b) is much more difficult and will be obtained in Lemma 4.8.

Property (c1) can be obtained by using the lower semicontinuity of V, stated in Remark 2.9, which gives

\begin{equation*} c_1 V(u,A)-c_2{\mathcal{L}}^d(A)\leqslant E(u,A)\quad \text{for every}\ u\in L^0({{\mathbb R}}^d;{{\mathbb R}}^k)\ \text{and}\ A\in\mathcal{A}_c({{\mathbb R}}^d). \end{equation*}

By inner regularity the inequality can be extended to $ \mathcal{A}({{\mathbb R}}^d)$ and to $\mathcal{B}({{\mathbb R}}^d)$, recalling (4.4). This completes the proof of (c1). Properties (c2)-(f) can be derived arguing as in [Reference Dal Maso and Toader20, Theorem 3.16].

Property (g) for E in $\mathcal{A}_c({{\mathbb R}}^d)$ is a consequence of Lemma 4.4 and 4.9, while (h) is trivial. The extension of (g) and (h) to all Borel sets can easily be obtained using (4.4).

Finally, by [Reference Dal Maso17, Proposition 6.8] for every $A\in\mathcal{A}_c({{\mathbb R}}^d)$ the functionals $E'(\cdot,A)$ and $E^{\prime\prime}(\cdot,A)$ are lower semicontinuous with respect to the convergence of $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$. Since $E=E'=E^{\prime\prime}$, we conclude that $E\in\mathfrak{E}_{\rm sc}$. This concludes the proof.

Proof. We prove (4.6). Without loss of generality, we assume that $E^{\prime\prime}(\psi_R\circ u,A)$ and $E'(\psi_R\circ u,U)$ are both finite. Let $(w_n)_n,(v_n)_n\subset L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ be two sequence of functions converging in $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ to $\psi_R\circ u$ and such that

(4.8)\begin{equation} \limsup_{n\to+\infty}E_n(w_n,A)=E^{\prime\prime}(\psi_R\circ u,A)\,\,\text{and }\,\, \liminf_{n\to+\infty}E_n(v_n,U)=E'(\psi_R\circ u,U). \end{equation}

We fix a subsequence $(E_{n_h})_{h}$ of $(E_n)_n$ such that

\begin{equation*} \lim_{h\to+\infty}E_{n_h}(v_{n_h},U)=\liminf_{n\to+\infty}E_n(v_n,U)=E'(\psi_R\circ u,U). \end{equation*}

Without loss of generality, we may assume that there exists a constant M > 0 such that $ E_{n_h}(w_{n_h},A) \lt M$ and $E_{n_h}(v_{n_h},U) \lt M$ for every $h\in{\mathbb N}$. Remark 3.13 then implies $(w_{n_h}|_A)_h\subset GBV_\star(A;{{\mathbb R}}^k)$, $(v_{n_h}|_U)_h\subset GBV_\star(U;{{\mathbb R}}^k)$. By property (g) of Definition 3.8 for every $h,m\in{\mathbb N}$ we can find $i_{h,m}, j_{h,m}\in\{1,...,m\}$ such that

(4.9)\begin{eqnarray} &\displaystyle E_{n_h}(\psi^{i_{h,m}}_{\sigma R}\circ w_{n_h},A)\leqslant E_{n_h}(w_{n_h},A)+c_4{\mathcal{L}}^d(A^{\sigma R}_{w_{n_h},0})+C\frac{M+{\mathcal{L}}^d(A)}{m}, \end{eqnarray}

(4.10)\begin{eqnarray} &\displaystyle E_{n_h}(\psi^{j_{h,m}}_{\sigma R}\circ v_{n_h},U)\leqslant E_{n_h}(v_{n_h},U)+c_4{\mathcal{L}}^d(U^{\sigma R}_{v_{n_h},0})+C\frac{M+{\mathcal{L}}^d(U)}{m}. \end{eqnarray}

Since $\|\psi_{\sigma R}^{i_{h,m}}\|_{L^\infty({{\mathbb R}}^d;{{\mathbb R}}^k)}\leqslant \sigma^{m+1}R$, $\|\psi_{\sigma R}^{j_{h,m}}\|_{L^\infty({{\mathbb R}}^d;{{\mathbb R}}^k)}\leqslant \sigma^{m+1}R $, $\|\psi_R\circ u\|_{L^\infty({{\mathbb R}}^d;{{\mathbb R}}^k)}$ $ \leqslant \sigma R$, and $w_{n_h}$ and $v_{n_h}$ converge in $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ to $\psi_R\circ u$ as $h\to+\infty$, for every $m\in{\mathbb N}$ we infer that

\begin{eqnarray*} & \displaystyle \nonumber \psi_{\sigma R}^{i_{h,m}}\circ w_{n_h}\to \psi_R\circ u\,\,\, \text{in }L^1({{\mathbb R}}^d;{{\mathbb R}}^k)\, \text{ as }h\to+\infty,\\ &\displaystyle \nonumber \end{eqnarray*}

\begin{eqnarray*} \psi_{\sigma R}^{j_{h,m}}\circ v_{n_h}\to \psi_R\circ u\,\,\, \text{in }L^1({{\mathbb R}}^d;{{\mathbb R}}^k)\, \text{ as }h\to+\infty, \end{eqnarray*}

(4.11)\begin{eqnarray} &\displaystyle\lim_{h\to\infty}{\mathcal{L}}^d(A^{\sigma R}_{w_{n_h},0})= {\mathcal{L}}^d(A_{\psi_R\circ u,0}^{\sigma R})=0, \end{eqnarray}

(4.12)\begin{eqnarray} &\displaystyle \lim_{h\to\infty}{\mathcal{L}}^d(U^{\sigma R}_{v_{n_h},0})= {\mathcal{L}}^d(U_{\psi_R\circ u,0}^{\sigma R})=0. \end{eqnarray}

For every $V\in\mathcal{A}({{\mathbb R}}^d)$ we set

(4.13)\begin{equation} \widehat{E}'(\cdot,V):=\Gamma\text{-}\liminf_{h\to\infty}E_{n_h}(\cdot,V)\,\,\text{and }\,\,\widehat{E}^{\prime\prime}(\cdot,V):=\Gamma\text{-}\limsup_{h\to\infty}E_{n_h}(\cdot,V),\\ \end{equation}

and observe that

(4.14)\begin{equation} E'(\cdot,V)\leqslant \widehat{E}'(\cdot,V)\quad \text{and }\quad \widehat{E}^{\prime\prime}(\cdot,V)\leqslant E^{\prime\prime}(\cdot,V). \end{equation}

By (4.8)-(4.12), we deduce that

(4.15)\begin{eqnarray} &\displaystyle\limsup_{h\to\infty}E_{n_h}(\psi^{i_{h,m}}_{\sigma R}\circ w_{n_h},A)\leqslant E^{\prime\prime}(\psi_R\circ u,A)+C\frac{M+{\mathcal{L}}^d(A)}{m}, \end{eqnarray}

(4.16)\begin{eqnarray} &\displaystyle \liminf_{h\to\infty}E_{n_h}(\psi^{j_{h,m}}_{\sigma R}\circ v_{n_h},U)\leqslant E'(\psi_R\circ u,U)+C\frac{M+{\mathcal{L}}^d(U)}{m}. \end{eqnarray}

By Proposition 2.3, we have that $\psi_{\sigma R}^{i_{h,m}}\circ w_{n_h}|_{A}\in BV(A;{{\mathbb R}}^k)$ and that $\psi_{\sigma R}^{j_{h,m}}\circ v_{n_h}|{U}\in BV(U,{{\mathbb R}}^k)$ for every $h\in{\mathbb N}$. Thus, for δ > 0 and for every $m\in{\mathbb N}$ we may apply Lemma 4.2 to obtain a sequence of functions $(u^m_{h})_h\subset BV(A'\cup U;{{\mathbb R}}^k)$ such that $u^m_{h}\to \psi_R\circ u$ in $L^1_{\rm loc}({{\mathbb R}}^d;{{\mathbb R}}^k)$ as $h\to+\infty$ and such that

\begin{eqnarray*} &&\displaystyle\liminf_{h\to+\infty}E_{n_h}(u^m_{h},A'\cup U)\\ &&\leqslant (1+\delta)\liminf_{h\to\infty}\Big(E_{n_h}(\psi^{i_{h,m}}_{\sigma R}\circ w_{n_h},A)+E_{n_h}(\psi_{\sigma R}^{j_{h,m}}\circ v_{n_h},U)\Big)+\delta\\ &&\displaystyle\leqslant (1+\delta)\limsup_{h\to\infty}E_{n_h}(\psi^{i_{h,m}}_{\sigma R}\circ w_{n_h},A)+(1+\delta)\liminf_{h\to+\infty}E_{n_h}(\psi_{\sigma R}^{j_{h,m}}\circ v_{n_h},U)+\delta. \end{eqnarray*}

These inequalities, combined with (4.15), and (4.16), yield

\begin{equation*} \liminf_{h\to\infty}E_{n_h}(u^m_{h} ,A'\cup U)\leqslant (1+\delta)\Big(E^{\prime\prime}(\psi_R\circ u,A)+E'(\psi_R\circ u,U)+\frac{S}{m}\Big)+\delta, \end{equation*}

where S is a positive constant independent of m. Recalling that $u^m_h\to \psi_R\circ u$ as $h\to+\infty$, by (4.13) and (4.14) we obtain that

\begin{align*} E'(&\psi_R\circ u,A'\cup U)\leqslant \widehat{E}'(\psi_R\circ u,A'\cup U)\\ &\leqslant (1+\delta)\Big(E^{\prime\prime}(\psi_R\circ u,A)+E'(\psi_R\circ u,U)+\frac{S}{m}\Big)+\delta. \end{align*}

Letting $m\to +\infty$ and $\delta\to0^+$, we obtain (4.6).

A similar, but easier, argument shows that (4.7) holds true.

Proof. Let us fix $u\in L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ and $w\in W^{1,1}_{\rm loc}({{\mathbb R}}^d;{{\mathbb R}}^k).$ Without loss of generality, we may assume that $E'(u,A) \lt +\infty$. Let $(u_n)_n\subset L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ be a sequence of functions converging to u in $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ and such that

(4.18)\begin{equation} \liminf_{n\to+\infty}E_n(u_n,A)=E'(u,A). \end{equation}

Fix R > 0 and $m\in{\mathbb N}$. For every $i\in\{1,...,m\}$, we set $v^i_n:=w+\psi_R^i\circ (u_n-w)$ and $v^i:=w+\psi^i_R\circ(u-w)$. Note that for every $i\in\{1,...,m\}$ the sequence $(v^i_n)_n$ converges to vⁱ in $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$, so that by Definition of E $^{\prime}$, we have

(4.19)\begin{equation} \frac{1}{m}\sum_{i=1}^mE'(v^i,A)\leqslant \frac{1}{m}\sum_{i=1}^m\liminf_{n\to+\infty}E_n(v^i_n,A)\leqslant \liminf_{n\to+\infty}\frac{1}{m}\sum_{i=1}^mE_n(v^i_n,A). \end{equation}

By property (g) and by (4.18) we get

\begin{align*} \nonumber \frac{1}{m}\sum_{i=1}^mE_n(v^i_n,A) &\leqslant E_n(u_n,A)+ \frac Cm\Big(\displaystyle E_n(u_n,A)+{\mathcal{L}}^d(A)+\int_A|\nabla w|dx\Big)\\ &\nonumber \end{align*}

\begin{align*} \,\,\,\,\,\, +c_3k^{1/2}\int_{A_{u_n,w}^R}|\nabla w|\,dx+c_4{\mathcal{L}}^d(A_{u_n,w}^R). \end{align*}

By (4.18), this inequality gives

\begin{align*} \nonumber \frac{1}{m}\sum_{i=1}^mE'(v^i,A)&\leqslant E'(u,A)+ \frac Cm\Big(\displaystyle E'(u,A)+{\mathcal{L}}^d(A)+\int_A|\nabla w|dx\Big) \end{align*}

(4.20)\begin{align} & \,\,\,\,\, +\limsup_{n\to+\infty}\Big(c_3k^{1/2}\int_{A_{u_n,w}^R}|\nabla w|\,dx+c_4{\mathcal{L}}^d(A_{u_n,w}^R)\Big). \end{align}

Since $u_n\to u$ in $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$, we have that

\begin{equation*} \limsup_{n\to+\infty}\chi_{A^R_{u_n,w}(x)}\leqslant \chi_{A^R_{u,w}}(x)\quad \text{for }\ {\mathcal{L}}^d\text{-}\text{a.e. } x\in A. \end{equation*}

Thus, from Fatou’s Lemma, we deduce that

(4.21)\begin{equation} \limsup_{n\to+\infty}\Big(\int_{A_{u_n,w}^R}|\nabla w|\, dx+{\mathcal{L}}^d(A^R_{u_n,w})\Big)\leqslant \int_{A^R_{u,w}}|\nabla w|\, dx+{\mathcal{L}}^d(A^R_w), \end{equation}

which combined with (4.20) yields (4.17).

Proof. It is immediate to check that E $^{\prime}$ satisfies property (c2), and by Lemma 4.4, we have that E $^{\prime}$ satisfies property (g) for $B=A_1$ and $B=A_2$. Since the functional $E'(\cdot,A_j)$ is lower semicontinuous with respect to the topology of $L^0(A_j;{{\mathbb R}}^k)$ for $j=1,2$ (see [Reference Dal Maso17, Proposition 6.8]), an application of Lemma 3.16 then proves the claim.

Proof. By [Reference Dal Maso17, Proposition 15.15] we may choose $A^{\prime\prime}\in \mathcal{A}_c({{\mathbb R}}^d)$, with $A'\subset\subset A^{\prime\prime}\subset\subset A$, such that

(4.24)\begin{equation} E^{\prime\prime}(v,A^{\prime\prime})=E'(v,A^{\prime\prime})=E(v,A^{\prime\prime}) \quad \text{for every }v\in L^0({{\mathbb R}}^d;{{\mathbb R}}^k). \end{equation}

Lemma 4.5 implies that there exists $R_m\to +\infty$ as $m\to+\infty$ such that

(4.25)\begin{eqnarray} &\displaystyle \lim_{m\to+\infty}E'(\psi_{R_m}\circ u,A^{\prime\prime})=E'(u,A^{\prime\prime}), \end{eqnarray}

(4.26)\begin{eqnarray}&\displaystyle \lim_{m\to+\infty}E'(\psi_{R_m}\circ u,U)=E'(u,U). \end{eqnarray}

Thanks to Lemma 4.3, we obtain that

\begin{align*} E'(\psi_{R_m}\circ u,A'\cup U)&\leqslant E^{\prime\prime}(\psi_{R_m}\circ u,A^{\prime\prime})+E'(\psi_{R_m}\circ u,U)\\ &\leqslant E'(\psi_{R_m}\circ u,A)+E'(\psi_{R_m}\circ u,U), \end{align*}

where in the second inequality, we have used (4.24). Since $\psi_{R_m}\circ u\to u$ in $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ as $m\to+\infty$, by the lower semicontinuity of E $^{\prime}$, (4.25), and (4.26) we obtain

\begin{equation*} E'(u,A'\cup U)\leqslant E'(u,A^{\prime\prime})+E'(u,U)\leqslant E'(u,A)+E'(u,U), \end{equation*}

concluding the proof.

Proof. The result follows from Lemma 4.6 and by standard arguments (see, for instance, the proof of [Reference Dal Maso17, Lemma 18.4]).

Proof. Let us fix $u\in L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$. By the De Giorgi–Letta Criterion for measures [Reference De Giorgi and Letta18] (see also [Reference Dal Maso17, Theorem 14.23] for the particular version of the theorem employed here) it is enough to show that for every $u\in L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ the function $B\mapsto E(u,B)$ is increasing, inner regular, subadditive, and superadditive. It has already been observed that the set function $E(u,\cdot)$ is increasing, while the inner regularity follows from the definition of $E(u,\cdot)$. The subadditivity is proved in Lemma 4.7, whereas the superadditivity is a consequence of [Reference Dal Maso17, Proposition 16.12] and of the fact that E_n satisfies (b) of Definition 3.8. This concludes the proof.

Proof. Fix $u\in L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ and $A\in\mathcal{A}_c({{\mathbb R}}^d)$. We first show that $E(u,A)=E'(u,A)$. Since $E(u,A)\leqslant E'(u,A)$, it is enough to show the converse inequality. Without loss of generality we may suppose that $E(u,A) \lt +\infty$. It is immediate to check that E $^{\prime}$ and E satisfy property (c2) in Definition 3.8 on $\mathcal{A}({{\mathbb R}}^d)$ and using Remark 2.9 we check that E $^{\prime}$ and E also satisfy (c1) on $\mathcal{A}({{\mathbb R}}^d)$. Hence $V(u,A) \lt +\infty$. This implies that, given ɛ > 0, there exists a compact set $K\subset A$ such that

(4.27)\begin{equation} c_3V(u,A\setminus K)+c_4{\mathcal{L}}^d(A\setminus K)\leqslant \varepsilon. \end{equation}

Consider now $A',A^{\prime\prime}\in\mathcal{A}_c(A)$ with $K\subset A'\subset \subset A^{\prime\prime}$ and let $U=A\setminus K$. Note that $A'\cup U=A$. Hence, by Lemma 4.6 we obtain

(4.28)\begin{equation} E'(u,A)\leqslant E'(u,A^{\prime\prime})+E'(u,A\setminus K)\leqslant E(u,A)+\varepsilon, \end{equation}

where in the last inequality, we have used (4.3), property (c2) for E $^{\prime}$ and (4.27). These arguments show that $E(u,A)=E'(u,A)$ for every $u\in L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ and $A\in\mathcal{A}_c({{\mathbb R}}^d)$.

We now prove that $E(u,A)=E^{\prime\prime}(u,A)$. Since $E(u,A)\leqslant E^{\prime\prime}(u,A)$ it is enough to show that $E^{\prime\prime}(u,A)\leqslant E(u,A)$. It is not restrictive to assume that $E(u,A) \lt +\infty$, so that $u\in GBV_\star(A;{{\mathbb R}}^k)$. Given R > 0, we can exploit the same argument used to obtain (4.28), replacing (4.23) of Lemma 4.6 with (4.7), to show that for every ɛ > 0

(4.29)\begin{equation} E^{\prime\prime}(\psi_R\circ u,A)\leqslant E(\psi_R\circ u,A)+\varepsilon=E'(\psi_R\circ u,A)+\varepsilon, \end{equation}

where we have used the equality $E=E'$. By Lemma 4.5, there exists a positive sequence $R_m\to+\infty$ such that

\begin{equation*} \lim_{m\to+\infty} E'(\psi_{R_m}\circ u,A)=E'(u,A)=E(u,A). \end{equation*}

Since $\psi_{R_m}\circ u\to u$ in $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ as $m\to +\infty$, exploiting the lower semicontinuity of E $^{\prime}$ $^{\prime}$ and this last equality, we deduce from (4.29) that

\begin{equation*} E^{\prime\prime}(u,A)\leqslant E(u,A)+\varepsilon. \end{equation*}

Since ɛ > 0 is arbitrary, we get $E^{\prime\prime}(u,A)\leqslant E(u,A)$, concluding the proof.

Proof. For every $m\in{\mathbb N}$ we set

\begin{equation*} \mu_m(B):=\frac{1}{m}\sum_{i=1}^{m}E(\psi_{R_m}^i\circ u,B) \,\,\, \text{for every } B\in\mathcal{B}(A) \end{equation*}

and note that this defines a finite Radon measure, being $u\in GBV_\star(A;{{\mathbb R}}^k)$.

Let $U\in\mathcal{A}(A)$. Thanks to property (g) of Definition 3.8, we have that

\begin{equation*} \mu_m(U)\leqslant E(u,U)+c_4{\mathcal{L}}^d(U_{u,0}^{R_m}) +\frac{C}{m}\left(E(u,U) +{\mathcal{L}}^d(U) \right). \end{equation*}

Taking the limsup for $m\to+\infty$ in the previous inequality, we get

\begin{equation*} \limsup_{m\to+\infty}\mu_m(U)\leqslant E(u,U). \end{equation*}

For every $m\in{\mathbb N}$, there exists $i(m)\in\{1,...,m\}$ such that

\begin{equation*} E(\psi^{i(m)}_{R_m}\circ u,U)\leqslant \mu_m(U). \end{equation*}

Since $\psi^{i(m)}_{R_m}\circ u\to u$ in $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$ when $m\to+\infty$ and $E(\cdot,U)$ is lower semicontinuous with respect to the topology of $L^0({{\mathbb R}}^d;{{\mathbb R}}^k)$, we have

\begin{equation*} E(u,U)\leqslant \liminf_{m\to+\infty} E(\psi^{i(m)}_{R_m}\circ u,U)\leqslant \limsup_{m\to+\infty}\mu_m(U)\leqslant E(u,U), \end{equation*}

so that (5.1) holds for every $U\in\mathcal{A}(A)$. From [Reference Dal Maso and Toader20, Lemma 4.4] we then deduce that (5.1) holds for every $B\in\mathcal{B}(A)$.

To show that (5.2) with b = a holds true, consider a set $N\in\mathcal{B}(A)$, with ${\mathcal{L}}^d(N)=0$, such that for every $B\in\mathcal{B}(A)$, we have $E^a(u,B)=E(u,B\setminus N)$ and $\mu^a_m(B)=\mu_m(B\setminus N)$ for every $m\in{\mathbb N}$, where µ ^a is the absolutely continuous part of µ with respect to ${\mathcal{L}}^d$. By (5.1), we have that

\begin{equation*} \lim_{m\to+\infty}\mu_m^a(B)=\lim_{m\to+\infty}\mu_m(B\setminus N)=E(u,B\setminus N)=E^a(u,B), \end{equation*}

which proves (5.2). Taking the difference of (5.1) and (5.2), we obtain (5.2) with b = s.

Finally, equalities (5.2) for b = c and b = j can be obtained arguing as in [Reference Dal Maso and Toader20, Proposition 4.3], replacing Theorem 2.2(d) with our Proposition 2.4(d).

Proof. Let $N\in\mathcal{B}({{\mathbb R}}^d)$ be a Borel set, with ${\mathcal{L}}^d(N)=0$, such that $E^s(u,B)=E(u,B \cap N)$ and $E^s(w+\psi^i_R\circ (u-w),B)=E(w+\psi^i_R\circ (u-w),B\cap N)$ for every $i\in\{1,...m\}$, and for every $B\in\mathcal{B}(A)$. Since (3.6) holds for every Borel set $B\in\mathcal{B}(A)$, we get

\begin{eqnarray*} &&\frac{1}{m}\sum_{i=1}^mE^s(w+\psi^i_R\circ (u-w),B)= \frac{1}{m}\sum_{i=1}^mE(w+\psi^i_R\circ (u-w),B\cap N)\\ &&\quad \leqslant E(u,B\cap N) + \int_{B^R_{u,w}\cap N}(c_3k^{1/2}|\nabla w|+c_4)\, dx\\ &&\qquad+\frac{C}{m}\Big( E(u,B\cap N)+ \int_{B\cap N}(|\nabla w|+1)\, dx \Big)\\ &&\quad =E(u,B\cap N)+\frac{C}{m}E(u,B\cap N)=\Big(1+\frac{C}{m}\Big)E^s(u,B), \end{eqnarray*}

which proves the claim.

Proof. Let us fix $m\in{\mathbb N}$. Arguing as in [Reference Dal Maso and Toader20, Lemma 4.16], one can construct a Borel function $\omega_m:{{\mathbb R}}^d\to{\mathbb R}^{k\times d}$ such that

(5.12)\begin{equation} f(x,\omega_m(x))\leqslant f(x,\xi)+\frac{1}{m} \end{equation}

for every $x\in{{\mathbb R}}^d$ and every $\xi\in{\mathbb R}^{k\times d}$. Note that (f2 $'$), (f3), and (5.12) imply that for every $x\in{{\mathbb R}}^d$ one has

(5.13)\begin{equation} |\omega_m(x)|\leqslant \frac{c_2+c_4+1}{c_1}. \end{equation}

Since $\omega_m\in L^\infty({{\mathbb R}}^d;{\mathbb R}^{k\times d})$, by the Lebesgue Differentiation Theorem there exists $N_m\in\mathcal{B}({{\mathbb R}}^d)$, with ${\mathcal{L}}^d(N_m)=0$, such that for every $x\in{{\mathbb R}}^d\setminus N_m$, $\nu\in\mathbb{S}^{d-1}$, and $\lambda\geqslant 1$ there exists $\rho^{\lambda,\nu}_{m,\xi}(x) \gt 0$ such that for every $\rho\in(0,\rho^{\lambda,\nu}_{m,\xi}(x))$, we have

\begin{equation*} \int_{Q^\lambda_\nu(x,\rho)}|\omega_m(x)-\omega_m(y)|\,dy\leqslant \frac{1}{m}\rho^d\lambda^{d-1}. \end{equation*}

We set $N:=\bigcup_{m\in{\mathbb N}}N_m$ and note that ${\mathcal{L}}^d(N)=0$. Using (f5) we get that for every $x\in{{\mathbb R}}^d\setminus N$, $m\in {\mathbb N}$, $\xi\in{\mathbb R}^{k\times d}$, $\nu\in\mathbb{S}^{d-1}$ and $\rho\in(0,\rho^{\lambda,\nu}_{m,\xi}(x))$ we have

(5.14)\begin{equation} \int_{Q^\lambda_\nu(x,\rho)}|\hat{f}(y,\omega_m(x))-\hat{f}(y,\omega_m(y))|\,dy\leqslant \frac{c_5}{m}\rho^d\lambda^{d-1}. \end{equation}

For every $y\in Q^\lambda_\nu(x,\rho)$ and $i\in\{1,...,m\}$ we set

\begin{eqnarray*} &\displaystyle \nonumber w(y):=\omega_m(x)(y-x)+\ell_\xi(x), \end{eqnarray*}

(5.15)\begin{eqnarray} &\displaystyle R:=d^{1/2}\Big(|\xi|+\frac{c_2+c_4+1}{c_1}\Big)\lambda\rho, \end{eqnarray}

\begin{eqnarray*} &\displaystyle \nonumber v^i(y):=w(y)+\psi_R^i (u(y)-w(y)). \end{eqnarray*}

Note that with this choice of R, we have $|\text{tr}_{Q^\lambda_\nu(x,\rho)}(u-w)|\leqslant R$ ${\mathcal H}^{d-1}$-a.e. on $ \partial Q^{\lambda}_\nu(x,\rho)$. Recalling (3.5), we obtain $\text{tr}_{Q_\nu^\lambda(x,\rho)}v^i=\text{tr}_{Q_\nu^\lambda(x,\rho)}\ell_\xi$ and

\begin{equation*} \|v^i-\ell_\xi\|_{L^\infty(Q^\lambda_\nu(x,\rho);{{\mathbb R}}^k)}\leqslant (\sigma^m+1) R=c_{\xi,m}\lambda\rho. \end{equation*}

We claim that there exists $i\in\{1,...,m\}$ such that (5.11) holds with $v=v^i.$ To this aim, for every $i\in\{1,...,m\}$ we consider the following partition of $Q^\lambda_\nu(x,\rho)$:

\begin{align*} & Q^i_{\rm in}:=\{y\in Q^\lambda_\nu(x,\rho)\colon |u(y)-w(y)|\leqslant \sigma^{i-1}R\},\\ & Q^i:=\{y\in Q^\lambda_\nu(x,\rho)\colon \sigma^{i-1}R \lt |u(y)-w(y)| \lt \sigma^iR\},\\ & Q^i_{\rm out}:=\{y\in Q^\lambda_\nu(x,\rho)\colon |u(y)-w(y)|\geqslant \sigma^iR \}. \end{align*}

Using (b) of Definition 3.8 we may write

\begin{equation*} E(v^i,Q^\lambda_\nu(x,\rho))=E(v^i,Q^i_{\rm in})+E(v^i,Q^i)+E(v^i,Q^i_{\rm out}). \end{equation*}

Recalling (3.5) and (5.15), from (5.10) we obtain

\begin{equation*} E^a(v^i,Q^\lambda_\nu(x,\rho))=\int_{Q^i_{\rm in}}\!\!\!\hat{f}(y,\nabla u(y))\,dy +\int_{Q^i}\hat{f}(y,\nabla v^i(y))\,dy+\int_{Q^i_{\rm out}}\!\!\!\!\!\hat{f}(y,\omega_m(x))\,dy. \end{equation*}

Taking advantage of (5.13)-(5.15), by (f2 $'$) and (f3) we get

\begin{align*} \int_{Q^i}\hat{f}(y,& \nabla v^i(y))\, dy\leqslant c_3k^{1/2}\int_{Q^i}|\omega_m(x)+\nabla \psi^i_R(u(y){-}w(y))(\nabla u(y){-}\omega_m(x))|\, dy\\ &+c_4{\mathcal{L}}^d(Q^i) \leqslant c_3k^{1/2}\int_{Q^i}|\nabla u(y)|\, dy+(2c_3k^{1/2}|\omega_m(x)|+c_4){\mathcal{L}}^d(Q^i) \\ &\leqslant \frac{c_3k^{1/2}}{c_1}\int_{Q^i}\hat{f}(y,\nabla u(y))\,dy+\Big(2c_3k^{1/2}\frac{c_2+c_4+1}{c_1}+c_2+c_4\Big){\mathcal{L}}^d(Q^i). \end{align*}

By (5.12) and (5.14), we have

\begin{align*} \int_{Q^i_{\rm out}}\hat{f}(y,\omega_m(x))\,dy&\leqslant \int_{Q^i_{\rm out}}\hat{f}(y,\omega(y))\,dy +\frac{c_5}{m}\lambda^{d-1}\rho^d\\ &\leqslant \int_{Q^i_{\rm out}}\hat{f}(y,\nabla u(y))\,dy+\frac{c_5+1}{m}\lambda^{d-1}\rho^d. \end{align*}

From the previous inequalities, we get

\begin{eqnarray*} E^a(v^i,Q^\lambda_\nu(x,\rho))\leqslant \int_{Q^i_{\rm in}}\hat{f}(y,\nabla u(y))\,dy+ \frac{c_3k^{1/2}}{c_1}\int_{Q^i}\hat{f}(y,\nabla u(y))\,dy\\ {+}\, C_1{\mathcal{L}}^d(Q^i) +\int_{Q^i_{\rm out}}\!\!\!\!\hat{f}(y,\nabla u(y))\,dy+\frac{c_5+1}{m}\lambda^{d-1}\rho^d, \end{eqnarray*}

for some constant C ₁ depending only on $c_1,c_2,c_3,c_4,c_5$, and k.

Summing over $i\in\{1,...,m\}$ and dividing by m, we obtain

(5.16)\begin{equation} \frac{1}{m}\sum_{i=1}^mE^a(v^i,Q^\lambda_\nu(x,\rho))\leqslant\Big(1+\frac{C}{m}\Big) E^a(u,Q^\lambda_\nu(x,\rho))+\frac{C}{m} \lambda^{d-1}\rho^d, \end{equation}

for some constant C depending only on $c_1,c_2,c_3,c_4,c_5$, and k, which we may assume to be larger than the constant of Lemma 5.3. Finally, using that lemma, we get

(5.17)\begin{equation} \frac{1}{m}\sum_{i=1}^mE^s(v^i,Q^\lambda_\nu(x,\rho))\leqslant \Big(1+\frac{C}{m}\Big)E^s(u,Q^\lambda_\nu(x,\rho)), \end{equation}

so that combining (5.16) with (5.17) we obtain that there exists $i\in\{1,...,m\}$ such that (5.11) holds with $v=v^i$.

Suppose now that $\hat{f}$ is continuous in ${{\mathbb R}}^d\times{\mathbb R}^{k\times d}$. Thanks to (f2 $'$) and (f4), we have that for every $x\in{{\mathbb R}}^d$ there exists $\omega(x)$ which minimizes the continuous function $\xi\mapsto\hat{f}(x,\xi)$.

Since $\hat{f}\in\mathcal{F}$, we also have that $|\omega(x)|\leqslant \frac{c_2+c_4}{c_1}.$ This inequality, together with the uniform continuity of $\hat{f}$ on compact sets of ${{\mathbb R}}^d\times{\mathbb R}^{k\times d}$, implies that there exists a $\rho^{\lambda,\nu}_{m}(x) \gt 0$ such that for every $\rho\in(0,\rho^{\lambda,\nu}_{m}(x))$ one has

(5.18)\begin{equation} |\hat{f}(y,\xi)-\hat{f}(x,\xi)|\leqslant \frac{1}{m} \end{equation}

for every $y\in Q^\lambda_\nu(x,\rho)$ and $|\xi|\leqslant (c_2+c_4)/c_1$. In particular, we have

(5.19)\begin{equation} |\hat{f}(y,\omega(x))-\hat{f}(x,\omega(x))|\leqslant \frac{1}{m}. \end{equation}

Exploiting these inequalities and the minimality of $\omega(x)$, we obtain that

(5.20)\begin{equation} \hat{f}(y,\omega(x))\leqslant \hat{f}(y,\xi)+\frac{2}{m} \end{equation}

for every $y\in Q^\lambda_\nu(x,\rho)$ and $\xi\in{\mathbb R}^{k\times d}$. Indeed, if $|\xi|\leqslant (c_2+c_4)/c_1$, we apply (5.18) twice to get

\begin{equation*} \hat{f}(y,\omega(x))\leqslant \hat{f}(x,\omega(x))+\frac{1}{m}\leqslant \hat{f}(x,\xi)+\frac{1}{m}\leqslant \hat{f}(y,\xi)+\frac{2}{m}.\end{equation*}

If $|\xi| \gt (c_4+c_2)/c_1$, recalling the minimality of $\omega(x)$, by (f3) and (f2 $'$), we have $\hat{f}(x,\omega(x))\leqslant\hat{f}(x,0)\leqslant c_4\leqslant \hat{f}(y,\xi) $, which, together with (5.19), implies (5.20). As in the previous part of the proof, for every $y\in{{\mathbb R}}^d$, we set $w(y):=\omega(x)(y-x)+\ell_\xi(x)$. We can replace (5.14) by (5.20) in the argument that we used in the case where $\hat{f}$ was not assumed to be continuous on ${{\mathbb R}}^d\times{\mathbb R}^{k\times d}$ and this leads to the existence of $i\in\{1,...,m\}$ such that

\begin{equation*} E(w+\psi^i_R\circ(u-w),Q^\lambda_\nu(x,\rho))\leqslant \Big(1+\frac{C}{m}\Big) E(u,Q^\lambda_\nu(x,\rho))+\frac{C}{m} \lambda^{d-1}\rho^d. \end{equation*}

In the case where $\hat{f}$ does not depend on x, the same is true for ω and (5.20) holds for every ρ > 0, since it is a direct consequence of the minimality.

Proof. Let $N\in\mathcal{B}({{\mathbb R}}^d)$ and $\rho^{\nu,\lambda}_{m,\xi}(x) \gt 0$ be as in Lemma 5.9. Consider a $v\in BV(Q^\lambda_\nu (x,\rho);{{\mathbb R}}^k)$, with $\text{tr}_{Q^\lambda_\nu(x,\rho)}v=\text{tr}_{Q^\lambda_\nu(x,\rho)}\ell_\xi$, such that

\begin{equation*} E(v,Q^\lambda_\nu(x,\rho))\leqslant m^E(\ell_\xi,Q^\lambda_\nu(x,\rho))+\frac{\lambda^{d-1}\rho^d}{m}. \end{equation*}

Since the function $\ell_\xi$ is a competitor for the minimization problem in the right-hand side of the previous inequality, we also get

\begin{equation*} E(v,Q^\lambda_\nu(x,\rho))\leqslant (c_3k^{1/2}|\xi|+c_4+1)\lambda^{d-1}\rho^d. \end{equation*}

Thus, we may apply Lemma 5.9 to obtain a function $u\in BV(Q^\lambda_\nu(x,\rho);{{\mathbb R}}^k)\cap L^\infty(Q^\lambda_\nu(x,\rho);{{\mathbb R}}^k)$, with $\text{tr}_{Q_\nu^\lambda(x,\rho)}u={\text{tr}_{Q_\nu^\lambda(x,\rho)}}\ell_\xi$ and $\|u-\ell_\xi\|_{L^\infty(Q^\lambda_\nu(x,\rho);{{\mathbb R}}^k)}\leqslant c_{\xi,m}\lambda\rho$, such that (5.22) holds, concluding the proof.

Proof. From Remark 5.8, we deduce that the limits in the right-hand side of (5.24) and of (5.25) exist and that $\displaystyle f(x,\xi)\leqslant \inf_{\delta \gt 0} f_\delta(x,\xi)$, $\displaystyle f^\infty(x,\xi)\leqslant \inf_{\delta \gt 0} f^\infty_\delta(x,\xi).$

We are left with proving that the converse inequality also holds. Let $N\in\mathcal{B}({{\mathbb R}}^d)$ be the ${\mathcal{L}}^d$-negligible set of Corollary 5.10, let $ x\in{{\mathbb R}}^d\setminus N$, $m\in{\mathbb N}$, $\xi\in{\mathbb R}^{k\times d}$, and $t\geqslant 1$. By (5.6), for ρ > 0 small enough, we have

(5.26)\begin{equation} \frac{m^E(t\ell_{\xi},Q(x,\rho))}{t\rho^d}\leqslant \frac{f(x,t\xi)}{t}+\frac{1}{tm}. \end{equation}

We can now apply Corollary 5.10, with ξ replaced by $t\xi$, to obtain a function $u\in BV(Q(x,\rho);{{\mathbb R}}^k)$, with $\text{tr}_{Q(x,\rho)}u=\text{tr}_{Q(x,\rho)}t\ell_\xi$ and $\|u-t\ell_{\xi}\|_{L^\infty(Q(x,\rho);{{\mathbb R}}^k)}\leqslant c_{t\xi,m}\rho$, such that (5.22) holds with ξ replaced by $t\xi$. By taking ρ > 0 small enough, we may also suppose that $\|u-t\ell_{\xi}\|_{L^\infty(Q(x,\rho);{{\mathbb R}}^k)}\leqslant 1/2$. Inequality (5.26) then yields

\begin{equation*} \frac{E(u,Q(x,\rho))}{t\rho^d}\leqslant \frac{f(x,t\xi)}{t}+\frac{C_{t\xi}+1}{tm}\leqslant \frac{f(x,t\xi)}{t}+\frac{C_\xi+1}{m}, \end{equation*}

where $C_{\xi} \gt 0$ is the constant given by (5.21). Note that in the last inequality, we have used the estimate $C_{t\xi}\leqslant tC_\xi$ for every $t\geqslant 1$.

We now compare $E_\delta(u,Q(x,\rho))$ and $E(u,Q(x,\rho))$. Since $\|u-t\ell_{\xi}\|_{L^\infty}\leqslant 1/2$ and $[u-t\ell_\xi]=[u]$, we have $|[u]|\leqslant 1\ {\mathcal H}^{d-1}$-a.e. on J_u. Hence, by (c1 $'$) we get

\begin{eqnarray*} E_\delta(u,Q(x,\rho))&=& E(u,Q(x,\rho))+\delta|Du|(Q(x,\rho))\\ &\leqslant& E(u,Q(x,\rho))+\delta \int_{Q(x,\rho)}|\nabla u| \, dx+\delta|D^cu|(Q(x,\rho)) \\ &&+\,\delta \int_{J_{u}}|[u]|\land 1\, d{\mathcal H}^{d-1} \leqslant \Big(1+\frac{\delta}{c_1}\Big)E(u,Q(x,\rho))+\frac{c_2\delta }{c_1}\rho^d. \end{eqnarray*}

Letting δ be so small that $\delta /c_1\leqslant 1/m$ and $\delta c_2/c_1\leqslant 1/m,$ we then obtain

\begin{equation*} \frac{E_\delta(u,Q(x,\rho))}{t\rho^d}\leqslant\Big(1+\frac1m\Big)\frac{E(u,Q(x,\rho))}{t\rho^d}+\frac1m\leqslant \Big(1+\frac1m\Big)\Big(\frac{f(x,t\xi)}{t}+\frac{C_\xi+1}{m}\Big) \end{equation*}

and, recalling that $\text{tr}_{Q(x,\rho)}u=\text{tr}_{Q(x,\rho)}t\ell_\xi$, we get

\begin{equation*} \frac{m^{E_\delta}(\ell_{t\xi},Q(x,\rho))}{t\rho^d}\leqslant \Big(1+\frac1m\Big)\Big(\frac{f(x,t\xi)}{t}+\frac{C_\xi+1}{m}\Big). \end{equation*}

Evaluating this last inequality at t = 1 and taking the $\limsup$ as $\rho \to 0^+$, we deduce

\begin{equation*} f_\delta(x,\xi)\leqslant \Big(1+\frac1m\Big)\Big(f(x,\xi)+\frac{C_\xi+1}{m}\Big), \end{equation*}

while taking the $\limsup$ as $\rho\to 0^+$ first and letting then $t\to+\infty$, we obtain

\begin{equation*} f^\infty_\delta(x,\xi)\leqslant \Big(1+\frac1m\Big)\Big(f^\infty(x,\xi)+\frac{C_\xi+1}{m}\Big). \end{equation*}

Finally, letting $\delta\to 0^+$ first and then taking the limit for $m\to+\infty$, we conclude the proof.

Proof. By (5.4) there exists $v\in BV(Q_\nu(x,\rho);{{\mathbb R}}^k)$, with $\text{tr}_{Q_\nu(x,\rho)}v=\text{tr}_{Q_\nu(x,\rho)}u_{x,\zeta,\nu}$, such that

(5.28)\begin{equation} E(v,Q_\nu(x,\rho))\leqslant m^E(u_{x,\zeta,\nu},Q_\nu(x,\rho))+ \rho^{d}\leqslant c_3k(|\zeta|\land 1)\rho^{d-1}+\rho^d, \end{equation}

where the second inequality follows from (c2 $'$) and the fact that $u_{x,\zeta,\nu}$ is a competitor for the minimization problem.

Let us choose $R=|\zeta|$ and for $i\in\{1,...,m\}$ let $u^i:=\psi_R^i\circ v$. Note that with this choice of R by (3.5), we have $\text{tr}_{Q_\nu(x,\rho)}u^i=\text{tr}_{Q_\nu(x,\rho)}u_{x,\zeta,\nu}$ and $\|u^i\|_{L^\infty(Q_\nu(x,\rho);{{\mathbb R}}^k)}\leqslant \sigma^m|\zeta|$. By property (g) of Definition 3.8 and (5.28) there exists $i\in\{1,...,m\}$ such that

\begin{eqnarray*} E(u^i,Q_\nu(x,\rho))&\leqslant& E(v,Q_\nu(x,\rho)) \, +c_4\rho^d +\frac{C}{m}\left(E(v,Q_\nu(x,\rho))+\rho^d\right)\\ &\leqslant& m^E(u_{x,\zeta,\nu},Q_\nu(x,\rho))+(2C+c_4+1)\rho^d+\frac{c_3kC(|\zeta|\land 1) }{m}\rho^{d-1}, \end{eqnarray*}

concluding the proof.

Proof. Thanks to Remark 5.8, the limit in the right-hand side of (5.29) exists and

\begin{equation*} g(x,\zeta,\nu)\leqslant \lim_{\delta\to 0^+}g_\delta(x,\zeta,\nu). \end{equation*}

We now prove that the converse inequality holds true as well. Let us fix $m\in{\mathbb N}$. By (5.7) for ρ > 0 small enough, we have that

(5.30)\begin{equation} \frac{m^E(u_{x,\zeta,\nu},Q_\nu(x,\rho))}{\rho^{d-1}}\leqslant g(x,\zeta,\nu)+\frac{1}{m}. \end{equation}

We can now apply Lemma 5.12 to obtain a function $u\in BV(Q_\nu(x,\rho);{{\mathbb R}}^k)\cap L^\infty(Q_\nu(x,\rho);{{\mathbb R}}^k)$, with $\text{tr}_{Q_\nu(x,\rho)}u=\text{tr}_{Q_\nu(x,\rho)}u_{x,\zeta,\nu}$ and $\|u\|_{L^\infty(Q_\nu(x,\rho):{{\mathbb R}}^k)}\leqslant \sigma^m|\zeta|$, such that