Assumption 3.1 (Admissible weights)

Let $p>1,$ $K\geq 1,$ and let $A_p(K)$ denote the Muckenhoupt class as in definition 2.1. A sequence of measurable weight functions $\lambda _k \colon \mathbb {R}^{n} \to [0,\,+\infty )$ is admissible if:

• • $\lambda _k \in A_p(K),$ for every $k\in \mathbb {N};$

• • there exists a cube $Q_0 \subset \mathbb {R}^{n}$ such that for every $k\in \mathbb {N}$ there holds

  (3.1)\begin{equation} c_1\leq {- \hskip-1.05em \int}_{Q_0}\lambda_k\,{\rm d}x \leq c_2, \end{equation}
  for some constants $0< c_1\leq c_2<+\infty$.

Proof. Let $(\lambda _{k_h})\subset (\lambda _{k})$ be the subsequence whose existence is established by proposition 2.4. Hence, in particular, $\lambda _{k_h} \rightharpoonup \lambda _\infty$ weakly in $L^{1}(Q_0)$ and $\lambda _\infty \in A_p(c_3^{p}K)$, for some $c_3=c_3(n)>0$.

Let $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ and let $A\in \mathcal {A}(Q_0)$. Therefore, if $F''$ is as in (4.5) with $k$ replaced by $k_h$, by (3.2) we readily get

(4.7)\begin{align} F''(u,A) & \leq\limsup_{h\to\infty}F_{k_h}(u,A)\leq\lim_{h\to\infty}\beta\int_A\lambda_{h_k}(|\nabla u|^{p}+1)\,{\rm d}x \nonumber\\ & \leq \beta\int_A\lambda_\infty(|\nabla u|^{p}+1)\,{\rm d}x, \end{align}

hence (4.6) is proven for every $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$.

Now let $u \in W^{1,p}_{\lambda _\infty }(Q_0;\mathbb {R}^{m})$; then there exists $(u_j)\subset C^{\infty }(\overline Q_0;\mathbb {R}^{m}) \subset W^{1,\infty }(Q_0;\mathbb {R}^{m})$ such that $u_j \to u$ in $W^{1,p}_{\lambda _\infty }(Q_0;\mathbb {R}^{m})$. Hence, thanks to (4.7), to the fact that $u_j \to u$ in $L^{1}(Q_0;\mathbb {R}^{m})$, and to the lower semicontinuity of $F''$ with respect to the strong $L^{1}(Q_0;\mathbb {R}^{m})$-convergence, we obtain

\begin{align*} F''(u,A) & \leq \liminf_{j\to \infty} F''(u_j,A)\\& \leq \lim_{j \to \infty} \beta\int_A\lambda_\infty(|\nabla u_j|^{p}+1)\,{\rm d}x \ = \beta\int_A\lambda_\infty(|\nabla u|^{p}+1)\,{\rm d}x \end{align*}

and thus the claim.

Proof. The proof of this lemma is classical and follows the line of, e.g. [Reference Braides, Defranceschi and Vitali8, lemma 3.5] with minor modifications. However, since we work in a different functional setting, we repeat the proof here for the readers’ convenience.

Let $\eta >0$ and $M>0$ be fixed. Let $(a_j)$ be a strictly increasing sequence of positive real numbers such that for every $j\in \mathbb {N}$ there exists a Lipschitz function $\varphi _j:\mathbb {R}^{m}\to \mathbb {R}^{m}$ with Lipschitz constant less than or equal to $1$ satisfying

\[ \varphi_j(y)= \begin{cases} y & \text{if } |y|\leq a_j, \\ 0 & \text{if } |y|>a_{j+1}. \end{cases} \]

For every $k\in \mathbb {N}$ and every $j\in \mathbb {N}$ set $w_k^{j}:=\varphi _j(u_k)$. We have

\begin{align*} \int_A f_k(x,\nabla w_k^{j})\,{\rm d}x& =\int_{A\cap\{|u_k|\leq a_j\}}f_k(x,\nabla u_k)\,{\rm d}x+\int_{A\cap\{|u_k|>a_{j+1}\}}f_k(x,0)\,{\rm d}x \\ & \quad+ \int_{A\cap\{a_j<|u_k|\leq a_{j+1}\}}f_k(x,\nabla w_k^{j})\,{\rm d}x \\ & \leq\int_A f_k(x,\nabla u_k)\,{\rm d}x+\beta\int_{A\cap\{|u_k|>a_{j+1}\}}\lambda_k\,{\rm d}x \\ & \quad+ \beta\int_{A\cap\{a_j<|u_k|\leq a_{j+1}\}}\lambda_k(|\nabla u_k|^{p}+1)\,{\rm d}x, \end{align*}

where to establish the last inequality we have used the nonnegativity of $f_k$, (3.2), and the fact that $\varphi _j$ has Lipschitz constant less than or equal to $1$.

Let $N\in \mathbb {N}$ be arbitrary; we now want to estimate $1/N\sum _{j=1}^{N} F_k(w_k^{j},\,A)$, for every $k\in \mathbb {N}$. To this end we start noticing that $(\{a_j<|u_k|\leq a_{j+1}\})_{j\in \mathbb {N}}$ is a family of pairwise-disjoint sets. Therefore, we get

(4.9)\begin{align} \frac{1}{N}\sum_{j=1}^{N} F_k(w_k^{j},A) & \leq F_k(u_k,A)\nonumber\\ & \quad +\frac{\beta}{N}\sum_{j=1}^{N}\int_{A\cap\{|u_k|>a_{j+1}\}}\lambda_k\,{\rm d}x +\frac{\beta}{N}\int_A\lambda_k(|\nabla u_k|^{p}+1)\,{\rm d}x. \end{align}

In view of (3.1) and (4.8) we can find a constant $C>0$ such that

(4.10)\begin{equation} \beta \int_A\lambda_k(|\nabla u_k|^{p}+1)\,{\rm d}x\leq C, \end{equation}

for every $k\in \mathbb {N}$. Moreover, thanks to proposition 2.5 there exist $c,\,\sigma >0$ such that

(4.11)\begin{equation} \int_{A\cap\{|u_k|>a_{j+1}\}}\lambda_k\,{\rm d}x\leq c\,c_2|Q_0|\left(\frac{|A\cap\{|u_k|>a_{j+1}\}|}{|Q_0|}\right)^{\sigma/(1+\sigma)}, \end{equation}

for every $k\in \mathbb {N}$ and every $j\in {1,\,\ldots,\, N}$.

Therefore, we define the sequence $(a_j)$ recursively by imposing the following condition on $a_1$:

(4.12)\begin{equation} |A\cap\{|u_k|>a_1\}|\leq \left(\frac{\eta}{2\beta c\, c_2}\right)^{(1+\sigma)/\sigma} |Q_0|^{-\sigma} \quad \text{for every }k\in\mathbb{N}, \quad a_1\geq M, \end{equation}

which is clearly possible thanks to the boundedness of $(u_k)$ in $L^{1}(A;\mathbb {R}^{m})$. Eventually, by choosing $N\in \mathbb {N}$ in a way such that $C/N\leq \eta /2$, gathering (4.9)–(4.12) we obtain

\[ \frac{1}{N}\sum_{j=1}^{N} F_k(w_k^{j},A)\leq F_k(u_k,A)+\eta. \]

Therefore, for every $k\in \mathbb {N}$ we can find $j(k)\in \{1,\,\ldots,\,N\}$ such that

\[ F_k(w_k^{j(k)},A)\leq F_k(u_k,A)+\eta, \]

hence the proof is accomplished by setting $\varphi _k:=\varphi _{j(k)}$. Finally, we note that $N$ is independent of $k$.

Proof. Let $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$; by [Reference Dal Maso14, proposition 8.1] there exists $(u_k)\subset W^{1,1}(Q_0;\mathbb {R}^{m})$ such that $u_k\to u$ in $L^{1}(Q_0;\mathbb {R}^{m})$ and

(4.13)\begin{equation} \limsup_{k\to\infty}F_k(u_k,A)=F''(u,A)<{+}\infty, \end{equation}

where the last inequality follows by lemma 4.3.

Let $\eta >0$ be fixed: by applying lemma 4.4 to the sequence $(u_k)$ with $M:=\|u\|_{L^{\infty }(Q_0;\mathbb {R}^{m})}$ we obtain a sequence $(w_k)\subset W^{1,1}(Q_0;\mathbb {R}^{m}) \cap L^{\infty } (Q_0;\mathbb {R}^{m})$ which is bounded in $L^{\infty }(Q_0;\mathbb {R}^{m})$, such that $w_k\to u$ in $L^{q}(Q_0;\mathbb {R}^{m})$ for every $1\leq q<+\infty$ and

(4.14)\begin{equation} F_k(w_k,A)\leq F_k(u_k,A)+\eta, \end{equation}

for every $A \in \mathcal {A}(Q_0)$. Then, taking the $\limsup$ as $k\to \infty$ in (4.14) and appealing to (4.13) we obtain

\[ \limsup_{k\to\infty}F_k(w_k,A)\leq F''(u,A)+\eta. \]

Eventually, the claim follows by the definition of $F''$ and the arbitrariness of $\eta$.

Proof. Let $\eta >0$, $A$, $A'$, $A''$, $B$ and $S$ be as in the statement. We start observing that by (3.1) there exists a constant $C>0$ such that

(4.15)\begin{equation} \int_S \lambda_k\,{\rm d}x\leq C \end{equation}

for every $k\in \mathbb {N}$. Let $N\in \mathbb {N}$ be such that

(4.16)\begin{equation} \frac{1}{N}\max\left\{\frac{3^{p-1}\beta}{\alpha},3^{p}\beta C\right\}\leq\eta. \end{equation}

Let $A_1,\,\ldots,\,A_{N+1}$ be $N+1$ open sets satisfying $A'\subset \subset A_1\subset \subset \cdots \subset \subset A_{N+1}\subset \subset A''$, and for $i=1,\,\ldots,\,N$ consider the function $\varphi _i\in C_0^{\infty }(A)$ such that $\mathop{}\!\mathrm{supp} \,\varphi _i\subset A_{i+1}$ and $\varphi _i=1$ on a neighbourhood of $\overline {A_i}$. Finally, define

\[ M_\eta:=\frac{1}{N}3^{p-1}\max_{1\leq i\leq N}\|\nabla\varphi_i\|_\infty. \]

For every $k\in \mathbb {N}$ and for $i=1,\,\ldots,\,N$ we have

(4.17)\begin{align} & F_k(\varphi_i u+(1-\varphi_i)\tilde{u},A'\cup B) \nonumber\\ & \quad=F^{*}_k(u,(A''\cup B)\cap\overline{A_i})+F^{*}_k(\tilde{u},B\setminus A_{i+1})+F_k(\varphi_i u\nonumber\\& \qquad +(1-\varphi_i)\tilde{u},B\cap(A_{i+1}\setminus\overline{A_i})) \nonumber\\ & \quad\leq F_k(u,A'')+F_k(\tilde{u},B)+F_k(\varphi_i u+(1-\varphi_i)\tilde{u},B\cap(A_{i+1}\setminus\overline{A_i})), \end{align}

where $F^{*}_k$ denotes the extension of $F_k$ to the Borel subsets of $Q_0$.

Denote by $I_{k,i}$ the last term in (4.17). For every $k\in \mathbb {N}$ and for $i=1,\,\ldots,\,N$, using (3.2) we obtain

\begin{align*} I_{k,i}& \leq\beta\int_{S_i}\lambda_k|\nabla(\varphi_iu+(1-\varphi_i)\tilde{u})|^{p}\,{\rm d}x+\beta\int_{S_i}\lambda_k\,{\rm d}x \\ & \leq 3^{p-1}\beta\int_{S_i}\lambda_k(|\nabla\varphi_i|^{p}|u-\tilde{u}|^{p}+|\nabla u|^{p}+|\nabla\tilde{u}|^{p})\,{\rm d}x+\beta\int_{S_i}\lambda_k\,{\rm d}x \\ & \leq 3^{p-1}\beta\int_{S_i}\lambda_k|\nabla u|^{p}\,{\rm d}x+3^{p-1}\beta\int_{S_i}\lambda_k|\nabla\tilde{u}|^{p}\,{\rm d}x \ + N M_\eta\int_{S_i}\lambda_k|u-\tilde{u}|^{p}\,{\rm d}x\\ & \quad +\beta\int_{S_i}\lambda_k\,{\rm d}x \\ & \leq 3^{p-1}\beta\int_{S_i}\lambda_k(|\nabla u|^{p}-1)\,{\rm d}x+3^{p-1}\beta\int_{S_i}\lambda_k(|\nabla\tilde{u}|^{p}-1)\,{\rm d}x\\ & \quad + N M_\eta\int_{S_i}\lambda_k|u-\tilde{u}|^{p}\,{\rm d}x+3^{p}\beta\int_{S_i}\lambda_k\,{\rm d}x, \end{align*}

where $S_i:=B\cap (A_{i+1}\setminus \overline {A_i})$. Therefore, by the growth condition from below (3.2) on $f_k$ we get

\[ I_{k,i}\leq\frac{3^{p-1}\beta}{\alpha}\left(F_k(u,S_i)+F_k(\tilde{u},S_i)\right)+NM_\eta\int_{S_i}\lambda_k|u-\tilde{u}|^{p}\,{\rm d}x+3^{p}\beta\int_{S_i}\lambda_k(x)\,{\rm d}x, \]

for every $k\in \mathbb {N}$ and for $i=1,\,\ldots,\,N$. Hence, there exists $i_0\in \{1,\,\ldots,\,N\}$ such that

\begin{align*} I_{k,i_0}& \leq\frac{1}{N}\sum_{i=1}^{N}I_{k,i}\leq\frac{1}{N}\frac{3^{p-1}\beta}{\alpha}\left(F_k(u,S)+F_k(\tilde{u},S)\right)\\ & \quad + M_\eta\int_{S}\lambda_k|u-\tilde{u}|^{p}\,{\rm d}x+\frac{1}{N}3^{p}\beta\int_{S}\lambda_k\,{\rm d}x \end{align*}

for every $k\in \mathbb {N}$; thus by (4.15) we get

\[ I_{k,i_0}\leq\frac{1}{N}\frac{3^{p-1}\beta}{\alpha}\left(F_k(u,A'')+F_k(\tilde{u},B)\right)+M_\eta\int_{S}\lambda_k|u-\tilde{u}|^{p}\,{\rm d}x+\frac{1}{N}3^{p}\beta C. \]

Eventually, in view of (4.16) and (4.17) the proof is accomplished choosing $\varphi :=\varphi _{i_0}$.

Proof. Let $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$; by proposition 4.5 there exist $(u_k)\subset W^{1,p}_{\lambda _k}(A'';\mathbb {R}^{m})$ and $(\tilde {u}_k) \subset W^{1,p}_{\lambda _k}(B;\mathbb {R}^{m})$ which are bounded in $L^{\infty }(Q_0;\mathbb {R}^{m})$, converge to $u$ in $L^{q}(Q_0;\mathbb {R}^{m})$ for every $q\geq 1$, and satisfy

(4.19)\begin{equation} \limsup_{k\to\infty}F_k(u_k,A'') = F''(u,A'') \text{ and } \limsup_{k\to\infty}F_k(\tilde{u}_k,B) = F''(u,B). \end{equation}

Let $\eta >0$ be fixed; then, in view of proposition 4.6 we can find a constant $M_\eta >0$ and a sequence $(\varphi _k)$ of cut-off functions between $A'$ and $A''$ such that

\begin{align*} & F_k(\varphi_k u_k+(1-\varphi_k)\tilde{u}_k,A'\cup B) \\ & \quad\leq(1+\eta)\left(F_k(u_k,A'')+F_k(\tilde{u}_k,B)\right)+M_\eta \int_S\lambda_k|u_k-\tilde{u}_k|^{p}\,{\rm d}x+\eta, \end{align*}

where $S=B\cap (A''\setminus A')$. Since the sequence $\varphi _k u_k+(1-\varphi _k)\tilde {u}_k$ converges to $u$ in $L^{1}(Q_0;\mathbb {R}^{m})$, by (4.19) we obtain

\begin{align*} F''(u,A'\cup B)& \leq\limsup_{k\to\infty}F_k(\varphi_k u_k+(1-\varphi_k)\tilde{u}_k,A'\cup B) \\ & \leq(1+\eta)\left(F''(u,A'')+F''(u,B)\right)\\& \quad +M_\eta\limsup_{k\to\infty}\int_S\lambda_k|u_k-\tilde{u}_k|^{p}\,{\rm d}x+\eta. \end{align*}

Now let $\sigma >0$ be the exponent as in theorem 2.2, using the Hölder inequality and recalling (3.1) we get

\begin{align*} \int_S \lambda_k|u_k-\tilde u_k|^{p}\,{\rm d}x& \leq\left(\int_S\lambda_k^{1+\sigma}\,{\rm d}x\right)^{1/(1+\sigma)}\left(\int_S|u_k-\tilde u_k|^{p(1+\sigma)/\sigma}\,{\rm d}x\right)^{\sigma/(1+\sigma)} \\ & \leq c\,c_2 |Q_0|^{1/(1+\sigma)} \left(\int_{Q_0} |u_k-\tilde u_k|^{p(1+\sigma)/\sigma}\,{\rm d}x\right)^{\sigma/(1+\sigma)}. \end{align*}

Therefore, since $\|u_k-\tilde u_k\|_{L^{q}(Q_0;\mathbb {R}^{m})}\to 0$ for every $q\geq 1$, we immediately obtain

\[ \limsup_{k\to\infty}\int_S\lambda_k|u_k-\tilde{u}_k|^{p}\,{\rm d}x=0. \]

Hence, (4.18) follows by the arbitrariness of $\eta >0$.

Proof. Let $(k_h)$ be the subsequence whose existence is established by proposition 2.4. Thanks to the compactness of $\Gamma$-convergence [Reference Dal Maso14, theorem 8.5], a standard diagonal argument gives the existence of a further subsequence (not relabelled), such that the corresponding functionals $F'$ and $F''$ satisfy

\begin{align*} & \sup\{F'(u,B)\colon B\in\mathcal{A}(Q_0),B\subset{\subset} A\}\\& \quad= \sup\{F''(u,B)\colon B\in\mathcal{A}(Q_0),B\subset{\subset} A\}=:F(u,A), \end{align*}

for every $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ and for every $A\in \mathcal {A}(Q_0)$. We note that the set function $F(u,\,\cdot )$ is inner regular by definition.

Moreover, by virtue of lemma 4.7 we can reason as in [Reference Dal Maso14, proposition 18.4] to deduce that $F(u,\,\cdot )$ is subadditive.

We now prove that (4.20), which will ensure that $F$ is the $\Gamma$-limit of $F_k$ on $W^{1,\infty }(Q_0;\mathbb {R}^{m})$.

Since by definition of $F$ we have $F\leq F'\leq F''$, to get (4.20) it suffices to show that

(4.21)\begin{equation} F''(u,A)\leq F(u,A), \end{equation}

for every $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ and $A \in \mathcal {A}(Q_0)$.

To prove (4.21) we consider the localized functional $H:W^{1,\infty }(Q_0;\mathbb {R}^{m})\times \mathcal {A}(Q_0)\longrightarrow [0,\,+\infty )$ defined as

\[ H(u,A):=\int_A\lambda_\infty(|\nabla u|^{p}+1)\,{\rm d}x. \]

Therefore, by lemma 4.3 we immediately obtain that $F''(u,\,A)\leq H(u,\,A)$, for every $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ and $A \in \mathcal {A}(Q_0)$. For every fixed $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ the set function $H(u,\,\cdot )$ defines a Radon measure on $Q_0$, hence for every $\eta >0$ fixed there exists a compact set $K_\eta \subset A$ such that $H(u,\,A\setminus K_\eta )<\eta$. Let now $A',\,A''\in \mathcal {A}(Q_0)$ be such that $K_\eta \subset A'\subset \subset A''\subset \subset A$ and let $B=A\setminus K_\eta$. By (4.18) we have

\[ F''(u,A)\leq F''(u,A'')+F''(u,A\setminus K_\eta). \]

Then by definition of $F$ we readily obtain

\[ F''(u,A)\leq F(u,A)+H(u,A\setminus K_\eta)\leq F(u,A)+\eta, \]

thus (4.21) follows by the arbitrariness of $\eta >0$.

Finally, the inner regularity and subadditivity of $F(u,\,\cdot )$ together with remark 4.2 allow us to apply the De Giorgi–Letta measure criterion (see, e.g.[Reference Dal Maso14, theorem 14.23]) to deduce that $F(u,\,\cdot )$ is the restriction to $\mathcal {A}(Q_0)$ of a Radon measure on $Q_0$, and thus to conclude.

Proof. Proposition 4.8 ensures the existence of a subsequence $(F_{k_h})$ of $(F_k)$ such that $F_{k_h}(u,\,A)$ $\Gamma$-converges to a functional $F(u,\,A)$ for every $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ and every $A\in \mathcal {A}(Q_0)$. Then, it remains to prove that the functional $F$ admits an integral representation as in (4.22).

We will break up the proof of the integral representation into a number of steps.

Step 1. Definition of $f_\infty$. Let $\xi \in \mathbb {R}^{m\times n}$ be fixed and set $u_\xi (x):=\xi x$. By the measure property of $F$ established in proposition 4.8, the set function $F(u_\xi,\,\cdot )$ can be extended to a Radon measure on $Q_0$. Moreover, thanks to lemma 4.3, $F(u_\xi,\,\cdot )$ is absolutely continuous with respect to the Lebesgue measure. For every $x\in Q_0$ define

\[ f_\infty(x,\xi):=\limsup_{\rho\to 0^{+}}\frac{F(u_\xi,Q_\rho(x))}{|Q_\rho(x)|}, \]

where $Q_\rho (x)$ is the cube centred at $x$, with side length $\rho >0$, and faces parallel to the coordinate planes. Then, $f_\infty$ is a Borel function and the Lebesgue differentiation theorem guarantees that

\[ F(u_\xi,A)=\int_A f_\infty(x,\xi)\,{\rm d}x, \]

for every $A\in \mathcal {A}(Q_0)$.

We now show that $f_\infty$ satisfies the growth and coercivity conditions as in (4.2). To this end, we start observing that the growth condition from above readily follows from lemma 4.3. In fact, choosing in (4.6) $u=u_\xi$, $A=Q_\rho (x)$, with $x$ Lebesgue point for $\lambda _\infty$, the estimate from above in (4.2) follows by dividing both sides of (4.6) by $|Q_\rho (x)|$, and eventually passing to the limit as $\rho \to 0^{+}$.

To derive the growth condition from below on $f_\infty$ let $u\in W^{1,1}(Q_0;\mathbb {R}^{m})$ and $A\in \mathcal {A}(Q_0)$ be fixed. By the Hölder inequality and by the growth condition from below in (3.2) we get

\begin{align*} \left(\int_A|\nabla u|\,{\rm d}x\right)^{p}& \leq\left(\int_A\lambda_k|\nabla u|^{p}\,{\rm d}x\right)\left(\int_A\lambda_k^{{-}1/(p-1)}\,{\rm d}x\right)^{p-1} \\ & = \left(\int_A\lambda_k(|\nabla u|^{p}-1)\,{\rm d}x\right)\left(\int_A\lambda_k^{{-}1/(p-1)}\,{\rm d}x\right)^{p-1}\\& \quad+\left(\int_A\lambda_k\,{\rm d}x\right)\left(\int_A\lambda_k^{{-}1/(p-1)}\,{\rm d}x\right)^{p-1} \\ & \leq\frac{1}{\alpha}F_k(u,A)\left(\int_A\lambda_k^{{-}1/(p-1)}\,{\rm d}x\right)^{p-1}\\& \quad +\left(\int_A\lambda_k\,{\rm d}x\right)\left(\int_A\lambda_k^{{-}1/(p-1)}\,{\rm d}x\right)^{p-1}, \end{align*}

therefore the following lower bound

(4.23)\begin{equation} \alpha \left(\int_A\lambda_k^{{-}1/(p-1)}\,{\rm d}x\right)^{-(p-1)} \left(\int_A|\nabla u|\,{\rm d}x\right)^{p} -\alpha \left(\int_A\lambda_k\,{\rm d}x\right) \leq F_k(u,A), \end{equation}

for every $k\in \mathbb {N}$. Now let $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ and let $(u_h)\subset W^{1,1}(Q_0;\mathbb {R}^{m})$ be such that

\[ u_h \to u \text{ in } L^{1}(Q_0;\mathbb{R}^{m}) \text{ and } \lim_{h\to \infty}F_{k_h}(u_h,A)=F(u,A). \]

Hence, by the lower semicontinuity of $u\mapsto \int _A|\nabla u|\,{\rm d}x$ with respect to the $L^{1}(Q_0;\mathbb {R}^{m})$-topology and by proposition 2.4, evaluating (4.23) in $(u_h)$ and passing to the limit as $h\to \infty$ we find

(4.24)\begin{equation} \alpha \left(\int_A\tilde\lambda_\infty^{{-}1/(p-1)}\,{\rm d}x\right)^{-(p-1)} \left(\int_A|\nabla u|\,{\rm d}x\right)^{p} -\alpha \left(\int_A\lambda_\infty\,{\rm d}x\right) \leq F(u,A), \end{equation}

for every $u\in W^{1,\infty }(A;\mathbb {R}^{m})$ and every $A\in \mathcal {A}(Q_0)$. Now let $x\in Q_0$ be a Lebesgue point for $\lambda _\infty$ and $\tilde \lambda _\infty$ and choose in (4.24) $u=u_\xi$ and $A=Q_\rho (x)$; then, dividing both sides of (4.24) by $|Q_\rho (x)|$ and passing to the limit as $\rho \to 0^{+}$ give

(4.25)\begin{equation} \alpha\big({\tilde \lambda}_\infty(x) |\xi|^{p}- {\lambda}_\infty(x)\big) \leq f_\infty(x,\xi), \end{equation}

for a.e. $x\in Q_0$ and every $\xi \in \mathbb {R}^{m\times n}$. Eventually, (2.6) entails the desired bound from below.

Step 2. Integral representation on piecewise affine functions. Let $A\in \mathcal {A}(Q_0)$ and $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ be piecewise affine on $A$; i.e. there exists a finite family of pairwise disjoint open sets $A_j$ such that $|A\setminus \bigcup _{j=1}^{N}A_j|=0$ and

\[ u(x)=\sum_{j=1}^{N}\chi_{A_j}(x)(u_{\xi^{j}}+z_j), \]

for every $x\in A$ with $\xi ^{j}\in \mathbb {R}^{m\times n}$, $z_j\in \mathbb {R}^{m}$ for $j=1,\,\ldots,\,N$. By remark 4.9 and step 1, taking into account the locality of $F$, we have

\[ F(u,A)=\sum_{j=1}^{N}F(u,A_j)=\sum_{j=1}^{N}\int_{A_j}f_\infty(x,\xi^{j})\,{\rm d}x=\int_Af_\infty(x,\nabla u)\,{\rm d}x, \]

that is, the integral representation (4.22) on piecewise affine functions.

Step 3. Convexity properties of $f_\infty$. For every $A\in \mathcal {A}(Q_0)$ the functional $F(\cdot,\,A)$ is lower semicontinuous on $W^{1,\infty }(Q_0;\mathbb {R}^{m})$ with respect to the strong convergence of $L^{1}(Q_0;\mathbb {R}^{m})$, thus, in particular, it is lower semicontinuous with respect to the weak$^{*}$ $W^{1,\infty }(Q_0;\mathbb {R}^{m})$-convergence. Therefore, the function $\xi \to f_\infty (x,\,\xi )$ is $W^{1,\infty }$-quasiconvex (and rank-$1$-convex) for a.e. $x\in Q_0$ (see, e.g. [Reference Braides and Defranceschi7, proposition 4.3, corollary 4.12]). Then, it is easy to check that the growth condition (4.2) together with the convexity property of $f_\infty (x,\,\cdot )$ yield the local Lipschitz continuity in (4.3) (see, e.g. [Reference Braides and Defranceschi7, remark 4.13]).

Step 4. Integral representation. For $u\in W^{1,p}_{\lambda _\infty }(Q_0;\mathbb {R}^{m})$ consider the functional

(4.26)\begin{equation} u \mapsto \int_Af_\infty(x,\nabla u)\,{\rm d}x. \end{equation}

We observe that the local Lipschitz condition (4.3) satisfied by $f_\infty$ ensures that, for every $A\in \mathcal {A}(Q_0)$, the functional (4.26) is continuous with respect to the strong $W^{1,p}_{\lambda _\infty }(A;\mathbb {R}^{m})$-convergence. Indeed, using Hölder's inequality we easily get

\begin{align*} & \int_A|f_\infty(x,\nabla u_1)-f_\infty(x,\nabla u_2)|\,{\rm d}x \\ & \quad\leq 3^{{1}/{(p-1)}}L'\left(\int_A\lambda_\infty(|\nabla u_1|^{p}+|\nabla u_2|^{p}+1)\,{\rm d}x\right)^{(p-1)/p}\\& \qquad\times \left(\int_A\lambda_\infty|\nabla u_1-\nabla u_2|^{p}\,{\rm d}x\right)^{1/p} \end{align*}

for every $u_1,\,u_2\in W^{1,p}_{\lambda _\infty }(Q_0;\mathbb {R}^{m})$. Moreover, arguing as in the proof of lemma 4.7 we can deduce that (4.26) is also continuous with respect to the strong convergence of $W^{1,q}(Q_0;\mathbb {R}^{m})$, for $q\geq p(1+\sigma )/\sigma$.

Let $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ and $A\in \mathcal {A}(Q_0)$ be given; then there exists a sequence $(u_j)\subset W^{1,q}(Q_0;\mathbb {R}^{m})$ strongly converging to $u$ in $W^{1,q}(Q_0;\mathbb {R}^{m})$ for any $q\in [1,\,\infty )$ such that its restrictions to $A$ are piecewise affine. Since $F$ is lower semicontinuous with respect to the strong topology of $L^{1}(Q_0;\mathbb {R}^{m})$, appealing to step 2 and to the continuity of (4.26) we then obtain

\[ F(u,A)\leq\liminf_{j\to\infty}F(u_j,A)=\liminf_{j\to\infty}\int_Af_\infty(x,\nabla u_j)\,{\rm d}x=\int_Af_\infty(x,\nabla u)\,{\rm d}x. \]

Hence, to represent $F$ in an integral form it only remains to prove the opposite inequality. To this end fix $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ and consider the functional $\widetilde F:W^{1,\infty }(Q_0;\mathbb {R}^{m})\times \mathcal {A}(Q_0)\longrightarrow [0,\,+\infty )$ defined as

\[ \widetilde F(v,A):=F(u+v,A). \]

We observe that $\widetilde F$ satisfies the same properties as $F$, hence there exists a Carathéodory function $h_\infty :Q_0\times \mathbb {R}^{m\times n}\to [0,\,+\infty )$ such that

\[ \widetilde F(v,A)\leq\int_Ah_\infty(x,\nabla v)\,{\rm d}x, \]

for every $v \in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ and every $A\in \mathcal {A}(Q_0)$. Note that the equality holds whenever $v$ is piecewise affine on $A$.

Let $(u_j)$ be the sequence of piecewise affine functions considered above. Then

\begin{align*} \int_Ah_\infty(x,0)\,{\rm d}x& =\widetilde F(0,A)=F(u,A)\leq\int_Af_\infty(x,\nabla u)\,{\rm d}x \\ & = \lim_{j\to\infty}\int_Af_\infty(x,\nabla u_j)\,{\rm d}x=\lim_{j\to\infty}F(u_j,A) =\lim_{j\to\infty}\widetilde F(u_j-u,A) \\ & \leq\lim_{j\to\infty}\int_Ah_\infty(x,\nabla(u_j-u))\,{\rm d}x=\int_Ah_\infty(x,0)\,{\rm d}x, \end{align*}

hence the equality in (4.22) holds for every $u\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$ and every $A\in \mathcal {A}(Q_0)$.

Proof. Without loss of generality, we can assume that liminf in the left-hand side of (5.4) is actually a limit. Moreover, we can also assume that $(u_k) \subset C_0^{\infty }(\mathbb {R}^{n};\mathbb {R}^{m})$, ${\rm supp}(u_k) \subset \subset Q_0$, and

(5.5)\begin{equation} \sup_{k\in \mathbb{N}}\int_{\mathbb{R}^{n}}\lambda_k|\nabla u_k|^{p}\,{\rm d}x<{+}\infty. \end{equation}

Indeed, thanks to (2.10) from (5.3) we have

\[ \sup_{k\in \mathbb{N}}\|u_k\|_{W^{1,p}_{\lambda_k}(A;\mathbb{R}^{m})}<{+}\infty, \]

then, since $A \subset \subset Q_0$, the extension result [Reference Turesson34, theorem 2.1.13] allows us to replace $(u_k)$ with a sequence of functions in $W^{1,p}_{\lambda _k}(\mathbb {R}^{n};\mathbb {R}^{m})$, whose support is compactly contained in $Q_0$, and such that (5.5) holds. Moreover, since for fixed $k$ the space $C_0^{\infty }(\mathbb {R}^{n};\mathbb {R}^{m})$ is dense in $W^{1,p}_{\lambda _k}(\mathbb {R}^{n};\mathbb {R}^{m})$ (see, e.g. [Reference Turesson34, corollary 2.1.6]), a diagonal argument provides us with a sequence $(w_k)\subset C_0^{\infty }(\mathbb {R}^{n};\mathbb {R}^{m})$, with ${\rm supp} (w_k) \subset \subset Q_0$, such that

(5.6)\begin{equation} \|u_k-w_k\|_{W^{1,p}_{\lambda_k}(\mathbb{R}^{n};\mathbb{R}^{m})}<\frac{1}{k}. \end{equation}

Then, we readily get

(5.7)\begin{equation} \sup_{k\in \mathbb{N}}\int_{\mathbb{R}^{n}}\lambda_k|\nabla w_k|^{p}\,{\rm d}x<{+}\infty, \end{equation}

and by the compact embedding of $W^{1,p}_{\lambda _k}(Q_0;\mathbb {R}^{m})$ in $L^{1}(Q_0;\mathbb {R}^{m})$, (5.6) also implies that $w_k \to u$ in $L^{1}(A;\mathbb {R}^{m})$.

Furthermore, we observe that $u_k$ and $w_k$ are close in energy so that once we establish the estimate (5.4) along $(w_k)$, the same estimate will hold true along $(u_k)$. In fact, (3.3) gives

(5.8)\begin{align} & |F_k(w_k,A)-F_k(u_k,A)| \nonumber\\ & \quad\leq 3^{{1}/{(p-1)}}L\,\left(\int_A\lambda_k(|\nabla w_k|^{p}+|\nabla u_k|^{p}+1)\,{\rm d}x\right)^{(p-1)/p}\nonumber\\& \times\left(\int_A\lambda_k|\nabla w_k-\nabla u_k|^{p}\,{\rm d}x\right)^{1/p}, \end{align}

hence gathering (5.6)–(5.8) yields

\[ |F_k(w_k,A)-F_k(u_k,A)|<\frac{C}{k}, \]

for some constant $C>0$.

Therefore, in all that follows, with a little abuse of notation, $(u_k)$ denotes a sequence in $C_0^{\infty }(\mathbb {R}^{n};\mathbb {R}^{m})$, with ${\rm supp}(u_k) \subset \subset Q_0$, and such that (5.5) holds.

For $k\in \mathbb {N}$ and $i \in \{1,\,\ldots,\,m\}$, let $u_k^{(i)}$ denote the $i$-th component of the vector-valued function $u_k$. By applying theorem 2.8 to $|\nabla u_k^{(i)}|\in L^{1}(\mathbb {R}^{n})$ we deduce

(5.9)\begin{equation} \int_{\mathbb{R}^{n}}\lambda_k(M|\nabla u_k^{(i)}|)^{p}\,{\rm d}x\leq c_4\int_{\mathbb{R}^{n}}\lambda_k|\nabla u_k^{(i)}|^{p}\,{\rm d}x, \end{equation}

for every $k\in \mathbb {N}$, every $i=1,\,\ldots,\,m$, and for some $c_4>0$. Hence, by combining (5.5) and (5.9) it follows that the sequence $(\lambda _k(M|\nabla u_h^{(i)}|)^{p})$ is bounded in $L^{1}(\mathbb {R}^{n})$, for every $i=1,\,\ldots,\,m$. Let now $\tau >0$, then lemma 2.11 ensures the existence of a measurable set $E_\tau$, with

(5.10)\begin{equation} |E_\tau|<\tau, \end{equation}

of a constant $\delta _\tau >0$, and a subsequence $(k_j^{\tau })$ such that

\[ \int_B\lambda_{k_j^{\tau}}(M|\nabla u_{k_j^{\tau}}^{(i)}|)^{p}\,{\rm d}x<\tau, \]

for every $j \in \mathbb {N}$, every $i=1,\,\ldots,\, m$, and for every measurable set $B$ with $B\cap E_\tau =\emptyset$ and $|B|<\delta _\tau$.

To simplify the notation we drop the dependence of the sequence on $j$ and $\tau$, thus we write

(5.11)\begin{equation} \int_B\lambda_k(M|\nabla u_k^{(i)}|)^{p}\,{\rm d}x<\tau, \end{equation}

for every $k\in \mathbb {N}$, every $i=1,\,\ldots,\, m$, and every measurable $B$ with $B\cap E_\tau =\emptyset$ and $|B|<\delta _\tau$.

By the Hölder inequality we deduce

(5.12)\begin{align} \left(\int_{\mathbb{R}^{n}}|\nabla u_k^{(i)}|\,{\rm d}x\right)^{p}& =\left(\int_{Q_0}|\nabla u_k^{(i)}|\,{\rm d}x\right)^{p}\nonumber\\& \leq\left(\int_{Q_0}\lambda_k|\nabla u_k^{(i)}|^{p}\,{\rm d}x\right)\left(\int_{Q_0}\lambda_k^{{-}1/(p-1)}\,{\rm d}x\right)^{p-1}, \end{align}

hence by (3.1), (5.5) and (5.12), since $\lambda _k$ belongs to $A_p(K)$ we get

(5.13)\begin{equation} \left(\int_{\mathbb{R}^{n}}|\nabla u_k^{(i)}|\,{\rm d}x\right)^{p}\leq C\frac{K}{c_1}, \end{equation}

for every $k \in \mathbb {N}$, $i=1,\,\ldots,\,m$, and some $C>0$. In its turn (5.13) together with (2.11) provide us with a constant $L_\tau \geq (\tilde c/\tau )(CK/c_1)^{1/p}$ such that for every $k\in \mathbb {N}$, and $i=1,\,\ldots,\,m$

(5.14)\begin{equation} |\{x\in\mathbb{R}^{n}\colon(M|\nabla u_k^{(i)}|)(x)\geq L_\tau\}|\leq\min\{\tau,\delta_\tau\}. \end{equation}

For $k\in \mathbb {N}$, and $i=1,\,\ldots,\,m$ define the sets

\[ H_{i,k}^{\tau}:=H_{i,k}^{L_\tau}:=\{x\in\mathbb{R}^{n}\colon(M|\nabla u_k^{(i)}|)(x)< L_\tau\},\quad H_k^{\tau}:=\bigcap_{i=1}^{m}H_{i,k}^{\tau}. \]

Then lemma 2.9 yields

\[ |u_k^{(i)}(x)-u_k^{(i)}(y)|\leq c_5(n)L_\tau|x-y| \]

for every $k\in \mathbb {N}$, $i=1,\,\ldots,\,m$, and every $x,\, y \in H_k^{\tau }$. Namely, the functions $u_k^{(i)}$ are Lipschitz continuous on $H_k^{\tau }$ with Lipschitz constant $c_5(n)L_\tau$, for every $k\in \mathbb {N}$ and every $i=1,\,\ldots,\,m$.

Appealing to McShane's theorem [Reference McShane27] we can extend $u_k^{(i)}$ from $H_k^{\tau } \cap Q_0$ to $\mathbb {R}^{n}$ keeping the same Lipschitz constant $c_5(n)L_\tau$. We denote this extension with $v_k^{\tau,(i)}$ and note that we can assume that $v_k^{\tau,(i)}(x)=0$ if $\mathop{}\!\mathrm{dist} (x,\,Q_0)>1$. We then have

\[ v_k^{\tau,(i)}=u_k^{(i)},\quad\nabla v_k^{\tau,(i)}=\nabla u_k^{(i)}\quad\text{a.e. in }\; H_h^{\tau} \cap Q_0 \]

and

(5.15)\begin{equation} \|\nabla v_k^{\tau,(i)}\|_{L^{\infty}(\mathbb{R}^{n};\mathbb{R}^{n})}\leq c_5(n)L_\tau. \end{equation}

Now, let $x'\in \mathbb {R}^{n}$ be such that $\mathop{}\!\mathrm{dist} (x',\,Q_0)>1$, then

(5.16)\begin{equation} |v_k^{\tau,(i)}(x)|=|v_k^{\tau,(i)}(x)-v_k^{\tau,(i)}(x')|\leq c_5(n)L_\tau(\mathop{}\!\mathrm{diam}(Q_0)+2), \end{equation}

for every $x\in Q_0$. Hence, gathering (5.15) and (5.16) entails

\[ \sup_{k}\|v_k^{\tau,(i)}\|_{W^{1,\infty}(Q_0)}<{+}\infty, \]

for every $i=1,\,\ldots,\, m$, and every $\tau >0$. Therefore, up to subsequences (not relabelled), for every $\tau >0$ fixed, we get that in particular

\[ v_k^{\tau,(i)}\rightarrow v^{\tau,(i)} \text{ in } {L^{\infty}(Q_0)} \]

as $k \to \infty$, with

\[ \|\nabla v^{\tau,(i)}\|_{L^{\infty}(Q_0;\mathbb{R}^{n})}\leq c_5(n)L_\tau, \]

for every $i=1,\,\ldots,\,m$. Finally, set

\[ v_k^{\tau}:=(v_k^{\tau,(1)},\ldots,v_k^{\tau,(m)}),\quad v^{\tau}:=(v^{\tau,(1)},\ldots,v^{\tau,(m)}). \]

Now, define the set $B_\tau :=\{x\in A\colon v^{\tau }(x)\neq u(x)\}$, then it must hold

(5.17)\begin{equation} |B_\tau|\leq (m+1)\tau. \end{equation}

To prove (5.17) we start observing that there exists a subsequence $(k_j)$ such that if

\[ E:=\Big\{x\in A\colon \lim_{j\to\infty}u_{k_j}(x)=u(x)\Big\} \]

then $|A\setminus E|=0$; hence, as a consequence, $|B_\tau \cap E|=|B_\tau |$. Moreover, since $v_k^{\tau }\to v^{\tau }$ in $L^{\infty }(A;\mathbb {R}^{m})$ as $k\to \infty$ we have that

(5.18)\begin{equation} \lim_{k\to\infty} v_k^{\tau}(x)=v^{\tau}(x) \end{equation}

for every $x\in A$ and hence, in particular, for every $x\in B_\tau$.

Assume by contradiction that $|B_\tau |>(m+1)\tau$, then by (5.14) we obtain

(5.19)\begin{equation} |B_\tau\cap E\cap H_{k_j}^{\tau}|=|B_\tau\cap H_{k_j}^{\tau}|>\tau, \end{equation}

for every $j\in \mathbb {N}$. Therefore, by (5.19) and lemma 2.10 there exists $(k_{j_h})\subset (k_j)$ such that

\[ \bigcap_{h\in\mathbb{N}}(B_\tau\cap E\cap H_{k_{j_h}}^{\tau})\neq\emptyset. \]

Thus, if $x$ belongs to the set above by (5.18) we get

\[ v^{\tau}(x)=\lim_{h\to\infty}v_{k_{j_h}}^{\tau}(x)=\lim_{j\to\infty}u_{k_{j_h}}(x)=u(x), \]

which is a contradiction in view of the definition of $B_\tau$. Therefore, (5.17) holds.

To conclude, it only remains to prove the energy estimate (5.4). Let $E_\tau$ be as in (5.10); by the nonnegativity of $f_k$ we have

(5.20)\begin{align} \int_Af_k(x,\nabla u_k)\,{\rm d}x & \geq\int_{(A\setminus E_\tau)\cap H_k^{\tau}}f_k(x,\nabla v_k^{\tau})\,{\rm d}x \nonumber\\ & =\int_{(A\setminus E_\tau)}f_k(x,\nabla v_k^{\tau})\,{\rm d}x-\int_{(A\setminus E_\tau)\setminus H_k^{\tau}}f_k(x,\nabla v_k^{\tau})\,{\rm d}x. \end{align}

By (5.14) we get

(5.21)\begin{equation} |(A\setminus E_\tau)\setminus H_h^{\tau}|\leq\sum_{i=1}^{m}|(A\setminus E_\tau)\setminus H_{i,h}^{\tau}|< m\min\{\tau,\delta_\tau\}, \end{equation}

hence invoking (3.2), (5.15), (5.14), (5.11), (2.3), proposition 2.5 and (5.21) we obtain

(5.22)\begin{align} \int_{(A\setminus E_\tau)\setminus H_k^{\tau}}f_k(x,\nabla v_k^{\tau})\,{\rm d}x& \leq\beta\int_{(A\setminus E_\tau)\setminus H_k^{\tau}}\lambda_k(|\nabla v_k^{\tau}|^{p}+1)\,{\rm d}x \nonumber\\ & \leq\beta m^{p-1}c_5(n)^{p}L_\tau^{p}\int_{(A\setminus E_\tau)\setminus H_k^{\tau}}\lambda_k\,{\rm d}x+\beta\int_{A\setminus H_k^{\tau}}\lambda_k\,{\rm d}x \nonumber\\ & \leq\beta m^{p-1}c_5(n)^{p}L_\tau^{p}\sum_{i=1}^{m}\int_{(A\setminus E_\tau)\setminus H_{i,k}^{\tau}}\lambda_k\,{\rm d}x\nonumber\\ & \quad +\beta cc_2|Q_0|\left(\frac{|A\setminus H_k^{\tau}|}{|Q_0|}\right)^{\sigma/(1+\sigma)} \nonumber\\ & \leq\beta m^{p-2}c_5(n)^{p}\sum_{i=1}^{m}\int_{(A\setminus E_\tau)\setminus H_{i,k}^{\tau}}\lambda_k(M|\nabla u_k^{(i)}|)^{p}\,{\rm d}x\nonumber\\ & \quad+\beta cc_2|Q_0|\left(\frac{m\tau}{|Q_0|}\right)^{\sigma/(1+\sigma)} \nonumber\\ & \leq\beta m^{p-1}c_5(n)^{p}\tau+\alpha_\tau, \end{align}

where $\alpha _\tau :=\beta cc_2|Q_0|({m\tau }/{|Q_0|})^{\sigma /(1+\sigma )}$; thus $\alpha _\tau \to 0$, as $\tau \to 0^{+}$.

Now let $A_\tau \subset A$ be an open set containing $A\setminus E_\tau$ and such that

(5.23)\begin{equation} \left|\int_{A_\tau}f_k(x,\nabla v_k^{\tau})\,{\rm d}x-\int_{A\setminus E_\tau}f_k(x,\nabla v_k^{\tau})\,{\rm d}x\right|<\tau. \end{equation}

We note that this choice is always possible thanks to the growth conditions satisfied by $f_k$ (3.2), to (5.15), and in view of proposition 2.5. Indeed, we have

\begin{align*} & \int_{A_\tau\setminus(A\setminus E_\tau)}f_k(x,\nabla v_k^{\tau})\,{\rm d}x\leq\beta(m^{p-1}c_5(n)^{p}L_\tau^{p}+1)\int_{A_\tau\setminus(A\setminus E_\tau)}\lambda_k\,{\rm d}x \\ & \quad\leq\beta(m^{p-1}c_5(n)^{p}L_\tau^{p}+1)cc_2|Q_0|\left(\frac{|A_\tau\setminus(A\setminus E_\tau)|}{|Q_0|}\right)^{\sigma/(1+\sigma)}, \end{align*}

moreover, $|A\setminus A_\tau |\leq |E_\tau |<\tau$.

Eventually, by combining (5.20), (5.22) and (5.23) we deduce

\[ \int_Af_k(x,\nabla u_k)\,{\rm d}x\geq\int_{A_\tau}f_k(x,\nabla v_k^{\tau})\,{\rm d}x-\alpha_\tau-\tau(\beta m^{p-1}c_5(n)^{p}+1), \]

and hence the claim follows with $\beta _\tau :=\alpha _\tau +\tau (\beta m^{p-1}c_5(n)^{p}+1)$.

Proof. In all that follows $(k_h)$ denotes the subsequence provided by theorem 4.10.

We divide the proof into two main steps.

Step 1: Lower bound. In this step, we prove that

(5.25)\begin{equation} F'(u,A)\geq F_\infty(u,A), \end{equation}

for every $u\in W^{1,1}(Q_0;\mathbb {R}^{m})$ and every $A\in \mathcal {A}(Q_0)$ with $A \subset \subset Q_0$.

To this end, let $u\in W^{1,1}(Q_0;\mathbb {R}^{m})$ and $A\in \mathcal {A}(Q_0)$, $A \subset \subset Q_0$ be fixed.

Substep 1.1: $u \in W^{1,p}_{\lambda _\infty }(A;\mathbb {R}^{m})$. By [Reference Dal Maso14, proposition 8.1] there exists $(u_h)\subset W^{1,1}(Q_0;\mathbb {R}^{m})$ with $u_h\to u$ in $L^{1}(Q_0;\mathbb {R}^{m})$ such that

(5.26)\begin{equation} F'(u,A)=\liminf_{h\to\infty}F_{k_h}(u_h,A). \end{equation}

We observe that lemma 4.3 guarantees that $F'(u,\,A)<+\infty$; therefore, $(u_h)\subset W^{1,p}_{\lambda _h}(A;\mathbb {R}^{m})$ and (up to possibly passing to a subsequence) by (3.2) we get

\[ \sup_{h\in \mathbb{N}}\int_A\lambda_{k_h}|\nabla u_h|^{p}\,{\rm d}x<{+}\infty. \]

Now let $\tau >0$ be fixed and arbitrary; theorem 5.1 provides us with $(\beta _\tau )$, infinitesimal as $\tau \to 0^{+}$, $A_\tau \subset A$, with $|A\setminus A_\tau |<\tau$, and $(v_h^{\tau })$, $v^{\tau }$ in $W^{1,\infty }(\mathbb {R}^{n};\mathbb {R}^{m})$, such that $v_h^{\tau }\to v^{\tau }$ in $L^{1}(Q_0;\mathbb {R}^{m})$, as $h\to \infty$. Moreover, by (5.26) and (5.4) we obtain

(5.27)\begin{align} F'(u,A)& =\liminf_{h\to\infty}\int_Af_{k_h}(x,\nabla u_h)\,{\rm d}x \nonumber\\ & \geq\liminf_{h\to\infty}\int_{A_\tau} f_{k_h}(x,\nabla v_h^{\tau})\,{\rm d}x-\beta_\tau\geq\int_{A_\tau}f_\infty(x,\nabla v^{\tau})\,{\rm d}x-\beta_\tau, \end{align}

where the last inequality follows by theorem 4.10, since $v^{\tau }\in W^{1,\infty }(Q_0;\mathbb {R}^{m})$.

Now let $B_\tau :=\{x\in A\colon v^{\tau }(x)\neq u(x)\}$ be as in the proof of theorem 5.1 and recall that

(5.28)\begin{equation} |B_\tau|\leq(m+1)\tau.\end{equation}

By (5.27) and the nonnegativity of $f_\infty$ we have

(5.29)\begin{equation} F'(u,A)\geq\int_{A_\tau \setminus B_\tau}f_\infty(x,\nabla u)\,{\rm d}x-\beta_\tau, \end{equation}

for every $\tau >0$. Now, since $|A\setminus A_\tau |<\tau$, using (5.28) we get

(5.30)\begin{equation} |A\setminus(A_\tau\setminus B_\tau)|\leq(m+2)\tau, \end{equation}

thus, thanks to (4.2) and (5.30), we can pass to the limit as $\tau \to 0^{+}$ in (5.29) and obtain

(5.31)\begin{equation} F'(u,A)\geq\int_Af_\infty(x,\nabla u)\,{\rm d}x=F_\infty(u,A), \end{equation}

hence the lower bound for $u\in W^{1,p}_{\lambda _\infty }(A;\mathbb {R}^{m})$.

Substep 1.2: $u \notin W^{1,p}_{\lambda _\infty }(A;\mathbb {R}^{m})$. In this case, from (5.1) we have $F_\infty (u,\,A)=+\infty$, hence to conclude we need to show that $F'(u,\,A)=+\infty$. Assume by contradiction that

\[ F'(u,A)<{+}\infty. \]

If this is the case, we may argue exactly as in substep 1.1 and get

\[ +\infty>F'(u,A)\geq\int_{A_\tau \setminus B_\tau}f_\infty(x,\nabla u)\,{\rm d}x-\beta_\tau. \]

By the Fatou lemma, (4.2) and proposition 2.6 this yields $u\in W^{1,p}_{\lambda _\infty }(A;\mathbb {R}^{m})$ and hence a contradiction.

Step 2. Upper bound. In this step, we prove that

(5.32)\begin{equation} F''(u,A)\leq F_\infty(u,A), \end{equation}

for every $u\in W^{1,1}(Q_0;\mathbb {R}^{m})$ and every $A\in \mathcal {A}(Q_0)$ with $A \subset \subset Q_0$.

To this end, let $u\in W^{1,1}(Q_0;\mathbb {R}^{m})$ and $A\in \mathcal {A}(Q_0)$, $A \subset \subset Q_0$ be fixed. We start observing that by the definition of $F_\infty$, if $u\notin W^{1,p}_{\lambda _\infty }(A;\mathbb {R}^{m})$ then there is nothing to prove. Therefore, we only consider the case $u \in W^{1,p}_{\lambda _\infty }(A;\mathbb {R}^{m})$.

Since $A$ is Lipschitz, by [Reference Turesson34, theorem 2.1.13] we can find a function $\tilde u \in W^{1,p}_{\lambda _\infty }(Q_0;\mathbb {R}^{m})$ with $u=\tilde u$ a.e. in $A$. Then, by density (see, e.g. [Reference Turesson34, corollary 2.1.6]) there exists $(u_j)\subset W^{1,\infty }(Q_0;\mathbb {R}^{m})$ such that $u_j\to \tilde u$ in $W^{1,p}_{\lambda _\infty }(Q_0;\mathbb {R}^{m})$. Then, by the locality of $F''$ and $F_\infty$, the continuity of $F_\infty$ in $W^{1,p}_{\lambda _\infty }(Q_0;\mathbb {R}^{m})$, and the $L^{1}(Q_0;\mathbb {R}^{m})$-lower semicontinuity of $F''$, invoking theorem 4.10 we deduce

\[ F_\infty(u,A)=F_{\infty}(\tilde u,A)=\lim_{j \to \infty}F_\infty(u_j,A) \ = \liminf_{j\to\infty}F''(u_j,A)\geq F''(\tilde u, A)= F''(u,A), \]

thus the upper bound.

Eventually (5.24) follows by gathering (5.25) and (5.32).

Proof. By theorem 2.2 we can deduce the existence of an exponent $\sigma >0$ and a constant $c>0$ such that

(6.2)\begin{equation} \left({- \hskip-1.05em \int}_Q\lambda_k^{-(1+\sigma)/(p-1)}\,{\rm d}x\right)^{1/(1+\sigma)}\leq c\left({- \hskip-1.05em \int}_Q\lambda_k^{{-}1/(p-1)}\,{\rm d}x\right), \end{equation}

for every cube $Q$ and for every $k\in \mathbb {N}$.

Now, let $A\in \mathcal {A}(Q_0)$ and $u\in W^{1,p}_{\lambda _k}(A;\mathbb {R}^{m})$ be arbitrary, and let $\delta >0$ to be chosen later. By the Hölder inequality we have

\[ \int_A|\nabla u|^{1+\delta}\,{\rm d}x\leq\left(\int_A\lambda_k|\nabla u|^{p}\,{\rm d}x\right)^{(1+\delta)/p}\left(\int_A\lambda_k^{-(1+\delta)/(p-1-\delta)}\,{\rm d}x\right)^{(p-1-\delta)/p}. \]

For $\delta :=(p-1)\sigma /(p+\sigma )$ it is immediate to check that

(6.3)\begin{equation} \frac{1+\delta}{p-1-\delta}=\frac{1+\sigma}{p-1}; \end{equation}

hence by (6.2) we readily get

\begin{align*} & \int_A|\nabla u|^{1+\delta}\,{\rm d}x \\ & \quad\leq c^{(p-1-\delta)(1+\sigma)/p}|Q_0|^{(p-1-\delta)/p}\left(\int_A\lambda_k|\nabla u|^{p}\,{\rm d}x\right)^{(1+\delta)/p}\\ & \quad \times\left({- \hskip-1.05em \int}_{Q_0}\lambda_k^{{-}1/(p-1)}\,{\rm d}x\right)^{(p-1-\delta)(1+\sigma)/p}. \end{align*}

Moreover, since $\lambda _k$ belongs to $A_p(K)$, by (3.1) we also deduce that

\[ \left({- \hskip-1.05em \int}_{Q_0}\lambda_k^{{-}1/(p-1)}\,{\rm d}x\right)^{p-1}\leq\frac{K}{c_1} \]

and therefore

\begin{align*}& \int_A|\nabla u|^{1+\delta}\,{\rm d}x\\& \quad \leq c^{(p-1-\delta)(1+\sigma)/p}|Q_0|^{(p-1-\delta)/p}\left(\frac{K}{c_1}\right)^{(p-1-\delta)(1+\sigma)/p(p-1)}\\& \qquad\times\left(\int_A\lambda_k|\nabla u|^{p}\,{\rm d}x\right)^{(1+\delta)/p}. \end{align*}

Eventually, gathering (3.1), (3.2) and (6.3) gives

\begin{align*} & \left(\int_A|\nabla u|^{1+\delta}\,{\rm d}x\right)^{p/(1+\delta)} \\ & \quad\leq c^{p-1}|Q_0|^{(p-1)/(1+\sigma)}\left(\frac{K}{c_1}\right)\int_A\lambda_k(x)(|\nabla u|^{p}-1)\,{\rm d}x\\& \qquad +c^{p-1}|Q_0|^{(p-1)/(1+\sigma)}\left(\frac{K}{c_1}\right)\int_A\lambda_k(x)\,{\rm d}x \\ & \quad\leq c^{p-1}|Q_0|^{(p-1)/(1+\sigma)}\left(\frac{K}{c_1\alpha}\right)F_k(u,A)+c^{p-1}|Q_0|^{(p-1)/(1+\sigma)}\left(\frac{K}{c_1}\right)c_2|Q_0|, \end{align*}

for every $k\in \mathbb {N}$. Hence, (6.1) immediately follows by choosing $C:=\max \{C_1,\, C_2\}$ with

\[ C_1:=c^{p-1}|Q_0|^{(p-1)/(1+\sigma)}\left(\frac{K}{c_1\alpha}\right)\text{ and } C_2:=\alpha c_2|Q_0|C_1. \]

Proof. Let $u\in W^{1,1}(Q_0;\mathbb {R}^{m})$ and $A\in \mathcal {A} (Q_0)$, with $A\subset \subset Q_0$ be fixed and let $(k_h)$ be the subsequence whose existence is guaranteed by theorem 5.2.

We divide the proof into two main steps.

Step 1: Lower bound. Let $(u_h)\subset W^{1,1}(Q_0;\mathbb {R}^{m})$ be such that $u_h\to u$ in $L^{1}(Q_0;\mathbb {R}^{m})$. In this step, we want to show that

(6.6)\begin{equation} \liminf_{h\to\infty} F_{k_h}^{\psi}(u_h,A) \geq F_{\infty}^{\psi}(u,A). \end{equation}

We note that we can always assume that

(6.7)\begin{equation} \liminf_{h\to\infty} F_{k_h}^{\psi}(u_h,A)<{+}\infty, \end{equation}

otherwise there is nothing to prove. Moreover, without loss of generality, we may also assume that the liminf in (6.7) is actually a limit. Then, by the definition of $F_{k_h}^{\psi }$ we have that $(u_h) \subset W^{1,p}_{0,\lambda _{k_h}}(A;\mathbb {R}^{m})+\psi$; while by theorem 5.2 we get that $u\in W^{1,p}_{\lambda _\infty }(A;\mathbb {R}^{m})$ and

\[ F_{\infty}(u,A)\leq \liminf_{h\to\infty} F_{k_h}(u_h,A)=\liminf_{h\to\infty} F_{k_h}^{\psi}(u_h,A). \]

Since $W^{1,p}_{0,\lambda _\infty }(A;\mathbb {R}^{m})= W^{1,1}_{0}(A;\mathbb {R}^{m}) \cap W^{1,p}_{\lambda _\infty }(A;\mathbb {R}^{m})$, to conclude it is enough to show that $u$ belongs to $W^{1,1}_{0}(A;\mathbb {R}^{m})+\psi$.

To this end, we start observing that thanks to (6.7), proposition 6.1 yields the existence of an exponent $\delta >0$ and of a constant $C>0$ such that

\[ \int_A|\nabla u_h|^{1+\delta}\,{\rm d}x\leq C, \]

for every $h\in \mathbb {N}$. Then, by Poincaré's inequality the sequence $(u_h)$ is bounded in $W^{1,1+\delta }(A;\mathbb {R}^{m})$. This readily implies that, up to subsequences, $u_h\rightharpoonup u$ in $W^{1,1+\delta }(A;\mathbb {R}^{m})$. Since $(u_h)\subset W^{1,1+\delta }_0(A;\mathbb {R}^{m})+\psi$ and this space is weakly closed, we immediately get $u\in W^{1,1+\delta }_0(A;\mathbb {R}^{m})+\psi$, and therefore the claim.

Step 2: Upper bound. We start by considering the case $u \in C^{\infty }_0(A;\mathbb {R}^{m})+\psi$.

By proposition 4.5 and theorem 5.2 there exists a sequence $(u_h)\subset W^{1,p}_{\lambda _{k_h}}(A;\mathbb {R}^{m})$ such that $u_h\to u$ in $L^{q}(Q_0;\mathbb {R}^{m})$ for every $1\leq q<+\infty$ and

(6.8)\begin{equation} \limsup_{h\to\infty} F_{k_h}(u_h,A)\leq F_\infty(u,A)=F_\infty^{\psi}(u,A). \end{equation}

Starting from $u_h$ we now want to construct a recovery sequence which also satisfies the boundary condition. To this purpose, let $\eta >0$ be fixed. By the equi-integrability of the sequence $(\lambda _{k_h})$ (cf. proposition 2.5) there exists a compact set $K_\eta \subset A$ such that

(6.9)\begin{equation} \int_{A\setminus K_\eta}\lambda_{k_h}|\nabla u|^{p}\,{\rm d}x\leq\|\nabla u\|_{L^{\infty}(A;\mathbb{R}^{m\times n})}^{p}\int_{A\setminus K_\eta}\lambda_{k_h}\,{\rm d}x<\|\nabla u\|_{L^{\infty}(A;\mathbb{R}^{m\times n})}^{p}\eta, \end{equation}

for every $h\in \mathbb {N}$.

Choose $A',\,A''\in \mathcal {A}(Q_0)$ such that $K_\eta \subset A'\subset \subset A''\subset \subset A$. Then, proposition 4.6 ensures the existence of a positive constant $M_\eta$ and a sequence $(\varphi _h)$ of cut-off functions between $A'$ and $A''$ such that

(6.10)\begin{align} & F_{k_h}(\varphi_hu_h+(1-\varphi_h)u,A) \nonumber\\ & \quad\leq(1+\eta)\left(F_{k_h}(u_h,A'')+F_{k_h}(u,A\setminus K_\eta)\right)+M_\eta\int_A\lambda_{k_h}|u_h-u|^{p}\,{\rm d}x+\eta. \end{align}

Set $w_h:=\varphi _hu_h+(1-\varphi _h)u$; then by definition $(w_h)\subset W^{1,p}_{0,\lambda _{k_h}}(A;\mathbb {R}^{m})+\psi$ and $w_h\to u$ in $L^{q}(Q_0;\mathbb {R}^{m})$ for every $1\leq q<+\infty$. Moreover, by (6.10) and (3.2) we get

(6.11)\begin{align} F_{k_h}(w_h,A) & \leq(1+\eta)F_{k_h}(u_h,A)\nonumber\\ & \quad+(1+\eta)\beta\int_{A\setminus K_\eta}\lambda_{k_h}(|\nabla u|^{p}+1)\,{\rm d}x+M_\eta\int_A\lambda_{k_h}|u_h-u|^{p}\,{\rm d}x+\eta. \end{align}

Hence, by (6.8), (6.9) and (6.11) we have

\begin{align*} \Gamma\hbox{-}\limsup_{h\to\infty}F_{k_h}^{\psi}(u,A)& \leq\limsup_{h\to\infty}F_{k_h}^{\psi}(w_h,A) \\ & \leq(1+\eta)\limsup_{h\to\infty}F_{k_h}(u_h,A)\\& \quad +(1+\eta)\beta(\|\nabla u\|_{L^{\infty}(A;\mathbb{R}^{m\times n})}^{p}+1)\eta+\eta \\ & \leq(1+\eta)F_\infty^{\psi}(u,A)+(1+\eta)\beta(\|\nabla u\|_{L^{\infty}(A;\mathbb{R}^{m\times n})}^{p}+1)\eta+\eta. \end{align*}

Therefore, by the arbitrariness of $\eta >0$ we conclude that

(6.12)\begin{equation} \Gamma\hbox{-}\limsup_{h\to\infty}F_{k_h}^{\psi}(u,A)\leq F_\infty^{\psi}(u,A), \end{equation}

for every $u\in C_0^{\infty }(A;\mathbb {R}^{m})+\psi$.

Now, let $u\in W^{1,p}_{0,\lambda _\infty }(A;\mathbb {R}^{m})+\psi$. We extend $u$ to $\psi$ outside $A$; we clearly have that the extended function (still denoted by $u$) belongs to $W^{1,p}_{0,\lambda _\infty }(Q_0;\mathbb {R}^{m})+\psi$. Now, let $(u_j)\subset C_0^{\infty }(Q_0;\mathbb {R}^{m})$ be such that $u_j\to u$ in $W^{1,p}_{\lambda _\infty }(Q_0;\mathbb {R}^{m})$, hence, in particular, $u_j\to u$ strongly in $L^{1}(Q_0;\mathbb {R}^{m})$. By the $W^{1,p}_{\lambda _\infty }(Q_0;\mathbb {R}^{m})$-continuity of $F_\infty ^{\psi }$, by (6.12), and by the lower semicontinuity of the $\Gamma$-limsup with respect to the strong topology of $L^{1}(Q_0;\mathbb {R}^{m})$ we get

\[ F_\infty^{\psi}(u,A)=\lim_{j\to\infty}F_\infty^{\psi}(u_j,A)\geq\lim_{j\to\infty}\Gamma\hbox{-}\limsup_{h\to\infty}F_{k_h}^{\psi}(u_j,A)\geq\Gamma\hbox{-}\limsup_{h\to\infty}F_{k_h}^{\psi}(u,A), \]

for every $u\in W^{1,p}_{0,\lambda _\infty }(A;\mathbb {R}^{m})+\psi$, and therefore the upper bound.

Proof. By (6.13) and by (6.4) we have $u_k \in W^{1,p}_{0,\lambda _k}(A;\mathbb {R}^{m})+\psi$, for every $k\in \mathbb {N}$. Then, arguing exactly as in the proof of theorem 6.2 we may deduce the existence of a subsequence $(u_{k_h})\subset (u_k)$ which weakly converges in $W^{1,1+\delta }(A;\mathbb {R}^{m})$ to a function $u\in W_0^{1,1}(A;\mathbb {R}^{m})+\psi$. Furthermore, by the compact embedding of $W^{1,1+\delta }(A;\mathbb {R}^{m})$ in $L^{1,1+\delta }(A;\mathbb {R}^{m})$ we have that, in particular, $u_{k_h}\to u$ in $L^{1}(A;\mathbb {R}^{m})$. Now, extend $u_{k_h}$ and $u$ by setting $u_{k_h}:=\psi$, $u:=\psi$ in $Q_0 \setminus A$. Then, clearly $u_{k_h}\to u$ in $L^{1}(Q_0;\mathbb {R}^{m})$. Hence, by theorem 6.2 and by (6.13) there holds

\[ F_\infty^{\psi}(u,A)\leq\liminf_{h\to\infty} F_{k_h}^{\psi}(u_{k_h},A)<{+}\infty, \]

thus by (6.5) we get $u\in W^{1,p}_{0,\lambda _\infty }(A;\mathbb {R}^{m})+\psi$.