Proof of lemma 3.4. Assume $0\not \in \Omega$, so that $\mathcal {O}$ is an open annulus centred at $0$ (the case where $0\in \Omega$ can be treated similarly). Let $0< r_1< r_2$ and $0<\delta < r_1$ be such that $\overline {\mathcal {O}}\subset B_{r_2}\setminus \overline {B_{r_1}}$ and $\pi$ is Lipschitz continuous in $\overline {B_{r_2+\delta }}\setminus B_{r_1-\delta }$ with Lipschitz constant $L$. Let $M\geq 0$ be the maximum of $\pi$ on $\partial B_{r_1}\cup \partial B_{r_2}$. We first define $\widetilde {\pi }: \overline {B_{r_2+\delta }}\setminus B_{r_1-\delta }\rightarrow \mathbb {R}$ as

\[ \widetilde{\pi}(y):=\begin{cases} \pi\left(r_1\frac{y}{|y|}\right)-K(r_1-|y|) & \text{ if }r_1-\delta\leq |y|< r_1, \\ \pi(y) & \text{ if } r_1\leq |y|< r_2, \\ \pi\left(r_2\frac{y}{|y|}\right)-K(|y|-r_2) & \text{ if } r_2\leq |y|\leq r_2+\delta, \end{cases} \]

where $K:=L\vee {M}/{\delta }$. It is easy to see that $\widetilde {\pi }$ is Lipschitz continuous in its domain; moreover, by construction it coincides with $\pi$ in an open neighbourhood of $\overline {\mathcal {O}}$. We now show that $\widetilde {\pi }\le \pi$ on $\overline {B_{r_2+\delta }}\setminus B_{r_1-\delta }$. Indeed, if $r_1-\delta \leq |y|< r_1$, by the Lipschitz continuity of $\pi$ we have

\[ \widetilde{\pi}(y)\le \pi(y)+L\left|r_1 \frac{y}{|y|}-y\right|-K(r_1-|y|)=\pi(y)-(K-L)(r_1-|y|)\le \pi(y), \]

and similarly if $r_2\leq |y|\leq r_2+\delta$. Finally, we note that, if $y\in \partial B_{r_1-\delta }\cup \partial B_{r_2+\delta }$, then

\[ \widetilde{\pi}(y)\le M-K\delta\le 0. \]

We now conclude by considering $\hat{\pi} (y):=\widetilde {\pi }(y)\vee 0$ for $y\in \overline {B_{r_2+\delta }}\setminus B_{r_1-\delta }$, $\hat{\pi} (y):=0$ otherwise in $\mathbb {R}^n$.

Proof. Let $\widehat {\mathcal {E}}_\varepsilon$ be the auxiliary energy defined as in (3.1) with $\pi$ replaced by the function $\hat{\pi}$ given by lemma 3.4. Since $\pi \ge \hat{\pi}$ everywhere and $\pi \equiv \hat{\pi}$ on $\Omega$, we have that

(3.7)\begin{equation} \widehat{\mathcal{E}}_\varepsilon(y)\le \mathcal{E}_\varepsilon(y) \qquad\text{for every } y\in\mathring{W}^{1,p}(\Omega;\mathbb{R}^n). \end{equation}

For the sake of brevity we introduce the notation

(3.8)\begin{equation} \Omega_\varepsilon^-:=\left\{x\in \Omega: \ |\det\nabla y_\varepsilon(x)-1|\le 1\right\},\qquad \Omega_\varepsilon^+:=\Omega\setminus \Omega_\varepsilon^-. \end{equation}

Since $\widehat {\mathcal {E}}_\varepsilon (y_\varepsilon )\le \mathcal {E}_\varepsilon (y_\varepsilon )\le C\varepsilon ^2$, we have that $y_\varepsilon$ belongs to $Y^p$, hence in particular $\det \nabla y_\varepsilon >0$ a.e. in $\Omega$. By (W5) and theorem 3.2 we infer the existence of $R_\varepsilon \in SO(n)$ such that

(3.9)\begin{equation} \int_\Omega g_p(|\nabla y_\varepsilon(x)-R_\varepsilon|)\,\mathrm{d} x\le C\int_\Omega W(x,\nabla y_\varepsilon(x))\,\mathrm{d} x. \end{equation}

By (2.8) we deduce that

\begin{align*} \int_\Omega W(x,\nabla y_\varepsilon)\,\mathrm{d} x& =\widehat{\mathcal{E}}_\varepsilon(y_\varepsilon)+\varepsilon\int_\Omega\left(\hat\pi(x)-\hat\pi(y_\varepsilon)\det\nabla y_\varepsilon\right)\,\mathrm{d} x\\ & \le C\varepsilon^2+\varepsilon\int_\Omega\left(\hat\pi(y_{R_\varepsilon})-\hat\pi(y_\varepsilon)\det\nabla y_\varepsilon\right)\,\mathrm{d} x\\ & \le C\varepsilon^2+\varepsilon\int_\Omega|\hat\pi(y_\varepsilon)-\hat\pi(y_{R_\varepsilon})|\,\mathrm{d} x+\varepsilon\int_\Omega\hat\pi (y_\varepsilon)(1-\det\nabla y_\varepsilon)\,\mathrm{d} x. \end{align*}

Since $\hat{\pi} \geq 0$ by construction, the integrand in the last integral above is nonpositive on $\Omega _\varepsilon ^+$. Thus, using the fact that $\hat{\pi}$ is Lipschitz continuous and bounded, and applying Hölder's inequality we deduce that

(3.10)\begin{equation} \int_\Omega W(x,\nabla y_\varepsilon)\,\mathrm{d} x\le C\varepsilon^2+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_1+C\varepsilon \|\det\nabla y_\varepsilon-1\|_{L^2(\Omega_\varepsilon^-)}. \end{equation}

By (2.6) this implies

\[ \|\det\nabla y_\varepsilon-1\|_{L^2(\Omega_\varepsilon^-)}^2\le C\varepsilon^2+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_1+C\varepsilon \|\det\nabla y_\varepsilon-1\|_{L^2(\Omega_\varepsilon^-)}, \]

which, in turn, by Young's inequality yields

(3.11)\begin{equation} \|\det\nabla y_\varepsilon-1\|_{L^2(\Omega_\varepsilon^-)}^2\le C\varepsilon^2+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_1. \end{equation}

By combining (3.10) and (3.11) we deduce

(3.12)\begin{equation} \int_\Omega W(x,\nabla y_\varepsilon)\,\mathrm{d} x\le C\varepsilon^2+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_1, \end{equation}

and, as a consequence of (3.4), (3.9) and (3.12), we obtain

(3.13)\begin{equation} \int_\Omega g_p(|\varepsilon\nabla u_\varepsilon|)\,\mathrm{d} x=\int_\Omega g_p(|\nabla y_\varepsilon\,{-}\,R_\varepsilon|)\,\mathrm{d} x\,{\le}\, C\varepsilon^2+C\varepsilon\|y_\varepsilon\,{-}\,y_{R_\varepsilon}\|_1=C\varepsilon^2(1+\|u_\varepsilon\|_1). \end{equation}

Using the definition (2.4) of $g_p$ this implies that

(3.14)\begin{equation} \int_{\{|\varepsilon\nabla u_\varepsilon|\le 1\}}|\varepsilon\nabla u_\varepsilon|^2\,\mathrm{d} x\le 2 \int_\Omega g_p(|\varepsilon\nabla u_\varepsilon|)\,\mathrm{d} x\le C\varepsilon^2(1+\|u_\varepsilon\|_1). \end{equation}

By Hölder's inequality we obtain

(3.15)\begin{align} \int_{\{|\varepsilon\nabla u_\varepsilon|\le 1\}}|\varepsilon\nabla u_\varepsilon|^p\,\mathrm{d} x& \le C\left(\int_{\{|\varepsilon\nabla u_\varepsilon|\le 1\}}|\varepsilon\nabla u_\varepsilon|^2\,\mathrm{d} x\right)^{{p}/{2}}\nonumber\\ & \le C\varepsilon^p(1+\|u_\varepsilon\|_1)^{{p}/{2}}\le C\varepsilon^p(1+\|u_\varepsilon\|_1), \end{align}

where the last inequality follows from the fact that $t^{{p}/{2}}\le 1+t$ for $t\ge 0$.

Again from (2.4) and recalling that $p\le 2$ we also have that

(3.16)\begin{align} \int_{\{|\varepsilon\nabla u_\varepsilon|> 1\}}|\varepsilon\nabla u_\varepsilon|^p\,\mathrm{d} x\le\int_\Omega g_p(|\varepsilon\nabla u_\varepsilon|)\,\mathrm{d} x\le C\varepsilon^2(1+\|u_\varepsilon\|_1)\le C\varepsilon^p(1+\|u_\varepsilon\|_1). \end{align}

By (3.15), (3.16) and the continuous embedding of $W^{1,p}(\Omega ;\mathbb {R}^n)$ into $L^1(\Omega ;\mathbb {R}^n)$ we deduce that

\[ \|\nabla u_\varepsilon\|_p^p\le C+C\|u_\varepsilon\|_{W^{1,p}}. \]

Since $u_\varepsilon$ has zero average, Poincaré–Wirtinger inequality finally yields

\[ \|u_\varepsilon\|_{W^{1,p}}^p\le C+C\|u_\varepsilon\|_{W^{1,p}}, \]

which implies $\|u_\varepsilon \|_{W^{1,p}}\le C$. This inequality, combined with (3.11) and (3.13), provides (3.3) and (3.5).

Finally, if $(R_\varepsilon ')$ is a sequence of rotations whose corresponding rescaled displacements satisfy (3.5), then

\[ |R_\varepsilon-R_\varepsilon'|\le C(\|\nabla y_\varepsilon-R_\varepsilon\|_p+\|\nabla y_\varepsilon-R_\varepsilon'\|_p)\le C\varepsilon. \]

This concludes the proof.

Proof. Let $(y_\varepsilon )$ be a minimizing sequence satisfying

\[ \mathcal{E}_\varepsilon(y_\varepsilon)\le\inf\limits_{\mathring{W}^{1,p}(\Omega;\mathbb{R}^n)}\mathcal{E}_\varepsilon +\varepsilon^2. \]

Using the fact that $W$ is nonnegative and arguing as in the proof of lemma 3.6, we deduce that

\begin{align*} \mathcal{E}_\varepsilon(y_\varepsilon)& \ge \varepsilon\int_\Omega\left(\hat\pi (y_\varepsilon)\det\nabla y_\varepsilon-\hat\pi(y_{R_\varepsilon})\right)\,\mathrm{d} x\\& \ge -C\varepsilon \|y_\varepsilon-y_{R_\varepsilon}\|_1-C\varepsilon\|\det\nabla y_\varepsilon-1\|_{L^2(\Omega_\varepsilon^-)}\\ & \ge -C\varepsilon^2, \end{align*}

where the last inequality follows from lemma 3.6. This proves the first inequality in (3.17).

The other inequality in (3.17) follows trivially by the fact that the energy $\mathcal {E}_\varepsilon$ is zero on the identity map.

Proof. By lemma 3.6 the sequence $(u_\varepsilon )$ is uniformly bounded in $W^{1,p}(\Omega ;\mathbb {R}^n)$. Hence, up to subsequences, $u_\varepsilon \rightharpoonup u_0$ weakly in $\mathring {W}^{1,p}(\Omega ;\mathbb {R}^n)$. We now show that $u_0$ belongs to $\mathring {H}^{1}(\Omega ;\mathbb {R}^n)$. We first introduce the set

(3.18)\begin{equation} G_\varepsilon:=\left\{x\in \Omega: \ \varepsilon^{1/2}|\nabla u_\varepsilon(x)|\le 1\right\}, \end{equation}

and we observe that by Tchebichev inequality

(3.19)\begin{equation} |\Omega\setminus G_\varepsilon|\le C\varepsilon^{p/2}. \end{equation}

We claim that

1. (i) $\chi _{G_\varepsilon }\nabla u_\varepsilon$ is bounded in $L^2(\Omega ;\mathbb {R}^{n\times n})$;

2. (ii) $\nabla u_0\in L^2(\Omega ;\mathbb {R}^{n\times n})$ and, up to subsequences, $\chi _{G_\varepsilon }\nabla u_\varepsilon \rightharpoonup \nabla u_0$ weakly in $L^2(\Omega ;\mathbb {R}^{n\times n})$.

Assertion (i) easily follows from (3.5) arguing as in (3.14) and using that $G_\varepsilon \subset \{|\varepsilon \nabla u_\varepsilon |\le 1\}$. To prove (ii) we first note that (i) ensures that $\chi _{G_\varepsilon }\nabla u_\varepsilon \rightharpoonup v$ weakly in $L^2(\Omega ;\mathbb {R}^{n\times n})$, up to subsequences, for some $v\in L^2(\Omega ;\mathbb {R}^{n\times n})$. On the other hand, by (3.19) we have that $\chi _{G_\varepsilon }$ converges to $1$ boundedly in measure. Since $\nabla u_\varepsilon \rightharpoonup \nabla u_0$ in $L^p(\Omega ;\mathbb {R}^{n\times n})$, we conclude that $\chi _{G_\varepsilon }\nabla u_\varepsilon \rightharpoonup \nabla u_0$ in $L^p(\Omega ;\mathbb {R}^{n\times n})$. Hence, $v=\nabla u_0$ and (ii) is proved. By Sobolev embedding we have that $u_0\in \mathring {H}^1(\Omega ;\mathbb {R}^n)$.

Since $SO(n)$ is a compact set, there exists $R_0\in SO(n)$ such that $R_\varepsilon \rightarrow R_0$, up to subsequences. To prove that $R_0\in \mathcal {R}$ we argue as in the proof of lemma 3.6 and deduce

\begin{align*} & C\ge \frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon(y_\varepsilon)\\& \quad \ge \frac{1}{\varepsilon^2}\widehat{\mathcal{E}}_\varepsilon(y_\varepsilon)\ge \frac 1\varepsilon\int_\Omega(\hat\pi(y_\varepsilon)\det\nabla y_\varepsilon-\hat\pi(x))\,\mathrm{d} x\ge -c+\frac 1\varepsilon\int_\Omega(\pi(R_\varepsilon x)-\pi(x))\,\mathrm{d} x. \end{align*}

Note that in the last integral we used that $\hat{\pi} \equiv \pi$ on $\mathcal {O}$.

Multiplying by $\varepsilon$ and then letting $\varepsilon \rightarrow 0$ we infer that

\[ \int_{\Omega}(\pi(R_0 x)-\pi(x))\,\mathrm{d} x\le 0. \]

This implies that $R_0\in \mathcal {R}$ since by assumption the identity matrix is an optimal rotation.

Uniqueness of $R_0$ is a straightforward consequence of (3.6). Uniqueness (up to an infinitesimal rigid motion) of $u_0$ follows by arguing as in [Reference Maor and Mora22, theorem 5.1], recalling that displacements have zero average in our setting.

Proof. We write

(3.21)\begin{align} & \frac1\varepsilon \int_\Omega \left(\hat\pi(y_\varepsilon)\det\nabla y_\varepsilon-\hat\pi(y_{R_\varepsilon})\right)\,\mathrm{d} x\nonumber\\ & \quad = \frac 1\varepsilon\int_\Omega\hat\pi(y_\varepsilon)(\det\nabla y_\varepsilon-1)\,\mathrm{d} x+\frac 1\varepsilon\int_\Omega(\hat\pi(y_\varepsilon)-\hat\pi(y_{R_\varepsilon}))\,\mathrm{d} x \nonumber\\ & \quad=:I_\varepsilon+II_\varepsilon. \end{align}

We start by considering $I_\varepsilon$. Let $\Omega ^-_\varepsilon$ and $G_\varepsilon$ be defined as in (3.8) and (3.18). Since $(u_\varepsilon )$ is bounded in $W^{1,p}(\Omega ;\mathbb {R}^n)$, property (3.19) still holds. Moreover, by (2.1)

\[ \det\nabla y_\varepsilon(x)=\det(I+\varepsilon\nabla u_\varepsilon(x))=1+\sum_{k=1}^n\varepsilon^k\iota_k(\nabla u_\varepsilon(x))\quad\text{ for a.e. }x\in \Omega. \]

Since by (2.2) we have that for $k=1,\,\dots,\,n$

\[ |\varepsilon^k\iota_k(\nabla u_\varepsilon(x))|\le C\varepsilon^k|\nabla u_\varepsilon(x)|^k\le C\varepsilon^{k/2}\le C\varepsilon^{1/2}\quad\text{for a.e. }x\in G_\varepsilon, \]

we deduce that $G_\varepsilon \subset \Omega ^-_\varepsilon$ for $\varepsilon$ small enough. Therefore, using the nonnegativity of $\hat{\pi}$ and (2.1) again, we obtain

\begin{align*} I_\varepsilon & \ge \frac 1\varepsilon\int_{\Omega_\varepsilon^-}\hat\pi(y_\varepsilon)(\det\nabla y_\varepsilon-1)\,\mathrm{d} x \\ & = \int_{G_\varepsilon}\hat\pi (y_\varepsilon)\mathrm{div}\, u_\varepsilon\,\mathrm{d} x\\& \quad +\sum_{k=2}^{n}\varepsilon^{k-1}\int_{G_\varepsilon}\hat\pi(y_\varepsilon)\iota_k(\nabla u_\varepsilon)\,\mathrm{d} x +\frac1\varepsilon\int_{\Omega^-_\varepsilon\setminus G_\varepsilon}\hat\pi(y_\varepsilon)(\det\nabla y_\varepsilon-1)\,\mathrm{d} x \\ & =:J^1_\varepsilon+J^2_\varepsilon+J^3_\varepsilon. \end{align*}

We first show that

(3.22)\begin{equation} \lim\limits_{\varepsilon\rightarrow 0}J_\varepsilon^1=\int_{\Omega}\pi(R_0x)\mathrm{div}\, u_0(x)\,\mathrm{d} x. \end{equation}

Indeed, since $\hat{\pi} =\pi$ on $\mathcal {O}$, we may write

(3.23)\begin{align} \left| J^1_\varepsilon-\int_{\Omega}\pi(R_0x)\mathrm{div}\, u_0(x)\,\mathrm{d} x\right|\le & \int_{G_\varepsilon}|\hat\pi(R_\varepsilon x+\varepsilon R_\varepsilon u_\varepsilon(x))-\hat\pi(R_0 x)||\mathrm{div}\, u_\varepsilon|\,\mathrm{d} x \nonumber\\ & +\left|\int_{G_\varepsilon}\pi(R_0 x)\mathrm{div}\, u_\varepsilon\,\mathrm{d} x-\int_{\Omega}\pi(R_0 x)\mathrm{div}\, u_0\,\mathrm{d} x\right|. \end{align}

Using the Lipschitz continuity of $\hat {\pi }$ and the definition of $G_\varepsilon$, the first integral at the right-hand side can be bounded as follows:

\begin{align*} & \int_{G_\varepsilon}|\hat\pi(R_\varepsilon x+\varepsilon R_\varepsilon u_\varepsilon(x))-\hat\pi(R_0 x)||\mathrm{div}\, u_\varepsilon|\,\mathrm{d} x\\ & \quad\le C|R_\varepsilon-R_0|\|\nabla u_\varepsilon\|_p+C\varepsilon\int_{G_\varepsilon}|u_\varepsilon||\nabla u_\varepsilon|\,\mathrm{d} x\\ & \quad\le C|R_\varepsilon-R_0|+C\varepsilon^{1/2} \|u_\varepsilon\|_p\\ & \quad\le C|R_\varepsilon-R_0|+C\varepsilon^{1/2}, \end{align*}

where the last term goes to zero, as $\varepsilon \rightarrow 0$. Since $\chi _{G_\varepsilon }$ converges to $1$ boundedly in measure, we have that $\chi _{G_\varepsilon }\mathrm {div}\, u_\varepsilon \rightharpoonup \mathrm {div}\, u_0$ weakly in $L^p(\Omega )$, hence the second term in (3.23) goes to zero, as well. This proves (3.22).

We now prove that both $J^2_\varepsilon$ and $J^3_\varepsilon$ converge to $0$, as $\varepsilon \rightarrow 0$. By (2.2) and (3.4), since $\hat{\pi}$ is bounded, we obtain

\[ |J^2_\varepsilon| \le C\sum_{k=2}^n\varepsilon^{k-1}\int_{G_\varepsilon} |\nabla u_\varepsilon|^k\,\mathrm{d} x = C \sum_{k=2}^n \varepsilon^{k-1}\int_{G_\varepsilon}|\nabla u_\varepsilon|^{k-p}|\nabla u_\varepsilon|^p\,\mathrm{d} x. \]

Since $|\nabla u_\varepsilon |\le \varepsilon ^{-1/2}$ on $G_\varepsilon$, we have that

\[ |J^2_\varepsilon|\le C\sum_{k=2}^n\varepsilon^{{(k+p-2)}/{2}}\|\nabla u_\varepsilon\|_p^p\le C\varepsilon^{p/2}\rightarrow 0. \]

To bound $J^3_\varepsilon$ we use (3.3) and deduce

\[ |J^3_\varepsilon| \le \frac{C}\varepsilon {\|\det\nabla y_\varepsilon-1\|_{L^2(\Omega_\varepsilon^-)}}|\Omega\setminus G_\varepsilon|^{1/2} \le C |\Omega\setminus G_\varepsilon|^{1/2}, \]

which vanishes by (3.19).

By combining the previous inequalities we conclude that

(3.24)\begin{equation} \liminf\limits_{\varepsilon\rightarrow 0}I_\varepsilon\ge \lim\limits_{\varepsilon\rightarrow 0} \frac 1\varepsilon\int_{\Omega_\varepsilon^-}\hat\pi(y_\varepsilon)(\det\nabla y_\varepsilon-1)\,\mathrm{d} x = \int_{\Omega}\pi(R_0x)\mathrm{div}\, u_0(x)\,\mathrm{d} x. \end{equation}

We now claim that

(3.25)\begin{equation} \lim\limits_{\varepsilon\rightarrow 0}II_\varepsilon= \lim\limits_{\varepsilon\rightarrow 0} \frac 1\varepsilon\int_{\Omega_\varepsilon^-}(\hat\pi(y_\varepsilon)-\hat\pi(y_{R_\varepsilon}))\,\mathrm{d} x=\int_{\Omega}\nabla\pi(R_0 x)\cdot R_0u_0(x)\,\mathrm{d} x. \end{equation}

Assuming this is true, the thesis follows by (3.21), (3.24), (3.25) and the divergence theorem, since

\[ \pi(R_0x)\mathrm{div}\, u_0(x)+\nabla\pi(R_0 x)\cdot R_0u_0(x)=\mathrm{div}\, (\pi(R_0 x)u_0(x)). \]

To conclude we only need to prove (3.25). We can write the integrand in $II_\varepsilon$ as

(3.26)\begin{equation} \frac 1\varepsilon(\hat\pi(y_\varepsilon)-\hat\pi(y_{R_\varepsilon}))=\frac1\varepsilon (\hat\pi(R_\varepsilon x+\varepsilon R_\varepsilon u_\varepsilon(x))-\hat\pi(R_\varepsilon x)) \end{equation}

and owing to the Lipschitz continuity of $\hat {\pi }$ we have

(3.27)\begin{equation} \frac 1\varepsilon |\hat\pi(y_\varepsilon)-\hat\pi(y_{R_\varepsilon})| \leq C|u_\varepsilon(x)| \qquad \text{for a.e. } x\in\Omega. \end{equation}

Since $(u_\varepsilon )$ is bounded in $L^p(\Omega ;\mathbb {R}^n)$ and $|\Omega \setminus \Omega _\varepsilon ^-|\leq |\Omega \setminus G_\varepsilon |\to 0$ by (3.19), we deduce that

\[ \lim\limits_{\varepsilon\rightarrow 0} \frac 1\varepsilon\int_{\Omega\setminus\Omega_\varepsilon^-}(\hat\pi(y_\varepsilon)-\hat\pi(y_{R_\varepsilon}))\,\mathrm{d} x=0. \]

Hence, proving (3.25) is equivalent to show that

(3.28)\begin{equation} \lim\limits_{\varepsilon\rightarrow 0}II_\varepsilon=\int_{\Omega}\nabla\pi(R_0 x)\cdot R_0u_0(x)\,\mathrm{d} x. \end{equation}

On the other hand, $u_\varepsilon \rightarrow u_0$ strongly in $L^1(\Omega ;\mathbb {R}^n)$ by compact embedding. Thus, by (3.27) and the generalized dominated convergence theorem, (3.28) is proved if we show that the integrand (3.26) converges a.e. to $\nabla \pi (R_0 x)\cdot R_0u_0(x)$.

To this aim, we first note that, up to subsequences,

\[ \lim\limits_{\varepsilon\rightarrow 0}\nabla\hat\pi (R_\varepsilon x)=\nabla\hat\pi (R_0 x)\quad\text{for a.e. }x\in \Omega. \]

Indeed, the convergence is actually in $L^1(\mathcal {O};\mathbb {R}^n)$. This can be easily proved by approximating $\nabla \hat{\pi}$ with functions in $C^0(\overline {\mathcal {O}};\mathbb {R}^n)$. Now, by Rademacher theorem (we point out that we are working with a countable sequence of rotations $R_\varepsilon$) for almost every $x\in \Omega$ we have

\[ \frac{\hat\pi(R_\varepsilon x+\varepsilon R_\varepsilon u_\varepsilon(x))-\hat\pi(R_\varepsilon x)}{\varepsilon}=\nabla\hat\pi (R_\varepsilon x)\cdot R_\varepsilon u_\varepsilon(x)+\frac1\varepsilon o(\varepsilon|u_\varepsilon(x)|). \]

Since $u_\varepsilon \rightarrow u_0$ a.e., up to subsequences, and $\hat{\pi} \equiv \pi$ on $\mathcal {O}$, we deduce the desired convergence. This concludes the proof.

Proof of proposition 3.10. Without loss of generality we can assume

\[ \liminf\limits_{\varepsilon\rightarrow 0}\frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon(y_\varepsilon)<{+}\infty, \]

so that $y_\varepsilon \in Y^p$ and, up to subsequence, $\mathcal {E}_\varepsilon (y_\varepsilon )\leq C\varepsilon ^2$. By lemma 3.6 and proposition 3.8 there exist a (possibly different) sequence $(R_\varepsilon ')\subset SO(n)$ such that, up to subsequences, $R'_\varepsilon \rightarrow R_0$, the corresponding displacements $u_\varepsilon '$ satisfy (3.5) and, up to subsequences, weakly converge to $u_0+Ax$ for some $A\in \mathbb {R}^{n\times n}_{\rm skew}$. However, by remark 3.11 we can assume, without loss of generality, that $R_\varepsilon =R_\varepsilon '$ and so, $u_\varepsilon =u_\varepsilon '$ and $A=0$.

Let $\widehat {\mathcal {E}}_\varepsilon$ be the auxiliary energy defined as in (3.1) with $\pi$ replaced by the function $\hat{\pi}$ given by lemma 3.4. By the properties of $\hat{\pi}$ we have

(3.30)\begin{align} \frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon(y_\varepsilon)\ge \frac{1}{\varepsilon^2}\widehat{\mathcal{E}}_\varepsilon(y_\varepsilon)& = \frac{1}{\varepsilon^2}\int_\Omega W(x,\nabla y_\varepsilon)\,\mathrm{d} x+\frac 1\varepsilon\int_{\Omega} (\hat{\pi}(y_\varepsilon)\det\nabla y_\varepsilon-\hat{\pi}(y_{R_\varepsilon}))\,\mathrm{d} x \nonumber\\ & \quad+\frac 1\varepsilon\int_{\Omega} (\hat{\pi}(y_{R_\varepsilon})-\hat{\pi}(x))\,\mathrm{d} x. \end{align}

Arguing as in [Reference Agostiniani, Dal Maso and DeSimone3, proof of theorem 2.4] one can prove that

\[ \liminf\limits_{\varepsilon\rightarrow 0}\frac{1}{\varepsilon^2}\int_\Omega W(x,\nabla y_\varepsilon)\,\mathrm{d} x\ge \frac 12\int_\Omega Q(x,e(u_0)(x))\,\mathrm{d} x. \]

Since condition (3.3) is satisfied by lemma 3.6, we can apply proposition 3.9 and we obtain

\[ \liminf_{\varepsilon\rightarrow 0} \frac1\varepsilon \int_\Omega \left(\hat\pi(y_\varepsilon)\det\nabla y_\varepsilon-\hat\pi(y_{R_\varepsilon})\right)\,\mathrm{d} x \geq \int_{\partial\Omega}\pi(R_0x)n_{\partial\Omega}(x)\cdot u_0(x)\,\mathrm{d}\mathcal{H}^{n-1}(x). \]

Finally, assumption (2.8) guarantees that the last term in (3.30) is nonnegative. This proves the desired inequality.

Proof. Let $(u_0,\,R_0)\in \mathring {H}^{1}(\Omega ;\mathbb {R}^n)\times \mathcal {R}$. By mollification there exists $(u_\varepsilon )\subset \mathring {W}^{1,\infty }(\Omega ;\mathbb {R}^n)$ such that

(3.31)\begin{equation} u_\varepsilon\rightarrow u_0\quad\text{strongly in }H^1(\Omega;\mathbb{R}^n)\quad\text{ and }\quad\varepsilon^{1/2}\|u_\varepsilon\|_{W^{1,\infty}}\le 1. \end{equation}

We define $R_\varepsilon :=R_0$, so that $y_\varepsilon (x)=R_0(x+\varepsilon u_\varepsilon (x))$.

We first observe that $y_\varepsilon \in Y^p$ for $\varepsilon$ small enough. Indeed, by (2.3) it has zero-average and by (2.1) it satisfies

(3.32)\begin{equation} \det\nabla y_\varepsilon(x)=\det(I+\varepsilon\nabla u_\varepsilon(x))=1+\sum_{k=1}^n\varepsilon^k\iota_k(\nabla u_\varepsilon(x))\quad\text{for a.e. }x\in \Omega. \end{equation}

Since by (2.2) we have that for $k=1,\,\dots,\,n$

\[ |\varepsilon^k\iota_k(\nabla u_\varepsilon(x))|\le C\varepsilon^k|\nabla u_\varepsilon(x)|^k\le C\varepsilon^{k/2}\le C\varepsilon^{1/2}\qquad\text{ for a.e. }x\in \Omega, \]

for $\varepsilon$ small enough we obtain

(3.33)\begin{equation} |\det\nabla y_\varepsilon(x)-1|\leq C\varepsilon^{1/2}\le \frac 12\text{ for a.e. }x\in \Omega, \end{equation}

and thus, $\det \nabla y_\varepsilon >0$ a.e. in $\Omega$.

By (3.31) we have that for $\varepsilon$ small enough the set $y_\varepsilon (\Omega )$ is contained in the neighbourhood of $\mathcal {O}$ where $\pi$ and $\hat{\pi}$ coincide. Therefore, using also that $R_0\in \mathcal {R}$, we can write

\[ \frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon(y_\varepsilon)=\frac{1}{\varepsilon^2}\int_{\Omega}W(x,I+\varepsilon\nabla u_\varepsilon(x))\,\mathrm{d} x+\frac 1\varepsilon\int_{\Omega}\left(\hat{\pi}(y_\varepsilon)\det \nabla y_\varepsilon{-}\hat\pi(y_{R_0})\right)\,\mathrm{d} x. \]

Arguing as in [Reference Agostiniani, Dal Maso and DeSimone3, proof of theorem 2.4], one can show that

\[ \limsup\limits_{\varepsilon\rightarrow 0}\frac{1}{\varepsilon^2}\int_{\Omega}W(x,I+\varepsilon\nabla u_\varepsilon(x))\,\mathrm{d} x\le \frac 12\int_\Omega Q(x, e(u_0)(x))\,\mathrm{d} x. \]

On the other hand, by (3.32) we have that

\[ |\det\nabla y_\varepsilon(x)-1|\leq \varepsilon|\nabla u_\varepsilon(x)| +C\varepsilon\quad \text{for a.e. }x\in \Omega, \]

hence condition (3.3) is satisfied and we can apply proposition 3.9. By (3.33) we deduce

\[ \lim\limits_{\varepsilon\rightarrow 0}\frac 1\varepsilon\int_{\Omega}\left(\hat{\pi}(y_\varepsilon)\det \nabla y_\varepsilon){-}\hat\pi(y_{R_0})\right)\,\mathrm{d} x=\int_{\partial\Omega}\pi(R_0x)n_{\partial\Omega}(x)\cdot u_0(x)\,\mathrm{d}\mathcal{H}^{n-1}(x). \]

This concludes the proof.

Proof. Let $(y_\varepsilon )$ be a sequence of almost minimizers. By corollary 3.7 we have that

\[ \inf\limits_{\mathring{W}^{1,p}(\Omega;\mathbb{R}^n)}\mathcal{E}_\varepsilon \le 0, \]

hence by proposition 3.8 there exist $u_0\in \mathring {H}^{1}(\Omega ;\mathbb {R}^n)$ and $R_0\in \mathcal {R}$ such that, up to a subsequence,

\[ u_\varepsilon\rightharpoonup u_0 \quad\text{weakly in }\mathring{W}^{1,p}(\Omega;\mathbb{R}^n)\qquad\text{ and }\qquad R_\varepsilon\rightarrow R_0. \]

We now show that $(u_0,\,R_0)$ is a minimizer of $\mathcal {E}_0$. To this aim let $(v,\,S)\in \mathring {H}^{1}(\Omega ;\mathbb {R}^n)\times \mathcal {R}$ and let $(v_\varepsilon,\,S_\varepsilon )$ be a recovery sequence for $(v,\,S)$, as in proposition 3.12. Let $z_\varepsilon (x):=S_\varepsilon (x+\varepsilon v_\varepsilon (x))$. By proposition 3.10 we have

(3.36)\begin{align} \mathcal{E}_0(u_0,R_0)& \le \liminf\limits_{\varepsilon\rightarrow 0}\frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon(y_\varepsilon)\le \liminf\limits_{\varepsilon\rightarrow 0}\bigg(\inf\limits_{\mathring{W}^{1,p}(\Omega;\mathbb{R}^n)}\frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon\bigg)\nonumber\\ & \le \limsup\limits_{\varepsilon\rightarrow 0}\bigg(\inf\limits_{\mathring{W}^{1,p}(\Omega;\mathbb{R}^n)}\frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon\bigg) \nonumber\\ & \le \limsup\limits_{\varepsilon\rightarrow 0}\frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon(z_\varepsilon)\le\mathcal{E}_0(v,S). \end{align}

This implies that $\mathcal {E}_0$ is minimized at $(u_0,\,R_0)$ and, as a consequence, (3.35) holds.

To conclude, it remains to prove that $u_\varepsilon \rightarrow u_0$ strongly in $W^{1,p}(\Omega ;\mathbb {R}^n)$. We adapt the argument in [Reference Agostiniani, Dal Maso and DeSimone3, theorem 2.5] to our framework. We claim that the following properties hold:

1. (a) $\chi _{G_\varepsilon }e(u_\varepsilon )\rightarrow e(u_0)$ strongly in $L^2(\Omega ;\mathbb {R}^{n\times n}_{\rm sym})$, where the set $G_\varepsilon$ is defined as in (3.18);

2. (b) the sequence $(({1}/{\varepsilon ^p})\operatorname {dist}^p(\nabla y_\varepsilon ;SO(n)))$ is equi-integrable;

3. (c) the sequence $(|\nabla u_\varepsilon |^p)$ is equi-integrable.

The thesis follows from (a) and (b), by using Vitali's convergence theorem together with Korn's second inequality, see [Reference Agostiniani, Dal Maso and DeSimone3, proof of theorem 2.5] for more details.

We now prove (a). By choosing $(v,\,S)=(u_0,\,R_0)$ in (3.36) we deduce

\[ \lim\limits_{\varepsilon\rightarrow 0}\frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon(y_\varepsilon)=\mathcal{E}_0(u_0,R_0). \]

By (3.30) and assumption (2.8) we have

\[ \frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon(y_\varepsilon)\ge \frac{1}{\varepsilon^2}\int_{\Omega} W(x,\nabla y_\varepsilon)\,\mathrm{d} x+\frac1\varepsilon \int_\Omega \left(\hat\pi(y_\varepsilon)\det\nabla y_\varepsilon-\hat\pi(y_{R_\varepsilon})\right)\,\mathrm{d} x. \]

Therefore, letting $\varepsilon \rightarrow 0$ and applying proposition 3.9 yield

\begin{align*} \limsup\limits_{\varepsilon\rightarrow 0}\frac{1}{\varepsilon^2}\int_{\Omega} W(x,\nabla y_\varepsilon)\,\mathrm{d} x& \le \mathcal{E}_0(u_0,R_0)-\int_{\partial\Omega}\pi(R_0x)n_{\partial\Omega}(x)\cdot u_0(x)\,\mathrm{d}\mathcal{H}^{n-1}(x)\\ & =\frac 12\int_\Omega Q(x, e(u_0))\,\mathrm{d} x. \end{align*}

On the other hand, by Taylor expansion of $W$ around $I$ and by the weak convergence of $\chi _{G_\varepsilon }e(u_\varepsilon )$ to $e(u_0)$ in $L^2(\Omega ;\mathbb {R}^{n\times n}_{\rm sym})$ (see property (ii) in the proof of proposition 3.8) we obtain

\begin{align*} \limsup\limits_{\varepsilon\rightarrow 0}\frac{1}{\varepsilon^2}\int_{\Omega} W(x,\nabla y_\varepsilon)\,\mathrm{d} x& \ge \limsup\limits_{\varepsilon\rightarrow 0}\frac 12\int_\Omega Q(x,\chi_{G_\varepsilon}e(u_\varepsilon))\,\mathrm{d} x\\ & \ge \liminf\limits_{\varepsilon\rightarrow 0}\frac 12\int_\Omega Q(x,\chi_{G_\varepsilon}e(u_\varepsilon))\,\mathrm{d} x\\ & \ge \frac 12\int_\Omega Q(x,e(u_0))\,\mathrm{d} x, \end{align*}

see, e.g. [Reference Agostiniani, Dal Maso and DeSimone3, Reference Dal Maso, Negri and Percivale8, Reference Maor and Mora22]. Combining the previous inequalities yields

(3.37)\begin{equation} \lim\limits_{\varepsilon\rightarrow 0}\frac 12\int_\Omega Q(x,\chi_{G_\varepsilon}e(u_\varepsilon))\,\mathrm{d} x=\frac 12\int_\Omega Q(x,e(u_0))\,\mathrm{d} x. \end{equation}

Since $\chi _{G_\varepsilon }e(u_\varepsilon )\rightharpoonup e(u_0)$ weakly in $L^2(\Omega ;\mathbb {R}^{n\times n}_{\rm sym})$ and the quadratic form $Q(x,\,\cdot )$ is coercive on $\mathbb {R}^{n\times n}_{\rm sym}$ by (W2), (W4) and (W5), equation (3.37) proves claim (a).

To show claim (b) one can repeat verbatim the proof in [Reference Agostiniani, Dal Maso and DeSimone3, theorem 2.5].

We now prove claim (c). Given $\alpha >p$ and $\eta >0$, we deduce by (b) that there exists $M_\eta >0$ such that, setting

\[ f_1^{\varepsilon,\eta}(x):=\operatorname{dist}(\nabla y_\varepsilon(x);SO(n))\chi_{E^{\varepsilon,\eta}}(x),\enspace f_2^{\varepsilon,\eta}(x):=\operatorname{dist}(\nabla y_\varepsilon(x);SO(n))\chi_{\Omega\setminus E^{\varepsilon,\eta}}(x), \]

where

\[ E^{\varepsilon,\eta}:=\left\{x\in\Omega: \displaystyle \frac{1}{\varepsilon^p}\operatorname{dist}^p(\nabla y_\varepsilon(x);SO(n))\ge M_\eta\right\}, \]

we have that

(3.38)\begin{equation} \left\|\frac{f_1^{\varepsilon,\eta}}{\varepsilon}\right\|_p^p\le \eta\text{ and } \left\|\frac{f_2^{\varepsilon,\eta}}{\varepsilon}\right\|_\alpha^\alpha\le |\Omega|M_\eta^{\alpha / p}. \end{equation}

Theorem 3.3 now ensures the existence of $\widetilde {R}_{\varepsilon,\eta }\in SO(n)$ and of $g_1^{\varepsilon,\eta },\, g_2^{\varepsilon,\eta }$ such that

(3.39)\begin{equation} \nabla y_\varepsilon=\widetilde{R}_{\varepsilon,\eta}+g_1^{\varepsilon,\eta}+g_2^{\varepsilon,\eta}\text{ and }\|g_1^{\varepsilon,\eta}\|_p\le C\|f_1^{\varepsilon,\eta}\|_p,\quad \|g_2^{\varepsilon,\eta}\|_\alpha\le C\|f_2^{\varepsilon,\eta}\|_\alpha. \end{equation}

Since $\nabla y_\varepsilon =R_\varepsilon +\varepsilon R_\varepsilon \nabla u_\varepsilon$, we deduce that

(3.40)\begin{equation} \frac{\widetilde{R}_{\varepsilon,\eta}-R_\varepsilon}{\varepsilon}=R_\varepsilon\nabla u_\varepsilon-\frac{g_1^{\varepsilon,\eta}}{\varepsilon}-\frac{g_2^{\varepsilon,\eta}}{\varepsilon}, \end{equation}

hence

(3.41)\begin{equation} \frac{|\widetilde{R}_{\varepsilon,\eta}-R_\varepsilon|^p}{\varepsilon^p}\le C\left(\|\nabla u_\varepsilon\|_p^p+\left\|\frac{g_1^{\varepsilon,\eta}}{\varepsilon}\right\|_p^p+\left\|\frac{g_2^{\varepsilon,\eta}}{\varepsilon}\right\|_p^p\right)\le C(1+\eta+M_\eta), \end{equation}

where the last inequality follows from Hölder's inequality, (3.38), and (3.39).

On the other hand, by (3.40) we can write

\[ \nabla u_\varepsilon=R_\varepsilon^T\left( \frac{\widetilde{R}_{\varepsilon,\eta}-R_\varepsilon}{\varepsilon}+\frac{g_1^{\varepsilon,\eta}}{\varepsilon}+\frac{g_2^{\varepsilon,\eta}}{\varepsilon}\right). \]

Thus, by (3.38), (3.39) and (3.41) we have that for every measurable set $A\subset \Omega$

\begin{align*} \int_A|\nabla u_\varepsilon|^p\,\mathrm{d} x& \le C\left(\frac{|\widetilde{R}_{\varepsilon,\eta}-R_\varepsilon|^p}{\varepsilon^p}|A|+\int_{A}\left|\frac{g_1^{\varepsilon,\eta}}{\varepsilon}\right|^p\,\mathrm{d} x +\int_{A}\left|\frac{g_2^{\varepsilon,\eta}}{\varepsilon}\right|^p\,\mathrm{d} x\right)\\ & \le C\left((1+\eta+M_\eta)|A|+\eta+\left\|\frac{f_2^{\varepsilon,\eta}}{\varepsilon}\right\|_\alpha^p|A|^{1-p/\alpha}\right)\\ & \le C \left((1+\eta+M_\eta)|A|+\eta+M_\eta|A|^{1-p/\alpha}\right). \end{align*}

Now, for every $\delta >0$ we can choose first $\eta =\eta (\delta )$ and then $\omega =\omega (\delta,\,\eta )$ in such a way that the right-hand side above is less than $\delta$ for every measurable set $A\subset \Omega$ with $|A|<\omega$. This proves claim (c) and concludes the proof of the theorem.

Proof. We consider only the case $q\in (1,\,2]$, being the case $q=1$ analogous and even simpler. Let $\overline {C}>0$ be a constant for which ($\pi$3) is satisfied and let

\[ h(y):=\big(\overline{C}(1+|y|^{{p}/{q'}})\big)\vee(1+\overline{C}). \]

Since $p/q'\in (0,\,1]$, the function $h$ is Lipschitz continuous in the whole of $\mathbb {R}^n$ and $\pi \ge -h$ by ($\pi$3). Hence we can apply lemma 3.4 to the function $\Pi :=\pi +h$. This provides us with a function $\widehat \Pi$. It is now easy to check that the function $\hat {\pi }:=\widehat \Pi -h$ has all the required properties.

Proof. The choice of $\varepsilon _0$ will be made throughout the proof. We follow the lines of the proof of lemma 3.6. By lemma 4.1 inequality (3.7) still holds. By (W5) and theorem 3.2 there exists a sequence $(R_\varepsilon )\subset SO(n)$ such that

\[ \int_\Omega g_p(|\nabla y_\varepsilon(x)-R_\varepsilon|)\,\mathrm{d} x\le C\int_\Omega W(x,\nabla y_\varepsilon(x))\,\mathrm{d} x, \]

and

(4.4)\begin{equation} \int_\Omega W(x,\nabla y_\varepsilon)\,\mathrm{d} x\le C\varepsilon^2+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_1+\varepsilon\int_\Omega |\hat\pi (y_\varepsilon)| |\det\nabla y_\varepsilon-1|\,\mathrm{d} x. \end{equation}

We denote by $P_\varepsilon$ the last term in the above inequality. In the following, $c_2$ is the constant in condition (W6). If $q=1$, the function $\hat{\pi}$ is bounded and thus, recalling the definition (3.8) of the sets $\Omega _\varepsilon ^\pm$, we have

\begin{align*} P_\varepsilon & \le C\varepsilon\|\det\nabla y_\varepsilon-1\|_{L^2(\Omega_\varepsilon^-)}+C\varepsilon\|\det\nabla y_\varepsilon-1\|_{L^1(\Omega_\varepsilon^+)} \\ & \le C\varepsilon^2 +\frac{c_2}4 \|\det\nabla y_\varepsilon-1\|^2_{L^2(\Omega_\varepsilon^-)}+ C\varepsilon\|\det\nabla y_\varepsilon-1\|_{L^1(\Omega_\varepsilon^+)} \\ & \le C\varepsilon^2 + \Big(\frac{c_2}2+C\varepsilon\Big) \int_{\Omega}g_1(|\det\nabla y_\varepsilon-1|)\,\mathrm{d} x, \end{align*}

where we used Cauchy's inequality and (2.5). If instead $q\in (1,\,2]$, the function $\hat{\pi}$ is Lipschitz continuous and satisfies a $p/q'$-growth condition with $p/q'\leq 1$, so that we obtain

\begin{align*} P_\varepsilon & \le C\varepsilon \int_{\Omega_\varepsilon^-}(1+|y_\varepsilon-y_{R_\varepsilon}|)|\det\nabla y_\varepsilon-1|\,\mathrm{d}x\\& \quad + C\varepsilon\int_{\Omega_\varepsilon^+}(1+|y_\varepsilon-y_{R_\varepsilon}|^{{p}/{q'}})|\det\nabla y_\varepsilon-1|\,\mathrm{d} x. \end{align*}

Using that $|\det \nabla y_\varepsilon -1|\leq 1$ on $\Omega _\varepsilon ^-$ and applying Hölder's inequality, from the previous equation we deduce

\begin{align*} P_\varepsilon & \le C\varepsilon\|\det\nabla y_\varepsilon-1\|_{L^2(\Omega_\varepsilon^-)}+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}} +C\varepsilon\|\det\nabla y_\varepsilon-1\|_{L^q(\Omega_\varepsilon^+)}\\ & \quad +C\varepsilon\int_{\Omega_\varepsilon^+}|y_\varepsilon-y_{R_\varepsilon}|^p\,\mathrm{d} x+C\varepsilon\int_{\Omega_\varepsilon^+}|\det\nabla y_\varepsilon-1|^q\,\mathrm{d} x \\ & \le C\varepsilon^2 + \frac{c_2}4 \|\det\nabla y_\varepsilon-1\|^2_{L^2(\Omega_\varepsilon^-)} +C\varepsilon^{q'} + \left(\frac{c_2}4+C\varepsilon\right) \|\det\nabla y_\varepsilon-1\|^q_{L^q(\Omega_\varepsilon^+)} \\ & \quad +C\varepsilon \|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}}+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}}^p \\ & \le C\varepsilon^2 + \left(\frac{c_2}2+C\varepsilon\right) \int_{\Omega}g_q(|\det\nabla y_\varepsilon-1|)\,\mathrm{d}x\\ & \quad+C\varepsilon \|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}}+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}}^p, \end{align*}

where we used Young's inequality, (2.5) and the fact that $q'\geq 2$. Combining (4.4) with the previous bounds on $P_\varepsilon$, we obtain that

(4.5)\begin{equation} \begin{aligned} \int_\Omega W(x,\nabla y_\varepsilon)\,\mathrm{d} x & \le C\varepsilon^2+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}}+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}}^p \\ & \quad + \Big(\frac{c_2}2+C\varepsilon\Big) \int_{\Omega}g_q(|\det\nabla y_\varepsilon-1|)\,\mathrm{d} x \end{aligned} \end{equation}

in both cases $q=1$ and $q\in (1,\,2]$.

Now, if $\varepsilon _0\le c_2/(4\overline {C})$, where $\overline {C}$ is a constant for which (4.5) is true, then by (W6) we deduce that

(4.6)\begin{equation} \int_{\Omega}g_q(|\det\nabla y_\varepsilon-1|)\,\mathrm{d} x\le C\varepsilon^2+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}}+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}}^p \end{equation}

for every $\varepsilon \in (0,\,\varepsilon _0)$. Combining (4.6) and (4.5) yields

(4.7)\begin{equation} \begin{aligned} \int_\Omega W(x,\nabla y_\varepsilon)\,\mathrm{d} x & \le C\varepsilon^2+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}}+C\varepsilon\|y_\varepsilon-y_{R_\varepsilon}\|_{W^{1,p}}^p\\ & = C\varepsilon^2(1+\|u_\varepsilon\|_{W^{1,p}}+\varepsilon^{p-1}\|u_\varepsilon\|_{W^{1,p}}^p). \end{aligned} \end{equation}

Arguing as in (3.14)–(3.16) and using Poincaré–Wirtinger inequality we deduce that

\[ \| u_\varepsilon\|_{W^{1,p}}^p\le C(1+\|u_\varepsilon\|_{W^{1,p}}+\varepsilon^{p-1}\|u_\varepsilon\|_{W^{1,p}}^p). \]

Up to choosing $\varepsilon _0$ smaller, if needed, we obtain

\[ \| u_\varepsilon\|_{W^{1,p}}^p\le C(1+\|u_\varepsilon\|_{W^{1,p}}), \]

which implies $\|u_\varepsilon \|_{W^{1,p}}\le C$.

Inequality (4.2) now follows easily from (4.6), while (4.3) is a consequence of (4.7) and (W5). The last two statements of the lemma can be proved arguing exactly as in the proofs of lemma 3.6 and corollary 3.7.

Proof. We follow the lines of the proof of proposition 3.9. The only difference is in the analysis of the term $I_\varepsilon$ in (3.21), which now can be written as

\begin{align*} I_\varepsilon& = \int_{G_\varepsilon}\hat\pi (y_\varepsilon)\mathrm{div}\, u_\varepsilon\,\mathrm{d} x+\sum_{k=2}^n \varepsilon^{k-1}\int_{G_\varepsilon}\hat\pi (y_\varepsilon)\iota_k(\nabla u_\varepsilon)\,\mathrm{d} x\\ & \quad + \frac1\varepsilon\int_{\Omega_\varepsilon^-{\setminus} G_\varepsilon}\hat\pi(y_\varepsilon)(\det\nabla y_\varepsilon-1) \,\mathrm{d} x+\frac1\varepsilon \int_{\Omega_\varepsilon^+}\hat\pi (y_\varepsilon)(\det\nabla y_\varepsilon-1)\,\mathrm{d} x \\ & =:\tilde{J}^1_\varepsilon+\tilde{J}^2_\varepsilon+\tilde{J}^3_\varepsilon+\tilde{J}^4_\varepsilon, \end{align*}

where the sets $\Omega _\varepsilon ^{\pm }$ and $G_\varepsilon$ are defined as in (3.8) and (3.18). Here we recall that $G_\varepsilon \subset \Omega ^-_\varepsilon$ for $\varepsilon$ small enough. The first integral $\tilde {J}^1_\varepsilon$ can be handled exactly as in the proof of proposition 3.9. To conclude it is enough to show that the remaining terms are infinitesimal, as $\varepsilon \to 0$.

By (2.2) and the Lipschitz continuity of $\hat{\pi}$ we obtain

\begin{align*} |\tilde J^2_\varepsilon|& \le C\sum_{k=2}^n\varepsilon^{k-1}\int_{G_\varepsilon}(1+\varepsilon|u_\varepsilon|)|\nabla u_\varepsilon|^k\,\mathrm{d} x\\ & =C \sum_{k=2}^n\left(\varepsilon^{k-1}\int_{G_\varepsilon}|\nabla u_\varepsilon|^{k-p}|\nabla u_\varepsilon|^p\,\mathrm{d} x+\varepsilon^k\int_{G_\varepsilon}|u_\varepsilon||\nabla u_\varepsilon|^k\,\mathrm{d} x\right). \end{align*}

Since $|\nabla u_\varepsilon |\le \varepsilon ^{-1/2}$ on $G_\varepsilon$, we have that

\begin{align*} |\tilde{J}^2_\varepsilon|& \le C\sum_{k=2}^n\left(\varepsilon^{{(k+p-2)}/{2}}\|\nabla u_\varepsilon\|_p^p+\varepsilon^{k/2} \|u_\varepsilon\|_p\right)\\ & \le C \sum_{k=2}^n\left(\varepsilon^{{(k+p-2)}/{2}}+\varepsilon^{k/2}\right)\le C\varepsilon^{p/2}\rightarrow 0. \end{align*}

Using (4.2), the boundedness of $(u_\varepsilon )$ in $W^{1,p}(\Omega ;\mathbb {R}^n)$, and recalling that $|\det \nabla y_\varepsilon {-}1|\leq 1$ on $\Omega _\varepsilon ^-$, we deduce that

\begin{align*} |\tilde{J}^3_\varepsilon|& \le \frac{C}\varepsilon \int_{\Omega_\varepsilon^-{\setminus} G_\varepsilon}(1+\varepsilon|u_\varepsilon|)|\det\nabla y_\varepsilon{-}1| \,\mathrm{d} x\\ & \le \frac{C}\varepsilon \|\det\nabla y_\varepsilon{-}1\|_{L^2(\Omega_\varepsilon^-)}|\Omega\setminus G_\varepsilon|^{1/2}+ C\int_{\Omega_\varepsilon^-{\setminus} G_\varepsilon}|u_\varepsilon|\,\mathrm{d} x\\ & \le C |\Omega\setminus G_\varepsilon|^{1/2}+C\|u_\varepsilon\|_p|\Omega\setminus G_\varepsilon|^{{1}/{p'}}\le C |\Omega\setminus G_\varepsilon|^{1/2}+C|\Omega\setminus G_\varepsilon|^{{1}/{p'}}, \end{align*}

where the last term goes to zero, as $\varepsilon \rightarrow 0$, by (3.19).

To deal with $\tilde {J}^4_\varepsilon$ we consider the two cases $q=1$ and $q\in (1,\,2]$, separately. If $q=1$, the function $\hat{\pi}$ is bounded and so, we have

\[ |\tilde{J}^4_\varepsilon|\le \frac{C}\varepsilon\|\det\nabla y_\varepsilon-1\|_{L^1(\Omega_\varepsilon^+)}\le C\varepsilon, \]

where in the last inequality we used (4.2).

If instead $q\in (1,\,2]$, by using the $p/q'$-growth of $\hat{\pi}$ we have

\begin{align*} |\tilde{J}^4_\varepsilon|& \le \frac{C}\varepsilon \int_{\Omega_\varepsilon^+}(1+(\varepsilon|u_\varepsilon|)^{{p}/{q'}}) |\det\nabla y_\varepsilon-1| \,\mathrm{d} x\\ & \le \frac{C}\varepsilon \|\det\nabla y_\varepsilon-1\|_{L^q(\Omega_\varepsilon^+)} |\Omega_\varepsilon^+|^{{1}/{q'}}+C\varepsilon^{{p}/{q'}-1}\|\det\nabla y_\varepsilon-1\|_{L^q(\Omega_\varepsilon^+)}\|u_\varepsilon\|_p^{{p}/{q'}}. \end{align*}

By (4.2) and the boundedness of $(u_\varepsilon )$ in $W^{1,p}(\Omega ;\mathbb {R}^n)$ we deduce

\[ |\widetilde{J}^4_\varepsilon|\le C \varepsilon^{{2}/{q}-1}|\Omega^+_\varepsilon|^{{1}/{q'}}+ C \varepsilon^{{2}/{q}+{p}/{q'}-1}\|u_\varepsilon\|_p^{{p}/{q'}}\le C |\Omega^+_\varepsilon|^{{1}/{q'}}+ C \varepsilon^{{2}/{q}+{p}/{q'}-1}. \]

It is easy to verify that ${2}/{q}+{p}/{q'}-1>0$. Moreover, since $G_\varepsilon \subset \Omega _\varepsilon ^-$ for $\varepsilon$ small enough, we have that $|\Omega _\varepsilon ^+|\to 0$ by (3.19). Therefore, we conclude that in both cases $\widetilde {J}^4_\varepsilon$ is infinitesimal, as $\varepsilon \rightarrow 0$.

Proof. The proof is the same of propositions 3.10 and 3.12, once we have at our disposal proposition 4.4. The only additional remark is that the deformations $y_\varepsilon$ in the recovery sequence are admissible since $\det \nabla y_\varepsilon \in L^\infty (\Omega )$ by construction (see (4.1) for the definition of $Y^p_q$).

Proof. As for the compactness statement, since $\mathcal {R}$ is a closed set, there exist $S_\varepsilon \in \mathcal {R}$ and $A_\varepsilon \in \mathbb {R}^{n\times n}_{\rm skew}$ such that $R_\varepsilon =S_\varepsilon {\rm e}^{A_\varepsilon }$ and

\[ \operatorname{dist}_{SO(n)}(R_\varepsilon;\mathcal{R})=\operatorname{dist}_{SO(n)}(R_\varepsilon, S_\varepsilon)=|A_\varepsilon|. \]

Since, up to subsequences, $R_\varepsilon \rightarrow R_0\in \mathcal {R}$, we have that $A_\varepsilon \rightarrow 0$. The remaining properties follow from the fact the sequence $({A_\varepsilon }/({|A_\varepsilon |\vee \sqrt {\varepsilon }}))$ is bounded by $1$.

We now give a sketch of the proof of the liminf inequality. By proposition 3.10 or 4.5 we have that

\[ \liminf\limits_{\varepsilon\rightarrow 0}\frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon(y_\varepsilon)\ge \mathcal{E}_0(u_0,R_0)+\liminf\limits_{\varepsilon\rightarrow 0}\int_{\Omega}\frac{\pi(R_\varepsilon x)-\pi (x)}{\varepsilon}\,\mathrm{d} x. \]

In fact, the last term above was always neglected in the previous computations, since it is nonnegative by (2.8). Using that $S_\varepsilon \in \mathcal {R}$, a Taylor expansion of $\pi$ and of the exponential map yields

(5.7)\begin{align} & \int_{\Omega}\frac{\pi(R_\varepsilon x)-\pi (x)}{\varepsilon}\,\mathrm{d} x \\ & \quad=\frac{1}{\varepsilon}\int_{\Omega}\nabla\pi(S_\varepsilon x)\cdot S_\varepsilon {A}_\varepsilon x\,\mathrm{d} x \nonumber\\ & \qquad+ \frac{1}{2\varepsilon} \int_{\Omega}\big( \nabla\pi(S_\varepsilon x)\cdot S_\varepsilon {A}^2_\varepsilon x+ D^2\pi(S_\varepsilon(I+t_\varepsilon(x) K_\varepsilon) x)S_\varepsilon K_\varepsilon x\cdot S_\varepsilon K_\varepsilon x\big)\,\mathrm{d} x \nonumber\\ & \qquad+ \frac{1}{\varepsilon}\int_{\Omega}|A_\varepsilon|^3\nabla\pi(S_\varepsilon x)\cdot S_\varepsilon H_\varepsilon x\,\mathrm{d} x, \end{align}

where $t_\varepsilon (x)\in [0,\,1]$, $H_\varepsilon$ is a uniformly bounded matrix, and

\[ K_\varepsilon:={A}_\varepsilon+\frac{1}{2}{A}_\varepsilon^2+|A_\varepsilon|^3 H_\varepsilon. \]

We now note that the first integral in (5.7) is equal to $0$ by (2.9); thus, by multiplying and dividing by $\alpha _\varepsilon :=(|A_\varepsilon |\vee \sqrt {\varepsilon })^2$ we obtain

\begin{align*} \int_{\Omega}& \frac{\pi(R_\varepsilon x)-\pi (x)}{\varepsilon}\,\mathrm{d} x \\ & =\frac{\alpha_\varepsilon}{2\varepsilon}\left[\int_{\Omega}\left(\nabla\pi(S_\varepsilon x)\cdot S_\varepsilon \frac{A^2_\varepsilon x}{\alpha_\varepsilon} + D^2\pi(S_\varepsilon(I+t_\varepsilon(x) K_\varepsilon)x )S_\varepsilon \frac{K_\varepsilon x}{\sqrt{\alpha_\varepsilon}}\cdot S_\varepsilon \frac{K_\varepsilon x}{\sqrt{\alpha_\varepsilon}}\right)\right. \,\mathrm{d} x\\ & \quad\left.+\frac{2|A_\varepsilon|^3}{\alpha_\varepsilon}\int_{\Omega}\nabla\pi(S_\varepsilon x)\cdot S_\varepsilon H_\varepsilon x\,\mathrm{d} x\right]. \end{align*}

We observe that the left-hand side is nonnegative, hence the term within square brackets is nonnegative, as well. Since $\sqrt {\alpha _\varepsilon }=|A_\varepsilon |\vee \sqrt {\varepsilon }\ge \sqrt {\varepsilon }$, $|A_\varepsilon |\rightarrow 0$, $S_\varepsilon \rightarrow R_0$, ${A_\varepsilon }/{\sqrt {\alpha _\varepsilon }}\rightarrow A_0$, letting $\varepsilon \rightarrow 0$ and using the dominated convergence theorem yield

\begin{align*} & \liminf\limits_{\varepsilon\rightarrow 0}\int_{\Omega}\frac{\pi(R_\varepsilon x)-\pi (x)}{\varepsilon}\,\mathrm{d} x\\& \quad \ge \frac 12\int_{\Omega}\big(\nabla\pi (R_0x)\cdot R_0 A_0^2x+D^2\pi(R_0 x)R_0 A_0 x\cdot R_0 A_0 x\big)\,\mathrm{d} x. \end{align*}

The liminf inequality follows now from the divergence theorem.

For the construction of the recovery sequence we proceed as in proposition 3.12 or 4.5, but choosing $R_\varepsilon :=R_0{\rm e}^{A_\varepsilon }$ with $A_\varepsilon :=\sqrt {\varepsilon }A_0$. Since $R_0$ is in $\mathcal {R}$, we can write

(5.8)\begin{equation} \begin{aligned} \frac{1}{\varepsilon^2}\mathcal{E}_\varepsilon(y_\varepsilon) & =\frac{1}{\varepsilon^2}\int_{\Omega}W(x,\nabla y_\varepsilon)\,\mathrm{d} x+\frac 1\varepsilon\int_{\Omega}\left(\pi (y_\varepsilon(x))\det\nabla y_\varepsilon-\pi(R_\varepsilon x)\right)\,\mathrm{d} x\\ & \quad+\int_{\Omega}\frac{\pi(R_\varepsilon x)-\pi (R_0 x)}{\varepsilon}\,\mathrm{d} x. \end{aligned} \end{equation}

The first two integrals at the right-hand side can be bounded by $\mathcal {E}_0(u_0,\,R_0)$, arguing as in proposition 3.12 or 4.5. Repeating the same computations as for the liminf inequality, one can show that the last integral converges to $(1/2)\mathcal {F}(R_0,\,A_0)$.

Convergence of almost minimizers and of infima can be proved exactly as in theorems 3.14 and 4.6. Finally, we observe that the functional $\mathcal {F}$ is always nonnegative by (5.2) and $\mathcal {F}(R,\,0)=0$ for every $R\in \mathcal {R}$. Hence, by minimality we deduce that $\mathcal {F}(R_0,\,A_0)=0$. This concludes the proof.