[proofplan] The lower bound follows from the facts that $f^{**}\le f$ and that convex integral functionals with $p$-growth are sequentially weakly lower semicontinuous in $W^{1,p}$. For the upper bound, one first proves a recovery construction for affine functions by replacing a constant gradient $\xi$ with rapidly oscillating scalar gradients whose volume fractions realize the convex envelope value $f^{**}(\xi)$. The affine construction is localized on cubes to handle piecewise affine functions, and then arbitrary Sobolev functions are reached by approximation in $W^{1,p}$, using the $p$-growth bounds to pass energies to the limit. [/proofplan]

[step:Record the inherited growth and convexity properties of $f^{**}$] Because $f^{**}$ is the convex lower semicontinuous envelope of $f$, it satisfies \begin{align*} f^{**}(\xi)\le f(\xi) \end{align*} for every $\xi\in\mathbb R^n$. Since $f\ge 0$ and the zero function is convex and lower semicontinuous with $0\le f$, the maximality of the convex lower semicontinuous envelope also gives $0\le f^{**}$ on $\mathbb R^n$. The lower $p$-growth bound for $f$ is itself a convex [continuous function](/page/Continuous%20Function) of $\xi$, up to the constant term. Define \begin{align*} g: \mathbb R^n \to \mathbb R,\qquad \xi \mapsto c|\xi|^p - C \end{align*} Since $g$ is convex and $g\le f$, the maximality of the convex envelope gives $g\le f^{**}$. Hence, for every $\xi\in\mathbb R^n$, \begin{align*} c|\xi|^p-C\le f^{**}(\xi)\le C(1+|\xi|^p). \end{align*} In particular, if $u\in W^{1,p}(U)$, then $f^{**}(\nabla u)$ is integrable over $U$ with respect to $\mathcal L^n$. [/step]

[step:Prove the lower bound by convex weak lower semicontinuity]Fix $u\in W^{1,p}(U)$, and let $(u_k)_{k=1}^{\infty}\subset W^{1,p}(U)$ be any sequence such that $u_k\rightharpoonup u$ in $W^{1,p}(U)$. Define the Caratheodory integrand \begin{align*} G:U\times\mathbb R\times\mathbb R^n\to[0,\infty),\qquad G(x,s,\xi)=f^{**}(\xi). \end{align*} The function $G$ is measurable in $x$, continuous in $s$, lower semicontinuous in $\xi$, and convex in $\xi$. Its upper $p$-growth follows from the previous step. Since $U$ is bounded and Lipschitz and $u_k\rightharpoonup u$ in $W^{1,p}(U)$, [citetheorem:8738] applies to $G$ and gives \begin{align*} \int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x)\le \liminf_{k\to\infty}\int_U f^{**}(\nabla u_k(x))\,d\mathcal L^n(x). \end{align*} Because $f^{**}\le f$, we also have \begin{align*} \int_U f^{**}(\nabla u_k(x))\,d\mathcal L^n(x)\le F[u_k] \end{align*} for every $k\in\mathbb N$. Therefore \begin{align*} \int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x)\le \liminf_{k\to\infty}F[u_k]. \end{align*} Taking the infimum over all weakly convergent recovery sequences in the definition of $\overline F[u]$ yields \begin{align*} \int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x)\le \overline F[u]. \end{align*}[/step]

[guided]We want a lower bound for the relaxed functional. By definition, $\overline F[u]$ is obtained by taking the best possible weakly convergent sequence $u_k\rightharpoonup u$ and then the lower limiting energy. Thus it is enough to prove the same lower bound for every admissible sequence. Fix $u\in W^{1,p}(U)$ and let $(u_k)_{k=1}^{\infty}\subset W^{1,p}(U)$ satisfy $u_k\rightharpoonup u$ in $W^{1,p}(U)$. The integrand $f^{**}$ is convex and lower semicontinuous by definition of the convex lower semicontinuous envelope. From the previous step it satisfies the growth estimate \begin{align*} c|\xi|^p-C\le f^{**}(\xi)\le C(1+|\xi|^p) \end{align*} for every $\xi\in\mathbb R^n$. Define the Caratheodory integrand \begin{align*} G:U\times\mathbb R\times\mathbb R^n\to[0,\infty),\qquad G(x,s,\xi)=f^{**}(\xi). \end{align*} We apply [citetheorem:8738]. The theorem requires $U$ to be bounded and Lipschitz, which is a hypothesis of the theorem statement. It also requires measurability in $x$, continuity in $s$, convexity in $\xi$, and an upper $p$-growth bound. These hold because $G$ is independent of $x$ and $s$, $f^{**}$ is convex and lower semicontinuous, and $f^{**}(\xi)\le C(1+|\xi|^p)$ for every $\xi\in\mathbb R^n$. Since $u_k\rightharpoonup u$ in $W^{1,p}(U)$, the lower semicontinuity theorem gives \begin{align*} \int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x)\le \liminf_{k\to\infty}\int_U f^{**}(\nabla u_k(x))\,d\mathcal L^n(x). \end{align*} The reason this proves a statement about $F$, rather than only about the convexified functional, is the pointwise inequality $f^{**}\le f$. Applying that inequality at $\xi=\nabla u_k(x)$ for $\mathcal L^n$-a.e. $x\in U$ gives \begin{align*} \int_U f^{**}(\nabla u_k(x))\,d\mathcal L^n(x)\le \int_U f(\nabla u_k(x))\,d\mathcal L^n(x)=F[u_k]. \end{align*} Combining the two inequalities gives \begin{align*} \int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x)\le \liminf_{k\to\infty}F[u_k]. \end{align*} Since this holds for every weakly convergent sequence $u_k\rightharpoonup u$, taking the infimum over all such sequences gives \begin{align*} \int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x)\le \overline F[u]. \end{align*}[/guided]

[step:Build affine recovery sequences from scalar laminates] We invoke the scalar convexification recovery theorem for integral functionals with continuous $p$-growth integrands. Applied to the continuous integrand $f:\mathbb R^n\to[0,\infty)$ satisfying the displayed $p$-growth bounds, it states that if $\xi\in\mathbb R^n$ and $Q\subset\mathbb R^n$ is a cube, then for every $\varepsilon>0$ there exists a sequence $(\varphi_k)_{k=1}^{\infty}\subset W^{1,\infty}_0(Q)$ such that $\varphi_k\rightharpoonup 0$ in $W^{1,p}(Q)$ and \begin{align*} \limsup_{k\to\infty}\int_Q f(\xi+\nabla\varphi_k(x))\,d\mathcal L^n(x)\le \mathcal L^n(Q) f^{**}(\xi)+\varepsilon \mathcal L^n(Q). \end{align*} The hypotheses of that theorem are exactly the continuity and $p$-growth assumptions on $f$, the scalar target space, and the fact that $Q$ is a cube. Applying this lemma to the affine map \begin{align*} a_\xi:Q\to\mathbb R,\qquad a_\xi(x)=\xi\cdot x, \end{align*} and defining $a_{\xi,k}:Q\to\mathbb R$ by $a_{\xi,k}=a_\xi+\varphi_k$, gives $a_{\xi,k}\rightharpoonup a_\xi$ in $W^{1,p}(Q)$. Define the localized functional \begin{align*} F_Q:W^{1,p}(Q)\to[0,\infty),\qquad F_Q[v]=\int_Q f(\nabla v(x))\,d\mathcal L^n(x). \end{align*} Then, since $\nabla a_\xi(x)=\xi$ for $\mathcal L^n$-a.e. $x\in Q$, \begin{align*} \limsup_{k\to\infty}F_Q[a_{\xi,k}]\le \int_Q f^{**}(\nabla a_\xi(x))\,d\mathcal L^n(x)+\varepsilon \mathcal L^n(Q). \end{align*} [/step]

[step:Localize the affine construction on piecewise affine functions] Let $v\in W^{1,p}(U)$ be piecewise affine with the following cubical exhaustion property: for every $\eta>0$ there are pairwise disjoint cubes $(Q_j)_{j=1}^{N(\eta)}$ compactly contained in $U$ and a measurable remainder set $R_\eta=U\setminus\bigcup_{j=1}^{N(\eta)}Q_j$ such that $\nabla v=\xi_j$ $\mathcal L^n$-a.e. on $Q_j$ for some $\xi_j\in\mathbb R^n$ and \begin{align*} \int_{R_\eta}\bigl(f(\nabla v(x))+f^{**}(\nabla v(x))\bigr)\,d\mathcal L^n(x)\le \eta. \end{align*} This is obtained by choosing the cubes inside the affine cells of $v$ and using absolute continuity of the [Lebesgue integral](/page/Lebesgue%20Integral), since the displayed integrand belongs to $L^1(U)$ by the $p$-growth bounds and $\nabla v\in L^p(U;\mathbb R^n)$. Apply the affine recovery construction on each $Q_j$ with error $\eta/N(\eta)$. Extending the perturbations by $0$ outside their corresponding cubes gives a sequence $(v_{k,\eta})_{k=1}^{\infty}\subset W^{1,p}(U)$ with $v_{k,\eta}\rightharpoonup v$ in $W^{1,p}(U)$. Since each perturbation has zero trace on its cube boundary, the pieces patch without creating jumps. Because the number of cubes is finite, choose one diagonal cubewise index so that all finitely many limsup estimates hold simultaneously. Summing the cube estimates and leaving $v_{k,\eta}=v$ on $R_\eta$, we obtain \begin{align*} \limsup_{k\to\infty}F[v_k^\eta]\le \sum_{j=1}^{N(\eta)}\int_{Q_j} f^{**}(\xi_j)\,d\mathcal L^n(x)+\int_{R_\eta} f(\nabla v(x))\,d\mathcal L^n(x)+\eta. \end{align*} Using $\nabla v=\xi_j$ on each $Q_j$ and the defining bound on $R_\eta$ gives \begin{align*} \overline F[v]\le \int_U f^{**}(\nabla v(x))\,d\mathcal L^n(x)+2\eta. \end{align*} Letting $\eta\downarrow0$ yields \begin{align*} \overline F[v]\le \int_U f^{**}(\nabla v(x))\,d\mathcal L^n(x). \end{align*} [/step]

[step:Approximate arbitrary Sobolev maps and pass to the limit] Fix $u\in W^{1,p}(U)$. Since $U$ is bounded and Lipschitz, the Sobolev piecewise affine approximation theorem on Lipschitz domains gives a sequence $(v_m)_{m=1}^{\infty}$ of functions satisfying the cubical exhaustion property used in the previous step and \begin{align*} v_m\to u\quad\text{in }W^{1,p}(U). \end{align*} The hypotheses of that approximation theorem are the bounded Lipschitz regularity of $U$ and the membership $u\in W^{1,p}(U)$. After passing to a subsequence if necessary, $\nabla v_m\to\nabla u$ pointwise $\mathcal L^n$-a.e. on $U$. The continuity of the finite [convex function](/page/Convex%20Function) $f^{**}$ gives pointwise convergence of the integrands. The strong convergence $\nabla v_m\to\nabla u$ in $L^p(U;\mathbb R^n)$ implies uniform integrability of $\{|\nabla v_m|^p:m\in\mathbb N\}$. By the upper $p$-growth estimate $f^{**}(\xi)\le C(1+|\xi|^p)$ and the finiteness of $\mathcal L^n(U)$, the family $\{f^{**}(\nabla v_m):m\in\mathbb N\}$ is uniformly integrable. Vitali's convergence theorem therefore yields \begin{align*} \int_U f^{**}(\nabla v_m(x))\,d\mathcal L^n(x)\to \int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x). \end{align*} For each $m\in\mathbb N$, the previous step gives a sequence $(v_{m,k})_{k=1}^{\infty}\subset W^{1,p}(U)$ such that $v_{m,k}\rightharpoonup v_m$ in $W^{1,p}(U)$ and \begin{align*} \limsup_{k\to\infty}F[v_{m,k}]\le \int_U f^{**}(\nabla v_m(x))\,d\mathcal L^n(x)+\frac{1}{m}. \end{align*} Let $(\Lambda_i)_{i=1}^{\infty}$ be a countable subset of the unit ball of the [dual space](/page/Dual%20Space) $(W^{1,p}(U))^*$ that is dense in the operator-norm topology; such a set exists because $W^{1,p}(U)$ is separable and reflexive for $1<p<\infty$ on bounded Lipschitz domains, by the reflexivity of Lebesgue and Sobolev spaces [citetheorem:8729] and separability of the corresponding $L^p$ components. By the [Rellich Kondrachov compactness theorem](/theorems/8731) [citetheorem:8731], the [weak convergence](/page/Weak%20Convergence) $v_{m,k}\rightharpoonup v_m$ in $W^{1,p}(U)$ implies $v_{m,k}\to v_m$ in $L^p(U)$ after taking the relevant tail of the recovery sequence; the hypotheses are the bounded Lipschitz regularity of $U$ and $1<p<\infty$. Choose $k(m)\in\mathbb N$ large enough that \begin{align*} \|v_{m,k(m)}-v_m\|_{L^p(U)}\le \frac{1}{m} \end{align*} and, for every $1\le i\le m$, \begin{align*} |\Lambda_i(v_{m,k(m)}-v_m)|\le \frac{1}{m}, \end{align*} and \begin{align*} F[v_{m,k(m)}]\le \int_U f^{**}(\nabla v_m(x))\,d\mathcal L^n(x)+\frac{2}{m}. \end{align*} Define $u_m=v_{m,k(m)}$. The energy bound and the coercive lower estimate for $f$ give \begin{align*} c\|\nabla u_m\|_{L^p(U)}^p-C\mathcal L^n(U)\le F[u_m]. \end{align*} The right-hand side is bounded because the integrals of $f^{**}(\nabla v_m)$ converge to the finite value $\int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x)$. The $L^p$ estimate $\|u_m-v_m\|_{L^p(U)}\le 1/m$ and the strong convergence $v_m\to u$ in $W^{1,p}(U)$ give boundedness of $(u_m)_{m=1}^{\infty}$ in $L^p(U)$. Hence $(u_m-v_m)_{m=1}^{\infty}$ is bounded in $W^{1,p}(U)$. The diagonal choice gives $\Lambda_i(u_m-v_m)\to0$ for every fixed $i$. Since the sequence is bounded in $W^{1,p}(U)$ and the set $(\Lambda_i)$ is dense in the dual unit ball in the operator-norm topology, every $\Lambda\in (W^{1,p}(U))^*$ is approximated uniformly on bounded subsets by some $\Lambda_i$. It follows that $\Lambda(u_m-v_m)\to0$ for every $\Lambda\in (W^{1,p}(U))^*$, which is exactly $u_m-v_m\rightharpoonup0$ in $W^{1,p}(U)$. Since $v_m\to u$ strongly in $W^{1,p}(U)$, we have $u_m\rightharpoonup u$ in $W^{1,p}(U)$. Therefore \begin{align*} \overline F[u]\le \liminf_{m\to\infty}F[u_m]\le \int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x). \end{align*} [/step]

[step:Combine the two inequalities] The lower bound proved above gives \begin{align*} \int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x)\le \overline F[u] \end{align*} for every $u\in W^{1,p}(U)$. The recovery construction gives the reverse inequality \begin{align*} \overline F[u]\le \int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x) \end{align*} for every $u\in W^{1,p}(U)$. Hence equality holds: \begin{align*} \overline F[u]=\int_U f^{**}(\nabla u(x))\,d\mathcal L^n(x). \end{align*} This is the claimed weak $W^{1,p}$ sequential relaxation formula. [/step]