[proofplan] We use the direct method in the [calculus of variations](/page/Calculus%20of%20Variations). First we check that the infimum is finite, using boundedness of $U$, the lower $p$-growth bound, and the upper $p$-growth bound. Then we take a minimizing sequence, use the lower growth bound together with the admissible coercivity inequality to obtain boundedness in $W^{1,p}(U)$, and extract a weakly convergent subsequence by reflexivity and [weak sequential compactness](/theorems/214). The weak sequential closedness of $\mathcal A$ keeps the limit admissible, and weak lower semicontinuity of convex integral functionals gives minimality. [/proofplan] [step:Show that the functional is finite on $\mathcal A$ and bounded below] Let $u\in\mathcal A$. Since $u\in W^{1,p}(U)$, the weak gradient $\nabla u:U\to\mathbb R^n$ belongs to $L^p(U;\mathbb R^n)$. The upper growth bound gives \begin{align*} f^{**}(\nabla u(x))\le C(1+|\nabla u(x)|^p) \end{align*} for $\mathcal L^n$-a.e. $x\in U$. Since $U$ is bounded, $\mathcal L^n(U)<\infty$, and hence \begin{align*} \int_U C(1+|\nabla u(x)|^p)\,d\mathcal L^n(x)<\infty. \end{align*} The lower growth bound gives \begin{align*} f^{**}(\nabla u(x))\ge c|\nabla u(x)|^p-C\ge -C \end{align*} for $\mathcal L^n$-a.e. $x\in U$. Let $\mathbb 1_U:U\to\mathbb R$ denote the constant indicator function of $U$, so $\mathbb 1_U(x)=1$ for every $x\in U$. The negative part is bounded by the integrable function $C\mathbb 1_U$. Therefore $\overline F[u]$ is a finite real number. Moreover, \begin{align*} \overline F[u]\ge \int_U (c|\nabla u(x)|^p-C)\,d\mathcal L^n(x)\ge -C\mathcal L^n(U). \end{align*} Thus $\inf_{v\in\mathcal A}\overline F[v]>-\infty$. Since $\mathcal A$ is nonempty and $\overline F$ is finite on $\mathcal A$, also $\inf_{v\in\mathcal A}\overline F[v]<\infty$. [/step] [step:Choose a minimizing sequence and bound its gradients] Let \begin{align*} m:=\inf_{v\in\mathcal A}\overline F[v]\in\mathbb R. \end{align*} By the definition of the infimum, there exists a sequence $(u_k)_{k=1}^{\infty}$ in $\mathcal A$ such that \begin{align*} \overline F[u_k]\to m. \end{align*} Let $\mathbb N=\{1,2,3,\dots\}$ denote the set of positive integers. Choose $M>0$ such that $\overline F[u_k]\le M$ for every $k\in\mathbb N$ after discarding finitely many initial terms. This does not change the minimizing property of the sequence. For every remaining $k$, the lower growth bound gives \begin{align*} c\int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)-C\mathcal L^n(U)\le \overline F[u_k]\le M. \end{align*} Hence \begin{align*} \int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)\le \frac{M+C\mathcal L^n(U)}{c}. \end{align*} Thus $(\nabla u_k)_{k=1}^{\infty}$ is bounded in $L^p(U;\mathbb R^n)$. [guided] We need compactness, and compactness in $W^{1,p}(U)$ requires control of both the functions and their weak gradients. The energy directly controls only the gradients, so we first extract the gradient bound from the lower growth hypothesis. Define \begin{align*} m:=\inf_{v\in\mathcal A}\overline F[v]. \end{align*} The previous step proved that $m$ is a finite real number. By the definition of infimum, we may choose a sequence $(u_k)_{k=1}^{\infty}$ in $\mathcal A$ such that \begin{align*} \overline F[u_k]\to m. \end{align*} Since this sequence of [real numbers](/page/Real%20Numbers) converges, it is bounded above after finitely many initial terms. Let $\mathbb N=\{1,2,3,\dots\}$ denote the set of positive integers. We discard those finitely many terms and choose a constant $M>0$ such that \begin{align*} \overline F[u_k]\le M \end{align*} for every remaining $k\in\mathbb N$. Now apply the pointwise lower bound for $f^{**}$ to the map $\nabla u_k:U\to\mathbb R^n$. For $\mathcal L^n$-a.e. $x\in U$, \begin{align*} f^{**}(\nabla u_k(x))\ge c|\nabla u_k(x)|^p-C. \end{align*} Integrating this inequality over $U$ with respect to $\mathcal L^n$ gives \begin{align*} \overline F[u_k]=\int_U f^{**}(\nabla u_k(x))\,d\mathcal L^n(x)\ge c\int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)-C\mathcal L^n(U). \end{align*} Combining this lower estimate with $\overline F[u_k]\le M$ yields \begin{align*} c\int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)\le M+C\mathcal L^n(U). \end{align*} Since $c>0$, division by $c$ gives \begin{align*} \int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)\le \frac{M+C\mathcal L^n(U)}{c}. \end{align*} The right-hand side is independent of $k$, so $(\nabla u_k)_{k=1}^{\infty}$ is bounded in $L^p(U;\mathbb R^n)$. [/guided] [/step] [step:Use admissible coercivity to bound the minimizing sequence in $W^{1,p}(U)$] Define \begin{align*} B_\nabla:=\frac{M+C\mathcal L^n(U)}{c}. \end{align*} The previous step gives \begin{align*} \int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)\le B_\nabla \end{align*} for every $k$. The admissible coercivity assumption gives \begin{align*} \|u_k\|_{L^p(U)}^p\le K_1\int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)+K_2\le K_1B_\nabla+K_2. \end{align*} Using the standard Sobolev norm \begin{align*} \|w\|_{W^{1,p}(U)}^p=\|w\|_{L^p(U)}^p+\int_U |\nabla w(x)|^p\,d\mathcal L^n(x), \end{align*} we obtain \begin{align*} \|u_k\|_{W^{1,p}(U)}^p\le (K_1+1)B_\nabla+K_2. \end{align*} Therefore $(u_k)_{k=1}^{\infty}$ is bounded in $W^{1,p}(U)$. [/step] [step:Extract a weakly convergent admissible subsequence] Since $1<p<\infty$, $W^{1,p}(U)$ is reflexive by [Reflexivity of Lebesgue and Sobolev Spaces]([citetheorem:8729]). The sequence $(u_k)_{k=1}^{\infty}$ is bounded in this reflexive [Banach space](/page/Banach%20Space), so by the weak [sequential compactness](/page/Sequential%20Compactness) theorem for bounded sequences in reflexive Banach spaces there exist a subsequence $(u_{k_j})_{j=1}^{\infty}$ and an element $u_*\in W^{1,p}(U)$ such that \begin{align*} u_{k_j}\rightharpoonup u_*\quad\text{in }W^{1,p}(U). \end{align*} Because each $u_{k_j}$ belongs to $\mathcal A$ and $\mathcal A$ is sequentially weakly closed in $W^{1,p}(U)$, the weak limit satisfies \begin{align*} u_*\in\mathcal A. \end{align*} [/step] [step:Apply weak lower semicontinuity to the convex gradient integral] The integrand $f^{**}:\mathbb R^n\to\mathbb R$ is convex and lower semicontinuous. Since it is finite-valued and convex on the finite-dimensional space $\mathbb R^n$, it is continuous, hence Borel measurable. Therefore $x\mapsto f^{**}(\nabla w(x))$ is measurable for every $w\in W^{1,p}(U)$. Since $u_{k_j}\rightharpoonup u_*$ in $W^{1,p}(U)$, the gradients satisfy $\nabla u_{k_j}\rightharpoonup \nabla u_*$ in $L^p(U;\mathbb R^n)$. Under the standard linear identification $\mathbb R^n\cong\mathbb R^{1\times n}$, the integrand is convex, hence quasiconvex by [Hierarchy Of Convexity Conditions]([citetheorem:8766]). The lower and upper $p$-growth assumptions are exactly the standard $p$-growth hypotheses, and the preceding finiteness argument shows that the integral is well-defined on $W^{1,p}(U)$. Therefore the [Acerbi-Fusco Lower Semicontinuity Theorem]([citetheorem:8749]) applies with $m=1$ to the [weak convergence](/page/Weak%20Convergence) \begin{align*} u_{k_j}\rightharpoonup u_*\quad\text{in }W^{1,p}(U). \end{align*} Consequently, \begin{align*} \overline F[u_*]\le \liminf_{j\to\infty}\overline F[u_{k_j}]. \end{align*} Since $(u_k)_{k=1}^{\infty}$ is minimizing, every subsequence is also minimizing, and hence \begin{align*} \lim_{j\to\infty}\overline F[u_{k_j}]=m. \end{align*} Thus \begin{align*} \overline F[u_*]\le m. \end{align*} Because $u_*\in\mathcal A$ and $m=\inf_{v\in\mathcal A}\overline F[v]$, we also have \begin{align*} m\le \overline F[u_*]. \end{align*} Combining the two inequalities gives \begin{align*} \overline F[u_*]=m=\inf_{v\in\mathcal A}\overline F[v]. \end{align*} Therefore $u_*$ is a minimizer of $\overline F$ over $\mathcal A$. [/step]