[proofplan] The proof converts the lower $p$-growth of the elastic energy into a lower bound for $I[u]$ in terms of $\|\nabla u\|_{L^p}^p$. The fixed-trace Poincare estimate then converts gradient control into control of the full $W^{1,p}$ norm. The subcritical loading term has order $q<p$, so it cannot destroy the resulting coercivity. Since a minimizing sequence has energy values bounded above, coercivity forces its Sobolev norms to remain bounded. [/proofplan]

[step:Convert the elastic lower growth into a lower bound for the functional]Define the Sobolev norm map \begin{align*} N:\mathcal A\to[0,\infty) \end{align*} by \begin{align*} N(u):=\|u\|_{W^{1,p}(\Omega;\mathbb R^n)} \end{align*} and define the gradient norm map \begin{align*} G:\mathcal A\to[0,\infty) \end{align*} by \begin{align*} G(u):=\|\nabla u\|_{L^p(\Omega;\mathbb R^{n\times n})}. \end{align*} Let $u\in\mathcal A$ satisfy $I[u]<\infty$. Since the lower bound for $W$ holds for $\mathcal L^n$-a.e. $x\in\Omega$, with $y=u(x)$ and $F=\nabla u(x)$, integration over $\Omega$ gives \begin{align*} \int_\Omega W(x,u(x),\nabla u(x))\,d\mathcal L^n(x)\ge c\int_\Omega |\nabla u(x)|^p\,d\mathcal L^n(x)-C\mathcal L^n(\Omega). \end{align*} By the definition of $G(u)$, this is \begin{align*} \int_\Omega W(x,u(x),\nabla u(x))\,d\mathcal L^n(x)\ge cG(u)^p-C\mathcal L^n(\Omega). \end{align*} Using the estimate $-\ell[u]\ge-|\ell[u]|$ and the growth hypothesis on $\ell$, we obtain \begin{align*} I[u]\ge cG(u)^p-aN(u)^q-b-C\mathcal L^n(\Omega). \end{align*}[/step]

[guided]The first task is to isolate the part of the energy that controls derivatives. We introduce two maps on the admissible class. Define \begin{align*} N:\mathcal A\to[0,\infty) \end{align*} by \begin{align*} N(u):=\|u\|_{W^{1,p}(\Omega;\mathbb R^n)} \end{align*} and define \begin{align*} G:\mathcal A\to[0,\infty) \end{align*} by \begin{align*} G(u):=\|\nabla u\|_{L^p(\Omega;\mathbb R^{n\times n})}. \end{align*} The value $N(u)$ measures the full Sobolev size of $u$, while $G(u)$ measures only the deformation gradient. Fix $u\in\mathcal A$ with $I[u]<\infty$. The hypothesis on $W$ says that for $\mathcal L^n$-a.e. $x\in\Omega$, every $y\in\mathbb R^n$, and every $F\in\mathbb R^{n\times n}$, \begin{align*} W(x,y,F)\ge c|F|^p-C. \end{align*} We apply this pointwise inequality with $y=u(x)$ and $F=\nabla u(x)$. Since $u\in W^{1,p}(\Omega;\mathbb R^n)$, the weak gradient $\nabla u:\Omega\to\mathbb R^{n\times n}$ belongs to $L^p(\Omega;\mathbb R^{n\times n})$, so the function $x\mapsto |\nabla u(x)|^p$ is integrable with respect to $\mathcal L^n$. Integrating the pointwise lower bound over $\Omega$ gives \begin{align*} \int_\Omega W(x,u(x),\nabla u(x))\,d\mathcal L^n(x)\ge c\int_\Omega |\nabla u(x)|^p\,d\mathcal L^n(x)-C\mathcal L^n(\Omega). \end{align*} The boundedness of $\Omega$ implies $\mathcal L^n(\Omega)<\infty$, so the constant term is finite. By the definition of $G(u)$, \begin{align*} \int_\Omega |\nabla u(x)|^p\,d\mathcal L^n(x)=G(u)^p. \end{align*} Therefore \begin{align*} \int_\Omega W(x,u(x),\nabla u(x))\,d\mathcal L^n(x)\ge cG(u)^p-C\mathcal L^n(\Omega). \end{align*} Now include the loading term. Since \begin{align*} -\ell[u]\ge-|\ell[u]| \end{align*} and the hypothesis gives \begin{align*} |\ell[u]|\le aN(u)^q+b, \end{align*} the definition of $I[u]$ yields \begin{align*} I[u]\ge cG(u)^p-C\mathcal L^n(\Omega)-aN(u)^q-b. \end{align*} Reordering the terms gives the lower bound \begin{align*} I[u]\ge cG(u)^p-aN(u)^q-b-C\mathcal L^n(\Omega). \end{align*} This is the main energetic estimate: the positive term has order $p$ in the gradient, while the negative loading term has only order $q<p$ in the full Sobolev norm.[/guided]

[step:Use the fixed trace to replace gradient control by full Sobolev control] The trace-Poincare estimate gives \begin{align*} N(u)\le KG(u)+D \end{align*} for every $u\in\mathcal A$. Hence, if $N(u)\ge 2D$, then \begin{align*} N(u)-D\ge \frac{1}{2}N(u). \end{align*} Since $N(u)\le KG(u)+D$, we have \begin{align*} KG(u)\ge N(u)-D\ge \frac{1}{2}N(u). \end{align*} Thus, whenever $N(u)\ge 2D$, \begin{align*} G(u)^p\ge \frac{1}{2^pK^p}N(u)^p. \end{align*} Substituting this into the preceding lower bound gives \begin{align*} I[u]\ge \frac{c}{2^pK^p}N(u)^p-aN(u)^q-b-C\mathcal L^n(\Omega) \end{align*} for every $u\in\mathcal A$ with $I[u]<\infty$ and $N(u)\ge 2D$. [/step]

[step:Show that the polynomial lower bound is coercive] Define the constants \begin{align*} \alpha:=\frac{c}{2^pK^p} \end{align*} and \begin{align*} \gamma:=b+C\mathcal L^n(\Omega). \end{align*} Then $\alpha>0$ and $\gamma\ge 0$, and the previous step says that \begin{align*} I[u]\ge \alpha N(u)^p-aN(u)^q-\gamma \end{align*} whenever $u\in\mathcal A$, $I[u]<\infty$, and $N(u)\ge 2D$. Because $q<p$, the scalar function \begin{align*} \Phi:[0,\infty)\to\mathbb R \end{align*} defined by \begin{align*} \Phi(t):=\alpha t^p-at^q-\gamma \end{align*} satisfies $\Phi(t)\to\infty$ as $t\to\infty$. Indeed, if $a=0$, then $\Phi(t)=\alpha t^p-\gamma\to\infty$. If $a>0$, then \begin{align*} \Phi(t)=t^p\left(\alpha-at^{q-p}\right)-\gamma. \end{align*} Since $q-p<0$, we have $t^{q-p}\to 0$ as $t\to\infty$, so there exists $T_0\ge 1$ such that \begin{align*} at^{q-p}\le \frac{\alpha}{2} \end{align*} for every $t\ge T_0$. Hence \begin{align*} \Phi(t)\ge \frac{\alpha}{2}t^p-\gamma \end{align*} for every $t\ge T_0$, and the right-hand side tends to $\infty$ as $t\to\infty$. [/step]

[step:Apply coercivity to an arbitrary minimizing sequence] Let $(u_k)_{k=1}^{\infty}$ be a minimizing sequence in $\mathcal A$ such that $I[u_k]<\infty$ for all sufficiently large $k$. Define \begin{align*} m:=\inf_{v\in\mathcal A}I[v]. \end{align*} By hypothesis, $m<\infty$. If $m\in\mathbb R$, then the definition of a minimizing sequence gives $I[u_k]\to m$ as $k\to\infty$, so the tail of $(I[u_k])$ is bounded above. If $m=-\infty$, then the definition of a minimizing sequence gives $I[u_k]\to-\infty$, so the tail of $(I[u_k])$ is again bounded above. In either case, there exist an index $k_0\in\mathbb N$ and a constant $M\in\mathbb R$ such that \begin{align*} I[u_k]\le M \end{align*} for every $k\ge k_0$. Suppose, for contradiction, that $(u_k)$ is not bounded in $W^{1,p}(\Omega;\mathbb R^n)$. Then the sequence $(N(u_k))$ is unbounded, so there is a subsequence $(u_{k_j})_{j=1}^{\infty}$ such that \begin{align*} N(u_{k_j})\to\infty \end{align*} as $j\to\infty$. For all sufficiently large $j$, we have $k_j\ge k_0$, $I[u_{k_j}]<\infty$, and $N(u_{k_j})\ge 2D$. The coercive lower bound therefore gives \begin{align*} I[u_{k_j}]\ge \Phi(N(u_{k_j})). \end{align*} Since $\Phi(t)\to\infty$ as $t\to\infty$, it follows that \begin{align*} I[u_{k_j}]\to\infty \end{align*} as $j\to\infty$. This contradicts $I[u_{k_j}]\le M$ for all sufficiently large $j$. Hence $(N(u_k))$ is bounded, which is exactly boundedness of $(u_k)$ in $W^{1,p}(\Omega;\mathbb R^n)$. [/step]