[proofplan] We prove boundedness of each energy sublevel. The gradient growth lower bound turns an upper bound on $I[u]$ into an upper bound for $\|\nabla u\|_{L^p(U;\mathbb R^n)}$. Writing $u=g+w$ with $w\in W^{1,p}_0(U)$, the Poincare inequality controls $\|w\|_{L^p(U)}$ by $\|\nabla w\|_{L^p(U;\mathbb R^n)}$, and the triangle inequality then controls the full Sobolev norm of $u$. Finally, we prove that this sublevel boundedness is equivalent to the sequential formulation of coercivity. [/proofplan] [step:Turn the growth lower bound into a uniform gradient estimate] Fix $M\in\mathbb R$ and define the sublevel set \begin{align*} S_M:=\{u\in\mathcal A_g:I[u]\le M\}. \end{align*} If $S_M=\varnothing$, then $S_M$ is bounded. Assume $S_M\neq\varnothing$, and let $u\in S_M$. Since $u\in W^{1,p}(U)$, the map \begin{align*} \Phi_u:U\to U\times\mathbb R\times\mathbb R^n,\qquad x\mapsto (x,u(x),\nabla u(x)) \end{align*} is Borel measurable after changing representatives on a set of $\mathcal L^n$-measure zero. Since $f$ is Borel measurable, the function \begin{align*} F_u:U\to\mathbb R,\qquad x\mapsto f(x,u(x),\nabla u(x)) \end{align*} is $\mathcal L^n$-measurable. The lower growth assumption gives \begin{align*} F_u(x)+a(x)\ge \alpha |\nabla u(x)|^p \end{align*} for $\mathcal L^n$-a.e. $x\in U$. Because $a\in L^1(U)$ and $F_u+a$ is bounded below by the nonnegative [measurable function](/page/Measurable%20Function) $\alpha |\nabla u|^p$, the integral defining $I[u]$ is never $-\infty$. Integrating the lower bound over $U$ with respect to $\mathcal L^n$ gives \begin{align*} I[u]\ge \alpha \int_U |\nabla u(x)|^p\,d\mathcal L^n(x)-\int_U a(x)\,d\mathcal L^n(x). \end{align*} Since $I[u]\le M$, we obtain \begin{align*} \int_U |\nabla u(x)|^p\,d\mathcal L^n(x)\le \frac{M+\int_U a(x)\,d\mathcal L^n(x)}{\alpha}. \end{align*} The right-hand side is finite for every $u\in S_M$. Define \begin{align*} G_M:=\left(\max\left\{0,\frac{M+\int_U a(x)\,d\mathcal L^n(x)}{\alpha}\right\}\right)^{1/p}. \end{align*} Then every $u\in S_M$ satisfies \begin{align*} \|\nabla u\|_{L^p(U;\mathbb R^n)}\le G_M. \end{align*} [/step] [step:Use the affine boundary condition and Poincare inequality to control the zero-trace part] Let $u\in S_M$ and define \begin{align*} w:U\to\mathbb R,\qquad w(x):=u(x)-g(x). \end{align*} Since $u\in\mathcal A_g$, we have $w\in W^{1,p}_0(U)$. By the [weak derivative](/page/Weak%20Derivative) linearity in $W^{1,p}(U)$, \begin{align*} \nabla w=\nabla u-\nabla g \end{align*} as elements of $L^p(U;\mathbb R^n)$. Hence the triangle inequality in $L^p(U;\mathbb R^n)$ gives \begin{align*} \|\nabla w\|_{L^p(U;\mathbb R^n)}\le \|\nabla u\|_{L^p(U;\mathbb R^n)}+\|\nabla g\|_{L^p(U;\mathbb R^n)}\le G_M+\|\nabla g\|_{L^p(U;\mathbb R^n)}. \end{align*} We now use the Poincare inequality for zero-trace Sobolev functions on bounded domains. Since $U$ is bounded and $1<p<\infty$, there exists a constant $C_P=C_P(U,p)>0$ such that every $v\in W^{1,p}_0(U)$ satisfies \begin{align*} \|v\|_{L^p(U)}\le C_P\|\nabla v\|_{L^p(U;\mathbb R^n)}. \end{align*} Applying this inequality to $w\in W^{1,p}_0(U)$ yields \begin{align*} \|w\|_{L^p(U)}\le C_P\left(G_M+\|\nabla g\|_{L^p(U;\mathbb R^n)}\right). \end{align*} [guided] The boundary condition is used through the difference between $u$ and the prescribed datum $g$. Define \begin{align*} w:U\to\mathbb R,\qquad w(x):=u(x)-g(x). \end{align*} Because $u\in\mathcal A_g$, the definition of $\mathcal A_g$ says exactly that $u-g\in W^{1,p}_0(U)$, so $w\in W^{1,p}_0(U)$. This is the point at which the Dirichlet condition enters the proof. The gradient estimate from the previous step controls $\nabla u$, but Poincare applies to the zero-trace function $w$, not directly to $u$. We therefore compare their gradients. By linearity of weak derivatives, \begin{align*} \nabla w=\nabla u-\nabla g \end{align*} in $L^p(U;\mathbb R^n)$. Applying the triangle inequality in the [Banach space](/page/Banach%20Space) $L^p(U;\mathbb R^n)$ gives \begin{align*} \|\nabla w\|_{L^p(U;\mathbb R^n)}\le \|\nabla u\|_{L^p(U;\mathbb R^n)}+\|\nabla g\|_{L^p(U;\mathbb R^n)}. \end{align*} Since $u\in S_M$, the previous step gives $\|\nabla u\|_{L^p(U;\mathbb R^n)}\le G_M$, and therefore \begin{align*} \|\nabla w\|_{L^p(U;\mathbb R^n)}\le G_M+\|\nabla g\|_{L^p(U;\mathbb R^n)}. \end{align*} Now we apply the Poincare inequality for zero-trace Sobolev functions on bounded domains. Its hypotheses are satisfied because $U$ is bounded, $1<p<\infty$, and $w\in W^{1,p}_0(U)$. Hence there is a constant $C_P=C_P(U,p)>0$ such that \begin{align*} \|w\|_{L^p(U)}\le C_P\|\nabla w\|_{L^p(U;\mathbb R^n)}. \end{align*} Substituting the bound for $\|\nabla w\|_{L^p(U;\mathbb R^n)}$ gives \begin{align*} \|w\|_{L^p(U)}\le C_P\left(G_M+\|\nabla g\|_{L^p(U;\mathbb R^n)}\right). \end{align*} This is the key mechanism: the growth condition controls only gradients, and Poincare converts the zero-trace gradient control into full $L^p$ control. [/guided] [/step] [step:Bound the full Sobolev norm on each sublevel] Let $u\in S_M$ and let $w=u-g\in W^{1,p}_0(U)$ as above. The triangle inequality in $L^p(U)$ gives \begin{align*} \|u\|_{L^p(U)}\le \|g\|_{L^p(U)}+\|w\|_{L^p(U)}. \end{align*} Using the estimate for $\|w\|_{L^p(U)}$, we obtain \begin{align*} \|u\|_{L^p(U)}\le \|g\|_{L^p(U)}+C_P\left(G_M+\|\nabla g\|_{L^p(U;\mathbb R^n)}\right). \end{align*} Together with \begin{align*} \|\nabla u\|_{L^p(U;\mathbb R^n)}\le G_M, \end{align*} this gives a uniform bound on the norm \begin{align*} \|u\|_{L^p(U)}+\|\nabla u\|_{L^p(U;\mathbb R^n)}. \end{align*} This norm is equivalent to the standard $W^{1,p}(U)$ norm, so there exists a finite constant $B_M>0$, depending only on $M$, $\alpha$, $a$, $g$, $U$, and $p$, such that \begin{align*} \|u\|_{W^{1,p}(U)}\le B_M \end{align*} for every $u\in S_M$. Thus $S_M$ is bounded in $W^{1,p}(U)$. [/step] [step:Deduce the sequential coercivity formulation] We have proved that every sublevel set \begin{align*} S_M=\{u\in\mathcal A_g:I[u]\le M\} \end{align*} is bounded in $W^{1,p}(U)$. Let $(u_k)_{k=1}^{\infty}$ be a sequence in $\mathcal A_g$ such that $\|u_k\|_{W^{1,p}(U)}\to\infty$. Suppose, for contradiction, that $I[u_k]$ does not converge to $+\infty$. Then there exist $M\in\mathbb R$ and a strictly increasing sequence of indices $(k_j)_{j=1}^{\infty}$ such that \begin{align*} I[u_{k_j}]\le M \end{align*} for every $j\ge 1$. Hence $u_{k_j}\in S_M$ for every such $j$, so the sublevel boundedness just proved gives \begin{align*} \sup_{j\ge 1}\|u_{k_j}\|_{W^{1,p}(U)}<\infty. \end{align*} This contradicts $\|u_k\|_{W^{1,p}(U)}\to\infty$. Therefore $I[u_k]\to+\infty$. Conversely, if the sequential formulation holds and some sublevel set $S_M$ were unbounded, then for each positive integer $k$ we could choose $u_k\in S_M$ with \begin{align*} \|u_k\|_{W^{1,p}(U)}\ge k. \end{align*} Then $\|u_k\|_{W^{1,p}(U)}\to\infty$, but $I[u_k]\le M$ for every $k\ge 1$, contradicting the sequential formulation. Thus the sublevel and sequential formulations are equivalent, and the theorem follows. [/step]