[proofplan] We first check that the variational problem has finite infimum by using the upper $p$-growth bound on one admissible competitor. We then take a minimizing sequence and use the lower $p$-growth bound, the affine Dirichlet condition, and Poincare's inequality on $W^{1,p}_0(U)$ to prove boundedness in $W^{1,p}(U)$. Reflexivity gives a weakly convergent subsequence, weak closedness keeps the limit admissible, compact Sobolev embedding gives strong $L^p$ convergence, and convex-integrand lower semicontinuity passes to the liminf. The limit therefore realizes the infimum. [/proofplan]

[step:Show that the infimum is finite] Because $\mathcal A$ is nonempty, choose $v\in\mathcal A$. By hypothesis, $v\in W^{1,p}(U)$ and $\nabla v\in L^p(U;\mathbb R^n)$. Since $f$ is a Caratheodory integrand, the composition \begin{align*} x\mapsto f(x,v(x),\nabla v(x)) \end{align*} is measurable by [citetheorem:8735]. The upper growth bound gives, for $\mathcal L^n$-a.e. $x\in U$, \begin{align*} 0\le f(x,v(x),\nabla v(x))\le C(1+|v(x)|^p+|\nabla v(x)|^p). \end{align*} Since $U$ is bounded, $\mathcal L^n(U)<\infty$, and since $v\in W^{1,p}(U)$, the right-hand side is integrable over $U$. Hence \begin{align*} I[v]<\infty. \end{align*} Also $I[u]\ge 0$ for every $u\in\mathcal A$, because $f$ takes values in $[0,\infty)$. Therefore the number \begin{align*} m:=\inf_{u\in\mathcal A} I[u] \end{align*} is finite and satisfies $0\le m<\infty$. [/step]

[step:Choose a minimizing sequence with uniformly bounded energy] By the definition of the infimum $m$, there exists a sequence $(u_k)_{k=1}^{\infty}$ in $\mathcal A$ such that \begin{align*} I[u_k]\to m. \end{align*} Since $m<\infty$, there exists $M>0$ such that \begin{align*} I[u_k]\le M \end{align*} for every $k\in\mathbb N$. [/step]

[step:Use the coercive lower bound and Poincare's inequality to bound the minimizing sequence in $W^{1,p}(U)$]For each $k\in\mathbb N$, the lower growth bound gives \begin{align*} I[u_k]\ge \alpha\int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)-\int_U a(x)\,d\mathcal L^n(x). \end{align*} Hence \begin{align*} \int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)\le \frac{M+\int_U a(x)\,d\mathcal L^n(x)}{\alpha} \end{align*} for every $k\in\mathbb N$. For each $k\in\mathbb N$, since $u_k\in g+W^{1,p}_0(U)$, define the map $w_k:U\to\mathbb R$ by $w_k(x)=u_k(x)-g(x)$ for $x\in U$. Then $w_k\in W^{1,p}_0(U)$. By Poincare's inequality on the bounded Lipschitz domain $U$, there is a constant $C_P=C_P(U,p)>0$ such that \begin{align*} \|w_k\|_{L^p(U)}\le C_P\|\nabla w_k\|_{L^p(U)} \end{align*} for every $k\in\mathbb N$. Since $\nabla w_k=\nabla u_k-\nabla g$ in $L^p(U;\mathbb R^n)$, the triangle inequality gives \begin{align*} \|\nabla w_k\|_{L^p(U)}\le \|\nabla u_k\|_{L^p(U)}+\|\nabla g\|_{L^p(U)}. \end{align*} Thus $(w_k)_{k=1}^{\infty}$ is bounded in $W^{1,p}(U)$. Since $u_k=g+w_k$, the sequence $(u_k)_{k=1}^{\infty}$ is bounded in $W^{1,p}(U)$.[/step]

[guided]The purpose of this step is to convert an energy bound into a compactness statement. The energy only controls the gradients directly, so we must also recover control of the $L^p$ norms of the functions themselves. The affine boundary condition is exactly what permits this. For each $k\in\mathbb N$, the lower growth assumption says that, for $\mathcal L^n$-a.e. $x\in U$, \begin{align*} f(x,u_k(x),\nabla u_k(x))\ge \alpha|\nabla u_k(x)|^p-a(x). \end{align*} Integrating over $U$ with respect to $\mathcal L^n$ gives \begin{align*} I[u_k]\ge \alpha\int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)-\int_U a(x)\,d\mathcal L^n(x). \end{align*} The integral of $a$ is finite because $a\in L^1(U)$. Since $I[u_k]\le M$, we obtain \begin{align*} \int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x)\le \frac{M+\int_U a(x)\,d\mathcal L^n(x)}{\alpha}. \end{align*} Thus the gradients $\nabla u_k$ are uniformly bounded in $L^p(U;\mathbb R^n)$. This is not yet a $W^{1,p}$ bound, because the Sobolev norm also contains the $L^p$ norm of $u_k$. We now use the Dirichlet structure. For each $k\in\mathbb N$, define the map $w_k:U\to\mathbb R$ by $w_k(x)=u_k(x)-g(x)$ for $x\in U$. Since $u_k\in g+W^{1,p}_0(U)$, we have $w_k\in W^{1,p}_0(U)$. Poincare's inequality on $W^{1,p}_0(U)$ applies because $U$ is bounded and Lipschitz; it gives a constant $C_P=C_P(U,p)>0$ such that \begin{align*} \|w_k\|_{L^p(U)}\le C_P\|\nabla w_k\|_{L^p(U)}. \end{align*} Since $w_k=u_k-g$, its weak gradient is \begin{align*} \nabla w_k=\nabla u_k-\nabla g \end{align*} in $L^p(U;\mathbb R^n)$. The triangle inequality in $L^p(U;\mathbb R^n)$ gives \begin{align*} \|\nabla w_k\|_{L^p(U)}\le \|\nabla u_k\|_{L^p(U)}+\|\nabla g\|_{L^p(U)}. \end{align*} The first term on the right is uniformly bounded by the energy estimate, and the second is fixed because $g\in W^{1,p}(U)$. Therefore $(w_k)$ is bounded in $W^{1,p}(U)$. Finally $u_k=g+w_k$, so adding the fixed function $g$ preserves boundedness. Hence $(u_k)$ is bounded in $W^{1,p}(U)$.[/guided]

[step:Extract a weakly convergent admissible limit] By [citetheorem:8729], $W^{1,p}(U)$ is reflexive because $1<p<\infty$. Since $(u_k)_{k=1}^{\infty}$ is bounded in this reflexive [Banach space](/page/Banach%20Space), there exist a subsequence, not relabelled, and a function $u_*\in W^{1,p}(U)$ such that \begin{align*} u_k\rightharpoonup u_* \end{align*} weakly in $W^{1,p}(U)$. Since every $u_k$ belongs to $\mathcal A$ and $\mathcal A$ is sequentially weakly closed in $W^{1,p}(U)$, it follows that \begin{align*} u_*\in\mathcal A. \end{align*} [/step]

[step:Upgrade the subsequence to strong $L^p(U)$ convergence] By the Rellich-Kondrachov [compactness theorem](/theorems/2748) [citetheorem:8731], the embedding \begin{align*} W^{1,p}(U)\hookrightarrow\hookrightarrow L^p(U) \end{align*} is compact on the bounded Lipschitz domain $U$. Since $(u_k)_{k=1}^{\infty}$ is bounded in $W^{1,p}(U)$, after passing to a further subsequence, still not relabelled, there exists $\tilde u\in L^p(U)$ such that \begin{align*} u_k\to \tilde u \end{align*} strongly in $L^p(U)$. The same subsequence also converges weakly to $u_*$ in $W^{1,p}(U)$, hence weakly to $u_*$ in $L^p(U)$ under the continuous embedding $W^{1,p}(U)\hookrightarrow L^p(U)$. Strong $L^p$ convergence implies weak $L^p$ convergence to the same limit, so uniqueness of weak limits gives $\tilde u=u_*$ in $L^p(U)$. Therefore \begin{align*} u_k\to u_* \end{align*} strongly in $L^p(U)$. [/step]

[step:Apply convex-integrand lower semicontinuity to identify the minimizer] The functions $u_k$ converge strongly to $u_*$ in $L^p(U)$, and their weak gradients satisfy \begin{align*} \nabla u_k\rightharpoonup \nabla u_* \end{align*} weakly in $L^p(U;\mathbb R^n)$ because $u_k\rightharpoonup u_*$ in $W^{1,p}(U)$. The upper $p$-growth bound and the inclusion $u_k,u_*\in W^{1,p}(U)$ imply that $x\mapsto f(x,u_k(x),\nabla u_k(x))$ and $x\mapsto f(x,u_*(x),\nabla u_*(x))$ are integrable, while nonnegativity gives $I[u_k],I[u_*]\in[0,\infty)$. The lower semicontinuity theorem for convex scalar integrands [citetheorem:8738] applies precisely to a nonnegative Caratheodory integrand with upper $p$-growth, convexity in the gradient variable, strong $L^p$ convergence of the scalar fields $u_k\to u_*$, and weak $L^p$ convergence of the gradients $\nabla u_k\rightharpoonup\nabla u_*$. Hence it yields \begin{align*} I[u_*]\le \liminf_{k\to\infty} I[u_k]. \end{align*} Since $(u_k)$ is minimizing, $\lim_{k\to\infty}I[u_k]=m$. Thus \begin{align*} I[u_*]\le m. \end{align*} Because $u_*\in\mathcal A$ and $m=\inf_{u\in\mathcal A}I[u]$, we also have \begin{align*} m\le I[u_*]. \end{align*} Consequently \begin{align*} I[u_*]=m=\inf_{u\in\mathcal A}I[u], \end{align*} so $I$ attains its minimum on $\mathcal A$. [/step]