[proofplan] We prove the result by the direct method. First we verify that the infimum is a finite real number, using nonemptiness of $\mathcal A$ and the upper $p$-growth bound. Then we choose a minimizing sequence, use coercivity to obtain boundedness in $W^{1,p}(U)$, extract a weakly convergent subsequence by reflexivity and [weak sequential compactness](/theorems/214), and use weak closedness to keep the limit admissible. Finally, sequential weak lower semicontinuity identifies the weak limit as a minimizer. [/proofplan]

[step:Verify that the functional has a finite infimum] Let $m\in[0,\infty]$ denote the infimum \begin{align*} m:=\inf_{v\in\mathcal A} I[v]. \end{align*} Since $f$ takes values in $[0,\infty)$, we have $I[v]\ge 0$ for every $v\in\mathcal A$, and therefore $m\ge 0$. Because $\mathcal A$ is nonempty, choose $w\in\mathcal A$. Since $w\in W^{1,p}(U)$, the representatives $w:U\to\mathbb R$ and $\nabla w:U\to\mathbb R^n$ are measurable. By [citetheorem:8735], the map $x\mapsto f(x,w(x),\nabla w(x))$ is measurable. By the upper $p$-growth hypothesis, \begin{align*} 0\le f(x,w(x),\nabla w(x))\le C(1+|w(x)|^p+|\nabla w(x)|^p) \end{align*} for $\mathcal L^n$-a.e. $x\in U$. Since $U$ is bounded, $\mathcal L^n(U)<\infty$, and since $w\in W^{1,p}(U)$, the right-hand side is integrable over $U$ with respect to $\mathcal L^n$. Hence $I[w]<\infty$, so $m\le I[w]<\infty$. Thus \begin{align*} 0\le m<\infty. \end{align*} [/step]

[step:Choose a minimizing sequence and use coercivity to make it bounded]Let $\mathbb N=\{1,2,3,\dots\}$ denote the set of positive integers. For each $k\in\mathbb N$, choose $u_k\in\mathcal A$ such that \begin{align*} I[u_k]\le m+\frac{1}{k}. \end{align*} The sequence $(u_k)_{k=1}^{\infty}$ is a minimizing sequence in $\mathcal A$, meaning that \begin{align*} \lim_{k\to\infty} I[u_k]=m. \end{align*} In particular, the sequence $(I[u_k])_{k=1}^{\infty}$ is bounded in $\mathbb R$. If $(u_k)_{k=1}^{\infty}$ were unbounded in $W^{1,p}(U)$, then a subsequence would have $W^{1,p}(U)$-norm tending to infinity, and coercivity of $I$ on $\mathcal A$ would force the corresponding energies to tend to infinity. This contradicts boundedness of $(I[u_k])_{k=1}^{\infty}$. Therefore $(u_k)_{k=1}^{\infty}$ is bounded in $W^{1,p}(U)$.[/step]

[guided]The purpose of this step is to turn the variational information $I[u_k]\to m$ into compactness information about the functions $u_k$. Let $\mathbb N=\{1,2,3,\dots\}$ denote the set of positive integers. Since $m<\infty$, for every $k\in\mathbb N$ we may choose $u_k\in\mathcal A$ satisfying \begin{align*} I[u_k]\le m+\frac{1}{k}. \end{align*} This is exactly the defining property of the infimum: $m$ is the greatest lower bound of the set $\{I[v]:v\in\mathcal A\}$, so values of $I$ can be found arbitrarily close to $m$ from above. Because $I[u_k]\ge m$ for every $k$, we have \begin{align*} m\le I[u_k]\le m+\frac{1}{k}. \end{align*} The [squeeze theorem](/theorems/627) for real sequences gives \begin{align*} \lim_{k\to\infty} I[u_k]=m. \end{align*} Thus the energies are bounded. Now we use coercivity. Coercivity on $\mathcal A$ means that no sequence in $\mathcal A$ can escape to infinity in the $W^{1,p}(U)$ norm while keeping bounded energy. Suppose, toward a contradiction, that $(u_k)_{k=1}^{\infty}$ were unbounded in $W^{1,p}(U)$. Then there would be a strictly increasing map $j:\mathbb N\to\mathbb N$ such that \begin{align*} \|u_{j(k)}\|_{W^{1,p}(U)}\to\infty. \end{align*} Since each $u_{j(k)}$ belongs to $\mathcal A$, coercivity gives \begin{align*} I[u_{j(k)}]\to\infty. \end{align*} But $(I[u_k])_{k=1}^{\infty}$ is bounded because it converges to the finite number $m$. This contradiction proves that $(u_k)_{k=1}^{\infty}$ is bounded in $W^{1,p}(U)$.[/guided]

[step:Extract a weakly convergent subsequence in $W^{1,p}(U)$]By [citetheorem:8729], the space $W^{1,p}(U)$ is reflexive because $1<p<\infty$ and $U\subset\mathbb R^n$ is open. Since $(u_k)_{k=1}^{\infty}$ is bounded in the reflexive [Banach space](/page/Banach%20Space) $W^{1,p}(U)$, the weak [sequential compactness](/page/Sequential%20Compactness) of bounded sets in reflexive Banach spaces gives 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*}[/step]

[guided]The compactness step takes place in the Banach space $W^{1,p}(U)$ with its usual norm. By [citetheorem:8729], this space is reflexive because $U\subset\mathbb R^n$ is open and $1<p<\infty$. The preceding step proved that the minimizing sequence $(u_k)_{k=1}^{\infty}$ is bounded in $W^{1,p}(U)$. In a reflexive Banach space, every bounded sequence has a weakly convergent subsequence by weak sequential compactness of bounded sets. Applying this compactness result to $(u_k)_{k=1}^{\infty}$, there are a strictly increasing index map $j\mapsto k_j$ 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*} This is the point where the restriction $1<p<\infty$ is used: it gives reflexivity of the [Sobolev space](/page/Sobolev%20Space), and reflexivity is the source of weak compactness for bounded minimizing sequences.[/guided]

[step:Use weak closedness to show the weak limit is admissible]Since $u_{k_j}\in\mathcal A$ for every $j\in\mathbb N$, since $u_{k_j}\rightharpoonup u_*$ in $W^{1,p}(U)$, and since $\mathcal A$ is sequentially weakly closed in $W^{1,p}(U)$, it follows that \begin{align*} u_*\in\mathcal A. \end{align*}[/step]

[guided]The subsequence found in the compactness step converges weakly in $W^{1,p}(U)$, but compactness alone only gives the limit $u_*\in W^{1,p}(U)$. We must check that this limit is still admissible. For every $j\in\mathbb N$, the element $u_{k_j}$ belongs to $\mathcal A$ because it is a subsequence of the minimizing sequence chosen inside $\mathcal A$. The hypothesis that $\mathcal A$ is sequentially weakly closed in $W^{1,p}(U)$ says precisely that whenever a sequence in $\mathcal A$ converges weakly in $W^{1,p}(U)$, its weak limit belongs to $\mathcal A$. Applying this hypothesis to the sequence $(u_{k_j})_{j=1}^{\infty}$ and the [weak convergence](/page/Weak%20Convergence) \begin{align*} u_{k_j}\rightharpoonup u_* \quad \text{in } W^{1,p}(U), \end{align*} we obtain \begin{align*} u_*\in\mathcal A. \end{align*}[/guided]

[step:Apply weak lower semicontinuity to identify the minimizer] The sequential weak lower semicontinuity of $I$ on $\mathcal A$ applies to the sequence $(u_{k_j})_{j=1}^{\infty}$ and its weak limit $u_*\in\mathcal A$. Hence \begin{align*} I[u_*]\le \liminf_{j\to\infty} I[u_{k_j}]. \end{align*} Because $(u_k)_{k=1}^{\infty}$ is minimizing and every subsequence of a convergent real sequence has the same limit, \begin{align*} \lim_{j\to\infty} I[u_{k_j}]=m. \end{align*} Therefore \begin{align*} I[u_*]\le m. \end{align*} On the other hand, $u_*\in\mathcal A$ and $m=\inf_{v\in\mathcal A}I[v]$, so \begin{align*} m\le I[u_*]. \end{align*} Combining the two inequalities gives \begin{align*} I[u_*]=m=\inf_{v\in\mathcal A}I[v]. \end{align*} Thus $I$ attains its minimum on $\mathcal A$ at $u_*$. [/step]