[proofplan]
The proof is the direct method. Properness and boundedness below make the infimum finite and allow us to choose a finite-valued minimizing sequence. The coercivity assumption bounds this sequence in $W^{1,p}(U;\mathbb R^m)$; reflexivity and [weak sequential compactness](/theorems/214) then give a weakly convergent subsequence. Sequential weak closedness keeps the limit inside $\mathcal A$, and the assumed sequential weak lower semicontinuity forces the limit to attain the infimum.
[/proofplan]
[step:Choose a finite-valued minimizing sequence with uniformly bounded energy]
Define
\begin{align*}
\alpha:=\inf_{v\in\mathcal A} I[v]\in[-\infty,\infty].
\end{align*}
Since $I$ is bounded below on $\mathcal A$, there exists a constant $b\in\mathbb R$ such that $I[v]\ge b$ for every $v\in\mathcal A$. Hence $\alpha\ge b>-\infty$. Since $I$ is proper on $\mathcal A$, there exists $a\in\mathcal A$ such that $I[a]<\infty$, and therefore $\alpha\le I[a]<\infty$. Thus $\alpha\in\mathbb R$.
For each $j\in\mathbb N$, by the definition of the infimum, choose $u_j\in\mathcal A$ such that
\begin{align*}
I[u_j]\le \alpha+\frac{1}{j}.
\end{align*}
Because $\alpha\le I[u_j]$ for every $j\in\mathbb N$, the sequence $(u_j)_{j=1}^{\infty}$ satisfies
\begin{align*}
\lim_{j\to\infty} I[u_j]=\alpha.
\end{align*}
Moreover,
\begin{align*}
\sup_{j\in\mathbb N} I[u_j]\le \alpha+1<\infty.
\end{align*}
[/step]
[step:Extract a weakly convergent subsequence in the Sobolev space]
By the coercivity hypothesis, the sequence $(u_j)_{j=1}^{\infty}$ is bounded in $W^{1,p}(U;\mathbb R^m)$. Since $1<p<\infty$, the [Sobolev space](/page/Sobolev%20Space) $W^{1,p}(U;\mathbb R^m)$ is reflexive by [citetheorem:8729], applied componentwise to the finite product of scalar Sobolev spaces. By the weak [sequential compactness](/page/Sequential%20Compactness) theorem for bounded sequences in reflexive Banach spaces (citing a result not yet in the wiki: weak sequential compactness in reflexive Banach spaces), there exist a subsequence $(u_{j_k})_{k=1}^{\infty}$ and an element $u_0\in W^{1,p}(U;\mathbb R^m)$ such that
\begin{align*}
u_{j_k}\rightharpoonup u_0
\end{align*}
in $W^{1,p}(U;\mathbb R^m)$ as $k\to\infty$.
[guided]
The purpose of the coercivity assumption is exactly to prevent a minimizing sequence from escaping to infinity in the Sobolev norm. We have already shown that
\begin{align*}
\sup_{j\in\mathbb N} I[u_j]\le \alpha+1<\infty.
\end{align*}
The coercivity hypothesis says that every sequence in $\mathcal A$ with this kind of uniform upper bound on the energy is bounded in $W^{1,p}(U;\mathbb R^m)$. Therefore there exists a constant $C\in[0,\infty)$ such that
\begin{align*}
\|u_j\|_{W^{1,p}(U;\mathbb R^m)}\le C
\end{align*}
for every $j\in\mathbb N$.
Now we need compactness, but not compactness in the strong Sobolev norm. The natural compactness available for bounded sequences in reflexive Banach spaces is weak compactness. Since $1<p<\infty$, [citetheorem:8729] gives reflexivity of the scalar space $W^{1,p}(U)$; applying it to each component and using that a finite product of reflexive Banach spaces is reflexive, we obtain that $W^{1,p}(U;\mathbb R^m)$ is reflexive. The standard weak sequential [compactness theorem](/theorems/2748) for bounded sequences in reflexive Banach spaces then gives a subsequence, which we denote by $(u_{j_k})_{k=1}^{\infty}$, and an element $u_0\in W^{1,p}(U;\mathbb R^m)$ such that
\begin{align*}
u_{j_k}\rightharpoonup u_0
\end{align*}
in $W^{1,p}(U;\mathbb R^m)$ as $k\to\infty$.
This is the central compactness move in the direct method: boundedness comes from coercivity, while existence of a weakly convergent subsequence comes from reflexivity.
[/guided]
[/step]
[step:Use weak closedness to keep the limit admissible]
For every $k\in\mathbb N$, we have $u_{j_k}\in\mathcal A$. Since $\mathcal A$ is sequentially weakly closed in $W^{1,p}(U;\mathbb R^m)$ and
\begin{align*}
u_{j_k}\rightharpoonup u_0
\end{align*}
in $W^{1,p}(U;\mathbb R^m)$, it follows that
\begin{align*}
u_0\in\mathcal A.
\end{align*}
[/step]
[step:Apply weak lower semicontinuity to attain the infimum]
Because $u_0\in\mathcal A$, the definition of $\alpha$ gives
\begin{align*}
\alpha\le I[u_0].
\end{align*}
On the other hand, the assumed sequential weak lower semicontinuity of $I$ on $\mathcal A$ applies to the weakly convergent subsequence $(u_{j_k})_{k=1}^{\infty}$ and gives
\begin{align*}
I[u_0]\le \liminf_{k\to\infty} I[u_{j_k}].
\end{align*}
Since $(u_j)_{j=1}^{\infty}$ is a minimizing sequence with $I[u_j]\to\alpha$, every subsequence has the same limit, so
\begin{align*}
\liminf_{k\to\infty} I[u_{j_k}]=\alpha.
\end{align*}
Combining the two inequalities yields
\begin{align*}
\alpha\le I[u_0]\le \alpha.
\end{align*}
Therefore
\begin{align*}
I[u_0]=\alpha=\inf_{v\in\mathcal A} I[v].
\end{align*}
Thus $u_0$ is a minimizer of $I$ over $\mathcal A$.
[/step]
