[proofplan]
We first show that the infimum cannot be $-\infty$: otherwise a sequence with energies tending to $-\infty$ would be bounded by coercivity, hence would have a weakly convergent subsequence, contradicting weak lower semicontinuity. Once the infimum is known to be finite, we choose a minimizing sequence. Coercivity gives boundedness, reflexivity gives weak subsequential compactness, weak closedness keeps the limit admissible, and weak lower semicontinuity identifies the limit as a minimizer.
[/proofplan]
[step:Rule out the possibility that the infimum is $-\infty$]
Let $\mathbb N=\{1,2,3,\dots\}$ denote the positive integers. Define
\begin{align*}
m:=\inf_{u\in\mathcal A}I[u].
\end{align*}
By hypothesis, $m<\infty$. Suppose, for contradiction, that $m=-\infty$. Then for each $k\in\mathbb N$ there exists $u_k\in\mathcal A$ such that
\begin{align*}
I[u_k]\le -k.
\end{align*}
In particular, $I[u_k]\le 0$ for every $k\in\mathbb N$. By coercivity with $M=0$, there exists $R_0>0$ such that
\begin{align*}
\|u_k\|_X\le R_0
\end{align*}
for every $k\in\mathbb N$.
Since $X$ is reflexive, every bounded sequence in $X$ has a weakly convergent subsequence. Hence there exist a subsequence $(u_{k_j})_{j=1}^{\infty}$ and an element $u\in X$ such that
\begin{align*}
u_{k_j}\rightharpoonup u
\end{align*}
weakly in $X$. Since $\mathcal A$ is sequentially weakly closed and $u_{k_j}\in\mathcal A$ for every $j\in\mathbb N$, we have $u\in\mathcal A$.
Sequential weak lower semicontinuity of $I$ gives
\begin{align*}
I[u]\le \liminf_{j\to\infty}I[u_{k_j}].
\end{align*}
But $I[u_{k_j}]\le -k_j$ and $k_j\to\infty$, so
\begin{align*}
\liminf_{j\to\infty}I[u_{k_j}]=-\infty.
\end{align*}
Thus $I[u]\le -\infty$, impossible because $I$ takes values in $(-\infty,\infty]$. Therefore
\begin{align*}
-\infty<m<\infty.
\end{align*}
[/step]
[step:Choose a minimizing sequence with uniformly bounded energy]
Since $m=\inf_{\mathcal A}I$ is finite, for every $k\in\mathbb N$ there exists $u_k\in\mathcal A$ 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, because
\begin{align*}
m\le I[u_k]\le m+\frac{1}{k}
\end{align*}
for every $k\in\mathbb N$, and hence
\begin{align*}
\lim_{k\to\infty}I[u_k]=m.
\end{align*}
Since $m+1\in\mathbb R$ and $I[u_k]\le m+1$ for every $k\in\mathbb N$, coercivity with $M=m+1$ gives a constant $R_{m+1}>0$ such that
\begin{align*}
\|u_k\|_X\le R_{m+1}
\end{align*}
for every $k\in\mathbb N$.
[guided]
The goal of this step is to convert the variational information $I[u_k]\to m$ into compactness information about the sequence $(u_k)$. Since $m$ is now known to be a real number, the definition of the infimum gives the following: for each $k\in\mathbb N$, there exists $u_k\in\mathcal A$ satisfying
\begin{align*}
I[u_k]\le m+\frac{1}{k}.
\end{align*}
The opposite inequality
\begin{align*}
m\le I[u_k]
\end{align*}
holds because $m$ is the greatest lower bound of all admissible values of $I$. Therefore
\begin{align*}
m\le I[u_k]\le m+\frac{1}{k}.
\end{align*}
By the [squeeze theorem](/theorems/627) for real sequences, this implies
\begin{align*}
\lim_{k\to\infty}I[u_k]=m.
\end{align*}
Now we use coercivity. The definition of coercivity used here says that an upper energy bound forces an $X$-norm bound. Since
\begin{align*}
I[u_k]\le m+1
\end{align*}
for every $k\in\mathbb N$, coercivity applied with the real number $M=m+1$ gives a radius $R_{m+1}>0$ such that every admissible $u\in\mathcal A$ with $I[u]\le m+1$ satisfies
\begin{align*}
\|u\|_X\le R_{m+1}.
\end{align*}
Applying this to each $u_k$ gives
\begin{align*}
\|u_k\|_X\le R_{m+1}
\end{align*}
for every $k\in\mathbb N$. Thus the minimizing sequence is bounded in the [Banach space](/page/Banach%20Space) $X$.
[/guided]
[/step]
[step:Extract a weakly convergent admissible subsequence]
The sequence $(u_k)_{k=1}^{\infty}$ is bounded in $X$. Since $X$ is reflexive, the Eberlein-Smulian [weak sequential compactness](/theorems/214) theorem says that every bounded sequence in $X$ has a weakly convergent subsequence. Hence there exist a subsequence $(u_{k_j})_{j=1}^{\infty}$ and an element $u_*\in X$ such that
\begin{align*}
u_{k_j}\rightharpoonup u_*
\end{align*}
weakly in $X$. Since $u_{k_j}\in\mathcal A$ for every $j\in\mathbb N$ and $\mathcal A$ is sequentially weakly closed, it follows that
\begin{align*}
u_*\in\mathcal A.
\end{align*}
[/step]
[step:Pass to the weak limit using lower semicontinuity]
Since $I$ is sequentially weakly lower semicontinuous on $\mathcal A$, and since $u_{k_j}\rightharpoonup u_*$ weakly in $X$ with $u_{k_j},u_*\in\mathcal A$, we have
\begin{align*}
I[u_*]\le \liminf_{j\to\infty}I[u_{k_j}].
\end{align*}
Because $(u_k)$ 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 $m=\inf_{u\in\mathcal A}I[u]$, so
\begin{align*}
m\le I[u_*].
\end{align*}
Combining the two inequalities gives
\begin{align*}
I[u_*]=m=\inf_{u\in\mathcal A}I[u].
\end{align*}
Thus $I$ attains its minimum on $\mathcal A$ at $u_*$.
[/step]
