[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 $k\in\mathbb N$.[/step]

[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: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]