[proofplan] We argue by contradiction. If a minimizing sequence were unbounded in $X$, then one can extract a subsequence whose $X$-norm tends to infinity. Coercivity forces the values of $I$ along this subsequence to tend to $+\infty$, while the minimizing sequence property forces the same values to converge to $\inf_{\mathcal A} I$, which is strictly below $+\infty$. These two conclusions are incompatible, so the original sequence must be bounded. [/proofplan]

[step:Extract a subsequence whose norms tend to infinity]Define \begin{align*} m:=\inf_{v\in\mathcal A} I[v]. \end{align*} By hypothesis, $m<\infty$. Suppose, for contradiction, that $(u_k)_{k=1}^{\infty}$ is not bounded in $X$. Then no tail of the sequence is bounded; otherwise finitely many initial terms together with a bounded tail would make the whole sequence bounded. We construct a strictly increasing sequence of indices $(k_j)_{j=1}^{\infty}$ as follows. Having chosen $k_1,\dots,k_{j-1}$, the tail $\{u_k:k>k_{j-1}\}$ is unbounded in $X$, so there exists $k_j>k_{j-1}$ such that \begin{align*} \|u_{k_j}\|_X\ge j. \end{align*} Thus $(u_{k_j})_{j=1}^{\infty}$ is a subsequence in $\mathcal A$ and \begin{align*} \|u_{k_j}\|_X\to\infty. \end{align*}[/step]

[guided]Let \begin{align*} m:=\inf_{v\in\mathcal A} I[v]. \end{align*} The hypothesis says $m<\infty$, so the infimum is not $+\infty$. Assume, toward a contradiction, that the minimizing sequence $(u_k)_{k=1}^{\infty}$ is not bounded in $X$. This means that the set of [real numbers](/page/Real%20Numbers) \begin{align*} \{\|u_k\|_X:k\in\mathbb N\} \end{align*} is unbounded above. We need a subsequence whose norms actually tend to infinity, not merely a sequence with occasional large terms. To get this, observe first that every tail must be unbounded: if there were an index $N\in\mathbb N$ and a number $R>0$ such that $\|u_k\|_X\le R$ for every $k\ge N$, then the whole sequence would be bounded by the finite maximum of $R,\|u_1\|_X,\dots,\|u_{N-1}\|_X$. Now construct the subsequence recursively. Choose $k_1$ so that $\|u_{k_1}\|_X\ge 1$. After $k_1,\dots,k_{j-1}$ have been chosen, the tail with indices larger than $k_{j-1}$ is unbounded, so we may choose $k_j>k_{j-1}$ with \begin{align*} \|u_{k_j}\|_X\ge j. \end{align*} The indices are strictly increasing, hence $(u_{k_j})_{j=1}^{\infty}$ is a subsequence of $(u_k)_{k=1}^{\infty}$. Since $\|u_{k_j}\|_X\ge j$ for every $j$, it follows that \begin{align*} \|u_{k_j}\|_X\to\infty. \end{align*} This is the exact form of unboundedness needed to invoke coercivity.[/guided]

[step:Apply coercivity to the unbounded subsequence] The subsequence $(u_{k_j})_{j=1}^{\infty}$ lies in $\mathcal A$ and satisfies $\|u_{k_j}\|_X\to\infty$. By the sequential coercivity of $I$ on $\mathcal A$, \begin{align*} I[u_{k_j}]\to+\infty. \end{align*} [/step]

[step:Use the minimizing sequence property on the same subsequence] Since $(u_k)_{k=1}^{\infty}$ is a minimizing sequence in $\mathcal A$, \begin{align*} I[u_k]\to m. \end{align*} Every subsequence of a convergent sequence in the extended real line has the same limit, so \begin{align*} I[u_{k_j}]\to m. \end{align*} But the previous step gives $I[u_{k_j}]\to+\infty$. Therefore $m=+\infty$ would be forced, contradicting the hypothesis $m<\infty$. [/step]

[step:Conclude that the minimizing sequence is bounded] The contradiction arose from assuming that $(u_k)_{k=1}^{\infty}$ is unbounded in $X$. Hence the sequence is bounded in $X$, meaning there exists $R>0$ such that \begin{align*} \|u_k\|_X\le R \end{align*} for every $k\in\mathbb N$. This proves the theorem. [/step]