[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*}
[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]
[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]
