Coercivity bounds every finite-level sequence

Bounded sublevels. Let $\mathrm{\Lambda }\in \mathbb{R}$. By [F1] the sublevel set $\{u\in A:I(u)\le \mathrm{\Lambda }\}$ is bounded in the norm of $X$; that is, there is $R\ge 0$ with $\mathrm{\parallel }u\mathrm{\parallel }\le R$ for every $u\in A$ with $I(u)\le \mathrm{\Lambda }$.

Sequences with finite sup of values are bounded. Let $(u_j)\subseteq A$ satisfy ${\sup }_{j}I(u_j)=:{\mathrm{\Lambda }}_0<+\mathrm{\infty }$. Then ${\mathrm{\Lambda }}_0\in \mathbb{R}$ and $I(u_j)\le {\mathrm{\Lambda }}_0$ for every $j$, so the whole sequence lies in the sublevel set $\{I\le {\mathrm{\Lambda }}_0\}$, which is bounded by step 1.1; hence $(u_j)$ is norm bounded.

Minimising sequences with finite infimum. Let $(u_j)\subseteq A$ be a minimising sequence, $I(u_j)\to {\inf }_{A}I$, with ${\inf }_{A}I<+\mathrm{\infty }$. By [F2] there are $N$ and a real $M$ with $I(u_j)\le M$ for every $j\ge N$. Step 2.1 shows that the tail $(u_j{)}_{j\ge N}$ is norm bounded. The finite set of initial vectors $u_1,\dots ,u_{N-1}$ is also norm bounded, so the entire sequence is norm bounded, even if some initial functional values equal $+\mathrm{\infty }$.