I am trying to prove the following:
Let $X $be a reflexive Banach space and $J : X \to \mathbb R$ be a functional such that
$\forall (u_m)_{m\in\mathbb N} \subseteq X$ such that $\lim_{m\to \infty} ||u_m|| = \infty$ it follows that $\lim_{m\to \infty} J(u_m) = \infty$ $\tag 1$
Then, $J$ is sequentially weakly coercive.
My try:
-The definition of weakly coercive is that $\forall \alpha \in \mathbb R \forall (u_m)\subseteq X$ such that $J(u_m)\le \alpha$ implies $\exists u_{m_k} \rightharpoonup u $
-The contrapositive of stament (1):
$\text {If } \lim_{m\to \infty} J(u_m) < \infty \implies \lim_{m\to \infty} ||u_m|| < \infty \tag{2}$
Therefore let the hypothesis of the exercise hold and let $\alpha \in \mathbb R$ and $(u_m)\subseteq X$ such that
$J(u_m)\le \alpha$
Taking the limit:
$\lim_{m\to \infty} J(u_m)\le \alpha$ which implies by (2), that $ L:=\lim_{m\to \infty} ||u_m|| < \infty $. This means that
$\forall \epsilon >0, \exists M(\epsilon) $ such that $\forall m>M(\epsilon)$, $| ||u_m|| -L|<\epsilon$
$ \implies ||u_m|| <\epsilon +|L|$, which means that $(u_m)$ is bounded. By a Corollary of Banach-Alaoglu, every bounded sequence in a reflexive Banach space has a convergent subsequence, so I have the thesis.
1. Is this correct? If not, please enlighten me.
2. I am specially suspicious about taking the limit of $J(u_m)$ because I don't know if that limit exists, so I should just take a liminf or limsup I guess? But then I wouldn't know how to proceed.
So, here a short answer. A priori it is not known if $\lim_{m\rightarrow\infty} J(u_m)$ actually exists. But since $J(u_m) \leq \alpha$ for all $m\in\mathbb{N}$, $(J(u_m))_{\in\mathbb{N}}$ is a bounded sequence in $\mathbb{R}$ and we may use Bolzano-Weiterstraß to extract a converging subsequence.
Regarding your last comment, note that the contrapositive of a statement is not the same as the negation.