People familiar with the direct methods of the Calculus of Variations probably know the following
Theorem: Let $X$ be a reflexive Banach space and let $J: X \longrightarrow \mathbb{R}$ be coercive and weakly lower semicontinuous. Then $J$ is bounded from below, and in fact the minimum is attained.
The proof of the boundedness goes as follows:
Consider a sequence $u_n$ in $X$ such that $J(u_n) \to -\infty$. By coercivity, the sequence must be bounded. Since $X$ is reflexive, it converges, up to a subsequence, to some $u \in X$. By weakly lower semicontinuity, $$ J(u) \leq \lim \inf J(u_n) = - \infty, $$ a contradiction.
I would like to have an analogous theorem for functionals constrained to manifolds where the constraint is something of the form $$ M = G^{-1}(0) $$ where $G \in C^1(X, \mathbb{R})$ and $G'(u) \neq 0 \ \forall u \in M$.
My question is:
Do we have the existence of the weak limit $u$ in the case that the sequence is restricted to the manifold $M$? How can we prove that?
I think I got this. The key fact is that we must impose that $M$ is weakly closed.
Let $(u_n) \subset M$ be a sequence such that $J(u_n) \to - \infty$. By coercivity of $J$ on $M$, we have that the sequence $u_n$ must be bounded. Hence, by reflexivity, it converges weakly to some $u \in X$. Since $M$ is weakly closed, $u \in M$. From here the argument by contradiction is the same as in the question.