Coercive functionals and minimizing sequences

Question

My course notes provide just a brief mention to a common technique used in calculus of variations, but I can't understand it.

Don't expect total rigor, since it's just a mention.

Let $G:X\to \overline{\mathbb{R}}$ be a function where $X$ is a sufficiently regular function space. My notes define $G$ to be coercive iff its sublevels are precompact (I think that this means that its sublevels have compact closure).

My notes then affirm that if $G$ is coercive and lower semicontinous then the accumulation points of minimizing sequences (i.e. sequences $\{x_n\}$ in $X$ such that $G(x_n)\to \inf \ G$) are minimum points. Is this trivial? Because I don't see it.

Moreove is it assured that at least a minimizing sequence has an accumulation point?

I have never studied calculus of variations, so please try to be basic.

Answer 1

Set $m = \inf_{x \in X} G(x)$. By the definition of infimum, we can select $x_n \to x$ such that $G(x_n) \to m$. Since it is coercive, so there exists a subsequence of $\{x_n\}$ such that it is convergent and we still denote this subsequence by $x_n$. Suppose $x_n \to x_0$.

Moreover, $G$ is lower semicontinuous implies $G(x_0) \leq \liminf_{x \to x_0} G(x) \leq \lim_{k \to \infty} G(x_k) = m$.

On the other hand, $m = \inf_{x \in X} G(x) \geq G(x_0)$. So $G(x_0) = m$.

Answer 2

Premise. While I was working on the following one, a previously given answer was accepted: nevertheless I hope that mine could still be useful to someone.

The result is apparently well known though not trivial: it basically is the consequence of two facts

1. First recall the following propertiy of lower semicontinuous functions:
   Theorem (theorem 4 in [2]). Let $X$ be a topological space. A function $G:X \to \overline{\Bbb R}$ is lower semicontinuous if and only if its sublevel $$ G^{-1}(]-\infty, a])=\{x\in X \mid G(x)\le a\}$$ is closed for every $a\in\overline{\Bbb R}$.
   The proof is a very simple consequence of definition of lower semicontinuity.
   This property, jointly with the precompactness of the sublevels $G^{-1}(]-\infty, a])$ due to coercivity, implies directly that these sets are compact (since the closure of a closed set is the set itself).

2. As @Ning wrote in is answer, due to the very definition of $\inf_{x\in X} G(x)$, we can define a minimizing sequence $\{x_n\}$ conveying to it and, due to the compactness of the sub level sets, we have that $\lim_{n\to\infty} x_n=x\in X$.

Final note: The route for proving this result I followed here was inspired by the informal notes given in [1], chapter 2, §2.1, pp. 9-12. The authors deal with this problem in a less general setting (they assume in the end that $X$ is a Banach space) but nevertheless they show how semicontinuity, the structure of sublevels and the coercivity, defined in the traditional way as it is customary for Banach spaces, i.e. $$ \lim_{\substack{\|x\|_{X}\to\infty\\ x\in X}}\frac{G(x)}{\|x\|_{X}}=\infty $$ of the given functional interact in order to prove an existence theorem.

References

[1] Claudio Baiocchi, António Capelo, Variational and quasivariational inequalities. Applications to free boundary problems, translated from the Italian by Lakshmi Jayakar, a Wiley-Interscience publication. Chichester-New York-Brisbane-Toronto-Singapore: John Wiley and Sons, pp. ix+452 (1984), MR745619, Zbl 0551.49007.

[2] I. A. Vinogradova (originator), "Semicontinuous function". Encyclopedia of Mathematics, ISBN 1402006098.