# coercive implies bounded, variational principle

I have a question about the proof of the variational principle, see below. Any help is much appreciated!

• How does it follow from coercivity that $(x_k)_k$ is bounded?

• Why is $\alpha_0 > - \infty$?

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### Prerequisites

Let $(X, || · ||_X )$ a normed real vector space, $\, M \subset X$, $\, F : M \rightarrow \mathbb R.$

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Theorem. (Eberlein-Šmulyan) Let $X$ be reflexive, $(x_k)_{k \in \mathbb N} \subset X$ bounded. Then there exists $x \in X$ and a subsequence $\Lambda \subset \mathbb N$ with $$x_k \overset{w}{\rightarrow} x \quad (\text{for } \, k → ∞ , \, k \in \Lambda) .$$

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Definition. The function $F$ ist weakly sequentially lower semi-continuous (w.s.l.s.c.) at $x_0 \in M$, if $\forall \, (x_k)_{k\in \mathbb N} \subset M$ with $x_k \overset{w}{\rightarrow} x_0$ (for $k \rightarrow \infty$) there holds

$$F(x_0) ≤ \liminf_{k \rightarrow \infty} F(x_k).$$

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Definition. $F$ is called coercive on $M$ with respect to $||·||_X$ , if for $x \in M$ $$F(x) \rightarrow \infty, \quad (\text{for} \, ||x||_X \rightarrow \infty).$$

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### Actual question

Theorem. (Variational Principle) Let $X$ be reflexive, $M \subset X$ nonempty and weakly sequentially closed, $F : M \rightarrow R$ coercive and w.s.l.s.c.. Then there exists $x_0 \in M$ with $$F(x_0) = \inf_{x∈M} F(x).$$

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Proof. Consider a minimal sequence $(x_k)_{k ∈ \mathbb N} \subset M$ with $$F(x_k)→ \inf_{x \in M} F(x)=:α_0 ≥ −∞, \quad (\text{for } k→∞).$$

Since $F$ is coercive, $(x_k)_{k \in \mathbb N}$ is bounded. (Why?) By Eberlein-Šmulyan's theorem $(x_k)_{k \in \mathbb N}$ has a weakly convergent subsequence $x_k \overset{w}{\rightarrow} x_0$ $($for $k → ∞, k ∈ Λ)$. Since $M$ is weakly sequentially closed, it follows that $x_0 ∈ M$, and $$F(x_0) ≤ \liminf_{k→∞, \, k∈Λ} F(x_k) = α_0$$ since $F$ is w.s.l.s.c., in particular we have $α_0 > −∞.$

## 1 Answer

Because if $(x_k)_{k\in\mathbb N}$ weren't bounded, we would have $\lim_{k\to\infty} F(x_k) = \infty$, by definition of the coercivity, thus contradicting the convergence to the infimum, no?

• how do you explain that $\alpha_0 > - \infty$? – cesare borgia Jan 22 '17 at 16:53
• Because $F: M \rightarrow \mathbb R$, so $F(x_0) \in \mathbb R$, and you show that $F(x_0) \leq \alpha_0$. – Roberto Rastapopoulos Jan 24 '17 at 11:33
• @RobertoRastapopoulos I know it's been a while since you answered this question, but I found your post very helpful and I was wondering if you could explain to me one more thing of the above proof. Why does $\;\liminf_{k→∞, \, k∈Λ} F(x_k) = α_0\;$ hold? Thanks in advance – kaithkolesidou Apr 13 '17 at 15:50
• That's just because $\lim_{k\to \infty} F(x_k) = \alpha_0$. The subsequence of a convergent sequence converges to the same limit, and $\liminf$ and $\limsup$ and $\lim$ coincide for convergent sequences. – Roberto Rastapopoulos Aug 30 '18 at 13:51