How do I prove these results involving the liminf and the inf ? ( part of a proof on sequential lower semicontinuity of lower semicontinuous envelope)

Question

I am trying to prove equation (6.3) in the lemma below. This is part of a course in calculus of variations, but that is irrelevant here, this is actually a question about the liminf and the inf of sequences

Definition Consider the definition Let (X,d) be a metric space, $F:X \to \overline{ \mathbb{R}}$.

Then $\overline F:X \to \overline{ \mathbb{R}}$ is the sequential lower semicontinuous envelope of F, defined as: $$\overline F(x):=\inf \{\liminf_{n\rightarrow \infty}F(x_n):x_n\rightarrow x\}\tag{6.1}$$

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Some hints. My lecturer explained it to me in the following way:

Consider an increment $\overline F(x) +\frac{1}{2n}$ and by definition (6.1)

$\exists (y_k^n)_k $ such that $\liminf_{k\rightarrow \infty}F(y_k^n)\le\overline F(x_n) +\frac{1}{2n}$ and $y_k^n\to x_n,$ for $k\to +\infty$

$\exists y_k^n$ such that $$i) d(y_k^n, x_n)< \frac{1}{n} $$ $$ii) F(y_k^n)\le \overline F(x_n)+\frac{1}{n} $$

then let $y_n:=y_k^n$

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But it is still not complete or clear for me, so I am trying to fill in the details to get these two equations and I have little experience with proofs in analysis

Now what I've tried:

By definition of inf in (6.1), we have that: $\forall \varepsilon >0, \exists (y_k^n)_k $ (and taking $\varepsilon=\frac{1}{2n}, n\in \mathbb{N}$) such that

$\liminf_{k\rightarrow \infty}F(y_k^n)\le\overline F(x_n) +\frac{1}{2n}$ and $y_k^n\to x_n,$ for $k\to +\infty$

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EDIT: the initial part of the prove in the picture in more detail:

Assume $F(x)>-\infty$. Take $x_n \rightarrow x$. Since I want to prove that $\overline F(x)\le \liminf_k\overline F(x_k)$, I evaluate $\overline F$ using the definition (6.1 ) at $x_n$:

$\overline F(x_n) = \inf \{ \liminf_k F(z_k): z_k \rightarrow x_n\}$ By definition of inf we have that: $\forall \varepsilon >0, \exists (z_k^\varepsilon)_k $ such that

$\liminf_kF(z_k^\varepsilon)<\overline F(x_n) +\varepsilon$ and $z_k^\varepsilon\to x_n,$ for $k\to +\infty$

Taking $\varepsilon=\frac{1}{n}, n\in \mathbb{N}$ and defining $ y_k^n:=z_k^\varepsilon$ yields

$\liminf_kF(y_k^n)<\overline F(x_n) +\frac{1}{n}$ and $y_k^n\to x_n,$ for $k\to +\infty$

Now I observe that to apply the property Anne Bauval proved below, I need $\overline F(x_n) +\frac{1}{n} \in \mathbb{R}$ which is true by the remark $(***)$

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Now by definition of $\liminf$:

$L:=\liminf_{k\rightarrow \infty}F(y_k^n)=\sup_{m\ge 1}\alpha_m \tag{&}$, with $\alpha_m= \inf_{k\ge m} F(y_k^m)\tag{*}$

$\alpha_m$ is increasing: $\alpha_m\ge \alpha_{m_{\tilde\varepsilon}}, \forall m\ge m_{\tilde\varepsilon}\tag{**}$

Now by definition of $\sup$ in (&), I get that $\forall \tilde\varepsilon>0, \exists m_{\tilde\varepsilon}$ such that $L-\tilde\varepsilon<\alpha_{m_{\tilde\varepsilon}} \le L$. Using $(*)$ and $(**)$: $L-\tilde\varepsilon<\alpha_{m_{\tilde\varepsilon}} \le \alpha_m ,\forall m\ge m_{\tilde\varepsilon}$

$L-\tilde\varepsilon<\alpha_{m_{\tilde\varepsilon}} \le \alpha_m=\inf_{k\ge m} F(y_k^m)\le F(y_k^m) ,\forall k\ge m\ge m_{\tilde\varepsilon}$

But that is not yet what I need, what am I doing wrong and how should I proceed? Please be detailed in your answers as I said I am not used to this proofs, so this would be an important example. You are also welcome to provide alternative ways to proof (6.3)

Additionally, since it looks like a long proof, I'd like to ask you how is one able to write directly (6.3) withouth going through all of this? Is there some lemma or anything?

Answer 1

1. Your attempt (by the time of first editing this answer, and before multiple edits in your post) started well but got lost from "Now by definition of $\sup$ in (&), I get that $\forall \tilde\varepsilon>0, \exists m_{\tilde\varepsilon}$ such that $L-\tilde\varepsilon<\alpha_{m_{\tilde\varepsilon}} \le L$." What you can write instead is:
   Since $\liminf_kF(y^n_k)<\overline F(x_n)+\frac1n,$ we have $F(y^n_k)<\overline F(x_n)+\frac1n$ for infinitely many $k$'s. Choosing moreover $k$ large enough, we also have $d(y^n_k,x_n)<\frac1n.$
2. Here is the same proof of property (6.3) but with alleviated notations. Given $x\in X$ and $\epsilon>0,$ we must find some $y\in X$ such that $$d(x,y)<\epsilon\quad\text{and}\quad F(y)\le\overline F(x)+\epsilon.$$ Since $\overline F(x)+\epsilon>\inf\left\{\liminf_{n\to\infty}F(x_n):x_n\to x\right\},$ there exists a sequence $x_n\to x$ such that $\liminf_{n\to\infty}F(x_n)<\overline F(x)+\epsilon$ and therefore, $F(x_n)<\overline F(x)+\epsilon$ for infinitely many integers $n.$ Choosing moreover $n$ large enough, we also have $d(x,x_n)<\epsilon$ and we can set $y=x_n.$
3. The main property used in the two previous points is that if $\liminf y_n<A$ then $y_n<A$ for infinitely many $n$'s. Here is a proof of this lemma.
   By definition, $\liminf_ny_n=\sup_m\alpha_m,$ where $\alpha_m=\inf_{k\ge m}y_k.$ If $\liminf_ny_n<A$ then $$\forall p\in\Bbb N\quad\inf_{k\ge p}y_k=\alpha_p\le\sup_m\alpha_m=\liminf_ny_n<A$$ hence $$\forall p\in\Bbb N\quad\exists k\ge p\quad y_k<A,$$ q.e.d. (With a little more care, you can prove that $(y_n)$ even has a subsequence which tends to $\liminf_ny_n$.)
4. There remains to justify $(***)$, which was used in the two first points above as well as in $(6.3)$ of your book (incidentally, their attempt to justify it with $(6.4)$ is misprinted – $y_n$ should be $x_n$ there – and at best incomplete).
   Assume by contradiction that $\overline F(x_{n_i})=-\infty$ for some strictly increasing sequence $(n_i)$ of indices. For each of these indices $n_i$, there would exist a sequence $(z^i_k)_{k\in\Bbb N}$ such that $d(z^i_k,x_{n_i})<\frac1k$ and $F(z^i_k)<-k.$ The diagonal sequence $(z_k^k)$ would then converge to $x$ and its image by $F$ would tend to $-\infty,$ thereby contradicting the hypothesis $\overline F(x)>-\infty.$
5. However, we could as well avoid that hypothesis and its consequence $(***),$ by a modification of the second inequality in $(6.3).$ The author of your book (and my point 1.) chose $\overline F(x_n)+\frac1n$ as a strict upper bound of $\overline F(x_n)$ if the latter is $>-\infty.$ I did the same in my point 2. with $\overline F(x_n)+\varepsilon.$ What we can do, instead, is naming $t_n$ some strict upperbound of $\overline F(x_n),$ chosen such that $\liminf t_n=\liminf\overline F(x_n)$ (which is always feasible, whatever the (in-)finiteness of the $\overline F(x_n)$'s), and perform the whole proof without making that $t_n$ more explicit.