Cauchy sequence of functions and the limit inferior I'm trying to understand a step in a proof. I don't get a special trick that is used several times in the book I am reading, so this does not get out of my head. I try to explain the prerequisites and what I don't understand:
Let $Y$ be a Banach space and let $S$ be a set. For a mapping $f:S\to Y$ let
$\|f\| := \sup_{x\in S}|f(x)|$. Let $B(S;Y)$ be the set of all mappings $g:S\to Y$ with $\|g\| < \infty$. I don't understand a step in the proof of the fact that this is a complete metric space.
Let $(f_k)_{k\in\mathbb{N}}$ be a Cauchy sequence in $B(S;Y)$. Let $x\in S$. Since $|f_k(x)-f_l(x)| \rightarrow 0$ for $k,l\rightarrow \infty$, we can define the pointwise limit $f$ of the sequence by $f(x) := \lim_{k\to\infty}f_k(x)$. Now here is the step that I don't understand:
The author of the book (and the proof) states that
$\lim_{l\to\infty}|f_l(x)-f_k(x)| \le \lim\inf_{l\to\infty}\|f_l-f_k\|<\infty$.
Why is that? What has the $\lim\inf$ to do here. And why is this finite? I think that it has something to do with subsequences of Cauchy sequences, but I don't understand it.
This $\lim\inf$-trick is used several times in the book so that it seems a bit important to me.
Thank you very much in advance. I'm glad for every help. This does not get out of my head.
 A: Assume we are give two sequences $\{a_r:r\in\mathbb{N}\}$ and $\{b_r:r\in\mathbb{N}\}$ such that for all $r\in\mathbb{N}$ we have $a_r\leq b_r$. Fix some $l\in\mathbb{N}$ and consider the last inequality for all $r\geq l$. After taking infimum on the left side over all $r\geq l$ we get
$$
\inf\limits_{r\geq l}a_r \leq b_r
$$
Then we take infimum on the right side over all $r\geq l$, and we get
$$
\inf\limits_{r\geq l}a_r \leq \inf\limits_{r\geq l}b_r
$$
Note that sequences $\{\inf\limits_{r\geq l}a_r:l\in\mathbb{N}\}$ and $\{\inf\limits_{r\geq l}b_r:l\in\mathbb{N}\}$ are non decreasing hence they have limits (finite or infinite). Lets take this limits, then we get
$$
\lim\limits_{l\to\infty}\inf\limits_{r\geq l}a_r \leq \lim\limits_{l\to\infty}\inf\limits_{r\geq l}b_r
$$
It is known that $\liminf\limits_{l\to\infty}x_l=\lim\limits_{l\to\infty}\inf\limits\limits_{r\geq l}x_r$ (sometimes this equality is taken as definition of $\liminf$), so
$$
\liminf\limits_{l\to\infty}a_l \leq \liminf\limits_{l\to\infty}b_l\tag{1}
$$
Another interesting fact: if a sequnces $\{x_l:l\in\mathbb{N}\}$ have a limit then it is equal to limit inferioir and limit superioir
$$
\lim\limits_{l\to\infty}x_l=\liminf\limits_{l\to\infty}x_l=\limsup\limits_{l\to\infty}x_l\tag{2}
$$
Now we proceed to your question. Fix $k\in\mathbb{N}$ and consider sequences
$$
a_l=|f_l(x)-f_k(x)|\qquad b_l=\Vert f_l-f_k\Vert\qquad\text{ where }\qquad l\in\mathbb{N}
$$
Since $a_l\leq b_l$ for all $l\in\mathbb{N}$ from (1) we get
$$
\liminf\limits_{l\to\infty}|f_l(x)-f_k(x)| \leq \liminf\limits_{l\to\infty}\Vert f_l-f_k\Vert
$$
By construction sequence $\{a_n:n\in\mathbb{N}\}$ is convergent. Indeed 
$$
\lim\limits_{l\to\infty} a_l=
\lim\limits_{l\to\infty} |f_l(x)-f_k(x)|=
|\lim\limits_{l\to\infty}f_l(x)-f_k(x)|=
|f(x)-f_k(x)|
$$
hence from $(2)$ we get
$$
\lim\limits_{l\to\infty}|f_l(x)-f_k(x)|=\liminf\limits_{l\to\infty}|f_l(x)-f_k(x)| \leq \liminf\limits_{l\to\infty}\Vert f_l-f_k\Vert
$$
A: Notice that
\begin{equation}
|f_l(x) -f_k(x)| \le \|f_l - f_j\|
\end{equation}
for all $l$. Then $(f_k(x))_k$ is a Cauchy sequence and
\begin{eqnarray}
\lim_{l\rightarrow \infty} |f_l(x) -f_k(x)| &=& \liminf_{l \rightarrow \infty} |f_l(x) -f_k(x)| \\
&\le& \liminf_{l \rightarrow \infty} \|f_l - f_j\|
\end{eqnarray}
We know that there exist $\lim_{l \rightarrow \infty}|f_l(x) -f_k(x)|$ because $Y$ is Banach and hence complete but, we don't know if $\lim_{l \rightarrow \infty}\|f_l -f \|$. Hence we use $\liminf$. Now we need that $k>>1$ for that $\|f_l - f_j\| < \infty$ because $f_k$ is a Cauchy sequence.
