# 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.

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Mentioning name of the book would be nice. (And, perhaps, useful.) – Martin Sleziak Jun 28 '12 at 15:38

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 $$\lim\limits_{l\to\infty}|f_l(x)-f_k(x)|=\liminf\limits_{l\to\infty}|f_l(x)-f_k(x)| \leq \liminf\limits_{l\to\infty}\Vert f_l-f_k\Vert$$

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Awesome answer! Thank you very much! – Sh4pe Jun 28 '12 at 15:45
You are welcome! – Norbert Jun 28 '12 at 15:45

Notice that $$|f_l(x) -f_k(x)| \le \|f_l - f_j\|$$ 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.

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Why does $Y$ Banach imply that $\lim_{l\to\infty}|f_l(x) - f_k(x)|$ exists? What do you mean by 'hence we use lim inf'. I want to understand these details... – Sh4pe Jun 28 '12 at 14:27
$Y$ Banack implies $Y$ complete and $F_k(x)$ is a Cauchy sequence in $Y$ – user29999 Jun 28 '12 at 14:56