Question on proof of completness of space of all complex valued functions

[oblixps]

in the proof of showing that the vector space of all complex valued functions with the norm |f|_u = sup(|f(x)|) over all x in the domain is complete, there was a step that was confusing:

let {f_n} be a Cauchy sequence in the normed space Z. We know that |f_n(x) - f_m(x)| \leq |f_n - f_m|_u. So {f_{n}(x)} is a Cauchy sequence in \mathbb{C} which is complete so f_n(x) converges to f(x) for every x. Letting n approach infinite on both sides of the inequality, we get |f(x) - f_n(x)| \leq lim \inf |f_n - f_m|_u.

my question is where did that lim inf come from?

 

[Opalg]

in the proof of showing that the vector space of all complex valued functions with the norm |f|_u = sup(|f(x)|) over all x in the domain is complete, there was a step that was confusing:

let {f_n} be a Cauchy sequence in the normed space Z. We know that |f_n(x) - f_m(x)| \leq |f_n - f_m|_u. So {f_{n}(x)} is a Cauchy sequence in \mathbb{C} which is complete so f_n(x) converges to f(x) for every x. Letting n approach infinite on both sides of the inequality, we get |f(x) - f_n(x)| \leq lim \inf |f_n - f_m|_u.

my question is where did that lim inf come from?

In general, if $a_n\leqslant b_n$ and $a_n\to a$, then $a\leqslant \liminf b_n$. In fact, given $\varepsilon>0$ there exists $N$ such that $|a_n-a|<\varepsilon$ whenever $n\geqslant N.$ But there exists $m>N$ such that $b_m < \liminf b_n + \varepsilon$. For that value of $m$, $$ a < a_m+\varepsilon \leqslant b_m+\varepsilon < \liminf b_n +2\varepsilon.$$ Since $\varepsilon$ is arbitrary, it follows that $a\leqslant \liminf b_n$.