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

oblixps

Member
May 20, 2012
38
in the proof of showing that the vector space of all complex valued functions with the norm [tex] |f|_u = sup(|f(x)|) [/tex] over all x in the domain is complete, there was a step that was confusing:

let [tex] {f_n} [/tex] be a Cauchy sequence in the normed space Z. We know that [tex] |f_n(x) - f_m(x)| \leq |f_n - f_m|_u [/tex]. So [tex] {f_{n}(x)} [/tex] is a Cauchy sequence in [tex] \mathbb{C} [/tex] 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 [tex] |f(x) - f_n(x)| \leq lim \inf |f_n - f_m|_u [/tex].

my question is where did that lim inf come from?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
in the proof of showing that the vector space of all complex valued functions with the norm [tex] |f|_u = sup(|f(x)|) [/tex] over all x in the domain is complete, there was a step that was confusing:

let [tex] {f_n} [/tex] be a Cauchy sequence in the normed space Z. We know that [tex] |f_n(x) - f_m(x)| \leq |f_n - f_m|_u [/tex]. So [tex] {f_{n}(x)} [/tex] is a Cauchy sequence in [tex] \mathbb{C} [/tex] 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 [tex] |f(x) - f_n(x)| \leq lim \inf |f_n - f_m|_u [/tex].

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 $a\leqslant \liminf b_n$.