Consider $l_{\infty} = \{f : \mathbb{N} \rightarrow \mathbb{K} : \sup_{n \in \mathbb{N}} |f(x_i)| < \infty\}$. Suppose that $\{f_i\} \subset l_{\infty}$ is a Cauchy sequence. Given $\epsilon>0$, there exists $N \in \mathbb{N}$ such that whenever $s,m \geq N$ we have $\sup_{n \in \mathbb{N}} |f_s(n) - f_m(n)| < \epsilon$. This mean for all $k \in \mathbb{N}$ $|f_n(k) - f_m(k)| < \epsilon$. Therefore the sequence $\{f_i(k)\} \subset \mathbb{R}$ is cauchy. By completeness of $\mathbb{R}$ we have $f_i(k) \rightarrow c_k$.

Define the function $f : \mathbb{N} \rightarrow \mathbb{K}$ as follows $f(n) = c_n$. Then our goal is to show that $f \in l_{\infty}$ and $f_i \rightarrow f$. First we will prove that $f \in l_{\infty}$. Consider the sequence $U = \{\sup_{n \in \mathbb{N}} |f_i(n)|\}$. By reverse triangle inequality $|\ ||f_s||_{\infty} - ||f_m||_{\infty}\ | \leq ||f_s - f_m||_{\infty} < \epsilon$ whenever $m,s \geq N$.

This mean that $|\ sup_{n \in \mathbb{N}}|f_s(n)| - sup_{n \in \mathbb{N}}|f_m(s)|\ | \leq sup_{n \in \mathbb{N}}|f_s(n) - f_m(n)|$. Therefore U is a cauchy sequence so we must have $\sup_{n \in \mathbb{N}} |f_i(n)| \rightarrow m$.

Consider the set $B = \{|c_k| \in \mathbb{R} : c_k \in f(\mathbb{N})\}$. For all $k$ and for all $i$, we have by hypothesis $|f_i(k)| \leq \sup_{n \in \mathbb{N}} |f_i(n)|$. Therefore $|c_z| \leq m$ for all $|c_z| \in B$. Therefore supermum exist as B is bounded above, so $f \in l_{\infty}$.

Finally we prove that $f_i \rightarrow f$. By Cauchy criterion we applied to $\epsilon = \epsilon/2$. There exists $N_f$ such that $s,m \geq N_f$ implies that $|f_s(k) - f_m(k)| < \epsilon/2$ for all k. taking $m \rightarrow \infty$ we have $|f_s(k) - f(k)| \leq \epsilon/2$ for all $s \geq N_f$ and for all $k$.

Therefore $\sup_{n \in \mathbb{N}} |f_s(k) - f(k)| \leq \epsilon/2 < \epsilon$. Hence every Cauchy sequence is convergent as we started with arbitrarily one. Therefore $l_{\infty}$ is complete.

  • $\begingroup$ Could you elaborate on what you mean when you say "By reverse triangle inequality and the fact that $\|\cdot\|_{\infty}$ is a norm we know $U$ is Cauchy in $\mathbb{R}$"? $\endgroup$ – Aweygan Oct 23 '16 at 4:23
  • $\begingroup$ Yes sure. I will add the details. $\endgroup$ – user111750 Oct 23 '16 at 4:34
  • $\begingroup$ @Aweygan I added the details. $\endgroup$ – user111750 Oct 23 '16 at 4:44

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