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I start by taking a Cauchy sequence $(a_i)$ in $\ell_\infty$. I denote the terms of $(a_i)$ as $f_1, f_2, f_3, \dots$ and so on.

For each $x \in \mathbb{N}$, the sequence $(f_1(x), f_2(x), f_3(x), \dots)$ converges. I define the sequence $L$ as $L(x) = \displaystyle \lim_{n \to \infty} f_n (x)$. I'm having trouble showing $L$ is bounded, but I have what I believe is the first step to an argument.

Since $(a_i)$ is Cauchy, it is bounded, so $d(f_n, f_m) \leq S$ for all naturals $m$ and $n$. (Where $d$ is the metric in $\ell_\infty$).

So let $x, n \in \mathbb{N}$. Then $|L(x)| \leq |L(x) - f_n(x)| + |f_n(x)| \leq |f_m(x) - f_n(x)| + |f_n(x)| \leq d(f_m,f_n) + \text{sup} f_n \leq S + $sup $f_n$. So $S + $sup $f_n$ bounds $\{|L(x)| | x \in \mathbb{N} \}$, implying $L$ is a bounded sequence.

My issue is choosing some $m \in \mathbb{N}$ such that $|L(x) - f_n(x)| \leq |f_m (x) - f_n(x)|$. Does such an $m$ always exist regardless of the choice of $x$ and $m$?

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Your notation is a little confusing...if your sequence is $(a_i)$, why are you using $f_i(x)$? $l_\infty$ is a sequence space, there are no functions. Unless you mean $L_\infty$? – icurays1 Apr 9 '13 at 4:32
Mostly it's to keep the indexes straight. They're sequences, but sequences are just functions with a domain of natural numbers. – Max Lipton Apr 9 '13 at 4:33
I see, okay, it makes sense now. Thanks. – icurays1 Apr 9 '13 at 4:35

1 Answer 1

First, bound the sequence $f_n$, then use continuity to show that this is also a bound for $L$.

Since $f_n$ is Cauchy, pick $\epsilon=1$ to get some $N$ such that if $m,n \ge N$, then $\|f_n-f_m\|_\infty \le 1$. In particular, $\|f_n\|_\infty \le \|f_N\|_\infty + \|f_n-f_N\|_\infty \le \|f_N\|_\infty +1$, for $n \ge N$, and so you have $\|f_n \|_\infty \le B=\max(\|f_1\|_\infty,...,\|f_{N-1}\|_\infty, \|f_N\|_\infty +1)$ for all $n$.

It follows that $| f_n(x)| \le B$ for all $n$, for all $x$. Since $L(x) = \lim_n f_n(x)$, we have $|L(x)| = \lim_n |f_n(x)|$. It follows from this that $|L(x)| \le B$ for all $x$, and so $L \in l_\infty$.

Addendum: Showing convergence is fairly straightforward. Note that if $|f_m(x)-f_n(x)| \le K$ for all $m,n \ge N$, then $|L(x)-f_n(x)| \le K$ for all $n \ge N$. Let $\epsilon>0$ and choose $N$ such that if $m,n \ge N$, then $\|f_m-f_m\|_\infty < \epsilon$. Then $|f_m(x)-f_n(x)| \le \epsilon$ for all $m,n \ge N$, and so $|L(x) -f_n(x)| \le \epsilon$ for all $n \ge N$. Hence $\|L-f_n\|_\infty \le \epsilon$.

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What do you mean by $||f_n||_\infty$? Is it the distance between $f_n$ and the sequence of all $0$'s? – Max Lipton Apr 9 '13 at 4:47
It is the norm on $l_\infty$, ie, $\|f\|_\infty = \sup_n |f(x)|$. (So, yes to your question, but I don't think of it that way.) Use $d(f,0)$ if you prefer. The $l_\infty$ distance is translation invariant, so you can write $d(f,g) = d(f-g,0).$ – copper.hat Apr 9 '13 at 4:48
So boundedness has been (very clearly established), but there is still the question of uniform convergence, which can be found in the "second answer" (the accepted answer) of… (look also at the comments of the answer) – user39158 Jun 22 at 9:02
@user39158: I added a few lines showing convergence. – copper.hat Jun 22 at 22:52

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