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 $m$?

  • $\begingroup$ 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 $L_\infty$? $\endgroup$ – icurays1 Apr 9 '13 at 4:32
  • $\begingroup$ Mostly it's to keep the indexes straight. They're sequences, but sequences are just functions with a domain of natural numbers. $\endgroup$ – Max Lipton Apr 9 '13 at 4:33
  • $\begingroup$ I see, okay, it makes sense now. Thanks. $\endgroup$ – icurays1 Apr 9 '13 at 4:35

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 $\|L-f_n\|_\infty \le \epsilon$.

  • $\begingroup$ What do you mean by $||f_n||_\infty$? Is it the distance between $f_n$ and the sequence of all $0$'s? $\endgroup$ – Max Lipton Apr 9 '13 at 4:47
  • $\begingroup$ 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 $d(f,g) = d(f-g,0).$ $\endgroup$ – copper.hat Apr 9 '13 at 4:48
  • $\begingroup$ 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 math.stackexchange.com/questions/71121/… (look also at the comments of the answer) $\endgroup$ – Noix07 Jun 22 '15 at 9:02
  • $\begingroup$ @user39158: I added a few lines showing convergence. $\endgroup$ – copper.hat Jun 22 '15 at 22:52

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