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$(l^\infty,\|.\|_\infty)$ is a Banach space. In the proof, $\mathbb{F}$ is either the field of complex numbers or the field of real numbers.


Proof: Let $(x^n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $l^\infty$, where $x^n=(x_k^n)_{k\in\mathbb{N}}$. Consider the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ of $k$th coordinates of the sequence $(x^n)_{n\in\mathbb{N}}$.

Let $\epsilon>0$. Since $(x^n)_{n\in\mathbb{N}}$ is Cauchy, there exists $n_0\in\mathbb{N}$ such that $$\forall m,n>n_0,\|x^m-x^n\|_\infty<\epsilon.$$ Therefore for each $m,n>n_0$ we have $$|x^m_k-x^n_k|<\epsilon.$$

So the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ is Cauchy. Therefore it converges to some $y_k\in\mathbb{F}\ $ ($\mathbb{F}$ is complete).

Let $y=(y_k)_{k\in\mathbb{N}}$. Since $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ is Cauchy it is bounded. Choose $M>0$ such that for each $n\in\mathbb{N}$, $|x^n_k|<M$. But since $$y_k=\lim_{n\to\infty}x_k^n,$$ we have $|y_k|\leq M$ for each $k$. Therefore, $y\in l^\infty$.

Fix $m>n_0$. Then we have $\|x^m-x^n\|_\infty<\epsilon.$ Therefore $\|x^m-y\|_\infty<\epsilon$ as $n\to\infty$.

Therefore for each $m>n_0,\ \|x^m-y\|_\infty<\epsilon$. Hence $(x^n)_{n\in\mathbb{N}}$ converges in $l^\infty$ and the proof is complete.


Could someone please tell me if the above proof is alright? Thanks.

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  • $\begingroup$ Looks alright to me. The readability was fine even before it was over-pedantically edited. $\endgroup$ Commented Aug 23, 2017 at 7:35
  • $\begingroup$ The general procedure seems right, however near the end $\|x^m - x^m\|_\infty$ is something you should fix. $\endgroup$
    – mlk
    Commented Aug 23, 2017 at 7:36
  • $\begingroup$ @uniquesolution Then you are just more skilled than I am in reading wall-of text style questions. But I hope even you will agree that the readability is much better now. $\endgroup$
    – 5xum
    Commented Aug 23, 2017 at 7:37
  • $\begingroup$ When I first looked at it it was properly Latex formatted. If it was text-style and you converted to Latex, then thank you. $\endgroup$ Commented Aug 23, 2017 at 7:39
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    $\begingroup$ Since $(x^0_k, x^1_k, \dots)$ is Cauchy, it is bounded.Choose $M > 0$ such that for each $n \in \mathbb{N}, |x_k^n| < M$. Instead of that, I think you should argue by the Cauchy and boundedness property of the original sequence, that is $(x^n)_{n \in \mathbb{N}}$. The $M$ that you find depends on the $k$ also. $\endgroup$ Commented Mar 22, 2019 at 12:04

3 Answers 3

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Your proof is very rigorous and very detailed all the way up to the point where you say

Fix $m>n_0$. Then we have $\|x^m-x^n\|_\infty<\epsilon.$ Therefore $\|x^m-y\|_\infty<\epsilon$ as $n\to\infty$.

Now I know the inequality stands, but as you were very thorough with all your other inequalities, I think it would be nice if you wrote a little more justification for this one as well - it is not entirely obvious how the right inequality follows from the left one.


Other than that, the proof is very well written and easy to follow.

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  • $\begingroup$ sorry, the proof is wrong. In addition to your issue, $M$ depends on $k$ and should be $Mk$ as @EugenR observes in another post and tries to fix , but makes a mistake too. $\endgroup$
    – Anon
    Commented Jan 14 at 6:08
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Rather a comment, but my reputation does not allow me to comment yet. You write

choose $M>0$ that for each $n\in\mathbb{N}, |x^n_k|<M$

I am curios whether you have to write $M_k$, since hypothetically $M$ can depend on the choice of the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$. But then we have difficulties with proving that $y\in l^\infty$.

EDIT: the proof is more subtle than I initially thought. First we have to settle the issue 5xum mentioned. We start with $$ \forall k\in\mathbb{N}\ \forall \epsilon > 0 \ \exists n_0:\forall n,p>n_0:|x^n_k-x^{n+p}_k|<\epsilon. $$ This is inequality in $\mathbb{F}$, so we can let $p\to\infty$. Thus we obtain $$ \forall k\in\mathbb{N}\ \forall \epsilon > 0 \ \exists n_0:\forall n>n_0:|x^n_k-y_k|<\epsilon. $$ That means $\|x^n-y\|_\infty<\epsilon$, thus $(x^n)_{n\in\mathbb{N}}$ is converging to $y$.

We still have to prove that $y\in l^\infty$. But $$ \|y\|_\infty \le \|x^n-y\|_\infty + \|x^n\|_\infty \le \epsilon + \|x^n\|_\infty < \infty $$

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  • $\begingroup$ Could you explain the last bit please? $\endgroup$
    – Janitha357
    Commented Aug 23, 2017 at 9:53
  • $\begingroup$ I am using triangle inequality and addition of zero to prove that $y\in l^\infty$: $\|y\|_\infty \le \|x^n - x^n+y\|_\infty \le \|x^n-y\|_\infty + \|x^n\|_\infty \le \epsilon + \|x^n\|_\infty < \infty$ $\endgroup$
    – EugenR
    Commented Aug 23, 2017 at 11:10
  • $\begingroup$ You correction is incorrect "That means $|| x^n - y ||_\infty < \epsilon$" is an incorrect consequence as your $n_0$ depends on k. For a correct proof, see this for $lp$ and modify it for $l\infty$ - math.stackexchange.com/questions/1276470/… $\endgroup$
    – Anon
    Commented Jan 14 at 6:11
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You said: Fix $m> n_0$. Then we have $\| x^m - x^n \|_{\infty} < \epsilon$. Therefore
\begin{eqnarray} \| x^m - y \|_\infty < \epsilon \quad \text{as } n \to \infty \end{eqnarray} I try to clean up this statement but then I have a new question after this. \begin{eqnarray} \lim_{n \to \infty} \| x^m - x^n \| &=& \lim_{n \to \infty} \sup_{i \in \mathbb{N}} | x_i^m - x_i^n | \\ &=& \sup_{i \in \mathbb{N}} | x_i^m - \lim_{n \to \infty} x_i^n| \\ &=& \sup_{i \in \mathbb{N}} | x_i^m - y_i | = \| x^m - y \| \le \epsilon \end{eqnarray} How do you justify interchanging the order of the limit and the sup? Note also that the limit is $\le$ and not necessarily $<$. Think, for example, about the sequence $(1/n)$ all terms are $>0$ but the limit is $=0$.

To justify the interchange between the sup and lim consider the following reasoning:

\begin{eqnarray*} \lim_{m \to \infty} \| x_n - x_m \| &=& \lim_{m \to \infty} \sup_{i \in \mathbb{N}} | x_i^n-x_i^m | \\ &=&\lim_{m \to \infty} \left ( \sup_{i \in \mathbb{N}} |x_i^n - x_i^1 | , \sup_{i \in \mathbb{N}} |x_i^n - x_i^2|, \cdots , \sup_{i \in \mathbb{N}} |x_i^n - x_i^m|, \cdots \right ) \end{eqnarray*} If such a limit exists it is given by \begin{eqnarray*} \sup_{i \in \mathbb{N}} | x_i^n - x_i^{\infty}| = \sup_{i \in \mathbb{N}}|x_i^n- y_i|. \end{eqnarray*} The case of $\le \epsilon$ is easy. Just asume $\epsilon/2$ above and then $\epsilon/2 < \epsilon$.

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