# Prove that $(l^\infty,\|.\|_\infty)$ is a Banach space.

$$(l^\infty,\|.\|_\infty)$$ 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 $$n\in\mathbb{N}$$, $$|x^n_k|. But since $$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.

• Looks alright to me. The readability was fine even before it was over-pedantically edited. Commented Aug 23, 2017 at 7:35
• The general procedure seems right, however near the end $\|x^m - x^m\|_\infty$ is something you should fix.
– mlk
Commented Aug 23, 2017 at 7:36
• @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.
– 5xum
Commented Aug 23, 2017 at 7:37
• When I first looked at it it was properly Latex formatted. If it was text-style and you converted to Latex, then thank you. Commented Aug 23, 2017 at 7:39
• 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. Commented Mar 22, 2019 at 12:04

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.

• 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.
– Anon
Commented Jan 14 at 6:08

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 $$\|y\|_\infty \le \|x^n-y\|_\infty + \|x^n\|_\infty \le \epsilon + \|x^n\|_\infty < \infty$$

• Could you explain the last bit please? Commented Aug 23, 2017 at 9:53
• I am using triangle inequality and addition of zero to prove that $y\in l^\infty$: $\|y\|_\infty \le \|x^n - x^n+y\|_\infty \le \|x^n-y\|_\infty + \|x^n\|_\infty \le \epsilon + \|x^n\|_\infty < \infty$ Commented Aug 23, 2017 at 11:10
• 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/…
– Anon
Commented Jan 14 at 6:11

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 $$\epsilon/2 < \epsilon$$.