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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 $$\forall m,n>n_0,\|x^m-x^m\|_\infty<\epsilon.$$ Therefore for each $m,n>n_0$: $$|x^m_k-x^n_k|<\epsilon.$$

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. $\endgroup$ – uniquesolution Aug 23 '17 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 Aug 23 '17 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 Aug 23 '17 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. $\endgroup$ – uniquesolution Aug 23 '17 at 7:39
  • $\begingroup$ @mlk fixed it, and thanks everyone. $\endgroup$ – Janitha357 Aug 23 '17 at 7:40
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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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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 Aug 23 '17 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 Aug 23 '17 at 11:10

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