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$l^{\infty} := \{x=(x_n) : \sup_{n\in \mathbb N}|x_n| \lt \infty\} $

$\|x\|_{\infty} = \sup_{n \in \mathbb N} |x_n|$

Show that $l^{\infty}$ is a Banach Space (Complete) with respect to > $\|x\|_{\infty}$

If I wish a space is Banach, I must show Cauchy Sequences are convergent in this space.

I cannot find clear answers for questions below :

1) I have prior problem about writing sequences in Cauchy definition. Do I have to use sequences of sequences in $l^{\infty}$ like $\|x_{n_k}-x_{n_l}\|_{\infty} \lt \varepsilon$ ,$\forall k,l \ge N_{\varepsilon}$ or can I write Cauchy definition for only elements of $l^{\infty}$ like $\|x_{n}-x_{m}\|_{\infty} \lt \varepsilon$ , $\forall m,n \ge N_{\varepsilon}$ and if I can how should I use it? (Please fix me if there is any mistake)

2) Let $y_k$ is limit sequence of $x_{n_k}$ (or $x_n$, I am not sure about writing sequences in Cauchy definition), how should I show $y_k \in l^{\infty}$, I think it is easy but I couldn't realize :(

Thanks a lot in advance

Community
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usereb
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  • I would recommend that you take a look into some book on measure theory or Lebesgue-Integrals. Most definitely, you will find a proof for the completeness of arbitrary $L^p$-spaces. – Peter Wildemann Mar 20 '18 at 21:53
  • See the discussion under this link: https://math.stackexchange.com/questions/2403181/prove-that-l-infty-infty-is-a-banach-space/3711653#3711653 – Herman Jaramillo Jun 08 '20 at 21:13

2 Answers2

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Hint:

Suppose $(X_i)_i$ be a Cauchy sequence with elements $X_i\in l^{\infty}$ that converges to $A$. We need to show that $A\in l^\infty$, i.e., $\sup_{n\in\Bbb N}|A_n|\lt\infty$

Then, given $\epsilon\gt 0$, there exists $N\in\Bbb N$ such that for all $i\geq N$, we have $||X_i-A||_\infty\lt\epsilon$

Now, $$||X_i-A||_\infty=\sup_{n\in\Bbb N}|X_{in}-A_n|\lt\epsilon$$

This implies that $|X_{in}-A_n|\lt\epsilon~\forall~n\in\Bbb N$

By the reverse triangle inequality, we have,

$$||X_{in}|-|A_n||\leq|X_{in}-A_n|\lt\epsilon$$

which shows that $|A_n|\in B_\epsilon(|X_{in}|)$ where $B_r(x)$ is the open ball of radius $r$ and center $x$

Can you now use this and the fact that $X_i\in l^\infty$ to conclude that $|A_n|\lt\infty~\forall~n\in\Bbb N$ and hence $||A||_\infty=\sup_{n\in\Bbb N}A_n\lt\infty$ ?

Prasun Biswas
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You seem to be confused with the indices.
You can bravely introduce any notation which is convenient for yourself, e.g. writing the sequences as functions: $x_n=k\mapsto x_n(k)$.

  1. The second choice is surely good, as this is the definition: $\|x_n-x_m\|$ can be stressed to be arbitrary small for big enough $n,m$ indices.

  2. From where this limit $y$ is coming from? It's just the aim to prove that $x_n$ will have a limit.
    Well, the $x_n$'s converge pointwise, as for each fixed $k$, the sequence $(x_n(k))_n$ is Cauchy.
    It will indeed give an $y(k)$, and you also have to prove $\|y-x_n\|\to 0$.

    To find a bound for $y$, fix an $\varepsilon$ and belonging $N$. Then $|x_n(k)|\le \|x_N\|+\varepsilon$ for all $n\ge N$ and all $k$.

Berci
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