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I cannot really find a good definition of $l^\infty$-convergence of a sequence. Can someone help me with this? Cheers!

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  • $\begingroup$ We can define a sequence as a function whose domain is $\Bbb N.$ We can say that a sequence $f_1,f_2,...$ of members of $l^{\infty}$ converges to $f$ iff the sequence of functions $f_1,f_2,..$ converges $uniformly$ to the function $f.$ $\endgroup$ Feb 19, 2018 at 4:00

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By definition $l^{\infty}$ is the set of all bounded sequences. Let $x=(x_1,x_2,..)$ be such a sequence, then the norm ||x|| is defined by $max|x_i|$. Convergence then means max of the differences between elements of the members of the sequence and elements of the limit ->0. In other words, the norm of the differences between members of the sequence and the limit ->0.

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  • $\begingroup$ Thanks for your answer! What do you mean by "elements of the limit" tho? Cheers! $\endgroup$
    – asn32
    Feb 18, 2018 at 22:26
  • $\begingroup$ It is best to say "sup" and not "max" in case there is no maximum achieved. $\endgroup$
    – GEdgar
    Feb 18, 2018 at 23:59
  • $\begingroup$ I think "and elements of the limit" is a kind of typo, and should be just deleted. And "max of the differences" should be replaced with "sup of the absolute differences". $\endgroup$ Feb 19, 2018 at 3:55
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Your question is not very specific, you should get used to develop it more here on math stackexchange.

$d$ stands for distance and $K=\mathbb{R}$ or $K=\mathbb{C}$, where $K$ is the scalar field on the vector space.

$l_{\infty}$Space of limited successions such that $l_{\infty}=\{(x_1,...,x_n,...)\:x_k\in K,k\geqslant 1\}$ with the norm $|x|_{\infty}=\sup\{|x_k|:k\geqslant 1\}$. So you get the biggest value of the sequence $(x_1,..,x_n,...)$ with the supremo norm. You can check this is a metric or a distance, the $|.|_{\infty}$ by noting that if $x,y,z\in l_{\infty}$, then $d(x,y)=sup|x-y|>0$,$d(x,y)=0\implies\sup|x-y|=0\implies x=y$ and the triangle inequality $d(x,z)\leqslant d(x,y)+d(y,z)$, $\sup|x-y|=\sup|x-y+y-z|\leqslant \sup|x-y|+\sup|y-z|$, apllying modulus properties.

Update:

Convergence You have got for example two sequences let´s say $x_n$ and $x_m$ each of the sequences converges to a point$ (x_n,1...x_n,2...x_n,k...)(x_0,1...x_0,2...x_0,k...)$ for$ k=1,2,3...$. You can measure the distance between the two sequences$ d(x_n,x_0)=\sup|x_n-x_0|$, we can define that for(using Cauchy ctieria) $n,m>N_k$ and $\epsilon>0$ we have $ d(x_n,x_m)=\sup|x_n-x_m|<\epsilon$, which means $x_n$ converges to a sequence let´s say $x_0$. Then you can go on and prove the space is complete.

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  • $\begingroup$ Hmm so what does convergence mean in $l^\infty$-space? $\endgroup$
    – asn32
    Feb 18, 2018 at 22:25
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    $\begingroup$ @AngghaNugraha You have got for example tow sequences let´s say $x_n$ and $x_m$ each of the sequences converges to a point$ (x_n,1...x_n,2...x_n,k...)(x_0,1...x_0,2...x_0,k...)$ for$ k=1,2,3...$. You can measure the distance between the two sequences$ d(x_n,x_0)=\sup|x_n-x_0|$, we can define that for(using Cauchy ctieria) $n,m>N_k$ and $\epsilon>0$ we have $ d(x_n,x_m)=\sup|x_n-x_m|<\epsilon$, which means $x_n$ converges to a sequence let´s say $x_0$. Then you can go on and prove the space is complete. $\endgroup$ Feb 18, 2018 at 22:34

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