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I want to prove $l_1$ space is complete. $(x_n)$ is a Cauchy succession in $l_1$. if $x_n$ converge to x in $l_1$ then x is in $l_1$? Can I use Fatou's lemma?

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    $\begingroup$ Well first of all you need to specify a metric $\endgroup$
    – Ant
    Commented Apr 6, 2016 at 8:27
  • $\begingroup$ @Ant I guess that the metric is the standard one. $\endgroup$
    – Siminore
    Commented Apr 6, 2016 at 8:51
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    $\begingroup$ See math.upenn.edu/~kazdan/508F08/completeness-l_1.pdf $\endgroup$
    – Augustin
    Commented Apr 6, 2016 at 8:57

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A direct attack. Let $x_n=(x_{n,i})_i\in l_1$ and let $(x_n)_n$ be an $l_1$ Cauchy sequence. For each $i$, the sequence $(x_{n,i})_n$ is a real ( or complex) Cauchy sequence because $\lim_{n\to \infty}\sup_{m>n}|x_{n,i}-x_{m,i}|\leq \lim_{n\to \infty}\sup_{m>n}\|x_n-x_m\|=0.$ So for each $i$ let $y_i=\lim_{n\to \infty}x_{n,i}.\;$ Let $y=(y_i)_i.$

(1) We have $y\in l_1.$ Proof: $(x_n)_n$ is Cauchy so $M=\sup_n\|x_n\|<\infty.$ Now, given $j$, let $n$ be large enough that $\forall i\leq j \;(\; |y_i-x_{n,i}|<1/(j+1)\;).$ Then $$\sum_{i\leq j}|y_i|\leq \sum_{i\leq j}|y_i-x_{n,i}|+|x_{n,i}|\leq 1+\|x_n\|\leq 1+M.$$ So $\sup_j \sum_{i\leq j}|y_i|\leq 1+M.$

(2).$\lim_{n\to \infty}\|y-x_n\|=0.$ Proof. Given $d>0$, let $n_1$ be large enough that $n \geq n_1\implies \|x_n-x_{n_1}\|<d.$ Let $j$ be large enough that $$\sum_{i>j}|y_i|<d \quad\text {and }\quad \sum_{i>j}|x_{n_1}|<d.$$ $$\text {Let } n_2\geq n_1 \text {where } \forall i\leq j \;(n\geq n_2\implies(\;|y_i-x_{n,i}|<d/(j+1)\;).$$ Then for $n\geq n_2$ we have $$\|y-x_n\|=\sum_{i\leq j}|y_i-x_{n,i}|+\sum_{i>j}|y_i-x_{n,i}|<$$ $$<d+\sum_{i>j}|y_i-x_{n,i}|\leq$$ $$\leq d+\sum_{i>j}|y_i-x_{n_1,i}|+|x_{n_1,i}-x_{n,i}|\leq $$ $$\leq d+\sum_{i>j}|y_i|+|x_{n_1,i}|+|x_{n_1,i}-x_{n,i}|\leq $$ $$\leq d+d+d+\|-x_{n_1}+x_n\|<4 d.$$

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    $\begingroup$ An indirect attack: Show that the space $S$ of sequences $(z_n)_n$ such that $\lim_{n\to \infty}z_n=0$, with norm $ \|(z_n)n\|=\sup_n |z_n|$, is complete. So $S$ is a Banach space. Show that $S^*=l_1.$ The dual space of a Banach space is also Banach, i.e. the norm on $l_1$ is complete. $\endgroup$ Commented Apr 6, 2016 at 11:47

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