Show that ($\ell^1$, $\|\cdot\|_1$) is complete

Question

Show that the vector space $\ell^1 : = \{(a_n) : \sum_n|a_n| < \infty\}$ with the norm $\|(a_n)\|_1 : = \sum_n|a_n|$ where $(a_n)$ are sequences in $\mathbb C$ is complete.

Thanks in advance.

Answer 1

Consider a Cauchy sequence $\{x^n\}$in $l_1$. Where $x^n = (x_1^n, x_2^n, \dots)$. For any $\epsilon >0$ there exist $k_1 \in \mathbb{N} $ s.t. $\|x^p - x^q\|_1 < \epsilon$ when $p, q > k_1$. So $\sum_{i=1}^{\infty} |x_i^p -x_i^q| < \epsilon$. Thus for any fixed subscript $i$ the sequence $\{x_i^n\}$ is Cauchy on $\mathbb{C}$. As $\mathbb{C}$ is complete the sequence is convergent and let it converges to $x_i$ . Consider the element $x = (x_1, x_2, \dots)$.

Now two points to be shown

$x \in l_1$.

$\{x^n\}$ converges to $x$.

The sequence $\{x^n\}$ is Cauchy and hence it is bounded. Thus $\sum_{i=1}^{\infty} |x_i^n| < C$ for some $C \in \mathbb{R}$. Thus $\sum_{i=1}^k|x_i^n| < C$ for any arbitrary $k$. Taking $n \rightarrow \infty$ we shall get $\sum_{i=1}^{k}|x_i| < C$. Now taking $k \rightarrow \infty$ we shall get $\sum_{i=1}^{\infty}|x_i|$ is bounded.

Now form the inequality from the condition of Cauchy Sequence taking one supercsript $\rightarrow \infty$ we can show the sequence $\{x^n\}$ converge to $x$. That is,

$$\lim_{q \to \infty}\sum_i^k|x^p_i-x^q_i| = \sum_i^k|x^p_i-x_i| \leq \epsilon$$

for all $k$.

Hence the proof.