Show that

A Normed Linear Space $X$ is a Banach Space iff every absolutely convergent series is convergent.

My try:

Let $X$ is a Banach Space .Let $\sum x_n$ be an absolutely convergent series .Consider $s_n=\sum_{i=1}^nx_i$. Now $\sum \|x_n\|<\infty \implies \exists N$ such that $\sum_{i=N}^ \infty \|x_i\|<\epsilon$ for any $\epsilon>0$

Then $\|s_n-s_m\|\le \sum _{i=m+1}^n \|x_i\|<\epsilon \forall n,m>N$

So $s_n$ is Cauchy in $X$ and hence converges $s_n\to s$ (say).

Thus $\sum x_i$ converges.

Conversely, let $x_n$ be a Cauchy Sequence in $X$. Here I can't proceed how to use the given fact.

Any help will be great.


For the converse argument; let $X$ be a normed linear space in which every absolutely convergent series converges, and suppose that $\{x_n\}$ is a Cauchy sequence.

For each $k \in \mathbb{N}$, choose $n_k$ such that $||(x_m − x_n)|| < 2^{−k}$ for $m, n \geq n_k$. In particular, $||x_{n_{k+1}}−x_{n_k}||< 2^{-k}$. If we define $y_{1} = x_{n_{1}}$ and $y_{k+1} = x_{n_{k+1}} − x_{n_{k}}$ for $k \geq 1$, it follows that $\sum ||y_{n}|| ≤ ||x_{n_{1}} || + 1$ i. e., ($y_{n}$) is absolutely convergent, and hence convergent.

| cite | improve this answer | |
  • 1
    $\begingroup$ Hey, I am facing the same problem and your solution was really good! However, I don't know why proving $(y_n)$ is convergent finished the problem... Now, for every $\varepsilon >0$ there exists $N\in \mathbb{N}$ such that for all $n\geq N$ we have that $\sum_{n=N}^\infty y_n <\varepsilon$, but what does conclude the proof? $\endgroup$ – user326159 Feb 21 '19 at 17:12
  • 1
    $\begingroup$ @user326159 The sequence $(y_n)$ is a convergent subsequence of a Cauchy sequence $(x_n)$. This implies that $(x_n)$ converges. $\endgroup$ – Thiago Alexandre May 1 '19 at 23:37
  • $\begingroup$ How does $y_n$ convergent mean the space is complete? (how do we know it converges to an $x \in X$) $\endgroup$ – sma Feb 3 at 22:19
  • $\begingroup$ @ThiagoAlexandre By assumption, $y_n$ is an absolutely convergent sequence that converges in the sequence. $\endgroup$ – kam Apr 4 at 17:20

Hint: show that there is a subsequence $y_n$ such that $\|y_n - y_{n+1}\| < 2^{-n}$.

| cite | improve this answer | |

A normed space is complete if and only if every absolutely convergent series converges.


We will prove this by proving absolutely convergent series is a Cauchy series.

We define a absolutely convergent series. Suppose $x_n\in E$ and $\sum_{n=1}^\infty||x_n||<\infty$ and denote $$ s_n=\sum_{k=1}^nx_k $$ Because the sequence of partial sums converges in E, for every $\epsilon >0 $, there exists $k>0$ such that $$ \sum_{n=k+1}^\infty ||x_n||<\epsilon $$ To show $(s_n)$ is a Cauchy sequence, $\forall \epsilon>0,\exists M,\forall m,n>M$ such that $$ ||s_m-s_n||=||x_{n+1}+x_{n+2}+\dotsm+x_m||\le \sum_{r=n+1}^\infty ||x_r||<\epsilon $$ (without loss of generality, we assume $m>n$)

Since E is complete, $s_n$ converges.


We need to prove if every absolutely convergent series in a normed space converges, then the normed space is complete.

Let $(x_n)$ be an Cauchy sequence in E and therefore $\forall \epsilon>0,\exists p_k\in N,\forall m,n>p_k$ such that $$ ||x_m-x_n||<2^{-k} $$ without loss of generality, we can assume $(p_k)$ is strictly increasing.

Then the series $\sum_{k=1}^\infty (x_{p_{k+1}}-x_{p_k})$ is absolutely convergent and therefore, convergent and therefore, the sequence $$ x_{p_k}=x_{p_1}+(x_{p_2}-x_{p_1})+(x_{p_3}-x_{p_2})+\dotsm+(x_{p_k}-x_{p_{k-1}}) $$ converges to an element $x\in E$

Then $$ ||x_n-x||\le ||x_n-x_{p_n}||+||x_{p_n}-x||\rightarrow 0 $$ Q.E.D.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.