I am trying to do an exercise in an introduction to functional analysis course:

1) Let $X$ be a normed space and $\{x_{n}\}_{n=1}^{\infty}\subseteq X$.

Prove that $X$ is a banach space iff every abselutly convergent series is convergent (that is $\sum_{n=1}^{\infty}||x_{n}||<\infty\implies\sum_{n=1}^{\infty}x_{n}$ exist)

2) Use part $1$ to shoe that if $X$ is banach space and $M$ is a closed subspace then $X/M$ is a banach space relative to the norm $$ ||[x]||=\inf\{||x-y||,y\in M\} $$

I have managed to do the first part of the exercise, but not the second one.

A start is to assume $$ \sum_{n=1}^{\infty}||[x_{n}]||<\infty $$

and to try and show that $$ \sum_{n=1}^{\infty}x_{n} $$


By definition I know that $$ \sum_{n=1}^{\infty}\inf\{||x_{n}-y||,y\in M\}=L<\infty $$

hence for every $\epsilon>0$ there is a sequence $\{y_{n}\}_{n=1}^{\infty}\subseteq M$ s.t $$ \sum_{n=1}^{\infty}||x_{n}-y_{n}||<L+\epsilon $$

I really don't see how to continue. Can someone please help me out and point me to the right direction ?

  • 3
    $\begingroup$ There is a $\xi_n \in X$ with $[\xi_n] = [x_n]$ and $\lVert \xi_n\rVert < \lVert [x_n]\rVert + 2^{-n}$. $\endgroup$ – Daniel Fischer Sep 16 '13 at 13:56
  • $\begingroup$ I have the impression that the notation $[x_n]$ is a bit misguiding. The $x_n$ have no special meaning, any element of $X$ with the same residue modulo $M$ is equivalent. So when you have a sequence $\eta_n$ in $X/M$ with $\sum \lVert\eta_n\rVert < \infty$, you can lift it to a sequence $\xi_n$ in $X$, with $\sum \lVert\xi_n\rVert < \infty$ (and $\pi(\xi_n) = \eta_n$, that's what "lifting" means here; $\pi \colon X \to X/M$ is the canonical projection). Now you know that $\sum\xi_n$ exists. And you know that $\pi$ is continuous, so ... $\endgroup$ – Daniel Fischer Sep 16 '13 at 14:17
  • $\begingroup$ @DanielFischer - Thank you for your comment, can you explain a little bit more ? I thought about it: I wanted to take $y_{n}\in M$ with $||y_{n}||<\frac{1}{2^{n}}$ and $\xi_{n}=x_{n}+y_{n}$. But that gives me $\sum_{n=1}^{\infty}||[x_{n}]||<\sum_{n=1}^{\infty}||x_{n}||+1$. But I don't think that this helps, since $0\in M$ I could of shown $\sum_{n=1}^{\infty}||[x_{n}]||\leq\sum_{n=1}^{\infty}||x_{n}||$ by taking $y_{n}=0$. It seems that I get the inequality in the other direction than the one I want. $\endgroup$ – Belgi Sep 16 '13 at 14:29
  • $\begingroup$ it seems that I don't see how to lift $\eta_{n}$ to $\zeta_{n}$ . I also don't understand how to use the continuity of $\pi$, it seems that I would of liked to used linearity and take $\pi(\sum\xi_{n})=\sum\pi(\zeta_{n})=\sum\eta_{n}$ and so it will follow that $\sum\eta_{n}$ exist. (I am not sure if $\pi$ is linear, at first I thought that no, but on second thought it seems that yes and a sum will be mapped to the sum + $M$) $\endgroup$ – Belgi Sep 16 '13 at 14:29

Let $\pi \colon X \to X/M$ be the canonical projection. Then we note that the norm on $X/M$ is given by

$$\lVert \eta\rVert_{X/M} = \inf \{ \lVert \xi\rVert_X : \xi \in \pi^{-1}(\eta)\}.$$

So if we have a sequence $(\eta_n)$ in $X/M$ with $\sum\limits_{n=0}^\infty \lVert\eta_n\rVert_{X/M} < \infty$, we can, for each $n\in\mathbb{N}$, choose a $\xi_n \in \pi^{-1}(\eta_n)$ with $\lVert \xi_n\rVert_X < \lVert\eta_n\rVert_{X/M} + 2^{-n}$, and that implies that

$$\sum_{n=0}^\infty \lVert\xi_n\rVert_X < \sum_{n=0}^\infty \left(\lVert\eta_n\rVert_{X/M} + 2^{-n}\right) = \sum_{n=0}^\infty \lVert\eta_n\rVert_{X/M} + 2 < \infty.$$

So since $X$ is a Banach space, we know that

$$x = \sum_{n=0}^\infty \xi_n$$

exists. Let $y = \pi(x)$. Then, by the continuity of $\pi$, we have

$$y = \pi(x) = \pi\left(\lim_{N\to\infty}\sum_{n=0}^N \xi_n\right) = \lim_{N\to\infty} \sum_{n=0}^N \pi(\xi_n) = \lim_{N\to\infty} \sum_{n=0}^N \eta_n,$$

and hence $\sum\limits_{n=0}^\infty \eta_n$ exists (and equals $y$). So $X/M$ satisfies the criterion, hence is a Banach space.


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