0
$\begingroup$

Proof. Let $X$ be a normed space with norm $|\cdot |$ and $(x_n)$ be Cauchy. Then for all $\epsilon \gt 0, \ \exists N : m,n \gt N \implies |x_m - x_n| \lt \epsilon$ is the standard definition of Cauchy sequence. But it's easy to show that $||x_n| - |x_m|| \leq |x_m - x_n|$ and thus the sequence $|x_n|$ in $\mathbb{R}$ is Cauchy. By completeness of the reals under the absolute value norm, we have that $|x_n|$ approaches a limit and thus $(x_n)$ is absolutely convergent.

$\endgroup$
  • $\begingroup$ Is your normed space complete? Or does your definition of absolute convergence not entail convergence as well? $\endgroup$ – Andrés E. Caicedo Oct 3 '13 at 4:55
  • $\begingroup$ No, not complete. This is a lemma to showing that a normed space is a Banach space iff absolutely convergent sequences converge. $\endgroup$ – Shine On You Crazy Diamond Oct 3 '13 at 4:56
  • 1
    $\begingroup$ Ah, I see. The argument you indicate is correct. $\endgroup$ – Andrés E. Caicedo Oct 3 '13 at 6:04
  • $\begingroup$ Do you have a reference for this lemma? $\endgroup$ – Jeremy Jeffrey James Mar 13 at 14:57
4
$\begingroup$

The proof is correct. Applied more generally, it shows the following:

If $X$ and $Y$ are metric spaces, $(x_n)$ is Cauchy in $X$, and $f: X\to Y$ is a uniformly continuous map, then the sequence $f(x_n)$ has a limit.

The particular statement uses $Y=\mathbb R$ and $f(x)=\|x\|$ (which is a Lipschitz function).


That said, I don't understand the bigger picture. Apparently "absolutely convergent sequence" here means a sequence $(x_n)$ such that $\|x_n\|$ has a limit. This is the first time I see this term used anywhere (and I kind of hope it's the last one. It seems designed to confuse people.) More importantly, this notion of "absolutely convergent sequence" does not imply usual convergence, e.g., consider $x_n=(-1)^n$ in $\mathbb R$. And since $\mathbb R$ is a Banach space, this disproves the claim made in a comment, "a normed space is a Banach space iff absolutely convergent sequences converge".

Perhaps the intended claim was "a normed space is a Banach space iff absolutely convergent series converge". That is indeed correct, but then the argument given in the OP is not really relevant.

$\endgroup$

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.