Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Define $\ell^1=\{x\colon\mathbb N\to\mathbb F: \|x\|_1~\mbox{is finite}\}$ where $\mathbb F$ is either $\mathbb R$ or $\mathbb C$. If $(x_n)$ is a Cauchy sequence in $\ell^1$, does that mean that $(\|x_n\|)$ is Cauchy in $\mathbb F?$

share|cite|improve this question

We have by triangular inequality,$$\lVert x_n\rVert\leq \lVert x_n-x_m\rVert+\lVert x_m\rVert$$ and switching $m$ and $n$ we get $$|\lVert x_n\rVert-\lVert x_m\rVert|\leq \lVert x_n-x_m\rVert.$$ Now, we can deduce that $\{\lVert x_n\rVert\}$ is Cauchy: for a fixed $\varepsilon>0$, let $N$ such that $\lVert x_n-x_m\rVert\leq \varepsilon$ whenever $m,n\geq N$. Then $|\lVert x_n\rVert-\lVert x_m\rVert|\leq \varepsilon$ whenever $m,n\geq N$.

Note that it's true in any normed space, not only in $\ell^1$.

share|cite|improve this answer


  • Let $u:E\to F$ be Lipschitz between two metric spaces $E$ and $F$. Then, if the sequence $(x_n)_{n\in\mathbb N}$ is Cauchy in $E$, the sequence $(u(x_n))_{n\in\mathbb N}$ is Cauchy in $F$.
  • For any norm $\|\ \|$ on any vector space $E$, the map $u:E\to\mathbb R$, $x\mapsto\|x\|$, is $1$-Lipschitz.
  • Conclude that the answer to your question is yes.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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