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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 $\mathbb F?$

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2 Answers 2

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 $\ell^1$.

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Hints:

  • 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.
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