Let $X = \ell^1$, the set of absolutely convergent real valued sequences and let $d_2(\mathbf{x},\mathbf{y}) = \left(\sum_{k=1}^\infty |x_k - y_k|^2\right)^{1/2}.$

This is the $2$-norm on the $1$ space.

Is $(X,d)$ complete? I don't think it is but I can't find an example of a non-convergent Cauchy sequence.


Consider the sequence $\mathbf x^{(n)}\in\ell^1\subset\ell^2$ defined by $$ x^{(n)}_k=\frac{1}{k}\textrm{ for } k\leq n,\quad x^{(n)}_k=0\textrm{ for } k > n. $$ Then $\mathbf x^{(n)}\overset{d_2}{\to} \mathbf x\in\ell^2\setminus\ell^1$, with $$ x^{(n)}_k=\frac{1}{k}\quad \forall k. $$ So $\ell^1$ is not $d_2$-complete.

  • $\begingroup$ I did think of that sequence, but then I couldn't prove that it's Cauchy with respect to the $d_2$ norm in the $\ell^1$. I know it must be because it converges in $\ell^2$, but is there any way of thinking about it without embedding $\ell^1$ inside $\ell^2$? $\endgroup$ Jun 8 '16 at 11:46
  • $\begingroup$ Well $d_2$-convergence (in the $d_2$-completion of $\ell^1$, that is $\ell^2$) implies the Cauchy properties in the space $\ell^1$. That is the fastest way to conclude. However, also the direct proof is quite simple: $d_2( \mathbf x^{(n)},\mathbf x^{(m)})^2\leq \sum_{i=m}^\infty i^{-2}=:R_m$, for any $n\geq m$. And $R_m\to 0$. $\endgroup$
    – guestDiego
    Jun 8 '16 at 12:06
  • $\begingroup$ Oh that is quite simple. Thanks a lot for your help $\endgroup$ Jun 8 '16 at 12:11

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