Recall the following theorem:

Theorem: If $X$ is Banach space and $A$ is a bounded linear map on $X$ with $\|A\| < 1$, then $\sum\limits_{i=1}^{\infty}A^{i}$ converges, and the limit is $(I-A)^{-1}$.

I am trying to thinking of a counter example in case $X$ is not Banach.

That is, I want to find a normed vector space $Y$ that is not Banach, and a linear map $B$ on $Y$ such that $\|B\| < 1$ but $\sum\limits_{i=1}^{\infty}B^i$ does not exist as a bounded linear map on $B$.

The non-banach space that I feel one could find an example in is the space $Y = \ell_{c}(\mathbb{N})$, which consists of of infinite sequences with only finitely many non-zero entries, together with the norm $\|a\| =\sum\limits_{i=1}^{\infty} |a_i|$.

Can I find a map $A$ on this space with $\|A\| <1$ but such that the series $\sum\limits_{i=1}^{\infty}A^i$ does not converge to a bounded linear map on the space?


  • 1
    $\begingroup$ One problem that would occur for non-complete spaces is simply that a Cauchy sequence of vectors need not have a limit in the space. But/and this is not a pathology. $\endgroup$ Jun 23, 2017 at 1:53
  • $\begingroup$ Hi, I'm familiar with examples of Cauchy sequences without limits in non-complete normed spaces (such as $\ell_{c}(\mathbb{N})$. Could you give me a hint on how I could explicitly construct a linear map $A$ with the desired properties? $\endgroup$
    – ttb
    Jun 23, 2017 at 2:15
  • $\begingroup$ btw, I still have in my drawer print outs of your abstract algebra notes that I studied from a few years back! $\endgroup$
    – ttb
    Jun 23, 2017 at 2:22
  • $\begingroup$ For $Ae_n=e_{n+1}/2$ on the finite vectors in $\ell^2$, your sum is $\sum e_n/2^n$, which converges in $\ell^2$ of course, but has infinitely-many non-zero components. (Algebra notes! :) $\endgroup$ Jun 23, 2017 at 17:22

1 Answer 1


Try the composition of shift and division by 2: $$A(a_1,a_2,a_3,\dots) = (0,a_1/2,a_2/2,a_3/2,\dots)$$


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.