I'm trying to prove the in-completeness of $\mathcal{C}[0,1]$ with the $L_{1}$ norm. To do so, I've considered the function $f_{n}(x) = |\sin{(2\pi/x)}|$ for $x \in [1/(n+1), 1/n]$ and $0$ elsewhere, as well as the theorem that completeness is equivalent to every absolutely summable series in a space being summable.

This seems intuitively to me like it should work as a counter-example, as simply plotting the sine waves and considering the intervals I don't see how the series of functions could sum to some continuous function on $[0,1]$ (it looks like it would be some step function to me), however it is clear that the series of functions is absolutely summable with $L_{1}$ simply by bounding $|\sin{(2\pi/x)}|$ by $1$ in the integral.

My problem is, I don't know how to rigorously show that the series of functions is not summable to a continuous function to complete my proof. I've taken the definition of summability as well as the definition of continuity, and tried to formulate some contradiction by supposing that $\sum_{n=1}^{\infty} f_{n} = f$ for some $f \in \mathcal{C}[0,1]$, but I'm not sure where the actual argument involving the function should come into place. I'm starting to think that this approach with a contradiction may not be working altogether.

How can I formalize this part of my argument to complete the proof?

  • $\begingroup$ Is there a reason you don't just provide a Cauchy sequence that doesn't converge? That seems far simpler. $\endgroup$ – Elliot G Nov 20 '16 at 22:50
  • $\begingroup$ @ElliotG I had already seen a few of the standard non-convergent Cauchy sequence arguments which complete the proof when I recently approached this question, so I guess I'm just interested in making this one work. $\endgroup$ – hermitude Nov 20 '16 at 22:55
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    $\begingroup$ The sum of the series is continuous on $(0,1]$. So look at how it behaves near $0$. $\endgroup$ – Daniel Fischer Nov 20 '16 at 23:02
  • $\begingroup$ @DanielFischer Ahh of course, thank you Daniel. $\endgroup$ – hermitude Nov 20 '16 at 23:03

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