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I’m trying to show that the functional sequence below is not Cauchy Sequence on $C[-1,1]$ space wrt $\|.\|_{\infty}$ which is equal to $\sup_{[-1,1]}|f_n(x)|$ but I cannot obtain something concrete for exact proof. Could you please fix my mistakes or warn me about unecessary things?

$$f_n(x) = \begin{cases} 0, & \text{if $-1\le x\le 0$} \\ nx, & \text{if $0\lt x\lt 1/n$} \\ 1, & \text{if $1/n\le x\le 1$} \end{cases}$$

I have used negation of Cauchy Sequence definition

$\exists \varepsilon_0 \gt 0$ $\forall N \in \mathbb N$ $\exists n,m \ge N$ $\|f_n-f_m\|_{\infty} \ge \varepsilon_0$

Let $m\gt n$ and $1/m \lt 1/n$

I’ve written

$\|f_n-f_m\|_{\infty}$ = $\max\{\sup_{0\lt x\lt 1/m}|f_n-f_m|$ , $\sup_{1/m\lt x\lt 1/n}|f_n-f_m|\}$ = $\max \{\sup_{0\lt x\lt 1/m}|(n-m)x|$ , $\sup_{1/m\lt x\lt 1/n}|nx-1|\}$

I cannot find a $\varepsilon_0$

With all due respect, I think there ara many mistakes. I need your help

Thanks

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    $\begingroup$ Suppose $n>m$, note $\|f_n-f_m\|_\infty ≥ |f_n(1/{m})-f_m(1/m)|$. What is the expression on the right? $\endgroup$ – s.harp Mar 23 '18 at 21:35
  • $\begingroup$ @s.harp Thanks for your concern. Isn’t it 0? $\endgroup$ – esrabasar Mar 23 '18 at 21:45
  • $\begingroup$ @Math1000 since $1/n<1/m$, $1/n\le1/m\le1$ hence $f_n(1/m)=1$ and $f_m(1/m)=1$ again. Why is it wrong? $\endgroup$ – esrabasar Mar 23 '18 at 22:03
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    $\begingroup$ Ok, to be correct it has to be evaluated at $1/n$. Then $f_n(1/n)=1$ and $f_m(1/n)=m/n$. If you keep $m$ fixed and make $n$ as large as you like this difference becomes close to $1$. $\endgroup$ – s.harp Mar 23 '18 at 22:09
  • $\begingroup$ @s.harp I have understood. I missed $f_m(1/n)=m/n$. For an instant, it has seemed as 1. Thanks a lot $\endgroup$ – esrabasar Mar 23 '18 at 22:14
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Let $m \ge n$. The functions $f_m$ and $f_n$ are equal on $[-1, 0] \cup \left[\frac1n, 0\right]$ so we have:

$$f_m(x) - f_n(x) = \begin{cases} mx - nx, & \text{if $x \in \left[0, \frac1m\right]$} \\ 1-nx, & \text{if $x \in \left[\frac1m, \frac1n\right]$}\\ 0, & \text{otherwise} \end{cases} $$

Now we calculate $$\|f_m - f_n\|_\infty = \max\left\{\sup_{x \in \left[0, \frac1m\right] } (m-n)x, \sup_{x \in \left[\frac1m, \frac1n\right] } (1-nx)\right\} = 1 - \frac{n}{m}$$

Therefore, if we set $m = 2n$ we get

$$\|f_{2n} -f_n\| = \frac12 \not\to 0$$

so the sequence $(f_n)_n$ cannot be Cauchy.


A more conceptual way to see that $(f_n)_n$ cannot be Cauchy is to recall that $C[-1,1]$ is a Banach space and that the uniform limit of continuous functions is itself a continuous function.

If $(f_n)_n$ were Cauchy, then by completeness of $C[-1,1]$ it would converge to an element of $C[-1,1]$. Since uniform convergence implies pointwise convergence, the only candidate for the uniform limit is the pointwise limit $\chi_{\langle 0, 1]}$. However, this is not a continuous function so the convergence cannot be uniform.

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  • $\begingroup$ Thanks a lot. Is $C[-1,1]$ Banach wrt sup norm? $\endgroup$ – esrabasar Mar 24 '18 at 9:06
  • $\begingroup$ @esrabasar Yes it is. $\endgroup$ – mechanodroid Mar 24 '18 at 11:36
  • $\begingroup$ thanks a lot :) $\endgroup$ – esrabasar Mar 24 '18 at 15:09

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