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


  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.