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I need help to prove that the following sequence is not Cauchy under the $C[0,1]$ norm.

$$x_n(t) = \begin{cases}0 & \text{for } 0\le t\le\dfrac12 -\dfrac1n \\ nt -\dfrac n2 + 1 & \text{for }\dfrac12 - \dfrac1n\le t\le\dfrac12 \\ 1 & \text{for } t\ge\dfrac12\end{cases}$$

in the Luenberger's Optimization textbook, it is said that this sequence is not Cauchy under $C[0,1]$ norm but when I tried to compute the norm $\|x_m-x_n\|=\max_{t \in [0,1]}|x_m(t)-x_n(t)|$ as $n$ and $m$ goes toward infinity, it seems to me that this limit does go to $0$ which means that it is Cauchy.

Thanks!

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If it is Cauchy it would converge to a continuous function. But the pointwise limit is $0$ for $x <\frac 1 2$ and $1$ for $x >\frac 1 2$. Hence the sequence cannot be Cauchy.

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Note that with $t_m={m-1 \over 2m}$, we have $x_m(t_m) = {1 \over 2}$. Now choose $m\ge n$ such that $t_m \le {1 \over 2} -{1 \over n}$.

Then $x_m(t_m) -x_n(t_m) \ge {1 \over 2} $ and so for any $m$ there is some $n \ge m$ such that $\|x_m-x_n\| \ge {1 \over 2}$ and so the sequence $x_n$ cannot be Cauchy.

Note that $\|x_m-x_n\| \ge |x_m(t_m) -x_n(t_m)| \ge {1 \over 2} $.

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  • $\begingroup$ how do you know the norm is greater than 1/2? thanks! $\endgroup$ – xiao jiang Feb 15 at 8:16
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    $\begingroup$ The norm $\lVert x\rVert_{C[0,1]}$ is the supremum of $\lvert x(t)\rvert$ over all $t\in [0,1]$. The above answer calculates a specific $t_m$ such that $\lvert x(t_m)\rvert \ge \frac12$, where $x(t)=x_m(t)-x_n(t)$. So naturally the supremem over all $t\in [0,1]$ of $\lvert x_n-x_m\rvert$ is at least $\frac12$. $\endgroup$ – Ingix Feb 15 at 11:49
  • $\begingroup$ why does the first inequality in the answer hold? seems to me that if m is greater than or equal to n, then it would be the other way round... $\endgroup$ – xiao jiang Feb 18 at 8:09
  • $\begingroup$ You are trying to show that it is not Cauchy. $\endgroup$ – copper.hat Feb 18 at 9:28
  • $\begingroup$ You need to show that for all $\epsilon>0$ there is some $N$ such that for $n,m \ge N$ we have $\|x_m-x_n\| < \epsilon$. I have shown that there is some $\epsilon>0$ (${1 \over 2}$ above) such that for all $N$ there exists some $n,m \ge N$ such that $\|x_n-x_n\| \ge \epsilon$. $\endgroup$ – copper.hat Feb 18 at 13:23

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