# Prove that a sequence is not Cauchy under the c[0,1] norm

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!

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.

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

• how do you know the norm is greater than 1/2? thanks! – xiao jiang Feb 15 at 8:16
• 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 $\frac12$. – Ingix Feb 15 at 11:49
• 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... – xiao jiang Feb 18 at 8:09
• You are trying to show that it is not Cauchy. – copper.hat Feb 18 at 9:28
• 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 $\|x_n-x_n\| \ge \epsilon$. – copper.hat Feb 18 at 13:23