# Cauchy Sequence in $X$ on $[0,1]$ with norm $\int_{0}^{1} |x(t)|dt$

In Luenberger's Optimization book pg. 34 an example says "Let $X$ be the space of continuous functions on $[0,1]$ with norm defined as $\|x\| = \int_{0}^{1} |x(t)|dt$". In order to prove $X$ is incomplete, he defines a sequence of elements in $X$ by

$$x_n(t) = \left\{ \begin{array}{ll} 0 & 0 \le t \le \frac{1}{2} - \frac{1}{n} \\ \\ nt-\frac{n}{2} + 1 & \frac{1}{2} - \frac{1}{n} \le t \le \frac{1}{2} \\ \\ 1 & t \ge \frac{1}{2} \end{array} \right.$$

Each member of the sequence is a continuous function and thus member of space $X$. Then he says:

the sequence is Cauchy since, as it is easily verified, $\|x_n - x_m\| = \frac{1}{2}\left|\dfrac1n - \dfrac1m\right| \to 0$.

as $n,m \to \infty$. I tried to verify the norm $\|x_n - x_m\|$ by computing the integral for the norm. The piecewise function is not dependent on $n,m$ on the last piece (for $t \ge 1/2$), so norm $\|x_n - x_m\|$ is 0. For the middle piece I calculated the integral, it comes up zero. That leaves the first piece, and I did not receive the result Luenberger has. Is there something wrong in my approach?

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Notice that the points where you break up the function depend on the index, so you will have to break up the interval into *four* subintervals: $[0, \frac{1}{2} - \frac{1}{m}], [\frac12 - \frac1m, \frac12 - \frac1n], [\frac12-\frac1n, \frac12], [\frac12, 1]$. Now the functions are exactly the same in the leftmost and rightmost subintervals; we should take care of the middle ones. –  Srivatsan Jan 7 '12 at 13:26
Sorry if this is a very elementary question but: What is the value of the function in each of those pieces you mentioned? And did you have to assume $m>n$, or $n>m$ while calculating these new subintervals? –  BB_ML Jan 7 '12 at 13:34
Ah, that was an oversight on my part; I should have mentioned which number is greater (although if you stare at it, you could figure it out as well). I assumed $m < n$. –  Srivatsan Jan 7 '12 at 13:39
stackexchange rules! :) –  BB_ML Jan 7 '12 at 14:56
That's really nice being able to provide a direct link to a book page from Google Books. Since such reads are surgical, GB will allow in all the time no doubt. –  BB_ML Jan 7 '12 at 14:58

It's relatively easy to see that for $m<n$ we have $x_n(t)\le x_m(t)$ for each $t$. Hence $$\|x_m-x_n\|=\int_0^1 x_m(t) \mathrm{d}t-\int_0^1 x_n(t) \mathrm{d}t.$$ We can disregard intervals $\langle 0,1/2-1/m\rangle$, since both functions are zero there. We can also disregard $\langle 1/2,1\rangle$, since $x_m(t)=x_n(t)$ on that interval. Therefore $$\|x_m-x_n\|=\int_{\frac12-\frac1m}^1 x_m(t) \mathrm{d}t-\int_{\frac12-\frac1n}^1 x_n(t) \mathrm{d}t=\frac1{2m}-\frac1{2n}.$$ The last equality can be shown by direct computation. You can also see this geometrically: If you draw the picture, the first integral is area of a triangle with base $\frac1{2m}$ and height $1$. The second is a triangle as well, the base is $\frac1{2n}$.

I used metapost to create the picture. In case someone is interested to see it, it is figure 6 in this source code: rapidshare, megaupload, pastebin.

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Hm. Slicker than my brute force computation thing. +1 –  Patrick Da Silva Jan 7 '12 at 13:33
@Martin, did you mean to use $x_n$ for the second term in $$\int_0^1 x_m(t) \mathrm{d}t-\int_0^1 x_m(t) \mathrm{d}t$$ –  BB_ML Jan 7 '12 at 13:41
You're right @user6786. I've corrected it. –  Martin Sleziak Jan 7 '12 at 13:49
There is the inequality that says $||x_m|| - ||x_n|| \le ||x_m-x_n||$. I guess in this case it is simply $||x_m|| - ||x_n|| = ||x_m-x_n||$. Is that because $x_n(t) \le x_m(t)$? –  BB_ML Jan 7 '12 at 15:35
@user6786 I've added link MetaPost source - usually I prefer ifile for sharing files; but it's not working for me at the moment. –  Martin Sleziak Jan 7 '12 at 15:49

A picture may help:

The area is $${1\over 2}\cdot 1\cdot ( {1\over 2}-a_m) - {1\over2}\cdot1\cdot({1\over2}-a_n) = {1\over 2}{1\over m} -{1\over2}{1\over n.}$$

I did not see Martin's answer as I was typing this. Should I delete this?

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Considering that answers have been added more or less simultaneously, I don't see why we can't keep them both. (If you have a look at edit history of my answer, you'll see that originally I've posted an answer without a picture and I've added it only later - it took some time to prepare a picture.) You're picture is nicer - I did not use colors... :-) –  Martin Sleziak Jan 7 '12 at 14:11
Did you guys use Latex or a WYSIWYG editor to draw these? –  BB_ML Jan 7 '12 at 14:30
@user6786 jsxgraph.org –  David Mitra Jan 7 '12 at 14:39
@user6786 I used metapost. I can add source code to my answer, in case it could be useful/interesting for you. –  Martin Sleziak Jan 7 '12 at 15:38
Please attach the code, that would be great. –  BB_ML Jan 7 '12 at 15:44

He does not "define" that $X$ is incomplete, he proves it.

The idea is that the function $|x_n - x_m|$ looks like this : assume $n < m$, so $$(x_n - x_m)(t) = \begin{cases} 0 & \text{ if } t \le \frac 12 - \frac 1n \text{ or } t \ge \frac 12 \\ nt- \frac n2 + 1 & \text{ if } \frac 12 - \frac 1n \le t \le \frac 12 - \frac 1m \\ (n-m)t - \frac{n-m}2 & \text{ if } \frac 12 - \frac 1m \le t \le \frac 12. \end{cases}$$ Computing the integral gives you $$\left( \left. \frac{nt^2}2 - \frac {nt}2 + t \right|_{\frac 12 - \frac 1n}^{\frac 12 - \frac 1m} \right) + \left( \left. \frac{(n-m)t^2}2 - \frac{(n-m)t}2 \right|_{\frac 12 - \frac 1m}^{\frac 12} \right) = \frac 12 \left( \frac 1n - \frac 1m \right).$$ The first parenthesis is the integral over the second part of the piecewise writing of $x_n - x_m$ and the second parenthesis is the integral over the third part. The integral over the first part is $0$. The sequence is Cauchy because of this.

Hope that helps,

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My bad, prove not define. I corrected it. –  BB_ML Jan 7 '12 at 13:38
Patrick, I tried to obtain the last equality, unfortunately I was lost in a sea of algebra. I am sure there is a simplifying trick unknown to me that'd make things a bit easier.. but anyways. –  BB_ML Jan 7 '12 at 16:47
There is ; look at Martin Sleizak's answer. –  Patrick Da Silva Jan 7 '12 at 18:00
No I meant for the algebra you mentioned. I guess there is none. –  BB_ML Jan 7 '12 at 18:47
Sometimes if one does not see no tricks, he has to dirty his hands. In this case we got lucky, but sometimes we don't. I thought I didn't. =) –  Patrick Da Silva Jan 7 '12 at 18:49