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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.2012-01-07
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    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?2012-01-07
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    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 $m < n$.2012-01-07
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    stackexchange rules! :)2012-01-07
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    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.2012-01-07
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    @user6786 Using Google Books link is IMO good for adding the context. (I use such links a lot.) But you should always keep in mind that if the page is displayed to you, it does not automatically mean that everyone else will be able to see it. So, to be on the safe side, all questions should be asked in a such way that they are self-contained and it's possible to understand them without using external links.2012-01-07
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    A closely related question: http://math.stackexchange.com/questions/21878/examples-of-function-sequences-in-c0-1-that-are-cauchy-but-not-convergent2012-01-07

3 Answers 3

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It's relatively easy to see that for $m 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 $\frac1{2n}$.

functions

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. +12012-01-07
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    @Martin, did you mean to use $x_n$ for the second term in $$\int_0^1 x_m(t) \mathrm{d}t-\int_0^1 x_m(t) \mathrm{d}t$$2012-01-07
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    You're right @user6786. I've corrected it.2012-01-07
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    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 $x_n(t) \le x_m(t)$?2012-01-07
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    @user6786 I've added link MetaPost source - usually I prefer ifile for sharing files; but it's not working for me at the moment.2012-01-07
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    I downloaded it. FYI, you can also use pastebin.com2012-01-07
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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.2012-01-07
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    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.2012-01-07
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    There is ; look at Martin Sleizak's answer.2012-01-07
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    No I meant for the algebra you mentioned. I guess there is none.2012-01-07
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    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. =)2012-01-07
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    Ok, just checking.2012-01-07
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A picture may help:

enter image description here

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... :-)2012-01-07
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    Did you guys use Latex or a WYSIWYG editor to draw these?2012-01-07
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    @user6786 http://jsxgraph.org/2012-01-07
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    @user6786 I used [metapost](http://en.wikipedia.org/wiki/Metapost). I can add source code to my answer, in case it could be useful/interesting for you.2012-01-07
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    Please attach the code, that would be great.2012-01-07