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?

  • $\begingroup$ 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. $\endgroup$ – Srivatsan Jan 7 '12 at 13:26
  • $\begingroup$ 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? $\endgroup$ – BBDynSys Jan 7 '12 at 13:34
  • $\begingroup$ 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 $m < n$. $\endgroup$ – Srivatsan Jan 7 '12 at 13:39
  • $\begingroup$ stackexchange rules! :) $\endgroup$ – BBDynSys Jan 7 '12 at 14:56
  • $\begingroup$ 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. $\endgroup$ – BBDynSys 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.

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

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,

  • $\begingroup$ My bad, prove not define. I corrected it. $\endgroup$ – BBDynSys Jan 7 '12 at 13:38
  • $\begingroup$ 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. $\endgroup$ – BBDynSys Jan 7 '12 at 16:47
  • $\begingroup$ There is ; look at Martin Sleizak's answer. $\endgroup$ – Patrick Da Silva Jan 7 '12 at 18:00
  • $\begingroup$ No I meant for the algebra you mentioned. I guess there is none. $\endgroup$ – BBDynSys Jan 7 '12 at 18:47
  • $\begingroup$ 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. =) $\endgroup$ – Patrick Da Silva Jan 7 '12 at 18:49

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?

  • 1
    $\begingroup$ 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... :-) $\endgroup$ – Martin Sleziak Jan 7 '12 at 14:11
  • $\begingroup$ Did you guys use Latex or a WYSIWYG editor to draw these? $\endgroup$ – BBDynSys Jan 7 '12 at 14:30
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    $\begingroup$ @user6786 jsxgraph.org $\endgroup$ – David Mitra Jan 7 '12 at 14:39
  • $\begingroup$ @user6786 I used metapost. I can add source code to my answer, in case it could be useful/interesting for you. $\endgroup$ – Martin Sleziak Jan 7 '12 at 15:38
  • $\begingroup$ Please attach the code, that would be great. $\endgroup$ – BBDynSys Jan 7 '12 at 15:44

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