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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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    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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    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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    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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    Ple$a$se attach the code, that would $b$e great.2012-01-07