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

Solution 1:

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.

Solution 2:

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,

Solution 3:

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?