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? 

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. 
