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? 

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,
