# Showing $(C[0,1], d_1)$ is not a complete metric space

I am completely stuck on this problem: $C[0,1] = \{f: f\text{ is continuous function on } [0,1] \}$ with metric $d_1$ defined as follows:

$d_1(f,g) = \int_{0}^{1} |f(x) - g(x)|dx$.

Let the sequence $\{f_n\}_{n =1}^{\infty}\subseteq C[0,1]$ be defined as follows:

$f_n(x) = \left\{ \begin{array}{l l} -1 & \quad \text{$x\in [0, 1/2 - 1/n]$}\\ n(x - 1/2) & \quad \text{$x\in [1/2 - 1/n, 1/2 +1/n]$}\\ 1 & \quad \text{$x\in [1/2 +1/n, 1]$}\\ \end{array} \right.$

Then $f_{n}$ is cauchy in $(C[0,1], d_1)$ but not convergent in $d_1$.

I have proved that $f_{n}$ is not convergent in $(C[0,1])$ since it is converging to discontinuous function given as follows:

$f_n(x) = \left\{ \begin{array}{l l} -1 & \quad \text{$x\in [0, 1/2 )$}\\ 0 & \quad \text{$x = 1/2$}\\ 1 & \quad \text{$x\in (1/2 , 1]$}\\ \end{array} \right.$

I am finding it difficult to prove that $f_{n}$ is Cauchy in $(C[0,1], d_1)$. I need help to solve this problem.

Edit: I am sorry i have to show $f_n$ is cauchy

Thanks for helping me.

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Just compute $d_1(f_n, f_m)$ ... –  martini May 31 '12 at 21:54
The sequence is Cauchy with $d_1$. That is why you are having problems showing it is not Cauchy. If $m,n > N$, then it is straightforward to get the estimate $d_1(f_m,f_n) < \frac{2}{N}$. –  copper.hat May 31 '12 at 22:05
It happens to most of us... –  copper.hat May 31 '12 at 22:15
Be careful when you say "$f_n$ is not convergent since it is converging to discontinuous function". Remember, "convergent" means something different here than pointwise or uniform convergence - for example, the functions $f_n(x) = n^{-x}$ converge to the 0 function, even though $f_n(0) = 1$ for all $n$! So be sure you're using correct reasoning to show that the sequence isn't convergent in the given metric space. –  MartianInvader May 31 '12 at 22:39

Suppose $m,n > N$. Then $f_m(x) = f_n(x) = -1$ when $x \in [0, \frac{1}{2}-\frac{1}{N}]$. Similarly, $f_m(x) = f_n(x) = +1$ when $x \in [\frac{1}{2}+\frac{1}{N},1]$. And $|f_m(x)-f_n(x)| \leq 1$ when $x \in (\frac{1}{2}-\frac{1}{N}, \frac{1}{2}+\frac{1}{N})$.
Hence $d_1(f_m,f_n) = \int_{0}^{1} |f_m(x) - f_n(x)|dx = \int_{\frac{1}{2}-\frac{1}{N}}^{\frac{1}{2}+\frac{1}{N}} |f_m(x) - f_n(x)|dx < \frac{2}{N}$.