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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:

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$

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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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 $d_1(f_m,f_n) = \int_{0}^{1} |f_m(x) – f_n(x)|dx = \int_{\frac{1}{2}-\frac{1}{N}}^{\frac{1}{2}+\frac{1}{N}} |f_m(x) – f_n(x)|dx < \frac{2}{N}$.

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