Show that space of continuous functions on interval $[a,b]$ with integral metric is not complete - question

Question

Show that space $C[a,b]$ with metric given by $\rho(f,g) = \int_a^b|f(x)-g(x)|dx$ is not complete.

I was considering a sequnece of continuous functions defined as follows

$ f_n(x) = \begin{cases} 0, & \text{if $x\in[a,\frac{a+b}{2}$]} \\ nx - n(\frac{a+b}{2}), & \text{if $x\in [\frac{a+b}{2}, \frac{a+b}{2} + \frac{1}{n}]$} \\ 1, & \text{if $x\in [\frac{a+b}{2} + \frac{1}{n}, b]$} \end{cases}$

But I have a problem about showing that this is indeed Cauchy sequence. Clearly, $\rho(f_n, f_m)$ is equal to $\frac{1}{2}\cdot |\frac{1}{n}-\frac{1}{m}|$ since its the area of the triangle between these two functions. The question I have is about making it infinitesimal. How can we estimate such value to make it lower than $\epsilon$?

Answer 1

Let $f(x) = 1_{({1 \over 2},1]}(x)$. Then $f_n(x) \uparrow f(x)$ (pointwise) for all $x$.

In particular, if $n >m$ we have $f(x) \ge f_n(x) \ge f_m(x)$ and so $|f_n(x)-f_m(x)| \le |f(x)-f_m(x)|$.

Hence $\rho(f_n,f_m) \le \int_0^\pi |f_n(x)-f_m(x)|dx \le \int_0^\pi |f(x)-f_m(x)| dx = {1 \over 2m}$.

Answer 2

Let $\epsilon > 0$ be given. Choose $n_0 > \frac{1}{\epsilon}$

Then if $n,m > n_0$,it follows that $\frac{1}{n}, \frac{1}{m} < \frac{1}{n_0} < \epsilon$

and hence:

$$\frac{1}{2}\left|\frac{1}{n} - \frac{1}{m}\right| \leq \frac{1}{2}\left(\frac{1}{n}+ \frac{1}{m}\right)< \frac{1}{2}(\epsilon + \epsilon) = \epsilon$$