The space of continuous functions $C([0,1])$ is not complete in the $L^2$ norm

I am trying to prove that under the $L^2$ norm, $C([0,1])$ does not give rise to a complete metric space. To do this I am trying to find a Cauchy Sequence which does not converge in $C([0,1])$. As a template (on $C([a,b])$) I am led to believe the following is Cauchy:

$$f_n (x) = \begin{cases} 1 &\mbox{if } 0 \leq x \leq \frac{1}{2} \\ 1 - 2n(x-\frac{1}{2}) & \mbox{if } \frac{1}{2}\leq x \leq \frac{1}{2n} + \frac{1}{2}\\ 0 & \mbox{if } \frac{1}{2n} + \frac{1}{2} \leq x \leq 1 \end{cases} $$

But I am struggling to show this is Cauchy, I have tried integrating from 0 to 1 but this is giving a very nasty integral and I was wondering if anyone has a better method?

Solution 1:

You can make things easy and approximate $$\Vert f_n- f_m\Vert_2^2\le 4\cdot |1/(2n)-1/(2m)|$$ Note $f_n-f_m$ is $0$ of the interval with endpoints $1/(2n)+1/2$ and $1/(2m)+1/2$.

-- David Mitra