# $C[0,1]$ with metric $d_2$ not complete [duplicate]

Possible Duplicate:
How to prove that $C^k(\Omega)$ is not complete

May i ask you for a little help about the following problem.

Consider $C([0,1])$ with the metric $$d_{L^{2}}:=\left(\int_{0}^{1}\left|f(x)-g(x)\right|^{2} dx \right )^{1/2}$$ What i have to show is that this metric space is not complete.

I have to find a Cauchy sequence that does not converge, but i have difficulties with finding an example of such function.

Thank you in advance.

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## marked as duplicate by Norbert, Cameron Buie, Henry T. Horton, rschwieb, Hagen von EitzenNov 12 '12 at 15:32

Try $f_n(x) = (-1)1_{[0,\frac{1}{2}-\frac{1}{n})}(x)+n(x-\frac{1}{2}) 1_{[\frac{1}{2}-\frac{1}{n},\frac{1}{2}+\frac{1}{n}]}(x)+1_{(\frac{1}{2}+\frac{1}{n},1]}(x)$.

Show it is Cauchy, and show that the 'limit' is not continuous, hence not in $C[0,1]$.

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For simplicity's sake I will use $C[-1,1]$ instead of $C[0,1]$.
Consider the sequence $(f_n(x))$ defined as $$f_n(x)=\begin{cases} -1 & \quad x\in \left[0, - \frac{1}{n}\right] \\ nx& \quad x\in \left[- \frac{1}{n},\frac{1}{n}\right]\\ 1 & \quad x\in \left[\frac{1}{n}, 1\right]\\ \end{cases}.$$ Then

1. $(f_n)$ is Cauchy:
If $n>m$ then $|f_n(x)-f_m(x)|=0$ if $|x|>\frac{1}{m}$ and $|f_n(x)-f_m(x)|\leq 1$ if $|x|\leq \frac{1}{m}$. Thus $d_2(f_n,f_m)<\sqrt{\frac{2}{m}}.$
2. $(f_n)$ is not convergent in $C[-1,1] \ldots$
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Thank you very much! I did it. –  Lullaby Nov 12 '12 at 12:19