Is $C[0,1]$ not complete with the $L^2$ distance?

Question

$$X=C[0,1]\qquad d(f,g)=\left(\int_{0}^{1}\vert f-g \vert^{2}dx\right)^{1/2}$$ The complete metric space is defined that every Cauchy sequence should be convergent on my book, so I think that the main goal is to either find at least one Cauchy sequence that is not convergent or to prove that all Cauchy sequences are convergent.
I have completely no idea how to deal with the given distance here, could anybody give me some idea?

Answer 1

$C[0,1]$ is not complete with the metric $d(f,g)$. To show it pick the sequence $(f_n)_n\subset C[0,1]$ defined by $$ f_n(x)=\left\{\begin{array}{ll}0,&0\leq x<\frac{1}{2}-\frac{1}{n}\\n(x-\frac{1}{2}+\frac{1}{n}),&\frac{1}{2}-\frac{1}{n}\leq x<\frac{1}{2}\\1,&\frac{1}{2}\leq x\leq1 \end{array}\right. $$ Draw the first ones before you keep reading! This is a Cauchy sequence as a consequence of the estimation $$ \int_0^1(f_n-f_m)^2dx\leq\int_{1/2-1/n}^{1/2}f_n^2dx\leq\int_{1/2-1/n}^{1/2}f_ndx=\frac{1}{2n},\hspace{.8cm}n\geq m, $$ but there are no $f\in C[0,1]$ such that $d(f_n,f)\to0$ because $f_n(x)\to\chi_{[1/2,1]}(x)$ for every $x$ in $[0,1]$.

Answer 2

Define $$ g_n(x)=\min(1,\max(-1,n(2x-1)) $$ Then $g_n$ and $g_m$ for $m,n\ge N$ only differ on an interval of length $\frac2N$ by at most $1$. Which means their $L^2$ distance is less than $\sqrt{\frac2N}$. Thus they form a Cauchy sequence, but the limit is obviously not continuous, there is a jump at $x=\frac12$.