TITLE: Easy example to illustrate that the following normed space is not complete
QUESTION [5 upvotes]: Consider the vector space $C([0,1])$ of real-valued continuous functions on $[0,1]$ endowed with the standard norm: $$ \Vert f\Vert_2 = \sqrt{\int_0^1 f(x)^2 dx}.$$ I know that this normed space is not complete. 
Is this because the function $f_(x) = x^n$ converges to the discontinuous function which is zero on $[0,1)$ and $1$ at $x=1$?

REPLY [6 votes]: The sequence $(f_n)$ converges in $(C[0,1],\|\cdot\|)$ to the function $0$ so this does not prove that $(C[0,1],\|\cdot\|)$ is not complete. 
An example is the sequence $(g_n)$ defined by $g_n(x)=0$ if $0\leqslant x\leqslant\frac12-\frac1{2n}$, $g(x)=nx-\frac12n+\frac12$ if $\frac12-\frac1{2n}\leqslant x\leqslant\frac12+\frac1{2n}$ and $g(x)=1$ if $\frac12+\frac1{2n}\leqslant x\leqslant1$. Then $(g_n)$ converges in $(L^2[0,1],\|\cdot\|)$ to the function $g$ defined by $g(x)=0$ if $0\leqslant x\leqslant\frac12$ and $g(x)=1$ if $\frac12<x\leqslant1$ but $g$ is not in $C[0,1]$ hence $(g_n)$ diverges in $(C[0,1],\|\cdot\|)$. But $\|g_n-g\|\leqslant\frac1{2\sqrt{n}}$ hence $(g_n)$ is a Cauchy sequence in $(C[0,1],\|\cdot\|)$.