Space of square integrable continuous function on $[-1,1]$

Question

I am confused if the space $L^2[−1, 1]$ of continuous functions on $[−1, 1]$ equipped with the $L^2$ norm (integration is by Riemann), Banach or not. I know with Lebesgue integration the space of square integrable functions is Banach with respect to the $L^2$ norm. But with the added continuity assumption and Riemann integration, I am just not sure how to go about proving or disproving this statement.

Answer 1

For each $n\in\mathbb N$, define$$\begin{array}{rccc}f_n\colon&[-1,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}1&\text{ if }x>\frac1n\\nx&\text{ if }x\in\left[-\frac1n,\frac1n\right]\\-1&\text{ if }x<-\frac1n.\end{cases}\end{array}$$Then, if $m,n\in\mathbb N$ and $m>n$,$$\bigl\Vert f_m-f_n\bigr\Vert_2=\frac2{\sqrt3}\sqrt{\frac1n-\frac1m}.$$Therefore, $(f_n)_{n\in\mathbb N}$ is a Cauchy sequence. However, it does not converge in your space, because if it did converge to some function $f$, then $f(x)$ would be equal to $1$ when $x>0$ and it would be equal to $-1$ when $x<0$. There is no such continuous function.