3

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.

Edumaths555
  • 535
  • SInce we are dealing with continuous functions here, there is no need to distinguish between the Riemann and the Lebesgue intgral, since they're the same in this context. – José Carlos Santos Oct 08 '17 at 18:49
  • @zwim Actually, it is, since it is the null function. What's your point? – José Carlos Santos Oct 08 '17 at 18:52
  • @JoséCarlosSantos yes, I was too hasty giving $x^n$, but a continuous step function works. e.g https://math.stackexchange.com/questions/336417/completeness-of-langle-mathscrc-0-1-cdot-1-rangle?noredirect=1&lq=1 It has to be adapted for $L^2$ norm but this is the same stuff. – zwim Oct 08 '17 at 19:13

1 Answers1

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.

José Carlos Santos
  • 440,053