Prove $C[a,b]$ with norm $\Vert f\Vert=\left(\int\limits_{a}^b\vert f(t)\vert^2\right)^{\frac{1}{2}}$ is not Banach space

Question

Prove $C[a,b]$ with norm $\Vert f\Vert=\left(\int\limits_{a}^b\vert > f(t)\vert^2\right)^{\frac{1}{2}}$ is not Banach space.

Banach space is complete normed space. We say complete normed space if every Cauchy sequence is convergent.

I have checked $\Vert f\Vert$ is norm. Now, I want to proof $C[a,b]$ is not complete under norm $\Vert f\Vert$, i.e. there exist Cauchy sequence which not convergent.

Now I confuse to choose the Cauchy sequence is not convergent. Can anyone give hint to choose Cauchy sequence is not convergent?

Answer 1

Consider, for every $k \in \{1,2,3,...\}$ the map $f_k \colon [-1,1] \to \mathbb{R}$ such that:

$f_k(x)=-1$ if $x\in [-1,-1/k]$,

$f_k(x)=kx$ if $x \in [-1/k,1/k]$ and

$f_k(x)=1$ if $x \in [1/k,1]$.

Verify that, w.r.t. this norm, it is the case that $f_k \to f$, being $f(x)=-1$ if $x\in[-1,0)$, $f(0)=0$ and $f(x)=1$ if $x\in (0,1]$.

Clearly $f_k$ is continuous for every $k \in \{1,2,..\}$ but $f$ is not. This in particular means that $\{f_k\}$ is a Cauchy sequence in $C([-1,1])$ with that norm, but it does not converge in $C([-1,1])$. Hence $C([-1,1])$ with that norm is not a Banach space.

Answer 2

Say, $[a,b]=[-1,1]$. Let $f_{n}(x)=0$ for $-1\leq x\leq 0$ and $f_{n}(x)=1$ for $1/n\leq x\leq 1$ and $f_{n}(x)=nx$ for $0\leq x\leq 1/n$, $f_{n}\rightarrow f$ in $L^{2}$, where $f(x)=0$ for $-1\leq x\leq 0$ and $f(x)=1$ for $0<x\leq 1$.