Let $L$ be the space of continuous functions $f(x)$ on $ 0≤x≤1$ and $||f|| = \max|x²f(x)|$, show that the space $L$ is not complete.
I have constructed few sequences of $f_n(x)$ such as $f_n(x) = nx$, when $0≤x≤1/n$, and $f_n(x)=1$ when $1/n≤x≤1$ so that way you get the function within the closed interval $[0,1]$. Also, $f_n(x) = x_n$ on $[0,1]$, then $f(x) = 0$, if $x=0$ or $x∈[0,1)$ and $f_n(x) = 1$ if $x=1$. However, after this I need to show that it is a Cauchy sequence in order to prove that the space is incomplete.
How do I prove that this is a Cauchy sequence? Can someone help me out with this?
$f_n(x)-f_m(x)=0$ if $ n<m$ and $x>\frac 1 n$. For $x \leq \frac 1 n$ we have $| f_n(x)-f_m(x)| \leq |f_n(x)|+|f_m(x)| \leq \frac 2 {n^{2}}$ because $f_n$ and $f_m$ take vales in $[0,1]$. Hence $| f_n(x)-f_m(x)| <\epsilon$ whenever $m>n>n_0$ where $n_0$ is such that $\frac 2 {n_0^{2}} <\epsilon$.
If this Cauchy sequence converges with the given norm then then limit function has to be $0$ at $0$ and $1$ at all other points, a contradiction to continuity. Hence the space is not complete.