Why isn't $\,\mathcal C[0,1]$ a Banach space in this unusual norm?

Question

I wish to ask the following question:

Let $\mathcal X$ be the normed space $\,\mathcal X=\mathcal C([0,1])$, with norm defined as $$ \|\,f\|= \max_{x\in[0,1]} x^2 \lvert\,f(x)\rvert. $$

Why isn't $\mathcal X$ a Banach space?

Answer 1

Take $$ f_n(x)=\left\{\begin{array}{ccc} \frac{1}{x} & \text{if} & \frac{1}{n}\le x\le 1,\\ n^2x & \text{if} & 0\le x\le \frac{1}{n}. \end{array}\right. $$ Then $\{f_n\}_{n\in\mathbb N}\subset C[0,1]$, and for $m\ge n$, $$ \|f_m-f_n\|=\sup_{x\in [0,1]}x^2\lvert\,f_m(x)-f_n(x)\lvert \overset{\ast}=\sup_{x\in [0,1/n]}x^2\lvert\,f_m(x)-f_n(x)\lvert\overset{\ast\ast}\le \frac{1}{n}\to 0, $$ owing $\ast\ast$ to the fact that in $[0,\frac{1}{n}$ all $f_n$s are positive and thus $f_n-f_m\leq f_n$, which in turn isn't greater than its value in $\frac{1}{n}$ which is precisely $\frac{1}{n}$, combined with $x^2\leq1$, and owing $\ast$ to the fact that $x^2\cdot\frac{1}{x}$, which is what we would be taking the supremum of in $[\frac{1}{n},1]$, is $x$, and values greater than its supremum there, which is 1, are attained bu $n^2x\cdot x^2$ in $[0,\frac{1}{n}]$, at least for $n$ sufficiently big. But the sequence does not converge is $C[0,1]$, since it is not bounded in the standard $C[0,1]$ norm.