Show that this linear operator is not continuous. This is a question of my last exam in Functional Analysis of my graduation:

Consider that norm: $|f|=\int_0^1 x^2 f(x)$ where $f\in C^0([0,1])$. Show that the linear operator $f(1-x)\in C^0([0,1])$ is not continuous.

MY ATTEMPT
I show that this operator is bounded with norm 1. So the operator is continuous and the teacher give me 0 in this question
MY FRIENDS ATTEMPT
Talking with my friends, they say that:
"This norm is more "refined" that the L1 norm, so the operator is limited."
MY TEACHER BLA BLA BLA
He say the operator is NOT limited but don't say the demonstration
I believe that all is wrong but I don't know how to show this... 
 A: Write $T$ for your operator. If you let $f_n(x)=n(1-x)^n$, then $|f_n|\to0$, while $|Tf_n|\to1$.
A: There are 3 relevant spaces here. You need to be careful where each element is and what the relevant notion of convergence is.


*

*The interval $[0,1]$

*The Banach space $C[0,1]$ of continuous functions on the interval. This space has the norm $\|f\|=\int_0^1 x^2 |f(x)|dx$

*The vector space $V$ of all automorphisms $C[0,1] \to C[0,1]$ with the operator norm. That is to say $\|A\|_{op} = \sup \{\|Af\| \colon \|f\|=1\}$.
Breaking down the definition of the operator norm we get. . .
$\displaystyle \|A\|_{op} = \sup \Big \{ \Big \|\int_0^1 x^2 |Af(x)|dx \Big\| \colon \int_0^1 x^2 |f(x)|dx = 1\}$.
What you want to prove is this value is undefined (infinite) for your given operator.
Your given operator is an element of $V$. That is to say a automorphism of $C[0,1]$ given by $Af(x) = f(1-x)$.
One strategy is to find a sequence of functions $f_n$ such that the RHS of the $:$ remains at $1$ while the LHS increases to infinity.
