Can someone take a look at my answer and tell me what I am missing, also, what do you think this problem is trying to teach me?
Problem
The functions $f_n=x^n$ are square integrable and continuous on $[0,1]$. Use this sequence of functions to show that $L=\frac{d}{dx}$ is bounded on $L^2[0,1]$ and unbounded on $C[0,1]$ with $sup\, norm$. (Hint: compute $lim_m \, ||Lf_n||$ for the two cases
Solution Attempt
Square integrable means $\int_{-\infty}^\infty |f(x)|^2dx < \infty$. But I don't see where to use this fact.
Also, calculating what the book says in the hint, $||Lf_n||= ||\frac{d}{dx}x^n||=||nx^{n-1}||=\sqrt{(nx^{n-1})^2}=nx^{n-1}$
I think $L^2[0,1]$ means this is a the $L^2$ norm for $x\in[0,1]$... if this is true, then the supremum of $||Lf_n||=nx^{n-1}=n$ when $x=1$, so $L$ is bounded by $n$ for this particular $f_n$, but that does not prove $L$ is bounded for any square integrable function.
For the second part, I think $C[0,1]$ is the set of complex numbers $x=a+bi\in[0,1]$. This might mean $nx^{n-1}=n(a+bi)^n$, taking the norm: $||n(a+bi)^n||=n\sqrt{(a+bi)^{2n}}$. So, for example, take $a+bi=0+\infty i<1$. Thus $n||(0+\infty i)^n||=n\sqrt{\infty^{2n}}$ which is not bounded.