$T:C[0,1]\to C[0,1]$ defined by $Tx(t)=\displaystyle\int_0^tx(s)ds,t\in[0,1]$ .
How to show $T^{-1}$ is unbounded.
First off all in the book the norm is not mentioned so I suppose (I think) the norm is supnorm.
My intuition says that find a function in $C[0,1]$ such that its derivative is not bounded because T is like definite integral on $[0,1]$
An example is derivative operator on $C[0,1]$ is not bounded is the following:
Take $f_n(x)=\sin(2\pi x)$ so since for $x=1/4n\in[0,1]$, $|\sin(2\pi x)|=1$
so $$||f_n||_\infty=1$$
but $$\left|\left|\dfrac{d}{dx}f_n\right|\right|=2\pi n\to \infty$$
and Can I really say $T^{-1}$ is a differentiation operator on $C^1[0,1]\subset C[0,1]$ ?
