I am reading A course in real analysis by John McDonald, on page 530, it says "it is easy to show $|||J|||=1$" where $J$ is the linear operation $J:C([0,1])\rightarrow C([0,1])$, defined by $J(f)(x) = \int_0^xf(t)dt\\$. I have $|||J|||:=\text{sup}\{||D(f) \ : ||f||\le1\}$
This is what I have and it might be way off, so any help would be great.
$|||J||| = \text{sup}\{\max_{x\in[0,1]}|\int_0^xf(t)dt| \ : \ \max_{x\in[0,1]}|f(x)|\le 1\}$
Suppose $\max_{x\in[0,1]}|f(x)|\le 1$, then $f(x)\le1$ for $x\in[0,1]$. Thus $\int_0^xf(t)dt\le\int_0^x1*dt=x\le 1$ for $\forall \ x\in [0,1].$ So, $\max_{x\in[0,1]}|J(f)|\le 1$ and thus $|||J|||\le1$
Now let $f(x)=1$, then $f(x)\in C([0,1])$ and $\max_{x\in[0,1]}f(x)=1$ and $\max_{x\in[0,1]}|\int_0^x1*dt$= 1. So $\text{sup}\{\max_{x\in[0,1]}|\int_0^xf(t)dt| \ : \ \max_{x\in[0,1]}|f(x)|\le 1\}\ge 1$. Thus we have that $|||J|||=1$.