Finding the norm of integral operator

Question

I have the following operator:

$A: (C^1[0;1];|||\cdot|||)\rightarrow(C^1[0;1];||\cdot||_{\infty})$

$Af(x)=\int_0^x f(t)dt$

where

$|||f|||= ||f||_\infty+||Af||_\infty$

What is $||A||?$

I wrote

$||Af||_\infty\leq L(||f||_\infty+||Af||_\infty)$

but don't know how to proceed.

Thanks for the help!

Answer 1

Since \begin{align} \left\|Af \right\|_{\infty} := \sup_{x\in [0,1]} \left| \int_{0}^{x} f(t) \, \mathrm{d}t \right| \le \sup_{x\in [0,1]} \int_{0}^{x} |f(t)| \, \mathrm{d}t \le \int_{0}^{1}|f(t)|\, \mathrm{d}t =: \left\|f\right\|_{L_1} \le \left\|f\right\|_{\infty}, \end{align} then \begin{align} \left\| A \right\| &:= \sup_{f \in C^{1}[0,1]} \frac{\left\|Af \right\|_{\infty} }{\left\|Af \right\|_{\infty} + \left\|f\right\|_{\infty} } \le \sup_{f \in C^{1}[0,1]} \frac{\left\|Af \right\|_{\infty} }{\left\|Af \right\|_{\infty} + \left\|Af\right\|_{\infty} } \le 1/2. \end{align} On the other hand, taking $f =1$ shows that $\left\|A \right\| \ge 1/2$.