Prove that $T$ is not contraction and $T^2$ is contraction

Question

$T$ is not contraction and $T^2$ is contraction">

For (a) $||(Tf)(t)-(Tg)(t)||= ||\int ^t_0 f(s)ds-\int ^t_0 g(s)ds||\\ =||\int ^t_0(f(s)-g(s))ds||\\=\sup_{0\le t\le 1}|\int ^t_0(f(s)-g(s))ds| \\ \le\sup_{0\le t\le 1} \int^t_0|f(s)-g(s)|ds $

From here how we contradict that $T$ is not contraction and how to we processed for (b) and (c)

thank you ...

Answer 1

For (a), you should try to find a counter-example. For example, for any given $g(t)\in C[0,1]$, we can set $f(t)=g(t)+1$. Then you can show that $T$ is not contraction.

For (b), you can use the mean value theorem. \begin{equation} ||T^2 f-T^2g|| = \sup_{t\in [0,1]} |\int_0^t \int_0^s f(u)-g(u) \mathrm{d}u \mathrm{d}s| = |f(\xi)-g(\xi)| \sup_{t\in [0,1]}\frac{1}{2}t^2 \leq ||f-g|| \end{equation}

For (c), I think we should first try to prove that C[0,1] is a Banach space and then apply fixed point theorem to get the answer.