Q. Let $T: C[0,1] \rightarrow C[0,1]$ be a mapping defined by $$ T f(x)=\int_{0}^{x} f(s) d s $$ Show that: $T$ is NOT a contraction but $T^{2}$ is a contraction.
It seems to me that $\begin{aligned}\left|\int_{0}^{x}(f(s)-g(s)) d s\right| & \leq \operatorname{sup}_{[0,1]}|f(s)-g(s)| \int_{0}^{x} d s \\ & \leq \sup _{[0,1]} |f(s)-g(s) |\end{aligned}$ What point I have missed here?