I have a function $H$ whose domain is the set $C$ of the continuous functions $[0,1] \to \mathbb{R}$. $H$ is defined by $$H(f)(t) := \int_0^t \phi(f(s))ds $$ for $t \in [0,1], f \in C$ and with $\phi: \mathbb{R} \to \mathbb{R}$ continuous.
What would be a nontrivial condition (imposed onto $\phi$) under which $H$ is a contraction? I know I have to show for $f,g \in C$ that $$d_{sup} (H(f), H(g))= \sup_{t \in [0,1]} \left|\int_0^t \phi(f(s)) - \phi(g(s))ds\right| \leq L \sup_{t \in [0,1]} {|f(t)-g(t)|} = L d_{sup}(f,g)$$ with $L<1$. But all I arrive at is $$d_{sup}(H(f), H(g)) \leq M-m$$ where $M, m$ denote the maximum of $\phi \circ f$/minimum of $\phi \circ g$ which both exist by the properties of continuous functions on compact intervalls. What am I missing here?
Also is it correct that $H$ maps to itself, i.e. that $H(f)$ is continuous? Thanks!
Supone $\phi$ is Lipschitz with constant $L_\phi$. Then, for any $f,g\in C[0,1]$ \begin{align} |H(f)(t)-H(g)(t)|&\le\int_0^t|\phi(f(s))-\phi(g(s))|\,ds\\ &\le L_\phi\int_0^t|f(s)-g(s)|\,ds\\ &\le L_\phi\,d_{\text{sup}}(f,g), \end{align} and $$ d_{\text{sup}}(H(f),H(g))\le L_\phi\,d_{\text{sup}}(f,g). $$