Consider the following mapping: $\phi(f)(x)=\int_0^xf(t)dt$
The mapping is $\phi:C[0,1]->C[0,1]$, i.e. over the continuous functions on $[0,1]$. I know that $\phi$ is not a contraction, but I am trying to show that $\phi\circ\phi$ is, i.e. the composition of $\phi$ with itself. Any ideas?
Unless I'm mistaken, for two functions $f_1$ and $f_2$ we would have $||\phi(f_1)-\phi(f_2)||_\infty=\sup|\int_0^x\int_0^t f_1-f_2 dt|$. But I'm not sure how to go from there in order to relate this to $||f_1-f_2||_\infty$
$$|\phi(\phi(f_1))(s)-\phi(\phi(f_2))(s)|=|\int_0^s\int_0^xf_1(t)-f_2(t)dtdx|$$ $$\leq ||f_1-f_2||_{\infty}\int_0^s \int_0^x dtdx$$ $$\leq||f_1-f_2||_{\infty}\frac{s^2}{2}$$ $$\leq \frac{1}{2}||f_1-f_2||_{\infty}$$
So by taking supremum you have the desired conclusion.