Consider $V = C([0,1],\mathbb{R})$ with the norm $\Vert \cdot \Vert_{\infty}$. Define the application $\Phi: V \to V$ by
$$\Phi(f)(x) = 1 + \int_{0}^{x}f(\tau)\operatorname{d}\tau.$$
(a) Show that $\Phi$ is not a contraction, but $\Phi \circ \Phi$ is a contraction(b) Show that $\Phi$ has a unique fixed point in $V$.
I have an idea how to prove item (b). Actually, I proved in a previous question that if $f: \mathbb{R}^{n} \to \mathbb{R}^{n}$ is such that $f^{k} = f \circ … \circ f$ is a contraction, then $f$ has a unique fixed points and I don't remember using that $\dim \mathbb{R}^{n} = n$.
For item (a), I think I didn't understand the function, maybe it's a notation problem. For show that $\Phi$ is not a contraction is sufficient to find a function $f$ such that $\max f > 1$, because $\Vert \Phi'\Vert_{\infty} = \Vert f \Vert_{\infty}$? Now, I was very confused about $\Phi \circ \Phi$. How is that composition?
Best Answer
Let $f=1, g=0$. Then $\phi(f)(x) = 1+x$, $\phi(g)(x) = 1$ and so $\|\Phi(f)-\Phi(g)\| = 1$ and $\|f-g\| = 1$.
$\phi^2(f)(x) = 1+ \int_0^x \Phi(f)(t) dt= 1 + x + \int_0^x \int_0^t f(s) ds dt$.
Hence $\phi^2(f)(x) - \phi^2(g)(x) = \int_0^x \int_0^t (f(s) -g(s)) ds dt$ and so $\|\phi^2(f)(x) - \phi^2(g)(x) \| \le \int_0^x \int_0^t ds dt \|f-g\| = {x \over 2} \|f-g\|$.
If $\Phi^m$ has a unique fixed point then so does $\Phi$. To see why, suppose $\Phi(x) = x$, then $\Phi^m(x) = x$ and hence $\Phi$ has at most one fixed point. For existence, suppose $\Phi^m(x) = x$. Consider the sequence $x, \Phi(x), \Phi^2(x), \Phi^{m-1}(x)$. Note that each of these points is a fixed point of $\Phi^m$ and hence $\Phi^k(x) = x$ for all $k$.
Aside: When proving the Picard iteration theorem for ODEs, there are two broad approaches: (i) Choose a small enough interval so that $\Phi$ is a contraction on that interval. (ii) Iterate $\Phi$ enough times so that $\Phi^m$ is a contraction.