Proving a function is not a contraction Let $\psi: C([0,1]; \mathbb{R}) \mapsto C([0,1]; \mathbb{R})  $, be such that if $x \in [0,1]$ then $$\psi(f)(x) = \int\limits_{0}^{x} f(t) dt $$
I am asked to show that $\psi$ is not a contraction but $\psi^{2}$ is. And then conclude that $\psi$ has a unique fixed point. 
I´ve managed to prove a statment that helps me conclude the last part but I´m having trouble proving that the function is not a contraction. This is what I've got. 
If $f,g \in  C([0,1]; \mathbb{R})$  then $$ |\psi (f) - \psi(g)| \leq \int\limits_{0}^{x} |f(t)-g(t)| dt \leq \sup\limits_{x\in [0,1]} |f(x)-g(x)| \int\limits_{0}^{x} 1 dt = d(f,g) x  $$ Then $|\psi (f) - \psi(g)| \leq d(f,g) x $,  and taking supremum on the left hand side allows me to conclude that $d(\psi (f),\psi (g)) \leq d(f,g) x$
Is there any problem with this procedure? Does anybody has an insight on how to procede with the $\psi^{2}$ part? 
Many Thanks! 
 A: $\psi$ is a contraction if there is a
constant $L$ strictly less than one such that
$$
d(\psi (f),\psi (g)) \leq L \, d(f,g)
$$
for  all $f$ and $g$.
This is not the case as you can verify with $f(x) \equiv 0$
and $g(x) \equiv 1$.
But
$$
\psi^2(f)(x) = \int_{0}^{x}  \int_{0}^{t}f(s) \,  dsdt
 = \int_{0}^{x}  \int_{s}^{x}f(s) \,  dtds
 = \int_{0}^{x} (x-s) f(s) ds
$$
so that
$$
 | \psi^2(f)(x) - \psi^2(g)(x)| \le 
\sup\limits_{x\in [0,1]} |f(x)-g(x)| \int_{0}^{x} (x-s)  ds
 = \frac {x^2}2 \sup\limits_{x\in [0,1]} |f(x)-g(x)|
$$
and therefore
$$
d(\psi^2(f),  \psi^2(g)) \le \frac 12 d(f, g) \, .
$$
A: $\psi(1)=x$, which gives $\|\psi(1)-\psi(0)\|=\|x-0\|=1$. So $\psi$ is not a contraction. You can rewrite $\psi^2$ using integration by parts:
\begin{align}
  \psi^2(f) 
     & =\int_{0}^{x}\int_{0}^{x'}f(t)dtdx' \\
     & = \left.(x'-x)\int_{0}^{x'}f(t)dt\right|_{x'=0}^{x}-\int_{0}^{x}(x'-x)\frac{d}{dx'}\int_{0}^{x'}f(t)dtdx' \\
     & = \int_{0}^{x}(x-x')f(x')dx'.
\end{align}
You can get the same result by drawing the triangular region and interchanging the orders of integration. Therefore,
$$
       d(\psi^2(f),\psi^2(g)) \le \int_{0}^{1}(1-x')dx'\cdot d(f,g)=\frac{1}{2}d(f,g).
$$
