Q:

Banach fixed point theorem application for continuous functions on [0,1]

The question asks to show there exists a unique f(x) on [0,1] such that this equation holds:

We define psi as such to show psi is a contraction and conclude it has a fixed point f(x), which is unique.
I get stuck with the algebra following the distance between the distance between two functions, f and g, under psi.
Thanks,

A:

The set $X=C([0,1])$ of continuous functions on $[0,1]$ with the metric
$$
d(f,g)=\sup_{0\le x\le1}|f(x)-g(x)|
$$
is a complete metric space. If $f\in X$, then $\psi\,f$ is a continuous function and $\psi\,f\in X$. Let $f,g\in X$. Then
$$\begin{align}
|(\psi\,f)(x)-(\psi\,g)(x)|&\le\int_0^1|f(x\,t)-g(x\,t)|\,t^2\,dt+\frac12|\cos f(x)-\cos g(x)|\\
&\le d(f,g)\int_0^1t^2\,dt+\frac12\,d(f,g)\\
&=\Bigl(\frac13+\frac12\Bigr)\,d(f,g).
\end{align}$$
We have shown
$$
d(\psi\,f,\psi\,g)\le\frac56\,d(f,g).
$$

