Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function in $[0, 1]$ , strictly monotonic , $f(0)=1$ as well as $f(f(x))=x$ forall $x \in \mathbb{R}$. Evaluate:
\(\displaystyle{f(f(0))=0\implies f(1)=0}\) and since \(\displaystyle{f}\) is continuous and strictly decreasing at \(\displaystyle{\left[0,1\right]}\)
we get \(\displaystyle{f(\left[0,1\right])=\left[f(1),f(0)\right]=\left[0,1\right]}\). Also, according to the
relation \(\displaystyle{f(f(x))=x\,,\,\forall\,x\in\mathbb{R}}\), we have that \(\displaystyle{f^{-1}=f}\)
at \(\displaystyle{\left[0,1\right]}\). There exists \(\displaystyle{a\in\left[0,1\right]}\) such that \(\displaystyle{f(a)\neq a}\).
Then, \(\displaystyle{\int_{0}^{1}(x-f(x))^{2016}\,\mathrm{d}x>0}\) and