function and sequence

[xaranton]

Let \(f:[{0,1}]\longrightarrow[{0,\,1}]\) a continuous function such that \(f(0)=0\) and for every \(x,y\in[{0,1}]\) holds: \[|{f(x)-f(y)}|\geq|{x-y}|\,.\] Let, also, the sequence \(\{{x_{n}}\}_{n=1}^{\infty}\), with \(x_1\in[{0,\,1}]\) and \[x_{n+1}=f({x_{n}})\,,\;n\in\mathbb{N}\,.\] Prove that:

1) \(f(x)\geq{x}\) for every \(x\in[{0,\,1}]\),

2) the sequence \(\{{x_{n}}\}_{n=1}^{\infty}\) converges to a real number \(\lambda\), such that \(f(\lambda)=\lambda\,.\)

3) \(f(x)=x\) for every \(x\in[{0,\,1}]\).

[Grigorios Kostakos]

For the continuous function \(f:[{0,1}]\longrightarrow[{0,1}]\) holds \(f(0)=0\) and, for every \(x,y\in[{0,\,1}]\): \[|{f(x)-f(y)}|\geqslant|{x-y}|\quad(1)\,.\] Because \(f\bigl({[{0,\,1}]}\bigr)\subseteq[{0,\,1}]\), for every \(x\in[{0,\,1}]\), holds: \[0\leqslant{f(x)}\leqslant1\quad(2)\,.\] 1) From \((1)\) and \((2)\) we have
\[f(x)\stackrel{(2)}{=}|{f(x)}|=|{f(x)-0}|=|{f(x)-f(0)}|\geqslant|{x-0}|=x\quad\Rightarrow\quad{f(x)}\geqslant{x}\quad(3)\,.\] 2) The sequence \(\{{x_{n}}\}_{n=1}^{\infty}\), with \(x_1\in[{0,\,1}]\) and \(x_{n+1}=f({x_{n}})\,,\;n\in\mathbb{N}\) is bounded because for every \(n\in\mathbb{N}\) holds: \[x_{n+1}=f({x_{n}})\in[{0,\,1}]\,.\] The sequence is, also, increasing because for every \(n\in\mathbb{N}\) holds: \[x_{n+1}=f({x_{n}})\stackrel{(3)}{\geqslant}x_{n}\,.\] So, it must be convergent to a real number, say \(\lambda\).
Because the function \(f\) is continuous at \(x=\lambda\), we have that \[f({\lambda})=\displaystyle\mathop{\lim}\limits_{x\rightarrow\lambda}{\,f(x)}=\mathop{\lim}\limits_{n\rightarrow+\infty}{f(x_{n})}=\mathop{\lim}\limits_{n\rightarrow+\infty}{x_{n+1}}=\lambda\,.\] 3) From \((1)\) and \((2)\) we have \[f(1)=|{f(1)}|=|{f(1)-0}|=|{f(1)-f(0)}|\stackrel{(1)}{\geqslant}|{1-0}|=1\quad\stackrel{(2)}{\Longrightarrow}\quad{f(1)}=1\,.\] Suppose that exist \(x_0\in({0,\,1})\) such that \(f(x_0)>x_0\). Then \begin{alignat*}{2}
0\stackrel{(2)}{\geqslant}{f(x_0)-1}>x_0-1\quad\Rightarrow\quad& 0\leqslant{f(1)-f(x_0)}=1-f(x_0)<1-x_0\quad\Rightarrow\\
& |{f(1)-f(x_0)}|<|1-x_0|\,.
\end{alignat*} Contradiction. From \((1)\) we have that for every \(x\in[{0,\,1}]\) holds \(f(x)=x\).