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 Post subject: Identity functionPosted: Tue May 30, 2017 11:54 am
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Joined: Mon Nov 09, 2015 1:36 am
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Location: Ioannina, Greece
Let $h\,:\mathbb{R}\longrightarrow\mathbb{R}$ be a function defined and valued in all real numbers, strictly increasing and continuous. We define the functions $h^{[n]}:\mathbb{R}\longrightarrow\mathbb{R}\,,\; {n}\in\mathbb{Z}$, as $$h^{[n]}(x)=\left\{{\begin{array}{ll} \mathop{(\underbrace{h\circ{h}\circ\ldots\circ{h}})}\limits_{n\;{\text{times}}}(x)\,,&{n}\in\mathbb{Z}^{+}\\ x\,,&{n}=0\\ \mathop{(\underbrace{{h^{-1}\circ{h^{-1}}\circ\ldots\circ{h^{-1}}}})}\limits_{n\;{\text{times}}}(x)\,,&{n}\in\mathbb{Z}^{-} \end{array}}\right.\,,$$ where $h^{-1}$ is the inverse function of $h$.
If
1. for every $x\in\mathbb{R}$, there exists an non-negative number $c_x$, such that, for every ${n}\in\mathbb{Z}$, holds
$$\left|{h^{[\rm{n+1}]}(x)-h^{[n]}(x)}\right|=c_x$$ and
2. there exists $x_0\in\mathbb{R}$, such that $h(x_0)=x_0$,

prove that $h(x)=x$, for all $x\in\mathbb{R}$.

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Grigorios Kostakos

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