functional equation (01)

[Grigorios Kostakos]

Find all bijective (1-1 & on to) and continuous functions \( h: \mathbb{R}\longrightarrow\mathbb{R}\), such that:
(a) there is \( x_0\in\mathbb{R}\), such that \( h(x_0)=x_0\) and
(b) \(h(x)+h^{-1}(x)=2x\,,\) for all \(x\in\mathbb{R}\) .

[Grigorios Kostakos]

For the bijective and continuous function \( h\,: \mathbb{R}\longrightarrow\mathbb{R}\) holds:
$$h(x)+h^{ - 1}(x) = 2x \quad (1)\,, \;{\text{for all}} \; x\in\mathbb{R}\,.$$ Because \(h\) is bijective and continuous is strictly monotonic and the inverse function \(h^{-1}\) has the same monotony. But both \(h\) and \(h^{ - 1}\) can't be strictly decreasing, because from (1) we have that the sum \(h+h^{-1}\) must be an increasing function.
So both \(h\) and \(h^{ - 1}\) are strictly increasing \((2)\,.\)
The functions $$ h^{[\rm{n}]}(x) = \left\{{\begin{array}{ll}
\underbrace{({h\circ{h}\circ\ldots\circ{h}})}_{\rm{n-\text{times}}}(x)\,,&{\rm{n}}\in\mathbb{Z}^{ + }\\
x\,,& {\rm{n}} = 0\\
\underbrace{({h^{ - 1}\circ{h^{ - 1}}\circ\ldots\circ{h^{ - 1}}})}_{\rm{n}-\text{times}}(x)\,,&{\rm{n}}\in\mathbb{Z}^{ - } \\
\end{array}}\right.$$ are bijective and continuous and because of (2), are strictly increasing.
By induction we have that for every \({\rm{n}}\in\mathbb{Z}\) holds: $$ h^{[\rm{n}]}(x) = {\rm{n}}\,({h(x) - x}) + x\,,\; x\in\mathbb{R}\,.$$ Suppose that there exist \(\alpha\in\mathbb{R}\), such that \(\alpha < h(\alpha)\).
For the intervals \(\bigl[{h^{[\rm{n}]}(\alpha),\,h^{[\rm{n} + 1]}(\alpha)}\bigr]\), \({\rm{n}}\in\mathbb{Z}\), we have

\(\bullet\quad {\rm{diam}}\bigl({\bigl[{h^{[\rm{n}]}(\alpha),\,h^{[\rm{n} + 1]}(\alpha)}\bigr]}\bigr)=h(\alpha)-\alpha>0\,,\)

\(\bullet\quad \mathop{\bigcup}\limits_{{\rm{n}}\in\mathbb{Z}}\bigl[{h^{[\rm{n}]}(\alpha),\,h^{[\rm{n} + 1]}(\alpha)}\bigr]\) includes the fixed point \(x_0 = h(x_0)\) and

\(\bullet\quad \mathop{\bigcap}\limits_{{\rm{n}}\in\mathbb{Z}}\bigl({h^{[\rm{n}]}(\alpha),\,h^{[\rm{n} + 1]}(\alpha)}\bigr) = \varnothing\,.\)
For the fixed point \(x_0 = h(x_0)\) holds \(h^{[\rm{n}]}(\alpha)\neq{x_0}\), for every \( {\rm{n}}\in\mathbb{Z}\,.\)
( If for some \({\rm{k}}\in\mathbb{Z}\) holds \(h^{[\rm{k}]}(\alpha)={x_0}\), then
$$\alpha=h^{[-\rm{k}]}\bigl({h^{[\rm{k}]}(\alpha)}\bigr)=h^{[-\rm{k}]}({x_0})=x_0\quad\Rightarrow\quad\alpha= h(\alpha)\,.$$ Contradiction.)
Then there exists \({\rm{k}}\in\mathbb{Z}\), such that $$ x_0\in\bigl({h^{[\rm{k}]}(\alpha),\,h^{[\rm{k} + 1]}(\alpha)}\bigr)\quad (3)\,.$$ But \(h\bigl({\bigl({h^{[\rm{k}]}(\alpha),\,h^{[\rm{k} + 1]}(\alpha)}\bigr)}\bigr) = \bigl({h^{[\rm{k + 1}]}(\alpha),\,h^{[\rm{k} + 2]}(\alpha)}\bigr)\) and $$ x_0 = h(x_0)\in\bigl({h^{[\rm{k} + 1]}(\alpha),\,h^{[\rm{k} + 2]}(\alpha)}\bigr)\quad (4).$$ From (3) and (4) we have $$ x_0\in\bigl({h^{[\rm{k}]}(\alpha),\,h^{[\rm{k} + 1]}(\alpha)}\bigr)\bigcap\bigl({h^{[\rm{k} + 1]}(\alpha),\,h^{[\rm{k} + 2]}(\alpha)}\bigr) = \varnothing\,.$$ So doesn't exists an \(\alpha\), such that \(\alpha < h(\alpha)\). Similarly, it can be proved that doesn't exists an \(\alpha\), such that \( \alpha >h(\alpha)\). Then for all \(x\in\mathbb{R}\) must be \( h(x) = x\,.\quad\square\)