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I need to find all functions with $g(1 - g(x)) = x$ and $x\in \mathbb{R}$. I must also provide a specific example of such function. My first attempt was to invert the expression yielding $g(x) = 1 - g^{-1}(x)$ and inverting again $x = g^{-1}(1 - g^{-1}(x))$. But this doesn´t bring me any further. I am not familiar with this type of problem and appreciate some guidance.

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  • $\begingroup$ What sort of solutions you are looking at? There is no continuous solution for this problem! $\endgroup$ – achille hui Jun 13 '18 at 15:12
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    $\begingroup$ Complex solutions : $g(x) = \pm \imath x + (1 \mp \imath)/2$ $\endgroup$ – Ronald Blaak Jun 13 '18 at 15:17
  • $\begingroup$ any solution (necessary discontinuous) satisfy $g(x) + g(1-x) = 1$. This implies $S = g(-100)+\cdots+g(101) = 101$. $\endgroup$ – achille hui Jun 13 '18 at 23:40
  • $\begingroup$ @Steve_Steve Let $h(x) = 1-g(x)$, it satisfies $h(h(x)) = 1-x$. We have $$h(1-x) = h(h(h(x)) = 1 - h(x) \iff g(x) + g(1-x) = 1$$ This leads to $S = \sum\limits_{k=-100}^{101} g(k) = \sum\limits_{k=1}^{100} (g(k) + g(1-k)) = \sum\limits_{k=1}^{100} 1 = 100$ $\endgroup$ – achille hui Jun 15 '18 at 22:54
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There is no such continuous function. Notice that if for all $\,x\,$ we have $\, g(1-g(x)) = x, \,$ then the inverse function $\, g^{-1}(x) = 1 - g(x) \,$ implies that $\,g\,$ is a bijection. Since it is defined on the reals this implies that the function is strictly monotone increasing or decreasing. In both cases the composition $\, g(1-g(x)) \,$ is monotone decreasing which contradicts it being equal to $\,x.$

If there are no restrictions on the function $\,g\,$ then it can be very arbitrary. We prove a general result. Suppose $\, t(x) \,$ is any involution and suppose $\, g(t(g(x)) = x \,$ for all $\,x \in \mathbb{R}.\,$ Thus $\,g\,$ is a bijection. Given any $\, a \in \mathbb{R},\,$ then let $\, b := g(a),\, c := t(b),\, \,$ and now, $\, a = g(c). \,$ Thus $\, b = g(g(c)) = t(c), \,$ because $\, t \,$ is an involution. Thus $\, t(x) = g(g(x)) \,$ for all $\, x\in \mathbb{R}.\,$ This is the only restriction on $\,g.$

In other words, this general result uses only elementary group theory. Suppose $\, t,g \in G \,$ a group such that $\, 1 = tt = gtg. \,$ Then $\, t(tg) = (tt)g = g = g(gtg) = (gg)(tg), \,$ and therefore $\, t = gg.\,$

Note that in our original question, if we define $\, f(x) = g(x+\frac12)-\frac12, \,$ then $\, -x = f(f(x)). \,$ Some good solutions to this are in question 312385 "Find a real function f:R->R such that f(f(x))=-x?" If you want concrete examples of functions $\, g \,$ then take a look at the solutions there. A nice solution is given as $\, f(x) = \text{sign}(x) (1 - (-1)^{\lceil x\rceil}) \,$ which is only piecewise continuous.

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  • $\begingroup$ Only continuous bijections need to be montone. But still a nice argument. $\endgroup$ – freakish Jun 13 '18 at 15:23
  • $\begingroup$ Of course. I forgot. $\endgroup$ – Somos Jun 13 '18 at 15:24
  • $\begingroup$ @Somos - very helpful. Much appreciated $\endgroup$ – Steve_Steve Jun 14 '18 at 6:06

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