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When solving functional equations it can be helpful to substitute another function, say, g(x) rather than x to the original functional equation h(x). Under what condition is this a permissible substitution that leaves the original solution intact? Is bijectivity sufficient and if so, why is it?

Example: h(1-h(x))=x. Can I substitute g(x) for x in this equation?

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Any substitution is possible whenever the necessary assumptions are fulfilled. They are often helpful to simplify the equation in question. Let $h\colon\Bbb R\to\Bbb R$. Put $1-h(x)$ instead of $x$. Hence $h(1-x)=1-h(x)$, so $$h(x)+h(1-x)=1,$$ which, maybe, maynot, allows us to gain some extra properties of the solution. For instance, for $x=\frac{1}{2}$ we get $h\bigl(\frac{1}{2}\bigr)=\frac{1}{2}.$ This is not so visible by the original equation.

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  • $\begingroup$ Thank you. But is h(x) + h(1-x) = 1 equivalent to the original equaition and if so, why is it? Because then I could make further elaboartions on the newly derived expression without losing information. $\endgroup$ – TaylorJames Aug 13 '18 at 22:51
  • $\begingroup$ I assumed that $h$ is a solution. This is an implication only. For example, the identity map fulfils the equation of mine, but this is not a solution of the original equation. Nevertheless, the addditional knowledge of a solution could be derived from here. This is a necessary condition. If you collect enough information, you could check whether or not the function you arrived at is a solution of the original equation. $\endgroup$ – szw1710 Aug 13 '18 at 22:53
  • $\begingroup$ Does this mean that any solution of the original function must also satisfy your function? And why is this? $\endgroup$ – TaylorJames Aug 13 '18 at 23:00
  • $\begingroup$ Does this mean that all solutions to the original function must satisfy your function? I must determine the values of the sum = f(-100) + f(-99) ... + f(100) + f(101) for all possible f which would lead to all sums being equal to 101 as the rest would cancel out $\endgroup$ – TaylorJames Aug 13 '18 at 23:09
  • $\begingroup$ The first question is whether there exists a function satisfying the original equation. Next, if we suppose there is such function, you are right, with my equation (implied by the original one) your task is trivial. Now we have what follows: if $f$ fulfils the original equation, then the sum in question is $101$. $\endgroup$ – szw1710 Aug 14 '18 at 0:59

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