This was posted to Math Stackexchange, but got no useful answers, and the more I think about it, the harder it seems.

I would like to know whether there exists a differentiable function from the (open or closed) unit interval to itself satisfying $$1-x-f(f(x))-f(x)f'(f(x))=0$$ for all $x$.

Ideally, I'd also like a list of all such functions.

This arose in the course of an economics problem whose description would be off-topic here, so lest this question seem too localized, let me pose a general question:

What techniques are available for solving functional equations of the above type?
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    $\begingroup$ Can you provide a link to the MSE question? $\endgroup$ – Benjamin Steinberg Oct 11 '13 at 1:11
  • $\begingroup$ "Unit interval" is $(0,1)$, right? $\endgroup$ – fedja Oct 11 '13 at 1:28
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    $\begingroup$ can you provide some background? $\endgroup$ – Maurizio Monge Oct 11 '13 at 1:53
  • $\begingroup$ @fedja: Yes, "unit interval" is (0,1). $\endgroup$ – Steven Landsburg Oct 11 '13 at 2:08
  • $\begingroup$ Maybe this could help: en.wikipedia.org/wiki/Carleman_matrix $\endgroup$ – Steve Huntsman Oct 11 '13 at 3:43

Write $$1-x=f(f(x))+f(x)f'(f(x)).$$ Note, that $f(x_1)=f(x_2)$ implies $x_1=x_2$ and thus the function is injective on $(0,1).$ Therefore, it is strictly monotone. This implies that $f(f(x))$ is strictly increasing.

If $f$ is increasing, then $f'(f(x))\ge 0$ and thus $1-x\ge f(f(x)).$ Letting $x\to 1$ leads to a contradiction.

If $f: (0,1)\to [0,1]$ and $f$ decreases, then $f(x)\cdot f'(f(x))\le 0$ and $1-x\le f(f(x)).$ Letting $x\to 0$ leads to a contradiction.

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  • $\begingroup$ "decreasing" is not the same as "negative". Besides, nobody said that $f$ is onto. $\endgroup$ – fedja Oct 11 '13 at 15:12
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    $\begingroup$ I am not sure I understand your point. It seems I have never used the fact that it is onto. According to OP, the range of $f$ has to be inside $(0,1).$ $\endgroup$ – Alvin Oct 11 '13 at 15:24
  • $\begingroup$ @LoïcTeyssier: I just claimed that if $f'(f(x))> 0$ at some point then $f'(x)\ge 0$ for all $x\in (0,1)$ which would imply that $f(f(x))$ has to be small around $1.$ $\endgroup$ – Alvin Oct 11 '13 at 15:30
  • $\begingroup$ @Safoura: the point fedja is making is that you can only obtain $f'(x)f''(f(x))\leq 0$ from the fact that $f'(f(x))$ decreases. Yet I agree with you: I don't see where the «onto» objection is relevant here. $\endgroup$ – Loïc Teyssier Oct 11 '13 at 15:32
  • $\begingroup$ Right, the onto part of the comment was my mistake (I just missed one trivial logical step in the argument) :-). $\endgroup$ – fedja Oct 11 '13 at 15:40

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