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Is there a strictly increasing function $f : \Bbb{R}\to\Bbb{R}$ such that $f'(x) = f(f(x))$ for all $x$?

I think the answer is no and my argument goes like this: If there were, $f'(x) = f(f(x))$ would imply that $f$ is linear on some interval $J$ and therefore has the form $f(x)=\alpha x$ for some real $\alpha>0$. But then $f(x)=f'(x)=\alpha$ for $x\in J$. Contradiction because $f$ is strictly increasing.

I think my solution is too simple to be true. What is wrong with it?

vilma
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2 Answers2

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My quick proof to show there is no such surjective function:

Suppose $f'(x)=f(f(x))$ and $f$ is strictly increasing and surjective. Since $f$ is strictly increasing it is invertible, and the surjectivity assumption implies $f^{-1}$ is defined on all of $\mathbb R$.

Then $0<f'(f^{-1}(x))=f(x)$ for all $x\in \mathbb R$, a contradiction.

Forever Mozart
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  • Did you find the contradiction when $f$ is not surjective ? I got that $f$ has a horizontal asymptote at $-\infty$, say $L$. I also got that $f(L)=0$, hence $f(0)>0$ and $L<0$. But I can't see any contradiction yet... – Gabriel Romon Jun 27 '14 at 11:25
  • yes I got those conclusions as well. And also $f'(L)=f(0)$. I think that we must have $-1<L<0$...Mean Value Theorem? – Forever Mozart Jun 27 '14 at 17:15
  • @ForeverMozart how did you claim $f$ is surjective? – K41 May 08 '25 at 04:23
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Such a function does not exist.

Suppose the contrary, i.e. $f(x): \mathbb{R}\to \mathbb{R}$ is strictly increasing and $f'(x)= f(f(x)) \Rightarrow f$ is differentiable $\Rightarrow$ $f''(x) = f'(f(x))\cdot f'(x)$. Since $f$ is strictly increasing, $f'(x)\ge 0$, and therefore $f''(x)\ge 0$, meaning that $f'(x)$ is non-decreasing $\Rightarrow f''(x)$ is non-decreasing.

  • If $f''(x)=0, \forall x\in \mathbb{R}$, then $\exists a, b\in\mathbb{R}, f(x) = ax+b$, plugging it into the original functions gives $f\equiv 0$, a contradiction.

  • If for some $x_0\in\mathbb{R}, c=f''(x_0)>0$, then from the non-decreasing property of $f''(x)$, $f''(x)\ge c>0, \forall x\ge x_0$, therefore $f'(x)\ge c(x-x_0)+f'(x_0)$, putting $x$ sufficiently large, $\exists M\in\mathbb{R}, \forall x\ge M, f'(x)\ge 2$, and following the similar argument $\exists N\in\mathbb{R}, \forall x\ge N$, $f(x)\ge x+1\Rightarrow f'(x)=f(f(x))\ge f(x+1), \forall x\ge N$.

    Now by mean value property, select $a\ge N$ such that $f(a)>0$ (which is clearly possible), then $\exists \xi\in [a,a+1]$ such that $f(a+1)-f(a) = f'(\xi)\ge f'(a)\ge f(a+1)$, a contradiction.

erpxyr2001
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