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Question 1: Study the existence of $C^1$ function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfying $\forall x\in\mathbb{R},\mbox{ } f\circ f'(x)=x.$

Question 2: Study the existence of differentiable function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfing $\forall x\in\mathbb{R},\mbox{ } f\circ f'(x)=x.$

Question 3: Study the existence of $C^1$ function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfying $\forall x\in\mathbb{R},\mbox{ } f'\circ f(x)=x.$

For question 1, such a function can not exist because : f' must be injective and since f 'is continuous, f' must be strictly monotonous. For example, If we assume that f 'is strictly increasing We can show that $\displaystyle \lim_{x\rightarrow -\infty}f'(x)=-\infty$ and $\displaystyle \lim_{x\rightarrow +\infty}f'(x)=+\infty$ which implies that f ' is surjective .

with a simple argument it shows that f is injective ( if $f(x)=f(y)$ by surjection of $ f'$ , we have $f'(a)=x $ and $f'(b)=y$ for some real $ a,b $ , thus implie $a=f(f'(a)=f(x)=f(y)=f(f'(b)=b$ so $ x=y$).

the continuity of f proves that f est strictly monotonous. For example, if we suppose f strictly inreasing, we must have $f'>0$. this contradicts the surjectivity of $f '$

For Question 2 , I need help

Pascal
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  • I do not follow your argument. Why are you saying that $f'$ must be surjective? The given equality implies that it must be injective, no? – Ted Shifrin Aug 03 '19 at 17:37
  • @ Ted Shifrin it's a typo, I correct – Pascal Aug 03 '19 at 17:45
  • I think you'd better rethink your argument. Why can't $f'>0$ everywhere and still be injective? – Ted Shifrin Aug 03 '19 at 17:48
  • I prove the injectivity of f by using the surjection of f ', do you see it? – Pascal Aug 03 '19 at 17:52
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    But I already told you that you know $f'$ is injective, not surjective. The given equation tells you that $f'$ is injective and $f$ is surjective. I don't see any contradiction yet. Perhaps you have to use more information. – Ted Shifrin Aug 03 '19 at 17:55
  • I said : We can show that $\displaystyle \lim_{x\rightarrow -\infty}f'(x)=-\infty$ and $\displaystyle \lim_{x\rightarrow +\infty}f'(x)=+\infty$ which implies that f 'is surjective . Are u ok? – Pascal Aug 03 '19 at 18:00
  • For 2, Darboux theorem tells that a derivative must possess intermediate-value property regardless of its continuity, and together with the injectivity of $f'$, we can conclude that $f'$ is continuous. Therefore by Problem 1, no such function exists. – Sangchul Lee Aug 03 '19 at 18:31
  • @ Sangchul Lee Thank's I don't see! I know if $f: \mathbb{R}\to \mathbb{R}$ is a monotone function that satisfies the Intermediate Value Theorem then $f$ is continuous [https://math.stackexchange.com/questions/1679289/if-f-is-monotone-prove-it-is-continuous] but if f' is injective, why f' must be monotone – Pascal Aug 03 '19 at 19:35
  • @Tina, For a function on an interval, intermediate-value property(IVP) together with injectivity implies monotonicity. Then IVP plus monotonicity implies continuity. To initiate this argument for $f'$ in the case of Problem 2, all we need is to guarantee the IVP for $f'$, which is exactly the statement of Darboux theorem. – Sangchul Lee Aug 03 '19 at 19:49
  • Thank you very much Sangchul Lee – Pascal Aug 03 '19 at 19:52
  • @Tina, Glad it helped :) – Sangchul Lee Aug 03 '19 at 19:52
  • @ Sangchul Lee Please , where is the mistake in my attempt for question 1 ? – Pascal Aug 03 '19 at 20:39
  • @Tina, Your argument seems perfectly fine to me; in fact, my answer for Problem 1 is exactly your argument with a bit of simplification. – Sangchul Lee Aug 03 '19 at 22:53
  • @ Sangchul Lee Thank you – Pascal Aug 04 '19 at 09:07
  • For those interested i link the fact TVI+injectivity implies monotonicity https://math.stackexchange.com/questions/7172/injective-functions-with-intermediate-value-property-are-continuous-better-proo – Pascal Aug 04 '19 at 09:19
  • @ Sangchul Lee can u help me here thanks in advance

    https://mathoverflow.net/questions/337607/existence-of-function-satisfying-ffx-x-almost-everywhere/337611#337611

     – Pascal Aug 06 '19 at 18:08

1 Answers1

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Problem 1. We show that no such $f$ exists.

It is clear that $f$ is surjective and $f'$ is injective. Since $f'$ is continuous, $f'$ is strictly monotone.

Now we show that $f'$ is surjective. We know that $f'(\mathbb{R}) = (\alpha, \beta)$ for some $-\infty \leq \alpha < \beta \leq +\infty$. Then we have to prove that $\alpha = -\infty$ and $\beta = +\infty$. If both $\alpha$ and $\beta$ are finite, then $(f\circ f')(\mathbb{R}) \subseteq f([\alpha, \beta])$, which contradicts the surjectivity of $f$. So it follows that either $\alpha = -\infty$ or $\beta = +\infty$.

  • Assume that $f'$ is strictly increasing and $\beta < +\infty$. Then $f'(\mathbb{R}) = (-\infty, \beta)$. This shows that $f$ is strictly decreasing near $-\infty$, and so, $(f\circ f')(\mathbb{R}) = f((-\infty, \beta))$ must be bounded from below, contradicting the functional equation.

Other cases can be treated in a similar way to yield the contradiction, hence we conclude that $f'$ is surjective.

Now since $f'$ is bijective, the functional equation tells that $f$ is also bijective and is the inverse function of $f'$. But this implies that $f$ is monotone, which then contradicts the surjectivity of $f'$.

Problem 2. Again we show that no such $f$ exists.

We make use of the previous problem. To this end, it suffices to show that the given condition implies the continuity of $f'$. The following general lemma will be sufficient for our purpose:

Lemma. Let $g: [a, b] \to \mathbb{R}$ be differentiable and $g'$ be injective. Then $g'$ is continuous.

Proof. By the Darboux theorem, we know that $g'$ possesses intermediate-value property. Together with the injectivity of $g'$, this implies that $g'$ is strictly monotone. In particular, $g'$ can only have jump-discontinuities. Then again by the intermediate-value property, $g'$ cannot have jump-discontinuities. So $g'$ is continuous everywhere.

Problem 3. Again no such function exists.

Assume that $f$ solve the functional equation. Then $f$ is injective and $f'$ is surjective. The former implies that $f$ is strictly monotone, which in turn tells that $f'$ is either non-negative or non-positive. But this contradicts the surjectivity of $f'$.

Addendum. Interestingly, there exists a $C^1$-function $f$ such that $(f' \circ f)(x) = |x|$. Indeed, we can make an ansatz that $f$ takes the form $f(x) = c |x|^{\alpha}$ for some $\alpha > 1$ and $c > 0$ and then plug this to the equation to determine $\alpha$ and $c$. Then we should have $c^{\alpha}\alpha|x|^{\alpha^2 - \alpha} = |x|$, and so, we may choose $\alpha$ as the golden ratio $\phi = \frac{1+\sqrt{5}}{2}$ and then $c = \phi^{-1/\phi}$.

Sangchul Lee
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