Does there exist a real function $f(x)$ that satisfies the following properties?

  1. its domain is $\mathbb{R}$

  2. $f'(x) > 0$ for all $x$

  3. $f''(x) + (f'(x))^2 < 0$ for all $x$

The log function $\ln(x)$ gives some idea about conditions 2 and 3. But for now, I did not find any example.

Besides, I want to find a non-linear differential function $f(x)$ defined on $\mathbb{R}$ and that is:

  1. strictly increasing

  2. (quasi)-concave

I think this is easier than the previous one. By looking at the graph, I guess this function exists, but I did not find one explicitly.


Thank you for all the comments, especially by @mihaild. I have found an example for these questions. An example for the 2nd question is $-e^{-x}$. And an example for the 1st question is borrowing the idea of the 2nd one, which is the following.

If $f(x)=-e^{g(x)}$, then $f'(x)=-g'(x)e^{g(x)}$, $f''(x)=-g''(x)e^{g(x)}-(g'(x))^2e^{g(x)}$, and $f''(x)+(f'(x))^2=-g''(x)e^{g(x)}$. So we just need $g(x)$ is defined on $\mathbb{R}$ and satisfies:

  1. $g'(x) <0$ for all $x$,

  2. $g''(x) >0$ for all $x$

Then simply we choose $g(x)=e^{-x}$. So an example is $-e^{e^{-x}}$.

This is a wrong computation. I will try to fix it. $(f’(x))^2$ should be $(g’(x))^2 e^{2g(x)}$.


After checking my previous example and reading all comments again. I see there does not exist a real function for the 1st question. (Thanks a lot for a remark in @mihaild comment.)

Indeed, assume that there exists $f(x)$ that satisfies the 1st condition. Then $g(x)=e^{f(x)}$ is a positive, strictly increasing and strictly concave function. But by its concavity $$g(x) \leq g(0)+ g'(0)x.$$ Since $g'(0)>0$, we have $\displaystyle\lim_{x \to - \infty}g(x) =-\infty$, which contradicts its positivity.

  • $\begingroup$ $\log(\log(x))$ almost works. It just doesn't satisfy the domain requirements... $\endgroup$
    – Joe
    Commented Jan 30, 2021 at 17:11
  • $\begingroup$ @Joe yeah domain is the most important condition, that why I put it the first. Btw your example is very interesting. Thanks a lot. ^^ $\endgroup$
    – N.Quy
    Commented Jan 30, 2021 at 17:20
  • 5
    $\begingroup$ Could you show it? It may be helpful. Thanks. $\endgroup$
    – N.Quy
    Commented Jan 30, 2021 at 17:28
  • 1
    $\begingroup$ Does $-e^{-x}$ works for the second part? Also, logarithm of any positive function for the second part will work for the first part, but unfortunately there are no positive functions for the second part... $\endgroup$
    – mihaild
    Commented Jan 30, 2021 at 21:16
  • 1
    $\begingroup$ @manooooh you’re right, I made a mistake in computation of $(f’(x))^2)$ $\endgroup$
    – N.Quy
    Commented Jan 31, 2021 at 16:53

3 Answers 3


Let $g=1/f'$ so $g'>1,g>0\,\forall x\in \Bbb R$ which is impossible so such an $f$ does not exist.

  • 1
    $\begingroup$ That is very neat. $\endgroup$
    – TonyK
    Commented Jul 3, 2021 at 10:18
  • 1
    $\begingroup$ Why is it impossible for $g$ to satisfy $g'>1$ and $g>0$? $\endgroup$
    – Joe
    Commented Jul 3, 2021 at 13:33
  • 1
    $\begingroup$ @Joe It will not be satisfied for all $x<-M$ for some $M>0$ since at every point, $g$ increases at least at fast at $x$ which means it is unbounded in both directions. $\endgroup$
    – TheSimpliFire
    Commented Jul 3, 2021 at 13:38
  • 1
    $\begingroup$ @TheSimpliFire: Oh, of course! This is a very elegant answer. $(+1)$. $\endgroup$
    – Joe
    Commented Jul 3, 2021 at 13:45
  • $\begingroup$ @TheSimpliFire. I have decided to set a bounty on this exellent answer. Oh yeah, once you see it, it's simple. But the problem is in seeing it :-) $\endgroup$ Commented Jul 4, 2021 at 11:02

Adding to TheSimpliFire's solution (because I cannot comment), one can prove the assertion that a differentiable function from $\mathbb{R}$ to itself such that $g>0$ and $g'>1$ cannot exist by applying the mean value theorem to $g$, giving \begin{equation} |g(x)-g(y)|>|x-y| \text{ for all x,y}. \end{equation} By definition, the function is injective. Being continuous and injective it takes open sets to open sets. So $f(\mathbb{R})$ is open. Also, taking a sequence $(x_{n})$ converging to $x$ in $\mathbb{R}$ we observe that $f(x_{n})$ goes to $f(x)$ so $f(\mathbb{R})$ is closed. Being both open and closed in $\mathbb{R}$, $f(\mathbb{R}) = \mathbb{R}$.

The function is surjective, contradicting the fact that it can only take positive values.


After TheSimpleFire's elegant solution. Is my reasoning simple and correct?:

Let $g(0)=N>0$. Then $$\int_{-N-1}^0 g'(x)dx>\int_{-N-1}^0 1\;dx$$ $$g(0)-g(-N-1)>N+1$$ $$N-g(-N-1)>N+1$$ $$g(-N-1)<-1$$ contradicting the positivity of $g(x)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .