The question is as in the title.

If $f : \mathbb{R} \to \mathbb{R}$ is smooth, has a smooth inverse and $\lvert f(x) \rvert \leq \lvert x \rvert$, can we say anything about the growth bound of $f''(x)$?

I am trying to use the L'Hospital formula, but it does not seem to work well..

Could anyone please help me?

  • $\begingroup$ Consider smooth functions approximating a "staircase" shape. $\endgroup$
    – Karl
    Jun 1, 2023 at 14:47

3 Answers 3


There is no growth bound.

Let's say $f$ is increasing. You have $0 < f'(x) < \infty$ with $f(x) = \int_0^x f'(t)\; dt \le x$ for $x > 0$. The graph of $f'$ could have arbitrarily tall "bumps" as long as they are narrow enough to not make much of a difference to the integral. In those bumps, $|f''|$ could be arbitrarily large. Examples are not hard to construct.


im not sure we can find a bound. See this example: f(x) = xsin(x), f'(x) = sen(x)+xcos(x),f''(x) = 2cos(x) - xsin(x). f'' will grow indefinitely

  • 2
    $\begingroup$ But $x\sin(x)$ doesn't have inverse. $\endgroup$
    – mihaild
    Jun 1, 2023 at 14:54
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jun 1, 2023 at 15:13

We can't.

Let $g(x)$ be any smooth function s.t. $g(0) = 0$, $|g'(x)| \leq 1$. Then $f(x) = \frac{x}{2} + \frac{g(x)}{4}$ satisfies your conditions, but it's second derivative is $\frac{g''(x)}{4}$, which can be very large.

For example, consider $g(x) = \int_0^x \sin(x^2)\, dx$. Then we have $g'(x) = \sin(x^2)$ and so $|g'(x)| \leq 1$, but $g''(x) = 2x\cos(x^2)$, so it's not bounded.


You must log in to answer this question.

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