# Growth estimate of $f''(x)$ for smooth $f: \mathbb{R} \to \mathbb{R}$ with smooth inverse and $\lvert f(x) \rvert \leq \lvert x \rvert$

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..

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

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

• But $x\sin(x)$ doesn't have inverse. Jun 1, 2023 at 14:54
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.