# Smallest $c$ such that $f'<cf$ holds for all $f$ such that $f,f',f'',f'''>0$ and $f''' \le f.$

Let $$f: \mathbb{R} \to \mathbb{R}$$ be a $$C^3$$ function such that $$f,f',f'',f'''>0$$ and $$f''' \le f.$$ What is the smallest $$c$$ such that we can guarantee $$f'? Since $$f(x)=e^x$$ works, we must have $$c>1.$$ On the other hand, I managed to show $$c = 1.5^{1/3}$$ works.

Proof: First, we note $$\lim\limits_{x \to -\infty} f'(x) = \lim\limits_{x \to -\infty} f''(x) = 0.$$ Now $$f''' \le f \Rightarrow (f''^2)' = 2f''f''' \le 2f''f < 2f''f + 2f'^2 = (2ff')',$$ so integrating yields $$(f''^2)(x) - (f''^2)(x_0) < (2ff')(x)-2ff'(x_0)$$ for any $$x,x_0.$$ Taking $$x_0 \to -\infty$$ gives us $$f''^2 < 2ff'.$$

We use $$f''' \le f$$ again, but multiply by $$f'$$ instead: $$(f'f'')' = f''^2 + f'f''' \le f''^2 + f'f < 3ff' = (1.5f^2)'.$$ Integrating, we get $$(f'f'')(x) - (f'f'')(x_0) < (1.5f^2)(x) - (1.5f^2)(x_0) < (1.5f^2)(x).$$ Take $$x_0 \to -\infty$$ again to get $$f'f'' < 1.5f^2 \Rightarrow (\frac{1}{3}f'^3)' = f'^2f'' < 1.5f^2f' = (0.5f^3)'.$$ Integrate and take $$x_0 \to -\infty$$ for the last time to get $$\frac{1}{3}f'^3 < 0.5f^3 \Rightarrow f' < 1.5^{1/3} f.$$

• $1.5^{1/3} \approx 1.1447$. In this answer the weaker bound $c \le 3/6^{1/3} \approx 1.651$ is obtained by a different method. In this answer to a Putnam problem it is shown that $c \le 2$. – My guess would be that $c$ can be arbitrarily close to one. Jan 29, 2021 at 6:04
• It can be shown that $$c>\frac1x\int_0^x\frac{f'''f'}{f^2}\,dt$$ Jan 30, 2021 at 12:00
• @MartinR It is not $1$, but $1.0189420865882\dots$. The worst function is $1$ for $x\le 0$ and $e^x+2e^{-x/2}\cos(\sqrt 3 x/2)$ for $x\ge 0$. The equation for the corresponding maximum is too ugly to be solved analytically, but the Newton method can be used to get it with any precision you want. Feb 1, 2021 at 5:46
• Sorry, should be "$3$ for $x\le 0$" to glue the values properly. Feb 1, 2021 at 12:40
• @fedja: Can you post your calculations as an answer? Feb 3, 2021 at 8:17