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 at 6:04
• It can be shown that $$c>\frac1x\int_0^x\frac{f'''f'}{f^2}\,dt$$ Jan 30 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 at 5:46
• Sorry, should be "$3$ for $x\le 0$" to glue the values properly. Feb 1 at 12:40
• @fedja: Can you post your calculations as an answer? Feb 3 at 8:17