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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 $f'<cf$? 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 $\frac{1}{3}f'^3 < 0.5f^3 \Rightarrow f' < 1.5^{1/3} f.$

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  • 1
    $\begingroup$ $1.5^{1/3} \approx 1.1447$. 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. $\endgroup$
    – Martin R
    Jan 29 at 6:04
  • $\begingroup$ It can be shown that $$c>\frac1x\int_0^x\frac{f'''f'}{f^2}\,dt$$ $\endgroup$
    – TheSimpliFire
    Jan 30 at 12:00
  • 2
    $\begingroup$ @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. $\endgroup$
    – fedja
    Feb 1 at 5:46
  • $\begingroup$ Sorry, should be "$3$ for $x\le 0$" to glue the values properly. $\endgroup$
    – fedja
    Feb 1 at 12:40
  • $\begingroup$ @fedja: Can you post your calculations as an answer? $\endgroup$
    – Martin R
    Feb 3 at 8:17

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