Let $f:[0,\infty)\rightarrow \mathbb R$ be a smooth function on $[0,\infty)$ ($C^3$ or even $C^2$ might suffice as well). Assume further that $f$ is positive, strictly decreasing and strictly convex on $[0,\infty)$ i.e. $f>0$, $f'<0$ and $f''>0$. Also, assume that $f\rightarrow 0$ as $x\rightarrow \infty$. Does that imply that the inequality $$2\cdot f\cdot f'' - f'^2 > 0$$ holds for all $x\in[0,\infty)$? Is there some function $f$ where it fails to hold for some values of $x$?


1 Answer 1


A counter-example, based on the error function. First define $$ \begin{align} g(x) &= 1 -\operatorname{erf}(x) = 1 - \frac{2}{\sqrt \pi} \int_0^{x} e^{-t^2} > 0 \, , \\ g'(x) &= - \frac{2}{\sqrt \pi} e^{-x^2} < 0 \, ,\\ g''(x) &= \frac{4x}{\sqrt \pi} e^{-x^2} \ge 0 \, . \end{align} $$ Note that $2g(x)g''(x) - g'^2(x)$ is negative for $x=0$, and therefore negative on some interval $[0, c]$.

Then for sufficiently small $\varepsilon > 0$ the function $f(x) = g(x + \varepsilon)$ satisfies all conditions $f > 0, f'<0, f''>0$ on $[0, \infty)$, and $2f(x)f''(x) - f'^2(x) < 0$ on $[0, c - \varepsilon]$.

  • $\begingroup$ Love you man <3 I was wondering what your though process was to coming up with it? How did you figure our from the inequality that you want something that decays very fast to form a counter-example? $\endgroup$
    – miniii
    Jun 27, 2019 at 18:40
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    $\begingroup$ @YousefKaddoura: I tried to find an example with $f(0)=1$,$f'(0)=-1$ and $f''(0)=0$. Then $f'$ must be increasing, stay negative, with $\int (- f'(x)) dx < 1$ so that $f$ stays positive. There are probably many choices for $f'$, but the error function was the first that came to my mind. $\endgroup$
    – Martin R
    Jun 27, 2019 at 18:47

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