# Differential Inequality for a Convex Strictly Decreasing Positive Function

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 $$x$$?

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 $$[0, c - \varepsilon]$$.
• @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. – Martin R Jun 27 '19 at 18:47