An exercise on convex decreasing function properties

Question

A function f$(x)$ defined for $x\geq0$. It is positive, decreasing, convex and log-convex: $\frac{d^2}{dx^2}\log[f(x)]>0$, $f(0)<1$. Can we prove that $f''(x)x+f'(x)>0$ for sufficiently large $x$?

Thanks in advance!

Answer 1

We just have to show that $$ x+\frac{f'(x)}{f''(x)}>0 \tag{1}$$ holds for any $x$ big enough.

We have that $g\triangleq \log f$ on $\mathbb{R}^+$ is negative, decreasing and convex, so $g'=\frac{f'}{f}$ is negative and increasing. Since $f'=f g'$, we have $f'' = f'g'+f g'' $. Now $(1)$ is equivalent to:

$$ (-g'(x))+\frac{g''(x)}{(-g'(x))}>\frac{1}{x}. \tag{2}$$ Let $h\triangleq -\frac{1}{g'}$. Then $h$ is positive and increasing, and we have to prove that for any $x$ big enough $$ h < (1+h')\, x \tag{3}$$ holds. By setting $v(t)\triangleq h(e^t)$, this is the same as proving that any positive increasing function $v:\mathbb{R}\to \mathbb{R}^+$ satisfies: $$ v(t)-v'(t) < e^t \tag{4}$$ for any $t$ big enough. Since the solution of the ODE $$ v(t) - v'(t) = K e^t $$ for $K\geq 1$ is given by $v(t) = (v(0)-K t)e^t$, the positivity of $v(t)$ gives that $(4)$ holds for any $t>v(0)$, hence $(1)$ holds for any $$ x > \exp\left(-\frac{f}{f'}(1)\right).$$

Answer 2

If $f=e^k$ is decreasing and log convex on $(0,\infty)$ we are asked to prove that there exists $x_0$ such that $$xk'^2(x)+(xk'(x))'\geq 0$$ for $x>x_0.$ To study this, we use the fact that $k'$ is increasing and negative. Therefore we can write $$k'(x)=-C-\int_{x}^{\infty}k''(t)dt$$ with $C\geq 0.$ Now $$I=xk'^2(x)+(xk'(x))'=x\left(\int_x^{\infty}k''(t)dt\right)^2+(xC-1)C+(2Cx-1)\int_{x}^{\infty}k''(t)dt+xk''(x)$$ which is positive if $x>1/C$ and $C>0$.

If $C=0$ then $$I=\left(x\int_x^{\infty}k''(t)dt-1\right)\left(\int_x^{\infty}k''(t)dt\right)+xk''(x).$$ It is negative if for instance $k''(t)=1/4t^2$ say for $t>1.$