Is $f(e^x)$ log-concave if $f$ is positive decreasing convex function on $[0,a]$?

Question

Let $f$ be a decreasing convex function which is positive on $[0,a)$ and $0$ on $[a,\infty)$. Furthermore, suppose $f$ is differentiable on $(0,a)$, for simplicity. Then, is $f(e^x)$ log-concave? Examples are $f(y)=(a-y)^k\,\,(k\ge 1)$, $f(y)=e^{-y}$, for which the conjecture is true. So far, I had tried the following. I tried to show the derivative of $\ln f(e^x)$ is decreasing.
$$ \frac{d}{dx}\ln f(e^x)=\frac{e^x f'(e^x)}{f(e^x)}=\frac{y f'(y)}{f(y)},$$ where $y:=e^x$. Define $$ g(z):=f^{-1}(f(0)(1-z))$$ This function is positive, increasing, and convex, and $g(0)=0$ and $g(1)=f^{-1}(0)$. Then we only have to show $$ \frac{g(z)}{(1-z)g'(z)}$$ is increasing. By geometrical intuition, this ratio seems increasing. Here, I got stuck.

Answer 1

If $a<1$, the function $f(e^x)$ is identically zero on $(0,a)$, because $e^x\geq 1$ for every $x>0$, and $f$ vanishes on $[a,\infty)$, hence on $[1,\infty)$, because $a<1$. Therefore the logarithm of the function is not defined on $(0,a)$, and in particular it is not log-concave.

If $a\geq 1$, then the function $f(e^x)$ is log-concave in $(0,a)$ if and only if the function $f(y)$ is log-concave in $(1,a)$. (Note that since $f$ vanishes on $[a,\infty)$, the logarithm of $f$ is not defined on $[a,e^a)$.)