Suppose we have a convex function $g(x)$ (i'm particularly interested in $g(x)$ decreasing but that's not a requirement for an answer)
Are there necessary or sufficient condition(s) that we can look at to examine whether $\log g(x)$ is concave?
Looking at some examples:
- an affine function is convex and log-concave
- $x^2, x^4$, etc are convex and log-concave
- $e^x$ is log concave
- $e^{x^2}$ is not log concave
So a crude guess might be that a convex function is log-concave if it does not increase "too fast". But I don't know if this is correct, and even if it is, I don't know what "too fast" is (is $e^x$ the limit?)
Let's assume $g$ is positive and twice differentiable. Then $(\ln g)'=\frac{g'}{g}$ and $(\ln g)''=\frac{g''g-g'^2}{g^2}$, so the condition of concavity is $g''g-g'^2\leq0$. Since $g$ is convex $g''\geq0$, so this is equivalent to $|g'|\geq\sqrt{gg''}$, the first derivative is greater by absolute value than the geometric mean of the second derivative and the function itself.
"Not increasing too fast" does not do it: $\frac1x$ does not increase at all, yet it is log-convex instead (taking $x>0$). There might be something to $e^x$ being the "boundary", it has to do with $x$ being the boundary between concave and convex among the power functions. Take $g(x)=\cosh x$, then $g'(x)=\sinh x$, $g''(x)=\cosh x$, and $\cosh^2x-\sinh^2x=1\geq0$. So $\cosh$, growing slightly faster than $e^x$, is log-convex. On the other hand, $\sinh x$, growing slightly slower, is log-concave.