Pretty much the title.
I am looking for a convex function that is not logarithmically convex. A simple reminder is that a function is logarithmically convex if and only if $\mathrm{log} f$ is convex. That is, if
\begin{align}
\mathrm{log} f(\lambda x + (1-\lambda) y) \leq \lambda \mathrm{log} f(x) + (1-\lambda)\mathrm{log}f(y).
\end{align}
Another tool someone has at his disposal is the famous second derivative criterion.
My thoughts are for the function $f(x) = x^2$ which as we know is convex, since $f''(x) = 2 > 0 \ \forall x \in \mathbb{R}$ but $\mathrm{log} f(x) = \mathrm{log}x^2$ has second derivative $(\log f(x))^{''} = \dfrac{-2}{x^2} < 0 \ \forall x \in \mathbb{R}$ implying that it is not convex but concave.
Is this true and sufficient or one would need something more?
Best Answer
That's good enough, I think (although the final verdict is up to whoever is correcting your exams).
I might personally choose $x^2+1$, just to avoid any and all issues with the logarithm not being defined. The second derivative is a little more complicated, and not always negative: $$ \frac{d^2}{dx^2}\ln(x^2+1)=\frac{2(1-x^2)}{(1+x^2)^2} $$ But this second derivative is negative on $(-\infty,-1)\cup(1,\infty)$, so $x^2+1$ is certainly not logarithmically concave.