Characterization of log-convexity

Fix real numbers $x < y$ and $0 < \lambda < 1$. Since $x \mapsto e^{ax}f(x)$ is convex, we have $$ e^{a(\lambda x + (1-\lambda) y} f(\lambda x + (1-\lambda) y) \le \lambda e^{ax}f(x) + (1-\lambda)e^{ax}f(y) $$ which is equivalent to $$ \begin{align} f(\lambda x + (1-\lambda) y) &\le \lambda e^{a(1-\lambda)(x-y)}f(x) + (1-\lambda)e^{a \lambda(y-x)}f(y) \\ &= \lambda C^{1-\lambda}f(x) + (1-\lambda)C^{-\lambda}f(y) \end{align} $$ with $C = e^{a(x-y)}$. This holds for all $a \in \Bbb R$, therefore we can choose $a$ such that $C = f(y)/f(x)$. This gives $$ f(\lambda x + (1-\lambda) y) \le f(x)^{\lambda} f(y)^{1-\lambda} $$ and that is exactly the convexity condition for $\log \circ f$.

Remarks:

• The opposite conclusion holds as well: If $f: \Bbb R \to (0, +\infty)$ is log-convex then $x \mapsto e^{ax}f(x)$ is convex for all $a \in \Bbb R$.
• This characterization can be used to show that the sum of log-convex functions is again log-convex..