# Characterization of log-convexity

I read that if we take $$f: \mathbb{R} \to (0, +\infty)$$, such that for all $$a \in \mathbb{R}$$, $$x \mapsto e^{ax}f(x)$$ is convex then $$f$$ is log-convex (meaning $$\log \circ f$$ is convex).

I did try to prove it, but I couldn't get to the result.

However, I also read that if for all $$a \in \mathbb{R}$$, $$x \mapsto e^{ax}f(x)$$ is convex then $$x \mapsto (f(x))^{a}$$ is convex for all $$a > 0$$. I couldn't find a proof either ... But I managed to prove that if this last condition is verified, then $$f$$ is log convex.

Therefore could you please help me prove one of these two statements (either the first one, which is in fact the result I want in the end, or the second one, which could lead me to the first result) ?

Thank you.

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 $$\log \circ f$$.
• 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 $$a \in \Bbb R$$.