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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.

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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:

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  • $\begingroup$ Thank you for the answer ! However, I'm looking for a proof without assuming that f is differentiable ... $\endgroup$
    – JackEight
    Sep 13, 2020 at 16:40
  • $\begingroup$ @JackEight: Check the new version! $\endgroup$
    – Martin R
    Sep 13, 2020 at 17:00
  • $\begingroup$ Thank you very much ! Simple but efficient ! $\endgroup$
    – JackEight
    Sep 14, 2020 at 5:20

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