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For a continuous function $f: \mathbb R^+ \rightarrow \mathbb R$ if:

$$f\left(\sqrt{xy}\right) \leq 1/2(f(x)+f(y))$$

for all $x,y>0$ then $f(e^t)$ is convex for all $t \in \mathbb R$.

What I've tried

I am going after showing the inequality:

$$f\left (\exp (\lambda t_1 + (1-\lambda) t_2)\right) \leq \lambda f(e^{t_1}) + (1-\lambda) f(e^{t_2})$$

for all $t_1,t_2 \in \mathbb R$ and $\lambda \in [0,1]$.

My plan was to rewrite the first inequality as:

$$f\left(\exp 1/2\log x + 1/2 \log y\right) \leq 1/2(f(\exp \log x)+f(\exp \log y))$$

and claim that since $\log$ is a bijection from $\mathbb R^+$ to $\mathbb R$ then inequality holds for $\lambda = 1/2$.

I next tried to show using the continuity hypothesis on $f$ that this holds for any $\lambda \in [0,1]$ but I am struggling with that argument and think I might be missing a simpler way.

dmh
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    this is an unusual phrasing of the result that midpoint convex and continuous implies convex; the proof is not "hard" but not "short" – Calvin Khor Mar 24 '21 at 14:06
  • I wasnt aware of such a result. I'll see if working with that general claim is easier to show. – dmh Mar 24 '21 at 14:06
  • iirc, the idea: repeatedly split [0,1] into halves so that you can get any $k/2^j$ – Calvin Khor Mar 24 '21 at 14:07
  • In other words, with $g:=f\circ \exp$, you have $g(\frac{x+y}2)\le \frac{g(x)+g(y)}{2}$, i.e., the convexity condition only for $\lambda=\frac12$ (and trivially for $\lambda\in{0,1}$) instead of for all $\lambda\in[0,1]$ – Hagen von Eitzen Mar 24 '21 at 14:12

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