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Consider a strictly increasing convex function $f(x)$ defined on the interval $[0,1)$ such that $f(0)=1$ and $\lim_{x\to 1^{-}}{f(x)}=+\infty$.

My question: Is the function $f(x)$ logarithmically convex (also called super-convex) in the interval $(1-\epsilon,1)$ for $\epsilon$ sufficiently small? In other words, is $\log f(x)$ a convex function in this interval?

Thanks!

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2 Answers 2

Some complicated answers have been posted! Here's a dead simple one. Let $f(x) = 1 + x$ for $x \in [0, \frac12]$. You can always extend it to go to $+\infty$ in a convex way over $[\frac12, 1)$; how you do it doesn't matter. The point is that over $[0,\frac12]$, $\log f(x) = \log(1+x)$ is a translated version of $\log$, and it's clearly not convex.

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I'd be happy to hear from whoever downvoted about what is wrong with this answer. –  Rahul Jan 27 '12 at 23:23
    
Undownvoted${}{}$ –  no identity May 15 '12 at 16:51
    
This function is not logarithmically convex in the interval $[0,1/2]$ but does not answers my question that is more related with the behavior near 1. –  ght May 22 '12 at 0:21

A counterexample is $$f(x) = \frac1{1-x} -x^2 =1+ x + \sum_{k=3}^\infty x^k.$$

It is easy to check that $\frac{d^2}{dx^2} \log f(0) = -1$.

(I hope it fulfills all the requirements)

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Thanks! My question is more related to what happens when $x$ is close to 1. –  ght Apr 10 '12 at 13:39

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