If the world is nice, let $f$ be a positive and strictly increasing differentiable function, i.e. $f>0$ and $f'>0$.

  1. Is the following statement true?

    There always exist $p\in\mathbb{R}$, such that $g = f^p$ is convex.

  2. How about we replace 'convex' with 'concave'?

  3. I came across this question in my study of submodularity, a notion of concavity in combinatorial optimization. If you are familiar with submodularity, is it true that for any monotonic function $f$, I can take some power of $f$ and make it submodular?


1 Answer 1


No, for both of them take $f(x) =\arctan(x) + \pi$,

And note that for a convex(concave) function $g$ defined on $R$ , we have $\limsup_{|x| \rightarrow \infty} |g(x)| = \infty $

  • $\begingroup$ That's a good counterexample. How about I change f'>0 to f' is bounded away from 0? $\endgroup$
    – ChubbyRuby
    Jun 10, 2017 at 5:52

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