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The following function $f(x)={1\over (1+e^{-x})}$ is non-convex but $\ln(f(x))$ is convex. Given a non-convex function $f$, can we find a function $g$ such that $g\circ f$ is strictly convex? If yes, how to find such a g? If not, what property should f meet when such a g can be found?

Tianqi Tang
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    What assumptions are you making on $f$, and what are you requiring from $g$ ? Note that e.g. for $f(x):=\sin(x)$, the function $g(\sin(x))$ is constant on $\pi\mathbb{Z}$, so it is convex only if it is constant. – Pietro Majer Apr 26 '15 at 16:09
  • To add to Pietro comment, if there is a point with three preimages by $f$, say $x_1$, $x_2$, $x_3$, then $g\circ f$ must take the same value at these three points, and thus is constant. –  Apr 26 '15 at 16:15
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    @PietroMajer For g·f, I would like it to be strictly convex. And for g, actually I would like it to be invertible, which implies that I can make a one-to-one mapping between f(x) and g(f(x)). – Tianqi Tang Apr 26 '15 at 16:17
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    Then, if g is increasing and g(f(x) is convex, all sub-level sets of f, {x : f(x)<c} = {x : g(f(x)) < g(c)} must be intervals. – Pietro Majer Apr 26 '15 at 16:45
  • So you'd like to take a function with multiple local minima and transform it to one with no more than one? Hmm. – Michael Grant Apr 26 '15 at 23:00
  • @MichaelGrant Yeap~ Does it make sense? – Tianqi Tang Apr 27 '15 at 01:33
  • It makes sense. I just don't think it is possible in any wide sense, – Michael Grant Apr 27 '15 at 05:14

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Таке any convex function, say $s(x)=\exp(x)$. Let $f(x)$ be monotone. Solve $g(f(x))=s(x)=\exp(x), f(x)=t, x=p(t)$ and so $g(t)$=s(p(t))=\exp(p(t))$ will be OK!