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If $g(x)$ is positive on $(0,\infty)$ and $g(x)\to \infty$ as $x\to \infty$. Is there a convex function $h(x)$ on $(0,\infty)$ such that $h(x)\le g(x), h(x)\to \infty$ as $x\to \infty$.

I didn't know how to solve this one. But I find a solution says that $h(x)$ exists iff $\liminf_{x\to \infty} \frac{\log(g(x))}{x} = \alpha \gt 0$. I don't know how to do it in any direction. Can somebody give me a hint on what I should do? Thanks.

There is a similar question in Rudin Real and Complex analysis on the interval $(0,1)$. But I think they are not same question. At least for that one the answer is affirmative but this answer suggests otherwise.

Harald Hanche-Olsen
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user236626
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  • @almosteverywhere Not a duplicate, I think. There is a big difference between $x\to0$ and $x\to\infty$. – Harald Hanche-Olsen Mar 19 '17 at 17:52
  • What is that condition $\lim_{x\to\infty}\inf\ldots$ supposed to mean? Should it be $\liminf_{x\to\infty}$, perhaps? (And where did you find the solution you mention?) – Harald Hanche-Olsen Mar 19 '17 at 17:55
  • Okay, saw your edit. (I fixed it up a bit more. Hoping that the Log with capital L wasn't essential …) Now that seems to say that $g$ has at least exponential growth. But if $g(x)=h(x)=x^2$ is a counterexample. So there is something funny about that “solution”. – Harald Hanche-Olsen Mar 19 '17 at 18:03

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Here's a hint for your question as stated in the first paragraph. Note that any convex function on $(0,\infty)$ that is increasing somewhere has at least linear growth. Now think about logarithms, which don't have linear growth.

Harald Hanche-Olsen
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