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The question is related to my previous post. We already know that $\log x\ll x\ll e^x$ as a growth of the function (how fast the function diverges). I wonder if there is a middle function $f$ such that $\log x\ll f\ll x$. More generally, if $f,g$ be a real function that diverges to infinity as $n\to\infty$ such that the growth $f\ll g$. Then can we find the function $h$ such that $f\ll h\ll g$?

I don't know what tag will be adequate. I hope someone add appropriate tag of this post.

Note: $f\ll g$ if $\lim_{x\to\infty}\frac{f(x)}{g(x)} = 0$.

Karl
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one potato two potato
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2 Answers2

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You can always take the average of $f$ and $g$, namely $h(x)=(f(x)+g(x))/2$ to get $f(x)<h(x)<g(x)$. For $\log x <f(x) < x$, you could take for instance $f(x)= (\log x)^2$.

Slugger
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The geometric average $\sqrt{fg}$ should work, since $\frac f{\sqrt{fg}}$ and $\frac {\sqrt{fg}}g$ are both equal to $\sqrt\frac fg$, which approaches $0$ if $\frac fg$ does.

This is of course dependent on how you define the $<$ relation on functions. If we defined $f<g$ to mean $\lim_{x\to\infty}(g(x)-f(x))=\infty$, then the arithmetic average (Slugger's answer) would also work. The $<$ defined in this way is finer than the ordering provided by $O$-notation but arguably just as suitable for studying the order structure of function growth rates. I asked a related question (about an even finer ordering) that got some interesting references in the comments/answers.

Karl
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