I asked this Question on [Math.StackExchange](https://math.stackexchange.com/questions/2243715/comparing-growth-of-f-circ-g-and-g-circ-f) without success. Then I learned, that this might be the better place to ask. So, sorry for crossposting. I would agree on deleting my old question.




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Let $\mathbb R\_0^+:=\{x\in\mathbb R\mid x\geq 0\}$. Further let $f:\mathbb R\_0^+\rightarrow \mathbb R\_0^+$ and $g:\mathbb R\_0^+\rightarrow \mathbb R\_0^+$ two strictly increasing continuous functions (this can be weakened if necessary).



> 
> Is there anything that can be said about which of the functions $f\circ g$ and $g\circ f$ grows asymptotically faster (in some sense) with only assuming something about the asymptotic growth behavior of $g$ and $f$?
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I want to discuss this in a very general context. So I am open for all kinds of useful definitions of "grows faster" and "growth behavior". E.g. one can consider the usual $\mathcal O$-notation and ask for whether


$$ f\circ g\in\Omega(g\circ f)$$


whenever $f\in\Omega (g)$ or $g\in\Omega(f)$. But neither seems to hold in general. For example, conider $$f(x)=\log(x)\quad\text{and}\quad g(x)=x^\alpha.$$ Which of $f\circ g$ and $g\circ f$ grows faster depends on $\alpha$, while $g\in\Omega(f)$ regardless of $\alpha$. Also possible: we can call $f$ growing faster than $g$ if $f(x)>g(x)$ for all sufficiently large $x$. 


I was not abled to proof anything remotely useful for the connection between the "growth" of $f$ and $g$ and the connection between the "growth" of $f\circ g$ and $g\circ f$. So this is a soft question because I hope for input from no specific branch of math.




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Further useful assumptions might be


* $f$ and $g$ are convex functions.
* $f(x)>x$ and $g(x)>x$.




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I tried to prove



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> If $f>g^n$ for all $n\in\mathbb N$ ($g^n$ means $g$ iterated $n$ times), it holds $g\circ f< f\circ g$.
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For example, use $g(x)=x^2$ and $f=\exp$. I have not succeeded for any definition of growth so far but it seems plausible to me, at least for convex functions $f$ and $g$ strictly greater than identity.



Comparing the growth of $f\circ g$ and $g\circ f$
Suppose that $g$ satisfies $g(x)>x$ for all $x$, and also that $f$ and $g$ lie in a Hardy field, as suggested by Todd Trimble. (A *Hardy Field*, called an order of infinity by Hardy is a collection of germs of differentiable functions at $\infty$ that is closed under differentiation and composition, as well as the field operations. In particular, this means that for any Hardy field, $H$, and any $f\ne g\in H$, $f-g\in H\setminus\{0\}$, so that $1/(f-g)\in H$. In particular, $f-g$ is eventually positive, or eventually negative. That is, there is a total ordering on a Hardy field.)


Then I believe there is an extension Hardy field containing $f$, $g$ and a function, $h$ satisfying $h(g(x))=h(x)+1$. That is, $h(x)$ counts how many times you have to apply $g$ to 0 to get to $x$ (think of $\log^\* x$). 


Assuming the existence of such a Hardy field, there is now an affirmative answer to your question. If you conjugate $g$ by $h$, you obtain $\tilde g:=h\circ g\circ h^{-1}$ is $\tilde g(x)=x+1$. Conjugating $f$ by $h$ also to obtain $\tilde f$, your question is equivalent to asking whether $\tilde f(\tilde g(x))>\tilde g(\tilde f(x))$. That is, whether $\tilde f(x+1)>\tilde f(x)+1$. Notice that both $\tilde f$ and $\tilde g$ belong to the Hardy field.


The condition that $f>g^n$ for all $n$ means that $\tilde f(x)>x+n$ for large $x$, that is $\tilde f(x)-x\to\infty$. In a Hardy field, any function satisfying this satisfies $\tilde f'(x)>1$ for all large $x$. By the mean value theorem, this implies that $\tilde f(x+1)>\tilde f(x)+1$, as required. 


