Is there any notion of preserving asymptotic equivalence by a real-valued function? Any facts known about such functions?

To clarify what I'm asking I'll introduce one formalization of the idea which has come to my mind.

$\mathbb R_+$ denotes $[0,+\infty)$.

Consider a class $\mathcal{A}$ of continuous functions $f\,\colon \mathbb R_+ \rightarrow \mathbb R_+$, satisfying following conditions:

  • $\lim \limits_{x \rightarrow +\infty} f(x) = +\infty$,
  • for any two positive sequences $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$, such that $$a_n \rightarrow +\infty \;\;\;\text{and}\;\;\; a_n \sim b_n \quad\text{as}\;\;\; n\rightarrow \infty,$$ we have $f(a_n) \sim f(b_n)$, as $n$ goes to infinity.

Some facts about $\mathcal A$:

  • $\mathcal A$ contains such functions, as $\log(1+x)$ and $x^\alpha$ for $\alpha>0$.
  • For continuous $f,g\,\colon \mathbb R_+ \rightarrow \mathbb R_+$, if $f \in \mathcal A$ and $f \sim g$ then we have $g \in \mathcal A$. In particular, in our class lie the functions $f$ analityc at infinity, which means $\frac{1}{f(1/x)}$ to be analytic at zero. Thanks Joel Cohen who helped me note it.
  • One can show that $\mathcal A$ is closed under composition, multiplication and taking linear combinations with positive coefficients. Moreover, this operations are well-defined on set $A$ of $\sim$-equivalency classes of $f \in \mathcal A$.

Then, we can state some problems:

  • Is $\mathcal A$ bounded in meaning of asymptotic growth?
  • Can we characterize $f \in \mathcal A$ in some other terms?
  • Is $\mathcal A$ closed under taking indefinite integral? If so, is this operation well-defined on $A$?
  • If we modify the definition by replacing equivalent sequences with such $C^k$-smooth functions on $\mathbb R_+$, would we get the same class? What if we restrict this sequences/functions to be strictly increasing?
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    $\begingroup$ If you denote $g(x) = \frac{1}{f(1/x)}$, then the limit of $g$ at $0$ is $0$. This allows you to reformulate every thing in terms of asymptotic behavior at $0$. In that case, I think you might be asking for $g$ to be continuous in a neighborhood of $0$. $\endgroup$ – Joel Cohen Feb 25 '16 at 2:20
  • $\begingroup$ @JoelCohen Continuity condition is too weak, actually it follows from the first part of my definition. For counterexample consider $f(x)=e^x$. $\endgroup$ – quartermind Feb 25 '16 at 2:46
  • $\begingroup$ yes sorry, I think it might be something along the line of being smooth in a neighborhood of $0$. $\endgroup$ – Joel Cohen Feb 25 '16 at 9:33
  • $\begingroup$ @JoelCohen The same example (or slightly modified one) shows that smoothness is also not an option, but being analytic at zero turns sufficient for $g$, as in this case $g(x) \sim C x^n$. Though, you see, it doesn't hold for every function of our interest: consider $g(x) = x^\alpha$ for non-integer $\alpha$, $g(x) = -1/\log(x)$. $\endgroup$ – quartermind Feb 25 '16 at 11:48
  • $\begingroup$ Yes my comment was just a rough idea. Essentially I think it works for meromorphic functions that don't have an essential singularity at $0$ (or $\infty$ for $f$). But there might be other functions satisfying your condition, and it is also plausible that your regularity condition doesn't have any "simpler" characterization. $\endgroup$ – Joel Cohen Feb 25 '16 at 21:10

The functions which preserve asymptotic equivalence have turned out to be one of a vast number of generalizations of Karamata's regularly varying functions. Paper "On some extensions of Karamata's theory and their applications" by Buldygin et. al. introduces pseudo-regularly varying (PRV) functions which are, in fact, exactly that measurable functions, not necessarily tending to infinity, which preserve asymptotics at $+\infty$ in my sense.

The paper contains many properties of PRV functions and conditions for a function to be PRV. For example, PRV fuctions are characterized as being eventually ($\forall x>x_0$) of the form $$ f(x) = \exp \bigg( a(x) + \int\limits_{x_0}^x b(t) \frac{dt}{t} \bigg), $$ where $a$ and $b$ are bounded measurable functions with $\lim\limits_{c\rightarrow 1}\limsup\limits_{x\rightarrow +\infty} \lvert a(cx)-a(x) \rvert = 0$.

So the reference request is fulfilled, I'll return to finer class $\mathcal A$ described in the question and answer the "mini-questions".

Besides the above, we can charactirize $\mathcal A$ in terms of uniform continuity, although this characterization is not so constructive.

Proposition. A continuous function $f\colon \mathbb R_+ \rightarrow \mathbb R_+$ with $\lim \limits_{x \rightarrow +\infty} f(x) = +\infty$ belongs to $\mathcal A$ if and only if function $F(t) = \log f(e^t)$, correctly defined on some ray $[T,+\infty)$, is uniformly continuous.

Proof. Note that $\log f(x) = F(\log x)$.

1) To prove the "if" part, pick $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ as in the definition of $\mathcal A$. Since $\lim\limits_{n\rightarrow\infty} \log\tfrac{a_n}{b_n} = 0$ and $F$ is uniformly continuous, \begin{align*} \left| \log \frac{f(a_n)}{f(b_n)} \right| &= \lvert \log f(a_n) - \log f(b_n) \rvert = \lvert F(\log a_n)-F(\log b_n) \rvert = \\ &= \lvert F(\log b_n+\log\tfrac{a_n}{b_n}) - F(\log b_n) \rvert \leqslant \sup\limits_{t \geqslant T} \lvert F(t+\log\tfrac{a_n}{b_n}) - F(t) \rvert \xrightarrow{\;n\rightarrow\infty\;} 0, \end{align*} so finally $f(a_n) \sim f(b_n)$ as $n$ tends to infinity.

2) To prove the "only if" part, suppose that, coversely, $F$ is not uniformly coninuous. That means there exists $\varepsilon > 0$ and two sequences $\{t_n\}_{n=1}^\infty, \{s_n\}_{n=1}^\infty \subseteq [T, +\infty)$, such that $$ \lim\limits_{n\rightarrow\infty} (t_n-s_n) = 0 \quad\text{and}\quad \forall n\in \mathbb N\ \; \lvert F(t_n) - F(s_n) \rvert \geqslant \varepsilon. $$ It's necessary that $\{t_n\}_{n=1}^\infty$ is unbounded, because by Cantor's theorem $F$ is uniformly continuous on each finite interval. Hence, taking subsequence if needed, we can assume that $t_n$ (and so $s_n$) goes to infinity.

Define $a_n := \exp(t_n), \; b_n := \exp(s_n)$. We have $\lim\limits_{n\rightarrow\infty} \tfrac{a_n}{b_n} = \lim\limits_{n\rightarrow\infty} \exp(t_n-s_n) = 1$ and $$ \bigg\lvert \! \log \frac{f(a_n)}{f(b_n)} \! \bigg\rvert = \lvert F(\log a_n)-F(\log b_n) \rvert = \lvert F(t_n)-F(s_n) \rvert \geqslant \varepsilon \;\;\; \forall n \in \mathbb N, $$ what contradicts with $f \in \mathcal A$.

From this characterization it follows that a functions from $\mathcal A$ grows at most like power function. (It's also follows from the exponent expression for PRV functions.)

Corollary. If $f \in \mathcal A$, then $f(x) = O(x^\alpha)$ for some $\alpha > 0$ as $x \rightarrow +\infty$.

Proof. Take $F$ like in above proposition. Since $F$ is uniformly continuous, $F(t) \leqslant \alpha t \;\;\; \forall t \geqslant T$ with some $\alpha,T \geqslant 0$ (see this question). Then $$ f(x) = \exp\big(F(\log x)\big) \leqslant \exp(\alpha \log x) \leqslant x^\alpha \quad \forall x \geqslant e^T. $$

And finally, we can indeed replace equivalent sequences in the definition of $\mathcal A$ with such $C^\infty$-smooth functions on $\mathbb R_+$ and get the same class. It's straightforward, we just need to construct smooth function with given values in, say, $\mathbb N$.

We may also restrict the sequences from the definition to be strictly increasing, hence each real sequence not bounded above admits a strictly increasing subsequence. And thus the smooth functions from the alternative definition can be all stricly increasing. It's bit more tricky and this is crucial for l'Hopital's rule application to prove

Proposition. If $f, g \in \mathcal A$, then $F(x)=\int_0^x f(t)dt$ and $G(x) = \int_0^x g(t)dt$ lie in $\mathcal A$ and, moreover, $F \sim G$ at infinity.

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