# What operations is this asymptotic relation closed under?

For all positive functions $f$ and $g$ of the real variable $x$, let $\sim$ be a relation defined by

$f \sim g$ if and only if $\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = 1$

Then if $f \sim g$, we have for example, that $f^2 \sim g^2, \sqrt{f} \sim \sqrt{g}, f+g \sim 2g$, but we do NOT have that $e^f \sim e^g$. What other operations preserve, or do not preserve the asymptotic relation $\sim$?

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There are some things to point out in your examples. Each demonstrates functional composition i.e. functions $h$ such that $$f\sim g \implies h\circ f\sim h\circ g,$$ but these are only a subclass of operators preserving the $\sim$ relation (or are they?). If you put extra restrictions on the functions, your class of operators may expand fortuitously - l'Hôpital's suggests we could include differentiation if each of $f$ and $g$ are asymptotically either $0$ or $\infty$ (and are differentiable for sufficiently large $x$). Finally, your last example puts $g$ on both sides of the relation... – anon Oct 11 '11 at 0:19
You may find more complete answer or/and reference in my post. – quartermind Mar 5 at 18:28

There seems to be a connection between this question and the beautiful Karamata theory of regular variation.

A positive function $F$ is regularly varying both at $0$ and $\infty$ if and only if $F(t)=t^pL(t)=t^qK(t)$ where $p$ and $q$ are real and $L$ and $K$ are slowly varying at 0 and $\infty$ respectively, which implies in particular (due to the uniform convergence theorem for slowly varying functions) that $$\lim_{t\to0^+}\sup_{|a-1|\le \delta}\left|\frac{L(at)}{L(t)}-1\right| \to 0 \qquad \text{and} \quad \lim_{t\to\infty}\sup_{|a-1|\le \delta}\left|\frac{K(at)}{K(t)}-1\right| \to 0,$$ for some $\delta>0$.

Now, it follows that whenever $f\sim g$ then $F\circ f \sim F\circ g$, provided that (i) $F$ is continuous, and (ii) $F$ is regularly varying both at 0 and $\infty$. The proof is a straightforward (but slightly tedious) $\frac\epsilon 3$ argument based on (a) the fact that $$f(x)=g(x)(1+h(x)) \quad\text{ where }h(x)\to0,$$ (b) uniform continuity of $F$ on any compact interval $[b,c]\subset(0,\infty)$, and (c) the limit relations above.

While conditions (i) and (ii) on $F$ are sufficient, (ii) is not necessary: The function $$F(t)=t(2+\sin\ln t)$$ is a counterexample. This function is not regularly varying since $\lim_{t\to0} F(at)/F(t)$ does not exist when $\ln a=\pi$. However, whenever $f\sim g$ it is true that $F\circ f\sim F\circ g$. This is due to the fact that $$\sin(\ln g(x)+\ln(1+h(x))-\sin\ln g(x)\to0 \qquad \text{ whenever } h(x)\to0,$$ by the uniform continuity of $\sin$.

Generalizing this example seems to yield a class that preserves $\sim$ consisting of all continuous $F$ such that $F(t)= t^p G(L(t))$ where $L$ is slowly varying at 0 and $G$ is bounded away from 0 and $\infty$ and is uniformly continuous, with a similar representation at $\infty$.

Naturally, the class of positive $F$ that preserve $\sim$ is closed under composition and products. I don't have any of the classic texts on regular variation handy, but it is conceivable that necessary and sufficient conditions are known.

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Suppose $F$ is a differentiable convex or concave function on $(0, \infty)$. Then the mean value theorem says $F(g(x)) = F(f(x)) + F'(t)(g(x) - f(x))$ where $t$ is between $f(x)$ and $g(x)$, thus $$|F(g(x)) - F(f(x))| \le |g(x) - f(x)| \max(|F'(f(x))|,|F'(g(x))|)$$ In particular, if there is a constant $k$ such that $|t F'(t)| \le k |F(t)|$ for all $t > 0$, since $|f(x) - g(x)| = o(|f(x)|$ and $o(|g(x)|)$, we have $|F(g(x)) - F(f(x))| = o(\max(|F(f(x))|, |F(g(x)|))$, and so $F(f(x)) \sim F(g(x))$. Thus powers $F(t) = t^p$ preserve the relation, but exponentials don't.

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